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Does heatsink transfer plate thickness matter? I'm designing a water block for a liquid cooling system. It will be acrylic on 5 sides and copper on the side that attaches to a CPU heatsink/fan. What I'm stumped on is the thickness of the copper side of the block. A thicker sheet of copper can hold more total heat energy, but once the system reaches equilibrium that heat still needs to go somewhere, so I'm limited by the thermal interface between the water and copper, and then the copper and heatsink. So my gut instinct is that thickness doesn't matter. However, I often see cooling blocks advertised online as having extremely thick copper transfer plates, making me wonder if this is actually an important spec. I understand that if my water block is larger than the contact area of the heatsink a thicker plate will help transfer more heat to the heatsink, but in this case assume that the water block, transfer plate, and heatsink interface are all the same size. Does transfer plate thickness really matter when sandwiched between coolant and heatsink? <Q> Let's consider what the thickness of the metal plate is doing a) <S> It's making it rigid, so the CPU-metal contact is good b) <S> It's spreading the heat sideways from the CPU, to areas in contact with the water c) <S> It's impeding the flow of heat from the CPU to the water (a) is saying that it has to be thick enough mechanically, say at least 10% of its width. <S> (b) and (c) taken in combination say that if it's any thicker than the space between the edge of the CPU thermal contact and the edge of water contact, then it's not going to improve things making it any thicker. <S> However, that's an upper limit, a thinner plate may still conduct an adequate amount of heat. <S> It's not a hard upper limit, as a thicker sheet will still impede the flow of heat very little, more an economic limit above which there's no performance gain. <S> However, there may be an economic incentive for the manufacturer to use a very thick plate, as then he can advertise that it uses 'a very thick plate' <S> (so it must be better, mustn't it!) <S> Where Ww is the width of the water contact, and Wc is the width of the CPU contact, the plate should be at least 0.1*Ww thick, with <S> no thermal need for it to be thicker than 0.5*(Ww <S> -Wc) thick. <A> A few factors to consider. <S> The thermal resistance would be greator with a thicker place, everything else being equal. <S> However, how well a plates contact with the heatsink matters. <S> Also a thicker plate can dissipate heat more evenly over a wider area within the plate, reducing thermal resistance. <S> .... <S> The reality is likely a compromise of those factors, and money, space, weight,.... <A> Draw a grid of copper squares. <S> Compute the thermal resistance of each square. <S> Define the input heat region (presumably some 10% or 20% of the total width). <S> Compute the #squares the heat must flow through, to get to fins or however else you will contact the cooling water. <S> Experiment with thickness. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Thicker plate is more than heat capacity. <S> Capacity is only important for initial conditions. <S> In steady state it provides short path for energy between hot element on one side and each fin on other side. <S> Sometimes it is what makes the difference between "all fins are effective" and "only 10%,the ones inside are effective, and others do nothing". <S> If your plate becomes too thick, you can use heat pipes, they transfer heat much better than even copper.
| A thicker plate can maintain better contact over a wider area, reducing thermal resistance.
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Understanding a variable 220v AC to DC Power Supply Can someone just please explain how and why the diodes D3 D2 and D4 are used?Why D3 is single while D2 and D4 are in series?How do we calculate the capacitance of capacitor required in a circuit like this? Please, be a little descriptve I am a newbie. Thanks to everyone in advance <Q> The circuit has too many errors even to serve as a bad example! <S> 100pF polarized capacitors shown with wrong polarity, cap in series with DC input to a regulator, confusing schematic.... <S> Why does it keep getting resurected? <S> (and I'm sure the OP is long gone) <A> What is with diodes, I don't know. <S> And also that variable DC voltage output you have there should be regulated with resistor which is also not included in your schematic. <S> Capacitors which prevent oscillations on the output are also needed on the variable stage of the power supply. <A> About capacitors, C1 (smoothing capacitor) used to filter out twice the input frequency generated from D1 (full wave rectifier). <S> Generally calculated based on how much ripple you would want to allow. <S> Rule of thumb less than 10% of output voltage is preferred. <S> This article tell you how to calculate. <S> Other capacitors C2 - C6 serves same purpose to decouple AC and provide stable DC output. <S> Those values are often suggested in the datasheet. <S> For 78XX series IC product datasheet says the following, <S> The diode simply blocks DC Voltage. <S> Diode D3 serves no purpose as even without it conducts only in forward direction. <S> D2 and D4 in series treated as single diode as it doesn't enhance or affect any electrical properties except when forward biased will drop twice compared to single diode. <S> Again in this schematic it serves no purpose at that location.
| Capacitance of the smoothing capacitors should be between 2000uF-3000uF for 1 Ampere.100pF capacitors are originally just fine - they are used for preventing unwanted oscillations/noise.
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why two pair of terminals are used in a shunt? The working principle of two pair of terminals of a shunt is required here. <Q> Imagine the shunt looks something like this, where R1~R4 represent connection and wire resistances, and are variable with tightness of the connections, temperature and so on. <S> R5 is the shunt resistance itself (usually made of a low temperature coefficient alloy such as Manganin or Constantan . <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Provided only that the meter is high resistance compared to the values of the resistors the voltage measured will accurately reflect the current I1 multiplied by R5 (only). <S> It is important to use the larger connections for the high current and the smaller ones for the measured voltage because the larger ones are (or should be) on the 'outside' of the smaller ones. <S> If you were to just parallel them all in my example there would be a large (and variable) <S> 5m\$\Omega\$ resistance on each side so the error would be a very large (and variable) 10% of reading! <S> Even if you calibrate that out, it would change if you tighten the connections or if they relax, and it would change more than the tolerance of most shunts for a few degrees C change in temperature. <S> It's possible to use Kelvin connections to surface mount resistors, for example. <S> Some have 4 terminals, but you can split the ends of something like an 0805 to improve performance (not quite as good, but better than nothing). <A> It's called a Kelvin connection , or four-wire resistance measurement. <S> The issue is that the resistance of the shunt is not so very different from the resistance of the wires connecting to it, so sharing the same connection points for connecting power, and for measurement, the connections themselves would add enough resistance to affect the measurement. <S> So a large (low resistance) connection is used to connect power at each end of the shunt, and a second connection closer to the shunt itself takes voltage to the measurement device. <S> (This can be relatively thin, as it doesn't carry significant current, without affecting measurement accuracy). <A> It's connect at two points of a circuit. <S> Hence it has two terminals or connection points. <S> To measure a voltage across it, you need to measure it across those two points. <A> In some shunt resistors resistance is so small, it is not very different from pcb trace resistance. <S> So taking measurement from there is a bad idea. <S> They arrange another two pads such way that current will not flow there, so you can take measurements of voltage drop on tbe exact resistance you know.
| A shunt is a part with a very small resistance (typically know, not always small).
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Is it better to keep LiOn battery charge on the lows or the highs for battery lifespan? I know that neither keeping battery full or discharged is good for battery life, but I want to know a basic answer: which one of the two bad alternatives is better? Full or empty. This is given that self discharge bellow 0% is not an issue as I would plug the devise in after it's off for some time. I have a bad habit of either keeping my phone battery full (by constantly looking for charger) or having it almost empty (by charging when it's dead for 10min and then going after business). So I want to know which of the two habits will wear my phone battery faster? Keeping constantly charged to near 100, or barely charged. Thanks. <Q> For storage purposes, a Li-ion battery should be charged to 60-70% capacity. <S> This reduces the damage done to the electrode material over time while still allowing enough charge so that it will hopefully not be completely discharged due to self-discharge when you get around to using it. <S> For charge cycle purposes, a Li-ion battery should be charged during its constant voltage cycle with a voltage about 150mV less than the voltage specified in the datasheet. <S> This will result in only about 85% to 90% runtime versus full voltage, but an increase charge cycle count of 20% to 40%. <S> Of course, if the battery is in a device then all this is moot since the device manufacturer decides what "full" and "empty" mean for the battery; you're only responsible for connecting the power supply to the device. <A> http://batteryuniversity.com/learn/article/how_to_prolong_lithium_based_batteries Since your phone will prevent the battery from over-discharging due to self-discharge and you're keeping it charged up, being on the low-side is the best way to keep your battery in tip-top shape. <S> The lower in voltage you go, the less strain you place on the structure within the cell. <S> Much of the data presented in the link doesn't extend down to the 0-10% range because batteries will virtually last forever down in that range since they're hardly being stressed at that point. <S> All of the tables and graphs there point to extremely long battery lifetimes if you keep the battery at a very low state of charge (soc). <S> Just storing your battery at room temperature at 100% soc results in 20% loss in a year vs 4% at 40% soc. <S> Table 3: <S> Estimated recoverable capacity when storing Li-ion for one year at various temperatures: Temp 40% charge 100% charge 0 <S> ° <S> C 98% 94% 25 <S> ° <S> C 96% 80% <S> Too be clear, there's nothing wrong with keeping a li-ion battery at 0% as long as you don't let it drop below 0% (3.0V or 3.3V depending on the manufacturer). <S> Manufacturer's state that damage starts to happen at 2.0 or 2.5V for lithium ions. <S> Your phone has a protection circuit on it that would prevent your battery from working at all if it got down to this voltage. <S> Since 0% is defined much higher than that, unless you let your phone sit idle for a long time at 0%, your battery will live a very plush life between 0-10% soc. <A> Neither. <S> I've wondered about that myself, so I talked to the battery expert friends I have: <S> In either state, you risk that the extreme concentration differences in the chemicals that make up the cell lead to a diffusion of elements to regions they shouldn't be in. <S> So, go for something like a 1/3 charge or so to maximize shell life. <S> , or I constantly overcharge it <S> No, you don't. <S> You're not the one in charge of the charging cycles, your phone is. <S> Let it do its job, it does protect the battery as far as possible in trade off with fast charging and long usage. <S> It's more likely your battery will deteriorate through normal thermal effects than due to the amount you charge it to. <A> According to your comments, if you only have full/empty options, I'd go for full . <S> That's because Li-Ion batteries usually have control/safety circuits that only work between full and empty. <S> So, if you store an empty battery, self-discharge may get you below this circuit's minimum operating voltage, which would then block battery charging. <S> As a consequence your battery would be unusable, even if the chemistry was still fine. <S> On a full battery, self-discharge wouldn't be that much of a problem, except that its chemistry wouldn't like it and would potentially damage your battery earlier. <S> That very circuit should stop charging before the chemistry is 100%, effectively setting the "battery full" below that. <S> After question edit: If you're not storing the battery, I don't really see why you'd keep an empty battery in a system. <S> It would have no purpose. <S> I'd still keep it full, so it'd serve as backup power, or I'd remove the battery from the system (infinite lifespan, since you'd never have to replace the battery). <S> That would also avoid battery malfunction problems and power waste due to occasional recharge to 1%.
| However, for bog normal Li batteries, it probably won't matter much, and you should most likely, avoiding the shorting->explosion risk, store them discharged.
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Why is there no current sharing between neutral and ground? In the first system, will lamp1 turn on? If yes, then why is there no current sharing between neutral and ground as shown in the second system? I always read that neutral is a current carrying conductor but the ground is not. What is the reason behind that? I think current should be shared equally between the two wires of neutral and ground (assuming they have the same resistance). If no, does that mean earth is not a reference and it does not have "zero" voltage? So there's no need to ground the neutral wire? Arrows are the direction of currents. simulate this circuit – Schematic created using CircuitLab <Q> In your first diagram the answer is no. <S> Not because earth is not a reference, but because there is no current return path to the single phase generator. <S> What happens in the first circuit is the top of the generator is effectively "grounded" through the lamp. <S> The bottom of the generator will show the inverted AC voltage. <S> But since there is no circuit... there is no current. <S> Redrawing it like this helps. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In circuit 2 there is NO current passing to ground . <S> All the current coming from the AC source <S> MUST return to the AC source. <S> Current sharing would ONLY occur if you grounded both ends on the bottom schematic. <S> Which is what happens with bad wiring, or if there is a short somewhere. <S> Which is why we install GFCIs in circuit breakers these days. <S> simulate this circuit Typically though, houses and especially rural farms are wired this way. <S> simulate this circuit <S> The service neutral line can be omitted to reduce costs. <S> If it DOES exist, and is rated to carry return current from all phases, the neutral should not be connected to the ground at the client end. <S> Interestingly your schematic #2 brings up an interesting argument: " Would the electrical system actually be safer WITHOUT all this grounding? <S> " If it were a closed loop isolated system, in order to be shocked, you would need to touch Live AND Neutral. <A> Your first circuit won't work since there is no closed conduction loop path. <S> Your second circuit, minus the diodes or whatever those arrows are, is how power systems in single houses work. <S> The neutral is tied to earth ground in one place, usually close to where the power comes into the house. <S> Earth is not "0 Volts" or anything else since voltage is a relative concept. <S> You would have to explain 0 Volts to where ? <S> Since the earth is a large and pervasive conductor and many things are electrically tied to it, even if not deliberately, it is common to consider it the 0 Volt reference to measure other voltages relative to. <A> You may be confusing two different things: Ground in an electronics context, as in Vss, as in the normal current return. <S> Earthing, aka Equipment Safety Grounding in a mains-electrical context, which is entirely a safety shield around what is effectively an isolated system of hot and neutral, with neutral being the usual current return. <S> If the fault current is excessive, it will trip the circuit breaker. <S> If not, it will hopefully flow enough to induce a GFCI trip, and not hurt any humans. <S> In mains wiring, conductors (hot and neutral) should be connected to Safety Ground absolutely never , with a solitary exception made out of necessity: an isolated system has nothing to keep the conductors from attaining an unexpected voltage. <S> For instance leakage in a supply transformer could bias the entire isolated system at 2400V above ground, which would fail insulation. <S> Or the isolated-system voltage could "rattle" due to capacitive coupling. <S> So to prevent floating, the conductors are pegged to the earthing system, and the bonded conductor is defined as "neutral". <S> This neutral-ground bond is done in one single place to prevent redundant current paths (which themselves would be considered a ground fault). <S> Before it was just good practice; now GFCI/RCD and AFCI protection devices absolutely require it. <S> Your first diagram is correct on the left and a code violation on the right: you didn't wire a neutral <S> so you bootlegged neutral from ground. <S> It will light only if somewhere off-diagram there is a neutral-ground bond. <S> However if the source is GFCI/RCD protected, it will trip instantly. <S> Your second diagram shows an ungrounded system, much like you might have had in a 1946 house, except with the aforementioned neutral-ground bonding present. <S> By modern standards you would want to extend that safety ground out to the loads, though GFCI/RCD protection will protect people if present. <A> No to this one Earth is a reference for zero volts and neutral is (usually) connected at one point to earth. <S> This ensures that any current that flows through the live wire but not the neutral can be detected by an RCD and save lives. <S> In the US, an RCD is called a GFCI. <S> If you don't ground the neutral wire then there is no means to detect if an unsafe single fault appears and RCDs won't work. <A> It's impossible to completely isolate a circuit from an AC voltage source. <S> Even with perfect insulation there is always capacitive coupling. <S> Since it takes only a few thousandths of an amp to kill someone, it's easier to ground anything that could be electrically charged than to isolate everything. <S> This is the basis of grounding practices in the electrical code. <S> If a wire goes through a piece of metal, that piece of metal has to be grounded. <S> If an alternative path to ground exists, like a water pipe, the electrical ground has to be connected to the water pipe system. <S> The exception is when double insulation is used. <S> This just means the appliance has a non-conductive housing in addition to the wire insulation. <S> Grounding is never perfect and in a fault condition the grounded item could be at a higher potential than other local grounds such as a water tap. <S> The greatest risk of this is in the kitchen and bathroom where the sockets need to be protected by ground fault circuit breakers. <S> These measure the current difference between the hot and neutral wires. <S> If the difference exceeds the lethal few thousands of an ampere, the power is cut.
| There is never any current on Safety Ground except during a wiring fault.
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Is forward biasing or reverse biasing an inherent property of a diode? I read a tutorial about transistors and I reached the following sentence."The Bipolar Junction Transistor (BJT) is a three layer device constructed form two semiconductor diode junctions joined together, one forward biased and one reverse biased." It made me think that diodes are inherently forward biased or reverse biased but from what I know this depends on what is the direction of the voltage applied to the diode, so the same diode can be forward biased or can be reverse biased depending on the direction of the electric field (voltage). Is this right? <Q> The tutorial is worded very badly. <S> The tutorial should read something like.. <S> "The Bipolar Junction Transistor (BJT) is a three layer device constructed LIKE <S> two semiconductor diode junctions joined together, cathode to cathode, or anode to anode." <A> The tutorial is plain wrong and you are right to question it. <S> A good call on your part. <S> Biasing is an action and a consequence of applying a voltage. <S> The BJT's PN junctions will conduct when forward biased and oppose conduction when reverse biased (until the reverse voltage is high enough to cause an electrical breakdown). <S> The effect of one junction on another in a transistor leads to its useful application as a current-amplifying device. <S> Can I suggest that you pass feedback to the tutorial's author, if possible, so others don't get confused and misled. <A> And to paint the picture showing that the 'diodes' of the BJT can be biased in any direction simply by changing the polarity of the voltage across them. <S> Note the 'normal' way of setting up the BJT is (3) which is what the article seems to be alluding to. <S> Vbe forward, Vcb reverse but failed to give the whole picture. <S> Also you can't make a functioning BJT by simply connecting two diodes connected back to back - they need to be constructed in the same crystal. <A> so the same diode can be forward biased or can be reverse biased depending on the direction of the electric field (voltage). <S> at any given point, a diode is either forward conducting and reverse conducting (shutting off ideally). <S> whether it is forward biased or reverse biased will depend on the design.
| Biasing is an external stimulus not a property of the device.
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How can I determine the safe width of the PCB traces so that they withstand the momentary short circuit current before the circuit breaker trips? Many lamps are supplied by 220v and a circuit breaker through a PCB as in the picture. The total current of this circuit is 1A. For this implemintation I am going to use 2A circuit breaker. I am afraid that a short circuit could happen at the load and the traces of the PCB may explode. How can I determine the safe width of the PCB traces so that they withstand the momentary short circuit current before the circuit breaker trips? Any tips would be appreciated. <Q> A fuse might be more suitable. <S> SMD Fuse : <S> Current Rating <S> 1A , Voltage Rating - 350Vac , $0.18 @ <S> 500pc or a Polyfuse 1A <S> $0.40 (500pc) <S> 10A max radial <S> The whole idea of a fuse or a breaker or a Polyfuse to choose protection that reacts quicker than the fragile path. <S> got it? <S> Which was my 1st answer. <S> Ensure the track resistance is much much less than the SMT resistance, by design of the tracks and additional solder if necessary or busbar. <S> A 2A fuse is 41 mOhm cold. <S> Otherwise if you don't then the track becomes the fuse. <S> Beware of high resistance <S> The whole idea of a fuse or a breaker or a Polyfuse to choose protection that reacts quicker than the fragile path. <S> If a short thin track heats up faster than a breaker or a polyfuse, then the only choice is a fast blow SMT fuse. <S> got it? <S> Which was my 1st answer. <S> Ensure the track resistance is much much less than the SMT resistance, by design of the tracks and additional solder if necessary or busbar. <S> A 2A fuse is 41 mOhm cold. <S> Otherwise if you don't then the track becomes the fuse. <S> Beware of high resistance contacts. <A> A cube of copper 35 microns on a side (35 microns is the standard thickness of CU foil at 1ounce/foot^2) <S> has 0.0005 (500 microOhms) resistance. <S> I'll use a specific heat <S> I know <S> well ---- that of silicon, 2 picoJoules/ <S> cubic micron/ <S> deg <S> C ---- to run through the math. <S> In that 35micron cube, there are ~~ 40,000 cubes of size 1micron. <S> Thus the energy storage is 40,000 * 2pJ/micron^2 <S> /deg C or 80 nanoJoules/35microncube CU. <S> (again, I'm using silicon, not CU value). <S> What is the energy generated in that 35 micron cube, at 1 amp? <S> P = I^2*R, or 1*1 <S> *0.0005, or 500 microJoules/second. <S> Lets just divide 500 microJoules/second by 80 nanoJoules/deg C, and we find the dimensions are "degC/second" and the scale factor 500,000nJ/80nJ = 6,000. <S> Thus the rate of heating is 6,000 degree C/second. <S> Now you need to compute the specific heat of a cubic micron of copper, and tweak that 6,000 to a more accurate number. <S> edit <S> If you want 60 degree C rise, over 1 second, you need 100X the width I used. <S> That 35micron cube is 1.4 mils, or 0.0014 inches. <S> Thus 0.14 inches seems safe. <A> you could use a calculator like: http://www.saturnpcb.com/pcb_toolkit/ . <S> Go to the "Fusing Current" Tab. <S> Here you could insert the width of your track and how much time do you want the track to hold the short circuit current. <S> The short circuit current that the given track width can Sustain for the given time is calculated by the program. <S> you have to be sure that your fuse, CB can disconnect the current in Shorter than the inserted time in the program. <S> For example, say you have a track width of 3.1 mm <S> and you want that this track holds a short cicuit current for 1ms. <S> The program will tell you that the maximum short circuit current that this track can Sustain in 1ms is About 1kA. <S> Based on this result you have to make sure that your CB/Fuse can disconnect a short circuit current of 1kA in less than 1ms <S> (based on its datasheet). <S> so Long your track width is known, you can then Play with the time and get the maxumum allowable short circuit current at each time, then see if your ciruit breaker can handle These short circuit currents in less time than you insert in the calculator.
| If a short thin track heats up faster than a breaker or a polyfuse, then the only choice is a fast blow SMT fuse.
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Why is the diode forward voltage constant? When you have a diode with a certain barrier voltage (e.g., 0.7 V for Si) and you apply a voltage higher than this barrier potential, why does the voltage across the diode remain at 0.7V? I understand that the output voltage across the diode will increase as a sinusoidal input is applied until it reaches the 0.7 mark, I don't seem to understand why it remains constant after that point however. It makes sense to me that any potential greater than this barrier potential will allow current to pass, and correspondingly, the potential across the diode should be the applied voltage minus the 0.7 V. <Q> The voltage across the diode does not remain at about 0.7 V. When you increase the current, the forward voltage also increases (here: 1N400x): <S> And when you increase the current even further, the power dissipation becomes too large, and the diode eventually becomes a LED (light-emitting diode) and shortly afterwards a SED (smoke-emitting diode). <S> So a larger forward voltage cannot happen in practice. <A> Voltage is what we can observe and measure, but what is also changing is resistance. <S> A diode starts out as a large resistance, as you apply voltage to it that resistance remains fairly constant until you approach the forward breakdown voltage. <S> At that point the resistance starts to drop. <S> Past the knee the resistance is very low. <S> Any further increase after knee causes little change in the resistance. <S> Since R has gone down, in order to maintain that voltage you have to increase the current... <S> a lot. <S> The diode has become a small resistor "switch" and can therefor be referred to as ON. <S> The full voltage current relationship of a diode looks like this. <S> The slope before the knee is the forward off conductance (1/R), the slope past the knee is the forward ON conductance. <S> The actual math is of course a lot more complicated than that, but I find this description helps folks understand. <A> why does the voltage across the diode remain at 0.7V? <S> It doesn't. <S> Most of the time, a constant 0.7 V is good enough, just as flat earth is good enough for driving around town. <A> Diodes have a logarithmic relation between current through the diode and the voltage across the diode. <S> A ten:1 increase in current causes 0.058 volts increase across the diode. <S> (the 0.058 V depends on several parameters, but you can see that number in lots of on-chip-silicon bandgap voltage-references]. <S> What if the current changes 1,000:1, either increasing or decreasing? <S> You should expect to see (at least) 3 * 0.058 volts change in V diode . <S> What if the current changes 10,000:1? <S> Expect at least 4 * 0.058 volts. <S> At high currents (1 mA or higher), the bulk resistance of the silicon starts to affect the logarithmic behavior, and you get more of a straight line relation between I <S> diode and V diode . <S> The standard equation for this behavior involves "e", 2.718, <S> thusly $$Idiode = <S> Is * <S> [e^-(q*Vdiode/K*T*n) <S> - 1]$$and at room temperature and ideal doping profiles <S> (n=1)$$Idiode= Is * <S> [e^-Vdiode/0.026 -1]$$ <S> By the way, this same behavior exists for bipolar transistor emitter-base diodes. <S> Assuming 0.60000000 volts at 1 mA, at 1 µA, expect 3 * 0.058 <S> V = 0.174 V less. <S> At 1 nanoampere, expect 6 * 0.058 <S> V = 0.348 V less. <S> At 1 picoampere, expect <S> 9 * 0.058 volts = 0.522 <S> volts less (ending up with only 78 millivolts across the diode); perhaps this pure-log behavior ceases to be an accurate tool, near zero volts V diode . <S> Here is Vbe plot over 3 decades of Ic; we expect at least 3*0.058 volts or 0.174 volts; reality for this bipolar transistor is 0.23 volts. <A> As the other answers have explained, the voltage is not constant at 0.7V, but based on the reference to barrier potential in your question, I suppose you realize this and are asking more about the semiconductor physics behind why this happens. <S> As you apply forward voltage, the depletion region becomes smaller. <S> With low voltage the larger depletion region restricts most current, and as the voltage increases, the reduced depletion region results in a reduction in resistance (and therefore increased current). <S> This continues until approaching ~0.7V where the depletion region is very small as well as the resistance. <S> This causes the exponential V-I relationship. <S> This article has some good diagrams and explanations, as does the Wiki page . <A> The point is that you can't "apply a voltage higher than this barrier potential", the diode doesn't let you. <S> That is, the marginal impedance of the diode in conduction mode is less than the source impedance of your voltage supply: your voltage source can't drive more than "0.7V" across a 0.7V diode, so "the voltage across the diode remain[s] at 0.7V". <S> Of course, the marginal impedance of a diode in conduction mode is not exactly zero, so there will be some rise in voltage if your voltage supply attempts to supply more than zero current. <S> And the marginal impedance of your voltage supply may be very low, comparable to a diode, so it may be able to boost the diode voltage up quite high before the diode fails. <S> Those are the second-order effects. <S> The simple model of a diode, conducting above 0.7V, is a device that limits voltage by accepting infinite current. <A> Once diode is switched ON with sufficient biasing, it acts a voltage source of 0.7 or 0.6(depends on material) with a small series resistor. <S> So if we increase the input voltage, current across the small resistor will also increase. <S> So as input voltage increases there is variation across output taken across diode. <S> Usually diode is considered to be ideal, so there is no resistor in series. <S> So o/p voltage across diode remains constant.
| The reason is that the depletion region of a diode (with zero voltage applies) creates the barrier potential, as you already noted, of about 0.7V (assuming a typical Silicon diode).
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Why a sinusodial signal expressed by a cosine function instead of a sin function? In the signals and systems course one of the first things that we see is a sinusodial signal wave. And they say its expressed like \$A\cdot\cos(\omega t + \phi)\$. But why we use a cosine function when we work on a sinusodial signal? why its not \$A\cdot\sin(\omega t + \phi)\$? for example: http://users.abo.fi/htoivone/courses/sigbe/sp_sinusoidal.pdf or watch the first minute of this video: https://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/video-lectures/lecture-2-signals-and-systems-part-i/ <Q> A cosine and sine wave are essentially phase-shifted versions of one another. <S> For instance, \begin{equation*}cos(x+\phi) = sin(x+\phi-90) = <S> sin(x+\theta)\end{equation <S> *} <S> where \begin{equation*}\theta = <S> \phi-90\end{equation <S> *} <S> Hence, a cosine and sine wave are essentially the same, and what differentiates their use is the value of the initial phase. <S> However, a primary reason as to why the cosine notation is preferred is because of the frequent occurrence of complex envelopes in the area of Signals/Systems. <S> An example of the use of complex envelopes is when a sinusoid of low frequency, say m(t) , is modulated onto a sinusoid of higher frequency ( fc ); the resultant modulated signal r(t) can be expressed as: \begin{equation*}r(t) = <S> Re\{m(t)e^{j 2 <S> \pi f_{c} t} \}\end{equation <S> *} <A> Yes, the most common reason given is that a cosine is the real part of a complex exponential in Euler's formula, and thus implies Euler's dentity. <S> But why is a cosine the real part? <S> A cosine allows a signal with a phase and frequency of zero to have a non-zero DC value. <S> There are several other interesting aspects. <S> One is that the cosine is symmetric around the origin (0), and has a 1st derivative of zero. <S> In practical electronics, that means that time is reversible, and that a small phase error from a starting phase of zero will have the minimal effect on some circuit behaviors. <A> The general expression for an oscillating system relies on the relationship $$e^{j{\omega}t} = <S> cos(\omega t) <S> + j\text{ } sin(\omega t)$$ <S> The cosine term is the real part, while the sine term is the imaginary. <S> Physical measurements can only directly measure real quantities (barring special devices), so such systems are generally characterized by variables which are expressed as cosines.
| As it turns out, the real part of a complex envelope is a cosine waveform (Refer to Euler's formula) and hence it is much more convenient to represent a signal, such as r(t) , using a cosine.
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Problem in understanding bridge rectifier In a bridge rectifier (full wave) why do the electrons not flow from the other two diodes during a half cycle? Pls help me in understanding the concept. <Q> Does putting some voltages and voltage-drops across the diodes help? <S> Note that the diode will only allow current flow when it is forward biased, or else it acts as an opened (unconnected) element. <A> Your confusion is not uncommon. <S> Students often have a problem with this. <S> My suggestion is to replace the ac input with a dc input which you can swap in polarity. <S> Draw it with the dc battery supply one way round and trace the dc current flow, then do the same with the battery supply the other way round. <S> AC is just DC that "swaps over" regularly. <S> I find most of my students "get it" after this. <A> If the source is an AC source as given in your circuit, the two diodes that are off won't remain off forever. <S> As soon as the source start to be in its negative part of the cycle the ones that are off will turn on and the ones that are on will turn off. <S> Interestingly, the current flowing through the load never changes its direction. <S> That way you will have only positive currents even input voltages are negative. <S> I know this can get you more confused, but it's worth knowing it. <S> Good luck!!! <A> In a bridge rectifier (full wave) why do the electrons not flow from the other two diodes during a half cycle? <S> Pls help me in understanding the concept. <S> First you have to understand the fundamental operation of a diode: it only conducts in one direction. <S> Then when the signal goes negative, the other two diodes conduct while the first pair goes into a reverse bias state <S> (stops conducting).
| In a full wave rectifier, there are two diode that conduct when the ac signal goes in a positive direction and the other two diodes are reversed bias.
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Can you put transformers in series to get higher voltage? I was looking for the answer and found this related question: Wiring transformer output in series to get twice the voltage possible? But the answers don't seem to address what I was imagining. I am thinking of connecting the secondary coil of one transformer to the primary coil of the next, but the only information I can find talks about hooking the secondary of both together either in parallel or series. Am I making any sense? <Q> You may chain two transformers, and if both have a 1:10 ratio, then the combination has a 1:100 ratio. <S> But beware that each transformer winding has a maximum voltage that it can withstand. <S> If you exceed it (too much), you will experience an electric breakdown that will destroy your transformer. <S> For example, you cannot use a 220V-12V transformer, with the 12V winding connected to 220V mains, to produce 4kV. <A> Transformers have a maximum voltage that they can output before they suffer internal breakdown, adding a second transformer before the second one does not increase this maximum voltage but it does increase the primary-to-secondary insulation <S> (there are not many applications where this is needed). <S> Transformers that are not designed for HV generally have much lower breakdown limits that make them unusable much beyond their designed range. <S> It will reduce the power that you can draw from it, I'm not sure how much by though, it is better to just use a single HV transformer and <S> if that isn't enough then add a ladder of voltage doublers. <A> The problem is that a transformer that has a primary winding rated for a voltages that is significantly higher than mains voltage and a secondary rated for a significantly higher voltage that that is usually available only for a power transmission line power level. <S> A transformer's voltage rating can not be exceeded by more that 10 of 15 percent without overheating due to magnetic saturation and the resulting increase in magnetizing current. <S> Insulation breakdown voltages generally have a higher safety margin that that. <S> So the problem will be saturation, not breakdown.
| If the primary of the second transformer is rated for the secondary voltage of the first transformer, there is no problem with connecting them together.
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why does high speed current follows the path of least inductance? and how are the shape of lines are different at low and high speed? Does anybody know why the lines of electromagnetic field at high speed are bent and exactly following the shape of line, while in the low speed device the electromagnetic lines are in the usual propagation shape. <Q> This answer is a simplification, but is the best description I have seen. <S> First of all, "return current follows the path of least resistance" is actually a bit misleading. <S> The current returns by any and all means that it can . <S> The path of "least resistance", or more accurately "path of least impedance", simply carries more current than any other path. <S> However that may not be the majority of the current. <S> That is, the path of least resistance may carry 25% of the current but all the other paths may carry 75% in total. <S> This is actually true regardless of the signal frequency. <S> However, each wire can be represented by the image below. <S> Each point along the wire has a capacitance to the ground plane (Or to ground or whatever else is in the vicinity.) <S> As you know, with a DC signal those capacitors have no effect and can be ignored. <S> However, lets imagine a rising edge on a signal running down this line. <S> As it passes each capacitor the voltage on the bottom end of the capacitor briefly jumps up to some fraction of the signal voltage. <S> Electrons are immediately drawn into the capacitor from the plane to charge the capacitor. <S> When the signal reverses, the opposite is true. <S> These capacitances cause signal loss of course, but also has the effect of creating a potential well, or gulley in the plane which effectively becomes the new path of least impedance. <A> 1) those are NOT the electromagnetic field lines. <S> Read the text, the lines show the current path meaning, the path which the returning current will take to close the current loop. <S> 2) low frequency and DC currents just take the path of lowest resistance. <S> They cannot take shortcuts by capacitive coupling nor are low frequency and DC signals affected by any inductance. <S> 3) high frequency signals follow the path of lowest impedance , this can be a different path than the lowest resistance path (it <S> can be different, it does not have to be). <S> High frequency signals can couple between two conductors as two isolated conductors behave as a capacitor . <A> As other answer have mentioned, those lines are not the electromagnetic fields lines but rather current density lines for the returning path of the current. <S> Generalized current will return through the path of least impedance. <S> In the DC cases the path of least impedance tends to be the least resistive path which is usually the shortest distance path <S> In high frequency cases you the current will follow the path again of least impedance which is the path that creates the smallest amount of impedance. <S> For inductive looking paths this path tends to be the path that creates the loop with the smallest area. <S> So for example, in the picture you have shown even the notice <S> the in the low speed case <S> even though there returning path in shorter the area of the loop is bigger than the high speed case and therefore more inductive. <S> At high frequency the shortest return path will have more impedance than going the long way around of following the resistor shape. <S> Additionally , in high frequency cases you will have capacitive coupling which would find shortcuts for the current in places that have a capacitance (actual capacitor like decoupling or parasitics) which will offer the path of least impedance. <A> Nature seeks to minimize the energy required. <S> At DC, all possible paths get explored. <S> At approximately 50,000 Hertz, even as all possible paths still get explored, the inductance and the ability of large enclosed areas to store energy in that inductance, and the need for physical systems to minimize the required energy, causes all of that concurrent exploration to prefer paths with minimal enclosed area. <S> The non-preferred (high stored energy) paths are still explored but minimal energy is stored thus minimal interference occurs. <S> Thus coax cables begin to be effective in herding the electrons, at 50KHz. <A> Another way to look at this is as follows: <S> At high frequencies, ground plane impedance( Z ≈ jwL ) and from the definition of loop inductance, the total loop inductance for two conductors with opposite directions is: Lt = <S> L1 + L2 - 2 <S> M <S> Where L1 and L2 are the partial self-inductance of the two conductors and <S> M is the partial mutual-inductance between them. <S> If the two conductors are identical then: <S> Lt = <S> 2(L - M) <S> Then, for the return current path to have the least inductance , it will try to have M as high as possible. <S> As a result of that, the return current path will try to follow and be very close to the signal trace as possible, hence, more M and less Lt will get.
| The reason for following the shortest path is that is will have the lowest inductance and therefore also the lowest impedance .
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Why does shorting/putting heavy load on a motor slow it/stop it? I am doing a simple circuit which includes using a motor in theory (I am using a stepper-motor) as a break. One of the random fun-facts that I have known for a long time is how shorting a stepper motor (or putting very high load ) will cause high current and slower rotation (or sometimes stopping rotations all together). However I do not understand how it works from an electrical/electromagnetic point of view. Where does that force come from? <Q> When a motor is free-wheeling it turns into a generator. <S> By shorting out the motor you are applying a LARGE electrical load to the motor. <S> If you want the generator to keep turning you need to apply a large torque. <S> Since all you have driving it is the inertia of the motor and whatever it was driving, this torque turns into a braking force. <S> You can think of this another way. <S> Suppose you hook up a generator to a handle and wind the handle with nothing connected to the generator. <S> The generator will turn quite easily. <S> Now attach an electrical load to the generator. <S> You will now find it is harder to turn the handle at the same speed. <S> Note: YOU have to supply the power to run the load with your arm. <S> When you short out the generator, you are affectively applying a VERY LARGE LOAD. <S> Turning it by hand becomes REALLY difficult. <S> You just built a brake. <A> Two concepts: conservation of energy, and reverse EMF (electro-motive force). <S> When you move a magnet past a suitably oriented inductor coil connected to a load, the changing magnetic field induces a current in the inductor. <S> The current in the inductor creates its own magnetic field. <S> That newly generated magnetic field is oriented such that it produces a force that tries to accelerate or eject the magnet in approximately the opposite direction of its present motion. <S> The kinetic energy absorbed from the magnet as it slows down will be proportional to the energy that heats the load on the inductor (and other system losses), thus conserving it. <S> Conversely, if you open the circuit, there is no place for current to flow, and thus no less reverse or back EMF produced. <A> All motors can operate as generators under the right conditions. <S> Shorting the terminals is one of those conditions for some motors. <S> A generator applies torque to whatever is turning it and mechanical energy from the source, rotating inertia in this case, is converted to electrical energy.
| If the motor is shorted, the energy is dissipated in whatever conductors are carrying the current including the motor windings.
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Why is the power factor always 0.8? If reactive power means losses, why don't we correct the power factor and make it 1? For example, on marine vessels that I have been working on, the power factor is always 0.8. We could add capacitors in parallel, or over-excite a generator. Why this is never the case? <Q> Power factor is not corrected to 1 because power the power factor is likely to change with changing load. <S> With correction to 1, the power factor could become leading and that is often more difficult for the source to accommodate. <S> It could cause the voltage to rise. <S> In your marine vessel example, the engine generator set may be designed for 0.8 pf. <A> Power factors are often "corrected." <S> Typically, line loads from electric companies are inductive loads (transformers, high power industrial motors, etc). <S> So the solution is to have some giant capacitor, which is not terribly uncommon. <S> Why don't we correct everything? <S> It's not ever possible, and usually it's not practical. <S> Inductance is not constant. <S> E.g. night time, there is less power going to industrial machines so the inductive load may go down. <S> If you correct the P.F. for these inductive loads, your P.F. will change again when those loads are turned off. <A> Large industrial main loads are required or incentivized by the utility company to reach a minimum power factor by installing capacitors or active PFC. <S> Active PFC is starting to show up in consumer electrics, as well, to improve the situation from the buttom. <S> Some "green" certifications, such as certain Energy Star certifications, specificy a minimum PF. <S> You can find off the shelf LED bulbs that feature active PFC in order to reduce the harmonic currents that are generally introduced by rectifiers. <S> For more information, see the following links: https://en.wikipedia.org/wiki/Power_factor#Importance_of_power_factor_in_distribution_systems <S> http://www.eaton.com/ecm/groups/public/@pub/@electrical/documents/content/sa02607001e.pdf <S> http://powerelectronics.com/power-management/power-factor-correction-justified-home <A> You list two alternatives: over excite generator. <S> power factor correction. <S> We could over-excite the generator and supply leading reactive power to cancel out lagging reactive power and make power factor 1. <S> But you have done nothing, since the genset still supplies the same apparent power. <S> No gain. <S> Power factor correction is viable, but is costly. <S> It can be done for total or on a load by load basis. <S> To date, ship owners have been reluctant to make this investment, because if you compare cost of running ships' auxiliaries to cost of propulsion, the extra cost of running with lagging power factors is negligible. <S> I say to date because with EEDI (Energy Efficient Design Index), designers of new ships have to achieve 10% (20% and eventually 30%) reduction in \$CO_2\$ emissions. <S> A portion of this (5% to 8% at 30%) would have to come from the electrical system, so things like power factor correction are viable. <S> But even there, there are better options. <S> So it is an option, but I do not see it being used in many vessels. <S> On land, over-exciting synchronous motors and power factor correction are done to reduce power factors to reduce a facilities power factor, so power companies do not charge a surcharge for each kWh consumed. <S> At sea, there is no benefit to improve the power factor since the vessel is the producer and consumer of power. <A> I have not heard of an over excited generator!Generator excitation controls the output voltage and is set to give the required output voltage. <S> Most large ships have an over excited synchronous motor driving a forced draft fan, which every large engine room will have. <S> These are switched on during commissioning of the vessel and remain on throughout the life of the vessel. <S> Being synchronous they are used for power factor correction and have been in use for decades.
| In fact, power factor is corrected in many cases to reduce line currents and the associated distribution losses. If the engine is sized to produce a certain KW and the generator is designed to produce a KVA that is 20% more than that, there is not much to be gained by adjusting to anything higher than 0.8.
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How to switch DMM between 10 different batteries? I'm trying to figure a way to switch a DPM from battery to battery in a 10 battery series string. At first I thought I'd do it manually using a small SMD 2 pole 10 position rotary switch but the only ones I've found at a reasonable price & size are coded (BCD). Then I thought it would be better to automatically scan. I bet there is a device out in chip land that could do the deed. I would prefer to not go smart (i.e. PIC) or relays if possible. Something simple, small & cheap. Discrete FET switches would take up a fair amount of physical space which I don't have. Is there a 20 FET ASIC? What is a circuit I could use to switch between multiple batteries? <Q> There are all kinds of solutions to automate this. <S> Two spring to mind. <S> If your accuracy is simply .1V <S> You could use linear optical isolator across each battery to translate the voltage on each down to the right level to feed via a multiplexer or slide switch into a meter or again into a micro/display combination. <S> Or if you only care about battery volts <10V or some such, maybe just a led indicating that for each battery. <S> Maybe even 10 bar-graphs of leds.... <S> However, since you mention trays of batteries, and unless you want to collect the data in a computer or something, the cheapest (counting your time), quickest, and least effort solution would probably be to just buy 10 cheap volt meters. <A> Three hurdles are presented here: the voltages are high (over 100 VDC), and potentiallyhazardous the display only gives a voltage, but that information is useless unless you also know which taps are being probed (i.e. which 12V section) the 'DPM' presumably is a low-voltage voltmeter, which requires its own logic and bias power and some kind of ground reference, and attenuation of the input voltages. <S> There are solid-state high voltage switches that can select one tap of theeleven in your series, <S> see CPC7601 , and which might be powered from theHV string itself. <S> The DPM can be replaced with a microprocessor andsmall display, which can give both a digitized voltage and indication ofthe battery section. <S> And, there are <S> operational amplifier circuitsthat can buffer the battery voltages down to manageable levels so thatthe microprocessor can handle the input. <S> It will take three or more ICs and a display, and some programming, to make thishappen. <S> It would be less expensive to label test points, and poke themwith voltmeter probes, of course, but that exposes the terminalsand is a shock hazard. <S> A mechanical switch, while appealing, is going to require mechanical mounting and a suitable enclosure; it putsones fingers near those high voltages. <A> Thanks for all the input. <S> I'm thinking my initial approach isn't practical due to availability/cost of the switch (mechanical or electrical). <S> What I'm trying to do is connect a battery health DPM (from China - where else?) <S> to individual 12v AGM batteries in large UPS systems. <S> There are 8-10 batteries in an enclosed tray with 4 to 32+ trays. <S> Currently, the health of individual batteries can't be determined online. <S> I'm thinking of a device that could mount on the front of each tray and scan each battery. <S> The DPM will show its health. <S> How to scan is the problem. <S> Below is a concept I cobbled. <S> I realize the 74HC4067 will let its smoke out with the voltage involved. <S> The CPC7601 is too much $. <S> Maybe I'll be ale to fit some SOT-23 packages in the space I have to work with.
| then you could use 10 resister voltage dividers going from the top of each battery to ground and multiplex that into a micro, measure the voltage sequentially on the top of each battery and do the math there and show the results on some suitable display.
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Altium Designer Rule: Between Silkscreen and Via I have not find a solution about between top overlay or bottom overlay and vias. Can you help me about this . I need identify a rule.I don't want the silkscreen to touch a via <Q> The following worked for me on Altium 17: <S> It did not work when the minimum clearance was 0 mil... <S> likely a bug. <A> If this is the case, the answer is pretty simple, simply set a rule in the "SilkToSolderMaskClearance" option. <S> See image below. <S> This is found under the manufacturing section of the rule list. <S> Then you can select to create a new rule, and you can set a custom query under the "Where The First Object Matches" to "IsVia" - you can leave this as "All" if you want the rule to apply to all exposed copper regions, and then "Where The Second Object Matches" -> <S> Layer -> Top Overlay. <S> You can then select whether the clearance is from the exposed copper, or the Solder Mask opening and set the distance that you require. <S> If it is for a straight overlap, set this to 0 <S> and then it will only detect if it overlapping, not if it is close to. <S> This will save you having to tent the vias if it is not something you want to do - I think some fab houses can get a bit funny about doing it as well. <S> Or at least, in my experience. <A> If you don't want the silkscreen to touch a via, you can use the following: Where first object matches: OnLayer('TOP OVERLAY') <S> Where second object matches: IsVIa <S> Then set your clearance rule. <S> EDIT: <S> I just tried this at work and to my surprise <S> it didn't work for me either. <S> I also tried queries such as IsDesignator , and even played around with the Silk To Soldermask rules and had no success. <S> This seems like it may be an Altium bug. <S> I suggest you call their customer support and see if they can give you an answer that works. <S> Otherwise I recommend doing what I suggested in the comments, and tent your vias with soldermask. <S> Then it won't matter if the soldermask passes over the vias or not, they should still be visible. <S> Most manufacturers can tent vias with up to a 12 mil (0.3mm) hole diameter.
| For this, if you do not wish to tent your vias, it is safe to assume that there will be a solder mask layer present on the vias and any other pad for that matter which I also assume you will not want silkscreen to cover.
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Resistor or low impedance path between common ground point and mains earth? A simple EPA ( ESD Protected Area ) consists of 3 things: wrist strap antistatic mat common point ground Wrist strap and antistatic mat both should grounded to the common point ground. ANSI/ESD S6.1 recommends a non-resistor ground cord [...] to ground worksurfaces. However, the cord may have a 1 megohm resistor for non ESD purposes. You might think: just put a resistor between Common Ground and mains earth... ANSI/ESD S6.1 , section 6.4.2 recommends: The resistance [...] from the common point ground to the AC equipment ground shall not be greater than 1 ohm. Both plans below don't follow ANSI/ESD S6.1, because i think the path between antistatic mat and mains earth should NOT be a low impedance path . And if there should be a resistor between an dissipative surface (like an antistatic mat) and mains earth for user safety, then there should be a resistor between a conductive surface (unpainted inside of a PC case) and mains earth too. In theory, is one plan more preferable over the other? Plan A: only 1MΩ between wrist strap and common ground.Thus discharging takes less time. Therefore (maybe?) less chance of a potential difference between wrist strap and the PC (with ESD sensitive hardware inside) and less chance of an ESD event. Possible disadvantage of plan A: because the user is constantly charging himself (by moving) and discharging to the common ground, the common ground will be bouncing more than it would in plan B. (Please correct me if i'm wrong.) Both plans are easy to execute. Plan A Plan B The yellow marked resistors are inside the grounding plug. <Q> If by plan B you mean to add a resistor to the mains ground, then no plan B is not an option. <S> If you add a resistor in series you will have two problems: 1) <S> It affects the saftey of the earth mains, if you size the resistor wrong in the event of a fault you could have an issue with the resistor blowing out or creating common mode voltage in the case 2) <S> You create common mode voltages and cause the whole chassis ground to bounce depending on the amount of current flowing throw chassis ground which could cause problems for the supply and circuitry (if your ground bounces up and down by 10's of mV it might create digital noise, which could be possible depending on the resistor on earth ground.) <S> You would also increase the potential between the PC and other devices creating noise problems. <S> You can put resistances in series with the wrist strap and the anti static mat, but not the PC. <S> Anti static mats ( <S> all the mats I've seen already have MΩ's of resistance in the mat material itself. <S> Most wrist straps also have a 1MegΩ resistor built in. <S> If your not playing around with anything over 30V you probably don't need to worry about any of this anyway. <S> Keep the mains ground as low resistance as possible as required, and put resistors (that are probably already there, you can tell with a good DMM (voltmeter)) in between the mat and ground and the strap and ground. <S> Shown here: <S> If look at this question <S> the diagram would look like this, the grounds would be connected in the building wiring: <S> In the above image the wrist strap is connected to the mat, if it's the kind with a 1MΩ resistor in it, it doesn't matter where you connect it, if you connect it to the chassis (shown as wrist strap1) , it still gets grounded. <S> If you connect the wrist strap to the mat it gets grounded, and if you connected it to the mats ground (shown as wrist strap2) it still gets connected, each time with at least 1MΩ to ground or any other potential that might affect those who might grab onto earth mains. <S> If you take a 1MΩ resistor and poke it into an electrical outlet it's not going to kill you, but a fork might. <A> The reason to ground the PC case has nothing to do with ESD: it's a measure against PC case going live. <S> If the case is grounded, a live wire touching the case will create a prompt short circuit, which will (hopefully) trigger the circuit breaker and instantly remove the power. <S> Grounding yourself via wrist strap has nothing to do with your safety: it actually makes things worse for you. <S> If you touch a live wire while not wearing the strap, the current you'll get will be limited by your body's capacitance and leakage via your shoes etc. <S> Should that happen while you're wearing a grounded wrist strap (without a resistor), that current will only be limited by your body's resistance, which is not that big at high voltage. <S> That's why you need a resistance in series with your strap. <S> Now, it is true that your PC case which is grounded without any resistance remains a shock hazard: if you manage to touch it with one hand, and a live wire with the other one, you'll get a bad shock. <S> But this is quite improbable and requires you to do something very bad (touch a live wire) in the first place. <S> A case going live, on the other hand, is not an impossible event: a simple bump on the side may deform it hard enough to touch something on the power supply PCB. <S> Should that happen, you'll have that same very bad stuff right in front of you and without any warning: a 1MOhm resistor can withstand 230V indefinitely. <S> Edit: after the exchange in comments it became clear that you don't need protective Earth. <S> In that case, I would go for Plan A, as it has more wiring behind the resistor and less wiring connected directly to Earth. <S> Those resistors are installed for your safety as explained above, so it's clearly better to have more touch-safe wiring on your desk. <A> Without using the mat case: There is no meaning for adding the R2 resistor. <S> It is just going to make the discharge slower <S> and it does not add any additional protection function. <S> When using the mat case: <S> According to that the discharge resistors of the mat and the wrist strap are already built in, the point of discussion should be around the PC's discharge resistor connected to the ground. <S> Refereing to plan B, in case you have the PC on your mat (which is the normal thing to do) <S> the PC is already connected to ground through the mat and you dont have to add the discharge resistor between the PC and Ground. <S> Look at the following figure. <S> Additional info: <S> The equivalent circuit of the setup in this video with roughly estimated values of the resistors is as following:
| If your PC is not on the mat you should not touch the PC's hardware unless the PC is grounded which leads us to the first case.
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CAN bus communication mechanism Does every node on a CAN bus just keep transmitting data no matter what or does it only transmit data when it is "requested" to by other nodes on the same bus? <Q> First of all the basic CAN transfer mechanism is: Any node can initiate a transfer. <S> If a node wants to transmit and the bus is busy it waits until the end of the current packet. <S> If two nodes start to transmit at the same time then the message with the lower ID will take priority, the message with the higher ID will abort, wait until the bus is idle and then try again. <S> Most nodes will be configured to generate an ACK pulse at the end of every valid packet they receive. <S> This doesn't mean that they will do anything with the packet, just that they read a valid packet and checksum. <S> Multiple nodes can all generate an ACK for the same packet without any issues (normally every device on the bus will). <S> If the transmitter doesn't see any ACK pulse it will indicate to <S> it's controller that there was a transmission error. <S> All the CAN bus requires are that the bus has at least two nodes, the correct bus termination, and that the nodes are configured such that at least one node acknowledge every packet. <S> There are other standard protocols that can be used on top of CAN (e.g. ISO 15765-2 ) to control message transmission in a request/reply structure but they are entirely optional. <S> Whether it is best to just blast messages out, use a standard protocol, use your own protocol, or a mixture of the above depends entirely upon the application in question. <A> Generally, Nodes are configured to broadcast on a time period that makes sense for that particular node. <S> Receivers are configured to listen for the messages they're interested in. <S> There is nothing stopping you from building a request/reply type node, but that is not a good way of scaling up messaging systems. <S> (This requires two transmissions for 1 piece of data: One request, one reply). <A> Every node can independently transmit data i.e. no master permission or request is required. <S> To prevent any crash on the bus, every node first "listens" the bus then transmits the data if the bus is ready. <S> If there's a transmitter then there should be a receiver, right? <S> One can request some info and the other can respond, or one node can periodically pump a specific data and the other(s) can process it. <S> One good example is a car: <S> A node placed at a door (say, front right) can transmit a data containing the status of the door (open or closed, window status etc) at a rate of, say, 10 times per second so that the main (maybe a master) node knows it and warns the driver with a message on display and/or a sound.
| There is nothing about a CAN bus that requires the system to keep transmitting no matter what or to only transmit when requested, that is entirely up to the software controlling the messages.
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How to get a high precision sine wave not available from standard crystal? I am considering for a RFID transmitter, which just send power with no data. However, my frequency is not standard 13.56MHz not 27.12MHz, it is 27.095MHz. Can I use the block diagram shown above? If yes, where can I get the 27.095MHz signal source? I have two ideas now, first use a MCU and a PLL chip to derive the accurate frequency. Second, use the programmable crystal oscillator(which is also based on pll, with factory programmed) like SG8101 from epson. However, the output from SG8101 is a square wave, so I have to low pass it before it is send to amplify? <Q> How to get a high precision sine wave not available from standard crystal? <S> Inevitably, when transmitting power across a gap to an RFID device, to obtain best efficiency it is likely that you will make the transmit coil resonant (using a tuning capacitor) and similarly, you will make the RFID coil resonant also using a tuning capacitor. <S> The effect of "tuning" means that if you fed a square wave to the drive coil it would largely transmit a sine wave so, the process of producing a magnetic field will convert your square wave into a sine wave. <S> There may be some other reasons to use extra filtering to improve the sine purity of course. <S> So, if you can make a square wave oscillator that runs at 27.095 MHz then your coil and tuning capacitor design should produce a reasonable sine wave. <S> How pure this sine wave must be is down to you. <A> Filtering the output of the digital programmable oscillator you have to obtain a sine wave would be simple and accurate reliable way. <S> I have not used your device but have used the DSC8001 programmable MEMS oscillator, which was low cost and accurate to +-10 ppm across full voltage and temperature. <S> For the filtering, have a look at www.ti.com/lit/an/sbfa003/sbfa003.pdf <S> , Burr-Brown app' note 'Simple Filter Turns Square Waves Into Sine Waves'. <A> To expand on Tony's and Andy's answers <S> : Let's do a prime factorization of your frequency: $$ 27095000\,\text{Hz} = <S> 5419 <S> \cdot 2^3 <S> \cdot 5^ <S> 4 <S> \,\text{Hz}$$ <S> Square waves have every odd multiple of their fundamental frequency as a spectral component. <S> Now, we only have two different odd prime factors here, and we surely won't be using the 5419th harmonic (since that will practically not exist), so what you'd most likely do if digitally generating this is taking the fifth harmonic of a $$ 5419 <S> \cdot 2 <S> ^3 <S> \cdot 5^3 \,\text{Hz} = <S> 5419 \,\text{kHz} <S> $$ square wave. <S> However, it's slightly questionable <S> you'll find an oscillator that will readily run at a multiple of 5419 Hz, so it's not like you could simply run counter on a microcontroller. <S> From the top of my head, 5419 is really an awkward prime factor, since it's far from any power of 10 or 2. <S> You're mostly stuck with frequency synthesis by other means. <S> fractional-N synthesis with a wide range of factors. <S> Again, it's pretty possible <S> you'll never be able to perfectly hit your target frequency going from a common nominal oscillator frequency, but you just need to get <S> close - nothing is perfect in this world, and an RFID system will never need atomic clock precision, or even assume a comparable short-term stability. <S> Another option: generate a \$5419\cdot 5^ <S> 3\,\text{Hz}\$ sine (that's ca. <S> 677.3 kHz) in software, by using sin/cos (or, depending on a lot of things, CORDICs etc) mathematically. <S> That would mean that the period of the digitally calculated tone wouldn't have to be a (small) multiple of the clock period of your MCU. <S> Convert to analog using some DAC (you'll need one with at least ca. <S> 1.5 MHz sampling rate) and get the 8th overtone of that; frequency doublers can be relatively simple to build (coupling C -> diode -> simple filter). <S> I think the most important takeaway from your question is, however: <S> In technologically "non-trivial" situations like these, it becomes very necessary to know one's requirements well. <S> You didn't say what kind of frequency precision you need! <S> I'd personally assume that something like 25ppm wouldn't be any problem, and then you'd just use a slightly different frequency with much much nicer prime factors. <A> Firstly, why the fascination with sine waves? <S> You have a very narrow band transmit coil with lots of Q if you want reasonable efficiency and you are not doing any tricky modulation, so I would be thinking in terms of a class E power stage which will be 80+% efficient and will typically need square wave drive at twice the transmit frequency (But only cmos levels)... <S> A SILabs osc, 74LHC74 as the phase splitter <S> (Use the other half to latch the thing off in the event of excess reflected power), two butch low side gate drivers, a couple of mosfets (IXYS for the fets and gate drivers would be my suggestion), some assorted R, L and C, and a hunk of 43 mix ferrite for the output transformer, job done. <S> DC rail for the power stage would probably be designed to be about 100V or so. <S> Enhancement mode GaN might be worth a look, but that is a whole other learning curve. <S> Either the SILabs part or a synth will be fine, DDS feels like overkill. <S> The biggest challenge in such a thing is the mosfet drivers and the cap that ensures zero voltage switching (Lots of fast current in that part). <S> RF power sources like this are NOT usually designed the same way you would a transmitter with modulation, they are far more like fast switch mode power converters.
| Easiest, and probably cleanest, would be having a clock synth IC that supports
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Driving a 5A current through a coil To many electronics engineers this might seem a stupid question but I was given by my boss the task to find a solution to drive approx. 5A through a coil.The coil's resistance (not considering inductance which would be ~15mH) is about 90 Ohms. The signal is sinusoidal with a frequency of about 3kHz. Since he made this sound pretty easy I am a little bit unsure of how to proceed. Considering Ohm's law, driving 5A through a 90 Ohms would result in 450V which doesn't seem so trivial to me. I have a feeling that I'm overlooking some fact but I cannot figure out which one. A possible idea would be to use a OPA549 which gets at least roughly into that area (bandwidth, output capability) but again, I think that I'm overlooking something. All power amplifiers I can find are intended for audio applications and typically drive a 4 Ohm or 8 Ohm load. I'd be really happy if you could tell me where I'm wrong or if this actually isn't such an easy task. <Q> The inductance of the coil at 15mH gives you a reactive impedance of about 280 johms at 3kHz. <S> That's a larger impedance than your resistance, so will dominate the voltage you need across the coil, 5A <S> * 280ohms = 1400 volts, before you add the extra voltage for the resistance. <S> Assuming the 5A is 5A rms, you will be dissipating <S> \$I^2R\$ = <S> 2250 watts in the resistance of the coil, no small amount, and way out of the league of anything like OPA549. <S> I would suggest that you reduce the VA you need to drive the coil by cancelling the series inductance with a series capacitor. <S> To resonate 15mH at 3kHz needs a capacitor of 180nF. <S> You will still need to supply the full 5A, but only supply the 450v needed to drive it through the resistance. <S> The alternative parallel tuned circuit connection ideally needs a current drive (or an inductor) and must supply the full 1500v resonant voltage, but at a lower current. <S> Obviously, the series connection is easier on two counts. <S> One option is to buy a 3kW audio amplifier, and use a transformer to match its output drive to the requirements of the tuned circuit. <S> Obviously this transformer will need to handle 2.25kW, but at 3kHz, you will be able to use a much smaller core than an equivalent rating mains transformer. <S> At 3kHz, it cannot be a conventionally iron-cored mains transformer, you will need to use ferrite. <S> At that low frequency, ferrite heating losses will be low, so you will be able to run up near the saturation field. <S> Another option that's just within reach is to use a 450v H bridge. <S> Here, although the voltage supplied to the resonant circuit will be a square wave, the current flowing, and the voltage across the coil, will look very sine-like. <S> If you can tolerate the waveform distortion, and engineer the high voltage H bridge, then this will be cheaper than a 3kW amplifier and a transformer. <A> It's not an easy task. <S> The voltage across the coil (driven DC, ignoring inductance) will be 450V, as you have calculated. <S> The power dissipated in the coil will be \$450\mathrm{V}*5\mathrm{A}=2.25\mathrm{kW}\$. <S> When you add the inductance of the coil in, things get even worse. <S> The impedance of the coil is \$ <S> z <S> = <S> j\omega L \approx 280\Omega\$. <S> To drive 5A through it requires 1400V and 7kVA. <S> The combination of inductance and resistance comes to about 1500V and 7.4kVA <A> The daunting 1500 VAC that Jack B stated as a requirement could come from the secondary of a ferrite transformer .This <S> ferrite transformer does not need to have an air gap .It is possible to use powdered iron or some tape wound arrangement because of the low frequency .A <S> toroid could also work <S> .Your choice of transformer should be dictated by what you have on hand and what you are familiar with .if <S> you tune the 1500VAC transformer secondary with a combination of quality film caps that total the 180 nF that Neil UK called for .It would be difficult to find a single cap that would do the ripple current and the voltage .The <S> use of multiple caps is normal for induction heating applications .Now <S> with your caps the rating needed for the proposed transformer is not much more than the 2.25 KW that Jack B stated .This <S> is much better than 7.4 KW .Remember that the caps handle the reactive power. <S> Now build say a ZVS royer osc with cheap 30 amp IGBTs on the proposed primary .Make <S> the primary some convenient voltage to suit Bucked down rectified mains <S> .The system should run close to 3KHz when you do a low power test on a lab supply ramping up the input volts to say 50VDC .You <S> can fine tune it with your caps .
| Finally, you need a voltage source.
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Efficiency of obtaining differential signalling from a single-ended manufactured transducers This will be a conceptual question. I'm sometimes dealing with data-acquisition of transducers, like strain gauges, accelerometers ect. sort of sensors.Most of these sensors have their own precision amplifiers. So what I mean by the transducer output is the amplified sensor signal.These signals then go to data-acquisition's input amplifier which is simply a differential amplifier ect.But most of the time the transducer outputs are single ended. Sometimes I encounter all sorts of noise, common mode noise ect. Since differential signalling is more immune to noise, I thought about converting a single ended signalling to a differential signalling as below(I want to implement Figure 2): So here are my questions. 1-) Some transducers are manufactured and sold as differential signalling transducers. So they are ready to be wired to a differential amplifier.But if one has a transducer and want to use it as differential signalling as in my Figure 2, would that be a wrong treatment?Im asking because If I invert the signal myself to obtain a differential signalling as in Figure 2, then I might introduce noise to the inverted input by interacting it with the inverting opAmp circuit,and that will not be common on both signals. So my first question is: is it a common practice to convert single ended signalling to differential ended signalling(in the aim of noise immunity) where the transducer was actually designed for single ended signalling? 2-) If this method makes sense. Here is the typical inverting opAmp configuration: I would choose R1 and R2 10k. How does the input impedance of the data-acquisition's differential amplifier have affect on choosing R1 and R2 here?I want the inverting as precise as possible. Is there an opAmp category for that, an example would be great? I dont want use LM741 for instance. <Q> Since differential signalling is more immune to noise <S> Any signalling is susceptible to noise - it's how your receive amplifier handles those received signals that determines how much immunity can be acquired. <S> However, you can have a perfect differential amplifier attached to a single ended source (via a properly balanced cable) <S> that has problems. <S> If the output impedance of the hot wire is several tens of ohms compared to the impedance of the 0 volt transmit reference you have what is known as "earth impedance imbalance ". <S> Note that I said imbalance . <S> If noise comes along and "hits" the cable, it will develop a larger signal on the hot output than that developed on the 0 volt reference signal. <S> Here's what I mean for a good scenario: - The signal source is "perfect" in that it presents the same low impedance for hot wire as 0 volt reference. <S> Clearly, if any noise comes along then it hits both wires in the cable and, because both wires have equal impedance balance to ground, the noise received by the diff amp is equal and can be quite easily cancelled. <S> If the signal source has an output impedance that isn't zero <S> then there could be a problem that can be overcome by this: - Now, the impedances are largely the same - the added resistors are chosen to be identical and "swamp" the difference in impedance between hot wire and 0 volt reference. <S> Earth impedance balance will be good and noise will be the same on both received wires (providing your input amplifier has good input earth impedance balance as well). <S> Adding an inverting stage can make things worse - keep the earth impedance balance at the sending end good and you minimize problems without adding an amplifier. <S> Of course, in extreme circumstances you have to transmit a bigger signal and this can be done (carefully) with a balanced buffer. <S> To keep "balance" (the same for both signals) use an inverting amplifier and a non-inverting amplifier - this largely ensures that the impedance at high frequencies will be equal. <S> You cannot achieve this using the "original" signal and a buffer amplifier because you have no way of controlling the impedances relative to each other. <S> If it works it's just luck <S> and that's not good engineering. <A> The important thing about balanced lines for interference rejection is that the source impedance is matched, not that the voltage be differential. <S> Due to this fact you can do nearly as well as an active differential stage simply by matching the impedance to ground between the two legs at the transducer. <S> A resistor connecting the -a leg to ground at the transducer, selected to match the output impedance of the sensor will cause the differential input amplifier to reject significant interference (And does not need any power at the transducer). <S> The way to think about this stuff is to see it as a Wheatstone bridge where the exciter voltage is the noise and the signal voltage <S> is series injected either into one leg or in anti phase in both legs, from which we can see that providing the resistances are matched <S> you get cancellation even if the wanted signal is only driven onto one leg. <S> Incidentally, high common mode impedance at the receiver will substantially reduce the impact of small mismatches in source impedance. <S> If you are working in audio sorts of bandwidths, THAT CORP have an excellent (patented) bootstrapped receiver that is really very good. <S> http://www.thatcorp.com/datashts/THAT_1200-Series_Datasheet.pdf <S> I would also suggest that reading a paper on this by Bill Whitlock of Jensen transformer may be interesting. <S> https://sound-au.com/articles/balanced-interfaces.pdf <A> I do not think you should unbalance it like that. <S> Whether you really need to is answered by others here. <S> If you are doing it yourself, you should use a single to differential converter like a LT6350 <S> However, you have to get the power supplies over there and filter them well . <S> Also check your signal range is within the range of the device. <S> If this is an industrial application, there may be such conversion devices, pre-amps, available on the market for a suitably overpriced cost.
| There is no way you will be able to match the impedance of both signals.
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detect VGA port (through pinout) and send a command to microcontroller How can i realize that a video project is connected to laptop through a VGA port??I think I can understand it from the VGA pinout; but i don't know which one??If I realize it then I can send a command to microcontroller and then do some functions... <Q> Which "side" are you on - the monitor side or the driving side? <S> Either way round the magic words are "EDID" and "VESA DDC" . <S> The protocol is basically I2C. Monitors will return a list of supported resolutions and timings. <A> project is connected to laptop through a VGA port <S> You seem to be building displaying device. <S> Then you detect connection by seeing if there're horizontal and/or vertical sync signals. <S> Some monitors will not get out of sleep if there's no defined level on SCL and ID0 lines, you may sense them, if they at some time become ground, it's time to turn your display device on (this is actually simplified treatment of Vesa DDC - see pjc50's answer). <A> ID pins <S> set-up4 11 <S> 12ID2 <S> ID0 <S> ID1n/c <S> n/c <S> n/c <S> no monitorn/c <S> n/c GND <S> Mono monitor which does not support 1024x768n/c GND <S> n/c Color monitor which does not support 1024x768GND GND n/c Color monitor which supports 1024x768 <S> Practically all modern displays will be in the last category, pulling pins 4 and 11 low.
| In addition to VESA DDC, you can find out that a VGA display is connected (and which one) by examining pins 4, 11 and 12 of the connector:
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Why transistor doesn't oscillate in this configuration? This probably sounds funny but I'm curious. simulate this circuit – Schematic created using CircuitLab Current will flow between base and emitter. This will switch the transistor on and current flows from collector to emitter. But this should create low potential at base probably? Because once transistor switches on all the current should flow through collector and emitter. Transistor should turn off due to no base current and current flowing through collector and emitter should stop. Again current will flow through base. Shouldn't this happen to transistor and wont transistor oscillate? <Q> Why would all of the current flow through Vce? <S> Even if the transistor is entirely on (dead short), you still have LED voltage drop <S> so you'd have LED voltage drop across 100k <S> so you'd still have at least some base current. <S> You are right to an extent that these two things will balance eachother out to some degree and find an equilibrium. <S> Nothing here appears to cause any sort of oscillation though. <S> Even if you removed the LED, instead of oscillating, the transistor would find some equilibrium point where the voltage Vc would be enough to support the transistor base being on to some degree. <S> Maybe you're having trouble with this idea because you believe transistors are like on-off switches? <S> They're not. <S> They're gradual turn-on/turn-off devices. <A> As others have answered, the transistor does not 'rush' to the <S> on-state, conduct too much LED current, drag its supply rail down and so rush back. <S> It travels towards a balance point, and stays there. <S> Risking yet another parallel, it's the same reason why pulling a spring apart in your hands sharply doesn't result in endless oscillations when you then hold your hands still and get them pulled together a bit. <S> The spring stretches to a balance between how hard it can pull your hands together against how strongly you are holding them apart. <S> Back to transistors, below is a circuit that acts as what someone once called to me a 'programmable diode'. <S> This shows an application of this 'just-on balance' idea. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Its normal application is with a voltage across Vce supplied through a resistor somewhere up above. <S> I'll avoid that to keep on this circuit. <S> If base voltage Vbe goes below (say) <S> 0.7 V, Q1 isn't on at all which means the rail isn't pulled down. <S> So Q1 finds a balance: just enough to hold its base just on enough. <S> This produces the function of a configurable voltage drop, with Vce kept at something like Vbemin(Ra+Rb)/Rb volts, where Vbemin is 0.6..0.7 V and particular to that transistor. <A> But this should create low potential at base probably? <S> No. <S> This is your basic misconception. <S> The base voltage doesn't really change as the transistor conducts collector to emitter. <S> Also, if the transistor really worked the way you think, it still wouldn't oscillator. <S> To get oscillation, there needs to be a loop gain above 1 at some frequency, but not at DC. <S> What you describe is negative feedback, which would cause the transistor to find the operating conditions where the competing effects balance out. <S> As a aside, that is not a good way to connect a LED. <S> If the transistor were to turn on fully, there would be nearly 5 V across the LED. <S> Something would have to give, like the battery dropping voltage, the transistor blowing up, or the LED blowing up. <S> If your LED lights without getting damaged, then it is only because of the limited gain of the transistor keeping the collector current low enough.
| Transistor Q1 turns on but the more current it conducts, the more it drags down the supply across the potential divider and so across its own base.
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What kind of IC is "ATMLU324" "16B 1" "Z8J0534B"? I recently bought what I thought it would be an ATtiny85. After a few unsuccessful tests, I discovered (guess what?) true ATtiny85 had "ATtiny85" printed on it, like this one: Mine is this one: The printing says: ATMLU32416B 1Z8J0534B I searched for a datasheet, but all I found was datasheet sites searching for it too. What is this chip? <Q> It looks for all the world like an Atmel AT24C16B , a 16kbit two-wire serial EEPROM chip. <S> In particular, page 14 of the datasheet has this diagram explaining the markings on the DIP version of the chip: <S> Seal Year <S> | Seal Week | <S> | <S> | <S> |---|---|---|---|---|---|---|---| <S> A <S> T <S> M <S> L <S> U <S> Y <S> W <S> W <S> |---|---|---|---|---|---|---|---| <S> 1 <S> 6 <S> B <S> 1 <S> |---|---|---|---|---|---|---|---| <S> * Lot Number <S> |---|---|---|---|---|---|---|---| <S> | Pin 1 Indicator <S> (Dot)Y = SEAL YEAR <S> WW <S> = SEAL WEEK 6: 2006 0: 2010 <S> 02 = Week 2 7: 2007 <S> 1: 2011 <S> 04 = Week 4 8: 2008 <S> 2: 2012 :: : :::: : <S> 9: 2009 <S> 3: 2013 :: : :::: :: 50 <S> = Week 50 52 = <S> Week 52 <S> Maybe you were simply sent the wrong part by mistake. <S> Or maybe the seller tried to scam you with a cheaper (or counterfeit) part. <A> First of all, I believe it should read ATMLU924. <S> http://panda-bg.com/en/products/Integrated-Circuits/Memories/Memories-EEPROM/011602/ <S> http://www.eca.ir/forums/thread33452.html <S> You could try hooking these up and see if you can still use them as memory. <S> However, if they were advertised as ATtiny85, it's likely that somebody scammed you, unfortunately. <A> Multiple pictures with that same code show up on websites like Alibaba, claimed to be an attiny. <S> It's a counterfeit of some kind, maybe it is an attiny inside and will work like one, maybe not. <S> Only way to find out is to try to program it.
| There are multiple sites online that suggest these are knock-off serial eeproms, see
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Mechanically, is a latching switch always a momentary switch? I am looking for a switch that (mechanically, not by circuit design) is first a momentary switch and then latches down when depressed far enough to be a maintained switch. Do all latching switches have a "momentary moment" before latching, or do I need to look for a specific sub-type of latching switch? I'd rather not buy a switch only to find out it doesn't have the momentary function. Edit: Sorry I'm not sure if my description was clear. What I am looking for is a switch that when you depress it to say 5mm connects the circuit (turns on an LED etc.). If you release it at this point it just springs back and turns off the LED, i.e., a momentary switch. However, if you depress it to 10mm it locks mechanically and keeps the LED on permanently. Then when you depress it to 15mm this unlocks it and it springs back up and disconnects the circuit and you are back where you started. I'm sure I've used these in real life before, I just wasn't sure if it was standard behavior. <Q> Most of the major manufacturers of industrial controls make what are referred to as "Alternate Action" or "Push-On / <S> Push-Off" or simply "Push-Push" buttons. <S> When you select those operators, you can then also get a contact block for it that is <S> N.O.E.M. <S> (Normally Open Early Make). <S> With that combination, the contacts will close BEFORE the button latches, then if you continue pressing hard enough, it will latch and stay closed, until you press the same button again to release it. <S> Allen Bradley Alternate Action Push Button operator Allen Bradley NOEM contact block <A> nope, that's not a common feature of latching switches. <S> In fact, I'd even consider it kind of an anti-feature. <S> Normally, you'd want a switch to be in either stationary state, and nowhere in between! <S> So, yes, you'll have to look for that in the switch datasheets. <S> With a relay (or a transistor), you could use the "upper" contact as momentary closing switch use the "lower" contact to power coil of a relay that powers itself and closes the circuit permanently. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Rather - it's a design feature that ensures that the latch position ON (down) is always connected even when manufacturing tolerances rear their ugly head. <S> However, this type of operation is standard in closed-circuit intercom belt packs used by professionals. <S> Operation is simple: a quick tap-and-release of the button flips the state of the latch that drives the circuit. <S> If the circuit was OFF, the quick tap toggles it ON. <S> If the circuit was ON, the quick tap toggles it OFF. <S> A sustained press of the button (more than about 1 second <S> ) disables / resets the latch and the circuit functions as Momentary ONLY. <S> If the circuit was latched ON, pressing and holding the button resets the latch. <S> Note that the circuit is always ON anytime the button is pressed. <S> Older gear used a CD4013 dual D-FF and a single transistor per button. <S> Modern gear uses a tiny microcontroller to do the same thing. <A> This is a side effect of the mechanical path the latch spring takes in the switch. <S> On some, it's easy enough to modify, removing the part that holds the button in. <S> IIRC <S> most often it's the opposite of what you described, as it normally takes a deeper press depth to latch, then a shorter press to unlatch. <S> The catch is pushed around by the physical shape of the path, and a spring provides the tension needed. <S> Its basically the same mechanism used in clicky pens! <S> The v looking grove is what provides the latch function. <S> Click in, it falls into the grove, click out, it gets pushed out and then pushed back into the ready position, like a pinball. <A> I can't find anything already in production like you described, but you may be able to create this yourself fairly painlessly. <S> What you want is a "dual stage" push button, but pretty much every implementation I've seen of a dual stage button has been done by combining two existing button types. <S> Examples being the Steam Gaming controllers triggers, which are just a plastic trigger than moves a pot, and at full pull presses a momentary pushbutton, and <S> this DIY example using two offset microswitches <S> So my best suggestion would be to mount a limit/microswitch for your momentary press since they allow for a travel distance past the contact point, and <S> a latching pushbutton next to it, then 3d-print a button cap/housing to make it look like a single button. <S> If done right you could make a custom pcb they mount on and add pins/connector to create a single button module out of the whole piece that could then be modeled as a single object for design and produced in larger quantities.
| Although I have seen push-button switches that behave as you describe, I'm not certain that this behaviour is specified in the datasheets. You could probably also find (possibly even easier) a switch that is "dual-momentary", with the first contact closing at a different depression than the second (think of digital camera shutter buttons, which you slightly depress to set the focus point, and fully to take a picture).
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Is the "antenna aperture" the same as the "hole in the shield"? Is the concept of the "Antenna Aperture" the same as a hole in a electromagnetic shield? The wikipedia article is quite vague, and I am a newbie in this field. For example if we have an electric device and we put it in a steel box, that has sufficient depth. But the steel box has 1 piece of 1 cm hole in it. That should block any signal up to 29.979 GHz, since the maximum wavelength that can leak out from that hole is 1 cm. Is this the same concept as the "Antenna Aperture", if not then what is the relationship? <Q> This is always smaller than the physical size of the antenna since no antenna can extract all of the RF power that passes through it. <S> It does not refer to physical aperture but is called that due to the similarity to the aperture of a camera, the larger the aperture the more power is received from a given intensity of radio waves/light and in both is basically a measure of the efficiency/sensitivity. <S> However a generalisation of the concept could be applied to what you are talking about, you could create a similar metric for the amount of power that passes through an aperture of a given size in terms of an area of the source field. <S> That would give you an "effective aperture" which would be useful in calculating the power that will leak in but except for very simple apertures it is something you would need to measure to have a accurate result. <S> Though when you are designing RF shielding you do not normally use anything like the accuracy used in designing antennas, since it is not very hard or costly to make something much more effective than is needed. <S> I googled some pages that might help, I only just skimmed them though to see if they looked useful: http://incompliancemag.com/article/the-basic-principles-of-shielding/ http://www.compliance-club.com/archive/keitharmstrong/design_techniques4.html <S> https://interferencetechnology.com/analysis-shielding-effectiveness-board-level-shielding-apertures/ <S> Even if it's not a problem you might <S> as well just stick some aluminium tape on it if you have some lying around. <A> Can mean different things. <S> Aperture like a speaker hole in a bass reflex design for woofer bass can be 10m wavelength but <S> a small fraction hole size still lets out energy. <S> Although reasons of cabinet resonance and dampening air flow it still escapes out the fractional wavelength of the hole size. <S> For this reason the grid size in microwave screen doors and Lingren walk-in Faraday cages are sized as small as necessary to achieve high attenuation and tuning holes for 3m high power transmitters use something like a <5mm hole. <S> However if we want a resonant antenna at 1/4 or 1/2 wavelength to emit with many elements to get more focus or a narrower beam angle, this achieves some gain like a parabolic mirror for flashlight <S> then we consider the aperture of the half power angle of the beam at a certain distance rather than the stray radiation 100 db down (or whatever) for shielding purposes. <S> It is this beam spreading or focus aperture that you read about in Wiki. <S> Keep in mind <S> a small pin hole still lets out light but not efficiently, unless the emitter is tiny and directly under the hole. <S> Like a 1mm chip inside a 5mm LED directly under a thin 1mm hole. <A> The word "aperture" has more than one meaning. <S> One meaning is a physical hole or opening cut into some material such as a shield. <S> And it appears that you are confusing that definition with antenna aperture , which are two totally different concepts. <S> In this context, the effective aperture of an antenna system is directly related to its gain. <S> Keep in mind that effective aperture is neither directly related to the actual size of an antenna (the possible exception being a parabolic dish) nor to the thickness of its elements. <S> Having said this, antennas DO exist consisting of a flat sheet with a slot in it, and the actual radiation comes from that hole ("aperture"). <S> The one I'm thinking of has balanced feedline connected in the center of a rectangular slot. <S> As for shielding, holes are permissible as long their size does not exceed a certain relationship to the wavelength. <S> (Google shield holes cutoff frequency .) <S> https://leadertechinc.com/blog/how-apertures-affect-emi-shielding/
| Antenna Aperture refers to the area of radio waves that is equivalent to the power that is obtained from a given antenna.
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2.5mm Audio jack to micro USB First of all, just a caveat: I am a complete newbie with electronics, computer software is more my thing! I have bought a (used) car this morning and it came with an old TrafficMaster Satnav installed which I have now removed. It seems that a power jack has been installed in the dashboard to power it. I usually have my phone in a dock and run a long micro usb charging cable to a cigarette lighter USB charger so I can charge my phone while driving. It would be nice to utilise the power jack that has been installed, to charge the phone as it will be a neater solution than running the long USB cable. The old Trafficmaster is powered through a 2.5mm audio jack, the other end is moulded to the unit. Therefore the power jack installed in the car is the female end. This all works and the Trafficmaster recieves power. The TrafficMaster is pictured here What I want to do if possible is make a cable that has the 2.5mm jack at one end and a micro USB on the other. Is it possible to take a spare USB charging cable cut the micro USB off, cut the 2.5mm off the trafficmaster and solder the ground USB wire to the ground audio jack wire and the 5V USB wire to the 5V wire in the 2.5mm connector? (Is there one!?) <Q> " <S> 2.5 mm jack" is a mechanical spec. <S> You also need the electrical spec. <S> What voltage does this jack provide? <S> At what maximum current? <S> Is it a regulated and de-glitched 5 V? <S> A direct connection to the car 12 V system? <S> Something else? <S> Measure what power comes out. <S> Only then can you decide what needs to be electrically between this jack and a USB charger output. <S> If the jack provides 5 V, then maybe you can just connect them directly. <S> If 12 V, then you have to assume all the nasty spikes on the car 12 V system come along with it. <S> You could then hack up the same charger you now plug into a cigarette lighter outlet to plug into this jack instead. <A> You'll have to figure out the pinout of that 2.5mm connector, it's impossible to guess for us here. <S> Measure the voltages between the different rings? <S> Does one give something like a stable 12V, 5V or so? <S> Note that you probably do not want to use some old car nav supply to charge your devices. <S> Modern smart phones can draw multiple Amperes, and there's so far no reason to assume that there's a DC/DC converter built into your dashboard and that instead of the car-typical 12V, you get 5V on that connector, that <S> if there is such a DC/DC converter built into the dashboard, it will give you exactly 5V, <S> that if it exists and gives you 5V, it will be powerful enough to charge a modern phone (which will probably draw a lot more current when charging than some old-school car nav). <S> So, while if you're already measuring, sure, look for 5V between any of the rings on the converter, then connect a 5V/2A = 2.5Ω resistor and see how much that voltage drops if you draw a 2A charging current. <S> But: If you'd start by walking into a store, buying a multimeter, trying to grab a lot of resistors to put in parallel so that you can even have such a dummy load: forget about it. <A> Not likely directly possible. <S> The jack for the traffic master most likely puts out +12V to match the vehicle power system.
| The micro USB to your phone is 5V. Note also that cigarette lighter phone chargers have an internal DC-DC converter that converts vehicle +12V to 5V.
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STM32 clamping diodes - what is the maximum input voltage? As you can see on the attached picture from a reference manual of STM32F7 , GPIO pins have internal clamping diodes to protect from overvoltage. But what maximum voltage can I put into the pin? I know that 5V is max, and 4V is max for VDD. But what would happen if I put, let's say, 10V into the pin? Shouldn't the diodes clamp this overvoltage? What parameter of a diode decides how many Volts can be clamped safely? If I want to use external clamping diodes, let's say 1N4148 , then what will be maximum voltage that I can put into the pin? <Q> The answer to "Shouldn't the diodes clamp this overvoltage?" is yes <S> and no. <S> It really depends on the output impedance of whatever is feeding it and the strength of the power rail. <S> If you happened to connect that pin to a 10V power supply, what do you think is going to happen? <S> Will the diode pull down the 10V supply, or will Vdd be pulled up to 10V minus a diode drop. <S> Or will the diode just burn out. <S> It really all depends on the rest of the circuitry. <S> But as crude as that example is, you can perhaps grasp the idea that the diode has a limit on how far it can go in the role of signal clamping. <S> The pins ALSO have a current limit. <S> The diode will only survive as long as you do not exceed that current limit. <S> Will adding an external diode help? <S> Sometimes the internal clamping diode is not actually a diode but an FET type circuit. <S> An external diode that clamps at less than the internal protection voltage will allow you to dump more current. <S> If you can't find a diode that clamps to less than the internal circuit, then there is no point. <S> The chip may fry before your diode ever really kicks in. <S> You still can't short to the other rail though. <S> Adding an appropriately sized series resistance is generally required. <S> Addendum: <S> The input protection diodes are really there for "just in case" protection. <S> They should not normally be relied upon to be a functional part of your design. <S> Proper signal preparation before injection into the pin is the better design method. <S> Generally I prefer this approach. <S> R1 needs to be chosen to limit the current to the zeners specified reverse current at the indicated zener voltage. <S> That is, the Zener may be rated at say \$3.1V @ <S> 5mA\$. <S> So \$R1 = <S> (V_{signal} - V_{zener}) <S> / <S> I_{zener}\$ <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Shouldn't the diodes clamp this overvoltage? <S> Yes, until it is shot by over-current. <S> 2) <S> What parameter of a diode decides how many volts can be clamped safely? <S> The maximum current rating, typically not provided. <S> But I would be worried if there is more than 20 mA going through those diodes and typically design for 1 mA. Say, STM32L082xx datasheet <S> I/ <S> O current injection characteristics tells most pins can withstand up to 5mA worth of injected current without damaging the device. <S> 3) <S> If I want to use external clamping diodes, let's say <S> 1N4148, then what will be maximum voltage that I can put into the pin? <S> It depends on the design. <S> If you just use two clamping diodes (i.e. they are parallel to the on-die clamping diodes), it depends on which of them kicks in first. <S> A typical design has at least one and often two resistors there, so it will depend on the maximum rating + resistors' size. <A> The important thing is the volume of current. <S> You cannot exceed that too. <S> So you definitely cannot use 10V source with higher current capability. <S> It's important to use input resistor to limit the current.
| In summary, when attaching non-device level signals care needs to be taken to ensure the source impedance is high enough to not over-rate the device and not swamp the rail you are clamping to.
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How to mount a DB-9 adapter to an enclosure I have a DB-9 connector that needs to be mounted to an electronics enclosure. This hole for the DB-9 connector will be manually cut out. Question: How can the DB-9 connector be secured to the enclosure? I am using the following DB-9 connector . I need a DB9-to-DB9 connector, not DB9 connector on one side, and some other pins on the other side. Is there a better variant of it that has mounting screws or something? <Q> Those look like they are designed to join two cables together, not in a box or anything, just cable to cable. <S> Look for a bulkhead or panel mount connector or feedthrough . <S> More usual is to have a DB-9 cable outside, and something cheaper, smaller and more flexible inside - such as a ribbon cable. <A> Perhaps there are some nuances I've missed here, but if not then:- <S> Look at your example and do the same - simply remove the supplied screws and replace with longer ones and bolt through the enclosure. <S> You can also get long versions of hex ended screws like these. <S> Don't use screws. <S> Glue it to the enclosure with something like epoxy or hot glue. <S> If the enclosure wall is too thick to allow the exterior connector to mate, glue the adaptor with the bulk of it on the outside rather than the inside. <S> Wacky idea. <S> Does it need a connector at all? <S> There is also the possibility of connecting a cable only to the internal DB-9 connector. <S> Then simply feed the cable through the enclosure wall. <S> Figure out some form of strain relief. <S> You're then free to do what ever you want at the loose end. <S> No one said that equipment can't have a flying lead. <S> Most of it used to, and this is sometimes a practical solution. <A> The devices you have pictured are gender changers and not intended for the use you are looking for. <S> They will work for your application but, you will also need to be creative, which is why I believe you should use a DB9 that has solder pins on one side. <S> Wire <S> what ever you have in your enclosure directly to the solder side of the DB9. <S> If that is not possible, make a short cable inside your enclosure. <S> Be mindful of your connecting cable, if it is a null cable then make sure your internal cable is a straight through of you will cancel the null of the external serial cable.
| You might also be able to buy a panel-mount connector and short serial cable in one piece, sometimes called a panel-mount serial extension or similar. You might have difficulty finding something at your usual supplier, as it's actually quite unusual to use a DB-9 cable inside the case of an appliance.
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Replacing Multimeters with Megaohmmeters Standard Digital multimeters can measure up to the ranges of 20 Megaohms. Then why should we use Meggers at all? <Q> From Wikipedia... Megohmmeter (sometimes referred to as a megger) is a special type of ohmmeter used to measure the electrical resistance of insulators. <S> Insulating components, for example cable jackets, must be tested for their insulation strength at the time of commissioning and as part of maintenance of high voltage electrical equipment and installations. <S> For this purpose megohmmeters, which can provide high DC voltages (typically in ranges from 500 V to 2 kV) at specified current capacity, are used. <S> Acceptable insulator resistance values are typically 1 to 10 megohms, depending on the standards referenced. <S> Meggers will measure into the GΩ and TΩ ranges. <S> Don't try using one on your logic circuit though... <A> An ordinary multimeter uses low voltage to measure the resistance. <S> The multimeter will show infinity, but as soon you use a megger with specified test voltage you find out that the resistance is very low - insulation breakdown. <A> There are plenty of components that show high resistance at low voltages but then suffer breakdown when subjected to higher voltages (ziener diode and low voltage capacitors being the obvious examples). <S> Meggers with their higher test voltage in the 1kV-2kV range are used to ensure the high resistance is still apparent in a high voltage situation.
| A megger has a high voltage power supply for measuring, indeed the measured resistance is different compared to multimeter, when you test the broken insulation.
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how to end an assembly code I'm trying to write a code for pic16f877 by translating a simple algorithm into assembly code. The instructions set doesn't have such an instruction. Here is what I wrote: MOVF 0X40,WMOVWF 0X50SUBWF 0X41,WBTFSS Status,CGOTO labelMOVF 0X41 ,WMOVWF 0X50label END My question is: is there a way to "end" an algorithm when programming a microcontroller using assembly code? <Q> Just to make it clear: in microcontroller assembly code there usually isn't an end. <S> There certainly isn't an end instruction. <S> There isn't an OS to return control to, and there's no way to just "stop". <S> Microcontroller code nearly always contains an infinite loop. <A> When you work with a µC, you have to make an infinite loop in the end of your code. <S> Otherwise, you won't control what the µC will do. <S> If you forget such a loop, the µC will trigger it and block the execution of the program. <S> (called ISR_TRAP in TI's MSP430) <S> In C, you will basically end your code with something like while(1) or for(;;) . <S> By doing so, the processor will do some idle task so what it remain active and still aware of interruption. <S> Same for assembly, you have to make your code loop so what it remains active. <S> loop: jmp loop <A> Stop and actually think about it. <S> What do you want the microcontroller to do after it is done with the little bit of logic you show? <S> Surely there is something . <S> Should perform the operation again? <S> Should it go back to some larger task? <S> Should it wait for something external to happen, then do it again? <S> Even if it's supposed to do "nothing" until power is removed, that's actually something you have to tell it to do. <S> It doesn't make sense. <S> Everything requires some action. <S> Even powering itself down via some external switch requires activating that switch. <S> On a separate topic, flow charts should really indicate the next level up logic, not the details of how the processor will perform that logic. <S> You are specifying details about things like the W register in your flowchart. <S> Instead, it should show something like a comparison block between A and B (or whatever), then two or more choices to take depending on the result. <S> When you code this in assembly, then you figure out the actual instructions to realize the higher level logic described by the flowchart. <S> It's also bad to refer to variables by their addresses. <S> Refer to variable symbolically, then let the linker decide where to stick them in memory. <S> That's really none of your business, other than perhaps the bank they should be in. <S> Your flowchart refers to three different variables at 40h, 41h, and 50h. <S> Give them names and use the names in the flowchart. <S> Good names also helps illuminate the logic. <S> Even when you code this in assembler, you still give variables names and refer to them by names in the code.
| The point is, that there is no "end".
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Amplifying a PWM signal with an op amp. Is the slew rate a problem? I need to amplify a PWM signal from 5V to 24V in order to drive a mosfet that in turn drives a small DC motor.The input signal has a frequency of 500Hz and comes from Arduino uno (pin 9). For amplifying the signal I thought of using a typical non inverting amplifier configuration simulate this circuit – Schematic created using CircuitLab If I use an op amp such as the TL071, the typical slew rate is 16 Volts/microsecond. This means that the op amp will take 24/16 =1.5 micro seconds to reach the high output of the PWM. This seems acceptable to me since with a 500 Hz PWM frequency, the PWM period should be 2000 micro seconds, therefore 1.5 over 2000 is negligible. Is there any other consideration I should do? For instance, should I consider the time the mosfet needs to charge up the gate?Is there a better way to amplify a PWM signal? Furthermore, suppose that I would like to increase the PWM frequency. For instance up to 2.5kHz. in this case the PWM period should be 380 micro seconds. considering 1.5 over 380 the the slew rate still seems acceptable to me. <Q> There are several things wrong or confused here: <S> 24 V is very high for a MOSFET gate. <S> These are usually specified to switch fully with 10 or 12 V. <S> 24 V might be the absolute maximum, not what it's intended to switch at. <S> A TL071 is totally inappropriate here. <S> Those need several volts of headroom from both supplies at both the input and the output. <S> Typical specs are meaningless. <S> Use a FET driver. <S> Driving a FET gate from a digital signal is exactly what they are for. <S> If the motor is powered from 30 V or less, then something like the IRLML0030 would work. <S> You just connect its gate directly to the digital output. <S> 500 Hz is probably fast enough for the motor to mechanically filter the pulses. <S> However, there will likely be audible whine, and the current will probably change significantly during the on and off time of each pulse. <S> Even if you don't care about the whine, having a steady current matters. <S> Think of the current thru the motor broken into its DC and AC components. <S> Only the DC component moves the motor. <S> The AC component does nothing useful, but still causes heating due to the resistive component of the motor coils. <S> In short, the less AC component, the more efficient the overall motor drive is. <A> Simple N-Channel one, low-side switching with a resistor going to the positive supply voltage: simulate this circuit – Schematic created using CircuitLab <S> The value of R1 depends on what you need to switch on the outside. <S> This is in fact an inverting circuit, but that really doesn't matter – most MCUs can simply set the polarity of the PWM, or you can just logically invert the duty cycle. <S> Which really raises the question why you think you'd have to raise the gate voltage of the MOSFET you're trying to drive! <A> What you have analysed and concluded looks fine, good work. <S> You should put a resistor between the op-amp output and your FET gate. <S> Without it, the op-amp has the capacitance of the FET gate on its output which can cause it to oscillate. <S> I can't say the resistor value without knowing the FET's gate capacitance. <S> However, you usually find values around 470 R or 1 K typically used <S> so I imagi <A> Is there any other consideration I should do? <S> Gate drivers are generally a function of the switchers you intend to drive. <S> The most important factors there are current capabilities, frequency limits, and drive topology. <S> Very rarely you see an linear amplifiers used here. <S> Google gate driver may help you get more insights.
| Depending on the motor power supply voltage and the motor current, you might be able to use a FET that switches nicely with just 5 V on the gate. For this kind of voltage amplification, you'd typically use ... a MOSFET.
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Connect panel mount switch to PCB I'm looking for more information on how to connect a panel mount switch to a PCB. Switch Diagram: I've read that it's generally bad to connect this kind of mechanical device directly to the PCB (since it's a more likely part to fail). Since there won't be a panel for the device I'm trying to make, how can I attach it to the PCB in a sane way? This is just a hobby project that I want to have some switches for easier testing. It's not a critical component. I've tried searching but I can't find something that fits this. Perhaps I'm using the wrong terminology. Specific questions: Should I use a "terminal block" for this purpose? If so, what specifications should I look for? I think it's 3 pin, 7.1mm pitch, but I don't know what to do with the tab width of 4.8mm. Am I just missing some alternative method that would be better for my purpose? <Q> I don't think you need to use a terminal block, if you don't need to be able to rapidly remove and replace the switch. <S> If you are having the PCB made by a PCB manufacturer that will provide plated through holes, then three of those spaced 7.1 mm apart will provide good mechanical strength attachments to the tabs on the connector. <S> Just specify round holes that are large enough for the pins to go through. <S> So, the finished diameter of each mounting hole will have to be at least... <S> $$d = \sqrt{0.2^2 + <S> 4.8^2} = 4.804 <S> mm $$ <S> Plus add a little bit extra for manufacturing variance. <S> Also note that the through-hole plating reduces the inner diameter of the hole a little bit, by perhaps 0.125 mm, although many PCB manufacturers will take this into account so that you don't have to when you specify a hole size. <S> There are of course switches designed for mounting right on a PCB, but if you want to use this one, then big through holes for the tabs to go through is probably the best way for a cheap hobby board. <S> A plated through hole rectangular slot would match the shape of the tabs more closely for perhaps better mechanical location of the switch, but for a hobby board, round holes are probably good enough for what you want. <A> Don't directly solder the terminals to your PCB - that's good advice. <S> You stand the chance of damaging the switch internally. <S> This type of switch is generally meant to be used with "Fast-On" contacts. <S> These crimp on to single conductor wire and are available at so-called "home improvement" stores like HD & Lowe's, and also auto parts supply stores. <S> They come in several widths. <S> Your 4.8 mm terminal will probably take a 1/4" size, you need the female type. <S> You crimp the contact onto the wire. <S> You probably don't have a crimp tool, so you can use a pair of hefty pliers, or press it in a bench vise. <S> Cut the wire to the desire length, solder the other end into your PCB. <S> Then push the Fast-On onto the switch terminal. <S> You obviously need 2 or three of these wire assemblies depending on how you intend to use the switch ( i.e. single throw or double throw). <A> If you don't need this specific component, but just a switch functionality, there are switches designed to be soldered directly onto the PCB. <S> Personally, I would choose another one, because these are kinda bulky and there are more compact solutions depending on the application. <S> RS Components, Mouser, Farnell are some good websites to look for electronic components. <S> If you want to solder this specific switch to the PCB, you can design three holes with a 7.1mm pitch. <S> I'd make the holes 4.9mm in diameter if circular or 4.9mm <S> x 0.9mm if rectangular. <S> Terminal blocks are normally used to connect wires to a PCB: the component is soldered with through-hole pins on PCB <S> and then the wires are pinched in the case with some screws. <S> I would not use this kind of component as an improvised adapter for your switch.
| The design of your switch suggests that it is not for direct mounting onto the PCB, but to be contacted through "Fast-on" connectors or direct soldering of wires to the tabs.
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how to convert the difference in voltages into a digital signal How can I convert the differences between two voltages into a digital signal? ie. if sensor 1 V > sensor 2 V there is a digital one, but if sensor 2 V >= Sensor 1 V, there is a digital 0. I understand that there are methods of converting a single analog signal into a digital signal, but my understanding of a digital signal is that it must always be either high or low. if you convert both into a digital signal, then wouldn't it be impossible to compare them. Or rather, so long as eithers analog voltage is always greater than zero, then wouldn't a converter always convert the voltages to digital 1s? <Q> The classic method of comparing two analog signals is to use an analog comparator. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is an op-amp like device that produces a high on the output if the Plus pin is at a higher voltage than the voltage on the Minus pin. <S> (Some, like an LM339, require a pull-up on the output to attain the required logic high level.) <S> However, there is an issue with comparators in that >= requirement. <S> Comparators have a minimum offset error between the two pins so = is a bit of a relative term that can be problematic if the signals you are comparing are frequently at or near the same voltage. <S> Even if the offset can be tuned out, if the signals are equal, the output is undefined, or can actually oscillate. <S> For such conditions is it more prudent to digitize the signals into a digital byte or word using an analog to digital convertor and compare those values mathematically. <S> However, in that instance you have to remember there is a minimum step size in the digitization, so again "equals" <S> is really a relative number ( = <S> +- Half a step). <S> More often, the requirement is <S> "I want a HIGH if Signal A goes higher than signal B, but I don't want it to go LOW again till <S> signal A goes lower than say .1V <S> Below signal B. <S> " This is called hysteresis and can easily be created by adding positive hysteresis feedback to the op-amp circuit. <S> For very small signals it gets more difficult. <S> In order to provide reasonable accuracy some amplification of the original signals is sometimes warranted to give you a signal scale that is sufficiently high to reduce the effects of the inherent errors and offsets in the system. <S> In some situations it is better to subtract the signals in an analog circuit before amplification to ensure both signals are treated equally. <A> An analog comparator will compare two analog voltages, A and B, and give a digital output - a "1" if A is greater than B, or a "0" if B is greater than A. <A> If you are just looking for a single bit representing if one signal is higher or lower than another, you would probably be best off using a comparator. <S> But you can also convert both analog signals into digital values, and then compare them mathematically, i.e. check if one value is greater or less than the other. <S> That can be done with discrete logic, but typically it would be done with a microcontroller. <S> So for example, a small simple MCU can take both voltages into to ADC pins, and convert and compare them in firmware. <S> No, if the voltage is greater than zero, it doesn't necessarily imply a digital 1 output.
| It will depend on the reference voltages for the converter, number of bits, etc.
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How to generate crosstalk between 2 wires? I'm trying to recreate a situation where crosstalk is occurring. I'm having a bit of trouble trying to inductively cross couple two parallel wires that don't have any shielding or insulation and aren't twisted together. I basically have 2 oscillator circuits that are connected to an inductor since I want to measure the two inductors frequency. <Q> Crosstalk is due to either capacitive or inductive coupling. <S> You seem to assume only inductive coupling, but gave no justification for that. <S> Capacitive coupling is the easiest to simulate. <S> Just connect a capacitor between the two wires. <S> Depending on the length of wire, their spacing, and what insulation is between them, this might take a few pF to a few 100 pF. For inductive coupling, loop a few turns of each wire around the same ferrite stick. <S> Make sure to use a ferrite material specified for your frequency. <S> In this case, the far ends of the wire need to be loaded as they would be in real life. <S> Inductive coupling is due to current, so you won't get any coupling without current flowing thru the wires. <A> What you can do to increase crosstalk: increase frequency increase output impedance of circuit that is feeding "secondary" wire increase voltage increase current through "primary" wire <S> increase length of those wires <S> decrease distance between those wires <A> Vinduce = <S> [MUo MUr Area/(2pi*Distance between Wire#1 and Loop)]*dI/dT <S> Thus wire carrying 1 amp at 60 Hz has dI/dT of 377 amps/second. <S> If one meter away from 0.1meter*0.1meter (4" on a side) <S> square loop, with Loop and Wire#1 in the same plane, with no other metallic structures nearby, we have the induced voltage of [2*1e-7H/meter * 0.1m <S> * 0.1m/1 <S> m] <S> * 377 = <S> 2e-9 <S> * 377 = 754 nanoVolts
| You can compute crosstalk between 2 wires, if you know the frequency and current in wire#1, and know a LOOP (wire+return_path) area of another wire.
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Easy way to close 4 contacts at the same time I'm building a smartphone rig to check the smartphone camera quality, focus and shutter speed. This is my RIG, I have 4 bluetooth remote paired and I solder to the buttons some wires. If I take a single couple of wires and I put them in contact the smartphone shoot a picture, but If I group the wires together they doesn't work. I need to close the 4 couple of wires at the same times keeping them isolated one from each other. I cannot find a 4 way button. Any solution? <Q> Google 4pdt switch (4 pole double throw) for more resultes. <A> Part 15401 and 15451 from MEC seems to match with your need. <S> Here is the datasheet : <S> http://docs-europe.electrocomponents.com/webdocs/051b/0900766b8051b322.pdf <S> According to manufacturer website, 15401 is a momentary switch and 15451 <S> a latching switch. <S> http://www.mec.dk/Unimec-v12-p-72.html#2 <S> However, I couldn't find the part 15401 on any web site but RS Europe. <A> I would have made this a comment, but still don't have that ability here yet, so this will depend on answers to a few questions I would have asked - namely regarding on-hand parts and availability. <S> Your mileage may vary. <S> That being said... <S> If you are comfortable with soldering (which it looks like you are), you can use a few components to create a 4-pole single throw momentary switch that may also cost less than a mechanical one. <S> What you'll need: SPST "Normally Open" Momentary Pushbutton or Tactile Switch 10k Resistor (or close to that value) <S> Breadboard, Perfboard or IC Prototyping Board <S> Batteries & Connector (two to four AAs or AAAs, or one 9v) <S> CD4066 Quad Bilateral Switch IC (pictured below) ICs may be intimidating to use if you are unfamiliar with them, but this solution really only relies on the ability to solder. <S> The 4066 is basically just 4 on/off switches that are individually controlled by an applied voltage, instead of manually. <S> Below, we will add a manual component to deliver the voltage. <S> IC Power <S> : Connect your positive battery lead to pin 14, and the negative to pin 7. <S> Control Switch: <S> Connect your positive battery lead to one terminal of your switch, and run 4 wires from the other terminal to the control pins (5, 6, 12, & 13) of your IC. <S> You can also do this with a single wire to one control pin, and run jumpers to the other three. <S> It can work without this, but it's common practice. <S> Trigger Interface <S> : Connect your trigger wire pairs to pins 1&2, 3&4, 8&9, and 10&11. <S> Whenever you press your control switch, you will activate all 4 switches on the chip - yielding a simultaneous triggering of all of the remotes. <S> EDIT: <S> Here is a quick video demonstration of the circuit in operation.
| You are looking for 4 pole 1 throw switch, and 4 pole double throw switch would work as well. Pulldown Resistor: Connect the 10k resistor between pin 5 or 6, and 7 (battery negative). There are many options over there. Also, allow me to disclaim this by saying: I know it is not as simple as a single 4-pole switch, but it's how I'd solve it with things I have immediately available, and at a cost below US$1. 15451 is nowhere to be found...
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How does a DC motor drive control the motor? AC motor drives are pretty well documented; topologies, inverters, frequency control, volts-per-hertz, space vector modulation. It's all out there in a dozen places. But what is a DC motor drive, and how does it control the motor? <Q> It doesn't drive the motor. <S> I assume you're talking about a brushed DC motor. <S> Brushed DC motors have the commutation built into the mechanics of the motor - the brushes are part of that. <S> You can just apply a voltage and it'll go. <S> But since voltage relate to speed and current relates to torque then you can start doing fancy things to control those variables to have the motor behave in a desired way. <S> Brushed DC motor controllers are there simply to help with one or both of those parameters but are not involved in commutating it. <A> I assume that "DC Motor" refers to a commutator motor and not a brushless DC motor. <S> The intersection of the motor's speed vs. torque capability curve and the load's speed vs. torque demand curve is the speed and torque operating point. <S> With a commutator motor, that can be done by changing the voltage supplied to the armature, the current supplied to the field, or both, either separately or simultaneously. <S> The most common method is to provide a constant field flux either by using a dedicated power supply for the field or by using permanent magnets. <S> The speed vs. torque curves shown below illustrate this method. <S> Note that only the curve for the lowest voltage intersects the Y axis at the locked-rotor torque point. <S> Curves for higher voltages result in current that would exceed the capability of the controller and perhaps also the motor. <S> Controllers for all but the smallest DC motors have a current limit feature that prevents the motor from operating on the dotted portion of the curve. <S> Since the current is limited by regulating the voltage, operation on a curve of a higher voltage is prevented until the motor has accelerated to a speed at which the current is below the limit. <S> Better Performance <S> For better performance, closed-loop armature current control can be used to control the torque. <S> The torque reference for speed control would be speed error from either an armature voltage controller or tachometer feedback. <S> If tachometer feedback is not used, a means for estimating armature IR voltage drop can be used to improve speed regulation. <A> DC motors' drivers use PWM (pulse width modulation) to Drive Motor and control their speed you can make PWM using 555 IC OR any microcontroller and a Mosfet,but If you want to control it to change the direction of the spinning shaft you need an H-bridge H-bridge <S> is some transistors working together to bias motor in forward direction and also backward direction by reversing the polarity <S> So if you use both PWM and H-bridge you will have a really good driver that helps you control the speed and also the direction of motor Spinning <A> On motors using 3 phase AC supply in the US of 5 hp or larger there are 2 D.C. Control approaches. <S> Smaller systems may use diodes to rectify the AC voltage then use SCRs or IGBT transistors to pulse width modulate this rectified supply to the motor. <S> The more common approach for larger horse power motors (20hp to 200hp) is to use a "stack" of SCRs. <S> Non regen drive systems can only supply torque in one direction and will have 3 SCRs one for each of the 3 phases. <S> Regen drives will have 6 SCRs, two for each phase. <S> The control of the SCRs will use some type of phase angle firing on the sine wave to control motor speed and torque. <S> Most D.C. Drives have another control board dedicated to field control. <S> Field control is usually in 2 flavors - either constant voltage or constant current- with constant current being more common as field weakening for operation above base speed uses reduction in field current.
| Most motor speed control is accomplished by providing a means by which the motor's torque vs. speed characteristic curve can be adjusted so that the speed vs. torque demand curve of the load can intersect it at any desired speed.
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Can I power ICs directly from a wall adapter? I am designing an audio circuit that operates at 5 volts. Is it reasonable to buy a wall adapter that outputs 5V, or should I buy a wall adapter that outputs a higher voltage and step the voltage down at the PCB with a linear or switching regulator? The audio circuit operates at 5V to achieve a volume level. The audio signal source will come from a microcontroller regulated at 3.3V. The wall adapter I am considering is medical grade; part # MDS-030AAC05 AB. <Q> The main distinguishing factor of audio circuits is the low noise level that must be maintained. <S> While powering a audio circuit from some wall wart will work, it may, and quite possibly will, result in unacceptably high noise. <S> Most wall warts today are switchers. <S> The switching frequency is usually well above the audible range. <S> However, depending on the control algorithm, there may be sub-harmonics in the audio range. <S> With a sufficiently strong and low impedance input signal, a little extra noise from the power supply might not matter. <S> You haven't said anything about the circuit you are powering. <S> The circuit may have some power supply rejection. <S> If the circuit is opamp-based, then this is probably true. <S> However, active power supply rejection only works up to some frequency. <S> Above that, the active circuit can't keep up with, and therefore compensate for, the power supply noise. <S> The switching frequency of a common wall wart may well be above that range. <S> You can't hear the 1 MHz or whatever switching frequency, but such high frequency noise inside a chip can mess up its normal operation on much lower frequencies. <S> The noise could get AM-demodulated and added to your signal, for example. <S> All in all, I think it's a bad idea for "audio" without knowing something more specific about the input signal, output signal, and the circuit. <S> If your circuit contains a power stage, then the worst power supply noise could be from the changing current demand of the power stage. <S> If this is not dealt with properly, then your circuit could oscillate. <S> It would be prudent to filter the supply to any sensitive front end circuit. <S> You can use the raw unfiltered supply for a power stage. <S> Note that sensitive front end circuitry generally requires little power current, <S> so a few ferrite chip inductors followed by capacitors to ground won't get in the way. <S> Medical grade or not has no bearing on audio use. <S> Medical grade means extra low leakage between power input and output, and possibly extra high isolation voltage. <S> It has nothing to do with noise on the output. <A> Yes you can, match the voltage of the rail and make sure that the total current of the circuit (and IC's) you are using is less than the current rating on the supply. <S> Realize each voltage source carries noise. <S> Be aware that for analog circuits the PSRR rating will tell you how much noise will transfer from the power supply rails to your signal (this number is usually well over 80dB). <S> If you have a direct pathway to the rail (like a pull up resistor or a DC path to the power rail) then this could inject noise in directly from the rail. <S> If the noise is to high for your application, find a new supply, use an LC filter or a precision linear regulator. <A> It just happens that I tried to power an audio transmitter last night using a few separate USB adapters. <S> Results varied but were never good, too much noise. <S> I ended up picking one that was more or less tolerable. <S> I think this can be improved by having a filter/conditioner <S> but I didn't have time. <S> In summary, if quality matters, you should invest in building a simple power source for your audio that would be free of these problems.
| Measure the noise on the power supply with a volt meter set on AC mode, this will tell you the RMS value of the ripple or noise, if its acceptable for your application then you can use it.
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Considerations when using Opamp as comparator For cost reasons, I want to use an opamp as a comparator. So far in terms of making sure the opamp is suitable, I have: Making sure the output can slew fast enough Making sure it can handle large differential inputs Making sure it can go rail to rail Ensuring that the current draw at maximum swing is acceptable Are there any other crucial points? <Q> This application note answers exactly your question and covers a few less obvious issues associated with incompatibilities that may arise due to the different optimization and expected usage of op-amps and comparators. <S> Give it a read: https://www.analog.com/en/analog-dialogue/articles/amplifiers-as-comparators.html <A> Yeah you can absolutely use operational amplifiers as comparator, however if you want to consider the frequency or time factors, then you should consider CMOS rather than opamps. <A> I have: <S> it is a lot more application specific than opamp vs. comparator specific. <S> Making sure it can go rail to <S> rail output R2R is not as important than input <S> R2R. input R2R is not as important as the ability to swing to ground (for inputs). <S> I would also put current drive up there as well.
| Making sure it can handle large differential inputs handling small differentials accurately is a lot more important than handling large differentials accurately.
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Outcome when logic output of IC shorts to VCC I wish to know how to determine what happens when the logic output of an IC becomes shorted to VCC (preferably without actually having to short it). I realize that there are a plethora of ICs out there, all with different properties. I have chosen to use the 74HC74PW so that I have an example datasheet to talk about. I also wonder about AND gates, shift registers, etc. Consider the example circuit below: simulate this circuit – Schematic created using CircuitLab SW1 is there to simulate a pin-short or other similar failure. Hypothetically speaking, If SW1 closes while Q is HIGH, then I'll assume no issue because VCC and Q are at equal potentials. If SW1 closes while Q is LOW, then what process should I go throughto determine the outcome? Brainstormed ideas (while Q is LOW): Q is low impedance, and V1 essentially shorts to GND. V1 and or the IC will be damaged. Q is high impedance, the IC is essentially unaffected, and the micro simply sees the wrong value (HIGH). Q is "medium/high" impedance. V1 is okay because it can supply ample current, but the IC is either damaged or effected in some negative way. Something else? The following table is from the flip flop's datasheet: Am I able to determine the failure outcome from that table? I have come across many ICs that don't disclose explicit output current or output clamping current though, so I'm wondering if there is a more general rule I could assume true when I the datasheet isn't helpful. (Perhaps I should assume all logic outputs are low impedance?) What specifically should I be looking for in an IC's datasheet, how should I analyze it, and what other rules can I use when the datasheet doesn't have said info? I imagine that relying on simulation software is risky for these kinds of conditions. I also don't want to rely on actually shorting the IC myself, because even if there appears to be no issue, it doesn't mean that there isn't one (especially over long periods of time). <Q> Philips NXP Nexperia has the HC(T) <S> User Guide , which shows in figure 33 what happens when you overload an output: <S> So the outputs will not go above a certain current. <S> But it's still too high. <S> This shows typcial values, not the possible maximum, but even so it certainly exceeds the I <S> O limit of 25 mA that can immediately damage the chip. <S> Actually, this graph is for 4.5 V. <S> There is another graph for 2 V, where the current stays below 10 mA, so you have to measure what happens at 3.3 V. <S> Anyway, the guide says in section 8.2: <S> The maximum rated DC current for a standard output is 25 mA and that for a bus-driver output is 35 mA. <S> These ratings are dictated by the current capability of on-chip metal traces and long-term aluminium migration, but it is expected that output currents during switching transients will, at times, exceed the maximum ratings. <S> A shorted output will also cause the maximum DC current rating to be exceeded. <S> However, one output may be shorted for up to 5 s without causing any direct damage to the IC. <S> maximum duration : <S> 1 <S> ms maximum duty factor : 10 % maximum V CC : 6 V Please note that shorting an output for 5 s will not cause "any direct damage", but this implies that there will be indirect damage that will shorten the life of the IC. <S> To avoid that, you'd have to keep the duration of the short below 1 ms. <A> CMOS is made from Nch and Pch devices with a designed RdsOn which was ~300 Ohms for CD4000 series at 15V and <S> 50 Ohms nominal for 74HC' series and 25 Ohms nominal for 74ALC' series with a wide tolerance. <S> This limits short circuit current and also affects transition current spikes with capacitive loads but generally optimized for speed and controlled impedance tracks at max speed using 50 Ohms approx. <S> Use the specs for Vol/Iol to get these for logic "0" and (Vcc-Voh)/Ioh for a logic "1" on P-channel. <S> There are some thermal and Vcc effects on RdsOn just like MOSFET switches, just as Vgs affects RdsOn on ~1V threshold switches. <S> From this understanding if you estimate effects of load impedance using impedance divider relationship. <S> Normal logic ignores since it is <S> high R but C values affect slew rate with RdsOn*C time constant, where C is internal, stray L,C in layout and input C of loads. <A> (If the datasheet specifies Iol(max) and Ioh(max) separately, you're interested in Iol(max) for pins shorted to VDD, or Ioh(max) for pins shorted to GND. <S> If the datasheet specifies Io(max) or similar, it's the same value for driving high or low. <S> I'll call the relevant one Iomax here.) <S> The output will deliver a current of at least Iom. <S> However, the datasheet value specified is the maximum current it can always deliver from every one of these ICs they make, so it's value <S> is really the "lowest of the maximums". <S> The actual current out of each individual IC (Ioactual) will be somewhat higher. <S> This causes a power dissipation in the totem-pole output's high or low transistor. <S> So this can be predicted to be (Iomax x VDD) watts but is really (Ioactual x VDD) watts for each IC. <S> It then depends on how long the design and structure of that IC can handle that very localised power dissipation in that transistor before it's damaged or destroyed. <S> As an aside, I used to use 74LS and other 80's logic chips and if I inadvertently shorted outputs or connected logic outputs together, the chip would almost always be damaged and need replacing. <S> Many modern logic ICs seem to be very tough, relatively speaking, and I've shorted out FPGA pins, memory chips <S> I/Os and such like for minutes before realising my mistake <S> and they've survived and lived long and happy logical lives. <S> Not what I'd bank on in production but from a lab' playing perspective <S> , they're pretty bomb-proof to me. <A> General rule is... Never attach an output pin to either rail.... <S> Clamping current is the maximum current the protection diode on the pin will withstand if you try to drive the pin with a voltage outside the supply range of the device. <S> You can ignore that in your current question. <S> If you tie a driven pin to a rail you will definitely exceed Io, and ultimately damage the device. <A> In this case, the maximum output current is ±25 mA which is basically how much current that output can withstand without damage. <S> If you short Q to VCC when it is high, nothing much will happen.
| If you short Q to VCC when Q is low, you will be sinking a pretty high current through the output devices on Q, which will cause damage to the IC. The life of the IC will not be shortened if not more than one input or output at a time is forced to GND or V CC during in-circuit logic testing (“back drive”) as long as the following rules are obeyed:
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May I use relay rated for 5A to switch 12A? I would like to use an electromechanical relay as an H-bridge, which is capable to drive 12A at 12V. Relays able to control this current are big and expensive. Is it ok to use a compact relay with nominal max switching current 5A if I promise to switch it over always at zero current? <Q> Most likley not. <S> The relay rating depends on both losses and actual breaking capability. <S> Check the contact resistance and calculate <S> I^2*R. <S> You'll end up with a big number. <A> Adding to other great answers: <S> It depends on the relay. <S> If the relays spec sheet only specifies a switching current, then it's not a great idea. <S> If however it shows a max current separately from the switching current, then change-over at zero or low current is acceptable. <S> HOWEVER: <S> You also have to take into account whether you can guarantee switching will never occur when current is flowing, for example, when the power goes out. <S> A one time event like that can fuse your contacts together. <A> Aging from Arrhenius effects of temp rise from contact resistance and excessive conduction current in addition to unspecified reactive loads can be predicted to last a very short time. <S> Normally Relays are rated for >=1 million mechanical switches and 50k at worst case ratings from the best suppliers (OMRON) <S> But with your application I could probably burn it out at 500Hz rate using NC contact to interrupt coil current in about 3 minutes. <S> Your mileage will vary. <S> For more learned Engineer, read this <S> This tells every aspect of Relay design for general purpose types. <S> Automotive Relays are designed to cycle perhaps every 5 minutes for a 15A radiator fan using a 20A breaker. <S> The contact resistance varies with material type. <S> Contacts may have with alloy layers Palladium, tungsten, nickel plated with silver or not. <S> Many variables . <S> Always derate with DC current and reactive loads !! <S> read OMRON detailed component specs for these values. <S> Many other less reputable companies may not have this data. <S> FYI <S> All relays rated <S> < 2A are gold plated for microcurrent due to oxidation and current required to wet contacts ( usually 10%) <S> This microplating would obviously burn up with high arc currents. <S> Power Relays are defined by ratings > <S> 2A are not gold plated but depending on vendor quality have temperature resistance materials with excellent contact design. <S> (See link) <S> Added info <S> But this is not a good design practise. <S> Switching delays cannot guarantee Zero current for AC loads even with a ZCS pulse triggered off. <S> Normally it is the switched arc current which determines additional contact temperature rise while steady current temp rise may be good at first. <S> Long idle periods with oxidation buildup can then aggravate the thermal rise. <S> So it depends why you are doing this compromise on integrity <S> when 25A 12V contact relays are cheap.
| If can you guarantee zero switching current, you may get longer life than minimal and have an exceptional circumstance.
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Are there any cases where single-cycle is better than pipelining? I've been asked by my professor When pipelining is better than single-cyle MIPS CPU's? I actually answered "always", but I'm not sure that's the correct answer. Excluding an increase in design complexity, and the added complexity in handling hazards, from the point of performances doesn't pipelining always give better ( or equal ) results? EDIT: I omitted I was talking about the MIPS architecture. <Q> It depends on what you mean by "performance". <S> Pipelining generally improves throughput, measured in terms of results per unit time, but it increases latency — the time that elapses from the beginning to the end of any particular computation. <S> Sometimes the latter is more important than the former. <S> In particular, if the code has lots of conditional branches with only small numbers of other instructions between them, the latency associated with deciding which way each branch will go makes it impossible to keep the pipeline filled with useful instructions. <S> For example, suppose that converting a single-cycle implementation to a five-stage pipeline allows you to double the clock speed, but that a conditional branch requires you to flush the pipe. <S> If 20% of the instructions are conditional branches, the pipelined implementation is already reduced to the same performance as the original implementation. <S> A single-cycle machine uses every clock cycle, so at 1GHz, it takes 5 ns to execute 5 instructions. <S> The pipelined machine runs at 2 GHz, so <S> it decodes the 5 instructions in 5 clock cycles, or 2.5 ns. <S> But it takes another 5 cycles (2.5 ns) until the result of the branch is known, during which no other instructions can be issued to the pipeline. <S> It is only after the 10th cycle that the next useful instruction can be decoded, which means that this machine also took a total of 5 ns to execute those same instructions. <A> Since interrupts cause immediate stacking and branching, what is currently in the pipe may need to be aborted and flushed, or the interrupt may be held off until the current pipe clears. <S> Neither is advantageous if the response time to the interrupt is critical. <S> In the worst case scenario, with random interrupts, a pipe-lined processor may actually cause the system to fail catastrophically. <S> Can a processor be designed to circumvent these issues? <S> Possibly, but that does not answer your question as asked. <A> One advantage of a single-cycle CPU over a pipelined CPU is predictability. <S> This is important if you're using the processor for timing-critical operations, such as low-level "bit-banging" or real-time processing. <S> An example is the Sitara processor used by the Beaglebone. <S> In addition to the main ARM processor, the chip contains two PRU (programmable real-time unit) microcontrollers, which have a single-cycle implementation instead of pipelined. <S> (Details in this presentation .) <S> Since each instruction predictably takes 5ns, you can ensure that real-time requirements are met. <A> Dave gets it right, but his examples are hard to follow imo. <S> Pipelining is the trade-off between throughput and latency. <S> If you pipeline, you've added latency. <S> A simple example: if you have a DIVIDE function that takes 4 pipeline cycles and an ADD function that takes 1 pipeline cycle, but all operations have to go through the entire pipeline to get to the output, ADDs would now take 4 cycles rather than 1. <S> The advantage of this pipeline is that if you were dividing lots of things, after 3 cycles, you could get 1 DIVIDE happening every clock cycle whereas without a pipeline, you could only get 1 DIVIDE every 4 clocks. <S> We've traded 3 clock cycles of latency in the ADD function for higher DIVIDE throughput after the pipeline gets going. <S> Note that this example is simplistic and any hardware architect would try and find ways around the latency penalty for the ADD function, but this is the basic idea of the tradeoff pipelines give.
| If you are designing a hardware control system that is heavily dependent on a myriad of interrupt signals from various sensors and timers, a pipelined processor may seriously degrade the obtainable performance of the system.
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Voltage divider with battery load I'm trying to find the current across all the components but am having trouble doing so because I don't know how to deal with the extra battery in this case. I've found that without the battery V(out)should be 4V, so is this cancelled by the battery? How does the battery add to current in this case? <Q> Since the battery will hold the voltage across the 4K resistor at 4 volts, and the voltage divider will also make the voltage across the 4K resistor 4 Volts, no current will flow to or from the battery. <A> Simply write down what the node voltages are, and use Ohms Law to get the resistor currents. <S> The battery currents will be determined by the resistor currents. <A> Wiki says: Superposition Theorem: <S> The total current in any part of a linear circuit equals the algebraic sum of the currents produced by each source separately. <S> Since you don't have any current source just replace once at a time one voltage source with a short circuit, and leave the other intact. <S> For example, if 10V is shorted first then the voltage across R2 would be 4V, hence a current of 4V/4koms = 1mA would flow through the resistor R2. <S> Now assume 4V is shorted and 10V is on, then R2 would be shorted too and no current would flow through it. <S> Hence the total current flowing through R2 becomes Itot = 1mA + 0A = 1mA.
| All nodes voltages are determined by the batteries.
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Any Hazards Associated with a High Current Conductor For a conductor carrying a low voltage (say 12V) and high current (say >500A), are there any hazards that we should beware of? I assume it is safe to touch the uninsulated conductor even when it is carrying a high current. I know when an inductor is involved, disconnecting the circuit could cause a high self-induced emf. But does that apply to any high current circuit? <Q> The inductive kickback phenomena is related to a rate of current change \$U_L=-L\dfrac{di}{dt}\$, therefor it applies to all voltage ranges. <S> The hazard of high currents involve risk of fire, burning. <S> The metal vapours are very dangerous since they have a high melting temperature, they are poisonous. <S> The metal vapour blowed in your face, results in almost cases in certain death. <S> The skilled electricians are killed by this vapour, not by electrocution. <S> Killed by arc: https://www.youtube.com/watch?v=J8auz60ma0A <A> The danger with high current circuits is short circuits. <S> Arcing and sparking is possible at low voltages (like 12 volts) when there is erratic contact within the circuit. <S> An example of low voltage arcing can be seen by disconnecting a cable from the terminal on your car battery and touching it intermittedly to the same battery terminal while the engine is being cranked, the horn is blowing or another high current load is switched on. <S> If you want a real convincing demonstration, short the battery terminals together with a wrench - ONLY KIDDING !!!! <S> (There's a good chance the battery will explode, blasting acid in every direction.) <S> Another often reported automotive electrical accident occurs when a mechanic wearing a finger ring gets that ring situated between the positive terminal of the battery, or <S> a between heavy positive cable which is carrying the battery voltage elsewhere in the car (e.g. t o the starter motor, or a distribution block), and chassis ground. <S> The ring stands a good chance of getting electro-welded to the chassis with the mechanic held in its grasp, and with a very burnt finger. <S> Moral to the story <S> : Respect both high voltage and high current! <A> Adding to already great answers. <S> It also depends if it is DC or AC. <S> If it's a wire carrying hundreds of amps of AC that can induce significant current in any metals brought close to the conductor, e.g. jewelry, rings and tools. <S> The latter can get hot very quickly. <S> If you are unfortunate enough to have a pacemaker or other metallic implants, stay away from high current AC lines. <S> BTW: That also applies, but to a lesser extent, to switching that 500A on and off too. <A> This video shows how to go blind/sunburned/dead with short-circuits on 230VAC. <S> Most of the effects would be the same at lower voltage if enough current/power is available. <S> Notice how the infrared blast from the arc causes the dummy's plastic facial shield to warp. <S> This one here is only 54V, but with enough batteries to fill a small truck... <S> nice fireworks, batteries catching fire, and don't stand near the fuses when they go boom. <S> It also has a few epic quotes, like " <S> the cable's insulation has fire retardant properties as long as it doesn't melt". <S> And this one is only 9V but stuff catches fire just fine.
| If the current should get redirected thru another conductive entity which is not entirely capable of carrying the resulting current, that entity could be damaged by the resulting resistive heating or by arcing.
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What is the modern way to do small scale programmable logic? I am designing a circuit for an electronic coil winder. It has a few binary counters, equality detectors, 7 segment decoders and flip flops. How it is possible to get all of this logic onto 1 programmable chip without resorting to totally retro technology? Is there something like a GAL that can is well supported for programming on a modern computer? I can easily make it from standard CMOS or TTL chips, but it would be a good learning opportunity to try programming my own logic. <Q> The way I'd recommend for your particular logic circuit is a CPLD. <S> Have a look at the Altera MAX10 family or the Lattic iCE40 devices. <S> You can buy a cheap demo' board like the iCEStick, download the free development tools and get something up to experiment with at home or work. <S> You will have to learn VHDL (my preference) or Verilog <S> but that's part of the learning you mentioned <S> you're interested in doing. <S> You could use a microcontroller to produce a similar result but by a quite different function... <S> but it doesn't address your question. <S> And you'd have to learn to programme that so there's work either way. <S> (I know Lattice call iCE40 an FPGA family, but across the logic chips market, they have more in common with CPLDs. <S> The name will do for the purpose here.) <A> Microcontrollers are the modern way, but if you have reasons to avoid them (safety, or you don't know C), you need to use low cost FPGA, from either altera (now intel), xilinx, lattice or microsemi. <A> Adding to TonyM's answer: <S> If you just want some programmable logic and don't want a sequential processor (a micro controller), here's a tiny iCE40 chip: http://www.mouser.com/ProductDetail/Lattice/ICE40UL1K-SWG16ITR50/?qs=XJu%252bLGjWfSDdbhkO3WpKug%3D%3D <S> There are also parts like this that give you little logic blocks to work with: http://www.ti.com/product/cd4048b
| The modern way to do such a digital logic circuit would be in an FPGA or CPLD.
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Does it matter if I connect an antenna via SMA or not? I am using this LoRa Radio shield and want to connect a simple ¼ wave wire antenna to it. There is a hole for soldering an antenna on close to the SMA jack, but then I would be skipping the 50 ohms of resistance the SMA connector provides. Would that be bad? I am very new to radio technology, so I am sorry if its quite a basic question... <Q> I would be skipping the 50 ohms of resistance <S> the SMA connector provides <S> There is no 50 ohms in that SMA connector, instead it has a 50 ohm characteristic impedance and that is something different. <S> You can just leave out the SMA connector and solder your antenna wire onto the board. <S> A SMA connector with a proper antenna might give better performance but your wire solution will also work. <A> Since you're a beginner, I won't go into details too much. <S> Basically, the 50 ohms is not the resistance of the connector, it's its characteristic impedance. <S> When dealing with "radio stuff", in order to maximize power transfer, you want to have all the parts with same characteristic impedance. <S> So you'd match the radio chip on the PCB to 50 ohms, use the 50 ohm connector, optionally use 50 ohm coaxial cable with 50 ohm connectors on it for the antenna, and finally, use a 50 ohm antenna, with appropriate 50 ohm connector matching the connector on the cable, to radiate out your signal. <S> Now the thing is that we don't know what type of antenna you want to use. <S> You should keep in mind that the impedance of the antenna is frequency-dependent, so antenna for one frequency might not have 50 ohms on another frequency band. <S> That's why you first need to figure out for which band is your board configured and then get an appropriate antenna. <S> Now, soldering a bit of wire to the via on the PCB might work as an antenna, but, since you're a beginner, you probably don't know how to cut it to appropriate length so that it works OK <S> and you probably don't have the test equipment needed to figure out if it's working OK or not. <S> Usually, it's not enough to just cut it to 1/4 wavelength, since the PCB and everything around it will also impact the impedance of the antenna. <S> So my advice would be to use the SMA connector and buy a factory-made antenna for your band of interest. <A> Unless you're absolutely certain that you'll never want to change the antenna, use a connector. <S> If you don't, then replacing the antenna means unsoldering the old one and soldering in the new one. <S> That's not going to work very well; it's a lot of work, and you'll eventually damage the board. <S> Imagine that in your house, instead of having light bulbs that screw in, you have them soldered in place: <S> whenever a bulb burns out you have to shut off the power, unsolder the old bulb, solder in a new one, and turn the power back on.
| The reason for using a connector is so that you can connect and disconnect the antenna easily.
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Choosing an optocoupler for a digital interface with old -ve polarity -5V electronics This is an amateur application - 'adding auto control to a manual motor control circuit' using an optocoupler to bridge opposite polarity circuits. The existing 20 yo motors x 2 (-12V) are controlled manually, forward and reverse, by push buttons. The push buttons switch -5V to motor control hardware (TTL presumably). I am adding auto control of the motors from a +5V (ST-4 guide camera) source. With help from Stackexchange experts, I have come up with a basic layout, below. Switch off auto to use manual. The push buttons will be physically guarded and wont get much use. This might need some modification to lockout inadvertent activation of the PBs. My questions... I need to choose an optocoupler (there are 4 channels). The optocoupler must be capable of a similar output as the push buttons (which I presume is logic level) and fast enough to be responsive to the ST-4 pulse guide commands. The circuit should not be overly complex, have few components and therefore be reliable - it's mission critical. I have perused a number of likely datasheets and to my inexperienced eye the FOD0721 look promising. It's fast and has an appropriate voltage rating. Is this a sane choice, overkill or just inappropriate for the application? What should I be looking for in an optocoupler for this type of application? simulate this circuit – Schematic created using CircuitLab This is the edited version based on discussion below. simulate this circuit <Q> I'd suggest that since you are unsure of the voltage and current levels on your switches (and you recognize that you don't have a lot of experience with electronics) you should use relays to activate your pushbutton circuits. <S> There are plenty of small relay boards designed for microprocessors such as Arduino that would allow yo to simply solder in voltage free contacts across your existing pushbuttons. <S> For example for about $5 you can get a 4 relay board with opto-isolated inputs that will run from 5 V. <S> You might not need opto-isolation, and those variants are available too. <S> Certainly this would seem to imply the absolute minimum of modification to your Telescope circuitry. <A> The optocoupler must be capable of a similar output as the push buttons (which I presume is logic level) and fast enough to be responsive to the ST-4 pulse guide commands. <S> The FOD0721 is not a good option for emulating a push-button switch. <S> Just use a standard opto-coupler which has an NPN bipolar transistor or MOSFET output (not Darlington or logic gates) and a good Current Transfer Ratio (eg. <S> Sharp PC812 ). <S> The opto-coupler is simply wired directly across the switch (in series with SW1, which I presume is for disabling it). <S> Don't worry about speed. <S> Any opto-coupler will be much faster than a pushbutton. <S> Mechanical switches 'bounce' for several milliseconds during operation, whereas the opto-coupler should switch cleanly in a few microseconds. <A> First your VDD2 is max 6 vdc, If you are going to send -12Vdc into the MOSFET <S> I think it's not going to like it, Second, are you sure of the motor current consumption. <S> What king of motor runs on less then 10mA. 10ma is the max current that the FOD0721 can handle, potential problem here. <S> Is V2 the actual motor, or just the input control of a motor. <S> Why the 750ohms ? <S> If SW2 does the job directly to the motor, why using a 750ohms for SW1-FOD721 series ? <S> Or it could be connected from -5TTL to the SW1 switch then to the R750 then to the motor... <S> Please clarify some of those enigmas so we can better help you. <S> Cheers... <A> I tested the circuit below on a breadboard (one side -ve polarity, the other +ve polarity and an ILCT6 bridging the gap). <S> A second look at the pushbutton wiring on the actual device confirmed ground side (low side?) <S> switching. <S> Input was a basic Arduino digital pin HIGH LOW sketch. <S> Emitter to ground (+ve) carried more current - brighter LED Irrespective of emitter to GND or collector to GND, the output voltage did not vary - on my multimeter. <S> Having also run this circuit in a spice program, I added a small cap across the emitter collector to supress voltage spikes in the order of 3 - 4 volts. <S> Great learning experience. <S> Circuit editor does not work with touch screen
| A standard opto-coupler should work just as well as a switch, so long as it can pass enough current with low enough voltage drop when turned on. ThePhoton is right, Your FOD721 should be connected between SW1 and the R750, not from GND to SW1, this way you are shorting the TTL signal to ground.
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Why there is no resistor in an AM Detector? Consider the simple AM detector below: Inductor L1 and capacitor C1 builds the "tank circuit", which purpose is to filter all frequencies, except the selecteded one. However, I'm having trouble to fully understand this circuit. In my understanding, the circuit above would not work, because the antenna is shorted with the diode D1. The diode is receiving the same voltage as the antenna, so the tank circuit is, theoretically, not interfering in the output. If I was trying to build a tank circuit to select a specific frequency, I would do something like this: simulate this circuit – Schematic created using CircuitLab With this configuration, the resistor acts like a "voltage divider", and the frequency plot would be similiar to this: (To select frequencies near 600kHz) Of course, if I try to simulate the same circuit, but without the resitor, Vout becomes the same as Vin, since there would be a short between them. But this is exactly the case in the AM detector! How it works, then? Any help is appreciated,Thanks! Obs.: The AM detector I showed is used in almost every AM received schematic I found. <Q> Perhaps you are misunderstanding the word 'detector', which is what rectifiers were called in the 'olden days'. <S> In your first circuit, the tuned circuit filters the incoming signal. <S> The antenna has a finite impedance, it's not a voltage source, so allows the tuned circuit to suppress the response at higher and lower frequencies. <S> Amplitude modulation is a way of transmitting a low frequency signal which is coded into the amplitude of an RF signal. <S> Mid level RF signal means zero signal. <S> Lower level RF means negative signal. <S> Larger amplitude RF signal means positive signal. <S> However, all this time, the RF signal is swinging positive and negative, with zero average. <S> D1 is shown without a load, but in reality it will be followed by a resistive load to ground. <S> When the incoming RF signal is larger, the peaks of the RF signal will be rectified by D1 and produce a large current in its load. <S> A small RF signal produces little current into the load. <S> The changing load current represents the signal modulated onto the carrier. <S> In your second circuit, you have the tuned circuit behaviour, but no rectifier, no detector. <S> For a big signal, the output swings a lot above and below ground, for zero average output. <S> For a small signal, the output swings a little above and below ground, for zero average output. <S> The average output then is always zero, regardless of the modulation. <S> You have 'filtered' the carrier, but you have not 'detected' it. <A> The tuned circuit is only loosely coupled to the transmitter by the antenna, this has a similar effect to the resistor. <S> You can model it by replacing the resistor (R1) in your simulation with a capacitor of value <S> say 10pF. <S> The diode will load down the tuned circuit and reduce its Q, often the inductor is tapped to provide a lower impedance drive to the diode that will reduce the loading effect. <A> The diode is receiving the same voltage as the antenna, so the tank circuit is, theoretically, not interfering in the output <S> An antenna always presents an impedance to the circuit it connects to therefore the tank circuit does become a selective filter. <S> A quarter wave monopole presents an impedance of about 37 ohms when it receives a carrier that has a wavelength precisely 4 times the length of the antenna. <S> See this: - When height is referred to in the above graph it means the length dimension of a vertical monopole and not the height that the antenna is placed above some altitude reference point. <S> Height in this graph means length. <S> At 1 MHz (wavelength = 300 m), a 75 m monopole presents about 37 ohms. <S> However, back in the days of crystal sets, nobody ever used an antenna that long so, practicality meant that the antenna was "short". <S> Look at the impedance presented by an antenna that is one fifth of the length of a quarter wave monopole (15 m in this example). <S> It has a theoretical impedance that is capacitive and about 1000 ohm reactive. <S> At 1 MHz that's about 160 pF. <S> The blue line in the graph indicates that at 0.05 wavelength the resistive part of the impedance is just an ohm or so (there are formula to get exact values of course). <S> The bottom line is that an antenna converts the impedance of free space (377 ohms) to an electrical impedance that can vary significantly and be complex in nature. <S> This forms the impedance by which your tuned circuit is allowed to resonate and block those frequencies not wanted and allow those that are wanted. <A> In my view, a very simplistic way to look at your circuit would be to interpret it as being two separate circuits. <S> The first part would be the Antenna: <S> Coil:Capacitor, which would represent your Resistor:Coil:Capacitor from your second circuit. <S> The Antenna in your initial circuit is just a converter of electromagnetic radiation into a source of alternating voltage, loosely coupled to the RF transmitter far away. <S> this is why we can consider that antenna like some kind of resistor between the RF source (transmitter) and the RF receptor (antenna). <S> Since the two are quite far apart, the coupling is low hence, resistor. <S> Now the second part of the circuit is the diode (can also be called the detector). <S> That diode will simply use the top half of the RF AC signal and turn it into an RF AC rectified signal, composed of only the top part of the signal, because the diode only conduct current in one direction (theoretically speaking). <S> Now, in that second part of circuit should be added another resistor, otherwise the circuit would not be completed and no current would circulate through it. <S> So if you add a resistor to the cathode of that diode and to ground, there will appear a current quickly increasing from zero to max, only in one direction and it will do that many times per second, based on the tuned frequency of the first stage of the circuit. <S> Sending this rectified signal to an acoustic transducer will then emit some sound pressure representing the transmitted modulation of the RF energy transmitted.
| The antenna may also couple too tightly to the tuned circuit as well and so may be connected by a small capacitor or to a tap part way down the inductor.
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Can a voltage divider circuit be used before regulator to reduce heat? For a project of mine I need to step down 12V to 5V and also 5V to 3.3V. However, I don't have any heat sinks for my voltage regulators. From the equation below, if I reduce the difference between the input voltage and output voltage of the regulator, I can reduce the power dissipation. If I use a voltage divider to step down 12V to 6V and then use the 6V as the input of the regulator, would that then decrease the heat significantly, or does the heat still generated but at the divider? Sorry for the noob question, it is my first time messing with power supply other than the 5V. EDIT: forgot equation for reference Power = (Vin - Vout) x Iout <Q> Your idea of using a voltage divider is not viable at all. <S> You have to dissipate the power in a linear regulation system somewhere, and typically would do this in the regulator. <S> .... <S> but you could add a series resistor to drop the voltage at your target load current ... <S> this would spread the dissipation between the regulator and the resistor. <S> You still have exactly the same power dissipation however, and you have to make sure you don't exceed the current limit you used to select the resistor. <S> You could also use a Zener diode or set up a pass transistor with a smaller resistor to allow you to spread the dissipation between multiple devices. <A> Voltage dropping through a resistor, or a transistor, from 12V to 5V, alwaysdelivers heat as the equation above expresses. <S> There is no exception forparts-added circuitry that only puts in more resistors, but there IS aless heat-producing scheme, using switchmode regulation and inductors. <S> If your plan was to use, for instance, a LM7805 regulator without a heatsink,suitable pin-compatible parts are available DC/DC converter thatare more efficient than that equation (and presumably don't need a heatsink). <S> Also possible, use a heatsink. <S> It can be ANYTHING, a dab ofglue and a big washer will do SOME heat dissipation. <S> There's anequation for the heat generated, and there's more equations for howthe package temperature rises, with and without heatsinking. <S> So,you can calculate. <S> The temperature of the regulator can be measuredunder load, so you can verify adequate heat shedding to keep the regulatorin its operating temperature range by testing with your load, whilewatching the thermometer. <S> Either by calculation or by experimentation, or by using different parts,three paths forward are open. <A> First of all, it is good to spread out heat dissipation across multiple components. <S> That's why for sure you should use some passive components for lowering voltage before your regulator. <S> Lets take the further example: First of all, you should know how to treat voltage regulator from aspect of power consumption. <S> Note that your regulator is taking some of your input current, so just for our current perspective, we can take next model for voltage regulator: <S> So, current for R1 and R5, wont be the same. <S> One part of R1 current will go through the R6 (consumption of regulator). <S> You can find this current in datasheet of the regulator you are using. <S> Now, if you know that you want to have 3.3V on output (Vout) of particular regulator, that means that next equation will do: R2*Iout = <S> 3.3V <S> On the other side of regulator, you want to lower your Vin voltage which will be: Vin = <S> V1 - <S> R1*I1 <S> Note also relations of current, where I1 = Iout + <S> Iu, where Iu is load current of regulator (according to simplified model, current through R6 resistor). <S> Now, when you know V1, you need to decide what should your Iout be (consult your datasheet for available range). <S> For example, you want it to be 500mA: R2*500mA = 3.3V1 - R1*(500mA+Iu) = <S> Vin <S> You would like Vin (in this case) to be something between 4 and 5V, so from two upper equations you can choose your R1 resistor <S> , presumably you know what your load (R2) is. <S> @dim <S> : Is this more suitable then my previous answer? <A> So the best solution would be the first, with low value resistor, but you will loose too much power in the voltage divider because of the current. <A> Like the above answers state, there's no getting around the physics of the problem: The amount of power you dissipate as heat from regulating the voltage is your current requirements times your voltage drop. <S> If this causes the regulator to reach a higher temperature than you are comfortable with or beyond the operating specifications of your part, then you can either heat sink your regulator, or cascade regulators, which in essence is just distributing the power dissipation onto multiple devices. <S> There are usually equations in good data sheets to help you calculate a) if you need a heatsink, and b) <S> The thermal resistance of the heat sink you require, given the maximum temperature of the part, and the ambient air temperature.
| This can work, but again you are simply spreading the power dissipation between devices instead of providing a heatsink for the regulator. Your voltage divider will consume current and if you try to put high value resistor to avoid this problem, your output voltage of 6V will decrease if you take too much current in output to supply the regulator.
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DC/DC converter, from 36V to 32V I have to do a DC DC conversion from 36V to 32V @1.5A. I found some switching regulators, but no linear regulators. I'm not sure using a switching regulator is the best solution (price, overkill, efficiency) to reduce voltage. What would you advise? <Q> Use a standard 3 pin adjustable linear regulator. <S> You can use those at much higher voltages since they do not directly use a ground reference. <S> Limiting factor is Delta Vin-Vout. <S> e.g. <A> Adjust you zener voltage accordingly. <S> You can stack several zener diodes in series to make 32+~0.7 V in total. <S> Choose a NPN which can handle more than your maximum input voltage plus margin, more than your maximum current and can dissipate (36-32)*1.5 W. <A> How about an op-amp voltage follower: - Plenty of op-amps will run above 36 volts so this shouldn't be a problem. <S> I might be tempted to add output current limiting as well if the 36 volt supply is capable of many amps. <S> The NPN transistor need to be a power transistor of course and requires a heatsink for the volt drop (4 volts) <S> x 1.5 amps of max power.
| A simple NPN-zener regulator would do the job.
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Driving Nokia TFT LCD with STM32 How do I increase the refresh rate of a Nokia LCD that I'm driving using the STM32F103 microcontroller? The LCD is 132*162 and takes a 9-bit instruction/DATA with RGB for each pixel in 565 format. Since SPI on STM32 can't do 9 bits, I'm using USART (running at maximum 4,000,000 bit/s max as I'm using HSPI) in 9-bits mode to drive the LCD. This works just fine, but the refresh rate is not that great. Right now it takes me 230 ms approx to write the whole screen. Doing rough calculations, (132*162*2*9)/4e6 = 97 ms. Mine is higher because of overheads + the fact that I need to mirror each byte before sending as USART does LSB first only. I am looking for some suggestions as to how can I reduce the refresh rate. Use DMA? I am not sure how useful it would be in this case as my microcontroller is just driving the LCD. Nothing else. Buffer the LCD display in RAM and send it out to the LCD in one go. I can't do as the microcontroller has only 16 KB RAM. Bit bang GPIO to drive the LCD. I am not sure if this would achieve a rate more that 4 MHz of USART. Plus with this I lose the ability to use DMA in the future if required. The LCD I'm using is used in the Nokia C100 series and use the controller SPFD54124B. <Q> You want something that looks like SPI, so I'd try to make SPI work. <S> Assuming your LCD (which you might want to specify) doesn't need some specific timing in between words, you should just use the word modes that the STM32F1 has to offer (8 or 16 bits, iirc), and push out 8 words at once – that'd be 72 bits, and that can be broken into 9 SPI transfers. <A> The SPFD54124B has a 4-pin, 8 bit SPI interface. <S> Simply use that instead of the 3-pin 9 bit "SPI-alike" interface. <S> See the datasheet page 118 following. <S> The four-pin SPI interface uses a "CSX": Chip Select (MCU->display controller) "SCL": Serial Clock (->) <S> "DCX": <S> Data||command (->) "SDA": Data (->) CSX, SCL, SDA can be handled fully automatically by SPI hardware on your MCU. <S> DCX would need to be a GPIO. <S> That's not a problem. <S> Really, I'm kind of sad <S> it took you so long to say which controller you use. <S> You could have had an answer much, much quicker. <A> You are using LCD with serial interface. <S> That has big impact on the speed. <S> You should use LCD with parallel interface if you want to speed up refresh rate significantly.
| You can just set the GPIO pin to data, stream out your high-rate pixel data, then catch the interrupt when that is done, toggle the DCX pin, send commands, catch the interrupt when it's done, and switch back to data mode.
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What is the difference between the servo track and the line track of a motorized potentiometer? I have this motorized potentiometer , which I would like to control with an STM32 microcontroller. As far as I understood there are several parts in this device: a potentiometer a DC motor a conductive knob for touch sense I need all of these. There are 2 connectors. The 2 pins connector is for the DC motor. Nothing special there. But the 7 pins connector is described this way: pins 1 2 3: line track (R1) pin T: touch sense pins 1' 2' 3': servo track (R2) The touch sense is straightforward. I guess 1 2 3 is the potentiometer. But what is 1' 2' 3' then? I measured with a ohmmeter, and it seems to be the inverse potentiometer, which doesn't make sense to me. The question is: what am I missing? I found this related question , but with no answer. <Q> I would suppose that pins 123 are what you use for the audio or other analog channel you're controlling, and 1'2'3' are what the motor/system uses to figure out where it is on the slide. <S> A similar device on the Sparkfun site explains: "The slide contains two separate 10k linear taper potentiometers so that you can use one as servo-feedback in order to read the position of the slider and use the other to control whatever your target is" <A> Servo track has a linear dependence of the resistance on the position and is used to monitor the position. <S> Depending on the last symbols marking three kinds of potentiometers characteristics are shown on the last page of the datasheet. <S> And about the touch sense track link . <A> Unfortunately the spec sheet is not very explanatory. <S> It appear obvious to me that the 1'2'3' servo track is to be used by the uControler to know the absolute position of the pot. <S> You could not do that with the 123 pins because those are used for audio signal. <S> Hence, 1'2'3' is repeating the same action as 123 but on a totally different pot, the Servo Pot. <S> I would use them for sending a DC voltage to an ADC input on the uControler. <S> See proposed schematic. <S> Furthermore, What puzzled me is the pin <S> T. <S> Is it only a one wire touching the slide metal ? <S> One could sense a 60hz buzzing signal at user touch and detect that someone is attempting to slide the pot? <S> A bizarre way of doing it but could work. <S> Any other idea on the T pin ? <S> Cheers :)
| Line track has a complex dependence of resistance on the position and is used to adjust the analog audio signals (main purpose).
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How to measure voltage in a PS/2 keyboard cable? I have an old PS/2 keyboard, I doing a future project with it, but for now I just want to measure the voltage in the cable. However I don't know how to do it, so let me explain. It's a PS/2 keyboard , and I have a P2/2 adaptor that transforms it into a micro USB cable that get's connected to the PC. It has 4 wires: Red: VCC power Black: Ground White: Clock Green: Data The circuit looks basically like this (apology for my drawing) The mini USB is connected to the computer, then it's transformed into a PS/2 cable via an adaptor. The adaptor acts as some kind of resistance. Then the wires are connected to the keyboard via a PS/2 port. I have basically drawn the 4 wires as they are there, I have removed the external insulator cable on that segment in the drawing so that I can measure the voltage. I have measured the voltages at the points illustrated on the picture. A1->A2 = 0.6 mV DC B1->B2 = 0.6 mV DC C1->C2 = 2.2 V DC D1->D2 = 5.22 V DC I have also measured the voltages between the wires, after the adaptor: A2->B2 = 5.24 V DC C2->B2 = 5.24 V DC D2->B2 = 5.24 V DC So I guess the (black) ground cable is like the return path of the circuit since A2,C2 and D2 have 0 Volts between them, they are only connected with B2 separately. Meanwhile I have also measured the voltages before the adaptor: D1 had 0 Volts between all other points A1->B1 = 5.25 V DC C1->B1 = 3.05 V DC A1->C1 = 2.2 V DC Question The question is what is the voltage in the "Yellow Area", meaning in the wires that go out of the adaptor and go into the keyboard. So my goal is to determine how much watts of power the keyboard uses, so I would need to multiply the voltage in each cable "going out" with the amperes going through each wire and add them together. How to measure exactly the voltage in the yellow area? So which measurement is the correct measurement from above, in order to determine how many volts there are in the wires in the yellow area. And use that volt to multiply with with the amperes flowing through each cable to determine the watts of current that the keyboard uses. <Q> I don't really care about the battery though, <S> what I am interested in is what is the minimum watts needed to use the keyboard. <S> What other way there is to decrease power consuption? <S> First thing you need to think about that traditional keyboards were not designed for low-power applications, thus if you think you are critical for power probably you need to use some special input device especially designed for these purposes. <S> Look here . <S> Power consumption will depend on what keyboard is currently performing. <S> If it is idle there will be small current, if you press 10 buttons on it and have all LEDs turned on, it will be another current and thus another, bigger power consumption. <S> In my opinion, the only way to properly decrease power consumption is to turn keyboard device off when it is not expected to be used. <S> When operator or application will need to use it, you should refer to maximal device's current per its datasheet. <S> I just took PS/2 keyboard I have, and on its rear side there's label saying that it requires 5 V DC. <S> No information about current though. <S> There might be information on the web about maximal PS/2 ratings, I found this one which says max current is 275 mA. <S> Also please keep in mind that you are currently thinking or talking about your particular case with particular keyboard. <S> Imagine you are designer of consumer product, which can be used virtually with any keyboard, thus you can not know which exact current attached device takes. <S> And thus you should refer to minimal and maximal definitions within the standard, planning your device operating cycle accordingly. <A> I'm not sure you understand your keyboard or your PS/2 to USB adapter <S> well yet. <S> Let's start with the PS/2 to USB adapter .... <S> there are two types: A Passive convertor with nothing but wires in it. <S> In this application the PS/2 keyboard itself can output the serial TTL or USB signals. <S> As soon as 5 V VCC is applied the keyboard tests to see if it is receiving a clock, and if not it goes to USB mode. <S> An active PS/2 to USB adapter. <S> This is actually a small MCU that talks serial TTL to the keyboard and USB to the host computer. <S> These adapters typically have both a serial PS/2 and Serial Mouse connector (Green and Purple) to USB 1.0. <S> You might look at the questions on this site ... <S> it might help you. <S> You could also Google for PS/2 to USB adapter schematics and get a clue as to their operation. <S> To get to your question. <S> While you could measure the current in each wire, the results would be adequate if you simply measure the 5 V current consumption of the keyboard. <S> It will vary from just a couple of mA (more modern keyboards) through to 100 mA or ( the classical 8031 processor based keyboards) so no matter whether the keyboard is feeding serial TTL or USB. <S> In addition, some keyboard processors are actually in a deep sleep mode until a key is pressed or data sent to the keyboard (you see this particularly in battery powered wireless keyboards, but I've seen it in many USB keyboards). <S> In these cases average current consumption will vary from sleep states around 10 uA to 40 mA bursts while scanning the keyboard array. <A> My goal is to determine how much watts of power the keyboard uses, If you have a DMM capable of 1mV accuracy, break/insert an R=10 Ohm resistor on Red wire (+5V) from PC side (how is up to you) <S> The voltage drop for this R displays current, as I=V/R so 50mV/10 = <S> 5mA. <S> We normally choose a current shunt in this range so the voltage drop is negligible yet readable on a DMM. <S> Without a current probe, this is the only way. <S> To see what ROOT USB HUB current is allocated, goto Device Mgr and try each USB ROOT HUB > Properties>Power and see what current is allocated to target. <A> Ps/2 and usb are both 5V bus protocols. <S> The data lines use minimal power. <S> It will most likely be under 500mA. <S> Most keyboards are about 100mA or less. <A> @JackCreasey has the correct explanation of how the USB/PS2 keyboards work. <S> If you just want to measure the voltage/current, I have a handy little device that does that- <S> it has a USB male at one end and a female at the other. <S> I don't remember the cost, but it was very cheap- <S> I bought a few of them to have around. <S> There are a few variations with different display technologies. <S> It has an LCD display inside and an MCU to measure the voltage and current, and it will integrate the current to give mAh. <S> The 80mA shown is from a small FPGA board- <S> I tried measuring a no-frills Dell keyboard (no LEDs, not even for caps lock), but it was too low to register with 10mA resolution.
| Simply measuring the current through the power lines is enough to accurately measure the power usage of the keyboard.
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Scope probe having ringing with 10x option I did a bit of a search in the forum but despite there being a few related topics related I couldn't find the specific answer for this question. Using a 50Mz scope I connected the probe to an FPGA output pin where there's an output of a 3.3Vp@25MHz square wave signal. I did make sure to connect the grounds as short as possible and close to that clock signal. Using 1x and 10x scope probe options I got the following results: 1x probe: 10x probe: Well, for the 1x I think that I'm getting attenuation given by the low pass filter of the 50MHz scope for a 25MHz signal, and that could explain as well why the signal is not square and the amplitude is lower... But the part I don't understand is the result of the 10x option: why is that ringing happening? And why is it so different to the 1x? Obviously that overshooting is causing that the signal to increase the amplitude to 5.39Vp... <Q> As @user2233709 said, first adjust your probe's compensation . <S> If the compensation is not set properly, you won't see a sensible trace on the oscilloscope. <S> Second, a 50MHz scope won't do a good job of showing a 25MHz square wave; it doesn't have enough bandwidth. <S> If you look at the Fourier series of a 25MHz square wave, it has all the odd multiples of 25MHz: 75MHz, 125MHz, etc. <S> A 50MHz scope will roll off those high frequency signals, and the result will look closer to a sine wave. <S> Third, since the signal with the 1X probe looks pretty much like a sine wave, it suggests that your circuit is getting loaded down by that probe. <S> That's why you switch to a 10X probe, but even then, it may be too big a load, and just looking at the signal may be distorting it. <A> Most likely you are ignoring the mismatched probe impedance with a long ground wire to the clip. <S> Any rise times < 50ns must be done without a ground clip and tip using the tip and barrel , otherwise the probe inductance is added to the signal and will ring at say 50MHz for a typical long probe ground. <S> This assumes that the Risetime and overshoot is already calibrated in the probe and this 1st order filter does not affect this ringing, other than slight gain change with Cadj. <S> A 1mm wire diam with length <S> 100mm is about 100nH and 2mm ( or 2% of length) <S> diam is about 10% less and 0.1mm <S> thick ( D=0.1% of length) is about 150nH. <S> While two parallel L's are always 1/2 inductance of same D/l ratio, of a single wire. <S> It is the cable capacitance and ground wire inductance and your test method at fault.. <S> Proper 200MHz measurement method for textbook waveforms from ideal source. <S> Twisted pair of about 120 Ohms is another method. <S> Impedance is the ratio of conductor diameter to gap ground or adjacent signal. <S> e.g. track width/gap can be around 50 Ohms with e=4 dielectric. <S> Of course long breadboard wires will do the same thing. <S> N.B. <S> It is NOT the probe differential capacitance at the tip with 10M , but the entire coaxial cable to scope at some 20pF/ft or 60pF <S> /m so a 1m coax probe cable of ~60pF and 100nH 10cm ground clip will ring about 60MHz <S> (mental calc, U check) <A> Good probes have a variable capacitor to compensate the internal capacitance of the oscilloscope’s input. <S> You should adjust it, using the test signal good oscilloscopes provide. <S> The oscilloscope capacitor behaves like a 1MΩ resistor paralleled with a few-pF capacitor. <S> But it also build a low-pass filter with the internal capacitance. <S> So a compensation capacitor is added, paralleled with the 9MΩ resistor, so that the voltage divider is frequency-neutral. <S> If that compensation capacitor is too low, you get a low-pass filter. <S> If it’s too high, you get a high-pass filter (what you get). <A> Is this a 1x/10x switchable probe? <S> If so, your probe likely doesn't get the rated bandwitdh in 1x mode, giving you only the fundamental, and a quite attenuated one at that. <S> See this video by EEVblog Dave L. Jones for more information: https://www.youtube.com/watch?v=OiAmER1OJh4 <S> For the 10x: Fourier analysis teaches us that there is a lot of high-frequency content in a square wave. <S> This means that your 25MHz square wave also have frequency components at 75, 125, 175, ... MHz. <S> This could also lead to a quite distorted signal - A quick glance seems to indicate that you ahve the first and third harmonic only (square wave has odd harmonics). <S> On top of that, make sure your probe is well adjusted. <S> Most probes have adjustable compensation capacitors. <S> You need to tweak this capacitor (ideally with a teflon or nylon screwdriver) using the compensation terminals on the scope - explaining <S> the entire procedure here would be outside the scope (hah, see what I did there) of this page. <A> Here is model of 10X scope probe (using just LRC model); <S> The IC has 31 Ohms Rout, providing the dampening. <S> The probe is 15pF and 100nH (that ground lead). <S> The ringing is about 300MHz. <S> In your case, I suspect the FPGA has VDD & GND ringing, <S> which the output driver Rout filters when loaded by a 200pF? <S> X1 probe. <S> Can you examine one of the outputs set to logic "0" to see GND, or set to logic "1" to examine internal VDD, as the FPGA continues its clocking? <A> The ringing you're seeing on the 10:1 probe mode is what the signal actually looks like. <S> The 1:1 setting is generally severely bandwidth limited, so it won't pass those higher frequency components to the scope. <S> If you want to learn more about why that would matter, you can check out the Keysight Oscilloscopes YouTube channel, I've talked about both of these issues in recent videos. <S> https://www.youtube.com/keysightoscilloscope
| A 10× probe is a 9MΩ series resistor that build a 1:10 voltage divider with the oscilloscope’s input.
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Does it matter which 2 nodes on a CAN bus you terminate? I know you need to terminate the 'ends' on a high speed CAN bus but what if you know nothing about the topology of the network or where the bus will go. For my application i have designed a data acquisition unit for a vechile and want the ability to add nodes to the network as i please. The location of the devices that may be used is unknown so in that sense im finding it difficult to define what the ends of the bus are. Can i terminate the bus on the same circuit and have connections to the bus through that device [node1] [node2] | | ------------------(connection to unit)---------------------- | | [Termination]===================[termination] | [internal node] <Q> Each beginning has its end. <S> This can't be answered, because the question is not correct. <S> What you have is the transmission line - a twisted pair that has two ends, and that's the place where the termination has to be done. <S> You can't choose the end of the bus, since there are only two of them. <A> It doesn't matter which nodes are at the ends, but it does matter that the terminators are at each end. <S> The bus is a transmission line. <S> To keep edges from reflecting at the ends of the cable, the cable has to be terminated with its characteristic impedance. <S> The common standard for CAN is twisted pair with 120 Ω impedance. <S> You therefore need 120 Ω at each end of the cable. <S> This 60 Ω load is considered in the drive levels and signaling levels. <S> In CAN, termination of the bus serves a second purpose, which is to passively hold the two lines together when nothing is driving the bus. <S> This is like a pullup resistor on a open collector bus. <S> The terminators are also a pull-together resistance. <S> You have another confusion. <S> You terminate the bus, not nodes. <S> Therefore your question of which to nodes to terminate makes no sense. <S> The bus is terminated at its ends, and the nodes can go anywhere along that bus. <S> The bus can extend past the last node at each end, for example. <A> Marko and Olin are both correct. <S> This is what the network should look like from a topographical sense: [Node1] <S> [Node2] <S> [Node3] | | |[Termination]-----------------------------[Termination] <S> The nodes are daisy chained and the ends of the bus are terminated. <A> I wanted a similar sort of topology (also for in vehicle daq) <S> a while back, and it isn't that easy to deal with. <S> If your customers are going to configure the bus then things get much trickier. <S> The best solution really is to slow the bus down (and importantly, get a CAN driver that can reduce its slew rate). <S> By doing this you reduce the high frequency components of the CAN bus signal and reflections become less of a problem. <S> This is essentially what low speed CAN and LIN bus do. <S> If you want to run differential high speed, then you could make it work if you keep stub lengths short, control slew rate properly, and configure the CAN sub-bit sampling times well, but you must be very very careful.
| If you are the one who will always configure the network, then you can just use T-junction style connectors and remember to always terminate the ends of the bus while keeping stub lengths short. If you go down to 125kbps and the slowest slew rate allowable at this speed, you can pretty much have bulk termination on the main module, and small terminators on the nodes out to about 3m stub/star lengths without any issues.
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Generic name for Molex(tm) Mini-Fit Jr(tm)? Standard definition? The PC industry uses a large variety of power connectors which were originally introduced by Molex, using their trademarked (and search-engine unfriendly) brand name "Molex Mini-Fit Jr". This started with the venerable ATX power supply connector shown below: And continued with a wide variety of other power connectors, notably the 6-pin and 8-pin PCIe auxiliary power connectors used by GPUs (pictures below). Question: What is the non-trademarked generic name for this family of connectors? These connectors are manufactured by several other companies (AMP, TE, FCI, Amphenol, etc) so it's not a Molex-proprietary technology, and I don't think it would have become so widespread if it were anyways. But the other manufacturers list their products under nothing more than a long sequence of digits, like TE's "1586041-6". I would like to be able to search for connectors in this family without limiting myself to Molex products, which are the only ones that can use the trademarked term "Mini-Fit Jr". There must be a generic term; I'm not a lawyer and don't want to devolve this into amateur legal debates, but my understanding is that in order to maintain a trademark ("Kleenex", "Xerox") there must be a generic term for the product ("Tissue", "Photocopier"); if one doesn't exist then Molex can't enforce their trademark on "Mini-Fit(tm) Jr(tm)" Additionally, I'm hoping to find the standard that specifies these connectors. After using and examining a lot of them, there appears to be a method to the madness of the keying. You'll notice that some of the pins are square and some are chamfered. Right off the bat that prevents upside-down insertion so long as the male connector (which has a female housing) contains at least one chamfered pin. However it also helps prevent insertion of incompatible connectors: for the vast majority of connectors in this family, each column of two pins has one chamfered and one un-chamfered pin. This means that for a two-row connector with n pins there are 2^(n/2) possible keyings. The ATX connector above is a great example of this. Another example is the 6-pin male PCIe connector shown below: However there are exceptions! I've come across GPUs before with the connector below that have two chamfered pins in the middle column: These require a cable that also has two chamfered pins (chamfered male requires chamfered female, but square male will mate with either chamfered or square female) like this one: Clearly there are a lot of subtleties here. Molex has an irritating habit of not fully describing the connector in the product name; for example the connector above is "Conn Power HDR 6 POS 4.2mm Solder ST Thru-Hole 6 Terminal 1 Port" which does not distinguish between the (at least) two possible incompatible keyings! All of this brings me to the second question: Is there a standard that names the possible keyings so that a particular keying can be searched for? A lot of great information on this topic (but, unfortunately, not the answer to my question) can be found at All about the various PC power supply cables and connectors <Q> The EPSV12 v2.0 specification actually names "Molex _____ or equivalent" in the specification so it's a bit more than a de facto but not quite de jure. <S> Sometimes politics gets in the way of those tables being complete (for example, DigiKey keeps track of drop-in replacements <S> but I've noticed they don't generally list any for Molex products). <S> Molex helpfully (/s) has a database that will tell you the equivalent Molex part number, but of course it doesn't work in the other direction. <A> If you want to find equivalent connectors, start with the pitch on any distributors website and start mulling over connectors. <S> Then get the mechanical documents and compare them. <S> More often then not there aren't equivalents. <A>
| Several suppliers keep track of equivalency tables for the various part numbers, but not all do. According to Wikipedia , the technical term is a pin and socket connector , although the vernacular (in my experience) tends to be either "molex connector" or "motherboard/CPU/whatnot power connector".
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Idea/solution for building target to detect nerf dart accuracy? I'm doing some modifications to a nerf gun, and have already done speed/FPS tests with a chronograph I built, and the next test is modifications that can be done to increase the accuracy. I'm trying to build a target that will sense the x/y position of a dart as it hits a target, or breaks a plane. I have a few ideas, but I'm trying to keep it as simple as possible. One idea involved a mesh grid of wires, and when the dart strikes the grid, it would cause wires to complete a circuit. Problem there is thats a lot of wires (to get any sort of accurate reading), a lot of points of failure (accidental wires touching), and what sounds like a huge pain in the butt to make. Another idea was an array of IR leds and sensors. The issue there is then I have to rig up an absurd number of IR receivers, and write an function to determine just which sensors were blocked by a dart. Sounds complicated and like a lot of work. Are there any sensors out there that react to light (IR, laser, visible, whatever) that are like a strip, and will provide an analog result depending on what section is lit or unlit? Or a sensor that uses a spinning laser that can detect it's position/angle when the laser bounces off something? Or am I making this a lot harder than it needs to be, and there's an easier solution than what I'm coming up with? I want to fire a dart at, or through, a target, and get an x/y coordinate reading on it. I'd like it to be moderately accurate (1/4" or 1/2" resolution?). Better is great too. <Q> A flexible transparent resistive touch panel could be the answer to your design problem. <S> This kind of panel might be able to cope with the dart impact force, and will give you a reading from which you can calculate the XY position of the contact point. <S> Resistive touch panels are 4-wire <S> analog sensors need to be interfaced through circuitry including ADCs, clocks and special-purpose controllers. <S> However, this circuitry is commercially available as breakout boards, not too expensive, and some of them come with USB interface for extra convenience. <S> The main challenge could be finding a panel with the size required by your application, but reasonably priced as well. <S> Only you know your budget constraints! <S> As an example: 17-inch panel with controller . <A> You could do this fairly easily with a computer and a webcam. <S> Set up the webcam to be focussed on the target <S> so it fills the image and have software to compare a before and after image to identify the projectiles position. <S> With a decent camera you ought to be able to able to measure to much greater accuracy than 1.4 inch. <S> but before it hits it. <S> That will remove any bounce / deflection effect of the impact. <A> How about a sheet of something with microphones or accelerometers in several places around the periphery, and measure the relative time of arrival of the collision impulse? <S> This is not an original idea, I recall reading a writeup somewhere, perhaps on hackaday. <S> On a cruder end we put an accelerometer on a breadboard, hung it from a DC power cable and used the breadboard itself as a nerf gun target in a hackathon project - but that was only a single sensing point. <A> You could use a distance sensor attached to a servo that sweeps a plane. <S> Using several of them would increase accuracy by averaging the distances. <S> Use the angle of the servo and the distance reported by the sensor to calculate the position on the plane the dart passes through. <S> This would be a simple Arduino project. <S> The two issues being the required accuracy and the chance of not "seeing" the dart. <S> Using multiple fast servos with LIDAR sensors would help with both problems. <S> They are not cheap though. <S> Here is a newest LIDAR from sparkfun <S> https://www.sparkfun.com/products/14032 <S> The accuracy is +/- <S> 2.5cm at distances greater than 1m, so you would want to keep it under 1m. <S> I also found a fast servo from hobbyking, rated at 0.08s/60° at no load. <S> You already calculated the speed of the dart at some distance from the gun so, given the length of the dart, will it be breaking the plane for at least 1/10 of a second? <S> If not, just add another sensor sweeping in the opposite direction. <S> I imagine a simple wood frame maybe 16" square with a servo/distance sensors attached at opposite corners.
| The ultimate would be to configure a system to fire the camera shutter as the projectile crosses a plane just in front of the target itself
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Indicating 3V3 and 5V with 2 LEDs I am building 3V3/5V source for my breadboard. The actual voltage will be selectable by the switch. I would like indication, which voltage is actually selected by lighting green LED for 5V and yellow for 3V3. Treshold for 2N7002 is like 2V, so my idea is like that: simulate this circuit – Schematic created using CircuitLab But there is problem, that both LED would be lit on 5V, can I somehow simply cut off the yellow (3V3) LED when the green one (5V) gets lit? Is this schema valid (maby after some resistor adjust)? Thank beforehands EDIT: it is just small part of testing environment, meant to anable by jumper 5V or 3V3 output on VCC (or none, if jumper missing). There will be more such device on one PCB and there will be jumper array to configure each, so I would like indicate, what is actually on which VCCxx. Mode of operation: set jumpers, switch power on, do something, switch power off, reconfigure all the thing, set jumpers ... It is also possible, that the jumper will be off (SW1 not connected here) and the VCC would be else not connected, or externally driven to 3V3 or 5V. Those internal sources would be on anyway (and would provide power). So desired result is: VCC around 5V (say 4.5-5.5) = green led on, yellow off VCC around 3V3 (say 3.0-3.5) = green led of, yellow on VCC unconnected or around 0V(say 0-0.5) = both leds off <Q> First, using the gate voltage of a 2N7002 as a voltage reference is not a good idea. <S> A TL431 could be used since it has a nicely defined threshold of 2.5 V. <S> You can use two. <S> One is set to turn on at 3.3 V (taking all errors into account), and the other at 5 V. <S> You could use a lockout so that when the 5 V circuit triggers, it turns off power to the 3.3 V LED. <A> A better way would probably be to use comparators instead. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I can't be bothered to do all the math for you, so left all the resistors as their default values. <S> But you just set it up so that when it is a 5V supply, OA1 turns its LED on, and when it is a 3V3 supply, OA2 turns its LED on. <S> Although, bear in mind I am still learning electronics myself so other people may well have much better solutions than me, but this is my suggestion! <S> I have done this myself before <S> so I know it works. <A> If I was doing a discrete implementation, I would not use mosfets as switches like that. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I think this circuit will "fill the bill." <S> The zener diode is set to 3.3V. <S> When the VCC voltage rises above about 4V, <S> the comparator output shifts HIGH, "turning on" M1, which allows current to flow through the green LED and shorts the gate of M2 to (one diode drop above) GND. <S> Conversely, when the voltage drops below 4V, comparator output shifts LOW, "turning off" M1, which raises the voltage potential at its source to nearly VCC, which passes to the gate of M2, allowing current through the yellow LED. <S> simulate this circuit – <S> Schematic created using CircuitLab
| I'd use standard transistors and a couple of zeners to set the thresholds and tie back the 5V Led drive to kill the 3.3V Led when on as shown below. Shouldn't be too difficult to do it like this.
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480V 3-Phase AC to DC 12V I am trying to figure out the best approach to convert a 480V 3-phase AC to a 12V DC. I am thinking of first converting the 3 phase to a single phase AC then using a transformer and rectifier shown in this link Transformer and rectifier to get 12V DC. Is this the right approach or can anyone suggest better approach? To Convert a 3 phase to single phase can we use one of the phases and a neutral wire in a 3-phase 4-wire system and give it as an input to the above transformer and rectifier or is this a bad idea? Thanks <Q> I am trying to figure out the best approach to convert a 480V 3-phase AC to a 12V DC. <S> I am thinking of first converting the 3 phase to a single phase AC ... <S> You cannot convert three-phase to single phase using transformers other than by dropping two of the phases as you suggest in your last paragraph. <S> To Convert a 3 phase to single phase can we use one of the phases and a neutral wire in a 3-phase 4-wire system and give it as an input to the above transformer and rectifier or is this a bad idea? <S> Yes, it's a bad idea. <S> It puts all the load on one phase when you could have a three-phase balanced load. <S> Is this the right approach or can anyone suggest better approach? <S> Figure 1. <S> 3-phase rectifier. <S> Source: if a standard three-phase 400V AC connection is rectified what DC voltage comes out of it? . <S> The DC voltage obtained will be very close (about 95%) to the transformer peak output voltage less the diode voltage drops. <S> Therefore to get 12 V out the transformer secondary voltage should be: $$ V_{IN} = \frac {1}{\sqrt 2} (12 + 2 \times 0.7) = <S> 9.5\ <S> \mathrm V $$ Figure 2. <S> 3-phase transformers are available in sizes from 100 VA or so up to magawatts. <S> DC power supplies are available similar to that of Figure 2 complete with built-in rectifiers. <A> Phase to neutral for 480 <S> V 4-wire is 277 V. That is outside the 85-265 V range specified for the linked power supply. <S> You need a power supply with an upper voltage limit of at least 305 V to allow for 10% over-voltage. <S> A 15% margin or more would be preferable. <S> I would get a 480:120 transformer to use with a purchased power supply. <A> I have the same setup due to crypto mining. <S> solution 1a) use a 480vac 3phase to 208 3phase PDU such as Liebert PDUs (ie 225kva Liebert)b) connect 208vac 3phase breakers to power strip 3phase <S> PDU (ie 60amps <S> HP/Dell PDU which gives you like 6 sockets in C19 socket)c) connect the PDU sockets to a PSU such as the 2880w IBM PSU which gives you a bunch of 12V PCIE cables for the miners or whatever you're doing with the 12VDC solution 2 which requires no step down transformer/PDU a) wire up a bunch of 277v breakers in your 480 3phase panel and connect them to a single phase PDUs with like 6 sockets, wire each socket to a 1400watts 277v PSU such as the AMP1400Y1 / Z1 PSU which will gives you a bunch of 12V PCIE cables for the miners or whatever you're doing with the 12VDC. <A> I would suggest to rectify all three phases, then use a flyback DC/DC. <S> There are plenty of them, i personally like ST viper. <A> To answer the second part, if you have a 4 wire 3 phase system your rated voltage is 277v between one leg and the neutral therefore the transformer in the link will not work <A> Or... just go buy a 480-24VDC power supply. <S> Lots of people make them. <S> Line side will be single phase 480V line to line, load side is 24VDC. <S> Why make this more complicated than it needs to be?
| A very simple and economical approach is to step the voltage down with a 3-phase transformer and rectify it.
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How to have a micro-controller detect doorbell? I live in an apartment were everyone has an individual doorbell on the front gate. When the button is pressed for my apartment, 21VAC rings a bell in my unit. I would like to have a micro-controller detect when this is happening. My initial thought was to put a reed switch next to the bell, but it seems the reed switch is too sensitive and would get triggered when my bell was not on. My guess is that this is due to interference on the line from other apartment's doorbells, but I have not been able to verify this. Now I'm thinking of using an AC relay, but I'm having some trouble finding one that would work with my voltages. I'm also looking into using an optoisolator, but know very little about them and am not sure of the pros/cons vs. a relay. So my questing is, should I use an AC relay, an optoisolator, or something else?And for whatever solution, which part would be best? Thanks. <Q> From what you've said, you'd want the detection circuit to reject false triggers. <S> Let's consider those to be voltages below around 15 Vac, to allow for voltage drops in the line. <S> The circuit below should do that. <S> It should result in pulses around 1 ms for every AC cycle that the switch is pressed. <S> Your MCU can do more false trigger rejection by only acting on (say) <S> 4 trigger pulses within (say) <S> 25 ms of each other. <S> Your final circuit values may vary a little but the principle should be sound. <S> R1 reduces the strain on D1 when the door switch is pressed in mid-cycle and it has an uncharged C1 loading it. <S> EDIT: <S> Added R4 to discharge C1 down below the D1/D2 cut-off of 21-odd volts promptly when a Vac pulse is removed. <S> Discharges C1 from 21 V to flat in something like 30 <S> ms. Made zener voltage visible. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> I would use something like this: simulate this circuit – <S> Schematic created using CircuitLab <S> It allows you to connect the bell with a small load and most optocouplers are DC anyway. <S> IMPORTANT: <S> The schematic is just a design example, all components values need to be selected by yourself to fit your situation. <A> A reed switch is a good solution because there is no impact on system doorbell. <S> The residual magnetism is a problem. <S> And only when it comes to series of pulses with a frequency determined by the doorbell.
| You must create signal processing microcontroller so that the operation was not then when the switch is on or off. You can also use a Hall sensor.
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do I need to have a common 3.3V line connection for the I2C communication to function properly between dev boards I'd like to connect two Sparkfun ESP32 dev boards (a 3.3V device) through I2C where one would act as a master and the other as a slave. Each board gets its own LIPO battery, however it is only the master that might be connected through USB; this allows for both wired and wireless operation. Currently I have 5 common connections bridging the two boards: GND SCL SDA V-usb (5V) V-lipo (4.7V) This is convenient because the 5 wires fit neatly into a 5 pin, micro-usb cable connection (and I'll be building a custom cable to connect both sides). My question is this: do I need to have a common 3.3V line connection for the I2C communication to function properly between dev boards , or am I ok with just the 5 common lines detailed above? <Q> I2C is open Drain so source/load can be a different 3.3V but good common ground. <S> CMRR is rated at 60dB @1kHz 1Vpp <S> so in noisy environments ( motors, relays solenoids) a Ferrite CM choke may be needed. <S> (Torroid or sleeve or clamshell or other) Pullup termination R depends on cable capacitance and data rate. <S> Termination Versus Capacitance <S> Cp and Rp effectively limit the maximum data rate which can be transferred over SDA and SCL. <S> A high Cp can be compensated with a low Rp and vice versa. <S> The following three pictures show the same part of an I2C transfer. <S> Compared to the first one, the second picture shows the signals with a modified Rp of 2 kΩ, the third one with a lower Cp of 150 pF. Note: Long wires increase Cp dramatically. <S> I2C connections should always be as short as possible and connected by a suitable wiring pattern (c.f. I2C specification, section 17.3). <S> SDA (above) and SCL (below) with Rp = <S> 10 kΩ and Cp = <S> 300 <S> pF. <S> The SCL clock runs with 100 kHz (nominal). <S> The same transfer as above, but this time with a reduced termination resistance (Rp = 2 kΩ, Cp = <S> 300 pF) <S> The same transfer as above, but this time with a reduced wire capacitance (Rp = <S> 10 kΩ, <S> Cp = <S> 150 pF) <S> Note that the I2C standard limits Cp to the maximum value of 400 pF. <S> However, with an appropriate termination resistance, it is often possible (although not recommended) to operate I2C buses with higher capacitance. <S> Credit Ref: http://www.i2c-bus.org <S> Twisted pair varies with wire type and 12pF/ft or 40pF <S> /m is common. <S> Shielded pair can be up to 100pF/m. <S> You can measure rise time with R and compute cable capacitance. <A> You need a common ground reference for both integrated circuits. <S> Otherwise those integrated circuits would not know what voltage the I2C clock and data lines are currently at. <S> Of course, light weight buses such as SPI and I2C were intended to run between integrated circuits usually on the same printed circuit board. <S> They were meant to reduce the pins on the integrated circuits lowering costs and failure points. <S> They are not meant to go any great distance. <S> Certainly not as far as USB can travel. <S> And absolutely not as far as 10/100/1000Base-T or RS488 can travel. <A> Warning <S> It may seem convenient to use USB Micro-B connectors because they happen to have 5 contacts but you are highly discouraged from doing that. <S> At some point someone will get confused and think your setup is a normal USB connection and will try plugging to it accordingly. <S> Of course it will not work and in the best case the person will just be confused. <S> In the worst cast the attached USB device will get damaged or vice versa your device will get fried. <S> Also take note that only two pins of a typical USB cable have wires robust enough to carry any amount of current. <S> The others are designed as signal wires.
| Some of the more modern "skinny" USB cables have such small conductors that even the power capable wires are going to cause a significant voltage drop across the cable.
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Why does microwave test equipment use N connectors? Almost all of the microwave test equipment I've ever seen and used has N connectors, with exception of some scopes that use either custom fancy BNCs (keysight does/used to do this)) or SMA. Exceptions that I know of are VNAs which seem to use special connectors (but the ones I've used are also rated up to 110 GHz and used 1mm-based connectors). Of course, there is nothing wrong with the N connector - in fact I like it, it feels rugged and like it can take a good bit of use (is this the reason?). However, every single occasion I've worked with the instruments, or seen someone work with them, the first thing they do is put on a N-to-SMA (or 3.4mm or 2.4mm) adapter. So why not just put those connectors on the equipment in the first place? Is it really just the ruggedness, some other reason I'm missing, or is this some silly "because it used to be that way and nobody likes change"? <Q> Two Reasons: Pout and S11. <S> N connectors have a large surface contact area for >1A power levels or >10dB and large diameter means better "potential return loss over a wide GHz range is possible with machining tolerances of 0.1% affecting this. <S> However SMA is more common for low power apps but quality varies with undocumented suppliers. <S> (greatly and not gauranteed) <S> But never try to pump >>1 <S> Amp thru 1u" flash gold plated SMA connector (crap... <A> N is simply bigger - as you said, ruggedness might be an important factor! <S> Now, most lab usage will happen at low power, but I've seen Spectrum analyzers that have inputs rated for peak voltages >= 100V <S> – you wouldn't want that to happen across the air gap between center and outer conductor in SMA. <S> So, in that aspect, I fully agree with Tony's answer. <S> Now, for things like network analyzers, where calibration is a time-intense, and often very costly, procedure, having a connector that you can very reliably fasten with a high, easy-to-measure torque, and that has large internal contacts leading to reproducible, and low, resistance, is relevant. <A> Ruggedness is huge (You would much rather break the adaptor then the connector on the instrument, as others have said), and actually a little imperfection makes very little different to transmitted power <S> (Run the numbers if you don't believe me), so for a scalar instrument N is good enough to at least 8GHz <S> or so. <S> Now a vector instrument (that is sensitive to phase and reflected power) needs something all together better altogether sooner, hence the wide assortment of APC and friends that you see <S> (These are smaller so they avoid waveguide modes to a higher frequency, but also often less robust). <S> I would note that good quality adaptors are worth the (not inconsiderable) money even for lowish frequency work, and that fitting M/F 'connector savers' is usually worth doing for the the smaller instrument connectors (And I usually keep something even on N types). <S> Finally, watch out for the 50/75 ohm trap, SOME connectors are intermatable, some are NOT (Including N type, which can take damage if you get this wrong!).
| Having one common connector to "go smaller" from is probably also a good thing – screwing a SMA adapter onto a N jack is much less likely to damage the N equipment than attaching a half-rigid BNC or a heavy-duty N connector to a cute SMA connector.
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Motherboard backside traces touching antistatic mat. What happens? On the backside of an ATX motherboard there are several traces. Clearly visible. A certain quantity of these traces is connected straight to the ground-plane. What happens when an unpowered board (and completely disconnected from any peripheral) is on a grounded antistatic mat? Will the ground-plane reach equipotential with the mat, because the GND traces make contact with the mat? Or nothing happens, because both the V+ traces AND the GND traces touch the mat? (Like a circuit: if there is 0V between V+ and GND, then nothing happens.) <Q> What happens when an unpowered board (and completely disconnected from any peripheral) is on a grounded antistatic mat? <S> All potentials on the board settle to zero (ok maybe not the battery and RTC circuit) but they would float to whatever ground is (which is close to 0V)It <S> would be almost like attaching a 1MΩ resistor to every point on the board and then to ground. <S> If there was air in between the board and the mat it would be even higher. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Will the ground-plane reach equipotential with the mat, because the GND traces make contact with the mat? <S> Eventually it will, depends on how much charge exists on the motherboard. <S> which should be really low. <S> The mat is 1MΩ (or more, I just measured mine and across the whole mat is 10MΩ to the button, <S> yes I realize its not super accurate, but its close) to ground, most mats are made of dissipative material <S> so this also won't affect the high speed signaling (noticeably anyway) if you run the motherboard right on the mat. <S> We do have ESD tables and they are 1MΩ also <S> but underneath that thin 1MΩ layer there is a conductive metal layer. <S> I would not run a motherboard on this because of a risk of capactivly coupling the signals to ground or to each other. <S> This probably wouldn't affect the operation of the board but it could <S> so I would not do it. <A> The mat will not short out any traces, and unless you have very, very high impedance signals it should be fine. <S> If you have a conductive surface, things change. <S> Some bags that are delivered with electronics are coated in a thin layer of conductor. <S> Some people suggest putting the motherboard on this bag during testing, but this is not advisable, as it can short out connections. <A> Remember that the mat has a pretty high resistance to earth, many megaohms. <S> Earthed wrist straps are typically connected to earth through 1 Mohm. <S> This is to leak static away rather than short it hard to earth. <S> Shorted static is likely to damage something its charged up on, such as when a board with a static charge on is dropped down on the mat. <S> It also makes people jump. <S> So the mat is not a sheet of earthed tinfoil. <S> This is also for safety, as the consequences of an electric shock from the mains or wherever would be much worse if people were connected straight to earth by their earth strap or the mat they're leaning on. <S> Therefore the tracks on your board that touch the mat are touching a large resistor to earth. <S> I would say that the impedance between V+ and earth on your board there is pretty low because of a lot of decoupling capacitors across the rail and a lot of conduction paths, some resistive. <S> So the effect of touching either rail to the mat produces a series circuit with the impedance to earth much higher than the impedance between the tracks. <S> Nothing dramatic happens.
| Most anti-static mats are static dissipative.
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Can I make a VCO full audio range with only one capacitor? I'm trying to make a VCO with a low voltage setup (3.3V), and i 'm using a mcp6002 op-amp. I'm controlling the frequency with an MCP4822 (dac). The problem is that i get a limited range of frequencies. I'm willing to make it full audio range (20-20000Hz) but till now I am only able to add an offset to the base frequency by using different values for the capacitor or changing resistor values. Is it possible to make it full range and my control signal to modulate the oscillator frequency to a wider range? I have tried using this circuit but my op-amp does not oscillate with this configuration. How should I proceed to adapt this configuration for my low-voltage op-amps? <Q> Substitute a CMOS-input RRIO op-amp with a low Vos guaranteed to work at 3V (and with GBW of at least 5-10MHz) and substitute a MOSFET such as the DMN5L06K for the BJT + 10K base resistor. <S> If 3V = 20,000Hz, then 3mV <S> = 20Hz, so keep the maximum offset spec of the op-amp at maybe a few hundred uV. Microchip probably has an inexpensive part that will suit. <S> It's a bit easier with a higher supply voltage such as 15V because the offset requirements are less stringent, so the BJT and op-amp offsets don't matter as much. <S> By the way, rather than coming up with V+/2, simply split the right-hand 51K resistor into two 100K resistors as a voltage divider between V+ and ground. <A> It's possible to design a voltage-controlled oscillator circuit with a range that extends down to zero, up to "infinity" (subject to the limits of the oscillator parts), or both, but such designs are apt to be very touchy when operated anywhere near the ends of their range. <S> For example, if a circuit charges a cap up to 4 volts and times how long it takes to discharge down to the control voltage, and such a circuit would output 1000.0Hz when given a control voltage of 2.0 volts, then in the absence of noise or device limitations, the voltage required to get any frequency would be 4 volts/2^(1kHz/freq)--a value which will always be in the range 0 to 4 volts for any frequency, but the response would be most helpful near the middle. <S> Starting at 1kHz and going down, for example: Freq 1kHz/freq <S> voltage1000Hz <S> 1 2.0v 500Hz <S> 2 <S> 1.0v 250Hz <S> 4 0.25v 125Hz <S> 8 <S> ~0.0156v62.5Hz <S> 16 <S> ~0.0000610v <S> Going up, things would again start nicely but get icky at the top Freq 1/kHz/freq voltage 1000Hz <S> 1 2.0v 2000Hz <S> 1/2 <S> 2.8v <S> 4000Hz <S> 1/4 <S> 3.36v <S> 8000Hz <S> 1/8 <S> 3.67v 16000Hz 1/16 <S> 3.83v 32000Hz 1/32 <S> 3.91v 64000Hz <S> 1/64 <S> 3.957v128000Hz <S> 1/128 <S> 3.978v <S> If one were to increase the RC time constant of the circuit by about a factor of six, the resulting range might be usable (a range from 0.0156v to 3.978v would handle frequencies from about 20Hz to about 20KHz) but the circuit would be more sensitive than ideal near the edges of its frequency range. <A> You can remove the series and just use the Log Pot or a 5 turn pot and get 4 decades. <S> However you compromise sensitivity greatly. <S> MY DESIGN AND SIMULATOR <S> To make a quasi Sine, one can use a Decade counter and R DAC <S> ( Johnson , Gray code or binary) <S> But to make a 4 decade VCO will have excessive Phase Noise due to high f/V noise. <S> Varicaps have a limited range even with 0~50V differential to 3 decades. <S> The most common approach is to use a TIA.
| Yes, you can get 20-20000 Hz with linear control voltage output from something very similar to this circuit with a single capacitor .
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Cat5 cable for low voltage underwater application I am setting up a design for a small submersible ROV and planning to try cat5 cable to power the various components. I have found the ratings for the cable, however I have a question about water vs heat dissipation. Apparently, I can get approx .5-.7 amps of power transmission safely for the motors, but would I be able to push it a little more to 1amp if the cable is submerged in cold northern lake water? The cable would be 100ft and I have factored in voltage drop due to resistance. Thanks for any advice. <Q> Nobody will give you any sort of guarantee. <S> It may or may not work, you have to try this experiment. <S> It is hard to say that the heat will not build up in the middle of the cable. <S> There is an isolation between water and wires itself. <S> So it will isolate heat transfer from the wires to the water. <S> Yes, cold water will help with cooling. <S> But will it be enough to keep the wires at acceptable temperature? <S> Hard to say. <S> But as I said, no guarantee. <A> Yeah, 1 Amp should be ok. <S> General rule about most maximum ratings <S> is that the manufacturer will actually set it lower than the actual safe rating of the device. <S> However, this is only the case for most devices, as there are exceptions to this rule when dealing with the extreme cases (for instance: HV step-up/down transformer sub-stations which actually do display their actual safe rating so they can get the most use out of it possible). <S> Given that you will be also using it in a freezing cold lake should remove any remaining risk of it overheating as well, so you basically have two thumbs up from me. <S> A message to all newbies reading this post: <S> Just because I said that Archaeus can use more amperage on his cable than the rated limits state is ok, there are a couple reasons why you should not try to do this yourself [yet] . <S> I took into account his current setup and situation <S> There are certain situations where you can and can not use more power than the rated safe limits. <S> In his case, there were a couple conditions that drove me to my conclusion: he was only using .3 <S> Amps more than the safe rating, he was using it in cold conditions, the item was a cable, etc... <S> Overloading a cable is not the same thing as overloading a multi-core CPU. <S> Moral of the story: Complexity is the enemy of above-rating loads. <S> We have safe rating limits for a reason. <S> Ok, i'm done. <S> Time to grab some more coffee. <A> As others have said, it might be a fairly safe risk - you are not wildly over the ratings. <S> A couple of points that do occur though. <S> This type of cable comes in various qualities, materials and gauges. <S> At worst perhaps, 26AWG aluminium - at best 24 AWG pure copper - which WILL handle 1.4A in multicore (7-24 cores) at RATED levels. <S> https://www.engineeringtoolbox.com/wire-gauges-d_419.html <S> What is the maximum voltage rating for the cable? <S> Have you considered using a higher transmission voltage and then using a simple DC>DC converter inside the sub. <S> If your cable has a working voltage of (for example) 90v then even at just 1A, you could then drop that to perhaps 12v onboard, and obtain the best part of 7.5A at 12v onboard using just a single twisted pair. <S> If you do not need THIS level of power and stick to 48vDC, then step down DC-DC converters can be bought online for about £2 / $3 <S> this will take the 48vDC and turn it into 5V, 12V - whatever you want basically. <S> So 48VDC > 12V 4A or 5V at very nearly 10A - all on a single pair. <S> My other point I am sure you have considered <S> but I mention in the 100-1 chance you had not, is of course make sure you use the cable designed for patchleads (multistrand) and NOT fixed wiring (single core conductors). <S> I am guessing you will NEED that cable to be resilient to being flexed on an ongoing basis.
| Applying more than the limit in general isn't a very sound engineering practice, although it can suffice in a pinch. Always try to find a solution that has both safety and efficiency in mind. My guess is yes, it will work.
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Can a multimeter measuring current damage low voltage devices? I'm using an arduino due (3.3V) which its power regulator can deliver 800mA of current. Can I connect a multimeter in series and get a reading of the current my circuit draws, and can i leave it connected so to measure the current in real time? (I regularly re-program the Due with the circuit connected is there a chance the multimeter can damage the device in this situation?) Will such a config affect the circuit? At last should I break the circuit and connect the multimeter to the positive or to the ground connection? simulate this circuit – Schematic created using CircuitLab <Q> If your DMM is of auto-ranging type, never use it for measurement of current in a low-voltage environment. <S> The auto-ranging DMM will change the measuring resistor (shunt) value in-flight, which will add a varying voltage drop affecting your circuit power supply. <S> Instead, use a 1% 0.1 Ohm resistor in-line with your power rail (between the regulator and Arduino) as a starting value, and use the milli-voltmeter mode of your DMM to measure the current, I (mA) = V(mV)*10 for a 0.1 Ohm resistor. <A> Using it in current mode between the wrong two points can however create a short so care must be taken not to have it in current mode when you are trying to measure a voltage to ground. <S> Also, as others have mentioned, avoid auto-ranging ammeter scales. <S> When in resistance or diode mode the meter actually generates a current in the probes. <S> Although the levels are normally small, use these modes with great care when testing in-circuit. <A> The "Output Short Circuit Current" of an MCP6002 operating at 5.5 V is in the region of 23 mA. Ref: Microchip datasheet . <S> It will be less at 3.3 V. <S> For a TL081, Output Current (Typ) is 10 mA. Ref: TI datasheet . <S> So your circuit as presented is not going to approach <S> 100 mA let alone <S> 800 <S> mA. <S> On an Arduino Due ( circuit diagram ), the 3V3 power output line comes through an LM2734Y with a max. <S> output current of 1 A to get 5 V <S> and then an NCP1117ST33T3G , also with a maximum output current of 1 A. <S> Both devices have thermal overload and over-current protection. <S> It appears that you do not need to worry about drawing too much current from the power supply circuitry both because your circuit won't draw much current and because the supply is protected. <S> If you expect to be drawing a high current, maybe more than 300 mA, (through the power supply ICs) for a sustained time, it would be prudent to either use a separate power supply instead or add small heatsinks to IC2 and IC4 on the Due. <S> It is my understanding that op amps may have better performance characteristics when their voltage supplies are in the higher part of their allowed range. <S> Adding a resistance to measure the current draw will have the effect of reducing the voltages on the power rails of the op amps, which is probably not desirable . <S> You have 5V available so it might be better to supply the op amps with that and translate the voltages, if appropriate, for a lower overall power draw and possibly improved op amp performance. <S> In summary, measuring the current with a multimeter will not physically damage your circuit, but it may affect its operation and does not appear to be necessary. <S> If the current measurement is important <S> then you could power your circuit from a bench PSU which has a voltage sense input. <S> Models with a voltage sense input are going to have an output current reading. <S> Connect the sense wire(s) <S> appropriately and connect the PSU ground to the Due ground. <S> If you determine that your circuit is never going to draw too much current for the Due, your job is done.
| Meters in current mode and voltage mode do not stimulate the circuit, so, when used appropriately should do no damage.
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How do switches respond when more than 10 dial pulses are sent? Forgive me if this is off topic but this is actually about circuitry design: My understanding is that electromechanical Strowger SxS switches have 10 "steps". What would happen if 11 pulses were dialed, either from a phone that happened to have 11 holes in the dial and could send 11 pulses or by tapping and depressing the switchhook/hookswitch rapidly 11 times in succession? Similarly, what about 12, 13, etc. pulses in succession? I'm not sure if 11 pulses would dial 10+1 or simply 10 and then stop listening. I've tried and failed to test this out, even though I have 2 PBX systems at home. I can dial the lower five digits easily using the switchhook but I fail more than half the time simply dialing 0 using the switchhook. And because the PBXs are electronic, they may automatically recognize 11 as invalid and default to a reorder signal. If there is a difference between how electromechanical switches and electronic switches would respond, please contrast them. (I don't know how many people just have a SxS switch at home to test with though...) <Q> The extra pulses will simply be absorbed, except for the final digit which will be lost. <S> The 1st, 2nd etc. <S> group selectors only accept a single digit, then hunt into the bank to find a free outlet to the next group. <S> Once the carriage steps up to the top level it can't go any higher, so it just hammers into the backstop. <S> Final selectors step up for the penultimate digit, then step into the bank for the final digit. <S> Another selector type was the Discriminating Selector Repeater (DSR) which was designed to drop out when reaching certain levels, in order to absorb digits that weren't need in the local exchange. <S> This monster had 29 relays, some with 2 or 3 independent coils. <S> Here's a schematic - see if you can figure out how it works! <A> Old SXS switching centers were electro-mechanical and in addition to the step switches used several different types of relays. <S> The rotary dial was designed to pulse on the release of the dial, the number of pulses was determined by which hole you placed your finger in. <S> Tapping out the digits is possible (I've done it) but tricky and doesn't work every time unless you have a very steady hand. <S> If you had the proper pause between the 10th and 11th digit, the 11th digit would become a one for the next digit in sequence. <S> Each step switch in sequence has a different function as you progress throught the digits, 1st selector, second selector, line finder and so on. <S> I'm trying to remember as much as possible as it's been since the early 1970's since I worked a SXS office. <S> A PBX will convert dial pulse to DTMF (not to be confused with MF). <S> The PBX/keysystem will also store digits and send them all at one time. <A> If you did this to some SxS switches, the wiper would fall off the top or right side of the bank of contacts depending on if you do it for the final number or somewhere before. <S> The selector was then out of service until someone at the switching office put it back. <S> A friend of mine asked himself the same question many years ago as a child in a small town. <S> He pulsed the hook 11 times and heard nothing more until the phone engineer who lived nearby scolded him for doing it. <S> It was very clear to the engineer who did it because he frequently answered my friend's many questions on how the phone system worked.
| An extra pulse in the final digit will cause the carriage to step all the way around the bank and out the other side, where it will 'drop out' back to level 0 and start stepping up again.
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Driver failure when driving IRF3710 mosfet using IR2101 driver I've made a driver circuit using IR2101 driver and a fairly large IRF3710 mosfet. The circuit works up to a certain current through the mosfet but then the driver mysteriously fails and pulls both mosfets high generating a short circuit on the output. I'm using 3x of these circuits to drive a BLDC motor with 0.5Ohm winding resistance. I'm able to reach currents of about 3A through the motor before one of the drivers fails. IRF3710 has ~3100pF input capacitance and 130nC total gate charge according to the datasheets. In terms of voltage used to drive the load I have only gone up to 30v so far. Regardless of whether I use 12V or 30V to drive the load, the driver seems to fail when the load draws around 3A. When the circuit fails, a short circuit occurs on the supply, but nothing gets hot (I have a current limit on the power supply so the short circuit current never exceeds 5A). Still I find it disturbing that failure of the driver causes both mosfets to switch on causing the short circuit. While I will have overcurrent protection in the final circuit, I still find it quite bad and would like to make sure that chances of this kind of failure are close to nonexistent. I have a few theories about what may be the problem but I have not been able to pinpint exactly what it is because I think it has something to do with a transient condition that happens very quickly not giving me enough time to see it on the scope. The IR2101 chips have failed repeatedly on 1 or 2 of the stages during my recent tests. The motor spins up to a certain point, then as I increase the motor speed (current increases as well) one or more of the drivers fail. None of the mosfets have failed so far though. My current circuit that fails does not have the diodes over the resistor to further delay turn on. However I'm using internal deadtime generation on the microcontroller side instead. I have also been using 0.1uF capacitors without any difference in the switching waveform. My pwm rate is 30Khz. Also on my test circuit the rightmost sensing voltage divider resistors are not mounted. I'm aware that the 2101 driver is perhaps not the optimal chip to drive these large mosfets because it can only provide 130mA of gate drive current. The datasheet also specifies 270mA negative current but I don't quite understand what that means. Is it the current generated when gate is discharged? Questions Could there be an issue with using smaller capacitor besides slower turn on times? Could it be an issue that the negative current is specified at 270mA and the gate discharge current happens to be higher than that and goes directly through the driver burning it? Would ir2110 be a better driver choice for these mosfets? Is there any other driver similar to the ones I use right now that can provide much higher currents (assuming current rating is the issue)? Can the above circuit work at all or do I have to completely redesign it? Should I maybe use a larger resistor in series with the gate to protect the driver against possible reverse currents? (if that is even a possible issue here). Edit: the problem of driver chip failure occurs as the motor speeds up and current through the motor is around 3A. Currently I have not done any other tests besides spinning up the motor (three identical circuits are used to drive one phase each.) Typically driver in one of the phases fails. At first it happened twice on the driver furthers away from 48v connector. However latest failure occured on two drivers at once - the one closest to the 48v connector and the one furthest away. The middle one continued to switch as normal - the broken ones went into locked state with both mosfets switched on causing a short on the power rail. Replacing the driver chips fixes the problem - only until next failure. edit: Current layuot (work in progress). I have added a single 47uF 63v cap in the middle at the top. Summary of working solutions So far adding bypass caps on 48v rail seems to be the solution (I added one 47uF 50V electrolytic and 2 0.1uF ceramics). I think it is early to say whether this really did make the issue go away completely but after more tests today no driver chips were burned. I was able to spin up the motor to speeds at which it draws 5A and do instant spinups from zero to full throttle as well as instant stops without burning anything so far. Added bypass capacitors Reduced gate series resistors from 100Ohm to 36 Ohm Added diode on path from 48v supply to bridge (prevents regenerative current going back to power supply) Added voltage clipping across the 48v rail (TVS/Zener) to let regenerative voltage spikes go to ground (provides a much needed path for high voltage spikes when current can not flow back to power supply) The application note AN-978 was very helpful as well: http://www.infineon.com/dgdl/an-978.pdf <Q> This can manifest itself in two ways: <S> When top FET turns OFF, current in the supply wires cannot go to zero fast enough due to wiring inductance. <S> This inductance creates a positive voltage spike on +48 rail, which blows the FETs and the driver. <S> The spike will be proportional to the current, which is why it only blows at high current. <S> Solution: decouple +48 rail. <S> If the motor is used as a generator (regenerative braking) then the power it generates will end up on +48V rail, causing voltage to rise. <S> How to decouple: <S> Add 100nF MLCC or larger, one per FET, as close as possible to the drains. <S> Add low-ESR cap(s), with ripple current rated for the motor current. <A> The gate resistance is way to high 100 ohms, the parallel resistors R17 and R18 are not supposed to be there, as they just spoil the performance of the bridge. <S> Not sure if the diodes D8 and D9 without a series resistor are OK. <S> You should read the datasheet and some application note, and not put elements "by the way" with the schematics you got in your dreams. <S> Why it has failed: Because the gate resistances are too big, the switching rise time is very big, therefore the MOSFETs are cross conducting at the same time making a short. <S> EDIT: <S> You can try (only my opinion, without calculations): <S> Remove R17 and R18 Remove <S> diodes D8 and D9 Replace resistors R15 and R16 with 10 ohm resistors. <A> I see a few things... <S> I think your gate R value are actually too low. <S> You need to drain or fill the gate capacitances as quickly as you can. <S> However one of those capacitances can go to the 48V value as the bridge switches. <S> As such the resistor needs to be closer to \$370\Omega\$ <S> The extra pull-downs are not needed. <S> The extra diodes across the gate resistors is only adding to the capacitance problem as well as adding a switching delay and a path from the switching transient voltages to get back to the device. <S> The diode you show is only rated for 40V. <S> It needs to be at LEAST 60V, especially D3 and D8. <S> You DO need to add some dead time to the signals you drive this circuit with. <S> You say you are already doing that <S> so it is good. <S> Hopefully you have measured the turn on and turn off times and are using at least double that value. <S> ADDITIONALLY <S> You have not shown how you are dealing with the fly-back currents in the above schematic, or shown how the motors are connected or your grounding system. <S> All of these can create issues.
| Solution: if you use motor as brake, you will need something to absorb this energy, like a comparator which turns on a FET and dumps the extra energy into a resistor if the +48V exceeds, say, +50V. OK, so seeing your layout, the reason it fails is lack of decoupling on the +48 rail.
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In theory, is the position derived from accelerometer absolute? I recently found multiple research papers that say that in theory double integral of accelerometer data gives a position of the device. Does that refer to absolute position (like the one given by a GPS)? or for example starting position and then relative position to the starting position?Thank you for helping me clarify this. <Q> In theory, is the position derived from accelerometer absolute? <S> Ok, so imagine you're a sensor. <S> All you can sense is acceleration. <S> You're an accelerometer. <S> Now you're at rest, or moving at a constant speed. <S> You can't tell the difference, since Newton's laws don't allow that – an object at rest or in linear motion experiences no acceleration. <S> Obviously, since you might be moving (or not), you can't tell where you are, and whether you'll be at the same position in 10 seconds. <S> So, that answers your question. <S> Mathematically, you'd have to differentiate twice to go from position to acceleration. <S> So you'd have to integrate twice to go back. <S> Each integration step adds an unknown "offset" to your result. <S> I'm a bit surprised you didn't come up with either approach! <A> Integration comes in two types: one type is the definite integral, whichtakes a function over a defined range of its independent variable. <S> Theother type is the indefinite integral. <S> The indefinite integral is only defined within an arbitrary constant. <S> So, twice integrating the acceleration of a body, you are in need oftwo constants of integration, before the formula is completeand can give a definite value which relates to reality. <S> The two constants needed, are an initial position and velocity. <S> Well, actually the position and velocity at ANY times covered by your acceleration data would do. <A> Imagine you have an acceleration sensor resting at fixed position. <S> The first integration will give the speed signal, another integration will give the position. <S> If the acceleration is zero, speed is constant. <S> But what if there is a small offset to the acceleration signal? <S> This offset will be integrated and the result is an increasing speed and the error of position will increase with time, faster and faster. <S> It is not possible to remove even the smallest offset to the acceleration signal, the result is the speed error and position error increasing with time. <S> Even if the absolute position was true at start, it will get lost over time.
| An accelerometer + signal processor can only tell the position relative to some starting position, and only if the starting speed is known.
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How does PCB traces length effects parasitic capacitance on the board How does length of trace effects parasitic properties of the PCB board? Is it longer the length, higher the parasitic? <Q> Capacitance between two plates is proportional to the area of those plates. <S> Therefore a longer PCB trace above a ground plane, for example, will have more capacitance to ground than a shorter one, assuming everything else is held constant. <S> Other factors that effect the capacitance is the separation distance and the dielectric constant of the material between the plates. <A> Posting this just to picturize Mr. Lathrop's answer: Imagine how a capacitor is formed: As you can remember, the capacitance of this capacitor is \$C = <S> \epsilon <S> A <S> /d\$ <S> where \$\epsilon\$ is dielectric constant of material, \$A\$ is surface area of the plates and \$d\$ <S> is the distance between plates. <S> In your example, \$d\$ is the thickness of the PCB and \$A\$ is the surface are of the trace (or plane). <S> Assuming one side is completely poured-copper for GND <S> , so the parasitic capacitance is proportional to the trace length. <S> The longer the trace, the more the surface area (A) <S> so the more the capacitance. <A> This closeness tends to intercept Efield aggressors, improving SNR, but does load the signal trace and require more charge to sustain the same AC voltage swing.
| Even long thin traces can have high capacitance, if GND is brought very close to the signal trace.
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EBike Controller - Does 'current' affect driving speed? I wanted to ask if a brushless controller Current actually affects electric bicycle speed. At the moment I have a 48V 250W brushless controller, rated current of 5A and maximum current of 10A. protection voltage is 42V. Here is a more detailed picture about my controller plugs, etc: So my main question is: if I switch to a same brushless controller but just this time with lets say a maximum Current of 17A, will it make any difference in any way? I also tried talking with the manufacturer and I didn't get a flat answer; they said it won't affect on the speed but I read in a few places that it might. My Battery: 48V 13Ah My Motor: 48V 250W ( model: bafang swx02 ) Also another question, is there any way to make my ebike drive faster than 25 km/h without switching the motor? <Q> To a first approximation (a flat, level road), increasing the current capability (not the current) will have no effect on your speed. <S> A BLDC motor is effectively driven at a certain AC frequency, and the motor turns at one revolution per cycle of the drive. <S> And you should note that, from a power perspective, it's clear that your battery and motor are well-matched. <S> The motor is rated for 5 amps @ 48 volts (peak 10 amps), for a total of 240 watts (480 peak). <S> The motor is rated for 250 watts, so the motor is the limiting factor. <S> So the first thing you would need to do is replace the motor. <S> However, you would also probably need to replace the controller as well. <S> I suspect there is a speed limiter somewhere in there. <A> The motor will draw whatever current it wants to unless the speed controller limits current for torque control or to protect itself. <S> If your motor would normally draw less than 5A at top speed then using a higher rated controller won't make a significant difference (it might be very slightly faster due to lower voltage drop across the FETs in the controller). <S> Motor speed is proportional to voltage. <S> If you want more speed (and can't gear it up) then use a higher voltage battery. <S> Just realize that it will draw more power, which may damage the motor or controller if their ratings are exceeded. <S> You will need a higher voltage controller as well as extra batteries, so you might consider replacing the entire drive chain. <A> Voltage correlates to raw (unloaded) speed, current correlates to torque and overall power (loaded speed). <S> The current handling of your motor is going to limit your speed. <S> There won't really be a way to go faster without changing out your motor. <A> open the controller, find the shunt and add a bit of solder it will increase amps output allowing more torque, as well as speed if the current was limited down low enough (which 5A sounds a little low to me) <S> this would cost a bit off of the battery range.
| As long as the motor gets enough current to produce the force necessary to overcome friction and drag, increasing the current available won't change anything.
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Removal of unused pads? I was reviewing a design earlier, and I noticed something interesting. The designer had removed unused pads on the chip. I have never seen this done before. Is this something that is good practice? Is it even okay? <Q> This is not a standard practice, and should be avoided. <S> First: Along with providing electrical connectivity, pins also mechanically anchor a chip to the board. <S> Each pad that's removed increases the stress on the remaining pins, which will increase the risk of the chip detaching from the board. <S> Second: All of the remaining pins have nothing but soldermask between them and a trace underneath them which they aren't supposed to be connected to. <S> Soldermask is not very thick, and it's not very durable either. <S> If the mask is breached -- from a pin vibrating against it, for instance! <S> -- the pin may become intermittently connected to something it wasn't supposed to be. <A> No, it's not good practice, but if done really carefully it can open more space for routing on the top layer. <S> This is only useful if that makes the difference in being able to drop layers and make the board cheaper. <S> This in turn means it only makes sense for high volume products where the price of the board actually matters in the overall product, and where it is significant relative to the engineering investment. <S> Otherwise, there are reasons not to do this: <S> Pads are for physical mounting too, not just for making electrical connections. <S> This matters more for some packages than others. <S> For a 64 pin TQFP like this, removing half the pads shouldn't reduce mechanical strength to where it matters for most cases. <S> Unless you are careful to remove the pads reasonably symmetrically all around the chip, the surface tension of solder can pull the chip off center during reflow. <S> This is a serious issue you have to think about carefully. <S> You have to be really sure that the unused pins done touch anything conductive. <S> The solder mask should take care of this, but stuff happens. <S> If there is a lot of vibration or thermal cycling, are you sure the unused pins won't eventually rub thru the solder mask? <A> It would ease the layout at the expense of making the assembly more fragile. <S> It's probably not okay <S> according to the IPC (using solder mask as an insulator)-- see for example <S> this <S> thread- <S> but I don't have specific chapter and verse to cite. <S> Further, the soldered leads will have a relatively large gap bridged with solder because the other leads are sitting on top of solder mask, which makes the assembly even weaker. <S> So I think this is amateur hour stuff <S> but it probably works okay enough and might be acceptable for a throw-away consumer product. <S> Definitely not acceptable for a high-reliability design. <A> He clearly needed that for routing. <S> Probably otherwise he would drill holes, make more layers... <S> As long as number of removed pads is low and it's safe to assume that the component will not fall, it's a brilliant idea for saving costs.
| You may also have to change the pad shapes for the remaining pads so that the center of the pull is where the chip is supposed to be.
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Can epoxy resin be reused? Does anybody know if this epoxy resin can be reused? I do not need all the resin in the package at this time and I want to store it for later use. I am asking because it says in the specifications that the useable life is 60mins. If it cannot be reused, is there a resin that can be reused? Or at least, in smaller packages? <Q> Yes and no. <S> In general, yes, epoxy resin can be re-used. <S> It's supplied in two containers, you measure and mix only what you need, taking great care to avoid cross-contamination when handling the two parts. <S> West Systems and other marine epoxy products are designed for this. <S> That pack however is intended for easy mixing. <S> If used as intended, that prevents reuse, as you mix it in the pack before opening. <S> You might be able to decant each part into separate containers, then measure and mix as normal, keeping the rest for future use, but it'll be a messier process than starting from separate containers, and you'd better not expect any guarantees from the supplier if you try. <A> After mixing the two parts should be used. <S> Not be reused. <S> You can mix the quantity you require keeping the proportion of parts. <S> Store tightly what remains. <A> Those packs are designed for clean and fast mixing of the whole pack for a single use. <S> You can buy epoxy with the two components in separate containers, and measure out and mix any quantity you need. <S> The pack often contains a spatula or something similar to use as a mixing and spreading tool - you can mix small amounts the components on almost any disposable material, e.g. a piece of paper. <S> Just be careful not to cross-contaminate the original containers, otherwise you may end up "glueing" the caps onto the containers, etc, which makes re-opening them rather difficult!
| If you want to "re-use" epoxy, don't buy it in "mix-in-the-pack" packaging like your link.
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Microphone sensor for picking up voice from across the room I'd like to create kind of a DIY Google Home using a Raspberry Pi and voice recognition software. I'm looking for a microphone sensor that will pick up my voice from across the room, but nearly all of the microphones I've seen look something like this: and are clearly designed to be close to someone's mouth/the sound source. (Please correct me if I'm wrong.) I've found a cheap condenser microphone that looks like this: and would it work? Or am I looking at the wrong thing entirely? <Q> Normal microphones are not very sensitive .Speak <S> in to them while monitoring the output volts on a scope <S> and you will see what I mean .The <S> Carbon microphone which was in my 1970s junkbox did have a high output level but lots of distortion .I have not tried a parabolic microphone .What <S> would always work was a loudspeaker backwards .I tried a horn type speaker backwards and that worked even better . <S> Most speakers are low impedence like say 4 or 8 ohms .What <S> I did in 1975 <S> was to use an output transformer backwards to provide a better match to the preamp .Mains <S> hum pickup was an issue and output transformers were becoming harder to find <S> so I used a simple common base transistor stage biased at about 1 mA and then fed it into a more conventional AF amp .I <S> could hear stuff 30 feet away with junkbox Ge 1960s <S> transisters .Things were easy in the mid 1970s .Modern <S> minature microphones do not put out much voltage .I <S> have used speakers from 4 inch dia to 8 inch dia with good results .I <S> have not used this approach with any crossover speaker systems because they were not in my junk box in 1972 .I have tried phillips 800 ohm Hi Z speakers directly and they work well but would be hard to find now . <A> As a kid, building high-gain bipolar ac-coupled amplifiers, the only signal source I had was a 2" transistor radio speaker. <S> Scratch on the cone, for strong signals. <S> Speak into the cone, for normal signals. <S> Eventually, I learned proper VDD filtering. <S> The first 2 or 3 bipolar stages had their own private VDD (local battery equivalent) with 5,000uF and 100 ohms. <S> The final 2 or 3 stages ran directly off the 9volt "B" size battery. <S> Output probably was to magnetic earphones, to prevent acoustic feedback. <S> That amplifier, with speaker pickup, easily monitored voices 10 or 20 feet away. <S> You should be able to do similar, today, with 2 or 3 stages of OpAmps. <S> Just arrange for private power to the first stage, to avoid VDD-based feedback oscillation. <S> Here is what Signal Chain Explorer suggests: 3 stage of opamp gain, 40dB/stage using Default models (UGBW = 1MHz); input is 1 microVoltPP; I had to edit the first opamp, reducing its noise density from 4nanoVolts (1Kohm) to 0.5nanoVolts (16 ohms); I also edited the gain-set resistors of that first stage: 5 ohms and 495 ohms. <S> Result? <S> 18dB SNR for 1uVpp input. <S> Nah---that is too easy. <S> Lets use 2 stages of bipolar. <S> We achieve 1,000 * 1,000 gain. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You do NOT need high gain on your microphone. <S> What you need is a high signal (voice) to noise ratio. <S> You will not get a a high signal to noise ratio just by amplifying the microphone signal. <S> That will amplify the ambient noise along with the voice - the signal to noise ratio will stay the same (or degrade a bit since the amplifier will add some noise of its own.) <S> What you need is a little gain - just enough that a loud speaking voice used close to the microphone <S> will get you within about half of full scale. <S> Gets you maximum range without distortion. <S> Next you will need several microphones, and an analog to digital converter with enough inputs for all the microphones, 16 bit sampling, and you will probably need at least a sampling rate of 22kHz. <S> Once you have the audio in a form that it can be processed, you will need software to pick out the voice(s). <S> Picking the voices out of the background noise is not trivial. <S> After you have the voice picked out and isolated, you can use an automatic gain stage to bring the voice up to a particular level to make things easier for the speech recognition section. <S> Finally, you get to decide how your gadget should react to specific words or phrases. <S> The Jasper project has already solved most of these problems for you if you are using the Raspberry Pi.
| The solution involves beam forming ("aiming" the microphones to pick out particular sources withou physically moving the microphones) and noise reduction.
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Is the diode in this circuit connected in parallel with the resistor? Using the example below my teacher started to explain the parallel clipping circuit. Then he said that the diode is connected in parallel with the resistor but I couldn't understand why, because I think they are connected in series. Could you tell me where I'm wrong? <Q> What we say isn't always precisely what we mean. <S> I think he meant to say 'the diode is connected in parallel with the signal at Vout'. <S> In this case, the signal Vin is generated by the AC source, and goes through a resistance R, before being read at the + and - terminals of Vout. <S> The diode/battery combination is in parallel to the signal at Vout. <S> There is an alternative clipping circuit, a series clipper, where the diode would be in series with the R, being placed in the line between R and Vout+. <S> Having said that it's a parallel clipper, it's 'obvious' to your teacher and any engineer that the diode is in parallel to Vout, and in parallel to where the resistor R connects to Vout. <S> This is probably the source of his momentary lapse in precision. <S> Of course, it's not obvious to somebody learning. <A> Actually it is both, in truth parallel and series is dependant on your viewpoint. <S> What appears to be in series form one viewpoint may appear in parallel from another. <S> Looking at your circuit that \$+ <S> V_O -\$ makes all the difference. <S> If you redraw your circuit like this perhaps you can understand. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The terminals <S> Vo+ and Vo- now have two circuits attached to them. <S> One circuit is a signal source with an output impedance. <S> The other circuit is a 10V clipper circuit. <S> So you can legitimately say the circuits are connected in parallel across the terminals. <S> Point of view makes all the difference. <S> Consider the following two resistors. <S> Obviously in series <S> right? <S> What about now? <S> Did you say still in series? <S> It depends on where you look from. <S> Looking from the left or right they are in series, but looking from the join between the resistors, they are in parallel. <A> Yes, it's in parallel with \$R\$ as far as the load at \$V_o\$ and the AC circuit are concerned. <S> DC analysis: <S> The -10 V DC source biases the diode, having the effect of "moving" its forward voltage 10 V down so that \$V^{\prime}_f = <S> V_f <S> -10\$. <S> When the sine input is superimposed to the DC bias, the diode will behave as a non-linear resistor \$R_D\$ <S> according to its V-I curve. <S> Note than the diode will clip when the sine voltage across the diode goes below \$V^{\prime}_f\$, and that we can also embed this behavior in the non-linear characteristic of \$R_D\$. <S> AC analysis: <S> Thus we end up simply with a voltage division between \$R\$ and \$R_D\$. <S> If we calculate the equivalent Thevenin resistance of the AC circuit, we find it to be \$R_{th} = <S> R || R_D\$, which is indeed a parallel of \$R\$ and the diode , and which will be highly non-linear as well. <S> Now you can load the output of the circuit with any load you want and calculate the output voltage with precision. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Actually the correct answer is both !.
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why does absolute encoder (even multiturn) have the option of an additional incremental output Many companies make absolute encoders for motor drives some can output an optional incremental track . my question is why is that needed since the encoder have a BiSS interface that can go as high as 10Mbit/s . for example : https://www.kuebler.com/usa/prod-sen-multiturn.html I have read on this site somewhere : Magnetic Encoders - Tips and Tricks Qoute :" I wouldn't recommend closing a velocity loop on absolute position feedback but rather use quadrature signals" If this is true then my initial guess is that the cascaded Closed loop drive use Absolute output for position loop and incremental for speed loop . But why ? It doesn't make any sense especially that absolute encoder have much higher single turn resolution (up to 17bit and even 21bit) which can be really helpful for controlling motor at very low speeds. <Q> I have mainly used optical encoders, but this should apply to all encoder types. <S> Your very carefully designed control loop can become unstable quite easily when there is a very small delay. <S> Even worse, you cannot model the delay as a finite linear system in an analog control loop. <S> I once had a D.C. Motor that became unstable when a \$1\ \mathrm{\mu s}\$ delay was introduced. <S> Latency is also important when you use controllers that are not based on PID control. <S> In a dead beat controller, for example, the transfer function needs to be controlled quite accurately. <S> Also keep in mind that for encoders that have higher resolution you cannot get the exact position immediately. <S> You can get the position in the past due to the latency. <S> For a linear encoder with resolution \$100\ <S> \mathrm{nm}\$ <S> you get 10,000,000 ticks per meter. <A> An industrial servo drive would usually consists of a motion controller and driver. <S> The motion controller calculates the position trajectory and has a position loop where the actual position is relevant. <S> It outputs the velocity setpoint, which is the input of the drive. <S> The drive is a speed controller where the velocity feedback is relevant. <S> A fully integrated servo drive would have a proprietary communication protocol between motion controller and the drive (VFD like), where the encoder signals are directly connected on drive electronics. <S> The actual position is then forwarded to the motion controller over this proprietary communication protocol. <S> Assuming that BiSS protocol is something new and open source, you have the ability to connect the encoder on your motion controller/PLC and connect the other quadrature or sin/cos output directly on the VFD. <S> You do the position loop on PLC/Motion controller and the VFD does the velocity loop. <A> One can certainly use high-res Abs. <S> position encoder or high-res Quadrature encoder <S> both which yield position and direction with simple signal processing. <S> In some cases the Quad detector may have logic to prevent dithering on one edge when almost still where as absolute detector requires averaging. <S> In the end it all comes down to specs for cost, Resolution, accuracy for use in position, velocity, acceleration and interface/processing costs, latency , noise rejection. <S> Without specs, I don't think you can compare apples and oranges to say which is better for you.
| The main reason you want to use incremental outputs for velocity control is latency.
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What makes the generator practical? "A student tried to make an AC generator consists of a coil of one turn, and cross sectional area of (7.54 × 10^-3 m²), and rotates in magnetic flux density (2.5 × 10^-2 T), and when it was taken out of the magnetic flux in 1.256 × 10^-3 sec an induced emf of 1.5V has been generated." That was the description of a generator, then it asked whether that generator is practical or not? The answer was "No" without an explanation, so could you please explain it for me? <Q> Otherwise it will not work. <S> even at 50MHz it won't work unless highly tuned with heavy copper thin hollow core conduit. <S> Then skin effects and RPM , inertial mass, imbalance etc make it impossible. <S> Thus the core in a generator must have a magnetization flux to couple the magnetic field thru the commutation DCR to generate any appreciable V*I= power. <S> otherwise the inductive impedance will just shunt or short out the magnetic field and generate very little voltage or current. <S> This can be done with permanent magnets, laminated iron core with an excitation field current to increase voltage etc. <S> Do an impedance ratio calculation with a 0.1uH single turn and see what Z(f) <S> you get at 60 Hz with conductor and current needed to get 1V <S> then conduction losses and magnet strength in Tesla to get that current. <S> You would need a cryogenic cooler and a superconductor. <A> I think you (and unfortunately, us) are taking this question far to serious. <S> This is a question of scale. <S> Spin a coil in a magnetic field or spin a magnet in a coil and all you get is 1.5V. <S> Although you can get into the theory, it is NOT practical because the generator does not have the capability of an AAA 1.5V battery. <S> A lot of trouble for a weak flashlight, which cannot move. <S> So no practical application, but it does demonstrate the principle. <S> $$V_{Ind} = <S> N B <S> l <S> v$$ <S> Significantly, increase all of those and now you have a practical generator. <A> It is not a description of a practical generator: <S> The magnetic field is weaker than the field of a typical refrigerator magnet. <S> There is no indication of a mechanism to sustain the magnetic field or maintain its direction. <S> The coil of wire is initially described as rotating, but then described as having been taken out of the field. <S> A generator requires continuous rotation or some other continuous motion. <S> There is no indication of a method of supporting, rotating or otherwise moving the coil of wire with respect to the magnetic field. <S> There is no indication of any means to connect any kind of electrical load to the coil. <S> There are practical generators that are only intended to detect how fast something is moving. <S> There may be some minimum voltage and current necessary for that function, but the minimum could be extremely low.
| A practical generator continuously supplies electrical power through electrical conductors to a load that can utilize the power. In order for it to be practical the frequency of current must yield an inductive impedance about 10x to 100x the DCR of the wire. It is not necessary to prove any minimum to say that a generator is practical.
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How to make a piezoelectric transducer buzz louder? I am trying to make a piezoelectric transducer buzz loudly, but have so far only been able to make it create a quiet hum. I have tried using a basic amplification circuit with an NPN, but this made very little difference. I am working with microcontroller that words at 3.3v. How would I get a loud buzz even for a short period of time, from a piezoelectric transducer similar to the below image. <Q> Such transducers work best with higher voltages, like 9V or more. <S> Depending on the project, you may already have that voltage available. <S> Another common option is to drive a small step-up transformer with the secondary winding connected to the buzzer. <S> I have seen piezo transducers in fire alarms which are driven by 1:10 transformers, and those are indeed pretty loud. <S> A push-pull configuration (which should have the same effect as doubling the voltage) can be implemented with a single NPN transistor as well: simulate this circuit – <S> Schematic created using CircuitLab <A> A larger vibrating surface moves more air and thus produces a louder sound. <S> Glue or fasten the transducer to the inside of the enclosure, at the center of the widest panel. <S> Attach a mass at the back of the transducer <S> , so it has more reaction mass to push against. <S> This will increase the force with which it pushes the enclosure wall, and therefore the loudness. <S> Sweep the frequency to find the resonance frequency of the system. <S> Some frequencies will be much, much louder than others. <S> Small piezos like this tend to work above 500 Hz. <S> Consider a loudspeaker not mounted into a speaker box. <S> Say you hold it in your hand. <S> When the membrane pushes air forward, it also sucks air in from the back. <S> Therefore, the air turns around the edge of the speaker, and this is called an acoustical short-circuit. <S> It works better at low frequencies, and this is why a loudspeaker standing in free air without a closed box behind it <S> has no bass. <S> Your piezo is much smaller, but it has the same problem. <S> Using the enclosure as I said above solves this. <S> Drive the piezo with a H-bridge, for example a 74HC six-inverter chip with 3 inverters per side. <S> This will give you a nice push-pull drive. <A> Put it in a tuned (acoustically) resonant cavity with a hole for the sound to get out. <S> You can find formulas for calculating the resonant cavity dimensions on the net. <S> Nothing should touch the element except a circular knife edge that is at the neutral bending point of the element (the nodal circle). <S> And give it more voltage. <S> Like <S> +/-15V Maybe you can use a 741 or something to drive it. <S> ;-)
| A straightforward solution would be to generate that voltage using e.g. a charge pump, then use the NPN transistor to amplify the MCU signal to that voltage. The solution is likely to be acoustical.
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When two transistors connected which one is on and which one is in cutoff from given circuit condition? here in this question's solution, given that upper BJT is on while lower one is in cutoff region for Vbb = 2.7 V and viceversa for Vbb = -2.7 V and both will be in cutoff for 0 V.While in other problem Q1 is on and Q2 is in cutoff. So my question is how to decide which one will be conducting and which one will be off and why? Is there any general procedure? <Q> Look at this circuit carefully and realize each transistor is being used in emitter follower mode. <S> However, as you say, at most one of them is on at a time. <S> Hint <S> : Remove any transistor that is off from the circuit, then analyze. <A> Your two questions are not complicated. <S> I suspect you are imagining they are worse than they really are. <S> Let's apply simple logic with the more recently added schematic: simulate this circuit – <S> Schematic created using CircuitLab First thing to do is to decide what the two transistors are doing. <S> Suppose <S> that \$Q_2\$'s \$V_{BE_2}\approx 700\:\textrm{mV}\$ and that it is therefore actively on . <S> Then it follows that \$V_X\approx -2.7\:\textrm{V}\$. <S> But if that were the case, then \$Q_1\$'s \$V_{BE_1}\approx 5.7\:\textrm{V}\$! <S> Given that the collector current goes up by a factor of about 10 for each \$60\:\textrm{mV}\$, this would imply a HUGE collector <S> current and probably a huge base current in \$Q_1\$. Not to mention that \$R_1\$ couldn't allow it, anyway. <S> Given that the two emitters share the same node, it's much more likely now that these details are exposed that \$Q_2\$ is actually off and \$Q_1\$ is the only transistor in an active mode. <S> Once you realize this, you then know that \$V_X\approx 2.3\:\textrm{V}\$, that \$Q_2\$ is off , \$I_O=\frac{+2.3\:\textrm{V}- \left(-6\:\textrm{V}\right)}{2\:\textrm{k}\Omega}=4.15\:\textrm{mA}\$, \$I_{C_1}\approx 4.15\:\textrm{mA}\$ <S> (slightly less than the emitter current because of a tiny base current), and that \$I_{C_2}\approx 0\:\textrm{mA}\$. <S> Another way to see that fact is to simply imagine the two transistors as both being completely off to start, and to then gradually activate \$R_1\$ by slowly pulling the \$-6\:\textrm{V}\$ rail downward, starting at the highest base voltage of \$+3\:\textrm{V}\$ and ramping it downward towards \$-6\:\textrm{V}\$ just a little bit at a time. <S> Note that as you do this mental step, \$Q_1\$ will definitely be the first BJT to turn on . <S> And once it turns on at around \$600\:\textrm{mV}\$ or so, when the bottom supply rail has only just reached about \$+2.4\:\textrm{V}\$, the emitter current will start rising upward by factors of 10 for each additional \$60\:\textrm{mV}\$ lower. <S> It doesn't take much to then realize that the emitter of \$Q_1\$ will never allow the base-emitter junction of \$Q_2\$ to become forward biased. <S> It just can't happen. <S> So \$Q_2\$ is off . <S> End of story. <S> Now you should be able to apply similar logic in your original case. <A> In cut off mode both junction of bjt (emitter-base and collector-base) will be in reverse biased condition. <S> It can happen only if base is biased negatively (in NPN transistor) and biased positively in (PNP transistor).
| To decide whether a transistor is on or off, look at the voltage across its B-E junction.
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Are laptop cells balanced while charging? I have little-to-no knowledge about how the laptop batteries work. I know there is a microcontroller to actually verify the balancing between cells (and in case blow up a fuse or something similar). I was wondering, does the recharge circuit balances cells as well while charging? Is charging procedure the same of the external chargers (diffent CV/CC phases) ? <Q> I don't have specific knowledge of any laptop battery charging circuit, but I'd be real surprised if the charger didn't do charge balancing. <S> I did work on a commercial battery backup unit once. <S> There were 8 lithium cells in series. <S> The voltage across each was measured separately. <S> Each cell also had a separate shunt resistor that could be turned on independent of the other cells. <S> During charging, we checked the cell voltages every second, and turned on the shunts around any cell that was above the median voltage. <S> During discharge there is little you can do. <S> There is no advantage to draining the high cells to keep pace with the low cells. <S> You do have to monitor each individual cell voltage, then shut down the whole pack when the lowest one gets to a threshold. <S> To really do this right, you should measure temperature too. <S> Adjust the full and empty voltage thresholds as a function of that temperature. <A> There are hundreds of examples on the web, and here is my own, ACER AL10B31 battery, 2P3S configuration, nameplate voltage of 11.1V <S> This particular battery pack uses BQ8050 battery pack manager with built-in cell balancing: <S> For more capable cells external FETs can be used, but it doesn't take much to balance internally. <S> The standard CC-CV charging, however, must be applied externally, which is typically located on laptop mainboard itself. <S> Many battery packs have protection switch which must be enabled from mainboard via I2C interface. <S> The interface is usually secured and non-disclosed to public, so some people might think that it has a "blown fuse". <A> Sorry for the late entry. <S> Based on my experience on cracking open a few laptop batteries, this is what I can tell:1. <S> Laptop batteries have NO balancing (HP, SONY, ASUS, etc.).2. <S> Yes, they have a fuse that, when is burnt <S> it's because the MicroControler told it to and, from my atempts to revive, there's no way to recover the pack.3. <S> Again yes, the Laptop charges with CC (first) and then CV just like any other LiIon charger.
| All modern laptop batteries (multi-cell ones) do have cell balancing circuitry built inside. If you want good life from lithium cells while getting good performance and not catching fire, you really need to manage each cell separately.
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Why isn't my battery charging? I have a YT7B-BS battery in my motorcycle. My motorcycle doesn't start because the battery voltage is only 12.05 Volts. The battery has the following inscriptions: Standard charge: 0.7 A x 5~10 hours Quick charge: 3 A x 1 hours 6.8 Ah (20 HR) / 110 A (CCA) 12 V, 6.5 Ah (10 HR) I tried to charge the battery with an external hard drive power supply rated as following: Switching adapter Model SYS1298-1812-W2E Input: 100-240 V ~ 1.0 A MAX, 50-60 Hz Output: +12 V continous 1.5 A Output power: 18W max I have plugged the positive terminal of the power supply to the positive terminal of the battery, and the negatives to the negatives. After a 2 hours, the battery voltage is exactly the same: 12.05 V. The power supply is a little warm but not too much. I plugged the motorcycle battery to my car battery for a few minutes and the tension went up to 12.5V, allowing my motorcycle to start (battery alone, no cables), so the battery doesn't seem to be dead. Why isn't the power supply charging the motorcycle battery? <Q> A 12V battery needs a higher voltage to charge. <S> No current will flow from the power supply to the battery. <S> If it was discharged even more however, you will be able to charge it up to 12V (Whatever the equivalent State of Charge corresponding to that voltage is). <S> Try using a 14V supply, and preferably with a current limitation (@3Amp MAX) feature so that you won't damage it. <A> Battery is near 50% SoC at 12V and >90% at 12.5V float after 1 hr and needs 14.2V <S> charge <S> Most likely the CCA has degraded from aging, temp or abuse (V<<1.5V for extended periods. <S> (Sulphation) <S> The CCA rating is about 30% lower at 0'C than at 20'C for the CA test. <S> You state CCA is 110A new which means 110A at 7.5V at 0'C . <S> If you cannot get 50A at 7.5V at 20'C , then the battery is < 40% start capacity of new and NG . <S> I suspect it is even worse. <S> If you cannot do a V reading during starter and measure I or any other load test to drop V and raise I by 50A <S> then I suggest you get a new battery. <S> First get a better charger for your battery with 14V. <A> Randomly connecting electrical things that aren't meant for each other when you have absolutely no clue what you are doing will usually end up with one of several outcomes : 1) It won't work but your stuff will be fine, 2) <S> It won't work and your stuff will be broken, 3) <S> It won't work and you'll end up causing yourself an electrocution or an explosion, 4) <S> You get lucky and it works. <S> In this case, you got lucky with outcome #1. <S> A real automotive battery charger will be labelled and sold as-such and will use a variety of voltage or current control methods while sensing the state of charge in the battery to adjust its delivery of power. <S> Improperly charging a lead-acid battery, in the worst case, can lead to excessive generation of hydrogen gas and explosion so this isn't really something that should be toyed with. <S> The other answers have given good detail about why your battery didn't charge, but fundamentally this didn't work because entirely the wrong tool was used for the job. <S> A commercial battery charger is not expensive and will generally be cheaper and do the job better (and more safely!) <S> than a DIY hack. <S> Designing and building a charger as a DIY project (that will actually work), while entirely feasible, is also most likely not going to be as cheap as simply buying a charger. <S> It sounds like your objective here <S> is to get your battery charged and not to embark on a didactic project to learn about electrical engineering so <S> , do yourself a favour, and just go buy an appropriate charger.
| Lead-acid battery chargers exist for a reason - because charging a battery isn't as simple as you might think (and, doubly, has the potential to be rather dangerous). The CA rating is the current shorted to 7.5V and is chiefly limited by plates ESR from aging which reduces the Ah capacity as well. You're using a 12V HDD power supply with outputs a constant voltage to charge a battery that already presents 12V across its terminals.
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LED lights wired in series will not light I have a pack of simple LED lights that I planned on wiring in my boat for cabin lights. I installed and wired them all in a simple series circuit. Power supply connected to switch, switch connected to positive side of first LED and negative to positive of the remaining LEDs and so on, then connected to ground. None of the lights will work. When I test each individual light with a 12V source they work fine but when I wire 2 or more in series they will not pass voltage through the LED to light the next light. What am I doing wrong? <Q> I think you are using 12v LED lights. <S> Thus, if you have 2 LED lights, 6v for each one. <S> If you have 3 LED lights, 4v for each one. <S> For more info have a look at this link . <S> Quouted from the previous link . <S> Finally, remember that for resistors in series, the current is the same for each resistor, and for resistors in parallel, the voltage is the same for each one. <A> If they individually work on 12V without immediately blowing up, they are probably LEDs with built in resistors designed to operate at a higher voltage. <S> If that voltage is 12V you can not connect them in series. <S> Verify they are actually 12V LEDS, and if they are, wire them up in parallel. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If they work one at a time perfectly witj 12 volts, either wire them all in pallel, or leave in series, and multiply the number of lights times 12...then get on amazon and get a step up converter, one with adequate max voltage that either meets or exceeds the sum you jist came up with, as well as current capacity. <A> You can run 6Sx12=72V or 3S2P =36V @ <S> 2x Irated or 2S3P @ <S> 24V <S> @ 3xI or 6P at 12V with 6xI but do not daisy chain more than 2 . <S> Also most 12V led-strips are designed with 9V threshold and rated at 12V or 14.2V for watt rating so you can increase V+ by 20% max but wire and supply must have ampacity to work well. <S> Confirm with valid specs. <S> 6S means 6 in series 2P means 2 strings in parallel . <A> Regarding using standard 12V lights (such as MR16) in series, it sounds right to use a higher voltage, but it might (will?) <S> cause problems. <S> I have a 36V electric bike which I made some lighting for. <S> Initially I used three 12V MR16 spotlights in series. <S> It worked fine for a while, but then I had a failure. <S> It may have been a one-off, so I replaced it. <S> Eventually, another failure. <S> I came to the conclusion that, although the bulbs are rated at 12V, they have chip inside to either step down the voltage or regulate the current and they didn't work in series. <S> I suspect one ended up going over voltage and failing. <S> The solution in my case was to buy a stepdown converter and to run them in parallel at 12V. Since then I have had zero failures of bulbs. <S> Two other things not mentioned: if in series to a higher voltage, one blows, they all go out and with standard led 12V bulbs, there are no issues about which way they are attached to DC as they rectify the current and work happily either way round on DC. <S> So, in a nutshell, do what others here recommend and go parallel. <A> There are following possible reasons: <S> May be <S> one of series LED is damaged. <S> As current is same but voltage is different in series circuit. <S> Therefore,there is a lot of chance that voltage is not properly supplied to each LED. <S> May sure the LED connections. <S> For more information about series circuit connection, please visit http://electronicslab4u.com/series-circuit-resistors-in-series/
| Accordingly, when you connect many LED lights in series, the 12v is going to be divided equally among the LED lights. In your case, you should connect them in parallel to get the job done.
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MCU - Minimum list of external components I can't find a complete answer for this anywhere, and I want to fill the gap in my knowledge of MCUs. In MCUs why do I need any other component except, maybe, a voltage regulator and an external clock? Why do PCBs with embedded MCUs look so complicated? If I were to design a minimal PCB with embedded MCU (say atmega324p), what is the minimum list of components I should consider, and why? <Q> Some microcontroller indeed only require a bypass cap to operate. <S> Others require some kind of external clock, like a crystal, resonator, R-C, or a digital signal fed in from elsewhere. <S> Newer micros tend to have internal clocks, but these are usually only good to a few percent accuracy. <S> For example, analog inputs may need to be scaled and conditioned before being presented to a A/D input. <S> Micros are usually used to control stuff. <S> That stuff can be complicated. <S> As for minimal parts list, see the datasheet. <S> Most micros need ground, power (with bypass cap, of course), maybe external crystal or clock, maybe a filter cap for internal voltage regulator, possibly something to drive the reset input, usually some connections to get the program in, etc. <S> For this information you have to read the datasheet , then also look at your overall system requirements. <S> Those dictate how you'll use the micro, and therefore what else you need to connect to it. <A> In MCUs why do I need any other component except maybe a voltage regulator and an external clock? <S> Some of the extra components are actually recommended, even for a minimum complexity design: <S> Protection devices like rectifier diodes, TVS diodes, etc. <S> OTOH, some MCUs will work without an external clock and/or voltage regulator at all, under certain circumstances (low-power, low-speed, battery-powered applications, for example). <S> Why do PCBs with embedded MCUs look so complicated? <S> Because they usually want to provide a minimum (or not so minimum) of functionality for the general purpose applications they're targeted to. <S> As an example: Signal LED's indicating whether the MCU is ON/OFF and/or any serial TX/RX communication is ongoing. <S> Interfacing circuits so that new code can be uploaded to the MCU. <S> This may include USB serial to TTL. <S> Connectivity as required (pin headers, USB, power barrels). <S> Several voltage regulators to provide dual 5V & 3.3V supply rails. <S> Power source selection circuits, to automatically select between DC power from barrel or from USB. <S> Of course, for a specific application, you could get rid of most of them. <S> If I were to design a minimal PCB with embedded MCU (say atmega324p) <S> what is the minimum list of components I should consider, and why? <S> It depends entirely on your application. <A> I assume you're talking about prototype boards here. <S> Those are made complex to make prototyping and debugging easier: they tend to include as many features as possible, limited by the fact that they have to remain generic enough, and by the price tag. <S> Then it comes to real projects, many basic MCU designs are much simpler than the corresponding proto boards. <S> There will be no LEDs or buttons, no hardware for unused features, etc. <S> As an example, consider this RFID tag made from a single MCU chip and a coil. <S> This is a bit extreme, but many commercial MCU applications have BoM lists which are finger-countable. <A> With some circuits even the external clock is not required. <S> In fact I have seen circuits with zero external components other than an LED for it to actually do something. <S> The complexity of the circuitry round it is a function of what you want the application to do more than the need of the micro. <S> Adding resets and crystals and whatever else is required to perform a particular application, increases the part count. <S> But most modern micros will be quite happy to start on their lonesome. <S> However, one could argue a micro with zero external components doesn't actually do anything. <S> "If a tree falls in a forrest and there is nobody there to hear it, does it make any noise?" <S> So a power supply would be optional too. <A> MCU doesn't need anything more that a supply source to works. <S> Unlike microprocessor, microcontroller have intern memory, build-in clock source (even if not accurate) Because MCU has lots of pins? <A> Why do PCBs with embedded MCUs look so complicated? <S> Depending on the application, there can be the need for: <S> Additional memory (SDRAM, EEPROM, flash) <S> Additional I/ <S> O devices such as temperature and humidity sensors, accelerometers, USB interface, WiFi module. <S> Additional I/ <S> O connectivity (Analog signal conditioning, GPIO level shifting, buffering and isoation.) <S> Power on reset circuitry. <S> Battery management
| Boards with micros can be complicated because the system needs to do additional things external to the micro. Typically, there will be at least some LEDs and buttons, proper clock and reset circuits, proper voltage references, plus the hardware required for communication protocols and debug interfaces that the MCU supports Under certain circumstances and design goals, the list can go down to just the MCU - nothing else required. Capacitors required by the voltage regulator for stability and load/line regulation performance.
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Thevenin Equivalent of RC circuit I'm trying to find the theveinin equivalent of this RC circuit but am a little confused. Since R2 and C are in parrallel, how does this factor into finding the thevenin equivalent resistance? Is it still \$R_{th} = \frac{R_2}{R_1+R_2}\$ or is there something more subtle here I'm missing? Does the capacitance remain the same here too? <Q> The Thevenin equivalent is found at a pair of nodes. <S> What it is connect to doesn't change. <S> In this case, the black dots are probably causing confusion. <S> The pair of nodes are either end of R2, between R2 and C. <S> If you are only interested in DC then the capacitor doesn't matter much. <S> If you are interested in the frequency dependent Thevenin equivalent then C will make a difference. <S> You still have the option of not including it in the circuit and get a Thevenin equivalent of the source and resistors. <S> Then the the circuit will look like the one on the right. <S> If you include the C in the equivalent, you would end up with a source and a complex impedance without C across the output. <A> You're on the right track, but your formula is wrong. <S> R1 and R2 are in parallel, so you need to use the formula for combining parallel resistors to get the Thevenin resistance. <A> No. <S> Your question is on the system's capacitance at it's input port , the thevenin capacitance. ' <S> C' is just a part of that!
| If you want the Thevenin equivalent made up of the source, R1, and R2 not including C then C doesn't change.
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Going 20 Volts over what a Fuse Holder is rated for I have a fuse holder that is rated for 32V DC, but I need to hook it up to some capacitors that are rated at 5 F and 50 volts. Will the fuse just blow right away even if the trip current isn't achieved, or will the fuse holder still function normally? If it will work normally, would Ohm's law apply and I just use a fuse rated lower than what I need? <Q> Below 300v or so, at normal air pressure, the voltage is too low to initiate a spark across any practical gap. <S> This points to the 32v rating of the fuseholder not being governed by breakdown issues. <S> The fuse itself is a different matter, as this starts to open with a current flowing, the perfect condition for maintaining current flow with an arc. <S> Are the fuses themselves rated at 32v, or some other voltage? <S> As 32v is in the range of voltages that various regulation bodies deem as 'safe to touch', I would at least consider the possibility that this fuseholder does not pass touch tests for high voltage, so is relegated to being used at touch safe voltages. <S> If this is a one-off for home use (so you don't need to get it insurance rated), and your 50v supply is isolated (so any accidental contact problems with the fuse holder are mitigated, however note that 50v is rated as touch-safe by some jusidictions), and the actual fuse in the holder has a voltage rating of 50v or above, then it should be OK to use. <S> Please add a picture of the holder, or a link to its specifications, to be sure. <A> The voltage rating of a physical component is specified such that if the maximum voltage is reached the component will not suffer catastrophic failure. <S> If the fuseholder rated for 32V is brought up to 50V <S> then the fuse may not matter since the current could either arc or creep at that voltage. <A> A fuse that has not yet blown will have close to 0 V across it; the voltage rating is irrelevant at this point. <S> The problem comes after the fuse starts to blow due to overcurrent.
| If the available voltage is greater than the fuse's voltage rating, it may fail to open, open more slowly than expected, or catch on fire.
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Side effects of using large resistances Are there any problems that can be caused by using resistors of large resistances (in the order of megaohms)? I'm designing a feedback network that is just a voltage divider, and I want the feedback to drain as little current as possible from the circuit. The only thing that matters is the ratio between the resistors. So my question is: is there any reason why one would pick, for example, resistors of 1 and 10 Ohms instead of 1 and 10 MOhms? <Q> There are many drawbacks to both low and high values alike. <S> The ideal values will fall in between very large and very small for most applications. <S> A larger resistor of same type will, for example, create more noise (by itself and through small induced noise currents) than a smaller one, though that may not always be important to you. <S> A smaller resistor will drain more current and create more losses, as you have surmised yourself. <S> A larger resistor will create a higher error with the same leakage current. <S> If your feedback pin in the middle of your resistors leaks 1 <S> μA <S> when the resistor feeding that leak is 1 MOhm, that will translate to an error of 1V, while a 10k resistor will translate to an error of 10mV. <S> Of course, if the leakage is in the order of several nA or less, you might not care much about the error a 1 MOhm resistor creates. <S> But you might, depending on what exactly you are designing. <S> It's all checks and balances, and if that's not enough information at this point, you might want to ask a more direct question about specifically what you are doing. <S> With schematics and that. <A> In addition to the issues that @Asmyldof mentions, when using high resistances in the megaohms (and especially at 10M and more) environmental contamination such as dust, skin oils, soldering flux residue etc can easily reduce the effective resistance in unpredictable and time-varying ways. <A> In addition to other answers, also consider thermal noise . <S> If you want very accurate measurements, this may be an issue. <A> It's not unusual at all to use high resistances in dividers and feedback circuits for the reason you mention - to reduce current consumption and loading, especially for high impedance sensors, for example. <S> A few precautions should be taken to ensure predictable operation though. <S> The board should be well cleaned before and after component placement to avoid contamination appearing as a parallel resistance. <S> A good quality flux cleaner followed by an isopropyl alcohol swab is good for this. <S> If the circuit is to be operated in an unpredictable environment (like where there may be moisture buildup or high humidity) <S> then a good conformal coating agent should be applied to the board and components, and baked out as per the manufacturers instructions to produce a sealed, high resistance moisture barrier. <A> First lets consider the problems using LOW resistor values, with opamps. <S> The biggest problem is the opamp's limited output current. <S> Often 20 mA is the maximum for accurate performance. <S> Yet, 1 ohm and 1 volt require 1 ampere. <S> It is not available. <S> Thus you must design with higher values. <S> Another problem with LOW values is the thermal distortion, as self-heating causes big temperature changes and big resistance changes. <S> Using 1 ohms and 9 ohms, to set gain in feedback loop of opamp, causes the 9 ohms to dissipate 9X the power. <S> At 1 millivolt input, the 1mA current may or may not cause detectable distortion. <S> Walt Jung discussed this, for Audio Power Amplifier feedback dividers. <S> Now for HIGH value resistances <S> :A problem with higher values comes with the capacitance on the -V IN pin of the opamp. <S> The phase shifts ---- 1 megohm and 10 pF have a Tau of 10 µS, thus a 45 degree phase shift at 16 kHz ---- <S> it leads to peaking, instability, and oscillation. <S> The cure is to use tiny capacitors in parallel with the high value Rfeedback resistors...another component to buy and install. <S> High resistances leave the circuit vulnerable to Efield interferers. <S> The capacitively injected charges will find a return path. <S> A 10Meg Ohm resistor, facing 160volt 60Hz wiring at 4", coupling into 14mm by 1mm PCB trace, induces 1.5 millivolts of 60Hz. <S> At the 1Kohm level, the interference is 10,000X smaller. <S> Lets also examine an LDO, providing regulated 2.5 volt output for any Vunreg over 2.7 volts, with standby current < 1uA per the datasheet. <S> What do we know about the output noise of that LDO? <S> simulate this circuit – <S> Schematic created using CircuitLab <S> We know this LDO has at least 60 microvolts RMS output noise, because of the 12Million Ohms (times 2) feedback resistors. <S> At least 60uV, because the internal opamp has high noise (at very low currents, expect high noise) and the 1.22 volt BandGap has high value resistors. <S> I recall an LDO, with 1uA Iddq, showing poor PSRR above 100Hz. <S> Turns out the Vin metallization was above the 12Meg Ohm voltage dividers. <S> Any trash coming into the LDO was directly injected into the servo-amplifier loop. <S> Learn to visualize these problems. <S> The original designer stated "the parasitic extraction did not show this as a problem." <S> Learn to visualize these problems.
| Smaller resistors in feedback systems, e.g. with inverting amplifiers using op-amps, may cause errors on the incoming signal if the incoming signal is relatively weak. As your resistance goes up, so does the noise.
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Is the maximum current of a wire related with its length? Well, I Know this seems to be a silly question.. Note that the voltage drop and the delivery efficiency is not under consideration, just the maximum current limited by the maximum dissipation power of the wire itself. Thanks. <Q> If the wire is very short, then the heat-sinking effect of how it is mounted must be taken into account. <S> Good heat-sinking will increase the amount of heat, that is the current flowing, to obtain any given temperature rise. <A> Short answer : <S> No <S> The current rating depends on how much power (Heat) your cable can dissipate, which in proportional to the cable surface area, which is in turn proportional to the cable's length. <S> To summarize, a longer cable will have a higher resistance (R=pL/A), and thus will will generate more heat (P = RI²), but this heat will be distributed on a broader surface, which will compensate. <A> I would also say, No. <S> Because each unit of cable dissipates its own portion of energy in the form of heat, it doesn't matter the length. <S> It then is a question of whether the heat can be conducted away from the wire at a rate greater than the heat being dissipated. <S> Power companies know this. <S> They can move a little more current down their transmission lines in the winter than in the summer. <S> They will even up the current further during ice storms prevent ice accumulation. <A> That depends if it's AC or DC. <S> For DC see the other answers. <S> For AC it is a little different. <S> Ignoring voltage drop along the line, i.e. a superconducting line, the alternating current in the line is still generating a magnetic field around the wire. <S> The longer the wire the bigger the field. <S> You can think of it as the primary of a very LONG transformer. <S> If it is infinite in length it's an ideal primary, and you will not be able to drive ANY current into it unless there is an accompanying secondary somewhere drawing a load current.
| For a long wire, long enough that no appreciable heat flows along its length to the mounting points, the length is irrelevant.
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Can cutting electrical steel alter its magnetic properties? Please forgive me in advance if this question is not appropriate here. If you think so - let me know and I'll take it down. However I am really hoping someone can share their thoughts! I want to make an experimental transformer using standard, off-the-shelf U I laminated electrical steel sheets. As a part of the experiment, I need to form a gap in the core of a certain specific shape. I want to cut the sheets myself, but I am just wondering if cutting the sheets will stress the electrical steel so that its magnetic properties might be changed, reducing the efficiency of the transformer. Can anyone comment on this? I was planning on cutting the sheet (0.5mm) with a hacksaw. Very low tech I know, but I don't have the money to get a custom core made...bad idea? <Q> So, yes and no: Of course, cutting steel is going to produce a lot of heat. <S> But: with a slow hacksaw, that effect is probably not really relevant at all. <S> What I'd worry much more about is that you still need to stack multiple of these sheets, and you will break the lamination close to your cut, leading to stray currents across the sheets, which might reduce efficiency slightly bend, wave the sheets, so that they won't perfectly stack, which will reduce efficiency <S> So, I don't think the hacksaw is the optimal tool here; I don't know where you are, but it's often pretty easy to find metal workshops in any city, and many of those have CNC cutters, and CNC drills. <S> They aren't really expensive, and you'll save yourself a lot of pain and get better results if you ask them to cut things for you. <S> It's generally a good idea to show up in person with your stock U's, and explain what you need to happen, and ask whether you can watch while they work on your metal. <S> Most craftspeople are kind of proud of what they do, and you might learn a lot about handling the metal the way they do it professionally – and that might save you a lot of headache, metal dust, finger damage and cost later on :) <A> Improper heat treatment can change the magnetic properties of the steel <S> yes. <S> However manually forming each sheet will not produce a great product. <S> A better solution may be to stack-and-mill. <S> That is, stack and fix your sheets to the thickness you require, and then mill the stack, or have some-one mill it, to your desired dimensions. <S> The milling process can be done using a coolant to ensure heat introduced is minimized and your resulting core will be much more uniform. <A> There might be a simpler way of doing that. <S> You can put a sheet of foil over the entire U-I joint surface. <S> Instead making a 0.5mm space, that would be very difficult to trim the saw to match exact cutting length, you can put a 0.25mm PE sheet on both sides. <S> Technically the gap will be the same 2 x 0.25mm = 0.5mm, also the characteristics of new transformer core.
| Heating changes the metal structure and will alter the properties.
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Using a home oven for solder paste I'm doing a project and I have a PCB which needs surface mount components attaching to it. One component has already been attached, and I have some 138deg solder paste in a syringe. This is possibly going to sound very stupid, but is it feasible to use my kitchen oven or a similar household device to heat up the solder? This is just a one off for me and I am not planning on doing any surface mount soldering in future, and I do not have easy access to a proper reflow oven. All the components will be mounted on one side if that's an issue. <Q> Lots of people use toaster ovens and electric fry pans for reflow. <S> It can get a little touch and go with lead-free solder, and I recommend leaded solder. <S> Large kitchen ovens likely don't have the oomph to bring things to temp fast enough, and your IC's may not tolerate the slow temperature profile. <S> I think whatever appliance you use, you should dedicate to reflow, and not use it for food. <A> Can you? <S> Yes. <S> Will it work every time... <S> Maybe not! <S> There are two reasons for this. <S> Temperature profile and timing. <S> Reflow is done using two temperatures, a pre-soak temperature and a reflow temperature as shown below. <S> Oxidation Commercial reflow is usually done in a nitrogen gas so that oxidation will not occur in the solder and joints during the molten phase. <S> That's rather hard to do in your domestic environment. <S> Some parts are MUCH more sensitive to being "cooked" for too long, especially plastic parts like connectors and the like. <S> If you are doing this at home it is prudent to identify those and manually add them after. <S> SAFETY CONCERNS <S> Other than the obvious... <S> "Don't burn yourself... <S> " it is not a healthy idea to solder using the same oven you plan on baking that apple pie in later in the day. <S> Lead and other noxious chemical fumes will permeate the oven. <S> Also, if you are married, severe tissue damage can occur in the rectal area from the insertion of your spouses footwear when she, or he, finds out what you did. <A> If you want to learn how to solder SMT components by hand in an unimpeachable fashion, I would recommend watching some of the Youtube videos by IPC <S> (J-STD-001) certified soldering teachers. <S> You will need pure IPA (isopropanol) not drugstore rubbing alcohol, cored solder in <S> appropriate size(s), appropriate liquid flux and preferably a decent stereo microscope or at least a magnifier. <S> And, of course, a clean well-lit work area, preferably with ESD precautions. <S> I would not recommend try to solder a one-off with solder paste. <S> I do it myself with a stencil but the oven parameters are known-good for most boards. <S> Even dispensing appropriate amounts of paste on each pad without a timed pneumatic dispenser is not easy (the solder paste behaves in a fairly unpleasant manner because it's not really a liquid- <S> it's a bunch of little balls of solder in a matrix of liquid flux). <S> Sn63Pb37 leaded solder temperature-time profiles are relatively forgiving, but unleaded solder means running some parts very close to the point where damage occurs so the parameters have to be well controlled. <S> For something like a resistor you clean the board with IPA, maybe the part if it is not pristine, then tin one pad and slide the part in to solder one side, then solder the other pad. <S> There are some subtleties as to where exactly you put the solder during the process- to get the part to wet, and then feeding it in at the junction of pad and part, while heating both with the iron. <S> Then you clean and inspect and move on to the next. <S> It takes a while but the results can be very good. <S> If you insist there are people who are successful using nothing more than a skillet but you will want to not use that for food afterward, especially if you are using lead-based solder. <S> Make sure you wash your hands after handling the paste <S> (as a paste it's worse than wire solder) and (if applicable) don't do it at all if there is any chance of pregnancy. <A> You probably wouldn't want to use a regular oven for three reasons: Its probably not a good idea to put lead where food is. <S> A larger oven has more uneaven heating because there are more convection currents <S> The temperature profile of the solder paste would be hard to duplicate in a larger oven Use a smaller toaster oven, get one from a second hand store. <S> Look up the temperature profile for the paste you have, you need to time the temperature profile to that of the solder, you will get better results. <S> If this is just a one off PCB and you won't do it long term, find a cheap assembly place. <S> Another option is skipping the oven all together. <S> A hot air rework station will probably do just fine in your case. <S> There are numerous videos and tutorials on the web that show how to solder or use a hot air rework station with QFN's QFP's and other components. <S> This is a good option if you have a low part count on the PCB or low quantity production <S> run ~(<3 <S> PCB's).
| Getting the oven right could result in damage to board. IN ADDITION: Not all components can withstand the same heat. As such, you REALLY need to use two ovens to do this right, quickly switching it from one to the other after the pre-soak time.
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How can I keep this MOSFET in cut-off when Vcc > 5.5V? This is a follow up-question for this question where I had troubles with a latch occuring at start-up. While solving the earlier issue I found another one which I had not been able to anticipate since this is the first time I'm playing with a P-channel MOSFET. Have a look at this schematic, where I have stripped away everything that's not relevant to the actual problem: Circuit explanation The circuit is a part of a voltage monitor. Nominal input is Vcc = 5V, and if the voltage goes higher than 5.5V (feel free to read the above mentioned question for further details) the outgoing supply line (Vcc_OUT) is cut-off to prevent damage. The opamp is configured as a comparator. It's unfortunately not rail-to-rail, so "high" is ~4.5V, "low" ~0.5V. When the opamp output is "low", it means that the voltage tripping has not occured. The PNP (Q2) is therefore saturated and pulls the Gate to ground, making the MOSFET go wide open and Vcc_OUT is around Vcc. This is expected, and also works as expected. When the opamp output is "high", it means that the voltage tripping has occured. The PNP is cut-off, and Gate is tied to Vcc effectively closing the MOSFET and cutting off the supply to Vcc_OUT. This also works as expected. Background info about Vcc Vcc comes from a linear regulated power supply. A 9V-transformer is rectified, fed and filtered into a 7805. Their design are beyond my reach. Sometimes error can occur and the Vcc gets raised slightly, and that's where my voltage monitor comes in. Absolute worst case disaster scenario would be if the rectifier diodes and the 7805 all gets shorted, meaning Vcc will be 9VAC (RMS), around 13VAC. This will instantaneously kill everything. The problem The opamp controls the MOSFET as I expect, but when I played around earlier with my lab supply I increased Vcc to the level of the worst case scenario described above; 13V. At this voltage, the MOSFET conducted (or whatever the word, it was not "cut-off")! It should definately not conduct at this Vcc-level! My test-LED that's connected to Drain was very, very lit. I turned my lab supply back to 5V, and the LED went dark. Increased to 6V, still dark. 7V, still dark. 8V: a tiny, tiny light can be seen in the LED. Increasing to about 10.5V made the Drain / Vcc_OUT have ~5V, and increasing to the disaster level 13V, Vcc_OUT was around 7V. The guess As stated earlier, this is my first try with a MOSFET so I'm unable to pin-point the error. My only hypothesis is that because Vcc is floating, and GND is not, the Vgs level is somehow affected at Vcc > 8V, making the MOSFET conduct (which it shouldn't). The question Using this setup, is there any way to keep the MOSFET cut-off at all times whenever opamp is "high", no matter what level Vcc has? Datasheets: FDD6637 , OPA2132 <Q> First of all I would not use an op-amp as a comparator, they are not really designed to work that way. <S> Instead find a suitable open collector true comparator of equivalent performance and wire it this way. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It may also be a prudent to add a little positive feedback to that circuit to add some hysteresis. <S> Otherwise, when the trigger is hovering around the reference voltage that output is going to turn on and off very rapidly due to any ripple or noise that is in the circuit. <S> simulate this circuit <A> If you read the data sheet for the opamp, page 6 tells you that the output cannot get to within about a volt of the positive rail. <S> This means that it can't really switch off the PNP BJT. <S> I expect as the supply voltage rises, the problem gets worse until the LED you have connected starts to glow. <S> Maybe make both 2k2 in value. <S> EDIT <S> For clarity, when I said base emitter resistor, I would employ a resistor from base to Vcc. <S> Sorry for any confusion. <A> I would recommend changing the Q2 for a P-channel MOSFET with a logic level gate drive. <S> What is most likely happening with your circuit is look in page 6 of the <S> OPA2132 the drive to positive rail is typical the rail voltage minus 0.9V which will cause Q2 to be turned on because there will be a 0.9V across the emitter to base junction. <S> Provided the datasheet shows the 0.9V is the value when the device is power from +/-15 <S> volts in the the rails. <S> This is difference between output voltage and the power supply probably is increases as the power supply increases leading to the behavior you are seeing. <A> Maybe I'm missing something here, but surely the most sensible thing to do is swap the PNP for an NPN, and swap the opamp inputs? <S> The 0.5V output of your opamp should be able to completely switch off the NPN, and if it doesnt you can just add a diode in series with the base to drop more voltage. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Since the base of the NPN is referenced to ground, the VCC voltage doesn't matter (up to a point), so it will work for your entire range (assuming you choose an NPN that can handle it). <S> I'll add a diagram later. <S> Edit: added diagram
| I might consider adding a diode in series with the base or, a base emitter resistor of the same value as the one feeding the base.
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Designing an Electromagnet for a Digital Clock I'm trying to design a digital clock that will use several electromagnets and iron filings to show the time. Each digit should have 7 electromagnets that turned on and off to attract the metal shavings through white plexiglass (I'm also considering using immersion oil to make it easier for the filings to reform). However, I'm stuck on how to design the electromagnet itself. I initially started with a solenoid, however, I quickly learned that the shavings will mainly attract two ends, leaving the middle of the digit line with no definition. I'm looking for advice on how this might be done. Will wrapping the wire vertically instead of horizontally be a better choice? Should I use a different shape altogether? Thank you, <Q> rotate your electromagnet by 90°, so that its axis is perpendicular to the plexiglass. <S> That's kind of obvious – it puts only one of the filings-attrackting ends close to the glass. <S> If you wind your magnet around a rectangular-crosssection piece of iron, then you can make "straight lines". <S> Notice that even these lines will be feathered – the filings will align along the magnetic field lines, and those form a convex shape from one end to the other <A> I see two workable options for accomplishing something close to what you're describing: <S> Build the electromagnets with a flattened (as opposed to round) core, then place one pole against the underside of the plexiglass. <S> This would attract filings fairly uniformly to that "segment." <S> This would form poles at either end, and gaussian lines connecting them, which may prove to give you better "resolution" in getting your iron filings to line up in a recognizable digit. <S> CAVEAT: <S> Both of these designs should work well for a 1-segment "test case," but there will be more problems to overcome when you start combining segments. <S> I think option #1 may have less "intersegment interference," so long as all poles are equal and synchronised in-phase with each other. <S> However, in practice there are a lot of ways for the electromagnets to interfere with each other when in close enough proximity to form a 7-seg. <S> display. <S> ... <S> In the end, you'll have to do a lot of testing & adjusting, and may end up having to use magnetically shielded segments of some form. <A> Filings are hard to manage, once you get them attracted to a pole piece they are actually hard to get off due to residual magnetic retention effects. <S> You may have to use a high frequency AC signal to demagnetize them. <S> You might be better off using a Ferro Fluid. <S> This is dense, but with very low magnetic retention. <S> Of course like all things novel..... <S> it's already been done quite effectively. <S> The researchers even wrote a paper showing the methodology and a good picture of the electromagnets. <S> I've always fancied making one of these since they showed the prototype...but never got around to it. <S> It looks like it would be fun manipulating the blobs of Ferro fluid to build characters. <S> Show pictures when you are done!!
| Build the electromagnets as small "horseshoe magnets" with both poles touching the underside of the plexiglass (and the windings around the middle).
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Measure diameter of a ball with Arduino I'm trying to design a hand-held device which would allow to measure the weight and a size of a fruit (orange/apple). For the weight I can use a load cell, but not sure how to measure the diameter. What I'm thinking at the moment is a cone shape. The bigger the orange the farther it will be from the end of the cone.What I can measure is the distance from the end of the cone to the orange or by measuring the distance from the end of the cone to where the orange touches the wall. For the first one I could use some kind of resistive rod which the orange will push out and then measure the resistance. For the second one I could use a touch bar sensor. Any ideas what kind of elements I can use for either solution? Can someone think of a better way to measure the diameter of an orange on the go with arduino? A little bit of a background: the quality of oranges in my local stores is not very good - you get a lot of dried out ones. The idea is that the heavier the orange given the same diameter the better it is (it contains more juice). I want to build a kind of hanging scale which you hold in your hand and put an orange in it - it will sense the diameter of an orange and it's weight and displays the "juiciness" of the orange. This way I can filter out the bad ones. <Q> Or you could even go about it like digital calipers do - a mechanism based on varying capacitive and measuring phase shift of PWM signals (i.e. probably more complicated than is worthwhile). <S> But, better yet, use a slide pot, which is resistive and dead-simple to integrate. <S> Sliders are relatively cheap and wouldn't necessarily require any external components - just reading voltage level through divider. <S> You'd just need to be sure to get one longer than the greatest diameter fruit you expect to encounter. <S> On the other hand, the manufacturers who produce calipers and linear encoders opt for other methods - capacitance, inductance, Hall effect, magnetoresistive, optical, etc. - for increased precision (resolution) and accuracy. <S> I expect that your application does not call for such precision, though. <S> In any event, you'd avoid the need for a cone, which adds a good deal of bulk to your portable device. <A> that's a nice question!I've got an Idea, which might help. <S> You can use an ultrasonic sensor, and place them in front of a board, based on how large you want it to be. <S> Measure the distance between the board and <S> the sensor(serial.print and then take 5-6 values get an average figure). <S> Put an object now, and subtract this reading from the initial distance(between the board and the sensor). <S> You got the DIAMETER! <S> Or if you don't want a board and want a hi-fi, cool looking thing that just measures the dia by just putting it there, you should take 2 sensors, put them at a desired distance and note that distance down(don't use serial.print here, use a scale/ruler/measuring tape) now put an object in between. <S> Take both the readings from the sensors and add them up. <S> Subtract this from the total distance between the sensors. <S> YOU HAVE THE DIAMETER! <S> ( with this you can also adjust the size of the whole setup) Or the last method: get a digital vernier calipers, and glue a gear to it and connect it to servo. <S> get it to the max distance initially, and make the servo go to as close as possible. <S> now google a method to take the value from the circuit inside the calipers, and use it as input and display it. <S> YOU HAVE THE DIAMETER!(I know the last one is shit) <S> let me know if it worked for you! <S> Good Luck! <S> :) <A> I would use an IR distance finder with your cone such as the Sharp GP2Y0A41SK0F that has an analog output and works off 5v supply so is quite easy to interface to. <S> It's very good at measuring distances close to (but needs to be above 4cm). <S> It is also fast, compact, light weight, low power, reliable, and has no moving parts. <S> Side note: a cone is bulky, perhaps would be more portable to just use a V-shape made from 2 rods - although then weighing would be tricky.
| You might also consider measuring diameter using a linear encoder, which would add precision and accuracy at the cost of...cost.
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How far can mains voltage arc in air? I wondered this while soldering a mains voltage circuit board and was surprised by how close together the traces were.It has obvious implications in design of electrical plugs, and the proximity of wires when doing anything to do with mains voltage. I've tried asking search engines sensible questions like "how far can 240V arc at 1 atmosphere" and "how far can electricity jump" but I haven't found any easy answer. This calculator states that it only takes voltages between 400 and 3000VDC. By asking this question, I hope future people will be able to find the answer quickly and simply. My research suggests that the arcing distance is dependant on the medium and the pressure, so let's assume air (~79% nitrogen, ~20% oxygen, ~1% argon and a few other things) at 1 atmosphere or 1.01325 Bar. An answer has also drawn my attention to the affect of temperature and humidity. Assuming that higher temperatures and higher humidities both increase the possible arcing distance, let's choose something harsh like 40 degrees Celsius and 95% humidity. Given a mains voltage of 230VAC in the UK, how close would two uninsulated copper wires (as an example) need to be before an arc could form between them? Is this different for traces on a circuit board, or pins in a plug? For bonus points, could answers be given for 120VAC too? Would 240V arc significantly farther than 230V? How about 110V compared to 120V? I'm looking for fairly concise answers, but perhaps the reason I haven't found a simple answer is because there isn't one... This question is just out of curiosity. I'm not going to start rewiring mains fixtures or designing 240V circuit boards any time soon. <Q> The breakdown voltage of air varies significantly due to changes in humidity, pressure, and temperature. <S> However, a rough guide is that it takes 1 kV per millimeter. <S> Since that's about where arcs happen, you want to be nowhere near that in a real circuit. <S> On a circuit board, you also have to consider conduction along the surface. <S> This is why you often see talk of clearance and creapage in the same discussion. <S> Clearance is the straightest path between two conductors. <S> This is where the rough guide of 1 kV/mm for arcing applies. <S> Creapage is the shortest distance between to conductors along a surface. <S> The breakdown gradient for creapage is lower than for clearance since dirt can accumulate on surfaces. <S> Some dirt is partially conductive on its own, but many things can provide leakage paths after soaking up some humidity. <S> Take a look at specs for medical power supplies, for example, and you'll see large minimum creapage requirements to guarantee low leakage currents. <S> For most ordinary consumer equipment, 5 mm clearance is good enough isolation between user-touchable parts and 120 V AC power. <S> However, you really should look at the relevant standards, especially if you are doing something out of the ordinary. <A> Given a mains voltage of 230VAC in the UK, how close would two uninsulated copper wires (as an example) need to be before an arc could form between them? <S> The answer is: it depends. <S> International standards boards (namely IPC, and IEC) have come up with minimum distances between insulated conductors. <S> Uninsulated conductors are not safe for use in products <S> so those distances are not provided. <S> Uninsulated conductors in PCB's or connectors are covered in the clearance section of the table. <S> These specs are to prevent arcing or any kind of fire hazard. <S> It should also be noted that to view the actual specs you'll need to buy them from IEC (like IEC 61010-1 ), but there is a lot of information regarding the content of these specifications available on the web. <S> Source: <S> http://www.pcbtechguide.com/2009/02/creepage-vs-clearance.html <S> It should also be noted that the distance changes depending on the environment (pollution degree), an environment that sees more dirt/humidity will have a shorter spacing. <S> The distances in the table above are for a pollution degree 2 which would probably cover most designs, if not, find a table (or buy the spec) for the pollution degree your designing for. <S> Source: <S> http://www.ni.com/white-paper/2871/en/ <S> Is this different for traces on a circuit board, or pins in a plug? <S> Yes. <S> In the first table, the distance essentially doubles for off PCB conductors. <S> For bonus points, could answers be given for 120VAC too? <S> Would 240V <S> arc significantly farther than 230V? <S> How about 110V compared to 120V? <S> In the table above, if only designing for 120V the distance is shorter. <A> At large gaps it depends roughly linearly on distance, and also depends on the composition, temperature and pressure of the gas. <S> For air at standard temperature and pressure it's about 3.3MV/m. <S> As the gap gets very small the voltage to create a spark actually increases again. <S> The spark is caused by free electrons that are accelerated by the voltage knocking other electrons off air molecules. <S> If the gap is too small they can't get enough of a run up to knock off another electron before hitting the positive electrode. <S> This means that there's a minimum sparking voltage of 327V at 7.5µm in normal air. <S> 240VAC has a peak voltage of <S> ~340V, so you might be able to get it to briefly spark near the peak with a gap close to 7.5µm. <S> 120VAC won't spark in air. <S> In the real world there can be transient overvoltages, contaminants, condensation, etc. <S> You shouldn't rely on the above for safety purposes.
| There are a variety of factors including air, pressure\elevation, humidity, and dirt from the environment all affect the distance that an arc can form between two conductors. There are various safety standards out there that require minimum clearance and creapage distances according to application, voltage, and sometimes environmental parameters. The minimum voltage required to initiate an electrical arc in gas is described by Paschen's Law.
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Controlling hardware with serial port I initially posted this in SuperUser SE Site, but it was recommended that I bring it here instead. So, here we go :) We have a 'dated' security system at the office. It is a model that was made (or at least rebranded) by a company named ATI Access. The board are stamped as: SA-2000-II V8.0. As I mentioned before, the company that made (rebranded) the board is "ATI Access", located 30 minutes away from me in Milwaukee, WI. As of a few weeks ago, they are now defunct. That means that all my tech support, etc, has gone out the window. With that being said, I am trying to have a go at a small project, but I am now flying blind. Here are the details: On the top left side of the board, there are 5 inputs that can be programmed to do various tasks such as trigger an alert, or open a door by pushing a button. Looking at the manual for the board, a simple 'dry contact' switch is all that is needed to trigger them. So, what I have done on one of my panels is connect wires to Input 2 and attached them to a doorbell attached to the side of a table in my IT Closet. I then programmed the input to open the door on my IT Closet when the button is pressed. It works great. Go me, right? Well, that is not good enough for me. I want to be able to control it with my PC as well. Now yes, I could just keep the security system open at all times, and alt-tab to it, right click on the door, and choose Open, but that is a lot of steps. I figured I would just write my own app that sits in the system tray that I can double click on, which would somehow emulate me pushing the doorbell. I am not looking to use the PC to control a motor to push the doorbell for me, although that would be kinda cool. I am thinking that perhaps the serial or parallel ports could act as a dry-contact switch (however I am pretty sure that's not a feasible idea), or perhaps using some sort of dry contact relay that can be controlled via the serial port. So, the sequence of events would then be: Click on the icon of my app in the system tray App sends a signal out via the COM or LPT port (this part I can handle, I love programming) That signal is picked up by [insert stuff I don't know here] that trips the relay momentarily to a CLOSED position, completing the circuit to the security panel input header. Security Panel works its magic and opens my door. As stated, I have this done the manual way utilizing a doorbell button, but I am looking to fancy it up and use my PC instead. Has anyone done here ever done something similar to this and is willing to point me into the right direction for [insert stuff I don't know here] ? For bonus points: Does anyone have experience with this type of security panel? I'd be interested in knowing if there are any alternatives to the software that I am currently using for it (StarAccess Pro is the software that is used). <Q> I recommend an Arduino board and any one of the many existing "relay shields" for it. <S> Arduino connected via USB to computer is easily programmed to present a virtual COM port where you can send commands. <S> Then you can close relay contacts by just switching an Arduino output pin from LOW to HIGH state. <A> A arduino nano or uno would work fine and connect over USB to your pc, emulating a serial port. <S> You would attach a relay with driving transistor or if your looking for an easy option, a relay shield. <S> This is how the code would run: initialisation stuff Wait for serial input and then... Match serial input to a command set- eg, activateswitch1 and then... <S> Do the corresponding actions eg, turn on pin 3 to activate relay <S> Then wait again for more commands! <S> This could cost off eBay from those Chinese sellers around $10 arduino clone- <S> $5 relay module- <S> $3 project box- $2 <S> some indicator leds as you want them- <S> $1 <S> You could use a hc-06 serial over Bluetooth module. <S> I have personally worked with these and are really easy to use. <S> I have even made them work with android apps <S> so I can send data to the arduino over Bluetooth. <S> Well that's to different routes and again, other people have talked about commercial options so the choice is yours. <A> You may be going a little OTT with your thinking. <S> Down the right hand side of your panel there seems to be an unused RS-232 port, so one option may be to use that to send a command to the panel to tell it to open the door. <S> There's also the RS-422 port you've labelled as going to the PC, presumably this is what the control software uses, but it may also provide a method of doing what you want without any extra hardware. <S> If you do need a separate bit of hardware there are lots of options. <S> I've used a bunch of Velleman K-8056 relay boards to control a whole load of motorised blinds in an office, which had an RS-232 port and accepted basic commands generated by a wall-mounted touch screen PC. <S> The one I used had 8 relays, which is a bit excessive for your application, but there are probably smaller alternatives or you could put one together yourself from a whole load of different options such as Arduino, Raspberry Pi etc. <S> and connect to it through serial, Ethernet, Wi-Fi or whatever you fancy trying - it depends how "dirty" you want to get your hands putting it together. <S> You could actually bypass the panel and provide a parallel connection for the door striker directly from your relay, though it won't make a lot of practical difference (although it would lose the recording of you opening the door from your PC that the software presumably provides at the moment). <S> As for alternative panels, there are loads on the market as a quick search online or browse through eBay will show.
| If you are looking for a easy to get up and running solution than an arduino would be a great choice.
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Why triodes for high power tubes The pros and cons of triodes vs screen grid tubes like pentodes and tetrodes are well documented elsewhere. When it comes to low and medium power tubes both triodes and screen grid tubes are readily available. At the really high power levels like kWs it seems to be an all triode show. Are high power say pentodes difficult to manufacture? Is it inconvenient to have more connections to the glass envelope when voltage clearances are considered? Are screen grid tubes more likely to flashover? <Q> Another consideration is that the screen grid connection (generally to a high and constant voltage) may consume 20-30% of the cathode current in a tetrode circuit. <S> In a triode, substantially all the cathode current reaches the anode. <S> I haven't done the math on high power tube amplifiers, but that seems likely to impact the power supply design, power dissipation, and overall efficiency. <A> Could be a matter of matching. <S> Pentodes have stiffer Rout. <S> To match to 50_ohm cables requires a smaller cap on plate and a larger inductor to resonate, moving energy into the large cap wired to the output cable. <A> The RCA808 power triode is designed for the grid to be driven positive (up to +300V), specified up to 210V, to get good saturation. <S> This is about the maximum saturation voltage of the tube, so it could be connected as the o/p of a darlington pair, where the grid current would come through the load & not be wasted. <S> Impedance matching wouldn't be an issue in a class C transmitter output stage, as the valve is being used as a switch and the tank circuit can be designed to transform the 1000 to 2000Vp-p down to aerial voltage. <S> Output stages are like power supplies. <S> They have a maximum load they are designed to handle, not the load that gives maximum power transfer. <S> If you whack a resistance of about 0.5 ohm across your electricity supply, you will get the maximum possible power out of it FOR A SHORT TIME.
| Triodes just may be more match-friendly.
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Can small wire gauge restrict current? powering a brushless motor The question is related to quadcopters and else.So my quadcopter (brushless dc motors and electronic speed controllers powere by 3cell lithium battery 2700mah 25C) has thin wires on the main power port, Im trying to figure out if it could affect current flow, I know that when gauge is too small it can overheat, burn or melt, but can it just restrict current flow without fires? I having hard time to figure it out because once I tried to jump start a truck with cables that were too thin and the starter motor just weren't crancking the engine, wires weren't getting hot or anything... connecting another pair of wires in parallel resulted in successful start of the engine, and it was matter of seconds, not like the battery had time to recharge. Can some one please explain this to me. <Q> Wire has resistance. <S> Thin wire has more resistance. <S> That's also why it dissipates more power than thick wire at the same current. <S> Assuming that your wires aren't getting hot enough to melt or cause other problems due to the heat, the issue is voltage drop. <S> Ohm's law tells us that the voltage across a resistor is the current thru <S> it times its resistance. <S> The wire has some finite resistance. <S> When there is current thru it, there is therefore some voltage across it. <S> When in series with a battery, this voltage across the wire subtracts from the battery voltage being delivered to the rest of the system. <S> How much this matters is up to what you consider matters. <S> In contrast, a 300 mV drop from a 6 V battery is a 5% reduction, which can matter in some cases. <A> There are two separate issues. <S> Will the wire get too hot? <S> That is purely a matter of current and the wire diameter. <S> The wires do not heat up instantly. <S> Brief overloads will not cause the wire temperature to rise much, and the wire may cool down pretty quickly after the current stops. <S> This is what happens with your starter motor in your car. <S> The second issue is, will the wire prevent power being transferred to the load? <S> This depends on current, system voltage, wire diameter, and wire length. <S> A 1V voltage drop in a 120V system is not a big deal. <S> But a 1V Voltage drop in a 3.3V system is probably catastrophic. <S> Note that wire length is equally important to wire diameter when considering power transmission. <S> Short wires will still allow power to get through, even if they are thin. <S> For all these reasons, low-voltage, high-current applications (such as quadcopters and such) often need to use surprisingly large diameter wires. <S> People also tend to underestimate how large wire should be when connecting a 12V inverter to a 12V battery. <A> So, the key thing that wire thickness changes is the resistance. <S> A thin wire has a higher resistance than a thick wire. <S> This can affect two related things, Power and Current. <S> If the resistance of a wire is higher, all else being the same, the current going through it will decrease. <S> This is what would have prevented your starter motor from turning - the resistance of the thin wire was too high, so not enough current could flow to make the motor turn. <S> At the same time, the resistance of the wire can affect the power. <S> Power, in this case, is the amount of energy per second that the wire is converting to heat. <S> This energy is being wasted as heat instead of doing whatever you want it to, like turning your quadcopter blades. <S> The equation for this lost power is \$\ P=I^2R\$, where I is the current flowing through the wire, and R is the resistance. <S> If you try to put current through the wire, a high resistance will mean a larger amount of the energy will be converted to heat. <S> If the power is too high, the wire will heat up so much that it melts. <S> It's slightly more complicated than that, as the voltage will also affect things, but what I wrote is basically correct. <A> This is similar to a water pipe, which can only fit a certain quantity of water down its cross-sectional area at any moment. <S> If you have a circuit that relies on more current flowing down the wire than it can carry, the circuit will produce a voltage drop across the cable. <S> This will dissipate power, and therefore heat, in the cable, following the law P = VI. <S> If the power is sufficient to raise the wire's temperature above its melting point, the wire will burn out. <S> At the other end of the scale, a smaller overamperage will cause the wire to restrict the current flow without noticeable effects on the wire.
| The wire has a certain resistance and also a certain capacity for carrying current, proportional to its cross-sectional area. For a 12 V battery, for example, a additional 100 mV drop is less than 1% and doesn't matter much.
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What is the "safe" amount of current I can apply to a DC motor? A motor I have ranges from 12 to 30 volts, but it says the current has to be at 0.5 amps (500ma) I am powering it from a 12.6v 20000mah battery. Basically I would to use that motor with 24 volts. There are step up voltage converters available, but the lowest current I could find with one steps it up to 3 amps, which is what I don't need. I've heard people say that the motor will only draw what it needs, but what is the limit before current starts being forced into it? I only need 500ma, not 3A. Will this damage my motor? <Q> You can only force one of current or voltage. <S> The load, which in this case is your motor, will determine the other. <S> For example, you can use a constant-current supply that drives 500 mA thru the motor. <S> You then have no control over the voltage. <S> When the motor is just starting, the voltage will be low. <S> As the motor speeds up, the voltage will go higher. <S> In your case it seems you want a supply that limits its output to 30 V or 500 mA, whichever is lower. <S> That is certainly doable. <S> It is only "forcing" one of current or voltage at a time. <S> When it's controlling the current to 500 mA, the voltage could be anywhere in the 0-30 V range. <S> When it's controlling the voltage to 30 V, the current could be anywhere in the 0-500 mA range. <A> Your specifications seem a little off. <S> Motors are usually defined with a tight voltage and a max current, which makes me wonder about the style of this motor. <S> A traditional DC motor will consume all the current it can take through it's stopped coil resistance. <S> The latter will be a low number <S> so start up currents can be large. <S> Once up to speed that current will drop to a much lower value depending on the mechanical load on the motor. <S> How FAST the motor can go for any given load is defined by the applied voltage. <S> So that leaves me wondering what exactly this motor is, and how fast you want it to run. <S> If you do not care about the speed, simply apply your voltage source and add a current limiter to stop the current from exceeding the 500mA value. <S> If you need a constant speed, you need to adjust the current anywhere up to 500mA, possibly using pulse width modulation (PWM), until that motor speed is reached and continue to control it to maintain that speed. <S> ADDITION: <S> When stalled, whatever voltage generates the 500mA current value is the max voltage you should apply to the motor. <S> Make sure the motor does not get overly warm during that testing. <S> Also verify that the motor functions as expected at that voltage. <S> Then design your driver not to exceed the numbers you measured. <A> Voltage is a measure of the ElectroMotive Force. <S> This is the potential which causes a current to flow. <S> Current is never forced. <S> It is the voltage which, by definition, is the forcing effect. <S> This is true for all circuits. <S> If you restrict the current in a circuit, a lower voltage is the result. <S> If you have a fixed voltage, the current depends on the load. <S> In this case, the more work the motor does, the higher the current. <S> The spec does not say that the current has to be 500 mA, it says the current will be 500 mA (as a result of the voltage). <S> This is clearly inaccurate - maybe it applies to full rated voltage at stall-load - maybe not.
| When you do not have a proper motor specifications, you really need to test it with a bench power supply that lets you vary and view the voltage and current that the motor requires under stalled and full load conditions.
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Mosfet Conventional current flow direction in the circuit I am working on a project to control heavy loads with Arduino up to 10 Amperes. I found the circuit which is made using p-channel Mosfet and a p-type transistor. I am confused in the flow of current through the circuit. I uploaded the diagram please see if the conventional current flow is right in the diagram? and what about the current through the red box (Gate of Mosfet) what will be IL=?.If the input current is up to 10 Amperes does it effects my arduino digital pin? Also if you have any recommendations regarding the circuit please share them. <Q> MOSFET gates are very high impedance, so no current (or almost no current) flows into them in steady-state conditions. <S> During switching on/off there is indeed current flowing to/from the gate as it charges/discharges and reaches its required Vgs level. <S> But this is only a transient condition. <S> If your load is switched only from time to time, its steady-state condition is no current flowing to/from the MOSFET gate. <S> Additional suggestions: If you plan to control inductive loads such as motors, use a flyback diode across the load terminals to avoid destroying the P-MOSFET due to inductive voltage spikes at its drain when the load is turned off. <S> Decouple the +12V supply rail with a big capacitor to avoid destroying the P-MOSFET due to inductive voltage spikes at its source when the load is turned off. <S> Due to the high currents involved, consider using an optocoupler instead of a BJT, to fully isolate the 12V circuit from the Arduino. <S> Consider using a logic-level N-MOSFET instead of a BJT for T1. <S> If you decide to keep the BJT, then add a base resistor lo limit the current into the base. <S> Also, add a pull-down resistor at the base to ensure the BJT is cut-off when the Arduino pin is at high impedance (something that can happen when the Arduino is off or when it is starting up, before the pin is configured as OUTPUT ). <A> The current through the red element is substantial only when the load current is switched on or off. <S> The current "?" exists during the state transition, because the mosfet is controlled by voltage. <S> The current is needed because there exixts a substantial internal capacitance between the gate and the drain & source. <S> Ic charges that capacitance when the loar current is turned to ON. <S> The capacitange gets discharged through R1 when the load current is turned to OFF. <S> T1 is not P-type, but NPN <S> The red element can be a wire. <S> Often a small resistor is used to damp unwanted radio frequency oscillations that are common in fast pulse circuits with no precautions. <S> If this circuit is properly realized, I1 is only a few milliamperes <S> , the major part of the Is goes to the load. <A> Those arrows make everything hard to decipher, but they look correct. <S> The gate of a MOSFET behaves pretty much like a capacitor. <S> So a current will flow into or out of the gate only when you are switching. <S> (The amount of gate charge should be found in the datasheet.) <S> The source/drain channel of a MOSFET behaves pretty much like a resistor. <S> (The resistance (R DS(on) ) should be found in the datasheet.) <S> In a MOSFET rated for large currents, this resistance typically is very low (milliohms), so it's usually ignored. <S> In other words, you can assume that the load behaves as if it were connected directly to +12V. <S> If the load is inductive (e.g., motor, relay, transformer), it can generate large voltage spikes when switched off, and you need to add a snubber to protect the rest of your circuit. <A> Since the voltages throughout the circuit weren't labeled or discussed, perhaps your question stems from a common misunderstanding. <S> This is the misconception that circuitry is based on electric current ...and that to understand circuits, we sketch in all the currents. <S> Actually, engineers and scientists view most circuits as voltage-controlled systems. <S> Everything is powered by constant-voltage supplies, and the signalling is voltage-based. <S> To understand a circuit, we sketch in all the voltages. <S> Then, using Ohm's law, we can determine the currents if necessary (or even ignore them entirely, and instead concentrate on input/output voltages and load wattage.) <S> For a nice animated view of voltages (and currents) inside circuits, try the little simulator at Falstad's site. <S> (a java applet) <S> For example, the current in the gate-wire of the PMOS remains zero, whether the transistor is on or off. <S> MOS transistors are voltage-driven devices, and their gate-current is usually irrelevant. <S> To analyze this circuit, notice that transistor T1 and resistor R1 form a voltage-divider between 12V and 0V. <S> When T1 is on, T1 forms a short circuit to ground, and it pulls down the PMOS gate to zero volts. <S> When T1 is off, it acts like an open circuit, and then R1 pulls the PMOS gate up to 12V. <S> In other words, T1 and R1 have converted the Arduino's small output voltage into a 12V signal. <S> This 12V signal then drives the PMOS transistor gate. <S> The PMOS transistor is wired as an inverter: when the voltage on the PMOS gate is zero, that transistor turns fully on, and when the voltage is 12V, it turns off. <S> (Yes, it should handle 10amps just fine. <S> If its on-resistance is low enough, it might not even need any heat-sink.) <S> Also, note that you'll need a resistor in series with the base of transistor T1. <S> The transistor's input acts as a diode to ground, and this diode would short out your Arduino's output pin. <S> (LEDs need a current-limiting resistor, and so does this transistor's base lead.) <S> The added resistor should be around 10x larger than the value of R1 (so if R1 is 10K, then add a 100K resistor to the connection between the Arduino and T1.)
| Your diagram is right (but difficult to read due the blocky arrows).
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Why do parasitic elements (e.g capacitance) appear on only high frequencies? We know that the impedance of a capacitor is inversely proportional to the frequency then why is it not negligible when at higher frequencies ? <Q> Shunt stray capacitances are problematic when their impedance is too low, as opposite to being too high, because they pull down signal lines and/or mismatches them. <S> When the stray capacitance value is much lower than the frequency, then the resulting impedance may be high enough so as not to heavily impair a line. <S> Everything might be reasonably under control. <S> However, as frequency goes up this may cease to be true, and then the resulting impedance may start to be low enough so as to pull down the line. <A> the assumption <S> Why do parasitic elements (e.g capacitance) appear on only high frequencies? <S> in question is false. <S> It appears at any frequency but impedance lowers with rising frequency. <S> Zc=1/2piCf <S> Thus 1pF at 60Hz is pretty high but compared to 10M resistor can be used to detect 50/60 Hz stray signals for a hand wave proximity power light switch to couple stray Efield into a comparator and toggle or click an up/down counter to dim up down with simple logic to control phase on a triac. <S> been there, done that 1970. <S> added 2nd question... <S> why is it not negligible at high frequency <S> If we use caps for coupling then we assume short circuit impedance. <S> But you don't want stray insulation to attenuate, cause crosstalk or reflections on signals, so stray pF is important. <A> If the capacitor is considered to be JUST a capacitor, then its impedance does approach zero as frequency increases. <S> BUT The key is in your question title - specifically, "parasitic". <S> You cannot consider a capacitor to be simply a capacitor. <S> Simply by having physical length it also possesses inductance. <S> So a simple model looks like simulate this circuit – Schematic created using CircuitLab <S> For non-RF components and low frequencies, L is negligeable. <S> As frequency rises the drop in the impedance in C greatly outweighs any contribution from L, but the impedance of L does rise. <S> At some frequency, called self-resonance, the impedance of L equals the impedance of C, and this forms a minimum overall impedance. <S> For greater frequencies the increase in impedance of L is greater than the decrease in impedance in C, so the total impedance rises with frequency. <S> And all of this, of course, ignores resistive effects. <A> One way to look at it: A very low impedance is a short circuit. <S> Short circuits can have a dramatic impact on circuit performance. <S> Another way to look at it: <S> Many times, a parasitic capacitor is placed so that you should be worried about its admittance, not its impedance. <A> You also need to consider the parasitic inductances: pcb traces(1nH/mm), pcb vias (~1nH, depending of height/width ratio), pcb planes (10% of what a trace would be), bond wires in the ICs (0.5nH each), leadframes (go with 1nH/mm), and the silicon inductances. <S> Assume every pin of every IC has to work in a 10nanoHenry environment, until proven to be smaller. <S> For fun and games, consider the parasitic onchip VDD---GND capacitance, often 100pF or 1,000pF or larger. <S> That resonates with the pin inductance. <S> Perhaps the bulk silicon resistances and well_contacts and substrate_contacts and metal resistance will adequately dampen: Rd = <S> sqrt(L/C) is good trial value. <S> Note 100pF and 10nH resonate at 50MHz. <S> Lots of small MCUs, with one set of VDD/GDD, will ring at that frequency. <S> Is 50MHz high frequency? <S> The parasitic inductance of a bypass capacitor --- labeled ESL --- prevents ideal filtering action. <S> 500uF and 20nH (20milliMeter size cap) <S> resonate at 50.3KHz. <S> Not a high frequency, eh?
| Parasitic capacitance is always frequency dependant impedance.
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Is 19V safe for charge Ni-Cd batteries? I have 5 Ni-Cd battery(1.2V 2200mAh). They connected in series. I have a 19 V 3.42 A adapter. Will it make a problem when I charge Ni-Cd batteries with my 19 V adapter? Is 19 V dangerous for charge Ni-Cd batteries? <Q> 19V would be too much for your NiCd. <S> I don't recommend doing this. <S> You can typically charge NiCd at about 1 C/hr; <S> in your case, 2.2 A max. <S> This is acutely dangerous. <S> You don't want fire mixed with cadmium. <S> Bad combination. <S> Get a charge controller. <S> Or simply, a charger. <A> If you were to fill a water ballon from a faucet, you wouldn't turn the faucet on all the way. <S> You also would want to stop when the balloon reached a certain size. <S> And you would not turn the water on and walk away. <S> These essential controls are the same needed for charging batteries. <S> You need to control how fast they charge, and know when to stop. <S> So, there is a piece missing from your proposal--a charge controller. <S> This is a good question, in the sense that everyone who gets into electronic design needs to understand what they are attempting before they start asking about how to do it. <A> You can do it, but you need to be very careful. <S> To start, the standard NiCd charge technique (for a fully discharged battery) is 16 hours at a 0.1 C rate. <S> C is the battery capacity. <S> So, for instance, if you have 2200 mAhr cells, you can charge at 220 mA. <S> The trick is knowing when to terminate charge. <S> You can try this. <S> Take your batteries and discharge them completely. <S> Discharge each cell seperately, not in series. <S> Now connect them in series. <S> Find the resistance needed to get the proper charge current. <S> Now start charging. <S> Wait 16 hours. <S> Measure the battery voltage and record it. <S> Now you can charge your batteries even if they are not fully discharged. <S> Just stop when the voltage reaches the cutoff voltage which you recorded.
| However, be aware that you'll need to check the battery voltage at least once per hour. With 19V-6V overvoltage, you'll potentially fry the batteries pretty quickly.
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