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What is the advantage of a tri-state output? I'm trying to understand what is the use of tri-state output and what is the advantage of using it. Below depicts both two-state and tri-state output: According to the above logic, it seems to me two-state output can only be ON or OFF.But in tri-state case, the output can be ON, OFF or floating. So to me the only difference is the floating state. As far as I read, floating signals are not useful. But it seems I'm ignorant at some points and want to ask what is the purpose of using tri-state output? <Q> Tri-state is essentially 3 different states as you have seen. <S> Instead of calling it ON and OFF, think of it as IN and OUT instead. <S> The third state is not floating, it is in a high impedance state, which is essentially disconnecting it from the circuit. <S> This means your 3 states are: Input Output High Impedance (disconnected) <S> This is very useful in circuits that have a common bus for multiple components. <S> With tri-state logic, components can either read from the bus, write to the bus, or be essentially disconnected so that it does not affect the bus, and will not be affected by the state of the bus either. <S> As an example of using tri-state logic is on a computer bus: <S> (Picture taken from http://www.lisbdnet.com/system-bus-functions-features/ ) <S> As you can see, multiple units are sharing the same busses. <S> You don't want the information on the bus to be read by every unit at the same time. <S> You also dont want them writing to the bus at the same time. <S> In this way, data can be sent back and forth and only the units that require the information will receive it. <S> Imagine, with the above diagram, you receive an input to access a memory address, which then needs to be sent to a register, then go through the ALU, and finally the data is sent to the output. <S> This can be done easily by changing the between the 3 states, so only the units that are needed will be connected to the bus at any time. <A> Suppose that instead of just one switch you have many switches connected to the same tristate wire. <S> That single wire can be used by all of these different switches to communicate information, as long as only one switch at a time uses the wire by connecting to '1' or '0'. <S> This kind of situation occurs in practice when you want to connect a processor to different memory <S> and I/O devices. <S> The processor only talks to one of them at a time, so the wires used for communication (the data bus) are shared. <S> Of course, the processor must have some mechanism for specifying which device is allowed to use the wires at any point in time... <S> that is usually done with a separate set of wires (the address bus and a handful of control signals) that is one-way from the processor to the devices. <A> Before TriState logic/bus_drivers appeared, logic systems used multiplexors (ORs of ANDs)or wired_ORs. <S> The wired_ORs depended on resistive pullups, and were slow. <S> National Semi provided additional on_silicon transistors (using all NPN, the fastest cheapest silicon of that time) to disable <S> both low_side and high_side driver of the TTL gate output circuits.
| With Tri-state logic, each individual unit can either read from the bus, write to the bus or just be disconnected. By using just one wire for all of the switches, rather than a separate wire from every switch to every other switch, we can greatly reduce the amount of physical wiring needed.
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How to measure the actual length of a long twisted pair? Sometimes we might need to know the exact length of the wire pairs inside a cable. One can be interested in calculating voltage drop or some other calculation regarding the exact length. Because the wires inside the cable are twisted, their actual(electrical) length is greater than the cable jacket length. Is there any practical method to calculate the actual length? I couldn't find a duplicate, but if there is I will delete this question. <Q> Of course you need to know the resistance of the cable. <S> The better and very precise way is to use a time-domain reflectometer (TDR). <S> This device sends an impulse into the cable which is reflected at the (open) end. <S> The time of the reflected signal is measured and because of constant wave propagation, the length of the cable is calculated. <A> From a purely geometric perspective you could calculate the length using the helical length equation. <S> Where H is the length of the twisted wire and <S> Where R equals the radius of the turns in the wire. <S> Basically from the center of the twisted pair assembly to the center of one of the wires. <S> So if the wire makes a complete rotation around the center in 10mm, and distance between the center of the twisted pair and the center of a wire is 1mm, then were you to untwist the wire and straighten it the length would be <A> Cut off a 1 meter piece, remove and straighten one of its conductors, and measure the actual length per meter. <S> Update: <S> Straighten out one of the conductors and measure it. <S> Say it's 1.05m long, or 5% <S> (made-up number - I've no idea whether it's realistic) longer than the cable it came out of. <S> Apply that extra 5% to the length of your cable to get the length of the conductors inside.
| One easy and obvious way is to connect the cable at the end and measure the loop resistance with an ohmmeter.
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Why have a smaller fuse on input & larger on output on a power supply This appears to be a common practice in automation control panels - for example - on the power supply it has a 10A fuse on 120VAC input side & a 20A fuse 24VDC on the output. That power supply ratings are: 100-240VAC~5.6A in & 24VDC~20A out. I am just tying to understand the reasoning for this. <Q> Fuses work on \$i^2t\$ heating. <S> It's the current which causes the fuse to melt. <S> If input requires less current and output gives more current, you will have to use a smaller current rated fuse at the input and a larger current rated fuse on the output. <S> This is the usual case with step-down or buck converters. <S> However, if you are working with a step-up or boost converter, these will draw higher current on input and deliver lesser current (with higher voltage) on the output. <A> Fuses are used to protect the wiring infrastructure. <S> So, if your incoming 120 VAC cabling is rated continuously at higher than 10 amps then using a 10 amp fuse makes sense. <S> By the same token if the cables used on the outgoing supply (24 volts) are rated greater than 20 amps continuously then use a 20 amp fuse if the output line can drive more than 20 amps. <S> Fuses prevent fires; they don't protect electronics. <A> Power is V*I, voltage multiplied with current. <S> At the input side you have a high voltage, so you only need a small current. <S> At the output side the voltage is reduced, meaning you need higher current to get the same power. <S> With the numbers from your supply: At the rated voltages the output can deliver a power of 24V * 20A = 480 W before the fuse may blow. <S> For the input fuse this is dependent on the supply voltage. <S> With 110V it will blow only when the power supply consumes 110V <S> * 10A = 1.1kW.
| In this case, you will find a larger current rated fuse on input and smaller current rated fuse on the output.
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Why the power supplies limit the amperage? I'm just learning electronics, and I can't understand something:If the amperes don't matter in terms of burning the circuit, only voltage, why do phone charges for example have a low current capability? I understood from the various explanations I've read, I've read that its ok to use a power supply with higher amper rating, but not with different voltage. If the amper rating is the maximum that device can supply, it only consumes what it needs, right? <Q> Three things: <S> Yes, for a given load that expects a constant voltage, you can attach a supply with a higher current capability* but the same voltage output, and everything works fine. <S> The supply can only give so much power (i.e., only so much current times voltage). <S> So if I try to draw one amp from a wall-wart that's only good for a quarter of an amp, then the wall-wart will either quietly shut down, or it'll burn up. <S> So the rule for most supply and load combinations is that you need to select a supply voltage that is within the range that the load can accept (or exactly on, by preference), and you need to select a supply current capability <S> that is the same as, or greater than, the load's rated current draw. <S> * With modern wall-warts and supplies. <S> Confusingly, older wall-warts would actually supply higher voltage if you asked less current of them. <A> If by "Burning the circuit" you mean "heating up the circuit", the heating is actually directly related to the current. <S> More current = more heat. <S> The current is determined by the device (the "load"), not the voltage source. <S> If by "burning" you mean "powering", that is running the device as it's supposed to, the power supply needs to be able to provide equal or more current than the device wants to consume. <S> A device such as a mobile phone can only consume 1-2 A of current, hence a bigger charger with higher current rating would be unnecessarily costly. <S> Achieving faster battery charging speed comes with a number of problems, such as how much current the battery can accept and how much current the cables safely can conduct. <A> Phone batteries can't charge at infinite current. <S> Cables and connectors can't handle infinite current either. <S> Therefore you only need about 1-2 amp charger for your phone. <S> If it tries to draw much more than that it may be broken and charger shuts down <S> so the phone or cable does not burn. <S> Phone will take only the current it needs, even if the charger can provide more, like for charging two phones at the same time. <A> One of the first things to understand is that there are constant voltage supplies, and constant current supplies. <S> As it turns out, most supplies are constant voltage supplies. <S> Therefore, you don't want the supply to exceed the voltage rating of your load. <S> For constant current supplies, it is the opposite. <S> The current is what the supply actually put outs, and the voltage is the capability of the supply. <S> The output voltage just ends up being whatever it needs (within the limits of the supply) to push that current through the load. <S> In this case, you do not want the current rating to exceed load's current rating.
| For constant voltage supplies, the voltage rating is actually what the supply put outs, while the current rating is the maximum capability. Current does matter in burning up the circuit.
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Why in the world would the tab on a TO-220 package have a connection to anything (but GND, if applicable)? I know this is in some way a different version of the prior question In i.e. a TO-220 package, why is mounting the (pin connected) tab to case safe? But I am not looking for ideas what to do to mount a TO-220 (in my case an SCR that has the anode connected to the tab. That's the input with the highest potential from ground!) What I am asking is why in the world would they have designed it that way? These things even have a hole to screw it to a heat sink and you'd want the best contact to the heat sink, not layers of plastic in between. Now I am certainly not the first one discovering that this is rather annoying. So why did the engineers so unanimously design things that way? Why not just leave it disconnected? <Q> Why not just leave it disconnected? <S> Because of the way the IC is made. <S> It starts off as a (relatively) large piece of doped silicon, and the features are deposited / etched / implanted into one side. <S> That leaves the back side of the die electrically connected to one node of the circuit. <S> Which node it is depends on what circuit is being made. <S> Generally for things like ICs, it's the most negative terminal. <S> It is whatever makes the circuit the most straightforward to design. <S> For a power device, the designer also wants the lowest thermal resistance from the active region to the case. <S> The most critical part of this path is the narrowest directly under the die, which is why a direct connection is used. <S> It's far better if an insulating layer is needed to put it between tab and heatsink, where the area available for heat conduction is orders of magnitude larger than the area under the die. <S> And remember not all applications will need this insulator, some can use an isolated heatsink. <S> The bottom line is cost/performance ratio. <S> An insulator in the package would increase cost and decrease thermal performance. <S> There are niche applications that would pay that premium, and there are some isolated tab power electronics out there to prove that if you look for them. <S> But most equipment manufacturers will do the extra bit of mechanical design to get the best overall cost and performance. <A> (in my case an SCR that has the anode connected to the tab. <S> That's the input with the highest potential from ground!) <S> What I am asking is why in the world would they have designed it that way? <S> Most power transistors (bipolar and FET) and some regulators are constructed in a similar manner, where the collector or drain is connected to the tab. <S> Unfortunately that means the heat sink will usually be above ground, unless an insulated mounting kit is used. <S> Construction of a thyristor (SCR) <S> It is possible to insulate the die from the tab, but this increases thermal resistance which makes it more difficult to cool. <S> Another advantage of having a direct connection is that it can be used as a terminal as well as a heat sink. <S> Some packages (for example, DPAK ) which are designed to have the tab soldered to the PCB, often don't have a separate lead for the collector or drain. <A> A transistor or any other semiconductor device is made from silicon. <S> Initially you get a slice of silicon. <S> Then you induce some N and P zones on one face of this slice. <S> There you will connect bondings (welded tiny wires), that connect the silicon chip to leads. <S> The opposite face of this slice you will need to solder to the case base - the tab in case of TO-220 or "exposed pad" of some SMD chips. <S> This solder holds the chip in place and guarantees the lowest possible thermal resistance. <S> There is no way to insert some insulation between the chip and the tab - the chip area is too small and when you will try to insert any insulation layer, you will dramatically increase the thermal resistance "junction-case". <S> In fact, you always want to insert insulation to connections with the maximal contact area. <S> Thermal resistance is roughly proportional to H/S where H is the thickness of the insulating layer and S is the contact area. <A> I am not looking for ideas what to do to mount a TO-220 <S> that has the anode connected to the tab <S> There are things like heatsink insulator pads for cases like yours. <S> Having the insulation as an optional external component enables you to have a design with direct connection between the IC and the heatsink and keeping the heatsink insulated, if you need to minimize the thermal resistance.
| Of course there is a penalty in thermal resistance compared to a direct connection to the heatsink, but having a layer of insulation inside the IC would result in a similar penalty as well. Because the anode is at the bottom of the die, which is bonded to the tab. But, on the other hand, it creates parasitic electrical connection to the chip's substrate.
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Meanwell LDH-45 LEDs Glowing When Controller is Off I'm using a Meanwell LDH-45B-350 Dali driver for currently 4 Bridgelux BXRE-30G0800-D-73 LEDs with forward voltage of 18v at 350mA. The problem is when I switch off the driver, the LEDs still glow a little bit. I saw in the datasheet today this paragraph in really bad English: At dim off,LDH output voltage will drop to the same level as input voltage.To get luminaires complete dark, A please make luminaires are light off when they are driving by the input voltage. The driver is fed by 24V, does this mean that with these LEDs there is no solution for this problem? I got the B version and 24v supply because it can drive my 7 LEDs together. The A version can only drive 4. <Q> The meaning of that statement is pretty clear, the minimum output voltage is the input voltage. <S> So to get less minimum brightness you can either use more LEDs in serie s or lower <S> the input voltage (or both). <S> The topology probably looks like this (from Wikipedia commons): <S> For minimum output, the switch is always open so the inductor and diode are in series to the output. <S> It's possible the diode is actually another switch. <A> You could throw in a zener to take up the additional voltage, but it should be as low leakage as you can get. <S> You'd be surprised how little current it takes to visibly light some LEDs. <A> This seems solved by replacing the cheap temporary chinese 24vdc power supply. <S> I now have a meanwell sdr-120-24 powering the LDH module and all led's are completely off!
| Too few LEDs and they might be damaged.
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What is the name of this current called in this regulator datasheet? Which parameter in this data sheet is the default current flowing from the input to the ground regardless of the load current? In my case an 12V input will be regulated to 5V by the TL780-05. So to estimate the total power dissipation I will use the following formula: Preg = (I_n × 12) + (I_load × (12-5)) My problem is I don't know how they name I_n in the regulator datasheets, so I got stuck. Where in the datasheet I can find I_n? And is that always named same way? <Q> This is the one: So typically 5 mA, <S> maximum 8 mA <S> How to quickly find that: I know it's a current <S> so I look for any parameter specified in A (or mA, uA) in the UNIT column. <S> Then I look at each and seeif that could be the one. <S> There will be a small increase with increasing load current, see the line under the red rectangle. <A> A more common term is "Quiescent Current". <A> It's called input bias current and is on page 3 of the data sheet you linked. <A> In most cases you can ignore this bias current as it is small enough. <S> What you should be concerned about is voltage drop on regulator multiplied by load current. <S> For example if you have 12V input, 5V output and expected load current 3A - the power dissipation on regulator <S> equial P = <S> (12-5)*3 = 7 <S> *3 <S> = 21W Load will take P = <S> 5V*3A <S> = <S> 15W (41.2%) with power loss of 21W (58.3%) <S> -- the circuit is very inefficient, produces a lot of heat and requires bulky radiator. <S> If your circuit is not critical to some power 'noise' <S> then solution based on LM2596 , LM2576 , MP1484 , MP1584 , <S> MP2307 is more desirable due higher efficiency. <S> Of course you are not limited by these ICs and can use any other DC-DC converter. <S> MPS , TI , ST RENESAS , ON , INFINEON DIGIKEY , MOUSER , 10 Best Electronics Suppliers
| In that particular data sheet, they seem to be calling it "Input Bias Current".
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Choosing a crystal for a long-running synchronizing application between two microcontrollers I learned from research that crystals (HC49/U style casing) experience drift which causes timing to be inaccurate in the long haul. My project is a laser tag system in which one microcontroller (AT89C4051) in the target continuously searches for specific laser patterns from a gun which also uses the same microcontroller. Both use the exact same crystal (22.1184 Mhz) from the exact same manufacturer and the circuit board traces on both boards from microcontroller to crystal leads are the same. The capacitors between the crystal leads and ground are all 33 pF NPO (5%) from the same manufacturer. Now if I write the software perfectly, start everything at the same time and run everything for about a minute or so, everything will fall in sync. I'm also running another unit that would cause everything to try to be in sync periodically (by the master sending sync bytes). But what about drift? Let's say I power each unit with a fully charged 6800 mAh battery for 10 hours. (Ok, I understand a real laser tag game would not last 10 hours, but I'm going for worst cases here) . Then one battery decides to die 1 hour later and I replace it. Since the crystal is almost 24 Mhz, the microprocessor would execute instructions about one every 540 nS. So if I had the micro running for say 10 hours, would instructions still run at 540 nS at that point? or would the time alter on every minute due to drift? Like, for minute one, would each instruction take 540 nS to execute then minute two 538 nS then minute three 541 nS? I'm just trying to understand the best crystal parameters regarding drift to choose for my application so the data doesn't get out of sync (just because of possible drift resulting from continuous operation). I mean I seen specs like frequency stability and tolerance measured in ppm when I looked at crystals on Digikey, but what are the best numbers to aim for and how do I convert those numbers to the amount of time I can have guaranteed instruction stability before major drift affects the operation of the synchronization? <Q> Garden-variety crystals only have an initial accuracy on the order of 100 ppm, before you even start to consider thermal and voltage effects. <S> You can spend more money and push things down to on the order of 10 ppm, or you can spend a LOT more money for temperature compensated or even ovenized crystals to do even better than that. <S> Two 22.1184 MHz crystals whose frequencies differ by exactly 100 ppm will have a phase difference of 2211.84 whole cycles after just one second. <S> Crystals 1 ppm apart will have drifted by 22.1184 cycles in that period. <S> So no, you can't rely on "blind" synchronization for any meaningful length of time. <S> You need to design your system so that the subsystems remain in sync despite the inaccuracies of their individual clocks. <S> Basically, they need to communicate with each other in some way on a continual basis. <S> Of course, it isn't at all clear why a laser tag system would require that kind of global synchronization in the first place. <S> Sounds like some sort of X-Y problem. <S> What problem does that sort of approach solve for you? <A> Any system that relies on instruction to instruction synchronization of two units is doomed to fail. <S> It is not needed, and nobody ever does it. <S> Using the same nominal frequency crystal will get you within +/- <S> 100ppm as built, which is completely adequate to get bit rates correct, so they can talk to each other. <S> +/- <S> 10ppm can be got with extra work and cost, but there is really little point for the system you want to run. <S> Once communicating, they should sync their activity to the exchange of messages. <A> A reasonably good crystal might have a spec of 10ppm <S> +/-15ppm/ <S> C. <S> In this example we are ignoring the temperature effects on the load capacitors and driving circuit that pull the crystal's frequency, and aging effects where it's frequency changes over it's lifespan. <S> You want to use 22.1184MHz, so 1ppm is ~22Hz, 10ppm is ~220Hz. <S> We can assume your crystals are not at exactly the same temperature, perhaps <2C between them. <S> So that's another 30ppm or 660Hz, ~880Hz total. <S> So, worst case, in one second, one crystal may have drifted from the other by about 800 ticks. <S> That is 800 / 22118400 or about 3ms. <S> After a day that would be 22 seconds. <S> Even the best commercialy available crystals that employ temperature compensation and sit in little ovens only manage in the order of ~0.01ppm <S> and they are expensive. <S> Any scheme where you have more than one timing reference and the timing relationship matters you need a syncronisation mechanism. <S> That could be some sort of locking mechanism (e.g. a PLL and VCXO). <S> Or external time reference like GPS. <S> Ok, so how to fix it in your case? <S> You have to ask yourself why do you think you need a precise shared time reference?
| The conclusion is that you can't rely on a crystal to keep good time autonimously.
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How does energy dissipates in a purely reactive circuit? Suppose in a home, power factor drops to zero. It means that the circuit becomes purely inductive. If the energy meter is connected at incoming power supply point,will the energy meter still keep on increasing? If yes, then where is this energy being consumed? We know that reactive power keeps on oscillating between source and load, so in this case, how does the reactive power fits into picture? <Q> Then, you wake up. <S> In the "real world" there are NO genuinely reactive components. <S> Inductors can resistance (and some capacitance). <S> Capacitors have ESR (resistance) and some inductance. <S> Wires have the same. <S> The energy meter shunt is often resistive, and even if it uses a Hall sensor it will not be absolutely lossless. <S> In the very very "best" case, it may take a wee while for things to die down. <S> But never anything like indefinitely. <S> You can work out the order of time for yourself by looking at eg published capacitor ESR values and inductor resistance, and wiring resistance. <S> These set a lower level on losses before you start to look at the other (usually) second order components. <A> Please note there is a difference between power factor (PF) and displacement power factor (DPF) <S> PF is associated with the harmonic content while DPF is associated with the fundamental DPF = <S> cos( \$\Theta\$ ) <S> PF = <S> \$\frac{DPF}{\sqrt{1+THD^2}}\$ <S> why is this important? <S> Because there are two scenarios where the real power can be zero. <S> 1) <S> Purely reactive load causing 90° phase shift between voltage and current 2) <S> Purely harmonic current draw, causing no current draw at the fundemental. <S> colloquially, PF is used interchangeably with DPF more freely than it should be. <S> In the scenario you are considering, the PF would equally the DPF as the THD = 0 <S> What would happen? <S> Well assuming you could acquire an ideal inductor and you had ideal cabling to your house, you would draw current equal to <S> \$ \frac{V}{j\omega L} \$ <S> while drawing no real power. <S> This current could potentially be quite high. <S> Domestic billing is measured and charged in WATTS so you would theoretically not be charged anything. <S> The grid and the generation however still need to source this current and this is where the problems are. <S> Industry are charged in VA so every harmonic or DPF they draw, they must pay for since the generation and distribution must manage this. <S> If you or an industry complex did draw excessive DPF or PF there would eventually be a representative from the local electrical board to investigate (how quickly depends on how much disturbances since this impacts the generation, the distribution and everyone at the PCC) <S> In practice you cannot have an ideal inductor and thus there would be real power being pulled to feed the resistive element of the inductor and the cableing while also drawing (potentially) excessive current and DPF <A> Not only do all (non-superconducting) reactive devices have a non zero resistance, but also, any circuit of finite size (larger than a single electron) has a loop area, and will radiate EM energy. <S> Both will heat up the universe, conserving all the added energy. <S> Or something will melt. <A> In a reactive, oscillating circuit energy is dissipated by: a) <S> Resistance b) <S> Electromagnetic radiation c) <S> Gravitational radiation from the accelerating charges (negligible) d) <S> Changes in dimensions of components resulting in mechanical heat losses and sound
| If there is GENUINELY ZERO resistive energy then energy will not dissipate, an ideal energy meter will not increment and the ideal reactive components will maintain the situation indefinitely.
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How is a series resistor limiting the voltage for a diode? I try to understand some basic concepts, but I'm stuck with this question.I have this schematic, with a power supply 9V, a resistor 450ohm and a LED.Assuming the LED consum 16mA, this resistor have an voltage drop of: 7.2VMy questions is why this resistor "limit" the voltage with this voltage drop? I learn that resistors limit the current, not the voltage, so my power supply can deliver 1A, only 16ma would reach to the LED, so why I have a LED power by 9-7.2v = 1.8V and 16mA, and not power by 9V and 16mA? <Q> Source: <S> Tutorialspoint.com <S> As you can see in the diagram, the voltage of a diode is kind of "fixed" around the specified forward voltage (around 0.7V for a normal diode, <S> 1.8V for a red LED). <S> If the voltage over the diode is increased further, the current is increasing rapidly . <S> Without a series resistor this is exactly what would happen - the current increases and the diode burns up (if the power supply delivers more current than the diode can handle). <S> But with a series resistor there is a kind of "negative feedback": If the voltage over the diode would increase, the current would increase a lot . <S> This higher current would create a higher voltage drop in the series resistor (U = I*R). <S> And a higher voltage drop in the resistor means less voltage for the diode. <S> Thus the resistor is effectivly limiting the current through the diode and only by that it is also fixing the voltage. <A> My answer is simple: the resistor limits the current and diode limits the voltage . <S> You can consider this network as a simple current source (a voltage source + resistor) loaded by a voltage stabilizer (the diode). <S> This is a relatively good combination. <S> The best combination would be a perfect current source driving a voltage stabilizer . <S> The worst combinations would be a perfect current source driving a current stabilizer and perfect voltage source driving a voltage stabilizer . <S> It is easy to explain how the linear resistor limits the current. <S> Just apply Ohm's law to the rest of the voltage across the resistor <S> I = <S> (V - VLED)/R. <S> But it is much harder to explain how the diode limits the voltage across itself. <S> Perhaps for this reason there are no intuitive explanations for this phenomenon. <S> I will try to explain it by using the concept of dynamic resistance . <S> In the (almost) horizontal part of the IV curve, think of the diode as an ohmic resistor with high and (relatively) constant resistance. <S> So, when you increase the voltage across the overall network, the voltage across diode increases proportionally to the current through it <S> - V = <S> I.R (see the top of the picture below). <S> But when you enter the vertical part of the IV curve, supprisingly the diode begins changing its resistance - you increase the current but it decreases its resistance with the same rate. <S> As a result, the product of them - the voltage across diode, stays constant (see the bottom of the picture above). <S> I have shared this explanation in the RG discussion of How many types of resistance can be represented by a variable resistor? <S> You can see there a graphical representation of another version of this clever trick where the overall voltage stays constant... the limiting resistor decreases its resistance RL... but the "diode" decreases its static resistance RST as well. <S> As a result, the two IV curves rotate and their intersection (operating) point moves along the new IV curve of the "diode". <S> It is a vertical line in this region; so the voltage drop across the "diode" stays constant. <A> consider this simulate this circuit – <S> Schematic created using CircuitLab <S> For normal silicon diodes, between a picoamp and many milliAmps, the current increases 10:1 for every 0.058 volts increase in diode voltage. <S> Thus the current is exponential function of the voltage, and the voltage is a logarithmic function of the current. <S> Thus there is no threshold voltage. <A> You have to sacrifice an amount of voltage to activate/open LEDs. <S> That voltage is known as forward voltage Vf. <S> In your case, Vf of the used LED is around 1.8 V, so it sucks 1.8 Volts and after that the LED is almost a short circuit. <S> Remaining 7.2 volts drops on the resistor. <A> Subtract LED volt drop from supply voltage. <S> Now, connect your resistor to the led and do the math. <S> If LED drop is 1.8v, 7.2V / 450 Ohms is 16.66mA.
| The resistor is not limiting the voltage for the diode, the voltage on a diode can be considered as constant:
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circuit for hot-pluggable device I have a scenario, where device A provides 24V, device B converts 24V to 3.3V and feeds it back to device A . Both devices communicate with TTL (@3.3V), so we have 2 signal lines as well. What circuit is required so that I can safely plug device B in or out while device a is powered and so that I'm not limited to connector types that are hot-plug-safe .Also there is no requirement to limit inrush currents, because all inputs have low capacitances and the 24V rail may drop. I know there are ICs for hot swap circuits, but they are quite overpowered and I prefer a circuit consisting only of passive components. Does something like the following circuits make any sense (using diodes like for esd protection)? From my understanding, the act of connecting two devices is simmilar to an electrostatic discharge in some way. Version 1: Version 2: Edit: as desired, the DCDC-circuit: Block diagram of TPS54240 : <Q> The single most-important feature that your connector must have is that the ground pin makes connection first and breaks last. <S> Sketch up your circuit and connectors and observe what happens if both of your power supply connections are made but the ground connection is open. <S> There is a very good chance that both your low-voltage rail AND the signal connections have 24V presented to those nodes. <A> To analyze what happens when you connect two powered circuit, without any special connector, try to imagine that you connect each wire individually in every possible order. <S> Usually the worst case scenario is when you connect GND last. <S> For example: assume that you connect the 24V first and then the 3.3V and try to imagine which path the currect would flow. <S> What would happen inside the DCDC when it does not have a GND connection? <S> Will it isolate the 24V from the 3.3V, or will it allow some current to flow through? <S> What will happen to this current in device A? <S> If it increases the 3.3V, will device A tolerate it? <S> The diodes will not do anything. <S> If the corresponding wire has already been connected, it is reverse biased, so it is closed. <S> And if the corresponding wire has not been connected yet, then the diode is not connected either, so it does nothing. <S> Please copy the schematics of the DCDC here, and I will try to help more. <A> The most common necessary part and <S> that is the hot-insertion type connector which connects ground first. <S> At the same time you must provide ESD transient protection. <S> The requirements to avoid SCR-latchup are that the signal lines must never go outside the power rails by more than 0.2V. <S> I doubt there will be any diode solution for this as transient currents can easily exceed 0.2V unless you shunt every I/O port until DC is OK. <S> That would require a DC OK signal and some handshaking to activate Reed Relays or solid-state switches to go from "protected disconnect" to connected mode.
| I would suggest you add 10k resistors in series to limit the current so that the diode protection you have on CMOS inputs can work better with power rails not connected.
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Getting a USB PD 3.0 port to output a constant voltage I am trying to use this AC-DC wall adaptor as a simple constant voltage source to power a compact adjustable DC-DC converter. www.amazon.com/Charger-RAVPower-Adapter-Compatible-MacBook/dp/B07PLR7T1M I want the high power density that these new GaN transistor converters have, but the only ones I've found have only a USB-like output rather than a simple constant voltage that I want to power my DC-DC converter. Is there a way to make this AC adaptor output just a constant 20V? I have tried to research how the PD 3.0 system decides what voltage and current limit to supply, but I have not understood an explanation for what I want since it is not what the USB PD system was designed for. I am thinking the solution would involve connecting certain value resistors to certain pins in the connector to signal the adaptor to output a constant voltage. Can any USB experts help me? <Q> From some searching it seems the STUSB4500 should fill your requirements. <S> It supports 20V mode and once the non-volatile memory is programmed it can work without support from a host processor. <A> USB C power delivery or quick charge technology is a bilateral communication protocol that is used to let a device request a specific voltage and current output. <S> It is much more complicated than the older standards which were just a signalling voltage or resistor voltage divider on the usb A data pins. <S> In order to allow a typically 5V device to exist on the same port as a 20V device, a complicated protocol was deemed to be the best chance of not frying the low voltage device. <S> In order to use it as you want, a microcontroller typically is needed to either directly implement the handshake or a middleman dedicated USB PD ic is used. <S> For a diy project, there are plenty of PD or QC trigger devices that will allow you to dumb down the communication to a simple jumper type selection. <S> Look for these on the usual import friendly sites. <A> That appears to be a software configuration with Quick Charge 4 compliant protocol with 20 mV increments to 20V. ref <S> https://en.wikipedia.org/wiki/Quick_Charge <S> Perhaps you may prefer the more reliable Laptop charger supplies that provide 19.5V on a 3.5mm connector.
| To get 20V out of a USB C power supply you need a USB power delivery controller designed to work with USB C (there was a version of USB power delivery for A/B connectors but afaict it never took off).
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What benefit are opto-couplers for relay switching providing, if both sides share the same ground? I wanted to connect some relays to my Raspberry Pi (for 12V, not 230V) and I found the waveshare RPi Relay Board , which has the following schematic: I like that they spend some thoughts on filtering the 5V supply, but what is the point of the opto-couplers? According to the schematic it appears like they won't provide any isolation. <Q> I think it's redundant. <S> That said, I have heard from someone of experience on SE that although relays provide galvanic isolation, they can't be relied upon to prevent arcs from jumping from the primary contacts to the coil. <S> I've run into more than one newbies online where they are using a relay to switch a large motor (always an AC motor if I remember correctly) and <S> every time it switches (both on and off) it causes their processor to reset. <S> It's a seemingly unsolvable problem that I can only attribute to the above thus far, but I never hear more experienced people having the same problem nor do I hear them talking about using optos with relays. <S> However, an opto whose output is not isolated from its input side (as yours is), would not help. <S> Your 5V is probably not non-galvanically isolated from your 3.3V supply and it needs to be for the opto do provide any protection. <S> Now that I think about it, the focus of the help was always on flyback diodes or RC snubbers across the relay coil and decoupling capacitors, none of which helped. <S> An AC motor would preclude from using flyback diodes across the motor itself ( <S> unless they were bidirectional TVS diodes) which is why it was never suggested (and unnecessary if the relay was as isolated as we assumed it to be). <S> However, placing a small RC snubber across the motor itself was never suggested either which should solve the issue if an arc jumping from primary contact to coil is the problem. <A> Step currents being full spectrum not only cause <S> a broadband radiated EMI as well as conducted EMI and DC ground shift since that ground is not isolated. <S> Thus the loop current radiates noise and the conductor path shifts by V=IR+L*dI/dt on the shared return ground path. <S> Opto isolation will NOT reduce this EMI if you neglect the radiated noise on cables NOR will it reduce and filter your ground if it is sharing a path to the source . <S> It is a newbie design error or overkill to use opto-isolators with an obvious shared ground from an RPi which uses a 5V to 3.3V LDO. <S> There is no advantage. <S> Thus it is important to focus on the radiated EMI reduction with shielding, or cable orientation, or CM choke or any combination of the above when using relays to abruptly switch inductive loads from low level logic. <S> To minimize conducted ground noise issues, you must minimize the impedance from the switched inductive load to the power supply. <S> e.g. PSU to relay board heavy AWG, short "Star Path", twisted pair, <S> it relies on your experience to physically isolate load wires from control side. <S> not only gap,and orientation but as well use twisted pairs for both. <S> opto isolator = <S> if you have a shared ground <S> this does not reject common mode voltage if the grounds are common. <S> it can perform OK in spite of shared common if you follow the above EMI rules of thumb for conducted and radiated EMI. <S> Conclusion Design documentation failure by omission gives false hope of better performance unless they specify using isolated RETURN PATH to power source. <A> IIRC, the reason that this particular board uses optos is to separate the logic and relay grounds. <S> In addition, they translate the common 3.3V logic to the 5V drive level needed to operate the relays. <S> It's not necessarily good practice across the board, but it can save a lot of problems and debugging time right from the start for a fairly low cost.
| It is to prevent ground bounce issues when switching the relays (as the inductive relay coil can cause quite the spike on both power and ground nets, even with the freewheeling diodes across the coil).
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Is there any reason as to why the schematic symbol of comparators is almost equivalent to that of op amps? They do similar tasks, but they are almost completely different because comparators output digital signals. In that case, I am confused as to why their schematic symbol is almost entirely the same. Is it implying something else that I am not realizing? <Q> They are not completely different. <S> They are much more similar than you think. <S> Being linear in the linear region is not a design priority. <S> An op-amp is a comparator optimized to operate in the linear region (to be very linear in the linear region where it will spend all its time due to negative feedback). <S> An op-amp used as a comparator will be slower since it is slow to get out of saturation. <S> Most opamps also have frequency compensation so that they can be easily used with negative feedback. <S> Comparators have no need for such a thing. <S> Those are the big ones but op-amps also tend to place higher priority on things like noise, offset, and input bias currents than comparators. <S> All that good analog stuff. <A> A comparator is often close to being a linear amplifier that is optimised to achieve rapid switching around the point where the voltage difference between two inputs are very close to zero. <S> A comparator MAY include hysteresis which makes it less linear in action, but this is not an essential component. <S> If you look at the block diagrams of the very old (and so simple) <S> LM324 opamp and LM339 comparator <S> you'll see that they are very very similar to each other internally. <S> _____________________________ <S> LM324 opamp at left. <S> LM339 comparator at right. <S> The opamp (operational amplifier) has added linear output stages - but shares the same longtailed-pair front end with the comparator. <S> An output stage of the users choice could employ the comparator as an amplifier - with potentially superior characteristics to the "linear" device. <A> Both op-amps and comparators are types of amplifiers. <S> The triangle symbol is used for these and many other kinds of amplifiers. <S> You will find that power amplifiers and digital NOT gates also have very similar symbols (but only one input pin each), for example. <A> The comparator is a mixed device with analog input and digital output; so it can be considered as a 1-bit analog-to-digital converter (just as one flip-flop can be considered as a 1-bit register). <S> Usually, the comparator output is implemented as an open collector stage. <S> This allows to make it compatible with various circuit inputs and to connect multiple outputs in "wired OR".
| A common practice is comparators to be single supplied. Technically, a comparator is just an op-amp optimized to operate in the saturation region (have the output saturate since it is always either LO or HI, but be able to leave saturation quickly for fast switching).
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How to wire a 12V DPST switch? I have this 12V DPST switch that I want to wire, to make or break both Negative and Positive cable to open or close a solenoid valve. I would say: 3. (+) to battery 4. (-) to battery 2. (+) to load 1. (-) to load Is that correct? Anything that I should know? simulate this circuit – Schematic created using CircuitLab <Q> That is an illuminated switch, not the simple DPST that you indicate in your question. <S> Since Vellman doesn't provide a datasheet, you will have to do some testing to find how the LED is connected. <S> You may find that one pair of terminals is for the LED, and the other is for an SPST switch. <S> If you ask Vellman, they may be able to supply more infromation on the switch, or may be able to direct you to the actual manufacturer for information. <S> A common recommendation here is "No datasheet, No Sale", as you often need information from a datasheet to properly use an item. <A> There does't seem to be a datasheet for this switch, but this is from a similar product from ZF <S> that illustrates the arrangement. <S> The LED in the rocker body is connected to the contact plates by the springs and plungers that bear on the top of them. <S> Whether you want to have the switch lit all the time or when the circuit is powered determines whether you connect the incoming power to the center terminals (second circuit from left), or to the right hand terminals - third circuit - respectively. <S> The arrangement of the terminals in the base is always the same, with the ones at the center being the fulcrum for the contact plate to pivot over. <S> The empty positions on your switch are or the double throw versions. <S> The blade along the center of the body separates the terminals of opposite polarity, and there's a wall inside the body that isolates the moving contacts. <S> What may not be the same is the polarity of the LED, which will obviously only illuminate with the right polarity applied. <S> You may have to swap terminals 3 and 4 to get the LED to illuminate. <A> It will tell you which terminals are common, which are normally open, which are normally closed, and in what position common terminals are connected to others. <S> When you come upon or salvage a switch with unknown contacts, you have to do some continuity and resistance checking <S> (you have a multimeter, right ?) . <S> Measure the resistance from one contact to another. <S> Operate the switch and make notes about whether the resistance changes. <S> Or use continuity mode and listen for the beep. <S> A lighted switch usually has separate contacts for its internal LED or bulb. <S> For those you will need to apply whatever voltage the light is meant to operate at. <S> If you don't know, start low and increase it gradually. <S> In the case of an LED it will have polarity, so for example you might try 3 volts from terminal X to Y, then from terminal Y to X. <S> Then try 5 volts from X to Y, and from Y to X. <S> It can be tedious. <S> From the link in your comments, I can make out "R13-244" on one of the diagrams. <S> Searching for that, it appears to be a SCI Parts switch . <S> However, I was not able to find the specific 12V LED version. <S> Other answers and comments address other points, such as: <S> whether the switch is actually SPST with separate terminals for the light why you don't need to switch both sides of the solenoid including a flyback diode when driving a solenoid why a datasheet is important <S> So I'll leave it at that.
| Knowing how to wire a switch is a matter of looking at the product datasheet. Solenoids generate a large voltage on interruption of the current through them, you should put a freewheel diode across it to prevent arcing at the switch contacts, again observing the polarity - cathode to positive side.
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How do I use multiple GY-521, I2C sensor with Arduino? I'm working on an Arduino robotics project that requires I find the position of 12 different angles (I'm making a walking scale model of the Star Wars AT-AT and need to measure the angle of three different parts of the four legs). I'm currently going to use GY-521 Gyroscope sensor, however, I need 12 (3 angles, 4 legs) and each sensor has the same I2C address. Any possible solution? My knowledge of Arduino is moderate; knowledge of programming is basic/moderate; my knowledge of electronics is moderate/advanced. Would you recommend using something else to find the position of the leg?ALL advice is welcome!!! <Q> Let's try removing some XY-problem aspects first. <S> The basic goal is to find angles of 2-joint system. <S> This can be done with GY-521 breakout boards, however there are many cons in this solution. <S> First, you cannot measure angle with gyroscope. <S> Second, while MPU6050 has built-in DMP, making it work is a pain. <S> Many hobbyists usually give up and implement fusion in software. <S> This is not hard, but considering the amount of sensors it will be quite a draw on CPU and memory requirements. <S> If you go this way, I would recommend getting breakout boards based on BNO055 sensor, which has nice fusion DMP that is much easier to use. <S> Third, unless you guarantee your model will be used on flat level surface only, you also would need additional MPU on the body itself, and then quite a lot of math to make sense of all of it. <S> All these troubles can be avoided by installing simple rotary pots in the joints and then reading them with ADC. <S> If you have enough analog inputs on your Arduino this would be all you need. <S> Otherwise you can multiplex feedback channels via MCU-controlled analog switch, like ADG731 to a single analog input. <S> Finally, if you planning on using your model on uneven or non-level surface, I'd recommend adding one MPU in the body and load sensors in the "feet". <S> With some (much simple than above) math you should be able to control leg movements with enough precision to scale minor obstacles. <A> The modules have selectable address so two modules can be connected per bus segment. <S> Split the single bus into multiple isolated segments for example with a 8 channel multiplexer. <A> You can use an I2C multiplexer to select which I2C network to communicate with, for example <S> https://learn.adafruit.com/adafruit-tca9548a-1-to-8-i2c-multiplexer-breakout <S> with 8 outputs. <S> This allows using 8 devices that all have the same I2C address, giving you 16 in this case where each sensor can be set to one out of two different addresses. <S> There are basic driver libraries available online. <S> However, do you really want to use gyros? <S> They will tell the rate of rotation, not the angle when immobile. <S> An accelerometer would give you the static angle towards the ground but would be thrown off when the body part accelerates or slows down. <S> It's common to combine the two for a full picture, but that takes some algorithms to do. <S> Maybe you instead want to measure the angle of each joint? <S> Be careful if using I2C over larger distance than 10-20 cm as it is more prone to interference than other comm links.
| For stationary system it can be done with accelerometers, but for mobile system you need a fusion of the gyro and accelerometer data.
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What is the difference between a tank circuit in an oscillator and a resonance circuit? I understand that if a charged capacitor is placed with an inductor (and then remove the power supply,) the capacitor and inductor will exchange energy back and forth, it will basically oscillate and then slowly die out. Similarly, a resonance circuit (as I know RLC circuit with AC source) oscillates. Then, what's the difference between these two circuits? Am I right thinking a resonance circuit sustains oscillation and a tank circuit oscillation dies out? <Q> Am <S> I right thinking a resonance circuit sustains oscillation and a tank circuit oscillation <S> dies out <S> A circuit is called in resonance if the frequency of the driving signal is same as the natural frequency of the circuit. <S> For an ideal LC circuit (shown below), the natural frequency is \$\omega_0 <S> = \frac{1}{\sqrt{LC}}\$ . <S> If this circuit is driven with a current source at this frequency then the circuit will be in resonance. <S> At resonance, the voltage across the capacitor (or inductor) is going to have infinite amplitude. <S> This is because the impedance of the circuit between node A and B, given by \$\frac{1}{sC+\frac{1}{sL}}\$ , is infinite. <S> Further, since there are no losses in the system the circuit can sustain oscillations at the natural frequency. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Practically, there cannot be circuit with infinite impedance. <S> And in reality, the inductors and capacitors have parasitic resistances which can be approximately modelled as a resistance parallel to the LC circuit close to the resonant frequency. <S> This ensures that at the resonant frequency the impedance is finite (and equals \$R\$ ). <S> But the presence of the R implies losses inside the system and consequently this circuit will not sustain oscillations which will eventually die out. <S> simulate this circuit To compensate for these losses usually a non-linear active element is placed in parallel with the tank which would compensate for the losses inside the lossy LC tank (due to R). <S> what's the difference between these two circuits? <S> Both the circuits are parallel resonant LC circuits, ideal LC has no loss while oscillator tank has some loss. <S> Hopefully, it is clear. <S> Let me know if not. <A> An LC tank (consisting of an inductor and a capacitor in series) can be considered a resonant circuit, as you describe. <S> And yes, in an ideal world if you place a charged capacitor across an inductor <S> they wil oscillate at their natural frequency forever. <S> But in the real world there is resistive losses turning the electrical energy into heat, causing the oscillation to die out. <A> A resonant circuit will absorb/transform almost totally or in a large part the energy that he has or is being given. <S> A LC tank and antennas are electromagnetic resonators, piezoelectric crystal suck as FBAR are electromechanical resonators. <S> A system can be defined as an oscillator when it creates it's own signal. <S> For instance Ring oscillators don't have any resonators. <A> in another interpretation simulate this circuit – Schematic created using CircuitLab
| Both the ideal LC circuit and the oscillator tank (as you call it) are resonant circuits.
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How to desolder an element from a PCB? I want to desolder an element from a PCB, I can see all the solder joints that are holding it on to the board and I can reach all of them easily. I am not very experienced in working with a soldering iron, tbh this is the first time I am working with one but I informed myself how to do it correctly and I also got myself a desoldering pump to remove the melted solder joints. The only thing that is giving me headaches is the fact that I read it can be difficult to desolder solder joints that are close or on a ground plane because the ground plane absorbs heat. I took a look at my board and this may be the case for 3 or 4 of them. My question is do I have to take special care of those solder joints, do I need a specific tool to remove those or can I remove them just like the others? <Q> Whenever you have to desolder something the first question you need to ask yourself is: do I need the component intact? <S> Because if you removing damaged component then often the easiest way is to separate its pins first. <S> For most but the smallest SMD components this can be done with wire cutter. <S> The plastic connector shroud can be broken or melted away etc. <S> Once you've done that removing pins <S> one-by-one becomes trivial. <S> If you do need the component, see if you can distribute heat to all pins simultaneously. <S> For 2-5 pins this usually can be done by adding solder to the pads, making a blob of molten solder. <S> For more pins you can cut a piece of heavy gauge copper wire and put it along the pads, then cover everything in solder and keep heating until all pins got loose. <S> Here is very good video demonstrating this technique. <S> Pump and/or wick can be used afterwards to clean the pads. <S> In either case having an iron suitable for the job is most important. <S> Big chisel or screwdriver type tips work best for this. <A> Tricky, especially if you are not experienced. <S> It's a connector? <S> What I would try is first go along the pads with a solder sucker, suck out solder, and check that each pin is physically free with the tip of a small screwdriver or similar. <S> Sometimes you have to refill the hole with solder and repeat before you can get enough solder out. <S> Even so it depends a bit on how big the hole is compared to the pin going through. <S> If the fit is too tight this won't work. <S> The other option is to try to heat all pads at once and then prise the component out. <S> A hot air gun (the type for surface mount soldering) is good, but you have to be careful not to overheat the board and the component. <S> It's not really a beginners job, to be honest. <S> Needs a bit of experience, the right tools, and some patience. <A> Depending on the geometry of the pads and how they are connected to the ground plane, they will require longer heating to let you remove the solder. <S> A tip is to use a bigger solder tip, which has more thermal mass. <S> It will help delivering enough heat before it is drawn away by the ground plane. <A> Decide if you want to keep the board or the component intact. <S> Keepingboth intact is a hairy problem even for experts. <S> Desoldering lead-free solder joints is tedious. <S> It's actually easierif <S> you first generously add leaded solder (SnPb38Cu2 or SnPb38Ag2,not SnPb40!). <S> Connect all the pins with solder. <S> Solder in a thick copper wirebridge as a helper. <S> Heat the whole thing up at once. <S> Pull the component. <S> Congrats, you are almost done. <S> Remove the wire bridge. <S> Use solder wick to remove the solder from the vias. <S> Don't use a pump,they tend to reap the copper eyelets from the board. <A> This can take a while. <S> Desolder pumps are nice, but desoldering wick can also be useful. <S> Where it's not contraindicated, the easiest way is to cut the piece, and take out the pins one at a time.
| Also use solder wick to draw solder out.
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External clock with Arty Z7 FPGA development board I have a digital audio source that is basically I2S. I need to run the FPGA at the exact clock as the I2S provides. I've got four such inputs required, but I am assuming that the clocks are the same. Is there an input on the Arty Z7 board that works as an external clock? If not, is there a dev board for zynq that does? Any pointers to how to configure that in the vhdl? Update: maximum sample rate is 384Khz. maximum bits per channel is 32. maximum channels is two. Consequently, maximum clock is 24MHz. Max z7 clock is ~100Mhz? I do not believe that this is a high enough multiple to avoid clock inaccuracies on the output (but would be happy to be wrong). This is an audiophile application and avoiding jitter is very important. <Q> If the frequency is low (which is usually the case for audio applications), generally what is done on FPGAs is to use clock enables. <S> The idea is to detect the edges of the input signal with a fast core clock and generate single-cycle pulses. <S> Then use those pulses to sample/latch in the input data and control internal operations. <S> With that setup, there is no need to use a specialized global clock capable input pin to supply the clock. <S> This also means you don't need to deal with clock domain crossing because all of the logic is in the same clock domain. <S> The internal clock can also be much faster, enabling pipelined processing and even time-interleaving processing of multiple channels using the same logic. <A> Yes, there are pins available on the Arty Z7 that can be used as native clock inputs to the PL fabric. <S> Page 9 of the schematic shows the pin names in each of the banks of the PL used in the design. <S> This Xilinx Forums question about SRCC and MRCC pins gets to the necessary information: <S> the pins labeled SRCC and MRCC are clock input <S> capable pins, some of which are routed directly to user-facing pins on the Arty Z7. <S> The PMOD pins are prefixed with JA and JB while the Arduino headers use the prefix CK_IO . <S> Assuming single-ended clock inputs for your I2S-like implementation, you have these pins available: <S> IO_L11P_T1_SRCC_13/ <S> U7 - CK_IO29 <S> IO_L13P_T2_MRCC_13 <S> /Y7 <S> - CK_IO37 <S> IO_L14P_T2_SRCC_13/ <S> Y9 - CK_IO41 <S> IO_L12P_T1_MRCC_34/ <S> U18 - JA3_P <S> Of course, other boards will have similar IO features. <S> Depending on your FPGA fabric, processor, and IO needs, a different board will be an option. <S> For more details from Xilinx, see: UG472 (v1.14) <S> July 30, 2018 section "Clock-Capable Inputs" pg 30 and onward, <S> explaining how to use these inputs UG475 (v1.18) <S> July 16, 2019 <S> "Table 1-12: <S> 7-Series FPGAs Pin Definitions (Cont’d), Other Pins" pg 30-31 providing the definitions for the pin name abbreviations <S> As for configuring this in VHDL, you'll be using this as one of the clock domains in your design. <S> Your pin assignment constraints will need to place the input clock pin in the correct location given the above information and your choice of how you're connecting. <S> As for the design, it will depend on whether or not your clock input and output rates are constant or dynamic and continuous or discontinuous. <S> If you're looking for the most flexibility, I would suggest having a constant-clock resampler/custom-DSP running on one of the board clocks with FIFOs to/from the different precise and ratio-frequency IO domains. <A> If you implement a digital circuit with two independent clocks, you will need to create a block which transfers data between clock domains. <S> Such transfers have an inherent problem known as metastability which essentially means if two independent clock edges arrive with a certain timing, they will violate the setup/hold times on your triggers, which can then output an uncertain logic level until the next clock pulse. <S> As a result, your circuit will never be 100% reliable, and you will have an MTBF (mean time between failures) <S> metric associated with your inter-domain data transfer. <S> To get a working design, people increase this MTBF to a high enough value: 1 failure per billion years sounds acceptable, doesn't it? <S> Increasing MTBF is done by chaining several synchronization stages of flipflops in a sort of FIFO, so it has a price in terms of resources used to implement the flipflops, and the latency in data transfers. <S> Sampling the clock pin (as @alex suggests) has the same metastability problem: if the edge arrives at an unfortunate time, you can end up missing your single-cycle pulse, or generating two pulses instead of one. <S> However, the problem will be much less visible in the first place, since only 1 bit is affected, and the cost of dealing with it (if needed) in terms of resources will be proportionally smaller: in your case, 4 times as small compared to the 4 data inputs.
| Since you're after external IO, you'll probably use the PMOD or Arduino headers on the Arty Z7. To elaborate on your external clock idea: of course it's possible to use an external input as a clock, it just makes very little sense doing so for audio, unless you can use this clock for the whole design.
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Can I use aluminum foil for a heat gun as a nozzle? Wanted to desolder a small chip with 48 pins on a PCB. Since my heat gun didn't come with a small nozzle included, can I use aluminum foil to create a ghetto nozzle for it? What can I do as an alternative option? <Q> Of course, you can use aluminum as the nozzle, even though it's melting point temperature is low, I think aluminum alloy covers the upper limit temperature a chip can bear. <S> But why your heat gun has no nozzle? <A> I don't see why not. <S> You just need to secure it well and play around til you get a good size nozzle at the end with the airspeed you like. <S> If its blowing too hard you could try cutting some vents in the side to take away some of the pressure. <S> Like the user in the comments said, it may work better to just shield the rest and use the heat gun as is if its just a one off type thing. <A> You don't necessarily need a nozzle, I occasionally reflow solder with it off (if I'm doing a panel by hand, for instance, and want to spend less time waiting for my board to heat up between component placements). <S> The main problem is that without one you'll spread heat over a wider area, so you might accidentally shift or desolder components if you're not careful. <S> I do recommend getting a real, temperature controlled rework tool if you're going to be doing more of this in the future but a heat gun with no nozzle should be fine for a one off. <A> I don't know how well it will work, but you need to consider that aluminium has rather low melting point temperature(660.32 °C). <S> You probably wont have that temperature near PCB, but temperature will be higher near heat gun.
| If you're using an actual heat gun, you might risk high temperatures that could burn your solder mask (although I've used them sucessfully in a pinch).
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Beginner soldering help - solder balls/doesn't liquefy and soldering tips blacken Very frustrated!! Have gotten into electronics recently and watched many videos about soldering and proper technique. I started out with the iron from this kit: https://www.amazon.ca/gp/product/B0009Z3JJA/ref=ppx_yo_dt_b_asin_title_o02_s00?ie=UTF8&psc=1 It's not variable temperature, just plug in and use. So I waited for it to heat up, tinned the tip with my solder and used as I had seen in all the videos I watched, cleaning it on a wet sponge after every few solders. I thought maybe I had left the iron on for too long as the tip became blackened and the solder wouldn't liquefy when making contact with the iron tip. So I clean it off on the wet sponge a bunch and it starts looking better, but I still can't tin it. So I follow the instructions in this video: https://www.youtube.com/watch?v=2VlvjdaMARM I purchase this flux: https://www.amazon.ca/gp/product/B01EYJJEEK/ref=ppx_yo_dt_b_asin_title_o01_s00?ie=UTF8&psc=1 Get my iron heated and sink it into the flux, but the solder still won't stick to the tip, it just kind of balls up if i bring it to the edge of the coated part of the iron (the edge opposite of the tip). I think ok maybe it got burned out or something because I left it on too long, also it was part of a $20 kit so maybe it was just a piece of crap. So I buy this guy here: https://www.amazon.ca/Elenco-SL-75-Soldering-Station/dp/B002KIYSHQ Variable heating, I check the manual it says to heat it just beyond the 4th notch to about 700 degrees for lead-free solder. I heat it right where it says to and wait, it tins and everything is fine and dandy. I start using it, this time cleaning between every solder because I'm trying to be extra good to not mess it up, but after 28 solders it stops working I can't tin it, even while trying after dipping it in the flux. The solder doesn't liquefy at all. The station came with two tips, the left one is the one I used that no longer works, on the right is what it looked like before: https://i.imgur.com/0o1Z2Vz.png This is the tip of the first iron I used that stopped working: https://i.imgur.com/GzHoOk4.png https://i.imgur.com/tGIWY87.png <Q> The soldering iron you bought is provably low quality. <S> A high quality soldering station I would recommend in the same price margin as your previous would be the 937D+ , also get solder with flux in it. <A> It is taxing even on a $1k temperature controlled soldering station if you are using the wrong tip size. <S> Not only are your irons not getting hot enough, they are not maintaining temperature and you are using tips that are too small. <S> Lead free does not let you get away with things like that. <S> I have a $1200 station and still use nothing smaller than a 2.4mm chisel tip (conical tips suck because they do not make enough contact) for thru hole with lead free. <S> Is it twice as big as the joint? <S> Yes <S> but I can still squeeze it in and <S> it maintains temperature. <S> Also lead free solder dissolves the iron in cheap soldering iron tips really fast. <S> I saw a new one get eaten away in about 3 months of soldeiing a few hours every week. <S> It had 4mm or 5mm just disappear. <S> This was a BIG 6mm wide chisel tip. <A> I'm probably being contrary to what you have been reading, but I think that both of the soldering irons that you have purchased are of adequate quality for your use. <S> Instead, I think that you may have two problems: iron temperature is far too high and your choice of solder. <S> I'm in Canada <S> and we don't have the RoHS restrictions that Europe has. <S> That's <S> good - most of our clients are Industrial and they prefer that we use lead-based solder if at all possible. <S> Let's go back to the first iron that you purchased. <S> Obtain a brass "brillo" pad and gently clean the tip while it is hot. <S> Amazon has many but the least expensive appears to be: Stanz (TM) <S> Soldering Iron Tip Cleaner with Brass Wire Sponge, No Water Needed. <S> Be sure the wire pad is brass and not steel. <S> There is a much less expensive version but the description says "steel". <S> Now try tinning the tip with lead-based solder. <S> My favorite solder is Kester 63/37 with type 44 Rosin flux. <S> You can purchase a small pack for not much money just to try it. <S> I use Amtech NC-559 which is also available from Amazon. <S> I've purchased this from several sellers but <S> the best bang for the buck appears to be this: Souked 100g AMTECH <S> NC-559-ASM <S> No Clean Lead Clean Solder Flux Solder Paste <S> Can <S> $18.99 <S> That jar of flux will last you years and years! <S> Get some copper wire and practice soldering wires together. <S> Old CAT-5 Ethernet cable is good and available for free much of the time. <S> Strip the insulation from the conductors, twist pairs of wires together into a pigtail configuration, solder. <S> It will take time and patience to get good at this. <S> Graduate to soldering components on circuit boards after you have mastered soldering wires together.
| Cheap irons dont work well with lead free. You may also want to invest in a decent flux.
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Crimp directly to PCB (or holes in PCB) I know there are certain SMD components which can receive a Dupont male connector. I know there are ways to solder male and female connectors to a board directly. As part of a manufacturing technique, I'm wondering if there is any way on the market to connect wires to a board, without soldering of any kind. So you might crimp some steel piece to a wire, then that steel piece would go into a through hole in the PCB, where it is ergonomically crimped a second time. Is there any sort of system in place for connecting wires to boards without soldering? I am looking to build a machine that can do this. <Q> You did not provide any requirements, and it does not look like you've done any research. <S> Nevertheless, there are several options available. <S> For example for low power you can use so called "card edge" connectors, like EdgeMate series from Molex. <S> For high currents there are Han-Fast Lock terminals from HARTING. <S> Note, that this is not the best option in general, I would still go for good ol' screw terminal. <S> A bit of soldering, but beats the rest in convenience and peace of mind. <A> press-fit connectors connect to plated circuit board holes without solder. <S> Bolts, or machine screws, are another alternative. <A> If you can work with thin flexible cable you could use conductive adhesives (like the edge connections used in commercial LCD video screens). <S> These are small flexible parts with rows of conductive layers sandwiched vertically along its length. <S> You place a fine PCB pattern at the location, then place the strip over the pattern, then a flex circuit lays over the strip. <S> A clamp holds the whole assembly together. <S> The vertical layers in strip makes the connections. <S> (Once popular with watches with small LCDs.) <S> There are more bulky conductive epoxies (often containing silver or carbon particles) that might better hold thicker wires to a PCB pad or hole. <S> Lastly, there are many versions of "Swage Pins" that might work. <S> You insert the pin into a plated PCB hole (and possibly the wire too) <S> then deform the pin with a press, (similar to using a rivet). <S> With many types available there are likely some that would work with the wire type you are using.
| But really, if you consider the cost of connectors, crimping tools and contact reliability, your best option will probably be a simple and dirt-cheap ring terminal either bolted or riveted to large circular pad on PCB. You could also look into what is called "Zebra Strips".
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Can I Use an Isolated DC/DC Converter to Step Down a Switching Dynamic Signal? I have a 24V device that takes a signal of either of 36V or 72V that will switch on and off at around 2Hz. This signal needs to be converted to 24V and I'm considering this isolated DC/DC converter which can convert 18-75V to 24V. The problem is that I don't know if the device is capable of consistant switching since these are normally used for providing constant power. I looked on the datasheet and it doesn't say anything about switching lifespan or capable number of power-ons. My concerns are about activation time, reduced lifespan, start-up delay, start-up voltage jump, etc. Is this a viable use of the converter or should I use another device or configuration? Edit: For clarification, the input signal can come in as either 0V, 36V, or 72V (DC). The output is either 0V and 24V (DC). A 36V and 72V input must translate into a 24V output, and a 0V input must translate to a 0V output. <Q> Both 24V and 36V inputs are the same output in this system, while a 0V input is still a 0V output. <S> So you just have to clamp the output to 24V. Here are two possible clamping approaches. <S> The top circuit will clamp at the Vz and all its variation with current. <S> The bottom circuit will clamp at 24V + Vd. <S> The resistor size is not critical. <S> It is simply sized to limit the current going through the diode and the current that the output must source. <S> Use the maximum expected input voltage in this calculation, 72V. <S> In the top circuit, with a 72V input, 48V will appear across the resistor. <S> You can use this to calculate the current. <S> In the bottom circuit, with a 72V input, 47.3V will appear across the resistor (because the diode forward voltage drop takes up some of the slack). <S> Note that the bottom does not clamp if the system is not powered. <S> This could fry your input depending on how your system works and is expected to be used. <S> However, the 2K input impedance and R2 may still save your system in this case since it forms a voltage divider. <S> With the shown value of 10K, it steps down 72V to 12V when unpowered. <S> If one fails, the others are still there to limit current. <S> Diodes normally fail short which is fine diode failures probably won't fry your input. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You can use a resistor and a zener in the following configuration (ignore shown voltage levels). <S> Image from: https://next-hack.com/index.php/2017/09/15/how-to-interface-a-5v-output-to-a-3-3v-input/ <S> In your case, the zener diode voltage should be 24V. <S> The main drawback using this configuration is that the signal (72V or 36V)must be able to drive some mA to the Zener. <S> With this configuration, if in your output you have 0V, your input will be 0V. <S> If in your output you have 36V or 72V, your input will be clamped to 24V. <A> If you could benefit from isolation, another option is to use an SSR plus an input resistor. <S> This one is recommended to have 3mA drive so a resistor of about 10K on the input would be about right for 30-77V. <S> The switching time is less than 10ms, so acceptable for 2Hz. <S> DC-DC converters might work okay, or they might not. <S> I have seen them take a good fraction of a second to fully start so that would be unacceptable in your application. <S> They will also have an output capacitor that will need a load to discharge it within a reasonable length of time.
| If you are worried about resistors failing and frying your 24V output then you can use multiple resistors in series to achieve your desired value.
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Convention for inverted signal Is there a standard convention for marking an inverted signal using only text? The package currently being used does not support having a bar over the signal in all instances. IE: NOT_Reset or Reset_Bar etc? when you can't use \$ \overline{Reset}\$ <Q> Often a tilde ~ is used to indicate negation <S> (just like the bitwise NOT-operator in the C-programing langage) if only text without special formatting, like overbars, can be used. <S> E.g. <S> ~reset <A> <A> I use NOT_ because it's clear what it is if I forget the convention or whoever is reading it doesn't know the conventions. <S> But I am also very verbose in my labeling. <S> Principle of least astonishment. <S> Another reason is that it's difficult to miss when reading and difficult to forget when typing. <S> I never liked things like a single character prefix (and a suffix is even worse) because it's way too easy to gloss over when reading it and interpret it as an uninverted signal, or when typing it out too (especially if it appears in code and you have the uninverted variant also floating around) <A> There are many languages and logical forms of expression. <S> Since schematics are also "logic diagrams" , they too express the standards adopted by the international or corporate community. <S> They must also beware of the limitations in character set and ability to draw vectors such as "overline" unlike "underline" which is available in some. <S> From a historical point of view for documentation on signals used in both limited character sets for say "Theory of Operation" and schematic drawings, the most common symbol I have seen since the early 80's was the ! <S> symbol placed either before or more often after the word to indicate "Negation" or the active state is low. <S> e.g. reset! <S> is active low. <S> Similarily I have also seen reset~ , /Reset A or A! <S> are the binary states for normal and inverted signal or in Boolean logic A=1 and A!=0 ! <S> is also called the Bang key . <S> Here are all the software variants which includes examples of "negation" <S> may not be relevant to schematics but may be familiar that use extended character sets. <A> I usually use a preceding exclamation mark: ! <S> RESET <S> The main reason for this is I can easily denote signals like this: UP/!DOWN <A> I've seen all kinds of conventions: <S> RESET- <S> RESET_L <S> nRESET <S> Just pick one and use it consistently. <A> Check you coding environment documentation to determine if you need to code active-low signals a certain way. <S> I typically code HDL using the convention reset_n or resetn. <S> However, one development suite I use regularly will sometimes get confused if I don't use <S> -only- resetn <S> ; it does not interpret reset_n as one might expect during its parsing and autogeneration stages.
| I often use "/RESET" using the forward slash to indicate active low
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does proportionality always works when analyzing DC motors? Let's consider this example:a Trolley that can reach 36kmph when there is lite traffic with a current of 700A, and when the traffic gets heavy, the speed decrease to 20kmph.now i want to deduct the current form these information. so my question is, would proportionality rules always apply for these kinds of analysis.for example the current would be: I2 = (20/36)*700 = 389A the problem with this approach, is that it contradicts with how DC motors work, which is when speed decreases, the current increases <Q> Motor voltage or EMF is proportional only if no load. <S> as RPM/V or kV/RPM Motor current is proportional to torque. <S> Thus current will rise as load increases and voltage drops from a light load condition. <S> Overall , power drain increases, so you can presume current rise is faster than voltage sag up to some point. <A> it contradicts with how DC motors work, which is when speed decreases, the current increases <S> That is only the case for a DC motor that is driven with a constant voltage (and either a constant or increasing field; I assume you're thinking about permanent-magnet motors with constant field). <S> would proportionality rules always apply for these kinds of analysis. <S> for example the current would be: I2 = <S> (20/36)*700 <S> = 389A <S> No. <S> Assuming you're thinking about permanent-magnet motors, motor current is more or less proportional to the motor torque, plus some losses. <S> The torque needed to drive your streetcar is going to depend on speed in a complicated way; there will certainly be components that are proportional to speed, but there will also be constant components, and components that depend on speed squared or have more complicated relationships to speed. <S> So, basically, it ain't that simple, and determining just what actually happens is difficult to deduce from first principles. <A> Voltage is mostly proportional to speed (but you have to include the current through the motors internal resistance which resistance reduces the voltage available for speed. <S> Your trolley problem is under-specified, the current will be whatever the trolley driver deems necessary at any instant.if heavy traffic <S> means more passengers then current goes up. <S> if it means lighter acceleration it goes down. <S> For the same mechanical load you have the same current, so hill-climbs at either speed will take the same current.
| Current is proportional to torque, if you slow a motor by increasing the mechanical load then current will increase.
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How does an inverting op-amp feeds back 0 volts when the Vout is not 0 So, I'm a Computer Science major, and have this class, Analogic and Digital Circuits. I'm learning the op-amp now. The inverting op-amp has something that I just can't comprehend . I understood the algebra that lead to Vout = (-R2/R1)*Vin. But, the feedback, Vout is going into V-. V- and V+ are aprox 0. How can Vout be 0V, and another bigger value at the same time? <Q> The voltage at \$V_{OUT}\$ changes so that the voltage difference between the inverting and non-inverting inputs will become very small, ideally zero. <S> When this happens, the current flowing through the feedback resistor will be just equal to the current flowing through the input resistor. <S> If the non-inverting input happens to be connected to ground then the voltage at the inverting input will also be essentially zero. <S> EDIT: <S> To clarify, I'm talking mostly about ideal <S> op amp behavior. <S> Check the specifications of any real op amp to ensure that the output swings all of the way to ground. <A> Look more closely at your statement V- and V+ are aprox 0. <S> V+ is not "approx 0" it is normally wired directly to 0. <S> However the feedback voltage V- is not zero. <S> Check the "open loop gain" of your opamp : <S> it isn't infinite; let's say it is about a million. <S> And let's say Vout <S> = -10V : to get that, V- must be 10 uV. <S> Unless you have a very expensive multimeter, that's close enough to 0V <S> you simply can't measure it : <S> but it's still not 0. <S> Consider the case where R1 = R2, so <S> gain = <S> -1.V- is simply the difference between Vin and Vout. <S> It's not precisely 0, and therefore Vout isn't precisely -Vin, but it's close enough. <S> If you used an opamp with an open loop gain of 1000 or so, then V- would be about 10mV for the same -10V output, and you'd be able to measure that. <S> So your initial statement is wrong <S> : it's a simplification that's near enough to make the analysis simple, but accurate enough to be within measurement limits. <A> simulate this circuit – Schematic created using CircuitLab <S> Consider the circuit above. <S> An Op-amp has the property that if voltage at the non-inverting '+' input higher than that at the inverting '-' input ( \$V_n\$ in this diagram) it will try to make the output ( \$V_{out}\$ in this diagram) as high as possible. <S> Similarly if voltage at the non-inverting '+' <S> input lower than that at the inverting '-' input ( <S> \$V_n\$ in this diagram) it will try to make the output ( \$V_{out}\$ in this diagram) as low as possible. <S> Op-amps have a high but finite gain. <S> The second property to consider is both inputs take and supply negligible current and in most cases we assume this to be zero. <S> Lets assume the gain to be near infinite and set Vin to 1 volt. <S> Since we have zero volts on the '+' input then if \$V_n\$ were <S> >0V <S> \$V_{out}\$ would go very low pulling <S> \$V_n\$ <S> negative which would cause \$V_{out}\$ to go very high. <S> There is a balance point however where \$V_n = 0 \text{ volt}\$ . <S> This gives you the current in \$R_1\$ , the same current must be following in \$R_2\$ <S> so it should be clear that. <S> $$V_{out} <S> = - V_{in <S> } \cdot \frac{R_2}{R_1}$$ <A> There are specs for maximum linear range for inputs (Vcm = <S> V_common mode)and outputs ( <S> Vo max,min) before Op Amps can work as analog devices. <S> This is especially true for bipolar junction transistor(BJT <S> ) type OP AMPs within 2V of the +/- <S> supply voltages where something is usually saturated or cutoff. <S> (CMOS Type OP AMPs tend to be Rail to Rail (RR) more ideal and lower input but also lower output current) <S> Inverting "Negative Feedback"(NFB) always tries to force the input difference voltage =0 thru the NFB ratio. <S> (e.g. Rfb/Rin) <S> When the output is in the analog or linear output voltage range, the input difference will always be 0 or defined by Vio, Iio spec (input-offset). <S> When any input changes, the Vout changes thru NFB ratio to keep the input error =0 <S> according to your "analogic". <S> e.g. <S> If Vin- goes lower, Vout goes higher. <S> if Vin+ goes higher, Vout goes higher, etc.
| In an inverting amplifier it is the current between the output and the inverting input that achieves negative feedback.
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Is there a cheap and accurate electromechanical method to provide a configurable counter-torque? Given a continuously rotating motor shaft, is there a cheap and accurate electromechanical method that would provide a configurable counter-torque? Rotation is in one direction only, RPM range is 0 - 12000, torque range from 0.001 to 1 Nm. edit: Before the edit I was asking for a constant counter-torque regardless of the motor's RPM. I have removed this as it just adds complexity, increases cost and is not strictly necessary IF the magnitude of the counter-torque is precisely known at whatever RPM the motor's shaft is currently spinning at. <Q> Do you understand the concept behind using the Prony Brake for measuring torque? <S> You can use electromechanical feedback (a simple servomechanism) to maintain constant torque at arbitrary speeds. <A> The cost will depend on the range of torque and speed you need and particularly the maximum horsepower, torque multiplied by speed. <S> The required steady-state accuracy and dynamic performance will also be factors. <S> You also need to consider how much of the system you are going to build yourself and how much you are going to purchase. <S> The prony brake may or may not be the least expensive option depending on the complete requirements. <S> Look at this Dynamometer information <S> That method of measuring torque should be the least expensive. <S> A prony brake can be arranged to serve as an electrically-controlled, trunion mounted absorber. <S> At the power level in question, wear and power dissipation should not be problems. <S> At the low end of the torque range bearing friction could become significant. <S> That would not interfere with measurement, but it is a residual torque that will set the lower limit of applied torque. <A> There are "Torque Motors" that provide approximately constant torque independent of RPM. <S> An example application would be a driving a take-up spool on a web process.
| If you use a trunion-mounted absorber, the torque can be measured directly as the force exerted on the lever-arm that prevents the stator of the absorber form rotating.
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Why does a thermocouple have two outputs (V+ and V-)? Does that mean that there is two different voltages? The potential is used to determine the temperature. (?) I understand it's the seebeck effect dictates how the thermocouple works. Apologies if I'm missing a simple concept. I have had no experience with thermcouples prior to my current thermopile project. Thanks! <Q> ALL voltage sources have at least two terminals. <S> Voltage is ALWAYS specified relative to some reference value (often "ground"). <S> The concept of a single terminal voltage source makes no sense. <S> So, a thermocouple is a voltage source with an output voltage measured on one lead relative to the other lead. <A> Really there is only one output, which is the voltage difference between the two terminals (V+, V-). <S> The V- is used as the reference going into the multimeter/circuitry. <S> On your diagram, Tv is simply V+ minus V-. <S> In other terms, you don't measure the voltage referenced to ground, something you'd normally do for a power supply or amplifier output. <S> Be warned that the voltage difference may be quite small (uV - mV range), so if you have a cheap multimeter you will likely need to amplify the voltage with an instrumentation amplifier. <S> Here is a quick reading: https://www.analog.com/en/analog-dialogue/articles/measuring-temp-using-thermocouples.html <A> It's effectively a voltage source, a bit like a battery (given fixed thermal conditions). <S> So, of course, it has two terminals. <S> The potential between the two terminals, measured at the meter, is a nonlinear function of the temperatures at Th and the temperature at T0, so if you want the temperature at Th you must measure the temperature at T0 using some method other than a thermocouple (or, as they did in the days of yore, place the two junctions at T0 in an ice-water slurry to control the temperature at 0°C). <S> Very roughly, the voltage output is proportional to the temperature difference between Th and T0. <S> To get a more accurate measurement you can calculate the thermocouple voltage equivalent to the temperature T0 if it was compared to a reference junction of, say 0°C, and then add that voltage to the measured voltage at the meter to get a new voltage, then find the temperature corresponding to that voltage with a 0 <S> °C reference junction (if T0 was 0 <S> °C rather than whatever it really is). <S> If T0 happens to be 0 <S> °C (perhaps due to that ice-water bath) the voltage will be 0uV.
| Yes the voltage difference is used to determine the relative temperature.
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How to suppress a short circuit to prevent fuses from blowing I have a circuit that is modular and one of the modules involves interfacing from an external relay. Now this sub-module has a fuse set to blow a bit below what the main module's max current. (why put a fuse when the main module has over current protection? So that when this sub module fails it wont take the whole system down with it) This sub modules can cater upto 24 external relays and in the image below i will be showing how one channel is setup What i would like to protect is the fuse from blowing from a false short circuit(not really false, since it is doing its job) in the cenario that the user might incorrectly wire the connection and shorts pin A and pin B. I need a current limiting solution so that the fuse wont blow up and the user of the sub module can rewire and try again. The circuit is designed to trigger one relay at a time, a signal will be sent by a portexpander(not shown in picture) to the bs138 mosfet to open. the signal typically last around 250ms - 1000ms. I have considered a polyfuse but it to slow to react the main module protection will trip first. Another is adding a resistor in series with the relay line but it is either a bulky high watt rated resistor or a high value resistor that might not make the relay trigger because of voltage drop. there is also the problem since there is a short the power to the sub-micro controller will be cut, so this protection circuit i am looking for would probably happen in either series or parallel of the connection for the external relay I need current limiting solution between pin A and pin B so that when A and B is shorted by the user and the mosfet turns on the fuse wont blow how do you recommend to solve this problem, Thank you. <Q> The fuse wouldn't be blowing by accident. <S> The fuse would be blowing because it's passing too much current. <S> I don't really understand what you're asking. <S> If you don't need it to do that, you should replace it with a wire. <S> If you don't want it to blow specifically from a short between A and B, you can move the fuse somewhere else, or put some sort of minimal load or current limiting device between A and B. <A> You have described an application for a slow-blow fuse or a Polyfuse to heat up faster than the load. <S> These tend to operate between 85’C and 125’C by design. <S> A better design would define that actual load V vs I and Fault current then define limitations for cost and area. <S> It might be feasible to use a 5.5V to 6Vdc input with a 3 terminal regulator with OCP,OTP and UVLO with say 0.4V dropout at 1A for only $0.80 (100pc) in a SOT-223 package e.g. <S> MCP1826ST-5002E/DB <S> Intelligent auto-resettable breakers need accurate specs for break and retry algorithms. <S> - But for thermal loads such as motors used to close car windows, a thermal switch is used that acts quicker than the motor to cut off power but takes just as long as the motor to cool off and retry power again. <S> this can be something like 1 sec OFF and 5 to 10 sec ON retry. <S> This usually uses a carefully selected polyfuse controlled relay. <S> Also replace the main fuse with a slow blow fuse. <A> I need a way to prevent a short circuit from ever happening on the outputs of the sub module no matter how the user mess up the connection to the external relays. <S> The only way to minimize the chance of this happening is to use polarized shrouded connectors instead of pins. <S> Any attempt to bypass or disable the fuse only defeats the purpose of having it. <S> There is no protection from the stupid and reducing the error coverage only increases chances for failures.
| The fuse SHOULD blow if you short it.
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KiCAD: Triaxial Connector Footprint from Drawing I am planning to use a PCB-mount triaxial connector, CBBJR79TL, in a design for a laboratory prototype. Making reference to the "recommended mtg hole pattern" section from the CBBJR79TL drawing , I have attempted to make a footprint in KiCAD. I am new to PCB design generally, as well as KiCAD specifically. There are 6 pads. The pads are located at (-79,0), (21,0), (129,129), (-129,129), (129,-129), and (-129,-129). Pad Number 1 corresponds to the center conductor, Pad Number 2 corresponds to the inner shield, and Pad Number 3 (technically 4 contacts) corresponds to the outer shield. All units are in mils. How to interpret the dimensions provided on the drawing? The pins themselves appear to be 50x40 mil (+-5mil in each dimension), and yet the recommended hole diameter is 62.5 mil. Suppose I add 22.5 mil of pad diameter to account for the annular ring -- that leaves a total pad size of 85 mil. The closest of the two pads (Pad 1 and Pad 2) are 100 mil apart (center to center). With a total pad size of 85 mil, this means that there will be only 15 mil of solder mask between (edge to edge) of Pad 1 and Pad 2. This distance is lower than recommended pad separation distances (50 mil) for through-hole parts that I have come across online. That being said, I am unsure what Pad Size and Hole Size I should be using. I have a feeling I am missing something – perhaps something related to the pre-tinning note (note 7) on the drawing? <Q> I'll try to answer your questions: <S> How to interpret the dimensions provided on the drawing? <S> I believe you have correctly identified the x-y coordinates of each hole. <S> The hole diameter of 62.5 mils is also correctly identified. <S> As far as I can tell, you have transferred the useful information from the mechanical drawing for an electrical connection. <S> You may consider adding additional silkscreen for indicating the body of the connector, but it's not mandatory. <S> The pins themselves appear to be 50x40 mil (+-5mil in each dimension), and yet the recommended hole diameter is 62.5 mil. <S> The diagonal of a 50x40 mil rectangle is 64 mil. <S> However, if they're recommending a 62.5 mil drill hole, I suspect the corners are rounded. <S> The paranoid might order the connector and get it in hand to verify this with calipers before fabbing the board. <S> With a total pad size of 85 mil, this means that there will be only 15 mil of solder mask between (edge to edge) of Pad 1 and Pad 2. <S> This distance is lower than recommended pad separation distances (50 mil) for through-hole parts that I have come across online. <S> I'll note first that by virtue of the hole diameter, there will be less than 50 mils separating pads. <S> However, what you're used to seeing in 100 mil pitch boards are typically 32 mil drills. <S> If this is a hand-made prototype, I'd probably be fine with those holes for signal and guard. <S> I would make the pad diameter for the ground pins much larger, maybe >100 mils, to help with ground plane soldering as well as heating up the big metal chunk of that connector. <S> ... <S> perhaps something related to the pre-tinning note (note 7) on the drawing? <S> I think this is a red herring. <S> Less than 2 mils isn't going to change much. <A> You can import the footprint into freecad using stepup and then check the dimensions in a powerful CAD environment (if you have access to freecad 0.17 then drawing dimensions workbench otherwise techdraw or export to dxf and then use librecad.) <S> More details: <S> https://forum.kicad.info/t/how-to-check-footprint-correctness/9279 <S> The tutorial <S> https://forum.kicad.info/t/tutorial-how-to-make-a-footprint-from-scratch/11092/4 has a section about how one can select the correct hole size if it is not given in the datasheet (or if you do not trust the datasheet) <A> The footprint I proposed in my original question worked without issue. <S> The part fit well in the PCB and soldered without shorting. <S> I had some trouble heating the connector body pins well enough for the solder to flow well, but I found that the trick was just to let the connector body heat up longer before soldering.
| I would go with the recommendation and not worry about it too much.
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2.4GHz WiFi module causing interference on 433MHz RF receiver module I have a prototype board which has two RF modules on it, a 2.4GHz Wifi module (ESP-01) and a low cost 433.92MHz RF receiver module (Wenshing RWS-371F-6).My 433MHz transmitter (a remote control) sends data encoded in manchester to be decoded by the receiver, this part is working well. The data rate is 2000 bits per second and I am using 433.92MHz SAW ressonator on the remote control PCB prototype. I have a frequency meter, and its transmission frequency is exactly at 433.9MHz constantly.When the ESP-01 module is not connected to the board, I can control the relay of the board from 50 meters distance with short presses on the buttons of the remote control, and it almost always work at this distance (to control the relay).When the ESP-01 module is connected to the board, the 433MHz RF performance is very poor. For example, from only 5 meters distance I have to do a long press on the button in order to control the relay, and sometimes it does not work. And with the ESP module connected to the board, It does never work doing short presses on the button like works when the ESP-01 module is not connected on the board... As can be seen on the photos, I have tested 433MHz module antenna with helicoidal and straight wire (17,3cm and 13cm), and with 13cm long wire the result is even worst. This board have 12V (100KHz - LNK626DG) power supply that is generated from the mains, then the 12V is reduced to 5V (7805) and 3.3V (LM1117IMP-3.3), two linear voltage regulators. The board has a ground plane also, shared by all them. The only thing connected to the 5V output is the 433MHz RF receiver. At the input of 5V regulator I have a 100nF ceramic, and at output 10uF ceramic 0805 16V, and at the two 5V supply pins of the 433MHz module I have a 100nF ceramic. The ESP-01 is being supplied by a regulated 3.3V (this power supply will be changed by a buck converter of 1.5MHz on the next prototype). At the output of this regulator I have one 22uF ceramic, and very close the the 3.3V pin of the ESP module I have 2x 22uF 0805 10V ceramic capacitors. In that board it is also possible to connect a bluetooth module like the HC-05 (bluetooth is also 2.4GHz) in place of ESP-01, and when the BT module is connected to the board and working, it does not cause interference on the 433MHz RF like the ESP module do, maybe due to the low-power RF of bluetooth. I have noticed that the 3.3V output is at one edge of the PCB and the ESP connector is at the other edge of the PCB... Do somebody know if is it possible in my board, to have a good performance on 433MHz RF and use the ESP module at the same time? Because in my board the ESP module seems to be causing interference on the 433MHz module's receiving capability. Why is this interference occuring? And what could be done to solve? Is it layout issue? Datasheet of the 433MHz receiver module: www.web66.com.tw/_file/C3/31776/Dfile/1402897191832file.pdf?t=2019091807 Regards. <Q> These have poor selectivity and so are prone to interference at other frequencies. <S> It also has no shielding, so a strong rf signal could 'break through' and be picked up by active components on the board. <S> The ESP-01's antenna is less than 10mm away from the 433MHz antenna, so the opportunity for interference is high. <S> First thing I would do is rotate the ESP-01 180° or mount it vertically, so its antenna is further away from the 433MHz receiver. <S> I would also consider using a higher-spec superheterodyne 433MHz receiver, oriented so its antenna is as far as possible from the ESP-01 and with shielding around it if not already shielded. <S> Your ground plane might be improved by moving it to the top of the board and routing more wires on bottom. <S> That way it could act as shield preventing rf from the ESP-01 from getting into traces going to other parts of the circuit. <A> As has been mentioned, regenerative receivers are notoriously terrible. <S> But even building something to receive OOK signals with a good sub--GHz data radio as Ti, SiLabs, etc sell, you still need to worry about local noise sources. <S> Decouple all power and signals between the radio and digital circuitry with low pass filters such as SMD ferrite beads. <S> If you have a debug serial cable especially decouple that. <S> Avoid using a switching power supply, if you must start with one follow it up with a linear regulator. <S> These are not theoretical guesses, but rather issues found in actually building a device of this type. <S> Conversely, with a reasonable protocol it is not all that likely that the wifi radio will be transmitting while you try to receive, nor <S> is that as likely (at least with a good sub GHz radio) to be an issue as one might guess. <A> In order to find root cause of interference, use the suggest methods given already to isolate Conducted noise with ferrite series chokes and Radiated clock noise with a dummy shield can over top to isolate coupling. <S> This could be a grounded <S> "can" put over a protective insulator just to attenuate radiated EMI such as from harmonics of the 26MHz square wave clock. <S> If it improves reception, then you know proximity and radiated EMI is the cause.
| Your 433MHz receiver module has a trimmer in the rf coil and no crystal, which suggests it is a superregenerative type.
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How to establish communication between microcontroller on a bread board and laptop using usb cable? I want to learn about microcontrollers and use them in some projects. But I do not want to buy an Arduino, although that is what would be easy for doing projects. This creates a problem in programming the microcontroller. Although a programmer can be used, I noticed that many microcontrollers have an SPI. I thought I could use this along with a USB to programming it serially. I may also read the data serially (using shift registers) into the computer. This way, I think, I can establish communication of some sort between the two. But I am not sure if my computer will allow me to write to a USB port without any identifiable device being connected. I am aware I would have to write some code to make this smooth since, both the microcontroller and the computer have different clock speeds. <Q> An Arduino does not have to be a pre-fabricated board - it can be as simple as a microcontroller IC and some support components. <S> An eg basic Arduino Nano clone from Asia for under $US3 incl shipping gives you a few basic ICs a voltage regulator, a USB to serial interface and some support firmware to allow bootloading from USB. <S> You gain nothing by not starting with a few parts on a board and the bootloader and free environment. <S> Once running you can pursue whatever purist directions that seem good. <S> Struggling with months to achieve what can be done in hours in setting up an Arduino will teach you much, but there are far far far better ways to use the same learning time. <S> I built my first microcontroller system on strip board about 45 years ago!!! <S> Based on a Nat Semi SC/MP. <S> NO O/S, Binary toggle switch program and data input. <S> (No keyboard or keypad initially). <S> LED I <S> /O (yes, we had LEDs :-) ). <S> No assembler (let alone an HLL). <S> Later we graduated to using Baudot code teleprinetrs - that we wrote our own code for. <S> So, yes, I know what you can learn the hard way. <S> Buy an Arduino to start :-). <S> You can learn far far far more quickly by starting with an Arduino and working DOWN once you have it running. <S> Arduino use in no way misses out on ANY computer architecture aspects. <S> You are doing yourself a massively large disservice by not starting with Arduino. <S> Arduino Bootloader: <S> What it does and why use it, & How to turn any appropriate AVR microcontroller IC into an Arduino. <S> An <S> Arduino-less Arduino here and Adding a bootloader with ICSP here and here <S> and here Stack Exchange discussion here <S> What it is and what it does here <A> USB and SPI are completely different and you cannot communicate between your computer and your microcontroller using SPI directly. <A> Answer is simple: Serial. <S> Get yourself a UART-USB converter like this , write some code in whatever language you please in order to use a COM port for communication. <S> Done. <S> Don't you worry about clock speeds and bit banging. <S> No need to reivent the wheel. <A> I noticed that many microcontrollers have an SPI. <S> I thought I could use this along with a USB to programming it serially. <S> These are wildly different protocols, and building your own adapter between them is definitely a project for after you've learned a lot about microcontrollers. <S> On the other hand, you can just buy one: <S> Totalphase <S> Aardvark Bus pirate <S> The other approach is to use the dedicated programming interface on the microcontroller, which may be "JTAG" or something like "ST-LINK" for "SWD" parts. <S> That lets you do all sorts of convenient things in a debugger, such as set breakpoints and inspect variables. <A> You need to avoid confusing Arduino the software environment and Arduino hardware. <S> There is nothing unique about the Arduino hardware and they are a very cost-effective way to obtain all the hardware for what you are requesting. <S> You don't have to build any hardware if you don't wish to. <S> I often use Arduino hardware (either Nano or Uno for ~3$) but write the program using AVR-GCC and program the processor using Avrdude and the Arduino boot-loader. <S> The software is available free (I use Homebrew to install on my Mac <S> but it also supports Windows and Linux). <S> The Arduino hardware incorporates a USB converter to allow serial communication with the host computer - just use printf() in your code and capture on the PC with your favorite terminal emulator (on a Mac I use "screen"). <S> I usually include an RTOS that I wrote with a command line interpreter in my software to allow debugging and control as well as feeding back data to the PC. <S> The fact that it is based on Arduino hardware in no way restricts what my software can do.
| However in most (to not say all) of the current microcontroller eval boards you will be provided with an UART-to-USB converter, and thus you will be able to communicate with your PC using an UART.
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Where do USB ports on laptops typically source their power when the laptop is charging? Suppose you are at a coffee shop and your phone battery is low. You decide to plug it in to your laptop via USB to charge it. At this point the laptop's battery serves as our energy source for charging the phone. But then you realize your laptop's battery is also low, so you plug your laptop in using the AC adapter it came with. So the coffee shop's electricity is charging the laptop battery. Now, do most charging circuits in laptops switch to source power (specifically on USB ports) from the AC adapter or not? Would the current draw from the AC adapter increase as a result of having the phone being plugged in, or would it just take longer to charge the laptop battery to full capacity? <Q> If the voltage at the bus connected to the battery is higher than the battery voltage (given by the State of Charge of the battery), current will flow into the battery, charging it. <S> In this case, current can obviously not flow from the battery at the same time. <S> Thus, the energy to supply the laptop has to come from the power supply brick attached to it. <S> This may change fast, as switch mode power supplies can regulate voltages fast. <S> It may go from charging to discharging the battery depending on cpu load, if the PSU is not able to supply the maximal consumption by the laptop, for instance. <S> When battery is discharging, the voltage of the battery is higher than the bus, even if there's other supplies connected the same bus. <A> do most charging circuits in laptops switch to source power <S> Yes, they all do the switch (specifically on USB ports) <S> Regarding USB ports, if a device (laptop) receives the power externally (via AC-DC adapter or else), it is called "high-powered device". <S> In this case a USB-compliant port must supply "at least 5 units of load", which would be 500 mA for USB 2.0 port, and 900 mA for USB 3 ports. <S> Would the current draw from the AC adapter increase as a result of having the phone being plugged in, or would it just take longer to charge the laptop battery to full capacity? <S> For most laptops there is a way to communicate the power capability of AC-DC adapter to the laptop, so the laptop knows its limit, and current won't increase beyond the adapter capability. <S> And, since you said your battery is "low", the adapter is likely already maxed out. <S> However, smart laptop will re-distribute <S> the internal power paths depending on design quality and selected power policy. <S> In some cases the system will limit CPU performance, so the battery might continue to charge at optimal rate. <S> But if your power policy is high-performance/presentation, the CPU will work at full power, and the battery might charge at slower rate. <S> It all depends... <A> Would the current draw from the AC adapter increase as a result of having the phone being plugged in, or would it just take longer to charge the laptop battery to full capacity? <S> As I understand it laptops will generally have an intelligent charge control system with a bypass path <S> *. <S> The charge controller will also monitor system power consumption so it can prevent overloading of the adapter. <S> So if the charge controller thinks the adapter has spare capacity and you increase system power consumption by plugging in your phone then the charge controller will just allow the extra current to be drawn from the power adapter. <S> On the other hand if the charge controller thinks the adapter is already maxed out then it will lower the charge current to the battery to keep overall input power levels within acceptable limits. <S> * The existence of this bypass path is why laptops can typically work without a battery present, unlike mobile phones.
| When the laptop is plugged in, power is routed from the AC adapter to the system via the bypass path and also routed from the mains input to the battery via the charge controller.
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Why are polarity assigned to resistors in the circuit in questions? I have confusion on questions that assign polarity to the resistors. Are these polarities arbitrary or are these polarities actually true? Do I have to consider these polarities while doing the circuit analysis? For example; if I consider drop of the potential as negative, I get this equation according to assigned polarity in the question. $$ 35-V_x-2V_x+V_0=0 $$ Also, by ohm's law, $$ V_x = 10*i $$ $$ V_0 = 5*i $$ This is my solution, but I don't know if it's correct or not. Basically, I have confusion on as to whether consider polarity as given by the question or not. Also, this one. Are these polarities provide only to confuse students? <Q> Assignment of polarities in typically introductory electronics questions is indeed possibly arbitrary. <S> You want to be able to answer a question with whatever sign convention the question or schematic pose. <S> However, when working with dependent loads/sources, the sign is important, as your work shows. <S> Working clockwise around the circuit using the passive sign convention, I get the same result for your scanned circuit. <S> No, the polarities aren't there just to be confusing. <S> They are there as a required sign convention and the results can be positive or negative, depending on the other circuit values. <S> You just need to do the math using that convention. <A> The polarities are important, and in this particular case the polarities are assigned specifically to see if you've been paying attention in class. <S> A more "normal" diagram would reverse the polarity on Vo. <S> That way, the three voltages other than the voltage source would sum to 35. <S> If you don't pay attention, you'll miss that and come up with a wrong result. <S> The intent is not to confuse students, but rather to make them entirely aware of what they are doing and what they are supposed to do. <S> Once you define the polarities it is straightforward to write the relevant equations in such a way as to preserve those polarities. <A> What the question is trying to teach you is that the polarity affects the sign in Kirchhoff’s voltage laws, which you have correctly identified. <S> Where you have made a mistake is in applying Ohm’s law. <S> source <S> So <S> \$v_0 = -5i\$ . <S> The voltage is flipped, since the polarity is the reverse of what you would normally expect. <S> If you flip the polarity, you get what you would normally expect $$v_1 = <S> -v_0 = 5i.$$
| Remember that the polarity of the resistor voltage is opposite the current flow.
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Is this DIN connector suited for MIDI devices? I'm having difficulties to find PCB mount "MIDI" connectors by a supplier like Farnell, no search with the keyword "MIDI" succeeds. I know a MIDI connector is a DIN5 connector, but there seems to be many variations of "DIN5". For example, do you think that a product like this is suited for standard MIDI cables (for electronic music devices)? The image looks ok but I wonder if the "angle" of each "line hole"/pin is not a little bit different than usual MIDI devices... Here is the datasheet . Here is a connector that I know working for MIDI, I wonder if the one I linked is similar. Edit : The angle of the "linear holes" seems to be very different from one item to another. Why? By the way, why is it a "linear hole" and not just a round hole like all other connectors? Linked to 5-pin DIN socket advertised as MIDI, not the same as on my MIDI gear but not a duplicate since here I have a more precise question about the angle of the linear holes (not present in the other question). <Q> While one might expect that the socket's contacts would be a round metal receptacle, there is another design that looks like a fork. <S> The two extensions of the fork make contact with the plug's pins by contacting the pin on either side. <S> This fork-contact has a flat profile, and during manufacture, it is inserted into the slot that you see as a "line". <S> The sides of the fork also flex inside this slot in order to provide a spring-loaded contact with the pin from the plug. <S> So these "lines" (slots), and the angle they are made at, have nothing to do with the pin arrangement in the socket. <S> Only the round hole where the pin actually enters is a factor in whether the "pins are in the right place". <A> Something may be adding to the confusion. <S> There are indeed 2 different types of 5-pin DIN connector. <S> In this image they are called 5-way 180º and 5-way 240º, and yes, there is a difference in angle between them. <S> 5-way 180º: <S> centerlines of the holes are across a 180º arc <S> 5-way 240º: <S> centerlines of the holes are across a 240º arc <S> The one you need for MIDI is the 5-way 180º. <S> As the other answers have indicated, the angle of the slit across each pole is not important. <A> If you look inside the socket you will see that the contact is a fork with two prongs. <S> They are angled to prevent interfering with each other. <S> The important point is that the circular pin entry points are in the required pattern, 3 o'clock, half-past four, 6, half-past seven and nine o'clock. <A> MIDI uses 5 pin 180 degree connector. <S> It seems that the datasheet you linked matches this specification, 45 degrees between pins.
| The datasheet has only one 5-pin connector and it does match up.
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Convert 1kHz PWM signal to 0-10V DC with minimal ripple For my application, I want to control the position of a valve which takes 0-10V input voltage. For the determination of the position I need to use a digitial 5V 1kHz PWM signal from an Arduino Nano. What is the best way to convert the 5V PWM signal to 0-10V DC? I have thought of a lowpass-filter and an OpAmp with gain 2, however, the DC voltage should change within <1.0s (better <0.5s) after a one-resolution-step of the PWM (which has 8-bit). On the other hand, the DC voltage should also be as stable as possible over larger times (>1s) once it reached its nominal value. I am not sure if both requirements can be met with a low-pass filter. The other option I thought of is a PWM-DAC and an OpAmp. What do I have to look for there? <Q> There are multiple ways to skin this cat. <S> I've used a double pole low pass filter set around 600Hz with the PWM set to 10kHz. <S> We achieve 0-10V with 10mV steps with that set up. <S> Or you can get a DAC chip that will do it as well. <S> The DAC chip will have fewer parts and better accuracy with simple parts, but will cost a little more $. <S> The filter method will require a few 0.1% resistors to make sure that you don't have offsets from board to board. <A> You could use a cheap 8-pin or 16-pin microcontroller with an internal DAC, driving a voltage-doubler op-amp circuit. <S> You may find a microcontroller with an internal oscillator. <S> The microcontroller can measure the PWM signal pulse width in software or using an internal timer. <S> Your overall software will be capable of allowing for the inaccuracies of the microcontroller clock. <S> The microcontroller can be a sub-£1 part cost <S> but there's a potentially large cost overhead to write and support the software, depending on your industry and situation. <S> If your application allows, you can use your Arduino's single output bit as a UART Transmit Data (TXD) instead of a PWM output and send the microcontroller a data byte value. <A>
| For minimum ripple, use a microcontroller with a decent timer/counter, time the PWM cycle and on-time, calculate the percentage, then use a (probably external) DAC to generate the 0-10V. Or bypass all that and just control the DAC directly with the Arduino.
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Rejection of particular frequency in a signal which contains many frequencies in its spectrum (fourier transform) How does a filter know which frequency is to reject? Detail: A signal may contain a lot of frequencies. How could we surely say that the result of linear combination of the result of applying these frequencies individually will produce the same result as applying the whole signal? At any point of time there will be a voltage level which will be applied to a circuit (linear combination of amplitude of contained frequencies.) Circuit only "sees" voltages, so how will it know that which frequency to reject? I mean the resultant signal amplitude values will be the value, after subtracting the value of rejected signal frequency amplitude from original input signal value at every time instant. <Q> I find the way you are phrasing your question a little confused <S> but I think I know what you are looking for. <S> I'll try to answer. <S> A circuit only sees (instantaneously) voltages - well yes, but there are elements on the circuit that are time-dependent. <S> If we are talking about an analogue circuit, this is L's and C's. <S> These elements store energy in one form or other, and release it with a rate which is variable depending on component value. <S> So now we have a circuit that has some "memory" characteristics. <S> It's instantaneous state is dependent on the sum of all previous states. <S> There are a number of ways of analysing this mathematically, but eventually we find that with certain circuit configurations we can arrange that the circuit's response varies according to the frequency content (Fourier spectrum) of the input. <S> (This is one form of analysis that works well for periodic signals - signals which are assumed to behave in the same way, over and over, through all of time. <S> In practice of course, nothing does this, but, say, a 1MHz waveform running for a second or two is close enough for the assumption to hold.) <S> Likewise a digital filter works by computing the instantaneous output based on some conmbination of previous (sampled) inputs. <S> Now the memory element is really memory, used to store some number of samples in a rolling buffer. <S> In the end the important fact is that circuit behaviour has a time (and therefor frequency) dependent element. <A> Circuits see more than voltages. <S> They can well remember something of the past voltage values of a signal and form current output with some math rule from current and past input voltage values. <S> That can attenuate different simultaneous sinewaves differently. <S> An example: A filter contains a 1 millisecond delay line and a summing circuit which creates the output by adding together the filter input and the 1ms delayed input signal. <S> The result: <S> 500 Hz sinewave or 500 Hz component of more complex signal dies totally, but DC and 1 kHz components get doubled. <S> A filter applies the same math rule to every input signal. <S> That rule is called "transfer function". <S> But we have taken a habit to present signals as a sum of simultaneous sinewaves. <S> As well we build filter circuits which do not smash signals in a too complex way <S> , our filters are built to retain the frequencies of the simultaneously inputted sinewaves. <S> We say our filtering circuits are linear. <S> They cause only amplitude and phase changes to the sinewaves. <S> The filter itself knows nothing of how we describe its behaviour. <S> As said, it applies the same math rule for every input. <A> But in the case of a signal we are applying voltages and signal is not periodic. <S> Two things. <S> First, periodic and continuous are not the same thing. <S> By definition, a frequency is periodic. <S> Also, the frequency components of a transient are periodic. <S> The characteristics of an RLC circuit vary as a function of the frequency of the signal going through it. <S> If the signal is a combination of two frequencies, the circuit appears to the signal have two different impedances or response curves, one for each frequency. <S> Second, for a long time, transient events were considered to be aperiodic - not having a frequency or frequencies. <S> The genius of Fourier's work was that he showed that transients are composed of an infinite series of frequencies, and it is the combinations of those frequencies that makes the signal die out or stay continuous. <S> This means that something that appears on a scope to be "not periodic", actually is from the point of view of the filter circuit. <S> Fourier's initial work was a mathematical model of how heat spreads out into a cold mass when it comes in contact with a hot point. <S> Doesn't get more aperiodic than that. <A> Both filtering and frequency are concepts that to be useful require time. <S> (A single point is an impulse that contains infinite frequencies, you can only remove them all by zeroing the point, or increase them all by boosting that point. <S> Not very useful.) <S> A realistic LTI filter has an impulse response. <S> It looks over a period of time to try and match the "shape" of a certain frequency or combination of spectral frequency bands and then afterwards produces some behavior. <S> (e.g. block one and allow the other). <S> The filter's result will modify a waveform over a period of time, as modifying just a single point will only produce a "glitch", or a startup transient. <S> And that result will look different to another filter that is long enough to detect the specific frequencies that were modified. <S> So, although a digital filter will modify signals point by point, you can only describe what they are doing after letting them run for a bunch of points, sometimes a very very large bunch of points for very narrow frequency band changes. <S> (assuming non-zero noise or any finite quantization)
| The longer the filter looks at the data, the better it can tell two spectral bands apart and perhaps treat them separately
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Best way to amplify 3.3V square wave at 1MHz to 10V square wave at 1MHz I am trying to build a bus transmitter complying to 1553 standards. Its input is a 3.3V square wave at 1MHz (CMOS data) and it has to be amplified to 10V or higher at the same frequency. The load will be a center tapped isolation transformer whose primary side inductance is around 5mH. My current implementation involves an LT1210 current feedback amplifier which is able to amplify square wave but the catch is high current drawn from the sources (500mA) which leads to high power consumption as the datasheet suggests. Following is my setup: Right now to the left of the 10 ohm resistor I get 8.5V peak to peak. What I need to know is if I am on the right track here. Are there any other operational amplifiers available that could amplify large square wave signal and use less current at the same time, or are there any modifications that I could make on my circuit to draw less current and achieve same levels of voltage? (I clearly don't seem to need 500mA.) P.S. This is a part of a project which is a 1553 bus transceiver analog front end. HI-1579 chip is very close to what I am intending to build. <Q> The signal will get inverted. <S> If this is a probem just feed it into another inverter (there is a total of 6 in the IC package). <S> Of course it can't provide the current you need. <S> That could be accomplished by a discrete CMOS inverter stage (discrete P-MOSFET/N-MOSFET-pair). <S> In addition if you use two drivers, one at each terminal of the inductor (= H-bridge configuration), you don't need a center tapped inductor to create pulses of both polarities and you don't need a negative supply rail. <S> Depending on which terminal is high and which is low you can provide the appropiate current direction. <S> See circuit diagram below: simulate this circuit – <S> Schematic created using <S> CircuitLab The inverter should be a 40106 operated at 15V. <S> Alternatively: There are also more specialized and still cheap level converter/driver ICs like the MC34152 , <S> DS0026 or ICL7667 <S> which could also fit for your application. <S> I think they all can drive up to 1.5A. <S> In any case I recommend to build and test it first on a prototype board before manufacturing a PCB. <S> That will be much more meaningful than any SPICE tests. <A> I fear you have misunderstood how op amps work. <S> You have overlooked the fact that zero times any gain remains zero. <S> That is, the portion of the input waveform which has a zero value will produce a zero output. <S> As a result, the op amp (which has a nominal gain of 4) will attempt to produce an output from zero to 13.2 volts. <S> Since Vdd is only 10 volts, the most you could possibly get is a 0 to 10 volt output, rather than a 0 to 13.2. <S> However, this op amp is not what is called a rail-to-rail op amp. <S> That is, its output will not reach Vdd or Vss. <S> Instead, if you look at the data sheet for "VOUT", "Maximum Output Voltage Swing", you will see that with VDD/VSS of <S> +/- <S> 15 volts, the op amp is only guaranteed to put out +/- <S> 10 volts. <S> The output swing is not specified at +/- <S> 10 volts, but something on the order of +/- <S> 5 to 7 volts is probably a reasonable projection. <S> In the event, you're getting 8.5 volts, so that's better than you deserve. <S> What you need to do is produce an offset, which will produce a 0 output with 1.65 volts in. <S> At the same time, you'll need to increase the gain from its current value of 4 (1 + 845/247) to at least a gain of 6. <S> You can try something like this: <S> simulate this circuit – <S> Schematic created using CircuitLab Note that the new op amp must be able to output about +/- <S> 10 mA at much greater than 1 MHz (why?) <S> in order not to distort the waveform, and another LT1210 would seem like an excellent candidate. <S> Furthermore, the output will now swing from about + 8.5 to - 8.5 unless you increase VDD and VSS, for exactly the same reason it is now only swinging to +8.5. <S> With this op amp and these power supplies, you simply cannot expect (or even hope for) a greater output swing. <A> I would look at using a low-side gate driver IC, which will give you level shifting and an output power stage capable of operating at 10V or more. <S> Also, they are inexpensive. <S> For example, the Microchip (nee Microsil) MIC44F18/19/20 will drive 100mA with a couple hundred mV drop, and typically draws less than 10mA idling at 1MHz.
| The cheapest way would probably be using a CMOS hex inverter 40106 which can be operated at up to 15V and biasing the input near below the switching threshold using a voltage divider and either AC-coupling the input signal via capacitor or DC-coupling through another resistor.
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LTspice, plotting a node voltage problem Edit: The question is answered in a such way that I am happy now, however for the interested ones, the mystery of the question is still on. These two are the parts from my circuit, related to the question. Now, I want to be able to plot V(pnp_c) , however it does not allow me to choose it. You can see from the above list that there is not such quantity as V(pnp_c) . If I remove -9v label, then it allows me to select it. You may say that "they are essentialy the same point, they are at equal potentials, why bother?". My answer to that is, while simulating the power of the pnp by ALT + left click , Ltspice formulates it so that it will depend the value of -9v label as follows: Please note that npn power expression is independent from V(+9v) instead it is related to npn_c which is nice. The problem starts here, If I want to change the power rail to another value, by directly connecting a voltage source, or by changing it with another label, then I also have to replot the power on the pnp. However, if it were be using V(pnp_c) instead of V(-9v) then this problem would have been eliminated. I wanted to think that it just don't accept two nodes with different names and it chooses the constant voltage one, however as you can see from the above picture, for npn, the arrangement is the same but it does not create a problem and recognises npn_c as a node. Things get more confusing at that point. So, why I am able to choose npn_c as a node but cannot choose pnp_c ? <Q> Like Michael Karas already answered and LTspice also states in its help file: <S> Each node in the circuit requires a unique name. <S> Still, playing around with labels on the same node seems to reveal that LTspice prefers the bottom labels above the top labels in a schematic (first rule) <S> next, LTspice prefers right side labels above left side labels (second rule) <S> LTspice picks the label that fits above rules best and uses that name as node name order of placement or alphanumerical order of naming is not relevant (what OP also already discovered) <S> you can still use the other (less preferred) label name to refer to its node (in the waveform viewer as well as in behavioral sources, etc) <S> So, in your case -9v is lower than pnp_c , so <S> -9v 'wins' becoming the final node name. <S> And npn_c is lower than +9v (first rule <S> , despite it is more to the right side (second rule)) , so npn_c 'wins' <S> becoming the final node name. <S> The ugly way to solve your issue is swapping -9v <S> and pnp_c in vertical order. <S> A better way is not connecting several labels to the same node, but insert a component between the labels instead. <S> I found that inserting a jumper will not show 2 different node names (despite the description suggests). <S> Inserting an e.g. 1mΩ resistor between the labels -9v <S> and pnp_c would work (same applies to other labels on the same net). <A> It may very well be due to the fact that you have labeled nodes with two different labels. <S> LTSpice probably refers internally to one of the two labels with priority. <S> It is hard to predict what determines the priority but it could be the order in which the labels were created. <S> There is a clean solution to this problem. <S> Do not use multiple names per node <A> I have posted a bug report to the LTSpice folks via the email link in the About Box on the Help Menu. <S> Below is the text of the mail sent. <S> I will try to update this when a response comes back from the folks at www.analog.com (formerly of www.linear.com). <S> From: mkaras@carousel-design.com <S> To: LTspice@analog.com <S> JUMPER Component <S> Not Working <S> Please reference this question/answer thread on the Electronics Stack Exchange Site: LTspice, plotting a node voltage problem <S> It turns out that the JUMPER component does not work correctly. <S> In fact placing the component into a schematic is no different than using a direct wire to connect two differently named nodes. <S> Please correct this as there is a true need for a JUMPER component. <S> In real life when a jumper (i.e. two header pins on a circuit board connected together through a push on shunt) is used it represents a small series resistance to the circuit. <S> Because of this it would be an acceptable fix if the JUMPER component was modeled as a very small valued resistor such as 0.001 ohms. <S> A separate model of JUMPER that had a control parameter to selectively make the jumper look open or closed would be handy as well. <S> I know that such behavior can already be modeled with the SWITCH component but the proposed controlled JUMPER could be simpler without the need to specify the open and closed impedances. <S> Michael Karas Carousel Design Solutions
| When you need to interconnect two nodes that have separate names then use the JUMPER component that can be found in the [misc] folder of the component symbol selector.
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Is EMC shielding of a grounding wire required? Consider the following 2 use cases: Fully electric vehicle with 2 different insulated voltage systems: 1 high voltage @1000V DC insulated from chassis and 24VDC system with minus pole (-) at chassis. Would PE cable (connected to chassis) need to be shielded? If yes, then why? Single phase house supply system. Would the PE cable need to be shielded? If yes, then why? In my opinion there is no need for shielding as these cables' sole purpose is to protect the people from indirect contact should any live part came into contact with metal exposed parts (that can be touched by people). In short, these conductors are not meant to carry any signal/current (if not for brief periods < 1s), I don't see any reason for EM shielding. Short clarification: In question 1, the the chassis is ground (as in "return path") only for 24V system while metal parts of chassis of high power loads are connected to the chassis with a PE cable (as in "Protective Earth"). In question 2, consider a TT system, the PE cable is grounded. <Q> The problem with shielding any sort of safety grounding is what do you shield it with? <S> I mean, you're out of groundy things. <S> You have vehicle chassis or your local earthing rods, where do you get a better ground than that? <S> You don't. <S> If your PE shield needs shielding, then what you really need is for the PE shield to be a better one than it is. <A> I see no reason to shield the grounding (return) that is connected to chassis. <S> As standard design practice, we twist the supply (hot) line with its return. <S> If there is a need for an overall shield around that twisted pair bundle, then the shield is connected to chassis at both ends. <S> The only reason for the shield is if you were to have some conducted emissions (CEs) flowing as common mode currents on your twisted pair supply+return cable, that would then become radiated emissions (REs) from an unshielded cable. <A> Fully electric vehicle with 2 different insulated voltage systems: <S> 1 high voltage @1000V DC insulated from chassis and 24VDC system with minus pole (-) at chassis. <S> Would PE cable (connected to chassis) need to be shielded? <S> If yes, then why? <S> Since the ground is tied to the chassis, the cable and chassis can be considered as being the same potential. <S> The chassis also acts like a shield because the impedance between cable and chassis will be very low and short out almost all emissions from high frequency electric fields inside of the vehicle. <S> Single phase house supply system. <S> Would the PE cable need to be shielded? <S> If yes, then why? <S> There is very little radiation emitted from grounding systems in a house.
| The reason for shielding cables is to prevent electromagnetic radiation from leaving or entering the cable. Not ever, grounding systems in houses are very low impedance and can be considered as ground and being at earth potential.
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Illuminated LED and push-button using only 2 wires I want to make a circuit with an illuminated LED push-button that is far away from the input board and I only have 2 wires available. The push-button is not something specific. The input board will have many buttons, but not all buttons will have LEDs. The input board will probably be a MCP23017 16-bit expander with an MCU. I came up with this circuit: simulate this circuit – Schematic created using CircuitLab The MCP23017 is powered with 5V. When the push-button is not pressed, the voltage at the MCP input pin is about 1V, low enough to register a 0 on the digital input. When the push-button is pressed, the voltage is 5.6V without the added 5.1V zener diode D3, and 5 volts with the zener in place - well above 0.8 * 5V VDD required by MCP23017. But because the circuit is just a voltage divider 220+220 ohm resistors over the 12V supply, it will draw ~28mA, resulting in a bit of heat from the resistors. Because the push-buttons will not be pressed for more than a few seconds, I can live with this. D2 LED was actually an opto-coupler to separate the long wire from the digital input, but I gave up on this protection for now. I searched a lot on the internet for other ideas about how to power an LED and have a button using only 2 wires and no complex circuitry, but could not find any relevant info. I built this circuit and it works, but one thing I cannot test is the influence of mains wires that run alongside my long button wires. They are separated and individually protected, but parallel to each other. Do you have any other suggestions on how to use a button and an LED on 2 wires? Later edit: The buttons can only be normally open. They are European type light push buttons available only NO. LED is added by me.Led should be lit all the time when the push-button is open. When the button closes there is no problem if the led tuns off or remains on. In this particular case, the led is used for night time finding only. One other idea I had recently was to use input-output feature of the IO expander chip. Using the output to turn on the led, and then switch to input to check for the push-button state. Doing this fast enough not to see a flicker. But I haven't drawn any circuit yet. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> A simpler circuit using two GPIO pins. <S> How it works: <S> IO2 toggles between high and low. <S> IO1 toggles between input and output high. <S> Table 1. <S> State Button <S> IO1 IO2 <S> LED <S> 1 Open <S> Out high Out lo ON <S> 2 Open Input <S> Out hi <S> Off IO1 reads low due to pull-down R2. <S> 3 <S> Pressed Out high Out lo <S> Off LED shorted out. <S> 4 <S> Pressed Input <S> Out hi <S> Off IO1 pulled high by IO2. <S> Light the LED when IO1 is high and IO2 is low. <S> Read the switch when IO2 is high and IO1 is input. <A> If you still want isolation, you just need a source of low-voltage AC power: simulate this circuit – <S> Schematic created using CircuitLab <S> Each switch controls the LED at the other end of the long 2-wire cable. <S> D2 and D4 are only necessary if the peak voltage out of the transformer exceeds the LED reverse voltage rating. <A> I'm planning to do the same thing as you. <S> (Use EU Push buttons with LED backlight added by me and read them with Multiplexer PCF8574 -> <S> MCU ) <S> I've found this solution on hackaday: <S> HariFun#133 <S> More I/ <S> O Madness <S> and I'm planning to use the same approach to solve the task. <S> It is implementation of your idea: Using the output to turn on the led, and then switch to input to check for the push-button state. <S> In the example code he's using bitbanging to read the inputs, but PCF8574 has also interrupt pin, so it's possible to go in that direction.
| Either LED could be part of an optoisolator.
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How do I read the state of AC switch, through a microcrontroller? I wanted to read the state of an AC switch using a microcontroller.S1 is an SPST switch and cannot be changed. What circuit will be required for doing this? I'm using this for my home automation project where I should have both manual and wireless control. This will also help me find the state of the load (on/off.) <Q> Welcome to the site! <S> If you really don't need to know the state of the physical switch, but instead know if there is current flowing through the switch, you could use a current sensor to measure that. <S> Something like an Allegro ACS724 . <S> As you are tying into high voltages, please be smart. <A> One way It would solve is if I also have Neutral going into my ? <S> Block <S> then I could use an AC optoisolator to see if the switch is closed or not. <A> That's something you don't generally want to DIY. <S> You are likely to create a hazardous situation. <S> Something like this . <S> You'd wire the physical switch to a circuit something like this: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The MCU could monitor that pin as an input to find out the state of the switch. <S> And if it wanted to, it could also dictate the state of that pin with as output. <S> That's just one possibility. <S> Deciding on a topology that satisfies a set of desired outcomes about precedence and what not is what designing something like this is all about, and that's where you come in. <S> UDPATE <S> #1 <S> As per the update to the OP's question, same idea but two separate connections to the microcontroller. <S> simulate this circuit <A> Get a 110V (or whatever voltage you're working with) relay and connect the coil from S1 to neutral. <S> Then connect C to uP ground, and NO to a uP input with a pullup. <S> When S1 is closed, that input will go low. <S> (If you want the opposite polarity, use NC instead of NO.) <A> If you just want to know whether the switch is in on/off state, you can try recreating the following circuit :- Source <S> The optocoupler mentioned here is PC817. <S> You can also use H11AA1 instead of PC817. <S> Whenever AC voltage is present at the input of optocoupler, output of it would be LOW and microcontroller will receive logic 0 as input. <S> And when AC voltage is absent at the input of optocoupler, output of it would be HIGH and microcontroller will receive logic 1 as input. <S> Here the H11AA1 is used.
| One way to do it relatively safely, assuming you have a way to separately power a microcontroller, is to introduce a solid state relay.
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Permanent circuit board jumper - I have a RTC chip that kills soldered battery before clock set If I wave solder in the battery, it begins draining rapidly until the RTC chip (clock) is set. So I installed a jumper. Now I manufacture without the jumper, and then in the bed of nails testing, I set the clock and install the jumper before I package it and put it in stock. I never EVER want the jumper removed, but it is happening. Is there a way to make a lockable jumper or other ideas for a solution? <Q> Only apply a solder bridge when required. <S> There are many land pattern designs out there if you google them. <A> Could you wave solder in a battery holder instead, and install the battery later in your manufacturing process? <S> Here's an app note from TI about this sort of thing using their parts: https://e2e.ti.com/blogs_/b/powerhouse/archive/2014/06/05/maximize-shelf-life-with-a-one-time-push-button-switch <A> More ideas: conductive paint dabbed onto a SMD resistor footprint use a simple graphite pencil to shorten two pads <S> (your RTC doesn't draw enough current, hopefully, to make these couple 100 Ω a problem) make through holes in the right distance. <S> Use an office stapler to close the circuit big fat REMOVING <S> THIS JUMPER DAMAGES THE DEVICE AND IS DETECTABLE silkscreen or neon sticker through-hole rivet <A> You could add an inexpensive SOT-23 or SC-70 MOSFET and a (say) 10M resistor and put the jumper in place to turn the RTC off. <S> There would be a drain of a few hundred nA, probably less than the battery self-discharge, until the jumper is removed. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If you can find a wire wrap gun somewhere, you can make a permanent no-solder jumper (assuming the current jumper is a standard 0.1" square pin). <A> The RTC datasheet says that after it has been totally unpowered, the device only needs the main supply shortly after connecting battery. <S> Apply the main power to board just after manufacturing so it stops consuming battery, even if it is being tested later.
| You could have a solder jumper land pattern that does not have solder on it normally (opposite side of board from wave soldering).
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Massive ultracapacitor banks, why don't they work? (400V 3.38F) This might be a stupid question, and I know that for high energy, quick discharge applications (like a railgun for example) supercapacitors don't really work, but why? If I were to take hundreds of ultracapacitors (let's say they are 2.7V 500F individually), and then hook them up in series to get a 400V 3.378F capacitor bank, shouldn't that work better than a 400V 63300µF aluminum electrolytic capacitor bank? <Q> There are a couple of reasons. <S> First is energy, <S> E = 0.5*C*V^2 <S> so capacitance is not as important as voltage in this regard. <S> Coil guns need an incredibly quick pulse of current to work effectively. <S> Think of this in terms of a time constant, RC. <S> Also, think of the ESR of a capacitor. <S> If you have hundreds of caps in series, that ESR will be through the roof compared to a bunch in parallel. <S> This will also increase your time constant. <S> Super caps are nice, don't get me wrong, they just have their own niche uses. <A> Let's do some math. <S> 500F 2.7V <S> supercap :- ESR = 1mΩ. <S> x 148 in series = 3.38F @ 400V, and 1m&ohm <S> ; x 148 = 148mΩ 4700uF 400V electrolytic capacitor :- ESR = 18mΩ. <S> x 13 <S> in parallel = <S> 61100uF @ 400V, and 18mΩ / 13 = 1.4mΩ <S> The 13 capacitor Al electrolytic bank has 105 times lower ESR than the 148 capacitor supercap bank. <A> This might be a stupid question, and I know that for high energy, quick discharge applications (like a railgun for example) supercapacitors don't really work, but why? <S> The difference in capacitors is the ESR (Equivalent Series Resistance), you can look up this value in most capacitor datasheets. <S> The series resistance is from the total physical resistance of the terminals and electrolyte, the resistance creates loss and a voltage drop internal to the capacitor. <S> Electrolytic capacitors have higher ESR in the 10mΩ to 100mΩ range. <S> So even if you were to create a bank of regular electrolytic capacitors, it would have a large internal resistance and this would make it less useful. <S> To make an equivalent bank from regular capactors would take much more volume and weight. <S> Also this: A supercapacitor is also known as ultracapacitor or double-layer capacitor. <S> A supercapacitor tends to differ from an ordinary capacitor due to its very high capacitance <S> Source: https://www.spscap.com/difference-capacitor-supercapacitor.html
| Super\ultra capactors have very low ESR in the mΩ and very high capacitance. If you have a larger capacitance, the bank will discharge slower and the field will be present longer which (if the projectile passes the coil) will actually slow it down.
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Why in Digital Design only Capacitive load? Why in designing of CMOS logic we only consider the capacitive load of fanout? There is some wire resistance. So, why don't we consider resistive load? <Q> When the wiring resistance becomes significant, then we do include it. <S> As long as the wiring resistance is much, much less than the \$R_{DS}\$ of the MOSFET then it can be ignored. <A> Wiring resistance is usually less than 1Ω and most wiring resistance being in the mΩ range. <S> With many digital logic families maxing out at 50mA, if you were to calculate the equivalent resistance of most push pull GPIO's, the source resistance would be somewhere in the range of 100Ω to 50Ω at DC. <S> Because of 50Ω is much bigger than mΩ's the wiring resistance can be neglected in allmost all cases. <S> However, if you have long wires or small wires that make the wiring resistance more than ~1Ω <S> then you should consider wiring resistance in your calculation. <A> We do. <S> Internal to the CMOS chips, the resistance of the deep-sub-micron metal layer wiring contributes significantly to the RC delays even a small distance across an SOC (less than 1mm) in or between logic blocks. <S> That analysis is a large part of timing closure to meet clock rate specs. <S> In PCB design, another consideration is inductance: mutual inductance can be a cause of crosstalk between long traces, reducing signal integrity. <S> Trace inductance can also contribute to ringing of steep transients, causing signal under/overshoot. <A> Why C and not R, what about Zo = <S> sqrt (L/C)? <S> Attenuation occurs when the source impedance is not zero or the load is not infinity. <S> Although 5V logic is around 50 Ohms is considered low relative to the expected gigaohm gate input resistance so we can neglect trace resistance as well as load resistance. <S> 5V (1Gohm)/(1 GOhm + 50 ohm) <S> = <S> 5V as a voltage divider on R <S> there is no attenuation , so we neglect load R. Impedance of CMOS transmission signals do matter when the 2~3ns rise time reduces to make reflection echos visible in the echos from mismatched Z load reflections. <S> These delays show as 35% typ overshoot & decayed ringing unless terminated with matched impedance. <S> However increased risetime and for longer traces , ringing can degrade signal quality. <S> Typical trace width, <S> W with a height, H over ground planes and trace length , L has an impedance based on ratios of these parameters. <S> When typical PCB FR4 trace geometry has a ratio H/W=2, Zo <S> = 83 <S> Ohms, L < 5 <S> nH/cm , C ~ <S> 0.7 pF <S> /cm with a propagation speed, v = < <S> 18 cm/ <S> ns <S> (Ref Saturn PCB.exe) <S> Rise time, Tr is defined as the 0 to 64% Vmax by a standard T=RC load capacitance (e.g.15pF) and the logic family standard source resistance. <S> For 5.5V logic this is 50 Ohms +/-50% and for 3.6V logic this is 25 Ohms <S> +/-50% typically. <S> Lowering the gate + supply voltage increases <S> RdsOn so 3.6V logic was designed to use a lower Vgs(th) and lower RdsOn to improve rise times <S> but then cannot be used at 5V as it would go into shoot-thru or cross-conduction with both hi & lo side drivers active. <S> Tr = RC is the rated at 15pF but your design may be more or less depending on trace width over ground planes, trace length , fan out of gate loads and logic family. <S> Ringing occurs when the prop delay approaches the rise time the mismatched load reflection causes an observable change with oscillation. <S> When the signals go on cables, we are either limiting the data rate and slew rate (e.g. RS-232) <S> or we terminate the cable with a load resistance matched to the cable ( e.g. 120 Ohms for ribbon or 50 typ. <S> for coax.) <A> CMOS logic input draws no DC current, so there is not much resistance or resistive load to consider when a CMOS output drives CMOS inputs. <S> The speed how fast a signal can transition between states depends mainly on output current drive strength and capacitances it needs to drive. <S> PCB wiring has very little resistance in comparison, though wiring does add capacitance easily. <A> Resistance is included, in the sense that the output voltage level of a CMOS device is specified at a given current. <S> From these two numbers, the output resistance of the device can be estimated, and from that you can assess the effect of an ohm or two of wiring resistance. <S> Here's an example from a 74C04 data sheet. <S> Note the Voh is spec'd to be at least 2.4 V at a current of -360 uA.You'll find a similar set of specs on almost any digital device. <S> It's a dynamic, switching, or AC specification. <S> This spec impacts how fast a circuit can switch (though it's not the only thing), and also impacts the dynamic power dissipation of the device.
| The capacitance has to do with how fast a device can change the voltage on a node.
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Circuit to limit a set of outputs so that only one is on Background: I have a set of irrigation control valve solenoids (24V ac) that I control via a set of mechanical relays with the GPIO pins of a microcontroller (ESP32). The power supply does not have a large enough current to operate two valves at once, which is fine because the system also does not have enough water pressure to run two zones at once. In theory , I should be able to limit the system to one valve at a time strictly in software, but I keep finding corner cases where I have not ensured that, and when I mess up the supply voltage to the IC drops below 3v and the system reboots, so debugging is a pain. Question: Is there a reasonably simple circuit I could construct that would prevent a second pin from going high if another one is already on? Bonus points if there is an output that I can use to trigger an interrupt when this happens. UPDATE: Several people have asked about a software solution. Yes, I did refactor my code so that pins were set in exactly one place. I missed one piece of code in the refactor. sigh I also had an interesting bug where an internal process still wrote to an external pin when writing to flash memory. The ESP32's GPIO matrix makes it pretty hard to ensure that a library never accidentally takes over a pin. So while I agree that the software solution is ideal, my lived experence is that I would benefit from some hardware fault protection here. <Q> If your relays have both NO and NC contacts, you could chain the power to the solenoids through the NC contacts of the previous relay. <S> Any time a relay activates, any relays further downstream do not receive power and cannot feed their solenoid valves. <A> This has the added advantage of needing fewer I/O pins. <S> eg. <S> 74HC154 <S> (active low outputs) or CD74HC4514 (active high outputs) 4 to 16 line decoders. <S> This won't stop errant software from trying to operate a valve when it is supposed to be doing something else. <S> As always, you should write code defensively. <S> Wrap your I <S> /O accesses in functions that don't let bad signals through, but instead return an error code. <S> That way you don't need a hardware interrupt to tell you something is wrong. <A> If you really want to hardware solution that disallows the uc from ever having both solenoids on you could rig up a simple inverter/enable circuit with some jellybean logic gates. <S> Could use a SN74HC08A for the AND gates and a SN74HC14N for the schmitt trigger inverter. <S> Then in software use the output enable pin to turn <S> allow the solenoid selection to work. <S> No matter what you do in software both solenoids cannot be on because the same selection signal is always inverted for the other. <A> A single output rule needs to be defined as a spec ... <S> It could be a "Priority Encoder" or a Contention Arbitrator in software when schedules selected by user collide with some round robin token ring that advances every xx minutes among the shared stack of contenders. <S> This would best be all be done in software. <S> bitcount(n) <S> : count = 0 <S> while n <S> > 0: <S> count = count + 1 <S> n = n & (n-1) <S> return count OR compute output <S> Hamming Weight and correct error <S> if not 0 or 1 <S> (if weight>1 <S> then fault recovery) <S> - Plan B <S> Cheap and dirty H/W IRQ <S> if more than 2 drivers are active assuming active high out and Vdd/2 Threshold. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Using direct output ports with a 10k Resistor array and comparator choose Vref just below when 2 outputs are high "1" V(2)=2/(6+2) <S> * <S> Vdd <S> so Vref < Vdd/4 <A> Software solution <S> A simple code solution would be to use a switch construction to write specific patterns to the output port. <S> // <S> There are 8 valves on port A.switch(valveSelect){ case 1: outputs = <S> b00000001; <S> break; case 2: outputs = <S> b00000010; break; case 3: <S> outputs = b00000100; break; case 4: <S> outputs = b00001000 <S> ; break; case 5: <S> outputs = b00010000 <S> ; break; case 6: outputs = b00100000; <S> break; case 7: outputs = b01000000; break; case 8: <S> outputs = b10000000 <S> ; break; default: outputs = b00000000;}writeOut portA, outputs; Now <S> your code can only output those values - provided you don't write to port A anywhere else. <S> Hardware solution simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Using relay changeover contacts to allow one device on only. <S> The electro-mechanical solution is to wire your relay contacts as shown in Figure 1. <S> The priority is top-down. <S> For any solenoid to be powered all the relays above it must be deenergised and all those below it have no power feed to the contacts. <A> Frame challenge here. <S> You need to simplify your code to make this not possible. <S> For instance, I suspect your "corner cases" are a result of spaghetti code slathered all over the place. <S> Have precisely one subroutine that turns a line on (or all off). <S> That will be the ONLY place in your code that accesses the valve hardware. <S> (you can do a simple find/grep to assure the hardware is mentioned only once.) <S> The one subroutine is kept VERY simple. <S> It receives a value from 0 to the number of valves. <S> (i.e. 0-6 for 6 valves). <S> 0 means off. <S> Loop through each valve, turning every one off. <S> Run down a 10 second timer. <S> Then, loop through each valve that exists, starting with 1. <S> If it's the requested valve, turn it on; otherwise turn it off. <S> Since no valve is numbered 0, that means calling this with a value of 0 turns them all off. <S> That is the only subroutine in the code that actually operates valves. <S> However, you should really re-assess <S> your power supply situation so the PC doesn't reboot every time this happens. <S> Is it a long cable run from the DC power supply to the valves? <S> Does the computer sit at the valves and tap the same power supply cable? <S> Most likely, voltage on that cable is dipping because of voltage drop. <S> Run a separate power cable just for the computer <S> so it's not affected by voltage drop on the valve power cable.
| If you only ever need to operate one relay at a time then you could send the valve number as a binary code, and use a binary to 1 of x decoder to operate the relays. In reality though, this sounds like it should be easily fixable in software by ensuring that the outputs can only be changed by one function.
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Why is inverting op-amp behaving like a low pass filter? can anyone explain why this circuit behaves like a low pass filter despite having no capacitor in the feedback loop? <Q> can anyone explain why this circuit behaves like a low pass filter despite having no capacitor in the feedback loop? <S> If there is 2 pF <S> present then it forms a low pass filter with the 400 kohm feedback resistor having a 3 dB point of about 200 kHz. <S> This problem would be improved by choosing an input resistor of 1 kohm and a feedback resistor of 40 kohm. <S> However, the 741 is pretty crappy so you will be unlikely to get a ten times improvement in frequency response. <S> Inside (virtually) every op-amp are compensation capacitors; these limit the frequency response in order to maintain closed-loop stability. <S> As Tony Stewart mentioned in his comment, this has a significant effect on frequency response when higher closed-loop gain is required. <S> See the picture below for the closed loop gain (40 dB) of a 741 and note that your circuit has a closed loop gain of 32 dB: <S> - So, the closed loop response starts to fall from 40 dB gain at around 50 kHz. <S> With your circuit (gain of 32 dB) you might see the fall in gain around 70 kHz. <S> Picture taken from this useful document entitled "Frequency response of op-amps". <S> See also my answer here on a very related subject. <A> basically, every physical system in this world has a finite bandwidth and hence exposes some low-pass behaviour, so in the larger scheme of things, this is to be expected <S> You use an ancient (LM)741 series opamp. <S> These things have very modest bandwidths – in fact, the current Texas Instruments <S> LM741 datasheet only claims <S> 0.5V/ms as slew rate, and doesn't specify a bandwidth for most device variants, at all. <S> You simply use a model of a very low-passy opamp. <A> Other answers are correct, but they miss the biggest point. <S> Operating a 741 with a single 5 volt power supply is so far beyond the requirements of the IC that I'm a bit surprised that you get anything out at all. <S> For that matter, it would not surprise me if most of the output is simply due to the resistors connecting the input to the output. <S> The 741 is only specified for a power supply range of +/- <S> 5 <S> to +/- <S> 15 volts. <S> Not only are you violating power supply level requirements, if you are applying a positive input signal, your output needs to be less than zero (this is an inverter, right?) <S> but the op amp CANNOT drive to less than the lowest power supply voltage, which is zero. <S> Additionally, look at the data sheet such as this one and find the Common Mode Input Voltage range. <S> Note that for +/- <S> 15 volt supplies it must be within the range +/- <S> 12 volts to guarantee operation. <S> In other words, at +/- <S> 15 volts the + and <S> - inputs must be about 3 volts lower than VCC and 3 volts higher than VSS. <S> Think about what this means for a VCC of 5 and VSS of zero. <S> So. <S> Replace the single power supply with a +/- <S> 5 to 15 volt supply. <S> Until you do that, you are wasting your time.
| All op-amp amplifiers will behave like low pass filters due to parasitic capacitance between the output and, in your circuit, the inverting input.
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What is "Supply Input for Analog Functions" on a buck converter? Regarding this datasheet for a DC Buck Converter, there is a VCC pin that on pg-2 shows an arrow pointing "out". What is this "out" connection arrow used for in practice? Further along on the datasheet (pg-12), they show a board layout with that pin connected to GND via a capacitor. Is that a viable way to connect this converter if I don't have a use for whatever that "out" arrow is for? <Q> What is this "out" connection arrow used for in practice? <S> It's on the first page: "The AOZ2253EQI-30 integrates an internal linearregulator to generate 5.3V VCC from input." <S> It means that they make some regulated power available with all the advantages and disadvantages of a LDO linear regulator. <S> I can't speak for anyone else, but if I used it at all, it would be for some low current consumption circuit that needed to come up before the switching regulator did. <S> Is that a viable way to connect this converter if I dont have a use for whatever that "out" arrow is for? <S> It's certainly what they call out. <S> You do need the cap, according to the datasheet -- LDO regulators, particularly those of a certain vintage, tend to need them. <A> The VCC pin is an output from the internal linear LDO regulator. <S> The supply is used internally by the chip functions, in general it is not meant to be used externally and the datasheet does not mention how much current could be used externally so officially, it can't be used externally. <S> And the datasheet ia clear on this; the capacitor on VCC pin is mandatory for stable operation, it can not be omitted, even if you don't use it externally, the chip does use it internally. <A> The Yellow copper pad area "Vout" on p12 is the MAIN output of this chip. <S> on the left is feedback input from Vout and the layout is critical to keep tight. <S> If you do not need Vout, all that is left is the 5.3Vcc LDO output which is a waste if that is all you use.
| The output of any LDO still needs a capacitor, so the only function here is to provide the capacitor externally for the internal LDO regulator, nothing else.
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LDO output voltage lower than expected I'm aware the same question has been posted before, but unfortunately, it has not led me to an answer. I have a small PCB in which I've added an LDO to convert input voltage to 3.3 V output voltage. The LDO in using is the AP2210-3.3TRG1 ( datasheet ).My understanding from the datasheet is that with an input voltage of 4.3 V - 13.2 V. (4.3 V because the manufacturer recommends Vout + 1 V).I'm supplying 9 V input voltage from a battery. My schematic is as follows: simulate this circuit – Schematic created using CircuitLab I'm expecting to get 3.3 V as output voltage.When I measure the output voltage, however, my multimeter reads only 0.5 V. I've gone through the datasheet to see if I've misunderstood or missed something, but I can not manage to figure it out. There's also some stuff I do not really understand, which doesn't help. So some concrete questions.Do I have some mistake in my schematic?Am I misunderstanding how to use this component? I hand soldered the part on a PCB (with hot air), is it possible I overexposed it to heat? UPDATE Thank you all for your answers! Some additional information: The load is an ESP8266 ( datasheet ). I've measured the current at the LDO output, which is ~135 mA. When I measure the voltage at the same spot I get 1 V now.In the original post, I reported reading 0.5 V, but I must've been mistaken since nothing has changed and now I get a solid 1 V (which is still lower than expected). In regards to "Power Dissipation"; I get the following results (as per page 23 of the LDO datasheet): (150°C - 25°C) / (0.135 A × 200°C/W) + 3.3 V = 7.93 V This is a lower voltage than I'm supplying. Can this be the issue?The datasheet says: Therefore, for good performance, please make sure that input voltage is less than ... I'm going to find a 5V power source and see what happens! Results I supplied ~4.8V and measure the following values at Vout: 85-90 mA 0.7-0.75 V I also feel the LDO is getting significantly hotter now, but I've not verified that. UPDATE Adding a screenshot of the board design. resistors on top layer are 10k ohm resistors. C1 is a 1 uF cap and C2 is a 2.2 uF cap. U1 is an ESP-12F <Q> I've been soldering PCB's for years and never had a component fail due to overheating. <S> You'll find this is the case for many others (see How to tell if chips overheat while soldering ). <S> If you are using the SOT-23-3 version of this chip, it's pretty much a guarantee you have a goof in your setup (dry solder joint, internally disconnected wire, burned out chip) since you only have three pins to connect. <S> There is no mistake in your schematic, so something else is going on. <S> Try checking the current draw with no load (if it's high, your chip is blown), and double check your pin mapping and continuity. <A> Update <S> Below I estimated 5.7V*109mA=600mW = <S> Pd loss for a 3.3V*109mA = 360mW load. <S> We do not know the linearity of your new results but the size of the chip without a copper heatsink suggests 600mW should be derated by at least 50%. <S> Reduce your load, or change the design to a part with a heatsink or much higher power rating than this. <S> Estimating the load impedance. <S> 0.75/90mA <S> ~ 8 ohms 3.3V/90mA ~ 37 ohms <S> You can experiment lowering the input voltage if you have no linear adjustable supply source ,do an experiment and add a 10 ~30 Ohm input resistor to reduce the temp rise in the LDO. <S> ** <S> hmm <S> ** <S> I am leaning towards internal load short cct. <S> and diode drop characteristic failure ESD suggestion 2) <S> 1) <S> Excess linear power drop? <S> \$\theta_{JA}=200 <S> ~ <S> ' C/W~~~~\Delta V_{IN-OUT}=5.7,~~~~ <S> T_J=150'C,~~~ <S> I_{max}=125_{'C_{rise}}~~/200'_{C/W}/5.7V = <S> estimated~ 109mA <S> ~~absolute~ maximum\$ 2 <S> ) Electro Over Stress (EOS) semiconductor handling environment (training/reading) 3) Excess liquid solder dwell time. <S> ( read solder profile for IC) <A> Pin2, "Vout", appears to be grounded on your pcb. <S> ... <S> and C2 is connected to Pin 1. <S> That's probably not what you intended. <A> The first thing I consider in these situations is load / power and the next is stability . <S> Parts with a PNP pass element (particularly earlier and less expensive parts) often require a minimum ESR in the output capacitor path for stability. <S> Here is the simplified schematic for your regulator: <S> I have added a note about what I do when I see a PNP (or any P type) <S> pass element: <S> Going to page 22 of the datasheet we do indeed see a minimum ESR requirement: <S> Given the small size of the parts on your PCB, it looks like they are ceramics which typically have an ESR below the recommended minimum. <S> That leads me to believe the device is possibly oscillating or otherwise unstable which would render it quite hot in very short order.
| It appears to be a diode drop for some reason
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External 0~10v controlled by a MCU Newbie question :I have an external power supply which gives me 10volts reference. The machine I need to command is reading an analog order from 0V (minimum speed) to 10V (maximum speed) and I use a STM32WB55. My solution is to add a DAC and then an op-amp. Do you see another solution? Thank you <Q> 1) PWM output and low-pass filter 2) digital potentiometer. <A> You can get a few DACs that will directly output 0-10V with 3V SPI or similar serial control, but they are mostly designed for process control and other high performance applications, <S> I doubt the cost differential is worth it if you have a low end application (and even for higher end applications the cost hit is significant). <S> Eg. <S> LTC1257, <S> DAC8771 etc. <A> You have two options from Alex. <S> Another two are:- Bit bang (emulate) a PWM type function. <S> The STM32WB55 runs at 64MHz. <S> That should provide enough clout to do PWM in software. <S> Then low pass filter, but be aware that getting a nice clean DC voltage may require experimentation. <S> Especially if the output levels are to change quickly. <S> There's some calculation details here . <S> Then buffer and amplify with the op amp, or other voltage amplifier (may need to cater for an offset though). <S> This is an example of someone's efforts, but it will depend on the bit depth you're after and how many I/ <S> O pins you have free. <S> Remember that all of your design will be <S> a little soft if there's no revs feedback from the motor shaft. <S> This may affect how much output resolution you want to build in.
| You can create a DIY DAC with a bunch of resistors in a configuration called a resistor ladder .
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Why do we have to discharge the capacitor before testing it in an LCR Meter? LCR meters charge and discharge the capacitor to find out the capacity value (by using the time constant formula.) I wondering why a charged capacitor (even few volts) can damage the LCR meter ?Why do we have to discharge the capacitor before testing it in an LCR Meter if the tester must charge it to find out the capacitor value? <Q> Both of the meters in question use some means of charging the capacitor under test by driving it with a controlled current and measuring the time it take to go between two voltages. <S> When you charge a capacitor with a constant current the voltage rises linearly. <S> (From \$ V = \frac {Q}{C} \$ we can differentiate to get \$ \frac {dV}{dt} = <S> \frac {1}{C} <S> \frac {dQ}{dt} = <S> \frac{1}{C <S> } <S> I \$ since current, I , is charge per second. <S> These circuits are typically powered from a 9 V battery and the actual test voltages will only be a couple of volts. <S> Capacitors, on the other hand, may be rated at much higher voltages - hundreds of volts - and if a charged capacitor, even if only a few volts higher than the test range, may destroy the test circuit. <S> In practice a small charge of one or two volts shouldn't cause a problem other than a reading error. <A> LCR meters do not use high voltages, it might only use up to 1V signals for testing. <A> Thinking about the dangers of a charged capacitor, I remembered my school years when "bad guys" charged capacitors to the network and then gave them to the unsuspecting girls to hold them a little in their hands:) <S> Others hid a small battery, an inductor and a switch in a "book" with metallic covers ... and again handed it to someone asking him <S> /her to open it... to read something interesting inside:) I tell these "funny" stories to my students to arouse their interest in the behavior of these exotic electrical elements... <S> Both they accumulate electrical energy and can be thought as of "rechargeable batteries": the capacitor as a kind of "rechargeable voltage source" and the inductor as a kind of "rechargeable current source". <S> It is interesting to see what of them is more dangerous (for people and devices). <S> The capacitor is not so dangerous if there are no high voltage sources around since it stores the same voltage. <S> But it keeps the voltage for a long time - enough to carry it elsewhere and discharge it there. <S> So a capacitor that is charged to high voltage is dangerous for a relatively long time. <S> In contrast, an inductor is dangerous even when it is charged with low voltage. <S> But it is dangerous only in the first moment when we disconnect it from the source. <S> That is why there is no warning for discharging an inductor... <S> because when we connect it to the meter, it will already be discharged... <A> If you use a cheap screwdriver to short out the terminals of a large-value charged capacitor (e.g. 10,000 µF at 50 V) the stored energy is enough to blast the end off the screwdriver. <S> I still keep the screwdriver I used to discover this by accident, several decades ago, as a reminder not to do it again. <S> Doing the same thing to an expensive meter is an even worse idea than doing it to a US$1 screwdriver.
| It must be discharged because it might be charged to high enough voltage to break the meter or high enough voltage to be hazardous to humans.
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Is it possible to generate a list of all Net names in Altium Designer If I have a schematics in Altium Designer and it is spread over 10 sheets then is it possible to generate a list of all net names that are used in it? How can I generate this list? I want to do this to double check if a certain Net name is miss-spelled at one or more places in any sheet. <Q> You can generate a net length report, which tells you all the net names and the length of all the tracks assigned to each one. <S> I tried to find this option in Altium v14.3 but cannot find it. <S> Can you tell how to generate this net length report in this version or any later version? <S> In the PCB view, open the PCB pane. <S> Select Nets, and from the net classes select . <S> Now right-click in the area where the individual nets are listed and choose "Report...". <A> and then you can choose the netlist format that us wish the report to come in. <S> The options for the netlist formats will be dependant on which netlist generators you have installed. <S> The report you generate you should be able to search the nets you are looking for or print out and review as you like ! <S> You could also use the Navigate Plane under the "Net/Bus" and also search for the net you are looking for - <S> the benefit here is that when you click on the net it will zoom into the location of the net on the schematic <A> For making net lists from schematic 1.Choose Parameter Manager Set Options - Nets and click OK 3.Select <S> All & Report <S> It will create Excel file with all nets in design
| You can generate a netlist report of the whole project via the menu option Design->Netlist for Project
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How to connect multiple power rail wires together? I'm working on a project that uses a 12 V supply, tower light, buck converter & Raspberry Pi. I am prototyping it on a breadboard, but want a permanent home for it in an enclosure. What would be a good way to connect all the positive rail wires together & all the ground rail wires together? I tried using the Wago splicing connectors , but they seem more suited for "electrical" circuitry rather than the finer electronic circuitry. Is there a best practice for connecting multiple rail wires together between boards? <Q> Terminal Strips are most common for pigtail wire bonding but in pairs demands a solid wire to connect up one side of the screw terminals by daisy-chain then to source._______|________|_________|___ Busbar connectors tend to be overkill $ REF 7 load sides per pole but this one is cheap $2.80 REF <S> $2.30 (5)Mount on FR4, wood or equiv. <S> and bridge all pins together with AWG 16 solid gnd wire. <A> I think you're looking for "choc block", a.k.a. terminal strip. <S> You can common all the terminals with one piece of wire on one side in the screw terminals, then attach your individual power wires to the other side: (Image from screwfix.com .) <S> There are holes between the terminals which you can use to fasten it to something. <S> Also, it is easily cut apart between the terminals using a utility knife. <S> It is cheap. <S> Best practice would be to use ferrules on the ends of the wires if they have stranded conductors. <S> Or you could do pretty much the same thing with tag connector strips. <A> For actually connecting 3+ wires/nodes together (and not just joining individual sets of two), you can use the common screw terminal blocks along with a "strip" thing that bridges the connections. <S> They can be easily cut to the desired numbers of terminals. <S> You can find block and strip sold together or separately. <S> (Image from ebay.com )
| You could also just crimp your own wires together with terminals to function like the red strip above.
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Can smaller trace width limit the current flowing through the PCB? I am a computer science student and I am a newbie in electronics. I have designed a PCB for SIM800L, and I forgot to change the trace width of the copper rails. By default, the routing width was set to 0.254mm. Now once my PCB arrived I have soldered all the components to the PCB, and I have found that my design failed. I suspected on the trace width of the PCB because since SIM800L module doesn't come with an onboard voltage regulator, an external power supply adjusted to the voltage between 3.4V to 4.4V (Ideal 4.1V) is required. The power supply should also be able to source 2A of surge current, otherwise, the module will keep shutting down. Unfortunately, this has happened to me as well. So as I have no prior experience in PCB designing I just wanted to confirm whether it is because of the routing width or else there is any other reason for that. <Q> This might not be the only issue, but is almost certainly is one of them: <S> the LM317 cannot supply 2A (its output transistor simply cannot pass that much current). <S> But even if it could, it probably would go into thermal shutdown. <S> It is a linear regulator and linear regulators work by turning excess voltage to heat. <S> That means the higher your input voltage is relative your output voltage, the more heat is produced for the same amount of current. <S> You have your regulator set to 4V and are feeding it 12V. <S> That is an excess of 8V At 2A <S> , that is 16W worth of heat which is a lot. <S> The thermal resistance of the larger packages in the datasheet indicate that it would be somewhere around 600C temperature rise. <S> The LM317 goes into thermal shutdown long before you ever get anywhere near that. <S> Probably somewhere around 125C which is only 300mA. <S> So if the current surge lasts too long and your capacitor is too small to provide enough reserve charge for the entire duration, the LM317 will try and help out, overheat, and protect itself by going into thermal shutdown. <S> When linear regulator datasheet says it can supply 2A, it means 2A under the best conditions, not all conditions. <A> Supply transient problems in GSM radios are common. <S> My recommendation is to beef up the traces carrying high current by soldering wires on the surface of the PCB parallel to the traces. <S> Another option is to scratch the solder mask off and add a thick layer of solder on top of the traces. <S> It's also possible that the problem is in your external power supply. <S> Your input capacitor C1 should help you in this case. <S> If you believe this is the problem try using a larger capacitance. <S> If you have access to an oscilloscope, look for transients on the supply voltage at the module pin and then upstream. <A> 100uF capacitor consumes lots of inrush current at its start-up <S> and I don't think LM317 can support with those short traces. <S> Secondly, LM317 is an very good linear regulator with tight voltage regulation at its output and so you can move those huge capacitors to its input side and keep low capacitance (~10-20uF) at its output side. <S> Irrespective of it, you need minimum 12mil trace thickness for carrying 1A of current. <S> So based on maximum current carrying capacity of GSM module you can setup trace width accordingly. <S> Maybe 20mil would be a ideal choice. <S> Someone suggested SMPS buck IC instead of <S> LM317 and <S> in case if you are going to replace then you can keep 100uF capacitor as it is. <S> Keeping SMPS IC will help in improving efficiency of the overall system but with additional cost. <S> It is up to you to choose. <S> If you are going to use SMPS BUCK IC, then please go through its datasheet parameters clearly. <S> Apart from those, I would say some more improvements on the schematic what you have given here. <S> For LED you have connected 300Ohm which corresponds to ~15mA for 5V. <S> We don't need that much for LED. <S> May be you can consider to reduce the LED current to ~5mA and increase the resistance accordingly (1kΩ-2kΩ). <S> This change improves your overall circuit efficiency. <S> For ADJ pin of LM317 you need minimum 100uA current only. <S> But at present you have huge current <S> (4.1V /320 Ω (220+100)). <S> So increase the resistance of 220Ω to 2.2kΩ and 100Ω to 1kΩ, which also helps in improving efficiency of the overall module. <S> Please comment back for any questions.
| And yes, thin traces cannot carry as much current because they have higher resistance so they heat up more and produce more voltage drop, which can either cause the trace to overheat or the voltage to droop too low.
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Why does nMOSFET keeps blowing up? I have designed a Boost PFC circuit, But upon switch on transistor is getting blown up and fuse is burnt. So I removed transistor and just connected source to ground. the circuit worked fine. Then I kept gate open and connected drain to boost inductor. The transistor again blown up. [Edit : Schematic is added] I am now confused because even I kept gate open and transistor being technically off. Why it is keep blowing away? <Q> To be absolutely sure that the FET is not supposed to conduct, you should've shorted the FET's gate to the "source". <S> If you leave it floating, god knows what can happen - it's a high-impedance input. <S> If your FET is indeed an SPP20N60, it is supposed to be an N-FET, in which case it should work in your circuit. <S> If you mistakenly inserted a P-FET, you know what would happen. <S> Is there any chance that your FET's are possibly fake? <S> Have you checked the function of the FET somehow in "safe conditions"? <S> I.e., small Vds and using a resistor or a lightbulb to limit the current (or a lab PSU with a constant current limit) and apply some voltage to the gate... <S> You could even try a breakdown test, if you short the gate to "source" and use a high-value series resistor in the "drain" against a full rail voltage... <S> With the PWM controller attached to the gate: is your D3 diode really present in the circuit? <S> (Have you checked for its forward voltage? <S> for cold joints?) <S> Is C3 in the circuit? <S> If you leave the PWM controller's output detached from the gate (and FET gate shorted, or FET removed from the circuit), does the PWM produce output pulses? <S> You don't want the FET to conduct all the time :-) <S> Is the L1 really the value you think it is? <S> The SPP20N60 appears to have 600V nominal voltage. <S> What's the rail voltage in your PFC circuit? <S> Around 400V ? <S> As a very general rule of thumb, in power electronics for 230V mains, I'm used to seeing 800V triacs or other silicon switches - and that's where the 400V PFC topology is NOT involved. <S> If you're working with a 400V PFC application, try using an even higher-rated FET, say 1000-1200V, to stay firmly on the safe side - at least as an experiment while you're debugging your issue. <S> Changing the MOSFET model may help you avoid a fake or a bad batch, and the higher-voltage silicon switch models (in a family of similar siblings) tend to have a slightly higher "sustainable thermal energy before destructive melt-down" - which may also be a factor in your prototype, for one reason or another :-) <A> I have found the answer. <S> Unfortunately transistor body is source and diode body is cathode which means technically <S> I have shorted VCC and GND by giving common heatsink. <S> Now that I have made them seperate, It is completely working fine. <A> In order not to blow a large number of transistors, it is advisable to connect some active load in series to the transistor's drain ( <S> back in the old times it was an incandescent bulb or two). <S> Sure, it will get the circuit out of spec, but nothing will blow and you can attach a meter or an osciloscope and look what happens. <S> The FET in your case (assuming it is not fake, not miswired and not killed by static before being mounted) blows because it stays in "switched on" state for too long and the current in L1 gets too high. <S> Using the method above you can check why. <S> It may be a wrong resistor or capacitor somewhere around the IC. <S> Keep in mind that the IC may be blown from the first experiment, too. <S> Never, ever power on a circuit where some MOSFET has it's gate unconnected. <S> It is perfectly capable because of the parasitic capacitance between the drain and the gate, to saturate itself just because of the drain voltage change. <S> Bipolars are (mostly) OK with floating base. <A> Although it is possible that static transients exited your MOSFET gate when you replaced it unconnected, this behavior seems more like an incorrect routing on the PCB, but I wouldn't know, as you have not posted your board layout, or even a photo of the burned MOSFET on the board. <S> MOSFETS have an internal parasitic body diode that renders them unable to block in the reverse direction. <S> Connect them in reverse, and you are asking for an instant short. <S> This is probably what gave your fuse a run for its money too.
| Your MOSFET is likely blowing up because it it connected incorrectly.
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How to know how much energy storage has powerbank? I just looked to this question and I don't understand how to know how much Wh this powerbank has (click "Specifications" tab), because it has different output voltages and I don't know how to calculate if it's 18Ah * 12V = 216Wh or 18Ah * 19V = 342Wh? <Q> Then put on a constant current discharge load. <S> Monitor the discharge voltage duration with a low cost USB voltage data logger to determine how long of time it takes till the unit cuts out. <S> From this you can directly determine the power bank energy density at the chosen load rate. <S> This procedure is valuable to learn the real capabilities on the power bank. <S> You cannot trust the numbers given by manufacturers because they are often fake inflated data in marketing claims. <A> It depends upon how many cells are in the unit. <S> There is probably only one with ~3.7V output so the energy is probably 18Ah * 3.7 <S> = ~66Wh. <A> The information provided on the vendor web site is not very clear or specific. <S> If you really care about this you should look for a different vendor, one that provides complete specifications. <A> There's waht <S> they're claiming <S> and there's what it is . <S> This pack is advertised as having output ranges from 12 to 19 volts. <S> Remember that ratio. <S> One clue is in the graphic: "18000/28800mAh polymer 3-port <S> " <S> That ratio seems familiar... <S> 18000 <S> mAh x 19V = 342,000 mWh 28800 <S> mAh x 12V = 345,600 <S> mWh <S> Those numbers are very close, aren't they? <S> These numbers jibe; they seem to be claiming 340 watt-hours at 100% efficiency , which they loudly disclaim that you won't get . <S> By contrast, Anker wants $60 for a 100 watt-hour Anker Power Core II 20000, so 60 cents per watt-hour. <S> The very large Goal Zero Yeti 3000 watt-hour, is $2300, or 80 cents per watt-hour. <S> So if you figure 80 cents per Wh, I would expect $80 to buy a 100-125 watt-hour unit from a reputable maker. <S> So golly, that unit off AliExpress is a great deal, eh? <S> snicker <S> I suspect a teardown of this unit will reveal batteries incapable of storing 340 WH... <S> and probably more like 100 <S> WH <S> if that ... <S> there's no free lunch...
| If you already have the power bank a good way to evaluate is to first fully charge the unit.
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How much voltage difference will lead to saturation, LM741? Which parameter should I look for? I want to know what should be ( Vin+ - Vin- ) to get a saturated output. <Q> The main one would be the large signal voltage gain. <S> This gives the output in V for a difference in input measured in mV. <S> You also need to look at the input offset voltage which in this case is 2 to 6 mV. <S> If you take the worst case, 6 mV, and maximum large signal voltage gain of 200 you can see that the output would be saturated without any difference between the inputs. <S> The 741 is ancient and should only be used to learn about why we don't use them anymore. <A> You are to be interested in the large signal voltage gain. <S> Be aware that the specs given are for a very specific set of operating conditions. <S> Voltage gain is specified as volts of output swing versus corresponding input differential voltage in mV. <A> Output saturation is defined in datasheet as Vout swing {Min,Max} for Vcc,Vee <S> If the output swing is say 10V and the gain is 1e6 then input change is 10uV <S> but if the Vio input offset is in mV level, it cannot be predicted nor is it important. <A> It sounds like <S> you're more interested in using the op-amp as a comparator. <S> Some more modern op-amps have even more unpleasant behavior than the antediluvian <S> LM741 when you try to use them as comparators. <S> If you need a comparator, a good idea would be to consider a comparator such as an LM393 . <S> There are many, many better comparators than the LM393 for individual characteristics but few that are as useful, inexpensive and ubiquitous. <S> It is well specified for the application, such as response time of 1.3us typically with a 100mV step and 5mV overdrive. <S> Offset is only +/-2mV max at room temperature and <S> the gain is typically 200,000 (min 50,000). <S> Op-amps used as comparators can have very long recovery times from saturation as well as slow output slew rate and can conduct large currents between the inputs if there is more than a few hundred mV of differential voltage present. <S> Occasionally, it is useful to use an op-amp as a comparator, for example if you wanted to compare two voltages that were a few tens of microvolts apart and were not interested in speed of response, though even then it might be better to amplify the difference with an op-amp and then perform a comparison to a reference voltage.
| What is important is the Vio Iio specs and requirements for input offset voltage and current and your passive impedance tolerances.
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Simple circuit to light a LED only if I have a square wave I have a simple circuit with a transistor that accepts an input 3.3V square wave (50Hz) and amplifies it to a 20V output inverted square wave. I'd like to have a LED that lights up if the square wave is active and goes off if no input signal. If I just put a LED on the output it doesn't work, it lights up when no signal and it lights slightly dimmed when the square wave is active. Is it possible to achieve this in a simple way? EDIT:The circuit power supply is 24V, the input square wave is 3.3V <Q> The circuit below could be considered to be a monostable multivibrator, as Spehro suggested, but is cruder than most such circuits. <S> Using an IC will give a "clean" result more easily, but this can be made to work well with suitable design and/or playing. <S> Component values shown are examples only !!! <S> This would work on input or output with suitable component values. <S> Loading of the waveform by C1 may need to be considered depending on the nature of the source. <S> A MOSFET could be used for Q1. <S> On the input side an "on voltage" (Vgsth) lowe enough to accomodate the 3V3 swing would be needed. <S> On the output side allowance for the high voltage would be needed as most MOSFETS are better operated with lower gate-source voltages. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> One gate of a hex Schmitt inverter (eg 74HC14) could be used in place of the transistor - leaving the other 5 inverters for "playing". <S> You could eg have an LED that flashed at a steady rate regardless of input frequency, flashed only initially then steadied, or vice versa or .... . <S> The amount of useful, and fun, things that can be done with a package of hex Schmitt inverters is astounding. <S> Just for fun - or usefulness Gate 1 serves the same function as the above transistor circuit. <S> Gate two provides a flashing LED at a user defined frequency when the square wave is present. <S> 4 square gates invite experiment. <S> The LED drive is limited by the gate drive capabilty. <S> For brighter LEDs a transistor buffer can be added. <S> Component values are "indicative". simulate this circuit <A> Yes, look at using a retriggerable monostable multivibrator to turn the 3.3V square wave into a sustained 3.3V logic level. <S> You set the timing to be longer than the period of the square wave. <S> For example, 74HC123 . <S> but I think the above method is more elegant and uses fewer components. <A> You don't specify what supply voltage you have available, which complicates things. <S> Let's say you have 5 volts. <S> Then simulate this circuit – Schematic created using CircuitLab R1 and R2 form a 4:1 voltage divider which will reduce the input to a 5-volt square wave and will not overdrive the 74HC04 which is supplied by 5 volts. <S> The 74HC04 will invert the square wave so that the inactive level is 0 volts and the LED is off. <S> R3 will depend somewhat on the color of the LED - it will provide 2-3 mA for a red LED, and much less for white. <S> Don't reduce R3 too much, since the 74HC04 can't supply more than a few mA. Or, you can use a transistor as an inverter: simulate this circuit <S> In this case, set R2 to V1/10 in kohm/volt. <S> That is, if V1 is 10 volts, use 1k for R2. <S> For larger voltages, say 10 to 20 volts, this will give about 10 mA through the LED regardless of color. <S> Regardless of color, the LED will noticeably flicker - this is because it is pulsing at 20 Hz, well below the frequency at which the eye smooths things out. <A> Why not just have the LED powered from the 3.3V incoming signal? <S> LED/series resistor between Base or Gate and Gnd. <S> Can the 3.3V source spare a couple of mA to drive an LED?
| You could also do it with a charge pump (capacitors, diodes) and a transistor + resistor
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Confused about op amp equation derivation I'm studying op amps and I'm a bit stumped on a part of the derivation of the difference amplifier. In equation 4-1, I understand that \$R_G\$ and \$R_F\$ form a voltage divider that divides \$ V_{REF} \$ into the voltage at the non-inverting terminal ( \$ =V_+ \$ ). I also understand \$ V_+ \$ is the input to an inverting amplifier circuit, hence the \$ -V_{in}\frac{R_F}{R_G} \$ term. However, I'm struggling to understand intuitively where the third term comes from: \$ (\frac{R_F+R_G}{R_G}) \$ <Q> You didn't explain how 4-1 arrived. <S> But 4-1 isn't at all intuitive. <S> A much better way to see it is to require (obviously) that the (+) and (-) inputs to the opamp are equal to each other. <S> Like this: $$\frac{V_\text{OUT}\cdot R_\text{G}+V_\text{IN}\cdot R_\text{F}}{R_\text{F}+R_\text{G}}=V_\text{REF}\cdot \frac{R_\text{F}}{R_\text{F}+R_\text{G}}$$ <S> The left side is the equation for the (-) input node. <S> The right side is the equation for the (+) input node. <S> Just set them equal to each other (on the assumption that the designer wasn't an idiot and didn't get the input nodes reversed.) <S> The denominators on both sides cancel, immediately, and the solution is: $$V_\text{OUT}=\left(V_\text{REF}-V_\text{IN}\right)\cdot\frac{R_\text{F}}{R_\text{G}}$$ <S> In case you are wondering (and you shouldn't be because it should be as automatic as breathing is to you), any pair of voltage sources separated by two resistors will have the middle node voltage determined by an expression of the exact same topology as the left side of the first equation. <S> To make this clear, examine: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> $$V=\frac{V_\text{A}\cdot R_\text{B}+V_\text{B}\cdot <S> R_\text{A}}{R_\text{A}+R_\text{B}}$$ <S> This always works. <S> (You should memorize it.) <S> Sometimes, one of the voltages (with reference to ground) is \$0\:\text{V}\$ . <S> In those cases, the above equation simplifies a bit. <S> But it is otherwise something to memorize. <S> So the topology simplifies into the right side voltage divider topology when either one of the voltage sources is assigned the value zero. <S> (In the right hand expression of the first equation, that source was "ground," which is why it looks simpler. <S> But in reality it is the same, more general equation as on the left side -- just that on the right side one of the sources was zero.) <S> So the above approach is obvious and very easy to understand. <S> It's just two pairs of voltage dividing resistor equations set equal to each other. <S> You are just lucky that one of four sources is ground. <S> There's nothing at all difficult to explain about it. <S> All the questions about odd factors just disappear. <S> I'll be happy to explain the factor you mentioned the moment you provide us with the logical flow used to develop 4-1. <S> (Barry already provided you with an explanation you may like, though.) <A> Applying superposition, set Vin = 0. <S> Then Vref is first input to the voltage divider consisting of Rg and Rf which yields the first part of the term for Vout. <S> Now the opamp is acting as a non-inverting amplifier for the divided term with a gain equal to (Rf + Rg)/Rg since with Vin=0, the left end of Rg is grounded. <S> This gives the second part of the term for Vout. <A> The term that you are asking for, is the gain of the OpAmp when the input is applied to the non inverting terminal (given there is ground instead of Vin in your circuit). <S> Now this voltage is amplified by ((Rf+Rg)/Rg) to get the output contribution of Vref . <S> Similarly there is contribution of output from Vin which equals, Vin*(-Rf/Rg). <S> Adding these two contributions (Superposition) you get the total output Voltage.
| The voltage at the non inverting terminal of the OpAmp is Vref*(Rf/(Rf+Rg)).
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Design of second order low pass filter using passive RC components and opamp I have an RC low pass filter connected to the non inverting terminal of the opamp. There is a feedback resistor from output of opamp to its inverting pin.Then an RC low pass filter is placed at the output of the opamp. So while designing the filter,should I consider these two low pass filter together as a second order filter and design or consider each RC low pass filter separately and design the filter.Which is correct to get the desired response? <Q> You have your sallen-key topology wrong. <S> Even if it was correct, I don't like the circuit because attaching a load will mess with the transfer function. <S> Ideally your load attaches to the output of the op-amp or a buffered node, as then attached loads <S> will not draw current from the filter but the op-amp itself, preserving the filter's characteristics. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> ^ <S> Sallen-Key two pole low-pass. <A> U1 completely buffers the two RC stages from each other. <S> This means you can calculate their frequency responses independently, and simply multiply the responses together. <S> This is not the ideal way to make a 2nd order filter, it will have a very 'soggy' response around the cuttoff frequency, which most people don't want. <S> While technically it is a 2nd order filter, most people would refer to it as 'two cascaded first order filters'. <S> Sallen Key is a reasonable topology, there are many cookbooks available on the net to help you design one. <S> Bear in mind that a SK lowpass filter fails in the stopband at high frequencies when the opamp output impedance rises. <S> Use a 3rd order Sallen Key, which uses the same single opamp, to mitigate (doesn't cure, but makes it easier to live with) this fault, and gives you more design freedom on the filter transition as well. <A> Deepa - it is correct that your circuit is a second-order lowpass. <S> However, as mentioned already, it is a rather bad one. <S> Two reasons <S> : It works like a passive filter only with a "bad" amplitude response (two real poles) within the passband and it does not provide a low-resistive output which is insensible to any load. <S> I have mentioned the "poles". <S> Note that we have various different second-order lowpass responses - and the most useful response <S> is the so called "Butterworth" response which has a maximally flat amplitude response within the passband. <S> This filter type has not two real poles but two conjugate-complex poles and a quality factor Qp=0.7071 <S> (in contrast, the quality factor of your circuit cannot be larger than 0.5). <S> Such a Butterworth filter can be realized only as an "active filter" - for example based on the mentioned Sallen-Key topology (other filter structures are aslo possible). <S> In any "active" filter the amplifier is not used as a buffer only (as in your case) but it has a feedback network which is part of the filter function. <S> However, calculation of the parts values based on a certain filter specuification is more involved... <S> in most cases, we make use of formulas readily available in textbooks or other sources (internet).
| For a better approximation to a flat passband, followed by a steep rolloff, design a proper 2nd order filter, using the opamp as part of the filter. If you used buffered RC/Active Lowpass, it would be each TF multiplied, as the op-amp will provide the current. To answer your question, for the sallen-key you need to analyze as a whole (or look up TF, it's somewhat complex).
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What is this schematic symbol that is a circled capacitor? I am trying to determine what this symbol is that looks like a circled capacitor and is labeled as "Z1" that I circled in blue below. Here is the full schematic with this component being near the top left at the input. In my search I found this question with a circled capacitor and a dot but I'm not sure if its the same. I also found websites saying Z is for zener diodes which this obviously is not or this website that says Z could be for a balun, general network, or phase shifter but I did not find circuit symbols that match this one when googling those. <Q> It's likely an MOV (Metal Oxide Varistor) used for surge suppression. <S> Context guides this answer - <S> the schematic is from the late 1990's when this was a common use and it is across the hot/neutral lines and in the earth ground path. <A> It could be anything from a spark gap to a gas-dicharge tube to a semiconductor device such as a MOV or TVS. <S> What does the parts list (bill of materials) say about it? <A> They are indeed varistors like others are (educatingly) guessing. <S> Varistors Z1 and Z2 have overvoltage protect function on the line input. <S> Source: <S> http://www.pavouk.org/hw/en_atxps.html <S> You can find this source by googling on ATX power supply using the website on the schematic. <S> Search term: ATX POWER SUPPLY site:www.pavouk.org
| The "Z" reference designator strongly suggests that it is some kind of voltage-clamping device.
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Why are Marx Generators built like that? Short introduction: A Marx Generator is a capacitor arrangement that transforms a DC input voltage of let's say 1 to 20 kV to a multiple of that. Capacitors get charged in parallel and when they reach the break down voltage, they discharge themselves over spark gaps and essentially get connected in series leading to a much higher output voltage (theoretically input voltage times capacitor amount). I built such an arrangement, but the sparks it produces are kind of small. The easiest thing would be to measure the output, but I don't have a multimeter for voltages that high. Additionally I also want to understand the theory behind it, which is why I wrote this question. This image is a circuit diagram of a Marx generator. During my research I found several diagrams for Marx generators. They differed a bit, but all of them had the problem described further down. The image will help me describing my problem: Let us assume that the DC input is 8 kV and the resistors (Rb and the 10 resistors without text) have a value of 1 Mohm. According to my knowledge, that would mean that the first capacitor is charged with 8 kV and a resistance of 1 Mohm which would be perfectly alright. However as far as I know, the 2nd capacitor isn't charged with the same resistance; it is charged with 8 kV and a resistance of 3 Mohm (Rb + 2 times resistor without text). The third would get charged with 8 kV and 5 Mohms and so on. My question is the following: Is my assumption about the charging behaviour correct? If yes, why would someone do that? Why would you connect the resistors in series instead of in parallel, just like the caps? Why would you want the caps to not get charged with the same current, so they achieve break down voltage at the same time?If no, what is my mistake? Thank you for reading this far. I am pretty new to the field of electrics, so please be patient with me. Edit: As I don't know how to post an image in a comment, I will put in here.Some of you mentioned that the resistors would be exposed to the full voltage if placed in parallel too. Have a look at the picture. Wouldn't that fix the problem? Edit again: New circuit diagram <Q> The main point of a Marx is to get a very high output voltage with modestly rated components. <S> When the Marx fires, each resistor still only sees the charging voltage. <S> If you used parallel resistors, the ones to the end stage would see the whole output voltage. <S> Eventually, all stages of the Marx will reach the same voltage. <S> You can improve on the charging time by using inductors instead of resistors. <S> This limits the length of the output pulse, but it's usually fairly short anyway. <A> If you wait sufficiently long all capacitors will charge up to the input voltage. <S> When the "diagonal" spark gap goes off all capacitors are connected in series through their spark gaps and the total voltage is the input voltage * capacitor stages. <S> The idea is that the spark gaps are at that time very low resistance. <S> To understand this, draw the diagram again, but with all capacitors charged and all "diagonal" spark gap replace with a short-circuit. <S> I built me one of these and learned not to let go off whilst your talking next to it on the phone. <S> In your second diagram you have a bottom resistor in series with the discharge circuit. <S> That would severly limit the output current. <S> I worked with one of these at the high voltage lab of the university of Delft. <S> It was 20 meters high and could produce several Mega volts. <S> They did not use your adapted scheme so I would assume there are some disadvantages <S> but I honestly can't see which. <S> Apart from this one: For safety you want the ground of your feeder voltage and the end spark to be the same. <S> By the way: before you were allowed to work on it, it had to be discharged. <S> You did that by turning a wheel which pulled a metal wire past the capacitors. <S> Each time it hit a capacitor you would hear the 'chack' of the discharge. <S> I did not want to be pedantic but your initial diagram is wrong. <S> The bottom capacitor does not add to the circuit. <S> The 'left' arrow should come from the bottom which makes that it is at ground level. <S> Which is important for my remark about safety and ground above. <A> The capacitors are in series when the device fires. <S> The resistors are too. <S> However, if they were fed directly from a source via resistors when charging then when the device fired the resistors would be across increasingly higher voltage points as you progress up the "cascade" and dissipate increasingly more power and increasingly load down the capacitors as the voltage points increased. <S> You COULD use parallel feed and increasing resistor values but you get the same effect this way. <A> to get the last capacitor more fully charged what you do is make all the other spark gaps slightly bigger than the last. <S> In that way the last capacitor has time to get fully charged before the spark gaps fire. <S> When a single spark gap fires all the rest are triggered by the voltage pulses transmitted through the capacitors, so it does not matter if the have slightly different gap distances. <S> if you use resistors to a common input bus the resistors see a very high voltage when the gaps fire, with the series resistors the voltage on the resistors is only about the same as the voltage on the capacitors. <A> All your observations are correct. <S> However ... <S> 1) <S> If you wait long enough, all capacitors will charge to more or less the supply voltage 2) <S> There are two common ways to get a Marx to erect. <S> One is to have one spark gap set less than the others. <S> When this breaks down, stray capacitance of the other nodes to ground will transfer some of the step to the other gaps, causing them to break down promptly. <S> If this smaller gap is the last one in the chain, it will ensure all the caps are charged to at least that voltage. <S> The other way is to use a triggered gap as the first gap. <S> You can wait until the requisite charging time has elapsed before triggering the gap. <S> 3) People build Marx generators to get verrrrrry high voltages. <S> This means space and components are at a premium. <S> The conventional layout of a Marx uses its length to good effect, adding the voltage withstanding of the space and the resistors in series. <S> Your parallel layout forgoes that advantage. <S> There is little point building a Marx for modest voltages.
| You could use a parallel arrangement of resistors, but you would need very high voltage resistors, and clearances round them.
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Can I replace a 12a fuse with a 15a fuse on a 12v - 120v automobile inverter? I'm hard wiring an inverter into my automobile. Tapping into a 15a fuse in the fuse box. To hard wire it I had to remove the "cigarette plug" and in so doing removed a 12a fuse. I have an inline fuse connector which only takes "bladed" automotive fuses and none of them appear to be 12a, everything I see online for this type of fuse seems to go from 10a to 15a. And I am having a very hard time finding simple 12a inline fuses designed for 12v systems and 12g wire. So, can I replace the 12a "cigarette lighter" connector fuse with a 15a inline fuse? Or could this endanger the inverter, since it's designed to be protected with a smaller fuse? <Q> Never replace a fuse with one rated for a higher current. <S> If you can't find a 12A fuse try a 10A fuse. <A> The fuses are to protect the wiring, not the downstream devices. <S> If you are sure that all the circuit wiring is #12AWG (on either side of the fuse), and you're sure that the wiring connectors and harnesses are rated for 15A, then you could feel safe about increasing the fuse size. <S> I personally wouldn't take the chance that I had missed something. <A> It's your car, but the whole point of fuses is to protect the wiring in the car. <S> If this were wiring in a house or structure , I'd be standing on the table yelling. <S> But I'll stand down since you're likely to confine the damage to the car itself. <S> But still, melting a fuse block or wiring harness sucks . <S> I would either get a tube-type fuse holder (which are cheap), or downfuse to 10A. <S> Your better plan, really, is to install a dedicated circuit for that inverter, with a better plug than a cigarette lighter. <S> That way you can pull more than 100 watts. <S> Try coming off an appropriate (provided) high-amperage hot-in-run terminal on your car. <S> (or from an always-hot terminal via a relay whose coil is hot-in-run).
| In the case of an electrical fault, you want the fuse to melt before the wiring does :) If the device is taking more current than it was designed for, the damage has already been done.
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Importance of EM to microwave design? I am curious, how important is a general understanding of electromagnetics to microwave circuit design? I have looked at Pozars book which provides some background on EM and transmission lines in the first chapter and then goes on its way. Would someone with little background in EM but a sufficient background in mathematics be foolish/stupid to start his book instead of doing an entire EM course beforehand? To what extent is a fullbackground in EM needed to understand a book like Pozars? <Q> If you want to do microwave circuit design (amplifiers, for instance), you should be comfortable with E&M theory. <S> If you end up dealing with interconnects, antenna elements such as notch or patch radiators, and antenna arrays <S> , then it's all about E&M. <S> Unfortunately, I'm not familiar with Pozar's book <S> so can't venture an opinion there. <S> To be honest, I just got by in my undergrad E&M course. <S> Now I wish I had paid more attention. <S> E&M theory is all based on basic principals of physics and never goes out of style or is overtaken by some new technology. <S> The tools may change, but the underlying principles remain the same. <A> I'm not familiar with that book, so I can only answer in general. <S> You can probably get part of the way by learning some rules of transmission line behavior (which is, I suspect, <S> what's covered in that first chapter). <S> But you'll never understand the "why" of those rules without knowing E&M, and as the field evolves you'll always be a bit behind, and the "slow guy in the room" when wacky new technologies are being discussed. <S> Then, too, you really can't begin to understand antennas without understanding electromagnetic wave propagation. <S> I'd learn E&M, if I were serious about the subject. <A> Having used Pozar's book as reference previously, I feel it is a very good book for understanding Microwave Engineering, but not so good for understanding the basics of EM <S> (Maxwell's equations/Boundary conditions etc) which is crucial for good understanding of Microwave Engineering. <S> Since you already have sufficient math background, I would suggest "Principles of electromagnetics" by Matthew N.O Sadiku to understand the fundamentals of EM. <S> It goes in depth of Maxwell's equations/Boundary conditions/wave propagation etc and does so in a thorough mathematical way. <S> This book is also used as text book for undergraduate EM course in several universities. <S> Doing Pozar after Sadiku is the way to go according to me.
| While a lot of the new tools out there (ADS, field solvers, etc) take a lot of the grunt work out of microwave circuit design, a good grounding (pardon the pun) in E&M theory helps understanding the results the tools give you.
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What to do with excess co-ax cable These days light aircraft come with lots of instruments that require thin co-ax cable for GPS antennas, transponders, collision avoidance systems, tracking and navigation systems.Like this GPS antenna as an example. Whats the best way (from the signal quality perspective) of handling extra co-ax? Space behind an instrument panel is often at a premium, manufacturers typically supply antennas with 1-2 meters of cable but in a small aircraft (microlight or glider) the antenna might be only 12 inches from the instrument. With 4 or 5 such antennas the space behind an instrument panel can quickly become a cluttered mass of wires. The people who install these instruments are often the pilots or engineers with little knowledge of electronics. There seems to lots of 'psedo-science' around how to handle the extra meter or two of cable. 1) the simplest (and most common) option seems to be to leave thecable excess in a neat loop (like in the photo) secured with a cabletie. However, some say this will cause inductance which will effectsignal quality (my view is that inductance is a current relatedphenomena, and the current flowing through the coax will benegligible) 2) the 2nd prevailing opinion is that the spare cable should belooped in a figure of eight fashion to prevent induction. 3) another suggestion I've heard is to loop the spare cable in thebiggest possible loop (i.e. a two meter cable looped twice into a 30cm diameter loop) the argument here is that fewer turns means less inductance. 4) the best solution is almost certainly to trim the cable tothe exact length, however this isn't without it's own problems, the SMA/SMC cables are so small that they can be really hard to crimp, and antenna often need to be moved about when searching for optimum positioning, so there's always the danger of getting the length wrong So back to the original question. From an electrical engineering perspective, with a view to getting the best signal quality without interfering with similar instrument nearby whats the best way of handling excess cable? Is it one of the options listed above? Or something else? Or maybe it won't really make any difference? <Q> Best would be to shorten the cables. <S> A shorter cable means less loss of signal in the cable, and a neater installation. <S> Doing so requires proper tools, new connectors, and some practice in attaching the connectors to the cables. <S> That's too much to expect pilots working on their own planes, or even mechanics who do the work for many pilots. <S> There are simply too many differing connectors, cables, and tools. <S> Doing it wrong is worse than just leaving the cable a bit too long. <S> Leaving the cables at the stock length is the safest way. <S> The cables were assembled at the factory by technicians (or machines) that can do the job properly and the cables can be tested for proper function. <S> Don't worry about induction losses in the cables. <S> They are shielded. <S> There shouldn't be any measurable inductance just from coiling them up. <S> Whether you just coil them or make them into a figure <S> eight shouldn't make any difference, either. <S> What you do need to be careful of is coiling them too tightly. <S> The cables are made of several layers of materials. <S> The proper function of the cable depends on the relationship of each layer to the other. <S> Here's a typical coaxial cable (from wikipedia. ) <S> In the cables you are dealing with, the layers are thinner and more easily damaged. <S> A kink in the cable can cause the center conductor to squeeze through the inner insulator and either directly short it or allow the signal to more easily cross through to the outer conductor. <S> Either is bad. <S> Make your coils large, don't kink the cables, and you should be fine. <A> The best thing to do is leave the cable the original length, and to gather the excess in whatever is the physically tidiest way. <S> Few people are equipped to shorten the cables and remake a connector properly, and it needs to be done properly. <S> If you have to ask, don't do it. <S> There are two aspects to 'cable inductance', differential and common mode. <S> Differential mode. <S> This is the inductance of the inner with respect to the return shield that the signal sees. <S> The whole point of coaxial cable is that this inductance is completely screened by the coax outer, and is totally unaffected by whatever physical configuration the cable takes. <S> Loop it small, loop it big, put it into a figure of 8, none of these will affect the signal handling at all. <S> Common mode. <S> This is the inductance of the outer with respect to the instruments at each end. <S> This is affected by the physical configuration. <S> The main effect of this inductance is to modify slightly the electromagnetic compatibility performance of the box. <S> Does RF leak out of the box onto the cables and get radiated, or is the box more or less susceptible to strong RF fields around the box <S> getting picked up on the outer and leaking into the box? <S> With a good design of the box, all of this shielding is provided where the coax enters the box. <S> However, if this aspect has been 'value engineered' to only just meet requirements, then cable looping might make a difference. <S> You often see cylinders of ferrite clamped onto computer video and USB cables, these make an order of magnitude more difference to the emissions performance than looping the cable around. <S> If you do have strange interactions between, for instance, your transmitter and other instruments, then moving the cables around might be a way to diagnose whether there might be any interaction from leaky boxes, but it would not be the way to get a cure. <S> A reliable cure would only be effected by buying higher quality instruments, or putting clamp-on ferrites on the cables close to the offending boxes. <A> Doing so does not cause or increase inductance as the coax is basically a shielded conductor. <S> Its inductance is determined by the geometry (conductor diameter, insulation thickness, etc) and not by how it is arranged physically. <S> Just make sure you follow manufacturer's or best-industry practice and don't bend the coax too much by trying to make a really small coil. <S> Our standard is to keep the bend radius at least 10 times the diameter of the cable. <S> The figure 8 arrangement will neither hurt nor harm the inductance of the cable, for the reason described above. <S> No need for the larger diameter loop. <S> This would be the best solution. <S> But you need to make sure the new connectors are properly attached to the coax. <S> And you also want to allow for a service loop on each end to aid in attaching and removing the coax from its jack on the equipment. <A> Inductance (of the whole cable) is unrelated to most usage scenarios of the coaxial cable. <S> The currents flowing in the core and the screen should exactly cancel each other so no magnetic field is expected outside the cable. <S> Thus, all inductance-related questions are pointless. <S> People usually fail on the crimping part and even if they don't, the signal level increase is usually minor, because manufacturers generally use adequate cables for the signals used. <S> Creating bigger loop is good up to the point of radius, say, 20x of the cable width. <S> Making smaller loop risks damaging the cable, making bigger loop is not any better. <S> Making 8's in the loop is not good - curvatures get sharper, increasing the possibility of damaging the cable.
| Related to signal quality, the best is to shorten the cable to the needed size AND crimp it adequately. Leaving the excess coax in a loop as shown in your picture is perfectly OK.
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Why do LED headlamp bulbs need active cooling? I decided to install LED headlamp bulbs in my car in the hope that I can avoid replacing bulbs every year. I was quite surprised to find that I can't fit replacement bulbs in my car due to the bulbs having a huge heatsink and, indeed, a fan to keep them cool. I don't have sufficient space in my lamp housing to incorporate the extra hardware. Looking on Wikipedia at the relative efficiencies of halogen and LED I see that they are both hugely less efficient than I anticipated: Halogen 3.5% (typical) LED 14.9% (worst case) My usual halogen bulbs use 55 watts, and higher-output LEDs use 20 watts. Therefore, the halogen bulbs are creating: 96.5% * 55 watts = 53 watts of heat The LEDs: 85.1% * 20 watts = 17 watts of heat. Why do the LEDs need active cooling when the halogen bulbs don't - yet they create over 3 times the waste heat? <Q> Here's an interesting video comparing a 25/25W LED headlight to a 90/100W halogen. <S> Philips Luxeon H4 Headlight Temperature Output vs Halogen <S> The passive heatsink at the rear of the LED lamp got up to 74 °C, while the halogen lamp reached 99 °C at the bulb holder. <S> Considering the difference in power consumption, the temperature difference doesn't seem that much! <S> The halogen lamp's quartz bulb has to reach a surface temperature of 250 °C in order to maintain the halogen cycle, and the filament inside runs at ~3000 °C, so why doesn't the outside of the lamp get a lot hotter? <S> One answer can be guessed from the measurements at the front of the lamps. <S> The LED lamp only got up to 33 °C, but the Halogen measured 60 °C. <S> An LED only emits visible light, so most of the heat produced has to be dissipated by the heatsink. <S> A halogen lamp dissipates a lot of heat too, but it also emits infrared light directly out the front along with the visible light. <S> Some of that infrared is absorbed by the front glass, but a lot of it passes through and so does not heat up the lamp. <S> In the graph below we see that a halogen lamp emits much more infrared than visible light. <S> If that light was useful to us the lamp would be considered far more efficient. <S> Halogen Lamp Spectrum <A> Because incandescents love heat, and LEDs hate it. <S> An incandescent in a 250°C environment is one happy incandescent. <S> That's why they work so well in oven lights. <S> An LED in a 250°C environment is dead. <S> So you need to keep the junction temp down to something like 115 <S> °C at the absolute max, though 85°C would be better. <S> That fits hand to glove with the car's cooling system. <A> Why do LED headlamp bulbs need active cooling? <S> 1. <S> Because they will quickly fail otherwise. <S> They might have the same lifetime as a cheap 100W bulb at 150'C <S> If you add thermal insulation to raise the temperature so that the mean life or MTBF= 1000 Hrs and then reduce the temperature, your MTBF will double for every 10°C drop towards room temp. <S> This is a Law of Chemistry called Arrhenius Law . <S> It just happens that Tungsten filaments operate around 3000 ~ <S> 4000 <S> °C and Semiconductors don't like to operate more than 200°C and preferably < 85°C junction temp . <S> Therefore due to Arrhenius Law Semiconductor LED's with very high power density require active cooling like CPU's. <S> For a test on a 100W light bulb, rated for 3000h, put a glass globe over the bulb and trap the convection heat exchange necessary for cooling and expect it to last 3 mos. <S> ( If I recall from memory) <S> Don't ask how everyone thought LED's should last 50kh= <S> MTBF that was when Mfg process quality was very poorly controlled <S> and no one did any real reliability analysis. <S> The 1000 and 5000 hour high temp life test follows this Law for all electrolytic capacitors. <S> You then derate your design stress to estimate thermal rise relative to design rise to estimate MTBF.
| The other difference is that halogen lamps are designed to run hot (which they must to work properly), but LEDs are less efficient at high temperature and may be damaged by going over 150 °C, so they need better heat sinking.
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What is reason of having 3 phase in the transmission in cities? What is the reason of having 3 transmission lines as three phases ? <Q> Let's say you are running 200A wires insulated for 5000 volts, for a nice round 1 MW capacity per wire one-way. <S> If you run single-phase, you need 2 wires to complete the loop, and you get 1.0 MW circuit capacity. <S> If you run 2-phase 90 degrees, you need 4 wires and get 2.0 MW circuit capacity. <S> If you run 3-phase "delta", you need 3 wires and get 1.732 MW capacity. <S> This is the most economical use of wire and terminations. <S> It simplifies life for motors and DC <S> loads <S> People who run motors love you, because 3-phase is the most efficient motor design, and it's fully able to self-start without burdensome compromises. <S> You love people who are driving DC loads (i.e. rectifying 3-phase) <S> because there's considerably less ripple between the most positive and most negative pole of 3-phase. <S> This means things that ultimately use DC have much better Power Factors with less harmonics and inductive heating. <S> (when you try to push a 300Hz harmonic through a 60Hz transformer). <S> That drawing is wrong <S> The power on the top 3 wires is 3-phase delta. <S> There is no neutral here. <S> That wire labeled neutral is conspicuous in its absence of any insulators. <S> It is a guy wire . <S> Note how the transformers are on the pole out of balance , their weight is pulling the pole over to the right. <S> Note also how the two down-guys are also pulling the pole over to the right. <S> Everything is teaming up to make this pole fall over, right? <S> No. <S> The upper conductors are probably ACSR <S> which means they have the dual role of also being in physical tension. <S> Those, and the lower cross-guy, are counter-balancing something that is happening on the pole to the left, off-picture. <S> Given the bottom-tier insulators, I presume the lower lines are 480 "wye" service, with 480V corner-corner and 277V corner-neutral. <S> "Why does neutral need insulation? <S> It's near ground." <S> No, it's only near ground if everything is working properly. <S> The rest of the time, lookout! <S> That is why the lower service conductors do in fact have the same insulator on neutral as hot. <A> Electricity is generated as three-phase, for very good reasons that I won't go into here, and if you distributed only one of those phases, then you have what's called an unbalanced load (in fact, it's about as unbalanced as you can possibly get), and power people hate those--they cause all kinds of problems with power distribution and--don't quote me on this--i'm pretty sure they can even damage generation equipment when they're really bad. <S> So instead, you distribute all three phases and just do single-phase (or split-phase) in each individual building. <S> (I'm a bit rusty on power transmission, so please feel free to correct any mistakes or misrepresentations in this answer!) <A> The key reason why 3 phases are generated and distributed is: to provide a solid rotational power for electric motors. <S> Two phases 90 <S> * shifted would theoretically be enough, but the 3-phase system probably provides for smoother operation of the motors. <S> Theoretically also more robust (fault tolerant) <S> but in practice the 3-phase motors don't like a missing phase very much. <S> The fact that when the load is evenly spread across three phases, the return wire gets hardly any load <S> , that's merely a "fringe benefit", a desired optimal state of things. <S> And it's true that with a 2-phase 90 <S> *-shifted system, this "balance around the neutral return" would not be achievable. <S> In this vein, I've read that HVDC transmission systems (when converting from AC to DC) and some railway power rectifier stations can employ phase-shifting transformers to produce 6 phases (shifted by 60 degrees) to produce an even smoother rectified output... <S> The three single-phase transformers are interesting. <S> I'm used to seeing pole-mounted 3-phase transformers .
| Another fringe benefit is that, when you need some non-trivial DC power, 3-phase rectification gives you smoother DC output, even without heavy capacitive filtering (though without PFC, the harmonic content is certainly ugly even that way). It's the most efficient way to transmit AC
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Audio being fed back into the circuit, beeping I am developing an audio amplifier with microphone, I have a problem that I tried in many ways to solve I could not, so I am here to ask you a second opinion. My problem is that this circuit is generating a lot of feedback is I can not eliminate this whistle, I tried to put the electret microphone in polarization with a resistor, I also tested capacitors of various values in parallel is not resolved (the sound does not come out or distorted ). I have already tried to lower the value of c17 to decrease the op amp gain from + -702 and the sound comes out noisy.what do I do. supply voltage: 12V. The sound captures the return from afar +- 1 meter <Q> U8 has a feedback resistor, but no series resistor between the - input and the microphone. <S> With that setup, U8 just amplifies the microphone signal with everything it's got. <S> Put a 100k resistor in the place of the connection from C17 to R39. <S> That will reduce the amplification to about 5. <S> If that's not enough amplification, use a resistor with a lower value than 100k. <S> This kind of circuit has a natural tendency to develop a feedback squeal. <S> You amplify the microphone signal, and send it back out through the speaker. <S> The microphone "hears" what comes out of the speaker, amplifies it again, and again it goes back out of the speaker. <S> It just keeps getting louder and louder. <S> You have to turn down the amplification or move the speaker and microphone so that the microphone doesn't "hear" the speaker as well. <A> Assuming it's not the case that your gain is too high and you have acoustic feedback (you can't noise cancel with phase, sound rapidly becomes non coherent)... <S> You have a number of noise sources: <S> Your amplifier+and- have one terminal driven with a potential divider and one driven with a capacitor. <S> It's therefore liable to oscillate driven by small noise on the psu line. <S> The microphone voltage excitation should be decoupled from the output stage voltage source, otherwise ripple on it will feed back into your transimpedance amplifier. <S> There should be no (DC) current through the microphone so you can use a slow RC network. <S> Your output stage may well not be very nice and linear and therefore producing large (in terms of a high gain trans impedance amplifier) power rail dips and spikes. <S> Verify first that you're not just getting acoustic feedback... <S> Use a resistor for the speaker as suggested by someone else and an oscilloscope. <A> You have negative feedback for each amplifier. <S> If you still have oscillations( sounds like you do), you can add system level negative feedback. <S> Connect from C9 back to U8 with a resistor. <S> That resistor will also set the overall gain of the system. <A> Your inverting amplifiers have a DC offset of Vcc/2 on their outputs. <S> The first one has a lot of gain <S> (A V1 = -500k <S> / R out-mic ... <S> where R out-mic is the output [internal] impedance of the microphone. <S> It's probably a few hundreds of Ohms. <S> So Av < -500) . <S> It can ridiculously amplify any noise on its inv input. <S> And also the output will have an offset of Vcc/2 (6VDC in your case). <S> The second one has a gain of -3. <S> The inv input has 6VDC offset + amplified signal coming from the previous stage. <S> Since the gain is negative, the DC offset coming from the first stage will not affect the output. <S> But the output will still have a DC offset of Vcc/2 due to the voltage divider consisting of R1 and R2. <S> This may result in saturating the output of the preamp stage if the input signal coming from the microphone is large enough (e.g. 10mV rms). <S> So you'll get a pure DC output, or a square wave at best. <S> You may want to build non-inverting amplifiers instead.
| Add capacitors to the lower resistors of your voltage dividers and filter the microphone excitation.
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Buffers output load in unpowered state Lets take under consideration a 3 state buffer, such as this one: Since the datasheet doesnt provide the information about the internal build of the buffer, I was wondering either it is safe to assume that in case the chip is not powered up (0V between VCC and GND), the output is in High Z state? Should one worry about any load at the outputs in unpowered state? I imagine there is a mosfet P + N half bridge at the output. <Q> I would assume quite the opposite: <S> all pins are clamped to Vcc through ESD diodes. <S> Since Vcc is 0V when not powered, that means the output pins are clamped to ground through these diodes, and you won't be able to apply a voltage greater than about 0.5V to them. <S> Having first made that blanket assumption, check the datasheet. <S> Some devices may be designed with different kinds of ESD protection. <S> However, the specific part you mention states pretty clearly in the datasheet, under "output clamping current" that the output pin may draw 20mA <S> if the voltage is 0.5V below ground or above Vcc. <A> Some devices such as the SN74AUP1G126 are designed to operate with outputs applied even without any power supply. <S> Datasheet <A> No, you cannot make that assumption. <S> Devices whose outputs are defined to be high-Z when not powered are sometimes referred to as being "cold spareable", which means you can connect something to them when they aren't powered (hence "cold"). <S> Such devices find use in systems with redundant hardware. <S> If devices have that capability, it's almost always called out on the data sheet. <A> From the datasheet limiting values it can be seen that output cannot be higher than VCC+0.5V unless the clamping current is limited to 20mA. <S> So output has protection diodes and it will try to back-feed the chip VCC from output if output is set externally high.
| No, it is not at all safe to assume that the output pins are floating when the device is not powered.
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Can we generate electricity efficiently from a peltier device by supplying heat? Is it possible to generate electricity from a Peltier device by only supplying heat (not ice)(something above 343 K )? How efficient can it be? Which is more efficient?- Using Peltier device or using the heat to convert the water to steam and then to electricity(heat is obviously not 343K in this case, it's something above 373 K)? <Q> Is it possible to generate electricity from a Peltier device by supplying heat (not ice)(something above 343 K )? <S> How efficient can it be? <S> Yes, that is possible. <S> The mechanism that generates electricity (seebeck effect) is dependent on the temperature gradient (temperature drop or \$\Delta T\$ ) across the module. <S> Typical peltiers are roughly 3% to 8% efficient, but the efficiency is dependent on many factors like temperature gradient, heat moving through the device, and the load on the peltier or current and voltage that the device is operated at. <S> Most steam conversion is 30% to 50% efficient, much more efficient than a peltier. <A> It's all about scale <S> Peltier devices are notoriously inefficient - only about 5%. <S> However, steam requires an army of skilled engineers present at all times the boiler is running (and hours before and after). <S> It doesn't matter if you're making 500 milliwatts or 500 megawatts <S> , you still need the army of engineers to avoid a deadly BLEVE (boiling liquid expanding vapor explosion). <S> Note that 343K is much too cold for a steam plant to operate efficiently. <S> You might be able to make it work with lower boiling point liquids like pentane if you can access an ultimate heat sink that is sufficiently cool. <S> However, most of those fluids are flammable. <S> At the lower ends of that range, you are much better off using gas turbines or plain diesels , which require a much smaller engineering staff, because they can't explode in the violent manner of a BLEVE. <S> These run about 30-40% thermally efficient, net. <S> If you have a gas turbine installation large enough, you can use the rather hot exhaust to boil steam and run a steam plant alongside. <S> This is called a combined-cycle plant. <S> In this case, the boiler management staff are managing a number of small boilers (at each turbine exhaust) feeding into one large steam turbine. <S> These combined-cycle plants have efficiency of 50% or even as high as 55-60%. <A> Heat engines with a low temperature differential will always be inefficient. <S> The theoretical maximum is the ratio of the absolute temperatures of the source and sink. <S> With a room temperature sink of 293K and 343K source, the maximum efficiency will be (343-293)/293 = 17%. <S> This is known as the Carnot limit ( Carnot cycle ). <S> Peltier devices will get much lower than that - a large source of inefficiency is the conduction of heat directly from the source to the sink side of the Peltier.
| You can get 30-40% efficiency from a well-optimized steam boiler/turbine. Even operating in efficient temperatures, steam is economically impracticable (depending on the cost of labor in your country) below a break point of 10 megawatts to 200 megawatts.
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Lead acid battery charging too fast I have an ebike. I usually charge my lead acid battery for 6 hours until the charger shows green light, and I can travel for 30 Km or so. But since last week, green light indicating full charge appears after only 2.30h, and I can only travel 10 Km max, on low speed. If I try to charge it with another charger, same thing happens. I tried everything, but I can't get the bike to charge after 2.30h. Does anyone know what could be the problem? <Q> It’s time for a new battery. <S> At least one cell in yours has lost capacity, probably the result of a broken plate or internal connection. <S> The vibration experienced in E-vehicle applications is very hard on batteries. <S> There's no reasonable way to repair it, so recycle it and get a new one. <A> The faster you charge your battery the shorter life you can get from your battery (generally). <S> When charging a lead acid battery, take into account the charging stages. <S> Some chargers tell you the current stage of charge weather be in bulk, absorption, or float. <S> Each stage requires CV or CC modes which your charger doesn’t have. <S> Try charging slowly, it might help( lower the current ) <S> ( I might be wrong but this all I can remember when working around lead acid batteries) <A> Unless it's a very old battery that requires maintenance, such as refilling cells regularly, the likely solution is to replace the battery.
| It's likely that a cell of your battery has failed with a high ESR, so that the overall voltage of your battery is lower when running, but higher when charging.
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Phone to TTL output? I am looking for a way to get my phone to output a TTL level signal, up to about 1200 baud. Any suggestions? The only thing I can think of is using an audio output with a square wave signal, and then amplifying it. Is that workable? Anything better? <Q> Use a non-linear amplifier to get TTL levels, in the form of a comparator with it's threshold set a the mid-point of the audio waveform level. <S> Possible devices might include an LM360. <S> This won't work for frequencies near zero, as audio output will likely stay close to the threshold and near the noise floor, producing garbage. <S> But will work for a high enough minimum toggle rate. <S> If you output a lower frequency, say 2400 Hz, to indicate TTL high, another possibility is to use an Arduino chip, audio to ADC input, and measure the time between threshold crossing to detect the tone frequency needed to output a high level on another IO pin. <S> But this lower audio frequency will limit the minimum duration of your TTL pulse transitions (and quantize them in time). <S> A higher audio frequency will also quantize the TTL output in time, but at a higher time resolution. <S> So an appropriate USB audio DAC might allow very high audio signaling frequencies by supporting a sample rate of 96k or 192k. <S> This would likely provide even better time resolution for your TTL output than signaling via USB (OTG) serial or ethernet packets, as a streaming real time audio driver on a mobile device (RemoteIO on iOS) has much more deterministic latency than either the USB or the network stack drivers on the mobile OS. <A> a way to get my phone to output a TTL level signal <S> I assume that this is just about adding a TTL output to a smartphone, not about transmitting data through a telephony channel, since you're not mentioning that anywhere. <S> The only thing I can think of is using an audio output […] <S> Anything better? <S> Modern phones have a USB on the go port, meaning that they can act as host to USB devices. <S> For example, USB-to-serial (TTL) adapters. <S> So, plug one of these into your phone, and write an app that has access to USB and interfaces with that adapter. <S> Usually, if you're using one of the more popular USB-to-serial adapters, it will be known to the Linux kernel on an android phone, and unless the manufacturer stripped out that kernel module, you'll see it appear as /dev/ttyUSBx, and can just be dealt with like on any other Linux machine. <S> Same would be possible via bluetooth, since that has a general-purpose serial profile, too. <S> Other options include: Blink the LED from software. <S> Photodiode or -transistor, bam. <S> send out bluetooth LE beacons with the desired bit sequence as payload and pick them up with a BLE-enabled microcontroller board. <S> display a "bar code" on your phone with say 17 bars, and place phototransistors on top of each bar, add a 16-bit parallel-in-serial-out shift register, each parallel input one photodiode, clock with the 17th. <A> That is how computers and synthesizers saved data to cassette tapes back in the day. <S> Some had analog audio interface to work with normal cassette recorders, and the ones that had TTL interface just had a similar analog interface stage in the tape deck. <S> You don't even need amplification. <S> Just AC couple the signal and feed it to a comparator circuit, preferably one that has some hysteresis built into it.
| If you need to get a TTL output toggling at any frequencies from 0 to 1200 Hz, then another strategy is to output a much higher frequency audio signal (say 12 kHz) for the duration of your TTL high, and use a bandpass filter on the output to filter and detect this tone (diode + cap, similar to a AM crystal radio detector) before the TTL output comparator (with its threshold set above the zero-output noise floor and below the bandpass filter ripple).
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Can I use a voltage regulator to bring high voltage down to 3.3V GPIO input for ESP32? I have a button connected to the power rail of my project and I want to know when it's pressed with an ESP32 chip. Since the input voltage (6.5V to 15V) is higher than the maximum rating of the ESP32 I need to bring it down. Like this video of a power switch circuit , I used a current limiting resistor of 10K ohm between the button and the chip, relying on the protective diodes of the GPIO pin. But when I measure the voltage it's 4.1V which is still higher then the maximum spec of the ESP32; so I'd like a safer way. Is this circuit using a 3.3V voltage regulator appropriate? I use a 1k ohm resistor to load the voltage regulator, and 2 10uF caps as suggested in the datasheet. The low response time of the button should be fine. simulate this circuit – Schematic created using CircuitLab <Q> That would still pull up to 5V via the 10k resistor. <S> Simply use a transistor, it will be cheaper than regulator and caps. <S> Pushbutton would drive the transistor and the transistor would drive the input pin. <A> Yes, it will work if you use an LM7833. <S> They're not that common . <S> You could also alter the design to substitute a 3.3V regulator that can easily withstand 15V input and alter the output capacitor to suit (often ceramic output capacitors are unsuitable). <S> Or use an LM7805 or 78L05 <S> (TO-92 package available) and add a 2K+3K divider on the output, with that exact circuit and it will work. <S> You can reduce the capacitors to 100nF. <A> Do you really mean "ESP32 chip" or rather "ESP32 dev board"? <S> Actually... it does not matter. <S> In either case there is 3.3V source somewhere powering ESP32 MCU. <S> All dev boards have VCC routed to some pin. <S> Simply use that one for your button, no extra parts except for one pull-down resistor, and 100% guarantee it won't be out of input range, even if VCC drops for some reason.
| There are a few places where it can go badly.. many fixed-output 3.3V regulators like to go unstable with the wrong type or value of caps or have a maximum input voltage that is not high enough for your application (sometimes as low as 5.5V maximum). I would also recommend about 100R resistor between the button and GPIO and 0.1uF capacitor in parallel to the pull-down.
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why use of 1uf capacitors in max232 ic, why does not use 10uf capacitor in max232 ic.? I have use 1uf capacitor in max232 circuitc1=1uf (pin 1 and 3)c2=1uf (pin 4 and 5)c3=1uf (pin 2 and vcc)c4=1uf (pin 6 and gnd)c5=0.1uf (pin vcc and gnd) why this type of values used? <Q> Because if datasheet says 1uF for the charge pump then that is the best value it works with the given frequency of the charge pump. <S> Higher capacitance charges more slowly so the charge pump would need slower frequency or it would take longer for the pumped voltage to settle. <S> There is a difference between MAX232 and MAX232A, the former needs 1uF capacitors while the latter only needs 100nF capacitors. <A> Because those are the values the datasheet says it was designed for: It doesn't say why, and it doesn't show the internal circuitry so you really have no choice but to follow its recommendation. <S> The capacitors are used to generate the voltages needed for true RS232 operation. <S> The positive output is boosted to 7 volts to make up for line and load losses. <S> The required negative voltage is generated from the given 5 volt supply. <S> It is inverted to give a negative output at about 7 volts. <S> The internal boost and inverter circuitry is intended for use with 1 microfarad external capacitors. <S> It might also work with higher (or lower values) but the manufacturer doesn't mention that option or give you any information you would need if you were to use a different capacitor value. <S> It might appear to function if you use different values, but then generate bad signals at random because the charge pumps couldn't hold the output voltage steady. <A> The correct capacitor for reduced conducted noise across Vdd,Vss is also 1uF . <S> This older technology also has higher switch Ron ~300 ohms . <S> If you upgrade to Max201 then they can all be 0.1 uF caps .
| Lower capacitances would charge too fast and can not transfer enough charge per clock cycle, so the voltage would droop under load.
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How do electricians calculate the maximum allowed on circuit breakers if the current is measured in AC? Correct me if I’m wrong but the majority of electronics circuit operate of a 3V3 constant DC voltage, which the current would be reasonably stable draw apart from charging up capacitors, chips loading up, so adding up current is relatively simple ie. i1 + i2 ... etc. =iTotal but most of appliances are powered from mains drawing an AC current which means a sin waveform, so the equation becomes more difficult i1sin + i2sin .... etc. , so to answer my own question would I be right in saying that circuit breakers are rated for total AC current? <Q> The breakers are chosen for fire safety and the standard practice is to have a current rating near the power entry on an imprinted label to define the RMS current for safety to calculate the branch current requirements. <S> The electrician simply adds all these ratings to ensure the wire and breaker ratings “meet or exceed” the local standards and use the sum total by a safe margin to prevent nuisance trips from high inrush motor surges. <S> So breaker ratings do not include inrush current, which for charging up caps or starting motors can be ~10x nominal , when starting at peak Vac, unless reduced by design. <S> These vary with input voltage range and within the range so a worst case rating is used on the label, not nominal. <S> The device must have its own protection if it a low power device. <A> RMS Current is what heats up wire or any ohmic resistance like RDs on in mosfets <S> .PVC copper wire has max safe temperature ratings so the RMS current is the most important parameter here. <S> In AC circuits the RMS current is generally greater than the Average Rectified current .With <S> pure DC <S> the Average current is the RMS current .For <S> a pure Sine wave the RMS current is 1.11 times the average Rectified <S> current .This <S> factor is called the form factor and can be regarded as a figure of merit where 1 is ideal .Bad <S> harmonic rich waveforms can have form factors of 2 or more .This <S> is one of several reasons why breakers have to be specified higher than one would initially think . <S> A real example of this is a 100 Amp single phase battery charger that despite having its average rectified current strictly limited to 100 amps blew a 120 amp test fuse . <A> Circuit breakers are regulated by electricity building norms. <S> The idea is to break the circuit soon enough, before nothing gets overheated, for ex. <S> the wires inside the walls. <S> The currents are in those norms said as RMS values. <S> That means an amperage of DC which would have the same thermal effect in wires and resistors. <S> If we say RMS AC current is 1 ampere, the sinusoidal current has peak value 1,41 A which is the RMS current multiplied by sqrt(2).
| Electricity building norms state maximum allowed circuit breaker ratings vs wire length, wire cross section and how the wires are insulated and placed inside walls, floors, assembly tubes etc...
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Offset when trying to convert square to triangular wave Hi I constructed a circuit to create a 1Hz square wave with LM741 Opamp. The first part makes that and it makes it perfectly. But I need to convert it to a 8Vpp triangular wave. Without any offset, I simulated the circuit below in LTSpice it was flawless. But in real life I got some weird offset from the output of the triangular wave converter. Figure I implemented, first part creates 1Hz 21 Vpp zero offset square wave,2nd part convert this into a triangular wave with 8Vpp: In real life it has 8.2 Vpp which is very good for me but it starts from -6 to +2 (it has -2 V offset) How can I get rid off this offset? Better one: <Q> Integration isn't memoryless, and can be influenced by initial conditions, in this case the voltage across the cap. <S> On integrators it's not uncommon to see a switch used to discharge the capacitor to make its initial state 0V. <S> Try shorting the cap just before operation and see if the offset is reduced. <A> If you are using a 10uF electrolytic capacitor, it won't be happy with more than a volt or so reverse bias and will create a large offset, which will be multiplied by 5 in the integrator. <S> Similarly, the 330uF will not be happy with reverse bias. <S> You can get around the first by connecting the capacitor (-) terminal to the negative rail, but for the integrator you would need a non-polar electrolytic, or to increase the resistor values to more like 130K/680K for R4/R5 and use two 10uF electrolytics in series (back to back). <S> Or use 68K/330K and a 10uF film capacitor. <S> 10uF ceramics are available but they have so much voltage coefficient <S> your results will be quite bad (very nonlinear triangle wave and much higher amplitude and frequency than expected for ideal caps). <A> Others have explained the need for nonpolar capacitors, but you are doing things backwards. <S> You don't try to convert a square wave to a triangle wave, you make a triwave oscillator of the right frequency and amplitude. <S> There is already a useable waveform on C1, you just need to change the ratio of R2 and R3 so it has the right amplitude. <S> This will change the frequency so recalculate the value of R4 to bring it back to 1Hz. <S> Make it closer to 1M <S> so you have a chance of finding a suitable capacitor. <A> That's the classic problem you get when trying to drive an integrator from a square wave. <S> The integrator's output will drift towards one supply or the other and may eventually saturate because:- 1) <S> Square wave is not perfectly symmetrical (square). <S> The saturation limits for the square wave generator are not symmetrical. <S> 2) <S> The integrator op amp has input offset voltage and input bias currents. <S> You can't use a polarised cap because it is being to charged to both polarities. <S> Use a non-polarised cap or keep cap value lower and use a non-electrolytic. <S> Below is a favoured way of generating a triangle wave. <S> It uses a non-inverting schmitt trigger driving an integrator. <S> Triangle wave output amplitude is set by ratio of R1 to R2. <S> Frequency is set by R3*C1 time constant.
| My guess is that there was some stray voltage on the cap that causes your signal to swing around that offset.
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Thermistor scavenge from digital medical thermometer To be more specific, there's this thermistor I scavenge from an old medical thermometer. Something like this Cut to the chase, I connected my multimeter to the two wire of the thermistor where my multimeter shows a resistance of 40.9 kilo ohm at normal temperature. Blasting it with the heat of my soldering iron and I got a resistance of 8.8 kilo ohm (I'm really sorry for the bad angle. It's hard holding it here while taking a shot) When trying to cool it off using my refrigerator, the resistance would become higher than 40.8 kilo ohm. The question here is, Is there a way to calculate the temperature from the resistance it gave? <Q> Those cheap home thermometers use a (relatively) precision NTC thermistor that was first developed in Japan in the late 1980s (IIRC). <S> Of course that only works if the unit is still functional. <S> If not you can try to create specific temperatures, maybe by mixing warm and cold water, and calibrate with another accurate thermometer. <S> A typical spec for the type of thermistor used in this kind of consumer application is 54K + <S> /-10% at 37°C and \$\beta_{32-42} = <S> \$ <S> 3700 +/-2%. <S> The thermistor will work over a wider range, but it's specified for measurements of human body temperature. <S> The loose reference resistance spec means that the measuring circuit has to be calibrated (each unit individually) with a trimpot or digitally, since that represents a few °C tolerance, which is totally unacceptable tolerance for the application. <S> Once you have the two parameters, as above, you can calculate the temperature from the resistance using the simplified Steinhart <S> –Hart equation (c = 0). <S> \$\frac{1}{T <S> } = \frac{1}{T_0} <S> + \frac{1}{\beta}\cdot\ln{(\frac{R_X}{R_0})}\$ where T is in Kelvin, R is in ohms. <S> So in the above example, T0 is 310.15K (37°C). <S> If you end up trying to calibrate the thermistor for a more general purpose application, more common reference temperatures used for the measurement of beta are 35°C and 85°. <S> You can mix hot and cold water in a large pot or whatever to get stable temperature readings on a second reference thermometer, (assuming you lack a laboratory dry block calibrator). <S> The resistances at lower temperatures are fairly high, so you need to keep moisture out of the probe. <S> If this sounds like a lot of trouble, you can buy a thermistor with known specifications for a dollar or so. <S> You might want to check and see if my values happen to be correct for your part (you can very easily check at 0°C and 100 <S> °C using ice-water slurry and boiling water- <S> the latter check will be less accurate) and if they don't pan out, toss the thing. <A> The relationship between temperature and resistance for a NTC thermistor is defined by the Steinhart-Hart equation. <S> For this equation, you need 4 constants that are provided by the thermistor manufactures. <S> The equation is: 1/T = a + b*(ln[Rth]) <S> + c*(ln[Rth])^2 <S> + d*(ln[Rth])^3 <S> -273.15 <S> T is the temperature in deg C, Rth is the resistance of the thermistor, and a, b, c, & d are the 4 constants from the manufacturer. <S> Depending on the accuracy needed, you could drop one or both of the higher order terms. <S> Like Spehro said, it's probably a lot easier to just buy a new part <S> and then you'll have the characterization data for it. <A> Is there a way to calculate the temperature from the resistance it gave? <S> Yes. <S> It is called calibration . <S> Measure the resistance for known temperatures (like ice water, boiling water, etc) and try to figure out the mathematical relation (it doesn't necessarily need to be linear). <S> Once you found the mathematical relation, you can relate measured resistance to temperature.
| The easiest way to figure out the thermistor specifications is to replace the thermistor with a variable resistor and find the values that give specific readings on the display.
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load of an outlet I have an incubator with an internal outlet that, according to the manual, "can provide a maximum of 1A". I'm trying to understand if this actually means that the maximum load is 120Vx1A=120W? In other words, is the limit set by the power or by the current? I ask because I would like to connect a shaker to the incubator's internal outlet, and the label on the back of the shaker says "19V, 4.74A", i.e. its load is 19Vx4.74A=90W. Will the internal outlet support the shaker? <Q> Read more of the label on the power supply <S> The power brick for my laptop says how much input current it needs, yours should too. <S> That is the number that will determine if that outlet is suitable. <S> The power supply brick for my laptop says "input 120-240V 1.3A" so <S> this power brick would not be suitable (this is a 45W power supply: 20V, 2.25A), so based on that it's not looking good for your 90W power supply being compatible with that outlet. <S> You my need to ask the vendor to read the details on the power supply.this page seems to have their contact info https://www.boekelsci.com/contact/general-info.html/ <A> I have an incubator with an internal outlet that, according to the manual, "can provide a maximum of 1A". <S> I'm trying to understand if this actually means that the maximum load is 120Vx1A=120W? <S> In other words, is the limit set by the power or by the current? <S> You should check how it is wired inside the device, but... <S> usually this means the outlet is controlled by a switch (in the case of an incubator, probably a thermostat) and the switch is rated for a 1 Amp current. <S> Either that, or they used very thin wires for cost. <S> I would like to connect a shaker to the incubator's internal outlet, and the label on the back of the shaker says "19V, 4.74A", <S> So your shaker is powered by a 19V power supply which probably looks like a laptop brick. <S> You have to use the power supply of course, you can't connect a 19V device to a 120V outlet, but you know that already. <S> Check the INPUT <S> current rating on the power brick label. <S> Something like "xxx mA 120V". <S> This will probably be less than 1A, so it'll be OK. <S> The output current rating like "19V 5A" does not concern the mains side of the adapter. <S> Since these supplies tend to draw a very large inrush current when plugged in, due to capacitor charging, I'd recommend checking if the incubator internal outlet goes through a switch or a thermostat. <S> If it does go through a thermostat rated for a low current, like a few amps, I'd rather avoid having it switch the 19V power brick, as the thermostat contacts could be harmed by repeatedly switching a load with high inrush current. <S> If the internal outlet is only switched on/off by a manual switch on the incubator, then this won't be a problem as the manual switch won't be operated very often. <S> Contacts only wear when they switch on/off. <A> The real key here is that IF the outlet provides 120 V power (you didn't say one way or the other), and the shaker requires 19 V. <S> You will damage or destroy the shaker if you plug it into a 120 V outlet. <S> What kind of plug does the shaker have on it's power cord? <S> I would be surprised if it's a standard 3-prong plug that's intended for 120 V use. <A> The circuit capacity is 1 amp x your line voltage, in VA <S> So if your line voltage is 120V, the socket supports 120 VA. <S> Watts is (roughly) <S> the part of the sinewave the appliance uses, but it doesn't tell the whole picture about the thermal heating that occurs in the supply wires. <S> Look at the thing that has an AC power plug <S> Your shaker doesn't plug into a 120V outlet, so its nameplate is irrelevant. <S> It plugs into a wall-wart or power brick; its nameplate is relevant. <S> That accounts for thngs like PCs with 850 watt power supplies that actually only have Intel on-board video instead of the dual 980s and six hard drives the power supply is sized for.
| If you don't agree with or believe the nameplate, you can get a tester like a Kill-a-Watt and actually measure the VA of the appliance at its most extreme operating mode, and go with that figure instead.
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Reverse polarity protection on 5-12V input and output Can someone help me with reverse polarity protection on 5-12V input and output using a using a P-Channel MOSFET? How can I calculate which MOSFET, resistor, zener diode I have to use? I use a 5V 10A power supply. <Q> A common rule of thumb for improved long-term reliability is to overrate electronic components by 100%. <S> Examples: <S> For a 12 V system, use a FET rated for 30 V or more. <S> For a peak load current of 3 A, use a FET rated for 6 A or more of continuous drain current. <S> Because your operating voltage range is relatively low (in FET terms), select a "logic-level" FET. <S> Check the datasheet, but they almost always are rated for 20 V Vgs, so you don't need a zener to protect the gate unless the reverse voltage might be way below -12 V. <S> A power MOSFET has an internal zener diode between the gate and source. <S> Pay close attention to the orientation of this diode in your circuit. <S> If your system can tolerate having the MOSFET switch in the GND rather than in the supply, an n-channel MOSFET will cost less and perform better. <S> Plus, the circuit is more simple. <A> You want a MOSFET that turns on with less than 5V on the gate, so "logic-level", for the zener it should be less than the max Vgs of the MOSFETyour MOSFET needs to be able to handle the max current draw of the device - perhaps 10A, perhaps less.the resistor is less critical 1K or 10K or 100 <S> K makes little difference. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> These are LED power supplies. <S> I don't know what application you have , but these power supplies do not need polarity protection. <S> It is the loads that need it. <S> Also 12V LEDstrings do not need reverse protection because they can handle it. <S> I suggest you use polarized connectors rated for 10A. <S> If you only need 1 , these are pretty common and can be paired and cut in half with soldered pigtails at one end and screwed into terminal block on supply. <S> Plan on using tiny tie wraps to tie holes to prevent stranded wire strain and breakage. <A> I wonder why this has not been suggested, but I think you could simply use a diode bridge (section "rectifier") as you would do to rectify AC voltage.
| I would use schottky diodes, because they have a very low forward drop and therefore also low power dissipation in a continuous current situation.
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How do I calculate the total wattage of a dc to dc circuit? I am trying to figure out the total wattage of the below assumed dc to dc converter circuit block diagram. Can I get it simply by just multiplying the total SUM of the system output voltages by total SUM of the system output Amps, like this: (3.3 + 3.3 + 6 + 6 + 9) X (.5 + .5 + 2 + 2 + 2.5) = (27.6V) X (7.5A) = 207W ??? Is that how we calculate it? <Q> No. <S> Those are the correct units, but the answer will be off. <S> This problem can be solved with simple physics plus common sense; specifically the conservation of energy plus the fact that no power conversion device is 100% efficient. <S> You need to calculate the power available from each output, by multiplying its current by its voltage (e.g. each one of your two 3.3V supplies needs 1/2 A, and so provides 1.75W). <S> Then you need to add all of those together. <S> The number you get is the maximum possible output power. <S> It's also the theoretical minimum of the maximum input power. <S> In reality you'll need to divide that by the efficiency of the supply (e.g. 80%). <S> If you really want to get down to details, each output will have its own efficiency, so you'd need to multiply each output power by the individual efficiency, then add those up. <A> Step and repeat for each sub-supply then sum all the wattages to get the total power but, don’t forget to take into account any inefficiencies in the individual voltage conversions. <S> Around 35 to 40 watts is your likely answer. <A> No. <S> It's V x A for each <S> , i.e. 3.3 V x 500 mA <S> + 3.3 V x 500 mA... <S> and once you understand that, you could add the currents for the two 3.3 V supplies and for the two 6V supplies. <S> Think of adding coins: 2 @ <S> $.05 <S> and 7 @ <S> $.25 <S> cannot be calculated by saying "That's 9 coins of $.30. <S> " <S> BTW, does the assignment mention the efficiency of the supply? <S> If you need 20 W out and the supply is 50% efficient, you'd need twice as much, 40 W, input (and get 20 W of heat).
| No, you multiply each individual voltage by the current taken by the load on that individual voltage to get the power on that individual circuit.
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Should I ground the heatsink? I am using an npn transistor in TO-220 package, and its collector is electrically connected to the metal part on it. I have checked it via a multimeter. I use a heatsink for that transistor, with isolator. So after mounting it to heatsink with isolators, it is no longor electricaly connected, I have checked again. However, when I inspect the signals in the circuit via oscilloscope, I see there is some noise on waveforms, while trying to understand where it comes from, I have touched the heatsink with a metal, to be precise with the tip of the following instrument: It becomes much smooter. I didn't understand how touching it would affect the stability. After some reading on the web, I saw some say that heatsink itself should also be grounded. I wanted to have your thoughts on this. By the way, transistors(npn&pnp) are used in a class B audio amplifier. p.s : This doesn't answer my question. <Q> By the way, transistors(npn&pnp) are used in a class B audio amplifier. <S> This is an important detail. <S> The transistor tab is usually connected to the Collector pin, and your standard insulator will add about 15pF capacitance between the tab and the heat sink, so this is about knowing whether this capacitance can cause problems or not. <S> If the heat sink is grounded, then the collector will have 15pF to ground capacitance in series with the connection inductance. <S> Then the heat sink will also capacitively couple to whatever traces or signals are around. <S> If your amp uses an emitter-follower output stage, then both power transistors collectors are connected to power supplies (or ground if it is a single supply amp). <S> Since power supplies are decoupled to ground, there is already a low impedance HF path between collectors and ground, therefore grounding the heat sink should make no difference. <S> If your amp uses a common emitter output stage, then the collectors are connected to the output, and the capacitance will be between output and heat sink. <S> Some amps omit the insulator for better thermal resistance and have the heat sink connected to the output voltage, but this requires it to be inside an enclosure and not accessible. <S> Anyway, if you ground it, then you have to consider if the extra 15pF to ground could make your amp unstable. <S> For an audio amp, this is unlikely. <S> when I inspect the signals in the circuit via oscilloscope, I see there is some noise on waveforms, <S> while trying to understand where it comes from, I have touched the heatsink (...) <S> It becomes much smooter. <S> If the output voltage appears fuzzy on the scope, your amp could have stability issues. <S> You can try single shot mode and see if you observe oscillations or just noise. <S> If the amp does have stability issues, they are probably not related to case-to-heatsink capacitance, the problem would come from somewhere else, like layout or decoupling. <A> The electrical insulator such as Mica or Thermoplastic becomes a capacitor (100pF ??) <S> but if you press your finger onto it your fingerpad, it could be 10x bigger (1nF) and shunt the crosstalk. <S> If the insulator is needed for safety, ground it. <S> If just for DC and not personal safety but to protect the IC, it doesn't matter. <S> You know what to do. <S> For HV and highspeed FETS, IGBT's its a different game and load capacitance should be minimized during high dV/dt. <S> AS long as you have an RC snubber on the speaker output to suppress parasitic oscillations... 10R+0.1uF <A> Your ungrounded Heatsink is actually a not too shabby <S> capacitively coupled Antenna which radiates into sensitive low level circuits .Ground your isolated heatsink and <S> the problems will go away .The time not to ground is say a SMPS that generates lots of noise at the drain of the switching transistor that you do not want to be injected into the ground causing an EMC fail. <A> seeing as the transistor is electrically insulated from the heatsink, grounding the heatsink should not do any harm. <S> but note that grounding the heatsing will change the capacitance to ground on the collector of that transistor. <S> that may effect the way your circuit operates.
| However, if the heat sink is external, or part of the enclosure, and the enclosure is already grounded at another point, grounding the heat sink on the amp pcb will introduce a ground loop.
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In a Class-C power amplifier, what does the resistor in the base do? Look at the picture, which shows a Class C power amplifier ( source ). What is the role of the resistance in the base of the transistor, and what happens if it's removed? <Q> This resistor provides discharging of Cin. <S> Without this, Cin will charge up to peak voltage of Vin (decreased by Ube junction forward voltage of Q1). <S> As the result the Q1 will remain open all the time, so the amplifier will simply not work. <A> Summary Since the input impedance is low when Vbe>0.6V <100 <S> Ohms <S> the base resistor should R1 pulls up more than down to 0V. R1 forces the conduction duty cycle or phase conduction angle to increase. <S> The concept is to pump as hard as possible for the shortest % of time so that the Bandwidth of the filter can reject the modulation noise when saturated. <S> Removing R1 only reduces the pulse duration when there is sufficient drive as it should be. <S> If there is insuff,drive , it turns off. <S> This shows with a slider for R1 on the right . <S> You can also remove it and vary the drive level. <S> other Class C means pumped LC resonator with practical efficiency 50% to 80% as it only conducts in saturation as a switch for low duty cycle to apply some modulation frequency to the carrier. <S> Conceptual only <S> It also means "low fidelity" modulation <S> so BP filtering is essential. <S> REF <S> What is the role of the resistance in the base of the transistor, and what happens if it's removed? <S> The peak current and pulse width is then increased, no problem. <S> The example in question is not really useful but just a proof of concept. <S> Whereas the above is more practical of a high efficiency RF AMP. <S> Cin and Vbe form a positive clamp circuit such that the Vbe = 0.6Vpk at 1mA and 0.7Vpk at >10mA <S> and also the input impedance drops with the base current. <S> Thus the dV/dt= <S> Ib/Rbe for the peak only resulting in a sharp narrow pulse on the peak input <S> if ~2Vpp Increasing the amplitude just saturates the Vce(sat) more and the current and carrier amplitude then increases. <S> Reference <S> Advantages of Class C power amplifier. <S> High efficiency. <S> Excellent in RF applications. <S> Lowest physical size for a given power output. <S> Disadvantages of Class C power amplifier. <S> Lowest linearity. <S> Not suitable in audio applications. <S> Creates a lot of RF interference. <S> It is difficult to obtain ideal inductors and coupling transformers. <S> Reduced dynamic range. <S> Simulation <S> Values not intended to be a useful design, just to show some characteristics. <S> It could be used for FM with a varicap to modulate the frequency and keep at constant envelope. <S> Crude example with 10KHz pump <A> A simple answer: <S> The capacitor Cin fully charges through the base-emitter junction during the positive half wave of Vin and slightly discharges through R1 during the negative half wave. <S> Thus it can be thought as of a "rechargeable battery" connected in series to Vin. <S> During most of the positive half wave, its voltage subtracts from Vin and it acts as an opposing voltage source that shifts down the base voltage with about 0.7 V. <S> Only at the top the effective voltage is enough to turn on the base-emitter junction and accordingly, the transistor. <A> In an rf amplifier that prevents parametric oscillation. <S> In a lot of Cb type amps it is omitted, but that isn't proper. <S> That is all. <S> The ARRL manual for the Radio Amateur discusses this in the rf amplifier section..... <S> At least it used to. <S> --Shane <S> KD6VXI
| The capacitor in series with the base is provided to keep DC off the input from rectification of the drive signal.
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Are 1uF NP0/C0G capacitors different than regular 1uF capacitors? I am reviewing an schematic and it contains several reference to 1uF 0402 SMD capacitors, some of them are marked simply as "1uF" and others as "1u, NP0/C0G" , I think both are the same, is there any difference? <Q> No not the same. <S> A 1uF C0G, NP0 capacitor (if you can even find one!) will be very large and expensive (definitely not 0402). <S> The default for a bypass capacitor or similar non-critical application would be something like X7R or X5R but that should be specified. <S> The voltage rating should also be specified. <S> C0G, NP0 refers to an EIA Class <S> I dielectric made with very low temperature coefficient and voltage coefficient, typically specified for capacitors in the pF <S> but now easily available up to many nF <S> , however uF levels of capacitance are (today) not easily available. <S> At higher capacitances, dielectrics such as X7R and X5R, X5S and so on EIA Class II diectrics are used because they allow a physically small capacitor to be manufactured inexpensively. <S> The designer must work around their many non-ideal characteristics, which include (sometimes extreme) temperature and voltage coefficient (the capacitance may drop by 80% or more at rated voltage) but also microphonics (they tend to generate a voltage from vibration) and aging (the capacitance decreases with time). <S> There are also Class III dielectrics such as Z5U <S> but they are significantly more horrible again in characteristics. <S> Graph below is from Kemet , a major manufacturer of this type (MLCC = Multilayer Ceramic Capacitors). <S> The graph emphasizes the temperature characteristics (and the relative values of the dielectric constants) but the voltage characteristics can be at least as important in design. <A> All ceramic capacitors have a different classification for their dieelectric. <S> Other dieelectric classes are X5R, X7R, Y5V, etc. <S> The dieelectric class only denotes their temperature range and capacitance stability within that range (DC bias stability not included). <S> C0G/NP0 means it is specifically calling for an capacitor whose capacitance is accurate (tight tolerance), stable with temperature and DC bias, and does not suffer from piezo-electric effects (microphonic noise). <S> Usually for timing or filters. <A> 1uF, 0402 size in NP0/C0G is not available: Per here <S> http://www.avx.com/products/ceramic-capacitors/surface-mount/c0g-np0-dielectric/ C0G (NP0) is the most popular formulation of the “temperature-compensating,” <S> EIA Class I ceramic materials. <S> Modern C0G (NP0) formulations contain neodymium, samarium and other rare earth oxides. <S> C0G (NP0) ceramics offer one of the most stable capacitor dielectrics available. <S> Capacitance change with temperature is 0 ±30ppm/°C which is less than ±0.3% C from -55°C to +125 <S> °C. <S> Capacitance drift or hysteresis for C0G (NP0) ceramics is negligible at less than ±0.05% versus up to ±2% for films. <S> Typical capacitance change with life is less than ±0.1% for C0G (NP0), one-fifth that shown by most other dielectrics. <S> yet: Features & Benefits C0G (NP0) formulations show no aging characteristics and have an operating temperature range from -55°C to +125°C <S> Voltage Range: 6.3V to 500V <S> Capacitance Range: <S> 0.5pF to 0.1µF <S> Sizes: 01005 to 2225 <S> More: <S> https://www.allaboutcircuits.com/technical-articles/x7r-x5r-c0g...-a-concise-guide-to-ceramic-capacitor-types/ <S> Class 1 Caps <S> As you may have noticed in the chart, C0G is extremely stable (note that C0G and NP0 both have a zero, not an uppercase “O”). <S> C0G is a Class 1 dielectric and an all-around capacitor superstar: <S> the capacitance is not significantly affected by temperature, applied voltage, or aging. <S> It does, however, have one disadvantage that has become particularly relevant in this age of relentless miniaturization: it is not efficient with respect to volume. <S> For example, if you go onto Digi-Key and search for a 0.1 µF C0G cap, the smallest in-stock part is a 1206. <S> In contrast, you can find a 0.1 µF X7R cap in the 0306 package, and with a voltage rating (10 V) high enough for 3.3 V or even 5 V circuitry.
| But unlike the other classes, C0G/NP0 is known to have additional characteristics and the label carries additional connotations.
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Does current still flow through ballast with fluorescent bulbs removed? I have a 3 fixtures each with 4 fluorescent tubes with ballast replaced about 8 years ago and these fixtures are connected to the same switch in the kitchen. This am there was an smokey smell and the fire department believes it originated from a bad ballast. I removed all tubes from the offending fixture of 4 and I am wondering if current will still flow through the 'bad' ballast if I turn on the lights to the other 2 banks (again on same switch). <Q> Yes, it could . <S> Image source <S> Here is a somewhat typical ballast. <S> As you can see, if one of the many components has failed short, current will flow from active to neutral. <A> If they are electronic ballasts, they're like power supplies, so even without tubes installed they will be powered. <S> However if there was smoke or "fried electronics" smell it is possible that some wires have melted insulation <S> so I'd really recommend opening up the fixtures to check. <S> Cut the power from the panel, you never know if the light switch really does cut the live... or if Protection Earth is actually connected... <S> Better safe than sorry. <A> Remove the ballast cover from the fixture, if the ballast is not already exposed. <S> This cover attaches with screws or inserts into slots cut into the fixture. <S> Replace the ballast if it appears burned, swollen or is leaking oil. <S> IMPORTANT: <S> They can even cause a starter to burn out! <S> Don't wait too long to fix the problem or you may end up with a bigger repair! <S> " Rapid Start" <S> use FL heaters in ends and are in series. <S> Which type do you have? <S> Tube p/n? <S> Modern "instant start" <S> passive Ballasts short both end pair pins in tubes and source and do not use a heater in tube. <S> " Instant Start" use high voltage strike with no heaters and are in parallel. <S> All Yellow are common. <S> They also have shunt ballasts so each other tube still works if any one fails or is removed. <S> Usually the failure is not so much the ballast but the tubes after 5 years when they get contaminated and draw excess current. <S> If the ballast is not protected and left too long , it begins to overheat. <S> Modern tri-phos tubes should last 10 yrs or 50kh. <S> Conclusion Smells like the series old tar type to me and not the new ones. <S> No worries for now. <S> No problem with tubes removed. <S> Recommendation Replace all ballasts and tubes with new parallel type, lower power <S> more efficient 86 LPW similar to many older LED types. <S> I prefer tri-phosphor true white only. <S> 4500'K
| If they are oldskool inductive ballasts, then removing the tube will open the circuit. Flickering fluorescent tubes can cause the ballast to overheat and fail prematurely! The diodes in the rectifier part of the circuit or one of the transistors can fail in this way.
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How to choose the appropiate analog power supply filter inductor I'm currently studying designs for a pulse oximeter for use with Arduino, I have seen two Schematic designs form the same company, https://www.protocentral.com/sensors/1113-protocentral-afe4490-pulse-oximeter-shield-for-arduino-v2-642078949425.html and https://www.protocentral.com/open-source-health/1252-healthypi-v3-basic-hat-kit-for-raspberry-pi.html . Both of them are identical and are base around the AFE4490RHAT IC from Texas Instruments. Both designs have 4 inductors : L1, L2, L3 , L4. 1) What are these inductors used for in the design? 2) In one of the designs they are using 4 CIG22L100MNE and in the other 4 LPS3010 (they don't specify the full part number) but the Samsung one seems to be discontinued and the other one seems to be not recommended for new designs, I'm trying to choose and equivalent I have compared the specifications between the two parts and a part from the inductance they seem to have very different properties regarding current rating, DC resistance and inductance frequency, what specifications are more important when choosing a replacement part? <Q> 1) <S> The inductors are used to filter noise from supply voltage. <S> As each supply node has its own inductor, noise caused by periodic gulping of current on one supply node cannot pass on to other supply nodes and cause problems taking the sensitive measurements. <S> 2) That is a good question and there may not be an answer. <S> It seems that the evaluation kit for the pulseoximeter chip used a 10uH inductor LPS3010-103MLB everywhere, most likely to reduce cost by using same part for everything, and that reference design got copied everywhere. <A> 1) What are these inductors used for in the design? <S> Each inductor forms a LC low pass filter (L1 & C19, etc), to filter out noise originating from the 3V3 source to which they are connected to. <S> The cutoff frequency is determined by $$ f_c = \frac{1}{2 <S> \pi \sqrt{LC} <S> } \approx 16 \text{ kHz} $$ 2) <S> […] what specifications are more important when choosing a replacement part? <S> If you don't know the current requirements of the loads (labelled as LED_DRV_SUP, etc) I would just pick the same current rating (or higher). <S> As shown above, the value of inductance is determining the cutoff frequency of a LC filter. <S> As I don't expect it needs to be exactly 16 kHz, you can change the value: 8.2 uH or 12 <S> uH will probably do as well. <S> It can be useful looking at distributor's websites for alternatives for an obsolete component. <S> E.g. Arrow gives as (one of the ) alternatives to the CIG22L100MNE <S> the CB2518T100KV which also has a 10 uH inductance, a 0.6A current rating, but just a smaller package shape <A> As others have noted, the inductors are part of the noise filtering. <S> They are extremely non critical - within reason "some inductance of about the right value" is liable to be suitable. <S> Pulse oximeters are liable to be moderately sensitive to noise but <S> not immensely so - the signal levels involved are small but not (usually) in the microvolt range. <S> Good layout and design - (PCB tracks, capacitor locations and quantity, ...) are liable to be as important as inductance values.
| So as long as the parameters are reasonably same almost any inductor could be used.
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Raspberry Pi with 74HC14 connecting to LED strip (WS2812B) This might be a stupid question, but I can't seem to figure out how to convert 3.3V to 5V with the 74HC14. I have a power source which can supply 5V (see the power adapter at the left), and use it to power the Raspberry Pi and the converter. The Raspberry Pi has GPIO at 3.3V, while the LED strip (WS2812B) operates at 5V. This means I would like to use a logic-level converter to convert the 3.3V from the RPi to the 5V for the LED strip. Unfortunately, I do not have a lot of experience in schematics, nor can I find any possible location where they use the 74HC14 the way I need it. I could use any help needed! Design 74HC14 (from this forum) EDIT 1: It appears that the 74HC14 does not support 3.3V input, but as suggested by @Justme, the 74LS14 seems to work well. I also found the TXB0108, should that also be useful, or is it better to use the 74LS14? EDIT 2: Thanks to the comments section, I stumbled upon the 74HCT125, which might work I've updated the design based on the suggestions, is there anything wrong with it or that could be improved? <Q> The 74HC14 does not work with 3.3V logic input when it is powered from 5V. <S> The original schematic uses 74LS14 which does work, and for example 74HCT14 would work too. <A> Thanks to @Erlkoenig <S> Down below is an updated schematic: <A> I use these light strips all the time. <S> They should work fine. <S> The 3.3v from your MCU is above the logic level '1' for the leds, which is all it needs. <S> The only precaution is to NOT supply a data line that is OVER the supply voltage to the leds. <S> Meaning you can't power them with 3.3v and run the data line from a 5v MCU. <S> As a side note, I have found that these light strips will work fine off a 1S LiPo battery without a regulator down to about 3.2 volts, at which point they will stop working. <S> This means you can feed the battery directly to the lights, and then run the battery though a regulator to power your project with 3.3v and connect the MCU data line directly to the lights. <S> That way you can get the maximum current to the lights without worrying about power dissipation from a regulator. <S> Hope <S> this helps some out...
| You can power them from the 5v source and used the data line directly from the MCU to drive them. I stumbled upon the 74HCT125N, which I can confirm works very well.
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Can LM339 comparator inputs go way above Vcc? The datasheet for the LM339 quad open-drain comparator shows the absolute max values: \$V_{cc} \le 36V\$ (Supply voltage) \$V_{I} \le 36V\$ (Input voltage range (either input)) I want to compare a (fairly weak, 100k impedance) \$0V\$ to \$+15V\$ input signal against a \$+3V\$ reference, while powering the device from a \$+5V\$ supply. According to the spec, that should be fine. However, I'm surprised and wary, because it's common that device inputs can't go much above the supply rail. Then again, I can imagine comparators being unusual in this respect, and if that were the case here, I'd expect to see the input defined as \$V_{I} \le V_{cc}+0.6V\$ , or similar. Have you used an LM339 like this, and was it fine? I've been looking for a circuit diagram of the internals of the device, but I'm struggling to find anything that makes it crystal clear. I don't want to find out in a year's time that we're gently destroying them! <Q> "Absolute maximum ratings" are about not destroying the device. " <S> Electrical characteristics" or "Recommended operating conditions" are about proper functionality. <S> For example, a supply voltage of zero volts is perfectly safe (well within maximum ratings!) <S> but of course the device won't operate... <S> So, your voltage seems allowed by max. <S> ratings, but will the comparator compare? <S> "Electrical Characteristics" in the datasheet: <S> There is fine print: <S> What's important is "will provide a proper output". <S> Some comparators/opamps will misbehave or reverse the output if input voltage is outside correct operating conditions (but still inside maximum ratings), for example. <S> In this case it will work. <S> Datasheet page 13 confirms: <S> Most likely the internal transistors (highlighted in yellow) have been designed to have quite high Vebo rating , so the b-e junction doesn't break down under such high inverse voltage. <S> Usually discrete BJTs have Vebo around 6V so the b-e junction would break down and avalanche around 6V when reverse biased. <S> Also there are probably ESD protection diodes from inputs to GND, but not from inputs to positive supply. <A> Figure 7.7 of the LM339 datasheet . <S> Note 3 says: (3) <S> The voltage at either input or common-mode must not be allowed to go negative by more than 0.3 V. <S> The upper end of the commonmode voltage range is V CC+ – 1.5 V; however, one input can exceed V CC , and the comparator will provide a proper output state as longas the other input remains in the common-mode range. <S> Either or both inputs can go to 30 V without damage. <S> Figure 6 of the datasheet shows the internal circuitry. <S> From Figure 6 we can see that the input transistors are protected against over-voltage by the base-emitter reverse protection diodes. <S> We can only surmise that the following transistors and base-emitter diodes will not break down provided the inputs do not exceed 30 V. <S> It may be worth studying section 9.2.2.1 also. <A> This is note (3) from Table 7.6 of the TI data sheet. <S> (3) <S> The voltage at either input or common-mode must not be allowed to go negative by more than 0.3 V. <S> Either or both inputs can go to 30 V without damage. <S> Based on this information, you should be OK with your intended use of the '339. <A> According to the spec, that should be fine. <S> (Edit) <S> I missed a note in the data sheet on this my first time around. <S> It appears that you can exceed VCC on one input pin (but not both). <S> If it's important not to load the 100k-ohm input, you may want to check how much current flows in this circumstance. <S> It's almost certainly different from normal, although it's not clear to me how much different. <S> I have ranted on the Internet innumerable times about reading the datasheet carefully , for every clause. <S> Clearly I failed in this case... <S> I want to compare a (fairly weak, 100k impedance) 0 to +15 input signal against a +3 reference, while powering the device from a +5 supply. <S> There's two issues here. <S> One is whether you can mess with the signal -- i.e., is it OK to load it down, or do other things depend on it not being disturbed. <S> The other is how accurate your measurement needs to be. <S> The LM339 has between 25nA and 300nA of bias current that flows out of the input pins, depending on temperature and luck. <S> Working against a \$100 \mathrm{k\Omega}\$ source impedance, that's 2.5mV to 30mV of error in your switching point -- can you stand that? <S> If not, you can correct for it with a resistor in the opposite lead (R2 in the schematic below). <S> That'll bring the error down to \$100 \mathrm{k\Omega}\$ times the input bias current, or 0.3mV <S> to 10mV. <S> If you want to avoid that you can use a rail-rail comparator with CMOS input, but you probably won't luck out on being able to pull an input higher than VCC, and you'll load the signal down. <S> If you can load the thing down, then the circuit below may work for you -- D1 won't conduct significantly until the input signal is positively above 3V, and it'll limit the non-inverting input of the comparator to around 3.7V. R1 <S> is just there in case you want to load the source less (but be careful of the LM339 input bias current <S> -- I don't know what it is, so you need to check to see how much it'll throw off your measurement, or you'll want to use a JFET or CMOS input device). <S> simulate this circuit – <S> Schematic created using CircuitLab
| The upper end of the common-mode voltage range is VCC+ –1.5 V; however, one input can exceed VCC, and the comparator will provide a proper output state as long as the other input remains in the common-mode range.
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Why do emitter-degeneration resistors linearize an emitter coupled pair? Is there a mathematical derivation of the following curve for an emitter coupled pair with degeneration resistors? I can't seem to find anything online <Q> The specific equation for that curve is probably not so easy to mathematically reproduce because it's based on Gm which is a measured parameter of a real transistor. <S> It is not a constant either. <S> It varies with operating conditions so would actually be represented by some function which would probably be very large and messy since it's measured from the real world and not something mathematically idyllic. <S> But the curve is nonlinear because transconductance is non-linear. <S> The following is the AC circuit (hence all the grounded supplies) for a common-source amplifier but the idea is the same for common emitter, and for the differential pair (it's just done twice, one for each leg). <S> Microelectronic Circuits 7th Edition, Sedra & Smith <S> You can see that as Rs (the source/emitter resistor) gets larger, that the transconductance's influence on the gain gets drowned out. <S> Since linear component is drowning out the nonlinear component, it gets more linear. <A> The way to obtain the curve is to realize that we can use the small signal model for the transistor. <S> This is because the base-emitter voltage varies very slightly when the differential input is changed due to the emitter degeneration. <S> To calculate how slightly is complicated because the tail voltage of the differential pair changes due to the constant current. <S> To acquire a more intuitive understanding of why this happens, consider the simpler common-emitter amplifier. <S> Then $$V_\mathrm{b} <S> = R_\mathrm{E} I_\mathrm{E} <S> + V_\mathrm{BE} = <S> V_\mathrm{BE} <S> + R_\mathrm{E} I_\mathrm{S} <S> (e^{V_\mathrm{BE} / V_\mathrm{T}} - 1).$$ <S> In this situation, a small change in the base-emitter voltage can result in a large change in the current due to the the exponential. <S> This results in a large change in the emitter resistor voltage. <S> So by increasing the base voltage the base-emitter voltage only increases slightly. <S> The majority of the change comes from the increase of the emitter resistor voltage, as can be seen from the plot of the base-emitter voltage as a function of the base voltage with a 1k resistor. <S> You can see that after the current has increased enough for the emitter resistor to develop enough of a voltage drop, the base-emitter voltage becomes a linear function of the base voltage. <S> Now back to the differential pair. <S> We know that the small signal base-emitter resistance taken at <S> \$v_1 = v_2\$ is $$r_\mathrm{e} <S> = \frac{V_\mathrm{T}}{I_\mathrm{EE} / 2}.$$ <S> Let <S> \$V_\mathrm{EE}\$ be the voltage across the current source, let \$i_1\$ be the emitter current, and let \$V_\mathrm{Q}\$ be the base-emitter quiescent voltage at <S> \$v_1 = v_2\$ . <S> Then $$v_1 = <S> i_1 (r_\mathrm{e} + R_\mathrm{E}) <S> + V_\mathrm{Q} <S> + V_\mathrm{EE} - V_\mathrm{CC}.$$ <S> Thus <S> $$i_1 = <S> \frac{v_1 - V_\mathrm{Q} - V_\mathrm{EE} + V_\mathrm{CC}}{r_\mathrm{e} <S> + R_\mathrm{E}}.$$ <S> Hence $$V_\mathrm{E1} = <S> V_\mathrm{CC} - \alpha i_1 R_\mathrm{C} = <S> V_\mathrm{CC} - \alpha R_\mathrm{C <S> } \frac{v_1 - V_\mathrm{Q} - V_\mathrm{EE} + V_\mathrm{CC}}{r_\mathrm{e} <S> + R_\mathrm{E}}.$$ <S> A similar equation can be derived for the other transistor. <S> The output is then $$v_\mathrm{O <S> } = V_\mathrm{E1} - V_\mathrm{E2} = - \alpha \frac{R_\mathrm{C}}{r_\mathrm{e} <S> + R_\mathrm{E}} <S> (v_1 - v_2).$$ <S> You can see that this is a linear equation. <S> Compare this with the case without emitter degeneration, which you can find in any standard reference. <A> The emitter resistances introduce a negative feedback--the negative feedback linearizes the transfer curve of the amplifying circuit-- <S> so the emitter resistances linearize the transfer curve of the differential pair... <S> Now a separate answer for each statement: <S> The negative feedback linearizes the transfer curve of the amplifying circuit since it decreases many times the enormous high open-loop gain of the amplifying stage. <S> So the variations of this initially high gain (because of various reasons including the non-linearity) will be also decreased many times. <S> Inspired by the H&H "transistor man" (The Art of Electronics, 1989, Page 64, Fig. 2.5), I have drawn a conceptual picture to show how the negative feedback makes the transfer characteristic of the common-emitter stage linear. <S> Here the output collector-emitter part of the transistor is emulated by a rheostat with variable resistance <S> \$R_{CE}\$ <S> so the stage is simply a triple voltage divider ( \$R_{C1}\$ , \$R_{CE}\$ and \$R_E\$ ) supplied by \$V_{CC}\$ . <S> "The little man's perpetual task in life" is to keep \$V_E = <S> V_{IN}\$ by moving the slider of the collector-emitter rheostat \$R_{CE}\$ . <S> For this purpose, he observes the base-emitter zero indicator and changes the rheostat resistance in the respective direction to zero the indicator reading. <S> Since the current of this circuit of three resistors in series is the same, there is the same proportion between the voltages and resistances - the ubiquitous \$V_{RC}/V_E = <S> R_C <S> /R_E\$ <S> and this is the gain of the stage. <S> It is constant, so the relation is linear. <S> By the way, this "man-emulated transistor" precisely equalizes <S> \$V_E\$ to <S> \$V_{IN}\$ <S> (there is no \$V_{BE}\$ here)... <S> and there is no current ( \$I_B\$ ) drawn from the input source (the indicator is bootstrapped). <S> Its only drawback is that it is slow...and of course, it engages a human being for this chore. <S> The power of this approach - to see the general idea behind the specific circuit, is that it is valid for various other implementations (FET, op-amp, etc.) <S> For example, the op-amp inverting and non-inverting amplifiers exploit the same idea.
| The emitter resistor introduces a negative feedback since the output current creates a voltage drop across it that subtracts (in a series manner) from the input voltage.
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LED ceiling light with mismatch driver I am new to the forum, so please forgive me if I am asking a stupid question or posting the wrong thing at the wrong place. I recently bought a Led ceiling light online that comes with a remote control which allowed me to change the light temperature and brightness. However, I have noticed that if I used the remote to turn off the light, the light still glow (dim, but enough to notice and bother me at night…) If I use the wall switch, this will not happen (but doing so kind of defeated the feature of the dimmable light remote control). Based on what I searched so far, I figured that this is probably a mismatch of the controller and the LED strip. My ceiling light is supposed to be a 24W light, but on the controller, it said (36-48w)x2. So I am guessing that they have gave me the light with the incorrect controller installed…. The thing is, If I return the item, the postage would probably cost me at least half the price of the light, so it is really not worth returning…. So I am looking at ways to fix this rather than returning it. I have tried to switch the L and N wire that goes into the controller (just in case the switch was on the wrong side), but it made no difference ☹ After reading online, I noticed that I might be able to put a resistor into the circuit to fix this, but I am not too sure where I should install it, and what would be a safe value of resistor to try out for. It would be great if anyone could shed some light on my problem. I have attached a picture of my ceiling light, and a close up picture of the controller (the writing is in Chinese though). Each of the led strip are 2835 2b4C x2. Many thanks in advance. <Q> Based on what I searched so far, I figured that this is probably a mismatch of the controller and the LED strip. <S> My ceiling light is supposed to be a 24W light, but on the controller, it said (36-48w)x2. <S> So I am guessing that they have gave me the light with the incorrect controller installed…. <S> No, that is the correct controller for the lights. <S> These are constant current LEDs (no resistors are present in the light strips), which means the driver generates the current needed to drive the LEDs. <S> If it was generating too much current, the diodes would burn out quickly. <S> Since this is a constant current driver, the "0.17A" is the current, and the 36-48 is probably the range of voltages the driver can produce. <S> So I am looking at ways to fix this rather than returning it. <S> I have tried to switch the L and N wire that goes into the controller (just in case the switch was on the wrong side), <S> You should never do this due to the risk of electrical shock, especially with cheaply made electronics like this. <S> Since this is a constant current supply, putting resistors in series will just cause the supply to put out more voltage to overcome the resistance. <A> Only do this if you are not scared of experimenting and can solder. <S> Solution #1 simulate this circuit – <S> Schematic created using CircuitLab <S> A CC generator might not be able to be programmed to 1% minimum and this is too bright. <S> In order to reduce 170 mA for both strings or 85mA /string <S> dummy loads of a power resistor must be shunt on A & B strings to absorb min. <S> CC drive. <S> Assuming CC minimum of 10% of 85mA or 8.5mA at 24V approx for string of 4P8S array *2 colorsDummy load is 24V/8.5mA=2.8k <S> - depending on how bright is DIM (Off) you can adjust R up to 10K or 24V/10k = 2.4mA of dummy load. <A> OK, you gave the link , unfortunately all the Chinese text is in pictures, which Google Translate can not parse. <S> Still... this confirms an important construction detail: <S> This is designed to be mounted on a ceiling. <S> Hot air rises <S> , so we can't count on convection inside the enclosure to carry heat away from the LEDs (ie, down) and through the plastic diffuser ; hot air will stay up. <S> Thus all the heat will be dissipated through the back of the luminaire. <S> Mounting it against a ceiling will result in zero airflow against the back of the metal enclosure. <S> So there is absolutely no way it can do 24 Watts (or even worse, 2x24W if both LED color temperature are on) without overheating... and the driver, which is rated for an ambient temperature of 45°C, is going to fry very soon. <S> A possible fix would be to mount it 50mm away from the ceiling by using longer screws. <S> But it's still gonna run quite hot. <S> I bet there is no thermal paste behind these MCPCBs... <S> In a better design, the LEDs would be mounted inside the aluminium profiles (they can be seen in the above picture) which would act as a heat sink in free air, just like in an edge-illuminated LED panel. <S> So don't feel too guilty about hacking this thing. <S> Now that wasn't an answer to your question, just advice on how not to have it self-destruct in a few months due to overheating. <S> I agree with Tony, you can put two resistors, one for each LED color, in parallel with the strips. <S> The most convenient place seems to be on the pins of the unused connector at the end: <S> One resistor between "A" and "+" and the other resistor between "B" and "+". <S> If the connector is female, you can stick the resistor legs into it to experiment. <S> If your soldering iron isn't powerful enough to heat the MCPCB you can always rip the plastic shell of the connector off and solder resistors to the pins. <S> Note Tony's resistor value which draws 8.5mA will result in about half a watt power draw when the light is off. <S> So you will probably want to replace the driver with a more "user friendly" one... <S> no hurry, you'll have this opportunity when it fries I guess
| After reading online, I noticed that I might be able to put a resistor into the circuit to fix this To solve this problem you probably need to replace the LED driver with a different one that is better designed.
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Why shouldn't two AC sources be connected in parallel? In data centers, servers usually have redundant power supplies, which makes it possible to plug them to two different UPS units. This way, the server continues to function correctly if one of the PSU goes down, or if one of the UPS units is shut down for replacement. Some servers don't have redundant power supplies. In order to be able to still connect them to two UPS units (in order to be able to replace one unit without shutting down the server first), special devices called transfer switches are used: they can receive input from multiple sources, and provide power for single-corded equipment. A transfer switch is usually expensive and contains lots of electronics inside. Basically, automated transfer switches provide power from one source, and when they sense that the source is down, they fallback to another source. What makes it impossible to have just a cable which would plug into two sources on one side, and to one server on another side? In other words, something like that: but the other way around, i.e. not a cable with one male and two female connectors, but a cable with two male and one female connectors? In other words, why is it possible to wire two DC power sources in parallel, but one cannot (or shouldn't) wire two AC sources in parallel? Is it because the waves may not be synchronized and could then cancel out each other? Or something else? simulate this circuit – Schematic created using CircuitLab <Q> When two DC sources are paralleled, blocking diodes prevent back feeding from one supply to the other. <S> The supply with the highest voltage will always supply current without being affected by the other. <S> There is no comparably simple and effective provision for paralleling two AC supplies. <S> Two AC supplies from the utility can be obtained with matching phases and voltages, but if one goes down, it must be disconnected to prevent back feeding. <S> Even dealing with small voltage or phase differences is not particularly easy. <A> Note: I am not trained in electrical safety. <S> Do not rely on this post for safety advice. <S> I know you are asking theoretically, but I feel like I should put this disclaimer here anyway. <S> Is it because the waves may not be synchronized and could then cancel out each other? <S> Or something else? <S> That is the main reason, but the consequences are worse than just having the voltage cancel out. <S> Let's imagine the worst case where the sources are perfectly unsynchronized (180 degrees apart - opposite voltages all the time) and that the voltage is at the maximum when you make the connection. <S> Remember that AC is just time-varying DC - at the instant when you connect them together, you have this circuit: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You've connected two voltage sources in a loop. <S> Forget the load - what happens to the voltage sources? <S> Well, V1 tries really really hard to make the voltage 240V, and V2 tries really really hard to make the voltage negative 240V. V1 makes lots of current flow from the bottom to the top, and V2 makes lots of current flow from the top to the button again. <S> This is basically a short circuit, but short circuiting two sources at the same time. <S> If V1 and V2 are alkaline batteries, they might get quite warm . <S> If they're NiCad batteries, they might get hot enough to melt plastic . <S> If they're lithium rechargeable batteries, they will definitely burn up . <S> If V1 and V2 are inverters (as in double-conversion UPSes) <S> the output circuit will probably blow up (literally). <S> If V1 and V2 are different phases of mains power, you'll get a big spark and pop the circuit breaker. <S> If enough power is available, the spark (known as an arc flash) can be big enough to cause serious injuries . <S> The main power feed to the datacenter probably <S> can do that. <S> I don't think you're likely to be seriously injured by plugging an IEC connector into a short circuit (even on two different voltage sources), but I still wouldn't try it. <S> If V1 and V2 are standby UPSes, running on the same phase of mains power, they should both be in phase - <S> until one of them switches to battery power, at which time it will blow up because its inverter is now in parallel with the mains power. <A> Danger to Wires <S> With this type of setup, while your intention may be that it would only ever be used with "one cable's worth of electricity", there is no guarantee. <S> So someone could use the exact same configuration to get double power - e.g., 20A from circuit 1 and 20A from circuit 2 in order to have 40A total. <S> There are a bunch of problems with this. <S> In particular: Each of these two cables will have hot & neutral. <S> In theory, the current on the neutrals and hots will be balanced - 1/2 <S> the current on one wire in each pair. <S> Actually there is no guarantee of that, but even if there is under normal circumstances , if one of the neutrals breaks due to a wiring problem somewhere, all of the hot - e.g., 40A in this example, will return on the other neutral. <S> But neutral wires are not normally protected from overload in the US, so this would result in the wire handling twice its nominal capacity and potentially start a fire. <S> Danger to People <S> Normally, one circuit break is flipped to turn off power to a device. <S> Let's say you have one of these 2 circuits also used for LED lights above the servers - just a few watts, so no big deal. <S> Someone comes in to work on the light fixture. <S> The servers are not on at the time. <S> He starts work by turning off the breaker and verifying it is off and now has exposed wires from the fixture. <S> You come in and turn on the server. <S> The server can backfeed power from circuit 1 into circuit 2, which will now make the rest of circuit 2 - including the exposed wires on the light fixture <S> hot and potentially fatal. <A> Your premise is incorrect for experts and only applies to newbs. <S> This is why an experienced Tech. <S> or Eng. should be consulted. <S> This is done all the time with the AC grid, DC battery boost start etc. <S> e.g. if ESR total is 10mOhms and voltage difference is 1V then connecting both batteries together results in a current of 100 Amps which may or may not cause damage. <S> Answer to your photo in question. <S> (wrong example) <S> This is just a Y splitter cable to go from 1 source to 2 loads with female receptacles being the source of power. <S> If you did the opposite from two outlets, with the exact same voltage, there is no advantage except current sharing or doubling up wire capacity <S> but you still might exceed the breaker rating before each outlet rating.
| When you apply voltage sources in parallel, the impedance, voltage and phase must be known before connection to know the resulting power being transferred , if it will be stable, safe and performs expected task.
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Cut 3D chunk out of PCB? I would like to cut a small square out of the middle of a large square PCB. The tricky thing is, I don't want to cut all the way through the board, just about half way. This would create a 1mm-deep square pocket in a 2mm pcb board. I could do this with my CNC machine easily, but I want to know if something like this could be achieved during circuit board manufacturing. I would hate to have to load hundreds of panels of boards onto a CNC machine; this would raise the cost of manufacturing significantly. One idea I had was to use fancy programming of the drill that drills holes in the pcb to act like a CNC drill. It would, of course, require changing the drill bit to a special one for milling. Can anyone who is familiar with these machines (already setup in most Chinese PCB fabrication facilities) foresee a (cheap) way to use them to cut pockets out of a PCB? <Q> Yes, it can be done. <S> It's (often) called Z-axis milling . <S> For my usual Chinese supplier it is a standard option and adds USD $100-$150 to a small order of 2mm thick PCBs. <S> You could also ask for a top layer to be routed through on a multilayer board which might give you better control over the surface finish and maybe thickness. <A> They can also make interior cut-outs. <S> So the tool to do this operation is standard equipment. <S> I've also had controlled-depth milling done but had to be selective about the vendor to be able to get the depth tightly controlled <S> (my project probably needed ~10 um tolerance on the depth). <S> And it will increase cost, because they won't be able to process multiple boards in a stack; they'll have to process each panel individually. <S> You'll probably need to call a few vendors before you find the right one for your requirements. <S> I also wouldn't count on getting this from a bare-bones Chinese vendor. <S> But you'll never know until you ask. <A> Not a direct answer, but aimed at solving your problem: can't you stack several thin PCBs on top of each other to get the desired effect? <S> E.g. if you want a 1.6 mm thick PCB and a 0.8 mm cut-out, you could just stack one 0.8 mm PCB with a cutout on top of a 0.8 mm PCB without one. <S> Ordering two standard PCBs is also cheaper.
| They use something very much like a router or end mill to cut the board outline.
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What is the minimum die area of a chip? Silicon chips are made by slicing up wafers; in general, the fewer transistors a design needs, the smaller the area it needs, so the more chips you get out of a wafer, and the lower the cost per chip. But there must be some limit to this. Maybe you can have a chip of one square millimeter area, but it seems unlikely that you could have a chip of one square micron area. What is the minimum... To be specific, I'm not asking what is the minimum physically possible area, but what is the minimum area below which chips stop getting cheaper ? The point at which you stop trying to shrink your design, because fewer transistors will no longer save money? And has this remained roughly constant over the decades, or has it changed with iterations of process technology, and if so, in which direction? <Q> The smallest cut defect-free with a kerf is roughly equal to the wafer thickness and the slotted diamond saw roughly equal to 1/2 of the wafer thickness. <S> Thus the question should be what is the cheapest way to process single diode junctions. <S> As the biggest demand for single diode junctions appears to be LED's , the question should be what is the maximum number of LED's per wafer? <S> Economy has driven the size of the LED up. <S> There is no benefit to going smaller. <S> For any given chip on a die and wafer size, this calculation limits the maximum number of Die Per Wafer. <S> This is not the limit but is a calculator to determine the mechanical yield of a wafer. <S> The trend now is to make small flip chips with bump pads stacked on a system in package (SiP) or system-in-a-package with a number of integrated circuits enclosed in a single chip carrier package. <A> Chip costs are not die only. <S> There is testing, bonding, packaging etc. <S> You would be surprised how much time on a big chip tester costs! <S> Thus a chip with a small analog die with larger amount of test-vectors can be more expensive then a big digital one with fewer test-vectors. <A> Even ignoring things like scribe keep-out margins, many IC designs end up being pad limited or bump limited given the IO required by the circuit (even capacitive IO coupling or antennas require some physical size). <S> This is true not only for small chips, but many older larger chips had empty space because any smaller would not have allowed enough perimeter for the required number of IO pads (thus allowing room for free "chip art" on the die.) <A> Many posters mention bond pad size as a limiting factor. <S> But some of our newer designs do not use wire bonds at all. <S> The chip is built with bumps on the bottom that allow for direct chip attachment and connectivity to the substrate (board). <S> I think minimum die size has got to be determined by wafer dicing capability, and fallout (bad die) from that operation.
| The minimum area of the chip is determined by the most cost effective solution not the smallest physical possible cut .
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How to solder to this glass pcb? I have this glass PCB I took from a calculator that I'd like to get working with my Arduino as a numpad, but it's a glass PCB. Is there any way I could solder wires to it? <Q> The transparent conductors are ITO (Indium Tin Oxide) deposited on glass. <S> They're not easily solderable (perhaps possible with an ultrasonic soldering iron <S> but even then they would likely be mechanically very weak). <S> You can use an elastomeric connector (zebra connector) as Dave Tweed suggests, but you'll need a closely matching pattern (preferably gold plated, or carbon printed) on a PCB. <S> It's also possible to use FPCs with conductive adhesive such as heat sealed types, but that's not practical for small quantities or one-off. <S> I believe conductive paint is sold in auto shops for repairing heated car windows. <A> You can't, it's a very thin layer of printed-on metal. <S> As Dave says, you need a Zebra strip or similar pressure connector. <A> You can try using anisotropic conductive tape . <S> Adafruit will sell you a 50mm x 150mm strip . <S> It's double-sided tape that only conducts through the thickness of the tape when you press it between two surfaces. <S> That way, you can make multiple independent electrical connections without shorting them. <S> Make a PCB with exposed contacts of the same pitch and geometry as that of contacts of the glass. <S> Then sandwich the tape between the PCB and the glass. <S> Obviously, be sure the conductive sides of the board surfaces face each other. <A> As mentioned by others these aren't meant to be soldered to. <S> In theory with indium based solder it might be possible, but it's really not worth the cost of the attempt. <S> If this originally had a thermally bonded flex cable, you can often non-destructively remove those from the original board with a hair dryer. <S> You can then stick them to a compatible PCB footprint by effectively "ironing" them down - a clothes iron would work if it can fit, perhaps a sealing iron as used for model airplane covering film. <S> I was able to get a solder station with a large curved tip to sort of work at low temperature, tested by touching against the side of the ribbon to make sure it would not melt it, then to make sure the temperature controller couldn't hit it with another bolus of heat <S> actually powered off the iron before using the residual heat to iron the ribbon onto the board. <S> If it had zebra strips and you didn't break those, then with suitable fixturing you could probably do the same thing with a suitable footprint on a custom PCB. <S> A 3d printer might help make something that would hold the display vs. board in the right place. <S> For an LCD display you'd also need to work out the column and segment mapping and figure out an appropriate driving circuit. <S> Something generating about a volt or two of AC across two probes can be handy for exploring. <S> You can also simply toggle GPIOs in opposite phase. <S> Edit: realized what was hiding behind your link was actually a keypad not a display, but much of the same still applies, except that you can probably sort out the connectivity visually. <S> That said, for a keypad you might do better just making your own conventional PCB...
| Silver-loaded or carbon loaded conductive paint might be a possible way to weakly attach to thin copper wires such as (stripped) AWG 36 magnet wire.
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2 / 4 quadrant switches - how can current have opposite direction to voltage? I am learning about 1, 2 and 4 quadrant switches, used in inverters etc. I am however struggling to understand how, in the 2nd and 4th quadrant, the voltage can have an opposite direction to current. Using this picture for example I dont understand here how you can have a positive voltage but current in the opposite direction. For e.g. On the left, current would flow clockwise, and on the right, current will not flow, but if it did (i.e. if diode was replaced with a resistor) it would be anticlockwise. However, the 2nd and 4th quadrants indicate current in opposite direction to voltage - but I can't imagine how this could happen since (conventionally) current goes from positive to negative - the electromotive force is acting in one direction; how could current ever flow the opposite way? Any help understanding this would be greatly appreciated; thank you. <Q> This answer is specific to thyristor / triac operation. <S> I can see from @Cristobol's answer that another meaning can be taken from your question. <S> Figure 1. <S> From Littlefuse's Fundamental Characteristics of Thyristors . <S> The Littlefuse diagram makes it clear that the voltage in question is the gate voltage relative to MT1. <S> The article goes on to say ... <S> The most common quadrants for Triac gating-on are Quadrants I and III, where the gate supply is synchronized with the main terminal supply (gate positive -- MT2 positive, gate negative -- MT2 negative). <S> Gate sensitivity of Triacs is most optimum in Quadrants I and III due to the inherent Thyristor chip construction. <S> If Quadrants I and III cannot be used, the next best operating modes are Quadrants II and III where the gate has a negative polarity supply with an AC main terminal supply. <S> Typically, Quadrant II is approximately equal in gate sensitivity to Quadrant I; however, latching current sensitivity in Quadrant II is lowest. <S> Therefore, it is difficult for Triacs to latch on in Quadrant II when the main terminal current supply is very low in value. <S> They're basically saying that if MT2 is positive then a positive gate voltage will work best (QI). <S> If MT2 is negative then a negative gate voltage will work best (QIII). <S> The application note is well worth reading. <A> As shown in the diagram posted with the question, four quadrant switches can block voltages of either polarity and allow current to flow either in the direction of supply to load or load to supply. <S> Blocking voltage in either direction is necessary to use the switch with an AC supply. <S> In other words allow current to flow in the direction supported or opposed by the supply voltage. <S> In terms of inverter usage, a four quadrant switch is necessary for a grid-tie inverter that can supply power to a load or accept power from the same system that includes an energy source that can supply more energy than is used locally. <S> Another use is to control an AC motor in four modes, forward motoring, reverse motoring, forward braking and reverse braking. <A> how could current ever flow the opposite way? <S> It could be pulsed switches between a battery, a charger and a load where current can travel in either direction in a positive voltage. <S> (half Bridge) <S> In order to have current in the opposite polarity of voltage, there just needs to be some external voltage that pulls the load current in the opposite direction. <S> It could be a switched reactive LC load, or motor to generate a back EMF which reduces the current with acceleration or it brakes (-ve acceleration) by design. <S> (Full bridge includes changing directions) <A> Current flows from positive to negative in a load which is dissipating power. <S> In order for current to flow from negative to positive you must be supplying power to the circuit. <S> Look at your voltage sources above... <S> the current is flowing from the negative to the positive, because it's supplying power.
| Allowing current to flow in either direction is necessary to use the switch allow the supply system to either supply power or absorb power.
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Why does inc R of a RL circuit increase dVL/dt. Here I have circuits and details Why does increasing R of a RL circuit increase the rate of change of voltage across an inductor with respect to time dVL/dt. Here a 50V 2k circuit and a 5V 200ohm circuit both create a change of current from 0 A to 25 mA t = 0 and that current is opposed by a back emf. Both circuits exhibit an increasing current because the inductors create a self-induced voltage proportional to the rate of change of magnetic field strength across the inductor with respect to time dϕM/dt. The way I understand back emf and VL is the current created by the voltage source Vs/R is passed to the inductor then the inductor receives current Vs/R. That current passes through the first loop, not exactly, I am just imagining this. Then the change of current 0 A to 25mA creates a magnetic field that increases in strength like the voltage across a capacitor increases when it is supplied a constant voltage x V. The rate of change of the magnetic field strength creates a self-induced voltage. The self-induced voltage creates the voltage drop across the inductor. So the voltage across the inductor is like the current passing through a capacitor that is proportional to the rate of change of voltage across the capacitor dVC/dt.Increasing inductance (L) decreases the rate of change of the magnetic field strength with respect to time dϕM/dt like increasing capacitance C decreases the rate of change of voltage across the capacitor with respect to time dVC/dt.Increasing resistance (R) decreases the rate of change of voltage across the capacitor with respect to time dVC/dt. So why does increasing resistance (R) increase the rate of change of magnetic field strength with respect to time dϕM/dt.Next, here, is a statement I am not confident about, continuing on resistance: The current passed to the inductor is 0 A to 25 mA in both circuits. The inductance (L) is the same.So how is the resistance (R) increasing the rate of change of magnetic field strength with respect to time dϕM/dt, dVL/dt, and dIL/dt. <Q> The thing which caused the speedup was the increase of voltage, not the increase of resistance. <S> If the resistance were zero ohms the current would grow infinitely with growth rate = <S> voltage divided by inductance. <S> That gives amperes per second if the inductance is in henries. <S> The series resistor gives a limit to for the current. <S> In both cases that limit is the same, but the initial growth rate, when the voltage loss in the resistor is small, is 50 amperes per second with 50V supply and 5 amperes per second with 5V supply. <A> Why does increasing R of a RL circuit increase the rate of change of voltage across an inductor with respect to time <S> dVL/dt. <S> The relevant formula is \$V= -L\frac{di}{dt}\$ . <S> This says the voltage across the inductor is always proportional to the rate of current change in it. <S> The minus sign indicates that the back-emf voltage opposes the current change causing it. <S> When voltage is first applied to the RL circuit, current is small so the resistance has neglible effect. <S> The inductor generates a back-emf equal to the input voltage, and current rises at the rate required to produce that voltage. <S> However as current increases the resistor drops an increasing voltage, leaving less voltage across the inductor. <S> This slows the rate of current rise to match the lower voltage across the inductor, continuing in an exponential curve as inductor voltage approaches zero and current approaches <S> Vin/R. <S> The higher the resistance <S> the faster this happens. <S> But the rate at which inductor voltage drops with higher resistance is not as great as your plots might suggest. <S> You have also raised the voltage, so the inductor must generate more voltage to match it, and the current must also rise faster to support that voltage. <S> This faster current rise also causes voltage drop across the resistor to increase faster, so inductor voltage drops faster. <S> However the time constant \$t = <S> \frac{L}{R}\$ <S> is the same no matter what voltage is applied, so the time taken to reach a particular fraction or percentage of the initial voltage does not vary with applied voltage. <S> Here is a simulation plot showing voltage across a 1H inductor with 200Ω on 5V (blue), 2kΩ on 5V (green), and 2kΩ on 50V (red):- With 2kΩ on 50V the voltage drop at any point is steeper, but reaches (eg.) <S> the same 10% of 50V (5V) as 2kΩ reaches of 5V (0.5V) in the same time. <A> When you look at the asymptote for the starting linear slope for this series RL current, you will find \$\tau = <S> 2\pi <S> \dfrac{L}{R}\$ <S> Thus reducing R increases the max current Imax= <S> V/R with a fixed voltage and also the increases asymptotic rise time. <S> This unlike charging a capacitor whose impedance (f) is inverse with C and has a time constant of T=RC.
| As the current increases, the growth rate decreases because there's less voltage left over the inductor, the resistor voltage loss = IR.
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How does an SR Latch get started Noob question, but I'm not seeing how an SR latch such as the one in the image can "get started". If each input depends on the output from the other, how can there be two inputs to each gate to set the thing in motion? In a rudimentary circuit consisting of exactly one NOR-based SR latch, what happens when the power is switched on (e.g. battery connected)? <Q> It is true that the latch will "wake up" in an unknown state. <S> In the real world, given a little time, the latch will have a valid state with either Q=0 or Q=1. <S> This will force the latch into a known state, regardless of whatever the previous state of the latch might be. <S> In other words, if you want to know the state of the latch then you must explicitly Set or Reset the latch. <A> When you power on a bi-stable element like a NOR SR latch (with inputs at 0) it will start at to some state. <S> That is not necessaraiy a 'binary' state, the the outputs can for instance be halfway between 0 and 1. <S> Most bi-stable elements are designed with a positive feedback factor, which means that any difference between the two outoputs (or, same thing but different viewpoint: any difference between an output value and the 'halfway' point) is amplified, so the element quickly 'drifts' towards one of the two stable binary states. <S> This 'drifting' time can be observable, and can in some circumstances involve osscilation. <A> This is a way of power-on-resetting a RS latch to a known state. <S> Problem with it, the speed of the latch is slowed down and so can only be used for low frequency applications. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> (lets assume that the two inputs of the circuit are tied to ground) <S> what happens when the power is switched on (e.g. battery connected)? <S> At an electrical level a logic gate is effectively a crude amplifier. <S> In the SR latch circuit the amplifiers are connected together in positive feedback. <S> This creates a circuit with an unstable equilibrium. <S> If the output voltage is above the equilibrium point then it will rise until it saturates at a logic high. <S> If it is below the equilibrium point then it will fall until it saturates at a logic low. <S> If there was no noise and the design was perfectly symmetric then by symmetry the output would sit at the equilibrium point. <S> In the real world there is always assymetry and noise, which will push the output away from equilibrium and then positive feedback will take over and push the circuit to a saturated state. <S> Which of the two states it ends up saturated in is unpredictable and may also be inconsistent. <S> How long this will take depends heavily on the particular logic family, with most oldschool logic it will be very quick, however some low voltage families may have an effective dead-band in the amplification resulting in much longer settling times. <A> When the circuit is powered on, its state is not defined. <S> Because of natural variations in physical hardware, the latch will eventually settle into either valid binary state; however, in a simulator, which assumes ideal (perfect) hardware, the latch will be bouncing back and forth or otherwise be in an invalid intermediate state (e.g. Q and Q' are both on or both off; Q' is omitted in your schematic). <S> Even with a real circuit, some oscillation may be visible on an oscilloscope when power is initially applied to the latch, and it may take an unknown amount of time for the circuit to settle into a stable state. <S> This is called metastability . <S> As a result, you cannot rely on the latch being in any particular state until you explicitly set or reset it. <S> See also: Metastable state when S = R = 1 in SR Latch? <A> Transistors of the latch are not ideal. <S> On start of the latch one shoulder will 'prevail' and the latch will switch into one of the two states. <S> As such situation has no predictable outcome on start, it is better to forcefully put the latch into known state.
| You make the latch "get started" by setting one of the inputs (R or S) to be a 1 while the other input is a 0.
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How low have mains frequencies gone? What areas once used 30 Hz, and why? This answer quotes from a book written by Jocelyn Bell Burnell who co-discovered the first pulsar in 1967 where the author mentions another astronomer who noticed a star was pulsing at around 30 Hz and recognized the frequency because The Crab pulsar, with a period of 30Hz, is sufficiently fast that many people cannot see it; however some people can see 30Hz and Sue Simkin knows she is one of them – as a child she lived in an area where the main power supply was at 30Hz. This page suggests that the area might be Detroit Michigan in the 1940's or 1950's. But here I'm asking a more general question about mains frequencies well below 50 Hz. Question: How low have mains frequencies gone? What areas once used 30 Hz, and why? <Q> The historical reasons for using 30 Hz and other frequencies is documented in the book Electrical Engineering Papers by Benjamin G. Lamme, published in 1919. <S> In the article 'Story of the Frequencies' he explains that large generators and those powered directly by reciprocating engines ran at lower rpm, and therefore preferred a lower output frequency to reduce the number of poles required in the generator. <S> Finding all the areas that once used 30 Hz would take far more intensive research than this question deserves. <S> My Google searches came up with nothing, however I did find some information that may be relevant. <S> In the 1880's electric power was used mostly for lighting, which preferred higher frequencies to reduce flicker, so frequencies such as 133⅔ Hz and 125 Hz were common. <S> These higher frequencies enabled the use of smaller transformers. <S> However the AC motors developed at that time required frequencies as low as 16⅔ Hz. <S> This was a problem because converting from one frequency to another was difficult. <S> So different frequencies were usually generated for residential and industrial use, with many factories having their own power plants free to run at whatever frequency they desired. <S> Around 1890 the 'compromise' frequency of 60 Hz was introduced, which eventually became a nationwide standard. <S> However 25 Hz was still used in some places until quite recently. <S> In the article Early Electrification of Buffalo , it is mentioned that Westinghouse had adopted 60 Hz for lighting and 30 Hz for power, but for the Niagara Falls project they settled on 25 Hz. <S> That service continued until 2006. <S> Several other hydro stations on the Niagara river also ran at 25 Hz. <S> I don't know if Detroit used Niagara power in the 1940's or 1950's, but it seems likely that at least some parts of Michigan did. <S> Sue Simkin might have lived an area which had a local power plant running at 30 Hz, perhaps attached to a factory. <S> This could have been anywhere in the US. <S> With so many private power plants involved I think it would be hard to rule out the use of 30 Hz in any area. <A> How low have mains frequencies gone? <S> What areas once used 30 Hz, and why? <S> Mains frequencies are optimized around the requirements and constraints of their specific environment. <S> The German railway for example uses 16.67 Hz (50/3). <S> It comes down to a trade off: cost and size of transformers vs inductive and/or capacitive losses of the actual power lines. <S> Some modern high power distribution system actually use DC https://en.wikipedia.org/wiki/High-voltage_direct_current <A> I have seen an antique synchronous motor electric clock designed to use 25Hz mains. <S> The Wikipedia article on Utility Frequency goes into some detail: 25 Hz origins <S> The first generators at the Niagara Falls project, built by Westinghouse in 1895, were 25 Hz, because the turbine speed had already been set before alternating current power transmission had been definitively selected. <S> Westinghouse would have selected a low frequency of 30 Hz to drive motor loads, but the turbines for the project had already been specified at 250 RPM. <S> The machines could have been made to deliver 16⅔ Hz power suitable for heavy commutator-type motors, but the Westinghouse company objected that this would be undesirable for lighting and suggested 33⅓ Hz. <S> Eventually a compromise of 25 Hz, with 12-pole 250 RPM generators, was chosen.[2] Because the Niagara project was so influential on electric power systems design, 25 Hz prevailed as the North American standard for low-frequency AC. <A> The technology of the prime movers probably influenced the frequency selection. <S> Low frequency would seem to be easier for hydroelectric generators. <S> Low frequency might also be better for AC motors for electric railways. <S> On the other hand, higher frequencies are better for eliminating flicker in incandescent lighting. <S> There are historical records of some studies and discussions about technical factors considered in selecting a standard frequency. <S> There were likely both commercial and technical influences that are now unknown.
| Lower frequency was also preferred for long distance power transmission.
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Due to overcurrent, shouldn't I be extremely cautious when setting an I/O pin as Output? The question can be applied to any microcontroller with I/O capabilities, but I’m currently working with the popular ATmega328p. Consider the following circuit: A simple SPST Normally Open switch with a 10k pull-up resistor and a ceramic capacitor attached, connected to any I/O port.The specific pin should be obviously configurated as INPUT in the respective DDRx. The datasheet describes, inside the DC Characteristics section, an Input leakage Current I/O pin \$ (I_{IL}/I_{IH})\$ of 1µA, while far in the horizon the DC current per I/O pin caps at 40.0mA. Not a problem at all. DDRD = 0x00; // Entire port as INPUT Now consider that I forgot the fact that the given pin has to be set as Input, I set it as Output and set the Port HIGH. DDRD = 0xFF; // Entire port as OUTPUTPORTD = 0xFF; // Entire port HIGH While the datasheet doesn’t provide an Output Impedance it can be estimated to be 25Ω , based on the given graphs.Now at the time the button is pushed the current finds its way through the microcontroller from the 5V source with a 25Ω resistor plus negligible resistance due to the copper trace. This theoretically produces an overcurrent of 160mA above the DC current per I/O pin ceiling. Could this fry the port and the device? Analogously, if I switch over to positive logic and set the port to LOW the same problem could be encountered: DDRD = 0xFF; // Entire port as OUTPUTPORTD = 0x00; // Entire port LOW Considering this type of circuit is avidly encouraged, how it doesn't seem to bring any troubles? Now, back to the original question, shouldn’t I be extremely cautious when setting an I/O pin as Output? Or at least attach a small series resistor as a primitive protection? Note: Not a native English speaker, feel free to edit the post if you see something awkward. <Q> As others have said, you shouldn't have a problem as long as you double check your code. <S> If you do get it wrong, by and large the ATMega IO pins will limit themselves to about ~80mA due to internal resistance of the MOSFETs <S> (value found by experiment). <S> This is not good for the chip, but as long as you don't leave it in this condition for an extended period, they tend to recover ok. <S> If you are worried, it can be a good idea to put a resistor in series with the inputs. <S> Something in the order of 330R for a 5V VDD, or 220R for 3.3V VDD. <S> This will ensure the short circuit current is limited to ~15mA which is comfortably within spec. <S> The resistor goes between the IO pin and whatever is driving it (e.g. button w/ pull-up, or CMOS). <S> As the ATMega has a pretty limited useful frequency range (IO freq < 10MHz), the extra resistor won't have any noticable impact on operation of the circuit as the inputs have very little in the way of leakage current and capacitance <A> You should be cautious but not terrified about it. <S> Setting a pin output and driving it to the other direction will cause a lot of current to flow, but it also does not blow up in smoke immediately. <S> You could for example power the device with current limited labotatory supply when bringing up a design so these kind of errors cause even less damage. <S> I'd be more worried about the pushbutton and capacitor getting damage over continuous use as pushing the button <S> effectively shorts out the capacitor and a spike of current is limited only by cap ESR, PCB wiring and button resistance, so it can exceed many amps. <S> And as there will be contact bounce at the pushbutton, the current has high frequency components and that combined with stray inductance of long wires will convert the current spikes into voltage spikes, protected only by IO pin protection diodes. <S> So in general, a resistor of even 330 ohms in series starts to be a pretty good idea to limit current through pushbutton. <A> Considering this type of circuit is avidly encouraged, how it doesn't seem to bring any troubles? <S> Because when engineers set up the I/O for their device, they are careful to ensure that they do not drive inputs as outputs. <S> For most <S> I/O, we will set pins as either inputs or outputs at startup and never change them later. <S> This gives us only one line of code we need to double-check carefully, which is not an exceptionally hard task. <S> There is no need for a protective resistor unless you think it is likely that you will be careless about your coding. <S> In that case, I strongly suggest you work on improving your coding practises, by doing better testing, or by reviewing your code before you run it. <S> As a similar example, if you connect a wire directly between your +5V and 0V supply lines, you will burn out your power supply. <S> This does not mean you need to protect your power supply - it just means you need to be careful not to make mistakes like that! <S> Now, back to the original question, shouldn’t I be extremely cautious when setting an I/O pin as Output? <S> Yes you should. <S> But then I should hope you are equally cautious about all the many other ways you could cause damage to your circuit, like shorting out pins, turning on both sides of an H-bridge, or other similar mistakes. <A> Can confirm, on an arduino uno, 328p chip, if you connect a driven output to ground or VDD, it will kill that pin. <S> I have accidentally done it several times in my early days with electronics. <S> On other chips I work with, I'm pretty careful not to do this but more modern ones do seem less susceptible to this failure mode. <S> For example, I accidentally set up a timer on an stm32 in pwm out mode and had it fighting an opto coupler for about 15 minutes... <S> The opto comprehensively won the tug of war... <S> But the STM chip survived. <S> The situation with the arduino 328 chips is sufficiently bad that people have made "ruggeduinos" with all kinds of protections to avoid common causes of death. <S> Edit: <A> Generally, shorting a logic output to ground doesn't immediately kill a chip. <S> It may not be good for it, but it's not so bad that mistakenly setting inputs to outputs cause things to burn up. <S> So -- go ahead and use the recommended circuits, and double-check your software. <S> If you have willing friends or colleagues, do code reviews. <A> You need to check "Absolute Maximum Ratings" in the datasheet like this one http://ww1.microchip.com/downloads/en/DeviceDoc/ATmega48A-PA-88A-PA-168A-PA-328-P-DS-DS40002061A.pdf , which declares that DC Current per I/O Pin . . <S> . . . . . . . . . . . . . . . . <S> 40.0mADC Current VCC and GND Pins. . <S> . <S> . . . . . . . . 200.0mA <S> Which means that if you don't comply with the above, the reliability of the part won't be assured. <S> But not realiable doesn't mean fail now or even fail someday later, this is all about the chance to fail. <S> For dev board in the lab, if it is expensive or not easy to get another one, then more care is needed. <S> You can use a special funcion to access the register which make it the single point of care unless memory is corrupted by other code. <S> But in most case it is not that expensive or time consuming to get another one <S> then this is not a big problem. <S> BTW ESD may have a larger chance to destroy your dev board.
| Adding a resistor, around 100 ohms in series with the pin, is in my experience enough to avoid the pin damage.
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Advice on BNC cables for beginners electronic lab I'm setting up an electronics lab so I can increase my knowledge of electronics. I've bought some equipment of ebay ('scope, power supply, signal generator etc.,) but don't know what sort of BNC cables I should get. Do I need to worry about impedance when connecting the equipment together and buy (for example) 50 ohm cables or would straight through work OK? <Q> Do I need to worry about impedance when connecting the equipment together and buy (for example) 50 ohm cables or would straight through work OK? <S> You seem to have the false impression (see my highlight in the above quote) that a 50 ohm cable will incorporate 50 ohm resistors. <S> A 50 ohm cable (coax or otherwise) is telling you about the impedance it will present when using RF signals (greater than 1 MHz) and has nothing to do with real resistance values measured using a multimeter (for example). <S> All cables have a characteristic impedance (including "straight through" cables) and its impedance is represented by the formula: - $$Z_0 = <S> \sqrt{\dfrac{L}{C}}$$ <S> Where L is the distributed inductance per metre and C is the distributed capacitance per metre. <S> If you looked at some cables in their data sheets you might see typical values such as: - <S> L = <S> 0.25 <S> μH <S> per metre C = 100 pF per metre <S> Take the square root of the ratio and you get 50 ohms (nothing to do with actual resistive components). <A> However, not all scopes have built-in 50 ohm termination so external BNC T junction and 50 ohm terminator may be useful. <A> A cable with a 50-ohm impedance will respond to transients that are fed to it the way <S> a 50-ohm resistor would during the time it takes a signal to pass down the cable and back. <S> If whatever's connected to the far end doesn't behave like a 50-ohm resistor, the cable will behave as though the cable injected as a transient the difference between the voltage change produced at the far end, versus the voltage change that would have appeared with a 50 ohm resistor. <S> There are three general ways one may use such a cable: <S> Drive a load that behaves like a 50-ohm resistor, with minimal impedance at the source end. <S> The far end will absorb all of the energy sent down the cable, resulting in a clean signal. <S> If the far end impedance isn't exactly 50 ohms, some of the signal will get reflected, and most of that will get re-reflected at the source. <S> Drive a load that behaves like a 50-ohm resistor, but also have a 50-ohm resistor at the source. <S> This will cause the signal to be attenuated by 50%, but even if some signal gets reflected because the load impedance isn't exactly 50 ohms, most of that signal will get absorbed by the source rather than being re-reflected. <S> Ignore the cable impedance. <S> This will result in signals getting reflected at both ends of the cable, but if the cable is short and the signals aren't too fast, the reflections may die down fast enough that they don't pose a real problem. <S> The second approach gives the cleanest signals, at the cost of 50% attenuation. <S> The last approach is simplest, and is often "good enough". <S> For many purposes, the first ends up being a good choice.
| Standard lab equipment such as signal generators usually use 50 ohm impedance so BNC cables between equipment should also have 50 ohm impedance.
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LM393 comparator only connects to GND I'm testing out the comparator LM393 but when I try and the comparator conditions meet it only outputs low and won't switch high. That or I don't understand how it works properly so if you can help me out it would be great. I have 2 power supplies (each independent.) One output a voltage of 12V, the second can output 10V or 13V. Output 1 would resemble V ref (12v) Output 2 would resemble V in (10-13V) In the picture below Vin is connected to the non-inverting input Vref is connected to the inverting input GND is connected for both supplies resistors are for current limiting The first diagram I've connected the led so when the comparator output switches low. When only V in > V ref it would light up. In the second diagram I've connected the led so when the comparator switches high the LED will turn on BUT no matter what the output won't switch high and the LED won't turn on.What am I missing here? <Q> The LM393 has what's called an open collector output, as shown here from the datasheet. <S> This means the output can be pulled LOW when the BJT is turned on, but requires an external pull-up resistor to go HIGH when the BJT is turned off. <S> The typical applications section shows this pull up resistor: <S> Because of this, I recommend using your first circuit implementation. <S> Let me know if you have any follow up questions! <A> At 25°C, 1.5V is just enough to guarantee operation. <S> From the datasheet : <S> You can make it work by adding voltage dividers on each input, say equal value to R so that you get ~6V on each input. <S> Also note that the output is open-collector so your top circuit will work (with dividers), but for the bottom one use something like this: <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Always observe IC specs for Input and Output range. <S> Comparators are often open collector output so you swap inputs and use active low. <S> Other Design details. <S> When making a lead acid 12V battery voltage OK indicator , you want to start with specs. <S> e.g. Consider a threshold of 12V and a tolerance + <S> /-5% or below. <S> 12.5 100% SoC <S> (after quick load or settling time) <S> 12.0V = <S> 50% SoC <S> 11.5V = 0% SoC <S> SoC= State of Charge <S> Then you need a diode or voltage reference that can give you 1% accuracy which might be hard so you can consider any diode, even an LED or precision Zener or LDO. <S> Here is one way using the same LED to scale down with a pot to adjust over the desire range only. <S> 1) used an LED for Vref= 1.566V <S> @1mA 2) <S> Vcm input is suitable for some Op Amps>1.5V 3) <S> Choose R divider to scale down to Vref with care 4) <S> Used 1k pot to reduce range control <2% 5) <S> Simulated with 2Vpp ripple 11~13Vdc 1Hz 6) <S> LED reduced to 18 mVpp ripple or <1% error 7) <S> slider interactive adjustment on right
| You need a supply voltage that is higher than Vref by a couple volts.
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