source stringlengths 620 29.3k | target stringlengths 12 1.24k |
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Is there an existing type of voltage regulator that holds a desired/specified voltage below the Vin? I'd like a relatively efficient way to regulate a voltage at say 6.5 volts less than the Vin to my circuit. The circuit requires a Vin range of 7 to 60 VDC. So for example if the Vin was 45 VDC the voltage regulator would output 38.5 VDC and if the Vin dropped to 12 the voltage regulator would output 5.5VDC. Ideally this regulator would be able to supply a maximum continuous current of say 150 mA but would not use that much when the circuit wasn't using it... Perhaps this device (and/or circuit) is common and I just need help with the jargon/terminology. <Q> You merely have to connect a negative voltage regulator of the desired "drop voltage" to the positive rail. <S> Then, connect your load between the output of this negative regulator and the "ground" or DC Common or negative rail, of your voltage source. <S> Look at negative regulator ICs such as 79M05 (-5 VDC), 79M12 (-12 VDC), etc. <S> for fixed voltage drops. <S> The 79Mxx series will work up to a certain supply voltage. <S> Eg. <S> from Vin = 6 up to maybe 35 volts or so. <S> For higher Vin voltages you will have to identify a more appropriate family. <S> There is a wide variety of both positive and negative IC voltage regulators in the marketplace. <S> Check DigiKey and Mouser. <A> simulate this circuit – Schematic created using CircuitLab Depending if you can accept reasonable tolerance values, this could be selected for 6.5V +/-0.2V <S> ~150mA <S> using 1.4V for Darlington and 4.1V with two yellow or Red 5mm LEDs for D1 D2 . <S> With 1W max (6.5V*0.15A) dissipating in the device , it should have a heatsink. <A> It is not a common requirement, and it is actually fairly difficult/complex to do accurately with small current draw and capable of 7-60V input. <S> Part of that reason is that it has to work to 0.5V out. <S> The power dissipation at 150mA out is also a bit high for a simple and easily available Zener diode. <S> Here is a simple (but not very accurate) approach: simulate this circuit – Schematic created using CircuitLab <A> What you need is a shunt regulator -- those kinds of regulators are designed to maintain a constant voltage drop, just like you want. <S> Two important things to keep in mind: They will likely require a minimal current. <S> If load draws 0 mA, the voltage may get too high. <S> You may have to add a load resistor if this is an issue. <S> If want to use IC, they are not very powerful. <S> The biggest regulator I found on digikey claimed 200mA, but it is in SOT23 case, so I am not sure it would actually be able to handle your requirements: https://www.digikey.com/product-detail/en/diodes-incorporated/AP432ASAG-7/AP432ASAGDICT-ND/1301734
| There are also adjustable negative voltage regulators which can make almost any (reasonable) drop you need.
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What to do when I have energy loss due to distance I have to connect a total of 15 12 V, 2 A motors in parallel. They are all going to be connected to a battery and I’m trying to bring a line to connect all 15 of them. The distance between them is 25 meters. What can I do to reduce the energy loss due to the distance? <Q> Everyone seems to be suggesting the thicker wire route here.. <S> but you do have a different option. <S> Your other alternative is to either use a boost convertor to convert your 12V into a high DC voltage or use an invertor to make high voltage AC and transmit that down cheaper cabling instead. <S> Then add DC-DC or AC-DC power regulators at each motor. <S> That may seem extreme, but you need to do a cost comparison of which is cheaper. <S> Big beefy hard to manage cables, or cheap wiring and low cost convertors. <S> The other benefit of this route is you separate the power transfer efficiency issue from the target voltage-drop issue. <S> That is, the system ought to be 100% functional despite the transmission costs. <A> Based on your comment your pulling 30 amps of current through a 12 awg wire. <S> Normally 20 amps is the limit for 12 gauge, but for some industries it has to do with how hot the wire gets. <S> Go up to 8 awg wire, coarse or fine stranded, and that should fix the problem. <S> To be safe <S> the (-) wire is normally black while the (+) wire is red. <S> Use 6 awg (gauge) only if the voltage drop is serious, as in you have a long distance from the battery to the furthest motor. <S> You could run 6 awg dedicated wire to each motor instead of motor-to-motor, but the wire cost goes up because you have the distance issue. <A> An online voltage drop calculator says the voltage drop for 30 amps dc current in a 25 meter run of copper sire is 7.82 V for No. 12 wire, 3.09 V for No. 8 and 1.22 V for No. 4. <S> If you run a pair of wires from each motor directly to the source and you have 375 meters to the furthest motor, you would need no. 4 wire for 10% voltage drop. <S> If the first motor is only 25 meters away, no. <S> 14 wire could be used for that one. <A> If you are loosing energy due to the length of the wire then buy a thicker wire. <S> Alternatively you could also use multiple smaller cables in parallel to create the equivalent of a thicker wire. <S> 15 motors that are 2A each is 30A total. <S> The wire gauge table... <S> https://www.powerstream.com/Wire_Size.htm <S> ...says you need at least at 6 AWG or larger wire for 30A. <S> Or if you are wiring each motor separately then 18 AWG wire for 2A each motor. <S> A 6 AWG wire has a resistance of 1.295 mOhms per meter. <S> I am not sure exactly how the motors are physically arranged but assuming the 30A went through all 25m <S> then the total voltage drop would be... <S> 1.295 mOhm <S> /m <S> * 25 m * <S> 30A = 0.972V drop going to the loads and another 0.972V coming back. <S> The total voltage seen by the motor would then be 12V - 0.972V <S> * 2 <S> = 10.05V. <S> If you need to reduce the losses further then use a 0 AWG or 1 AWG cable, or more smaller cables.
| You could go up to 6 awg wire but the cost starts to go up quickly. If the wiring needs to run 25 feet between motors, you need to add the voltage drops between each segment between from a given motor to the source. I think the main issue here is wire size, since you did not mention any issues with battery life.
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Electric BLDC motor voltage and power limit factors Preliminary:I know basic model of DC motor which depends of two electric parameters such as voltage/rpm and current/torque. There are two models electrical - resistance, back EMF, and optional inductance of coils. On mechanical side there are torque, speed and optional moment of inertia. Question:What is the real limit for maximum voltage/frequency applied to BLDC motor if motor will be powered by high range current source ?My presumption is that motor's mechanical construction is limited mostly for currend and torque and when higher rpms are needed then should be able to increase voltage/frequency of motor. How it is with both case permament magnet or induction asynchronous motors?Why some datasheets of similar motors have different both nominal/maximum rpm and voltage specified? To summarize last talk thanks to Gregory Kornblum Some motors have ratings 24 V and could operate for higher voltages.My assumptions: Controller current source has filtered signals before output rails such that only envelope is driving motor's phases. Bearings and mechanical construction of the motor is strong enough to handle high rpm. There are fast mosfets to handle phases currents. Any further explanation/ideas or confirmation to the subject? <Q> You are correct in assuming that the basic mechanical construction of a motor limits current and torque more than it limits voltage and speed. <S> Speed is limited by the friction losses that the bearings can withstand. <S> The aerodynamic drag (windage) is also a factor. <S> Another factor is balancing of the shaft and rotor to minimize vibration. <S> At some speed, centrifugal force stress on the rotor may be a factor. <S> Voltage is limited only by the winding insulation. <S> None of those factors have really "hard" limits. <S> For most motors, those items are conservatively designed so that motors operated within the voltage and speed rating will have quite a long service life expectation. <S> Exceeding the published ratings will reduce the expected motor life, but the ratings can often be exceeded by 50% and even 100% without very rapid failure. <S> Note that the aerodynamic drag can increase quite rapidly, add significantly to the torque load on the motor and increase the current. <S> However if the current is held within the rating, tee motor will not overheat. <S> In effect, motor's torque rating and efficiency will be reduced. <S> Note also that load vibration and other load forces effecting the motor bearings can also increase with increased speed. <S> The bearings are definitely the leading candidates for the first thing to fail. <S> If you are designing a race car and can afford to replace the motor after every race, it makes sense to push the limits a long way. <S> If you are designing production equipment and have to cease production of expensive products if something fails, you may not want to operate everything below the ratings. <A> How it is with both case permament magnet or induction asynchronous motors? <S> Permanent magnets are not as robust as the copper bars embedded in the rotor of an induction motor. <S> They are brittle and not very strong mechanically, so at high rpm they may disintegrate due to centrifugal forces. <S> If the magnets are glued onto the rotor then at higher temperature the glue may soften, allowing the magnets to move around or even completely detach from the rotor (not <S> just theoretical - I have repaired several motors that this happened to). <S> They can also be permanently demagnetized by high temperature and/or high current. <S> Operating the motor at higher voltage increases the potential peak current draw, and also often increases operating temperature. <S> Magnetic losses increase at higher rpm, so even if mechanical forces don't destroy it there is a hard limit above which the motor will dissipate more power than it can handle even without a load. <S> Most BLDC motors use Neodymium magnets which have a relatively low Curie temperature , so they are usually the first to go when the motor overheats. <S> Why some datasheets of similar motors have different both nominal/maximum rpm and voltage specified? <S> Motors with 'similar' external dimensions and electrical specifications may have quite different internal construction. <S> A motor designed for higher rpm may have its rotor wrapped in Kevlar to hold the magnets on. <S> It may have better bearings and be finer balanced to reduce vibration. <S> To reduce magnetic losses it might have thinner stator laminations, reduced magnet coverage, larger air gap, different number of poles and/or winding pattern, or even a totally different core (slotless, ironless). <S> Specifications may also vary depending on the expected application. <S> A motor designed for continuous duty in an enclosed environment may be rated lower than a similar motor used for traction. <A> Frequency of what? <S> But it's about really high frequencies, you drive will limit you much lower by switching losses, power for gate drivers, etc. <S> Voltage- <S> again, normal motor can stand voltages that are orders of magnitude highet than its ratef voltage for sane operation. <S> This is why in many applications they use a 310V drives to move small 24V motors. <S> Bearings will fail much earlier than with BEMF you will reach anything close to the physical voltage limit. <S> Bottom line is that motor is a so complex and diverse creature that to understand its limits you really need to have a good view of the system and the motor itself.
| If you are talking about PWM, then you the only limiting factor i can think about and is related to the motor is capacitance between coils and the enclosure, so too high frequency would leak to the cage.
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Can grid-tie inverters charge UPS batteries if connected to the UPS output? If you have a UPS powering some load in an off-grid situation, and you have a solar panel with a grid-tie microinverter, is it possible to connect the grid-tie inverter to the UPS output, such that the microinverter assists with powering the load plugged into the UPS, and if there's an excess of power available, actually charge the UPS batteries? I've read that it's possible with certain kinds of inverters, and that Tesla's "Powerwall 2" is apparently one of them. It accepts power from grid-tie inverters on its output and uses it to charge its batteries, or if they are full to feed back into the grid. I thought this kind of function would require additional circuitry to extract the extra incoming power, but from what I've read, it sounds like there are some inverter designs that inherently have this capability. Is this true? What sort of designs would work this way? Is there any quick way you could tell whether a given UPS is capable of this, without getting hold of a schematic? <Q> I thought this kind of function would require additional circuitry to extract the extra incoming power, but from what I've read, it sounds like there are some inverter designs that inherently have this capability. <S> Yes, As long as the voltage criteria are satisfied a mosfet will happily allow current flow in either direction. <S> So it's not difficult to design an inverter that can feed power in both directions. <S> Indeed APC made some UPSes which they called "delta conversion" that explicitly exploited this, the battery bank was connected to the output through a single converter, then a separate "delta converter" was used to push power from the input side of the UPS to the output side. <S> Having said that though, the devil is in the details. <S> I would be concerned about the potential for bad interactions between the control systems of the two converters, and about the possibility for excessive voltages on the DC bus of the UPS. <A> UPS devices are not designed to be used this way. <S> They are designed to be powered from the input, not from the output. <S> I can see a number of safety hazards in your connection: If somebody turns UPS off, the grid tie inverter starts to free-run. <S> If somebody then turns the UPS back on, the UPS may not be synchronized to the grid-tie inverter. <S> Most UPSs have a power on/off button. <S> Edit: <S> Perhaps this may not be an issue after all, as the inverter may be designed to avoid danger when the power turns on. <S> The UPS may turn off, noting the lead-acid battery is below a given voltage. <S> When the lead-acid battery has rested, its voltage may rise, and if the UPS is poorly designed, the previous scenario (UPS turning off and then back on) can happen automatically due to the increased rest voltage! <S> The electrical cables needed to connect the inverter to the UPS output, not to the input, may be easily misused. <S> A permanent installation instead of detachable cables could solve this issue, but most UPSs are not designed for permanent installations. <S> After all these safety hazards, you probably find that you cannot charge the UPS battery from the output. <S> UPSs are designed to charge batteries when there is voltage on the input. <S> It would be foolish to design UPS to power its charging circuitry from the output, as it most use cases it would be just a circular energy transfer, reducing UPS efficiency. <S> I wouldn't do what you are planning. <S> Do the proper thing and convert the inverter to the UPS input, not to the UPS output. <A> The Actual qestion may be as below shown in Picture.
| I would not connect a grid-tie inverter to the output of a UPS that was not explicitly designed to support it.
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How can I prevent my soldering tip oxidising while I'm using it? I have a temperature-controlled soldering iron. When I use it above 250°C, the tip quickly dulls and won't take solder. By "quickly" I mean "within 20 seconds". I have to clean the tip with copper wool or a damp sponge literally between every joint . I am using leaded solder, and I only solder at these higher temperatures when the wires/contacts are large and difficult to heat. Or when using desoldering wick, which amounts to the same thing. Is my tip just of poor quality? I have tried different tips, but maybe they just all suck. Should I be using a more expensive tip? EDIT: I'm using a MarkEthan SMD Rework Station, with generic Ebay tips, and "AlphaMetals 8-Sn60Pb40" leaded solder. <Q> Most tips have a protected layer when new <S> and you don't want to damage that. <S> Make sure to use A LOT <S> (Note: capitals AND bold) of solder to tin it. <S> I make a puddle and keep it in there for some time, regularly refreshing the solder or start a new puddle as the solder will 'burn/oxidize' after a while. <S> In the beginning you will see that the solder does not 'stick', like water on a fat surface <S> it stays away form the tip. <S> Only stop after the solder sticks to the tip like water does to glass. <S> I had the very bad habit of cleaning the tip when done with a solder joint. <S> Wrong! <S> The best thing is even to put some more solder on the top when you are done. <S> Make sure to get rid of it before you start a new joint again as the solder will have gone 'bad' (oxidizes). <S> So the procedure is: Clean tip <S> Solder using new solder Put solder on tip Put iron away for a while. <A> You're probably not still looking for an answer, but if I were seeing this with one of my irons and I trusted I'd cleaned it properly <S> I'd start to worry that the tip wasn't achieving the target temperatures and was actually hotter than intended. <A> I had the very same trouble with soldering tips going brown (or black), and then losing the ability to pick-up solder - and remain clean. <S> If the tip temperature is too high, oxidation occurs. <S> How I overcame this problem - was to place a variac in series with the electric soldering iron. <S> in this way, you can adjust the variac control, and get a 'feel' for the precise temperature that will create a good solder flow, and yet maintain a reasonably clean iron tip. <S> I only use cotton cloth to wipe the iron tip, as steel wool removes the fine coating of iron. <S> This seems to be much more critical on any soldering iron that does not have any form of temperature - control feature. <S> I have several soldering irons, and they all get plugged-in to a variac.
| The problem here, I believe, lies with the soldering-tip temperature. Some tips which may help: Do not use abrasive cleaning methods on a new tip.
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Comparator, inverting/non-inverting terminal voltage is more than supply voltage I am using Comparator (schematic shown below). I have connected output with Vcc via 2k ohm resistor, as it has open collector output. In this configuration output of the opamp should be 3.3V. But the output is coming 0V. But if I change the supply voltage to 5V, then the output is showing 5V. (schematic below) Is there any rule "voltage at the terminal (inverting/non-inverting) of the comparator should not be higher than supply voltage of the opamp. Regards, <Q> It's a comparator, not an op-amp. <S> The question does not arise with op-amps because they generally do not have open drain outputs. <S> The answer is maybe. <S> You will have to read the datasheet for the particular comparator (assuming it has open drain/open collector output). <S> In the case of the LM339 the maximum output voltage is 36V regardless of Vcc. <S> Some comparators may have protection diodes on the output, so the datasheet would likely state a maximum such as Vcc+0.3V. <A> Look up the common mode input range in the datasheet. <S> This most likely doesn't exceed the positive supply. <S> If it doesn't, then your first circuit is running the part out of spec, and you have no guarantee what it will do. <S> Use resistor dividers on the input signals to bring them down to the common mode input range. <S> As always, you have to actually read the datasheet . <A> For almost all ICs, the voltage on the inputs must be somewhere between the supply voltage and Ground for proper operation. <S> For many linear parts, the inputs must be a significant voltage below the positive supply, or above the negative supply, for proper operation. <S> Read the datasheet for your device to learn its requirements.
| Possible solutions: Run the comparator from a supply voltage that is high enough so that the expected input signals are within the common mode input range.
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Is a stage box with an iron grid safe using this way? See also my previous question: How should I bypass switches Where I made a circuit directly by mains, and as proposal by very useful comments to use a separate 'control' circuit at 9 or 12V. However, according to this video: How to Build On-Stage Light Boxes for Your Band the switch under the 'plateau' is directly connected to 220V. However, in my case I want to make a very important change: I don't want to use plexi glass, but a metal grid (like this): I assume it would not safe to make a switch directly under this iron grid that is connected to 220V? Or is it actually not really safe for the plexi glass solution as in the YouTube video neither? (Btw, I think according to the related earlier question, I'm going for a transistor/relay/triac solution anyway). And a side question: another band I seen, uses a foot switch instead of a switch under the grid. I'm almost sure that foot switch is connected directly to mains (meaning 220V goes through it). Now this is seen quite often with living room lamps etc... but I wonder if this is safe on a music stage (what if someone would drop beer over it, and the foot switch is not 100% 'closed' somehow) ? <Q> There absolutely is no issue in using a correctly rated momentary switch to turn on and off a lamp directly. <S> After all, our houses are full of wall switches, foot switches, and even not so safe kinda sloppy switches from non controlled markets. <S> If you can understand what a proper environment is, and enforce it, then go ahead with whatever design comes to your mind. <S> Using a low voltage switching circuit is certainly a hassle, because you need a transformer, a PCB, whatever: it's a mess. <S> But consider this: you can buy an off the shelf relay, connect the mains voltage to it, put everything in sealed box like what they use to make electrical connections in gardens, and just run around your low voltage control wires. <S> With low voltage control you can do a number of things that are more difficult and costly to do with the mains directly: you can add crappy foot switches everywhere, and enjoy the light going on and off when your drunken guitarist drops a pint on it while smiling to that girl that was "definitely looking at him". <S> You can add alwasy on, always off switches at no cost. <S> You can parallel foot switches so that you can turn the lights on and off from various points in the stage. <S> You can add a microswitch to your drummer pedal and have cool, perfectly timed, blinking lights. <S> You can even program an arduino or whatever to produce super cool effects in the future, if you want. <S> Be safe on stage, if you have to ask if it is safe assume it is not safe. <A> The light box video has design flaws but works with the weight of stepping on one side only. <S> But at least it is rugged and insulated. <S> The choice of the >=20A switch and it’s height are critical. <S> Safety of using a metal grid instead of 1/4” plexiglass depends on your common sense of insulation and keeping it dry. <S> Grounding it may or may not be a wise idea with unknown guitarists vacuum tube amps , leather shoes and leakage hum. <S> ( some vacuum tube guitar pre-amps may have unsafe leakage leakage current available to go thru the strings. <S> Leather boots are known to be good for small current leakage from sweat absorption, good for anti-stat, but may induce hum if there is leakage to metal grid. <S> Then again, if grounded , it may also increase or reduce hum. <S> and impossible to predict if it is safe) <S> Enclosing a 500W Halogen heat lamp is not a great idea but for pulsed flood light, it does the job OK <S> .. Consider cheap LED flood lights, faster response time without the tungsten PTC time constant. <S> But I think the stomp timing effect gives it a smoother power effect with the one foot stomp control on the corner switch. <S> Plexiglass vs metal grid? <S> I think you could make the metal grid work, if you guarantee the switch to be insulated under all conditions ( coat exposed contacts with silicone) and ground the grid. <A> Yet, it will be safer with 12V. <S> As Vladimir Cravero said, with 12V you have more flexibility for further sophistication without worrying too much about safety.
| It will be safe if the grid is grounded and of course the switches insulated and designed for this purpose (strong enough mecanicaly). But there's a catch: there is no issue if you can ensure that the switch will operate in a proper environment at all times.
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(how) can I use 250v fuse in a 9v circuit? I bought a nice big assorted set of fuses ranging .1A to 20A. All of them have 250V on them. The question is can I use them in circuits with 3, 6 9 or 12 V? If yes, does lower voltage changes the rating amp of the fuse? <Q> You can use them at lower voltages (they will be safe) but you may see excessive voltage drop for your application. <S> You did not specify whether your proposed application is AC or DC. <S> In general you can use AC-rated fuses for DC voltages beyond 12V (maybe up to 32V), but read the datasheets. <S> High voltage DC (such as 125VDC) is a horse of a different color- <S> generally not safe at all . <S> This also assumes that your fault current cannot exceed the interrupting capacity of the fuse. <S> For the small glass fuse <S> it's 35A. <S> For example, a bog-standard 100mA 5x20mm glass fuse from Bel drops about 1.63V <S> so you'll lose more than half your voltage at full current. <S> At 1.6V across them they may never open at all. <S> This is not a consequence of the fuses being rated for high voltage- <S> it's because they actually require enough power (mW) to heat the fuse element to melting. <S> but there will always be some voltage drop. <A> The design of a fuse conductor material gives rise to the PTC characteristic which causes a thermal runaway effect above rated current so that resistance rises rapidly with temp until it melts the conductor. <S> The thin shape ensures that when this occurs, it explodes the material with sufficient gap that at least 1 mm air gap exists so that ~1kV spikes can be blocked on a 250Vac line. <S> It has no effect on low voltage circuits other than being relatively high resistance when conducting. <S> You also might want to consider getting an assortment of polyfuses. <S> - Why? <S> - Fuses do not provide great protection for low voltage circuits, because of the thermal mass of the transistor junction is much smaller than the fuse link <S> so the fuse is too slow to protect semiconductors. <S> It is effective for some low voltage applications such LiPo battery protection. <S> Thus 100mA fuses of this type rated for 250Vac are relatively useless for low voltage due to the maximum voltage drop possible when operating below the specified fusing limit. <A> I have used 250V fuses on 12V circuits without relevant voltage drop (less than 0.5V). <S> But the quality of the fuses may differ. <S> I can confirm that they were working as some of these fuses blew instantly when short circuited. <S> The best way is to measure the voltage before and after the fuse and see if it's acceptable. <S> However, for less than 12V, or if you need a very precise voltage level use fuses specificaly adapted. <A> Yes, you can use them, <S> The amp rating does not change. <S> A 1V power supply that can deliver 25A will blow a 20A fuse if shorted (a 25A fuse would probably glow (light up)). <S> The 250V rating means that they are safe to use in 250V circuit, but may arc at higher voltages.
| There are one-time (even SMT one-time) and self-resetting 'fuses' that may be more appropriate for low voltage
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Why do push button telephones use dual-tone for signalling? Here is a related information from wikipedia: For touchtone service, the signal is a dual-tone multi-frequency signaling tone consisting of two simultaneous pure tone sinusoidal frequencies. Above shows that if one pushes number 1 he sends the mix of 697Hz and 1209Hz to the telephone station/center through a wire. My questions are: What is the practical reason or advantage to mix two signals instead of a single pure tone? Is there a reason to use such frequencies like 1209Hz which do not belong to any music tones(modern western twelve-tone equal temperament)? <Q> The two reasons are simple: <S> Eight frequencies are easier to discriminate with simple analog electronics - banks of bandpass filters or even vibrating reeds - than sixteen frequencies. <S> The equally tempered scale is too close to the natural scale, which has simple fractional relationships between the frequencies. <S> Consider that a phone line may be highly distorted : the second harmonic is one octave above the fundamental, and the third harmonic of a note is an octave plus a fifth. <S> If you used more musical intervals between dialing tones, harmonic distortion could result in dialing the wrong number. <S> The tone frequencies, as defined by the Precise Tone Plan, are selected such that harmonics and intermodulation products will not cause an unreliable signal. <S> No frequency is a multiple of another, the difference between any two frequencies does not equal any of the frequencies, and the sum of any two frequencies does not equal any of the frequencies. <S> The frequencies were initially designed with a ratio of 21/19, which is slightly less than a whole tone. <S> The frequencies may not vary more than ±1.5% from their nominal frequency, or the switching center will ignore the signal. <A> If you used single tones instead of dual tones, you'd need 16 of them instead of only 8 as in the DTMF system. <S> Given that they have to be spaced far enough apart for reliable detection, and that you only have the range from 300 to 2700 Hz to work with, you'd probably find it difficult to decode that many different tones reliably using the technology available between 1950 and 1963 <S> (when touch tone dialing became available.) <S> Two tones also helps to reduce false detections. <S> If you only pick up one tone, you ignore it. <S> If you pick up two valid tones at once then it is much more likely to be an intentional dial command. <S> As to why they didn't use musical notes, I'd guess because they were more concerned with getting tones in a range and spacing that was good for detection and easy to generate. <S> Also, you don't want a harmonic of one tone to be the same as one of your higher tones. <S> If you used musical notes, that would be the case. <S> Look at the frequencies used. <S> None of the higher tones are a multiple of the lower ones. <A> It is because phone systems were designed with in-band signalling due to technical limitations at the time. <S> The signaling worked over the existing telephone infrastructure that previously was operated by hand, aka hello operator, connect me to 4562 . <S> The dual tones required allow to distinguish the tones from normal speech that could have the same frequency, but would not have both at the same time. <S> The frequencies used were spread out to make it easy for the signaling system to distinguish them even when noise was present over the line. <S> I doubt it had anything to do with musical tones. <A> Less frequencies needed = <S> > simpler to produce digitally, analog base timing oscillator can be accurate enough because the allowed error is bigger, less tuned detectors needed: 8 vs 16. <S> 2 frequencies at the same time (carefully selected for preventing distortion to make false mixing products) = <S> > not highly probable to be produced accidentally in ordinary sounds. <S> A legend I have heard : Single tone coding has been in use as internal telephone system signaling method until a boy with capable brain, criminal mind and perfect pitch <S> ear learned how to whistle transcontinental free calls and how to put the central down.
| The frequencies were chosen (citation needed, no doubt - here for example ) to reduce or eliminate the possibility of harmonic or intermodulation distortion between tones being mis-detected as the wrong number.
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How do I wire this 3 way switch? I have this switch it goes 3 ways, I am a noob and just started learning about electronics. I have a small oled Display connected to my nodemcu which is displaying information. The middle switch is off right? So I want the middle switch to turn the display off, the left switch to show specific information and the right switch to make the display show other information. How exactly do I wire it? <Q> I would call that a single pole, double throw (SPDT) center-off toggle switch. <S> Electrically, it looks like: simulate this circuit – <S> Schematic created using CircuitLab <S> With the handle in the center position, there is no connection between any terminals. <S> suggested connection: <S> simulate this circuit <A> Your device has two switches inside, connected to three external terminals. <S> The center terminal is common to both switches, and the two ends are at the other ends of each of the switches. <S> When the toggle is in the center, both switches are open. <S> That means all three terminals are disconnected from each other. <S> When the toggle is to one side, the center will be connected to one of the two outer terminals. <S> When toggle is to the other side, the center will be connected to the other terminal. <S> The rest is up to you. <S> This device has three states it can be in. <S> You can certainly detect with of the three states. <S> What you do with that information is up to you, and it's not clear what exactly that is anyway. <A> Since the paddle is in the middle, I think that it is a 3 position switch (SP3T)(single pole tripple throw). <S> Here is a very rough diagram of the switch <S> The red slider moves when you tilt the paddle to either side and shorts out two of the blue contacts. <S> In center position, the slider does not touch either of the side contacts.
| To confuse newcomers, when the handle is to the left, the center terminal is connected to the right terminal, and when the handle is to the right, the center terminal is connected to the left terminal.
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Missing component on the Usb to RS485 converter I have a Usb to RS485 converter that was dead on arrival. I opened it up to see if I could fix it to find out that it was missing a component on the top left corner. This is a very common converter. Can someone who has the same device tell me what the missing component is? Thank you! <Q> There must be one at each end of the buss. <S> So if this is just point to point, then it may be added externally to the screw terminals. <S> (kudos to GK for the nudge) <S> I would use a logic probe, scope or DMM on the MAX chip pin 1,2,6,7 and check status. <S> If RE is low then RO = voltage(Pin 7-pin 6) <S> "1"=+ve, "0"=-ve <A> Look up the max485 datasheet and look at the typical application schematic. <S> Then trace out and compare the board you have to it. <S> Most likely is a resistor or cap. <S> Then again, it might not be missing. <S> Unpopulated components are common. <A> It might work okay on a clear day with a very short cable, but what you have there is a bit of a steaming pile of something. <S> The main chip is a CH340G, which is a standard Chinese-origin USB-Serial chip. <S> This chip should have kernel support by any recent version of Linux. <S> If it's not enumerating then there's something likely wrong with the USB chip, the crystal or the connector. <S> It's definitely not the resistor position you point out, and adding a 120\$\Omega\$ resistor there won't help and may make things worse. <S> You will find that they do not actually drive the bus actively in one direction and depend on resistors to the supply to passively drive the bus. <S> The drive impedance is thus both asymmetrical and incorrect. <S> You can try inspecting it for bad connections and replacing the crystal if you feel like it. <S> In particular, wiggle the connector while looking at the SMT pins ( preferably under a magnifier in good light ) and see if one of the connections is broken- <S> since that's a likely path for damage.
| This is one of those stupid crappy cheap USB-RS485 modules from China that do not properly terminate the output nor do they properly enable the fake RS485 driver. Looks like it is missing a resistor with the marking 121 on it (120Ω)
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How does this transformerless 5V power supply work? I recently took apart a crock-pot cooker that stopped working correctly. It had two boards: One that dealt with the high-voltage for controlling output to the heating element The other had a 4-bit microcontroller and a few buttons/LEDs. I was surprised to see no transformer anywhere, and very few parts, considering there was a signal labeled +5V for powering the microcontroller. I reverse-engineered the schematic: simulate this circuit – Schematic created using CircuitLab Is it me or is this a really crappy design? I was very surprised to see that the DC GND was connected directly to the 120VAC L ! Is the 5V DC basically just a rectifier with a voltage divider? <Q> It's you. <S> The 1.5uF film capacitor acts as a reactive (lossless) dropping element. <S> D21 shunts negative current away and D22 conducts positive current to the filter capacitor C23 and it is limited by the 5.1V Zener diode D23. <S> The 1M bleeder resistor (across the 1.5uF) prevents a shock from touching the plug pins after it's pulled. <S> R21 limits the peak current to about 7A if plugged in at an AC peak. <S> Any breach in that insulation (including you opening up) could lead to a potentially fatal shock. <S> As in the comments, your heater should go directly to the N side of the mains. <S> There's probably an overtemperature fuse cutout in there somewhere too, maybe buried near the heater, and probably an overcurrent fuse somewhere. <A> So if the triac is going to switch the L side of the load it needs a control signal relative to mains L. <S> As such it greatly simplifies the design to use mains L as the "ground" voltage for the electronics. <S> Obviously this means that the electronics are live but that can be dealt with by insulating the control buttons and knobs. <S> The 22 ohm resistor limits the effect of spikes on the power supply. <S> Without it a fast spike (including a badly-timed connection of the mains power) could cause extreme currents to flow. <S> The capacitor has a reactance of about 1.8K <S> so it limits the current to about 67mA. <S> The 1 Megohm resistor ensures that the capacitor is discharged when the mains is disconnected. <S> The Diodes convert the AC current to DC. <S> We need both diodes as there must be a path for current to flow in the capacitor during both half-cycles of it's AC waveform. <S> Finally the Zener diode and capacitor control and smooth the output voltage. <S> This type of power supply does have some drawbacks <S> The draw from the supply is basically constant and has to be designed around the maximum load the power supply will run. <S> The power factor is terrible. <S> The DC output current available is less than half the RMS AC input current. <S> But overall for providing a little bit of current at 5V for a control circuit that needs to be mains referenced it's not a terrible option. <A> I've seen this approach used in PowerLine Metering. <S> In that case, 0.68uF was used. <S> The circuit came to my attention because of small errors in metering, especially at the ZERO load point. <S> Turned out the magnetic field of the 80mA thru the capacitor was magnetically coupling into the +- current sense wires across the 100 microOhm shunt. <S> Theerror voltage was approximately 100 nanoVolts and caused big errors when monitoring the power consumption of minimal loads, such as 10 watt bulbs.
| This kind of circuit requires all elements that could come in contact with a human to be isolated from the mains for safety, including any kind of display, switches etc. The triac needs a control signal relative to the power line it is controlling.
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What causes a faulty Linear Voltage regulator to output wrong voltage I made a circuit where I needed DC 8v supply. I was using a KA7808 to regulate the voltage. While testing the voltage regulator unit, I identified that the Regulator IC might be faulty because I was getting an output voltage of 4.8v - 4.9v for input voltage 18v. I am curious to know what might cause a voltage regulator IC to output wrong (not the rated) voltage. My assumption was, a faulty one would either passthrough the input voltage as is, or burn out, or output 0v. I'm using the KA7808 with simple config, input pin directly connected to input voltage, Gnd pin connected to ground and measuring output voltage directly from output pin, no other passive component used. And I measured the output voltage at no load condition. My questions are: For the given circuit (see below) can we expect a Linear Voltage regulator to fail in a way that it generates a wrong output voltage which it is not rated for? If answer to first question is 'yes', does it mean we should check every regulator for expected output voltage before using it in a circuit? I mean, do professionals do that as an 'always do' in everyday work where these regulators are used in large numbers? This is not a yes/no question, but I was also looking for an explanation which might explain what could have caused the regulator to fail in this way, if the external circuitry supporting the Linear Regulator to work as expected, is correct. Can this be expected if any of the building blocks (the internal Error Amplifier, Thermal Protection, Current Generator, Series Pass Element etc) of the regulator fails? Or there is a chance of simple 'wrong labeling' by the manufacturer? (I thought, wrong labeling is impossible, because that is what an Electronics Professional would blindly believe in!) Hope this clarifies my exact questions. See the circuit diagram below: simulate this circuit – Schematic created using CircuitLab <Q> LM/KA78xx regulators are almost impossible to get to oscillate (unlike many other types such as LM1117, LM7905 etc). <S> Others are either repeating folklore or are much (un)luckier in this regard than I have been (and I've tried to get them to ring). <S> We shipped approximately 30,000 units with no output capacitor, for example, and zero problems. <S> And some 78xx datasheets explicitly state that the output capacitor is not required for stability. <S> One of the advantages (along with price) of non-LDO positive regulators. <S> Usually you want a cap on the output for transient response on a digital circuit, but typically nothing bad will happen other than poor regulation if you omit it. <S> The internal voltage divider is more than sufficient minimum load at normal temperatures. <S> If the ground pin connection is not in place your output will go too high, so it's definitely not that. <S> You could have a (possibly fraudulently) remarked 7805. <S> But, I suspect <S> your problem is that you have a mains-frequency transformer followed by a rectifier and either no capacitor or insufficient capacitance. <S> This is not a stability issue, but a filter capacitor/ripple voltage issue. <S> if the input voltage is above about 9.5V you'll get 8V out, but if it dips down to say 3V, you'll get very little out, so the output would be a clipped version of the input waveform, plausibly with an average that was 4.8V. simulate this circuit – <S> Schematic created using CircuitLab <A> All these linear regulators require capacitors around them to keep them stable. <S> If you do not include those, the device will not operate as you expect. <S> See this answer for more information. <A> As Trevor suggested, the IC could be oscillatin. <S> Or the Iload is too high. <S> Or the input ripple is tool high. <S> Or Vin is too low. <S> Or the GND pin is not connected; with many of the internal circuits needin a path to GND or wild behaviors result, you need to ensure GND is indeed 0 volts. <S> Or the IC put itself into foldback protection, because the Cload is too high and the self-heating during powerup caused thermal overload and the part shutdown.
| You could have a defective part.
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Reading dynamic resistance values over a wide range I'm trying to measure a material which might fluctuate between 500-100k Ohms with an ADC (e.g. an Arduino or Raspberry Pi with something like the MCP3008 ). I'm a novice with electronics, and my go to strategy was simply what I knew already: voltage dividers . The problem I'm running into is the wide range. My setup is like so, where I want to know R1 : 3.3V -- R1 -|- R2 -- GND | ADC I conduct the following simple calculations to get it: v_out = ADC * 3.3/1024R1 = R2*(3.3-v_out)/v_out The problem as many of you likely know, is that highly mismatched values for R1 and R2 make for poor resolution, so the output looks very "steppy." Here's a simulated plot of a range of R1 values (what we're measuring), and the value calculated using the formula above using some common resistor values for R2 : The 1k and 4.7k are great for the low end, but they really flatten out at the high end where a 10k would be much better. Is there a reasonably simple and low cost circuit/method to read a dynamic resistance that might swing several orders of magnitude? Dreams/wish list: lost cost (say, $25) works with hobby-level hardware (Arduino or RPi) measurement error of <= 1% Ideas considered Given the above, the dream seemed to have a variable resistor! I learned that digital pots exist and thought I could idea to use one . With some great answers there, it turns out I can, but the accuracy is poor (~5% vs. <1% with fixed resistors) and somewhat unrepeatable. I also thought of having several R2 resistors connected to separate pins on the ADC. As the value of R1 changes I could switch which pin I read from. With this circuit, all candidate R2 's will be connected to the output of R1 and analog inputs... I don't know what that will create with respect to a circuit. Are they floating? Can I sort of disconnect the unused pins so I don't get a weird multi-voltage-divider? That might be what this question gets at, just for the purpose of stopping power drain. Sorry if this is a dumb question. I can find plenty of confirmation that this problem is real, such as from this article on making an Arduino Ohmmeter : The accuracy of the Ohm meter will be poor if the value of the known resistor is much smaller or larger than the resistance of the unknown resistor. I just don't find much on what people actually do in this situation in the real world. Many thanks. <Q> In an Arduino, pins are high impedance when set to 'Input', and a good low output impedance when set to 'Output', capable of sinking <S> >20mA. <S> This enables you to split up the range, and measure each optimally. <S> This applies to both digital AND analog pins. <S> For instance, let's the split the 500ohm to 100k range, which is a resistance ratio of 200 end to end, into 3 sub-ranges, by taking the cube root of the ratio. <S> Then we can handle it in 500-3k, 3k to 17k, 17k to 100k ranges, each with a ratio of 6 end to end. <S> We get the best centring of the range when we measure it against the geometric mean of its end points, so use 1200ohms, 7k and 41k as the 'other leg'. <S> Connect up the Arduino as follows. <S> I've rounded the resistors to preferred values, as the actual values are not important, just that they're known. <S> Measure the voltage on pin A0, using the ADC with VCC as the reference. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> With all pins set to Input, the effective resistor to ground is about 40k, for the range 7k to 100k. <S> Setting pin A2 to output low, and making sure A1 is high impedance, by pinMode(A1, INPUT);pinMode(A2, OUTPUT);digitalWrite(A2, LOW); pulls R3 down to ground. <S> Now the resistance to ground is about 7k, for the range 3k to 17k. <S> Set it back to high impedance like this pinMode(A2, INPUT); When you set A1 low in the same way, the resistance to ground is 1200, for the lowest range. <S> Note the maximum current that A1 has to sink is about 2mA in this case, well within the capability of those pins. <S> You can of course split the range into more and finer ranges, by doing the appropriate sums and using more pins and resistors. <S> You may be able to substitute digital pins for A1 and A2. <S> While they too are high impedance on input, I've not yet waded through the documentation to check whether intermediate voltages are OK on a digital input. <S> They aren't on usual CMOS logic families, the inputs can oscillate and draw excess power, but Arduino may be different. <S> Analog inputs are (of course) happy with any intermediate voltage on the pin. <A> One simple way is to just use a better external ADC. <S> For example, an ADS1115 gets you 16-bit conversion with an on-chip PGA and reference, for around five dollars in singles. <S> I2C interface. <S> Pretty decent for non-critical applications. <A> I would look into two ways of doing this: Voltage dividers where the known resistances are multiplexed. <S> (This idea is already hinted at in the O.P.) <S> If you want to try this on the cheap, connect the resistors to digital pins on the microcontroller. <S> When the digital pin is low, the resistor is connected to ground. <S> When the digital pin is floating, the resistor is disconnected. <S> Otherwise, an analog mux like a CD4051 or ADG728 should do the job. <S> OpAmp current source <S> The current through the unknown resistance is supplied by an OpAmp current source, and this current source is adjustable. <S> (Look up voltage controlled current sources. <S> My favorite is modified Howland current source , very versatile.) <A> One could convert R to frequency then convert f to some other parameter like force or displacement. <S> There will likely be hysteresis or backlash effects. <S> I would use the resistance as a negative feedback value to shunt cap in an inverting Schmitt Trigger inverter. <S> Then use a reference Xtal clock to measure the frequency over a time interval. <S> Impressions: <S> I have yet to find such a material that has 1% accuracy and doubt such a material repeatability exists with such a wide dynamic range. <S> The accuracy of the frequency measurement can be made to better than 1% <S> but I suspect the material measurement after repeated motions to the original position will most likely change > 50% in the relaxed position which may require a minimum tension to reduce dynamic range but improve repeatability/accuracy. <S> Thus the question needs to be re-written to include all the properties (specs) of the material and not just a simple how to measure R? <S> Recommendation: <S> The best test method is not to try to design an ADC on an unknown material but rather use a calibrated RLC meter with at least 3 frequency values and characterize the material properties properly for Rs, Rp, C values @f and repeatability and perhaps variance with temperature, Vdc bias etc. <S> You are a novice and you have much to learn.
| Material capacitance is also of great importance when the resistance changes this much since AC impedance measurements will measure both at one frequency and one value, but introduce great errors when any variable changes.
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What sensors to use to read gym machine selected weight I'm trying to find a set of sensors (to build a solution) to determine how many weight blocks an athlete is lifting on the weight machine like this: I already tried this solutions: Put RFID tag on each weight block and set a RFID reader a little bit above the top block. The idea was that when blocks are moving up, the RFID reader would know what blocks were lifted just by comparing how many unique RFIDs were detected in a sequence. The problem with such approach was that RFID reader was skipping some of the RFID tags (probably because weight blocks were lifted too fast so that RFID reader couldn't keep up). Attach ultra sonic sensor (Sensor_1) ( https://www.sparkfun.com/products/13959 ) on top of the block weights stand and other ultra sonic sensor (Sensor_2) on the side as shown in the picture: Sensor_1 is facing down. Sensor_2 is attached to a helper stand as is facing horizontally to the side of the block weights stand. The idea was that when user starts lifting the weights, at some point Sensor_2 detects a weight block (that is horizontally at the same height), Sensor_1 remembers the distance (Distance_1) to the top block weight. When Sensor_2 shows there is no block at the same height anymore, Sensor_1 checks the current distance (Distance_2) to the top block again and then it calculate the number of blocks by the following formula (Formula_1): number_of_blocks = (Distance_1 - Distance_2) / height_of_1_block_weight, where height_of_1_block_weight = 4 cm, Distance_1 ~= 115 cm, Distance_2 < 115 cm The problem with such approach was that ultrasonic doesn't measure distance to the point (like a ray), but there is a wave with a 15 degree angle instead and therefore we have issues with wave reflection from other objects (block weights stand or other stuff) We tried the same approach as #2, but with IR sensors ( https://www.pololu.com/product/1137 ). It doesn't give accurate distance because the sensor measurement error sometimes is more than 5 cm so the Formula_1 doesn't give expected output. Also, the sensor has 25 Hz limitation so if you lift the weights quick, it will skip distances > 3.2 cm (80 cm / 25 measurements a second). The only way left that I see is to attach a camera and analyze the output image on the software layer... Please, let me know if you have any other ideas or if you see a gap in the solutions that I specified. Thank you! <Q> 3 come to mind. <S> An IR encoder as RoyC mentioned. <S> Paint the back or sides of the weights alternating white and black. <S> A camera as you already mentioned. <S> OpenCV would make quick work of this. <S> And You can do it at a distance without touching anything on the machine. <S> And the last option is an actual weight sensor. <S> A load cell placed under the weights. <A> I would look at it from a different angle and try to detect in which slot the metal pin is placed. <S> This is an easier task and could be done in several ways such as using proximity sensors or micro switches. <S> These could be wired to a micro controller and thus you could detect the amount of weight blocks lifted. <S> This is just an idea for a possible way forward. <S> Designing circuitry would be a different question. <A> Put a small infra red reflector on each block and arrange these in a vertical line. <S> Your sensor set up is an IR illuminator and a directional IR sensor (photo-diode or transistor in a tube) mounted to line up with the reflectors just above the top block at rest. <S> You need to amplify the output from the photo detector and set the detection threshold so that it gives an output only for the reflectors. <S> This will give a chain of pulses which you can count to give you the number of blocks passing. <S> This can be made to work at very high frequencies <S> so no problem with speed. <A> Small cheap magnets on the weight block and two hall sensor facing the magnets. <S> The first sensor needs to be a little above(the gap between the blocks) <S> the height of the top magnet and 2nd one exactly facing the top magnet. <S> Count trigger in the first one to check how many weights have been lifted. <S> And when the 2nd sensor is trigger and first one is not, the weights are at rest. <A> Looking at the top arm, I'd suggest a strain gauge . <S> It's a very simple component, which you can glue directly on top. <S> You may need to smooth the measured values a bit, as the strain depends on the force bending the top bar. <S> And while that's exactly the gravitational force in the stationary situation, when exercising the weights accelerate up and down <S> so you need to use the average force. <A> I'm not sure what kind of time/effort <S> you're willing to invest into this, or how elegant of a result you're lookign for, but here's a cheap/simple simple approach, which amounts to creating a switch between every pair of plates. <S> : <S> Connect each plate's bottom wire to a common wire. <S> Connect each plate's bottom wire to a separate digital IO pin. <S> Count how many plates there are by how many wires are connected to the common. <S> To save on IO pins, you could even use a priority encoder to use ceil(log_2(number_of_plates)) <S> pins. <S> On the downside, this will look janky, will only be able to tell the weight while it's in motion, and it might break with use. <S> But on the bright side, it's cheap, simple, and you probably already even have all the parts you need to get a working prototype
| Tape aluminum foil contacts on the top and bottom of each plate, in a consistent location so that each contact touches the one of the plate next to it. When the selected weights are lifted, the load cell will show a negative change equal to the weight that was removed upwards.
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Influence routing on power consumption in FPGA What does influence of routing in static/dynamic power consumption in FPGA design? I want to know that, different routing results different power consumption? <Q> There will definitely be effects. <S> Generally the place and route tools will attempt to make the wires as short as possible in an effort to minimize delay. <S> There could be variations in the static power consumption of routing components as well, though this would be extremely architecture-dependent. <A> The problem is FPGA's have a lot of gates that switch at the same time <S> (even amps of current can be drawn from some FPGA's), this results in the load increasing and the demand for current increasing on very short timescales. <S> This causes a voltage drop at the power pins on the FPGA. <S> If the voltage drop is to great, the gates will not be powered and output 'noise' instead of the correct voltage for a 1 or a 0 which will result in an error in a Boolean calculation. <S> PCB traces are simply conductors, every conductor has resistance (which causes a voltage drop) and inductance (which prevents current from moving from one place to another immediately). <S> This contributes the voltage at the FPGA to drop. <S> What can you do about it? <S> There are two ways to combat this problem: 1) <S> Make sure you have sufficient power filter capacitors. <S> Power filter capacitors provide a short term storage next to the FPGA to combat line inductance. <S> 2) Make sure the voltage drop through PCB traces is small enough for the current moving through it. <S> The first thing you will need to do is find a PCB trace calculator that can estimate the resistance of the trace. <S> Then estimate the current needed by the FPGA, most provide a tool or a spreadsheet (like Xilinx ) that can estimate the current. <S> If your FPGA needs at most 1A, and it the pcb trace is 0.100Ω <S> then this would result in a 0.1V drop <S> (V=I*R). <S> This may not be acceptable for a 1.2V or 1.5V design, check the FPGA datasheet for limits on power. <S> If you increased the trace size by 5x then you would only have 0.020Ω and a 0.02V drop. <A> An FPGA, at least the Xilinx ones I've used, has a variety of routing resources. <S> Clock distribution is optimised for high speed clocking, and covers the entire chip. <S> This is used preferentially for clocks by the PAR. <S> Data has a number of different lengths of line. <S> Some go LUT to LUT for very local connections, fast carry chains for instance. <S> There are so-called 'long lines' that cross the die, which can be broken into small sections to limit dynamic power consumption, or routed right across. <S> There are also several types of intermediate length line. <S> All of these will have different dynamic power consumption. <S> The PAR will select from the available resources to meet timing constraints. <S> With a sparsely populated die, it might be possible to anticipate what will get used for what. <S> With a dense design, where the PAR struggles, it will be difficult to guess what it's going to use for what. <S> This means that only the post-layout reports will give you accurate estimates. <S> There is only so much you can do to control power dissipation from interconnect, once you have done the obvious things of arranging adders in columns, putting logic next to its IO so you don't cross the chip, using the dedicated clock buffers and so on, the obvious stuff. <S> Basically if the PAR is to meet timing constraints, then it has to use what it uses.
| Yes, with each trace on the PCB there will be parasitic resistance and inductance . Longer wires have more capacitance that will need to be charged up on every transition, increasing power consumption.
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Need some help building a TL072 preamp circuit I have an early 70's Magnavox turntable/radio console that I want to be able to plug my smartphone into so that I can use it as an external speaker. I'm plugging into the Tape In input, which is an RCA cable. If I go straight in, the volume is too low--I have to crank the phone and the console to get a decent level. I suspected an impedance mismatch and asked for advice on the VintageAudio subreddit. I got this response: "The output of the phone is probably a modern "line out" level of 150mV. The output of the tuner circuit is more likely around 1-2v. The function switch in equipment of this era is almost ubiquitously a literal switch with no additional amplification adaptations between the various input selections. Also, the output of the phone is likely 8 ohm for headsets, vs. 47k ohm the input expects. You're correct that this would be an impedance mismatch. What I've done in these cases is build a small preamp module using a TL072 dual-channel opamp. They're less than a buck each and I just use the simple preamp circuit provided by the manufacturer in the data sheets for the IC. Hang a half dozen capacitors and resistors off that 8-pin chip, feed it 12V and it can take care of all the level and impedance mismatches like magic." So I bought a TL072 and figured I'd build the "AC amplifier" circuit that is listed on the datasheet (schematic shown below). I have all the necessary parts but I have a few questions as I am a total noob when it comes to building circuits: 1.) Is this in fact the correct circuit (or a correct circuit) that will do what I want? If not, is there a better or simpler circuit that will accomplish the same goal?2.) Notice the line that the red arrow is pointing to. What does that connect to? All of the other pins on the IC are labeled (VCC+, N1, OUT, etc.) but I don't understand what that line, which is connected to ground, is supposed to connect to on the IC.3.) I'm assuming that I would connect IN+ and IN- to the mini jack that I'm going to plug my phone into. And on the output, I would connect the OUT to one of the L/R leads of the RCA cable, and the ground to the other RCA lead. Are those assumptions correct? I guess I'm trying to figure out how a single output connects to a stereo Tape In. Please enlighten me. :) <Q> I have used the TL072 op-amp a lot because it is a quiet amplifier, good enough for at least 16 bit audio. <S> But it is designed for +/- <S> 12 volt to +/- <S> 15 volt power supplies. <S> The trick is to insert a 10 K resistor from the output pin to Vcc. <S> This causes an offset current that stabilizes the op-amp. <S> Replace your 1 Meg resistor with a 100 K resistor. <S> Delete your 50 ohm resistor. <S> Insert a 10 uF capacitor with the <S> (+) pin connected to your IN- input. <S> Do not use your IN+ input as this is not an instrument amplifier. <S> Now you have a single ended input with a gain of 10, and a stable op-amp. <S> Because your output has a voltage equal to 1/2 Vcc on it, I would insert a 47uF capacitor at the output with the (+) pin going to the op-amp output. <S> The pin your pointing at with a red arrow is the Vee or V- pin, which normally goes to a -12 volt to -15 volt rail. <S> You can just ground it if you use the trick I mentioned in the first paragraph. <S> Without that 10K pull-up resistor the op-amp may produce severe distortion. <S> Adjust the 100K resistor if you need more gain. <A> Op Amps designed for bipolar supplies are referred to as Vcc and Vee or V+ and V- and ground is your choice of external 0V or midpoint connection. <S> Since you are using a single supply <S> your input and output DC levels are both at Vcc/2. <S> The offset adjust pot is not needed, but you may or may not need a series Dc blocking cap for the output. <A> Turns out I posted the wrong schematic-- <S> the IC in my schematic is a TL071 instead of a TL072. <S> A user on Reddit pointed me to the following schematic that uses the TL072 <S> and that should do what I need. <S> This schematic shows only one channel, so I'll need to duplicate all the components (except the IC itself and capacitors C3 through C6) for the second channel.
| For stability and ripple filtering there should be a 47 uF capacitor from Vcc to the ground (Vee/V-) pin.
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Power supplies in series, can you ground the middle leg? Let's assume I have two 15V DC power supplies. These power supplies are allowed to be connected in series (per the user's manual). I have a device that requires a signal from +15V to -15V (motor driver circuit). Typically, I would put both my power supplies in series, I would then provide 15V (relative to ground) to the motor driver circuit to use as ground. So from potential earth perspective, the motor driver circuit has 0V as its -15V, 15V as its ground, and 30V as its 15V. However, what would happen if I were to put both of my power supplies in series and then ground (potential earth) the middle leg from image below (i.e. ground negative terminal on power supply 1/positive terminal on power supply 2). Would I need to worry about power supply 2 shorting to potential earth? I would guess that since each power supply can be used in series, that means the + and - terminals are isolated from ground. Thus, it would be possible to connect the middle leg to ground. This would require that power supply 2 be able to maintain a truly negative voltage relative to potential earth. Am I correct in this assumption? <Q> As long as the outputs of both supplies are floating (not connected to Ground, or to each other in some other way), connecting the two supplies in series, and calling the mid-point "Ground" is perfectly normal. <A> This will cause a problem. <S> The reason you can get away with them in series <S> it is because of galvanic isolation, which explicitly removes the continuity between the input and output via a transformer. <S> You can, however, treat the tap between the two as ground as far as your motor driver is concerned -- assuming it's not itself referenced to earth elsewhere. <A> (The outputs are always isolated from Earth ground in my experience for cases like this.) <S> Let's see how this might actually work <S> : simulate this circuit – Schematic created using CircuitLab <S> On the left side, the load is grounded on one end, with the other end attached to the \$-15\:\text{V}\$ rail. <S> Here \$V_1\$ isn't doing anything. <S> But \$V_2\$ is just working normally . <S> It sees the load and does the usual things you expect from a power supply. <S> So \$V_1\$ is inactive and \$V_2\$ is active (sourcing current into the load.) <S> On the right side, the load is grounded on one end, with the other end attached to the \$+15\:\text{V}\$ rail. <S> Now \$V_2\$ isn't doing anything. <S> But \$V_1\$ is now just working normally . <S> It also sees the load and does the usual things you expect from a power supply. <S> So \$V_1\$ is active and \$V_2\$ is inactive (sourcing current into the load.) <S> Nothing particularly troublesome, so far. <S> So here's the last case I want to cover: simulate this circuit <S> In this case, the load isn't attached to ground. <S> So the current through \$R_{LOAD}\$, sourced from \$V_1\$'s (+) terminal, must be somehow sunk by \$V_2\$'s (-) terminal. <S> This then means that there must be a path by which \$V_2\$'s (+) terminal can allow such currents to arrive at \$V_1\$'s (-) terminal. <S> When they say that the power supplies can be stacked, they are telling you that this path exists without impairing the function of the power supply regulation features. <S> So everything should work fine, I think.
| If the power supplies make it explicit that their outputs can be connected in series to stack up the voltage, it's been my experience that this works fine.
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How to determine the saturation level when an opamp schmitt trigger starts worknig? The Opamp inverting schmitt trigger is constructed as shown. The output signal is So I ask why is the output at the start of operation saturated to positive? Also, if the input voltage is moved from the negative to positive terminal the starting output will saturate to negative. Why? Is there a rule to follow? If so, what causes this rule? What will happen if R1 is replaced with a capacitor, as in a monostable multivibrator?What will be its starting saturation level that will charge the capacitor. What will be the UTP and LTP then? <Q> The Schmitt trigger works similar to a comparator, so imagine you have a comparator with a Vref in the non inverting input. <S> When the input voltage falls below that threshold, the output of the comparator saturates high. <S> If the input voltage then goes above the threshold, the output goes low. <S> If you were to put the Vref into the non inverting input, the same rule applies, but backwards, so when the input goes below the threshold, the output goes low, and if it then goes above, the output goes high. <S> The general rule is if the non inverting input is less than the inverting input, the output will be high. <S> The Schmitt trigger works in almost the same way, except it has upper and lower limits. <S> As you can see from your diagram, the input voltage (sine wave) starts at 0, which means the inverting input is lower than the non inverting, hence the output starts high. <S> Once it goes above the high threshold, the output goes low. <S> With a Schmitt trigger, the output will stay low untill the lower threshold is met. <S> As with the comparator, if you were to swap the input voltage to the non inverting input, it would be the other way round. <S> I believe this is what you were asking, but please correct me if I am wrong as the English was rather poor, as Olin pointed out. <A> How to determine the saturation level when an opamp schmitt trigger starts workig? <S> With the "theoretical" circuit you have shown and the input at 0 volts, upon power being applied, there is no real way to determine what the output state will be. <S> The output anomaly will resolve itself when the input signal passes one of the two thresholds. <S> Once this has resolved, the output is unambiguous. <A> For the sake of theory, we assume an initial state of output. <S> Here in your circuit for inverting schmitt trigger, it is assumed to be \$+V_{sat}*R_2/(R_1+R_2)\$ at the non-inverting terminal, at t= 0, and the circuit waveforms are analysed. <S> Even if you assume it to be \$-V_{sat}*R_2/(R_1+R_2)\$ at t =0, you will eventually end up in the same waveform, as the outputs are resolved itself. <S> The same happens in real circuits too.
| Many applications will want the output to be unambiguous at power-up (in the absense of a significant input signal) and circuit measures can be taken to force a high or low situation at the output but, on a simple theoretical circuit like the one shown, it is impossible to determine.
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Which NXP ARM processsors have LIN capability? I am looking for potential alternatives for my current STM32F091 uC (they currently can provide a very lousy stock). Searching through datasheets and reference manuals, I can't pinpoint where to look when I want to find NXP uC-s that have LIN protocol capability. DigiKey suggests that only MK22FN 1M0 AV LK 12 has LIN but I couldn't confirm that in the Datasheet . I am more interested in low pin count and/or small footprint uC, 64...80 WLCSP, QFN... So any ARM uC family will do. ARM and KEIL support is more crucial. Thanks a bunch! <Q> I have no idea about what specifications other than NXP and LIN bus that are important for you (and neither has anyone else, seeing as that is not specified in the OP), but what I can say is that it is very easy to get a rough estimate of what processors to look for, and which of them that are in stock at what distributors. <S> Go to octopart.com, choose ICs, filter on NXP, filter on Processors and Controllers and filter on LIN interface, and choose to view only "In Stock" <A> Just grab a transceiver. <S> You're need to do that in most cases any way to comply with the physical layer requirements. <A> Atmel (now Microchip) has ARM parts with LIN. <S> In general I would put their commitment to long-term availability of parts as better than NXP (just designing an NXP part out now, so it does have an upside for some of us). <A> Getting answer from NXP support forum, one should look for Reference Manual and search for "break character transmission / detection " parameter under U(S)ART - which would tell if the uC has LIN capability (not all give results for keyword LIN). <S> So Determine other parameters first <S> (CAN, USB, ... etc) <S> and then check if the uC also has these conditions under UART. <S> I mentioned it was a bit tedious to mention LIN capability like <S> so - I didn't know that "break condition" would determine LIN capability (silly me?). <S> Having "LIN" directly in the manual would vastly ease the search and confirmation process. <S> So that's that.
| Realistically, you can use any micro that has a UART.
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Using a 5vdc solar charger to trickle charge 6 vdc reachargeable 4 AA battery pack I am looking to purchase a remote water tank level monitoring system an Aquatel D110 Wireless Tank Level Monitor ( http://www.rainharvest.com/aquatel-d110-wireless-tank-level-monitor.asp ) - it uses 4 AA batteries.- Can I use a 5v solar charger (power bank) for smart phones. Can I wire direct from the USB output of the 5v solar smart phone charger to the 4 AA battery pack of the water tank level monitoring system (minus the batteries)? such as a https://www.amazon.com/YOUNGFLY-20000mAh-Waterproof-Dust-Proof-Shock-Resistant/dp/B01IH304ZM/ref=sr_1_11?ie=UTF8&qid=1515518609&sr=8-11&keywords=solar+charger+for+cell+phone <Q> The charging source must have a higher voltage than the battery being charged - exact details of the charging procedure depends on the battery chemistry. <A> A NimH batter requires about 1.45V/cell to fully charge at charge current of C/10 or less. <S> For a 2500 mAh cell that's about 250 mA. <S> As I charge rises Vfully-charged also rises BUT you then run the risk of overcharging. <S> 4 cells thus need 4 x 1.45 = 5.8V. <S> With 5V available 4 cells get 5V/4 <S> = <S> 1.25V each BUT 3 cells get 5V/3 = <S> 1.66V each. <S> SO a 5V power bank WILL charge 3 cells but not 4. <S> A fully charged NimH cell at light loads (< to << say C/10) delivers about 1.3V initially dropping rapidly to 1.25V and then settling down to around 1.2V for a good part of the discharge. <S> So, 3 cells gives 3.9V initially dropping to 3.75V and then 3.6V. Whether 3 cells will operate your equipment is "TBD". <S> The D110 operates from 4 AA cells. <S> If it is well designed this means it should work when they are almost depleted = <S> 4V or fully depleted = <S> ~ <S> 3.6V. <S> When new Alkaline cells produce > 1.6V so 4 gives 6.4V SO 5V from a powerbank will operate the D110 IF available current is adequate. <A> As peter has said, you won't be able to charge a 6v battery with your 5v supply. <S> However, as you may be able to run the Aquatel D110 directly from the solar power bank(with no batteries). <S> Its designed for 6v, but 5v may be sufficient. <S> The Aquatel will not be damaged by too low a voltage, just make sure not to hook it up backwards.
| A 5 volt power bank cannot be used to charge a 6 volt battery (or four 1.5 volt cells forming a 6 volt battery).
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Why is the resistance of an ideal voltmeter infinite? Consider an analog voltmeter. I understand that the internal resistance of the voltmeter should be high enough to not lower the actual voltage across the load connected to the voltmeter, but, at the same time, if the resistance was infinite, there will be no current flowing through the voltmeter which means no magnetic field will be produced and the pointer wouldn't deflect. <Q> Your reasoning is correct if you take “voltmeter” to mean galvanometer (a device which deflects a needle using a coil in a magnetic field). <S> But “voltmeter” just means a device which measures voltage, and we can imagine taking that measurement without disturbing the circuit. <S> The whole idea of an ideal instrument of any sort is to ignore the limitations of a physical implementation of the concept; there does not have to be a possible way to do it. <S> That said, there are ways to measure DC voltage with a nearly infinite input impedance. <S> The electrostatic voltmeter or mechanical electrometer . <S> An electrostatic voltmeter has a needle and pivot like a galvanometer, but instead of a coil, has a pair of shaped plates, much like a variable capacitor, which are electrostatically attracted by the voltage across them. <S> The electrostatic voltmeter electrically resembles a capacitor, so its input impedance is infinite at DC (if you ignore leakage across insulators). <S> If you remove it from a circuit, the needle will keep its reading unless you short the meter terminals! <S> In this type of meter the force is very small compared to a magnetic type <S> and so it only works well for voltages in the tens to thousands of volts, and takes a long time to settle. <S> Compared to a galvanometer, it is less practical to be adapted to measure current or have multiple voltage ranges. <S> A null voltmeter or (obsoletely) potentiometer uses the bridge principle to measure voltage. <S> In manual operation, an adjustable reference voltage is connected through a galvanometer to the circuit being measured, then the reference voltage is adjusted until the galvanometer is not deflected in either direction ("nulled"). <S> The same principle can be implemented electronically without manual controls. <S> This type of voltmeter will either draw or supply (!) <S> power from/to the circuit under test, until the nulling is done; the effective input impedance depends on how precise the null is. <S> (One of the original applications of these devices was measuring the voltage generated by thermocouples.) <A> An ideal voltmeter isn't something you can go out and buy. <S> It's a conceptual idea. <S> A moving coil voltmeter can't be ideal since it requires some current to operate the coil. <A> An ideal voltmeter would not draw any current at all. <S> A good voltmeter would draw as little as possible. <S> The less current a voltmeter draws, the less error it generates when drawing current from a real circuit. <S> In the bad old days when I started electronics, a reasonable meter would take 20uA for FSD (full scale deflection), also rendered as '50k/volt', so you could work out the series resistor needed to get any voltage range. <S> These days, amplifiers in ordinary meters take much less current than that <S> , my present £10 DVM will read 200mV with 10Meg input impedance, that's 20nA for FSD, and specialist low input current (electrometer) amplifiers can be used to draw orders of magnitude less than that. <A> Current has to go through the coil of an analog voltmeter to get the needle to move (against the torque of hairspring), however there is no requirement that the current come from the voltage being measured. <S> Devices like vacuum tubes, JFETs, and MOSFETs draw almost no current on their grid or gate and can control usable amounts of current. <S> Most of the current from the voltage being measured in meters that use such devices is typically the input divider, and it is usually in the M\$\Omega\$. <S> Some meters have the ability to switch the divider out, yielding input resistance in the G\$\Omega\$. Special meters designed to have the highest possible input resistance are in the hundreds of T\$\Omega\$. Keithley, for example, claims "reliable measurements down to 10aA" ( <S> that represents about 63 electrons per second). <S> The 10A range on your multimeter is eighteen orders of magnitude greater full-scale than that sensitivity. <S> Roughly the ratio between the size of a hydrogen atom and the circumference of the earth.
| If you had an ideal voltmeter, you could connect it to any circuit and measure the voltage, without having any effect whatsoever on the circuit you're measuring.
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Why bother with even parity? I am using an SPI peripheral in my application. The peripheral returns packets containing 15 data bits, plus an Even Parity bit for error detection. Therefore all zeros, and all ones both pass the parity check. This means that my microcontroller cannot detect the most common type of error: the peripheral being disconnected! In this case, the received bits are all zero, which passes the parity check. Presuming that it would have been just as easy for the manufacturer of the peripheral to implement Odd parity, my question is: Why would they have chosen to use even parity in this case ? Is there some other advantage of Even Parity in this case to make up for the fact that it's unable to catch the most common type of error? <Q> A single parity bit can only check for the presense of single or odd numbers of bits in errors so expecting it to detect when a peripheral is disconnected is probably expecting too much. <S> However, many systems will produce a continuous series of 1's when a peripheral is not present and this can be achieved with a simple pull-up resistor on the returning data line. <S> If there was actual 8 bit data being returned by a connected peripheral then the parity bit would be zero for decimal 255 being transmitted. <S> So even parity can detect when a peripheral is disconnected under these circumstances. <S> If odd parity were used, 8 high bits (decimal 255) would result in a high parity bit so rendering odd parity useless as a means of detecting loss of peripheral chip. <S> Horses for courses. <A> Parity, or any block error detection, is intended to detect errors within a data transmission itself. <S> Parity is not designed to detect whether or not data transmission is taking place. <S> Given a transmission line, there are several different kinds of concerns. <S> The two which are relevant here are: 1) outright failure of the line itself, and, 2) block data errors within a particular transmission. <S> Others less relevant are, for example, incorrect line voltages, protocol errors, or security errors. <S> Parity helps with 2 but not 1. <S> For a subsystem on either end of a transmission line to cope with 1 (outright failure of a connection), another protocol feature is required. <S> The error detection rate of a single parity bit is often higher than 50%. <S> Exactly what that rate is depends on the heuristics of the data segment in the protocol. <S> Say you have a packet, (MSB) 1011010111011110, and there is an single bit error in the last transmitted bit, the parity check would fail and correctly reject it packet. <S> Similarly, if you had a data error in the first bit (the parity bit), the packet would be rejected. <S> Performing this check in hardware is extremely simple and requires no complicated processing. <S> It is useful in applications with relatively low bit error rates to weed out things like clock skew or clock signals generated by processors running garbage-collected software stacks. <S> SPI is a physical link protocol that designed for short electrically connected lines where the single-bit error rate doesn't much depend on the loss of the line. <S> If you're running something across a lossy line, you're going to need something way more robust than parity. <S> This isn't really what SPI does. <S> To check whether a device is still connected, try something higher in the stack. <S> By comparison, TCP/IP (IP, specifically) doesn't specify parity bits while many of the 802.x Ethernet specifications do. <S> IP does, on the other hand, have a complicated, "are you there? <S> " protocol. <S> What are you running on top of SPI? <S> The answer to data link management is probably there. <A> In communication and storage schemes, the parity polarity (odd or even) should be selected to trap the most likely or highest-occurring failure modes. <S> As you say, an unresponsive target or broken data receive wire may well result in a MISO line stuck high or low. <S> When communicating even numbers of bits, such as bytes over SPI, an odd parity bit would detect a fault in this all-1 <S> 's or all-0's data but even parity wouldn't. <S> However, there's no such clear winner when communicating an odd numbers of bits, such as in your application with 15 bits over SPI. <S> Even parity would detect a fault in the all-1's case but miss the all-0's case. <S> Conversely, odd parity would detect a fault in the all-0's case but miss the all-1's case. <A> There is little difference in benefit with even or odd parity. <S> One can be converted to the other with a single invert gate. <S> The main purpose of the parity bit is to check only the 15 bits in that value. <S> It is not its purpose to do any other thing. <S> That one or the other could detect a missing, faulty, or disconnected chip is not a consideration. <S> You mention that being disconnected is the most common type of error in your case. <S> It does not matter. <S> The parity bit is not there to detect that type of error. <A> You are right to question this, I have the same criticism of even parity. <S> With an odd number of data bits before addition of the parity bit, as in your example, and as is common, even parity allows all 0s and all 1s as valid transmitted words, which is useless in detecting a dead link or dead chip. <S> The prior answer by Tony M is wrong in this regard. <S> See the 7bit data example table here for proof:- https://en.wikipedia.org/wiki/Parity_bit <S> Odd parity however would insert an opposite state bit in the all 0s or all 1s case, thus proving that the link and chip are alive, and would be a far better choice in this case.
| There's no obvious benefit of even parity over odd.
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Toggle between two colors in a common-cathode RGB LED using a SPST switch I'm working on a switching circuit for a guitar amplifier that uses one DPDT relay which is powered by 12V DC supply. A SPST switch (panel mount or footswitch) will power the relay and switch to a second amplifier channel. Is it possible to wire a common-cathode RGB LED to glow one color (red) when relay is off and other color (blue) when relay is on? The closest circuit I could find is this Alternating between two LEDS using BJTs but it has common anodes. Is it possible to rework the circuit to use common cathodes? If that's possible, I could bring a positive voltage to R4 when I turn the relay on. <Q> If you switch high side you can use a P-MOSFET as shown below. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As others have mentioned, a simpler solution is simply to mix the colours. <S> simulate this circuit <A> If all you want is to have two different colours depending on whether a SPST switch is on, the simplest possible way is to have one on permanently and one switched, giving you either red or yellow. <S> This answers the question as written. <A> This trick is based on differences in the forward voltages of the diodes, which are mostly unknown so far. <S> If the forward voltage of D3 is less than the combined Vf of D1 and D4, then when the switch closes, D3 will come on and D1 will go off. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> This trick is based on differences in the forward voltages of the diodes, which are mostly unknown so far. <S> If the forward voltage of D3 is less than the combined Vf of D1 and D4, then when the switch closes, D3 will come on and D1 will go off. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Trevor's answer would work to give one on or the other on (without mixing colours), which may be what you meant, however requires more complexity and components.
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Impact of increased pad length in PCB I was wondering why I cannot edit footprint of LED 1206 to take care of LED 0805 as well so that I can mount one of them depending on the availability. To do this, I have to increase the pad size of 1206 footprint on the inner side to give a clearance of around 1.2 mm between two pads. I want to know the impact of increased pad length in the PCB. <Q> If you increase the pad, you will probably increase the amount of paste too. <A> As long as you follow the board design rules that will not cause any problems. <S> You will also slightly enhance the heat dissipation for the 0805 part. <S> During smd assembly the parts self align using the solder surface tension the larger pads will allow the smaller part to move around more. <A> You don't have to worry about not been able to mount it. <S> If you use 1206 footprint you will be able to mount 0805 component. <S> SMD_SIZES <S> In the link you can see different sizes. <S> Try to create new copy. <S> My recommendation however is not to do this as this increses the posibility of creating a connection using only the paste(short circuit) if you are going to solder by hand.
| It will be in it's internal size limit but you will be able to mount it without problems. Too much paste will increase the possibility of the component tombstoning as the solder melts. You need to watch out for misalignment of the 0805 part. Also maybe you can't edit the component as it's protected against writing.
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A basic question about the meaning of dB of an antenna radiation pattern I know that dB is always a power ratio . And to obtain dB there must be a reference power or voltage measured with respect to another. Below is an antenna pattern which shows a -3dB point for instance: What does that mean literally? Can it be explained clearly? -3dB obtained by dividing which units? And what is relative to what? Basically I'm asking the idea of antenna pattern and the dB associations. And how is that obtained. Edit: Edit 2: I marked two points A and B in X, Y, Z coordinates. A is on the plot surface B is not. Red and pink share the same XY point and different Z. Are the powers at A and B same? If not is the power pattern only valid for that plotted surface? How about the rest of the sapce? <Q> See the point at the end of the lobe - it's called the "peak output power point". <S> The point called "-3 dB from peak" tells me that at this point, if there was an RF power density of (say) <S> 100 watts per sq metre at the peak point <S> then there will only be 50 watts per sq metre at the -3 dB point. <S> -3 dB when converted to a power ratio is \$10^{-3/10}\$ = <S> 0.50119 <S> (or basically a half if pin-point accuracy isn't required). <A> In this kind of plots you draw a function r = <S> f(θ), where r is the distance from the origin and θ is the angle measured with respect to a reference axis <S> (in your case it is the horizontal axis pointing right). <S> Note <S> : θ in your diagram is not the θ <S> I'm referring to. <S> Your f is, in this case, the radiated power from the antenna. <S> Therefore the antenna pattern contour show the power transmitted in each direction <S> (direction identified by θ) as points of the curve that are farther from the origin as the power in that direction increases. <S> This kind of plot is useful for quantities that depends on some angle representing a direction in space, for example, since the "lobes" of the contour show clearly in which directions you have a peak of the emission. <S> The main lobe, if it exists in the pattern, is the lobe whose "peak" lies farthest from the origin and hence identifies the direction of maximum emission. <S> As your diagram states, the -3dB points are relative to the peak of the main lobe. <S> This means that the power emitted by the antenna in the directions identified by those two points is half the power emitted in the direction where the main lobe peak lies. <S> To be more precise, a radiation pattern like that is just a 2D section of a 3D surface, known as the radiation solid of the antenna (or 3D radiation pattern) . <S> If the radiation solid is has a cylindrical symmetry around the 2D polar axis, then a single section may be sufficient to give the idea of the whole radiation characteristics of the antenna. <S> Otherwise you could need of other sections, across different planes (see also this link ): <A> Isotropic antenna or radiator is a hypothetical (not physically realizable) concept, used as a useful reference to describe real antennas for their peak gain in one direction. <S> • <S> For example the spherical Sun is an Isotropic radiator • <S> If you can imagine integrating all the energy in a sphere, then every ideal antenna "could always" have the same spherical output power or 0dBi average... <S> • <S> BUT antennae are defined as having a peak gain in one direction only of something greater than 0dBi with a some relative polar pattern. <S> • <S> the ideal dipole has a peak gain of 1.15 dBi since the field loses strength along on the axis yet the total integrated surface area <S> * gain = <S> 0dBi <S> average <S> • <S> If an antenna cannot be perfectly matched at some frequency then subtracted from the diversity gain are the mismatch losses and the result is the net peak gain. <S> The same concept is used for light magnification with a parabolic reflector or lens where the beamwidth and gain are ideal tradeoffs while losses naturally occurring are added.
| The antenna pattern you see is an example of a so-called polar plot , i.e. a plot of a function in a polar coordinate system .
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How to introduce delay to a signal I need to introduce a delay to the output of an astable, it needs to be in the order of 10 micro seconds. I have read that optical cable is good for delays, which makes sense for small delays, but I have no access to optical cable for the project I am working on. The ideal solution would be some components / IC's. I am working with a square wave, so the signal can be assumed to be digital <Q> simulate this circuit – Schematic created using CircuitLab <S> There is a tolerance on Vt+ and Vt- that shifts withtemperature that will make the delays asymmetric. <S> Also if the waveform is not repetitive, it will take 20% longer for the 1st edge. <S> This is my approach if the delay tolerance is adequate. <S> Since the Schmitt trigger thresholds are 1/3 to 2/3 <S> each delay is 2/3 of V+ which is very close to linear approximation of the RC exponential decay. <A> (Similar to @oldfart 's suggestion...) <S> Look at the 74HCT595 (5 V) or 74LV595 (3.3 V) 8-bit shift register. <S> This gives a serial input and serial output with an 8-CLK delay in between. <S> You can select your clock frequency to get the delay you want, where the total delay is 8 / fclk with 1/fclk of jitter. <S> If you want to increase the precision and reduce the jitter, you could cascade several 74x595 in series. <S> For example, using three of these gives you 24 fclk-periods of delay for less than a pound. <A> My first approach would be a set of shift registers plus an oscillator. <S> You can get 8 registers in a package. <S> The input goes in on one side the delayed signal comes out at the other side. <S> The uncertainty is about 1 clock cycle thus for 800KHz that would be 1.25us. <S> (Sample point at the input as your signal is asynchronous to the shift clock). <S> With the latter you also influence your uncertainty. <S> simulate this circuit – <S> Schematic created using CircuitLab Post edit: <S> Sorry <S> : corrected my numbers!
| You can change the delay by adding more registers or change the clock frequency.
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2.5 to 4.5 volts I need to level shift to 0 to 10 volts I have an output of 2.5 to 4.5 volts I need to level shift, and amplify it to give me 0 to 10 volts output. A rail-rail op-amp, using the non-inverting op-amp appears to be the best way to do it. I've noticed a few different ways of doing it. Got very confused with the methods shown. No AC coupling just a DC voltages in, and out. I would like to use a Texas Instruments TLV2370IP , I have any regulated voltage that the device will work on (16v maximum). Any help will be much appreciated. <Q> If you have access to a negative rail of 2.5 volts you can use a 1:1 potential divider between the input and -2.5 volts. <S> The centre point of the voltage divider will now be 0 volts when the input is +2.5 volts <S> and it will be +1 volt when 4.5 volts is applied to the input. <S> If your negative rail is say <S> -5 volts (or something else) follow <S> this link to a calculator that can help: <S> - The picture above shows an input at +2.5 volts and the output at 0 volts. <S> Gain is always going to be 5 because that is the output range divided by the input range i.e. 10/(4.5 - 2.5). <S> You can double check that the output rises to 10 volts when the input is set at +4.5 volts (with no change in R2, R3 or R4). <A> Ok, so we need to multiply by 5 and subtract 12.5 <S> An amplifier with a non-inverting gain of 5 has an inverting gain of -4. <S> So we need to bias the feedback divider at 3.125V. BTW <S> be-aware that "rail to rail" op-amps often have considerablly worse performance as they get close to the rail. <S> This may or may not be a problem for you. <A> You can do it like this: <S> The +5 Vref directly affects the output voltage- <S> a 1mV change in the referencerepresents a 2.5mV change in the output voltage. <S> You can use a series reference chip or a shunt reference and resistors. <S> Rail-to-rail i/ <S> o is unnecessary. <S> You only need an input common mode range that extends to +4.5V with whatever supply you are using, single supply or rail-to-rail output , and an output range that goes to +10 with the supply you are using. <S> Relaxing that constraint allows you to use cheaper and/or more robust op-amps. <S> Note that the output has to sink 2.5V/R1 with 2.5V in. <S> Regardless of op-amp type, it won't get all the way down to 0V without a negative supply, but it may get close enough for your purposes. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If you want to use a different reference voltage or change other parameters, just scratch down the equations and solve them. <S> With the case of a 5V reference, we can simply write down by inspection: <S> R3 = <S> R1||R2 <S> (zero balance with 2.5V in) <S> and R1 = 4*(R3||R2) <S> (gain of +5) <S> The remaining degree of freedom is scaling so we pick something reasonable.
| So follow this with a gain of ten op-amp and you have a solution.
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How well does telephone delay correlate to geographic distance? I have a friend who lives on the other side of the world, and every time we speak over the phone, there's roughly a second of delay between one of us saying something and the other hearing it. I've tested this by having her put her phone on speaker so my own voice is detected and returned back to me over her phone as feedback, and the delay over that particular distance is quite consistent. While surely there are a number of other factors contributing to phone delay beyond propagation time, many of these factors are constant regardless of distance. So if the distance between us were halved, would the delay be as well? <Q> It used to be that the telephone delay was due to the signal being sent through a geosynchronous satellite. <S> These are a long way up and the signal took that long to get there and back. <S> Today delays through routing devices are more likely to be the cause of the pause. <S> Half way round the earth is about 20 million meters with light or signal travelling at 300 million m/s direct transmission delay would only be about .066 seconds. <S> Of course your wiring distance may be a little different, <A> 40,075 km is the circumference of the earth. <S> ( 4e7 meters ) <S> Assuming 2 second delay around the earth's surface in the shortest path implies a minimum velocity of 2e7 <S> [m/s] or about 10% of c thru most controlled impedance submarine and above ground cables (2e8 m/s), or the equivalent distance traveled is actually 10x more and may include Satellite Hops and TCPIP router zigzags. <S> Considering the lowest cost now is VOIP via TCPIP, I suspect your phone service provider is routed in a spider web of global routers. <A> In the same way there are delays with an internet connection, satellite transmissions, etc, the current technology has these limitations. <S> The data being sent over the phone is processed and routed through several different devices, from several different operators, which prevents this communication from being instantaneous. <S> The devices I'm referring to are not necessarily on Earth, they can be satellites also. <A> I have worked in a (fixed) telephone company, and electrical signals are almost instantly. <S> However, signals are routed through devices which each takes some processing time. <S> Like signals (from different calls) are multiplexed, demultiplexed, error corrected, enhanced etc. <S> Also it is likely one signal is handled by different hubs/points where signals are received/transmitted each adding their processing time. <A> Nowadays, almost all telephone calls travel through the internet, so geographical distance does not play such an important role in signal propagation time. <S> In this case, the propagation time of the information is mainly related to the amount of Hops in the network between source and destination and the bandwidth of each one of them. <S> Wikipedia: <S> Almost all In computer networking, a hop is one portion of the path between source and destination. <S> Data packets pass through bridges, routers and gateways as they travel between source and destination. <S> Each time packets are passed to the next network device, a hop occurs. <S> The hop count refers to the number of intermediate devices through which data must pass between source and destination. <S> Since store and forward and other latencies are incurred through each hop, a large number of hops between source and destination implies lower real-time performance.
| The bigger the distance, the more "nodes" it passes through, so the higher the delay.
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Easy way to get 12v DC from 32v DC bus? I've got a 32 volt DC regulated bus. I also need about 30mA at 12 volts (DC). Is there an easy way to tap some power from the 32v bus? My first thought was just put in a 12v linear regulator. But those in my parts box have absolute ratings close to (or below) 32v, and I'm not one to push the envelope. Some divider make up using hairy arsed power resistors might work, but that's not too efficient and generates HEAT. If the solution gets complicated (e.g. build a buck down converter), I could find a 12v wall wart and tie across the mains supply to my 32v power supply. I know about voltage multiplier circuits. Is there a neat voltage divider trick somebody can teach a nubie? <Q> this is typically handled with a "switching regulator". <S> A "buck converter" is a type of switching regulator designed to regulate down from one voltage to another. <S> There is any number of prebuilt modules you can find that accept 32V and regulates to 12V. Cheap ones start at a dollar and a month shipping from China, but quality costs more. <A> Other answers cover that. <S> I want to suggest a simpler brute-force method that you may have overlooked from the sound of your question: <S> The good old LM317 . <S> This is a very common linear regulator, but the "trick" compared to its fixed-output 7812 relative is that it's floating . <S> It has a maximum voltage difference of about 40 volts. <S> Your input-to-output difference is 20 volts. <S> It will dissipate 20 <S> * 0.3 = 6 watts of heat, so it requires a reasonable heat sink, but since you have not given any more details about the project, this may not be a problem. <S> If it is, you need a buck converter. <A> Have a look at these buck converters from TI, they're a tested, complete buck converter on a module which meet your input and output voltage and current specifications, with a lot more efficiency than a linear regulator. <S> They're quite expensive, and you could design and implement your own buck converter for less, but they do provide a lot of convenience. <S> There are other parts which cost less, and are pin compatible with a TO220 linear regulator such as a 7812, but they have more restrictive maximum ratings, so you would need something else as well to lower the input voltage a little. <S> An example of those is this module from TI. <S> Other manufacturers are available and found by searching for buck converter module. <A> The Art of Electronics , 3rd ed (Horowitz & Hill) <S> 9.13.2 suggests using a depletion mode FET follower to hold the LDO's Vin a couple of volts above the regulator's required input voltage. <S> Vgs is about -1.5v. <S> Higher values can be obtained by connecting the gate to a resistor divider between the regulator output and FET source terminal. <S> Here's a portion of the circuit: <S> simulate this circuit – <S> Schematic created using CircuitLab Resistors and caps required by the LDO are not shown. <S> I assume, but have not bench checked, the strategy would extend to other regulators.
| The linear approach A buck converter is the most efficient way to go, and should be easy to find.
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Low pass filter and diode question 1.In this circuit the R3, C1 and R6, C2 are said to act as low pass filters. The R3, C1 combo is easy to understand, however I don't get the R6, C2 configuration as they are parallel to each other, how can they act as a low pass filter? Also whats the function of D1? 2.The exit delay is achieved via C3,R7. The time constant RC determines the delay period, during which the gate input is held low. Does D2 act as some sort of protection? <Q> R3 and C1 are a low pass filter and are present to probably debounce the N/O contacts at the input. <S> D1 and C2 is a kind of controlled latch - when IC1A goes high, C2 will rapidly charge to a logic 1 level and, if IC1A's output goes low, D1 prevents C2 from being disharged i.e. C2's terminal voltage remains high and is slowly discharged via R6. <S> In other words C2/R6 is NOT a low pass filter. <S> C3 is charged slowly via R7 (470 kohm) when S3-5 are open. <S> If all of S3-5 are closed then D2 and R2 act as a reasonably rapid discharge path for C3. <A> R6,C2 tries to ensure some minimum alarm activation period length in case someone succeeds to reconnect the sensor chain ultrafast. <S> C2 gets charged fast, but discharging is through R6. <S> D1,C2,R6 is a pulse stretching circuit. <S> D2 lets C3 to get empty in case S7 is turned OFF. <S> This ensures the exit period starts from the beginning. <S> This circuit applies several RC timing circuits where a CMOS logic gate is used as a voltage level detector. <S> This can cause problems because around Vcc/2 input voltages the gate can behave unexpectedly (oscillations, high current consumption, even overheating if Vcc is as high as 12V) <A> If you imagine the highest frequency you can have, all the capacitors will become a short-circuit. <S> So the signal will go straight to the ground. <S> For the lowest frequency you can have (it's just DC current) <S> the capacitor become an open circuit. <S> So, the signal will go to the resistor. <S> The diode is just a protection. <S> The IC doesn't want to receive any current. <S> The capacitor will get charged. <S> You don't want the capacirot do discharge in the IC but in the rest of the circuit. <S> A second tought would be appreciate here.
| The diode D2 seems to be a protection to be sure the voltage between C3 and R7 doesn't get higher than the power source.
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Negative voltage for Vce - BJT NPN I was wondering what happens when VCE voltage becomes negative? suppose I am forwarding both junctions (no resistors). I know that saturation happens when VCE is between 0.1V- 0.3V but what happens below that range (V1 > V2) as in the following scheme? simulate this circuit – Schematic created using CircuitLab <Q> The definition of saturation is that both junctions be in forward bias. <S> So, with appropriate voltages for V1 and V2, your transistor would be operating in the saturation region. <S> It would be important for the voltages to be not too large, otherwise very large currents would flow and destroy the device. <S> The circuit is difficult and confusing because of the way it is drawn. <S> For example, you chose to label the common positive node of the two supplies as GND. <S> It may have been better to leave GND out of it. <S> Depending on the specific voltages of V1 and V2, net current flow might be from emitter to collector or collector to emitter. <S> Because of the differences in the two junctions, detailed predictions about current and voltage are somewhat difficult. <A> If the voltages are high enough (and that doesn't have to be much above 0.6V) you will just destroy the transistor. <S> BTW: <S> the title of the question "Negative voltage for Vce - BJT NPN" doesn't fit to the schematic: Vce would only be negative if V1 <S> > V2. <A> simulate this circuit – Schematic created using CircuitLab Redrawn with proper conventions for orientation in a logic diagram. <S> In your question, Both BE and BC are forward conducting diodes, not the same, with different Vr breakdown characteristics when reverse biased due to doping differences. <S> ... <S> meaning yours is not logical. <S> below correct application as a switch shows both junctions with forward voltage <S> but Vbc has a forward drop <S> is due to transistor function with collector current. <S> We consider both junctions saturated and thus Vce=Vce(sat) saturated. <S> simulate this circuit <S> A transistor does not behave like 2 diodes yet the forward voltage differences indicate Vce{sat} at rated current and often rated at Ic/Ib=10. <A> When we apply a positive voltage between collector and emitter a tiny depletion region forms between the np junction of collector and base and the pn junction between base and emittor is biased directly and electrons are driven from emitter to the collector. <S> I think that in case of applying the reverse voltage the depletion region between collector and base disappears allowing the electrons of n side of collector to move down toward emittor, on the other hand the electrons of the heavily dopped n , in the emittor, is flowing up side to the collector! <S> Therefore two currents want to flow inversely. <S> I think the sum would be a really small current which is not considerable. <S> So, I think no current will flow!
| Assuming that V1, V2 > 0: Both internal diodes (from B to E and from B to C) are forward biased and will act just like discrete diodes.
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Relation between transformer core material and application frequency Regarding the following excerpt from a text: The text mentions that at RF air core transformers are adequate, but for 50/60Hz low freq. applications iron core needed to obtain a low reluctance. The text does not explain the reason. What is the relation between the frequency being low and the need for low reluctance medium? Low reluctance intensifies the magnetic field but why particularly at low frequencies it is needed? <Q> It is difficult to explain in an intuitive way. <S> But I like to think of it in terms of primary inductance. <S> When you apply a voltage to the primary, even if the secondary is open-circuit, a certain amount of current flows in the primary, because of the primary inductance. <S> In other words, when the secondary is open circuit, the primary just looks like an inductor. <S> You want that current to be small, which means that the inductance needs to be large. <S> But when a low reluctance core is used, the primary inductance is much higher, and the transformer size is more manageable. <S> The cross section area of the core and the number of turns are usually fine tuned so that the magnetic field strength in the core is within a range that is suitable for the transformer core material. <A> At high frequencies, hysterisis and eddy current losses come into play if a ferromagnetic material like iron is used a core. <S> Hence Air core is preferred at high frequencies, despite its higher reluctance. <S> But at low frequencies, hysterisis and eddy current losses are nullified and hence Iron core is a better choice. <S> Way better flux-linkage and better efficiency than air core ones. <A> The volts per turn of the transformer are the product of core flux, and operating frequency. <S> If you have a very high frequency, you don't need much flux, so can get away with an air core. <S> This is just as well, as a material core is expensive and lossy at RF. <S> If you have a very low frequency, you need the higher B field from the high permeability core. <S> Fortunately, the losses of very high permeability materials are low at low frequency, so they're OK to use. <A> Inductor impedance is Z(f)=ωL where L is a product of the core permeability μ. <S> In order for power transformers to be efficient at low frequencies, the no load impedance must be > 10x rated load <S> Z otherwise the excitation losses at no load would be excessive. <S> As frequency rises the trend to lower relative permeability towards air=1 is due to the non-linear eddy current losses. <S> Thus the result is a limited useful range \$µ_r/ω\$ for any give material. <S> You may look up the various relative mu <S> values from 1 for air to 600 for ferrite to 4000 for transformer steel to 100k for special mu-metal foil used to raise the impedance and bandwidth of submarine cable surrounded by high k=80 water.
| As it turns out, for 50 or 60 Hz, it is not really practical to do an air core, because the inductance will be too low for practical core sizes.
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brushless for outboard of a 4 meter boat SK3 - 6364-190kv Brushless Outrunner Motor ... this motor has a nominal voltage of 37 V with a maximum current of 65 A, Can I supply this motor with a voltage of 24 V, with a current of 30 A? how much couple do I lose? thank you all for the clarifications. In your opinion, to move a 4-meter boat at a speed of 5 nodi, is a torque greater than 10nm needed? <Q> Torque (couple?) is proportional to current. <S> With only 30 amps available, the motor should not be loaded with more than about 45% of rated torque. <S> Limiting the speed will probably reduce the torque in proportion to the square of speed 0.65 squared is 0.42 or 42%. <S> That should be enough for the motor to handle with the power supply limit. <S> That is just a guess based on the limited details given. <S> If you can find data on the motor, controller and propellor, a better estimate could be made. <A> Specs. <S> Turns: <S> 18T Voltage: <S> 10S Lipoly (37 Vmax)RPM/V: 190KV Internal resistance: 0.028 <S> Ohm Max Loading: <S> 65A <S> ( surge = 37V/0.028Ω = <S> 1321A <S> start , ideal voltage source) <S> Max Power: 2450W = ? <S> 37 <S> *65A=2405W <S> hmm Shaft Dia: 8.0mm Bolt holes: 32mm Bolt thread: <S> M4 Weight: 697g <S> Motor Plug: 4mm Bullet Connector <S> What this means is if your battery is limited to 30A peak current then start torque <S> will be 30/2405 = 2% of spec. <S> But if it is average current then with 24V it is 720W/2405= 30% of rated power and only 24V/37V=65% or rated RPM. <S> p.s. <S> I hope it has water cooling. <S> However @ 30A <S> it will run cooler with Pd=30^2 <S> *0.028Ω=25 <S> Watts <A> With 24V you can still supply 65A with duty cycle below 35%. <S> So for low speed no problem. <S> Because of the voltage your speed will also be limited.
| For higher speed you will loose just above 50% of torque - it is proportional to current. Limiting the voltage to 24 V (65% of rated) will probably limit the speed to 65% of rated speed.
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Use potting compound to secure solder joint I would like to solder a wire to these 2 pins. I was wondering whether I should protect the solder joints, as the cables may shift over time (and tension). Should I put glue on the around it to make sure it won't move (and of course keep the solder)? <Q> I was wondering whether I should protect the solder joints, as the cables may shift over time (and tension). <S> This is a PCB-mount switch. <S> Its contacts aren't intended for soldering discrete wires directly to them. <S> If you actually wan to achieve long term robustness, you should make a small PCB with: a footprint for the switch, a wire-to-board connector of your choice, and maybe mounting holes too. <S> Such is the proper solution to your technical issue. <A> From the pictures, those look like quite small pins. <S> For something like that, I would be inclined to use stranded wire, not solid core, to reduce the chance of snapping. <S> Then I would slip a length of an appropriate diameter heat shrink tube over each joint, and heat it with the side of a soldering iron. <S> That provides a bit of strain relief, as well as insulation. <A> In comments there are plenty of useful suggestions. <S> If for some reason you cannot add a pcb and you must solder wires, use thin multi-strand wire and do not allow the solder to rise inside the insulation. <S> Now you have long stiff levers which sooner or later break the joint if any vibrations exist. <S> Hold the wires in pliers when you solder to prevent the solder to creep under the insulation. <S> Twist the wire around the contact pin before soldering or insert a ferrule around the pin and wire. <S> You must have some strain relief clamp, only few, say 3 or less, centimeters of wire can be freely. <S> Even those few centimeters should be binded together to reduce the freedom of vibrations. <A> With solid wire being very brittle and prone to breaking after a few extreme bends, I agree a strain relief is necessary if this is intended to function as mobile switch at the end of a wire. <S> Depending on your needs for cosmetic appearance, I would suggest something that is firm but not as hard as epoxy. <S> White Polyurethane is used in power supplies to bond heavy parts to the board but this is a special item. <S> My favorite polyurethane adhesive is something like this. <S> Allow a few days to cure. <S> The former volatile sub-floor adhesives were much faster drying but now obsolete.
| Hint: if you're using multicore wire, or if the other ends of the wires aren't accessible, slip the heat shrink up the ends of the wires before you solder them!
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Is audio signal passing through headphone differential? I'm confused about how small audio signals are immune to interference.Headphone wires seems twisted but are they also differential signals? As far as I know differential/balanced signals are carried by two wires and they are mirrored as shown here: https://upload.wikimedia.org/wikipedia/commons/thumb/e/e7/DiffSignaling.png/1200px-DiffSignaling.png So the common mode noise is rejected. But in headphone case the little headphone which goes to our ear is not a differential amplifier hence it is just an impedance. It DOES NOT subtract two incoming signals. I accept my ignorance on the issue but couldn't find a clear explanation about this. Lets say you have a smartphone or laptop and you plug your headphone to the device, what kind of signal is that audio signal passes through the vibrating earphones? And how come the wires reject the noise? Can this be shown with a simplistic circuit model? EDIT: I would be extremely glad if someone can reply these two point confuses me: Premise: In below diagrams grounds are meant as analog signal grounds there is no earth involved. Imagine the circuits are in space to neglect confusion. Sorry for the above quick hand drawn diagrams. Question 1-) I. represents the original question. As you see the phone's headphone driver circuit has a GND which is connected to the one terminal of the headphone coil. So some say "no way to tell which, if any, of the lines is grounded". But there is ground. All phone circuit's analog ground is tied to the one terminal of the headphone. (?) Question 2-) This is regarding diagram II above. in A and B why B cannot reject the common mode noise? Isn't B similar to the headphone scenario in I ? <Q> A speaker is a two wire transceiver, as such it is differential by nature. <S> The speaker reacts to the difference between the lines, it does not matter if one line is ground or an anti-phase signal, the speaker can not tell the difference. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Looking at the image above you can see that the signal current that is driving the speaker forms a loop shown by the green arrows. <S> These arrows flip back and forward in unison as the applied voltage changes polarity and excites the speaker. <S> Common mode noise on the other hand forms currents that go in the same direction in both lines, as such they form no voltage difference across the speaker terminals relative to each other. <S> As such any AC hum that may be picked up on the cable does not actuate the speaker. <S> RE <S> YOUR EDIT <S> So some say "no way to tell which, if any, of the lines is grounded". <S> But there is ground. <S> Yes <S> but the difference is the ground on your headphone, or phone handset, comes from the same place as the audio. <S> If you had an extra wire coming from one terminal of your headphones tied to a local ground it would be different, your would hear all kinds on noise. <S> in A and B <S> why B cannot reject the common mode noise? <S> Isn't B similar to the headphone scenario in I ? <S> In this case B has local ground tied to one side of the amplifier. <S> That means the common mode current coming in each line sees a different impedance. <S> As such a different voltage is generated on each pin and noise is introduced and amplified. <S> Case A each signal sees the same impedance, generates identical voltages, and those cancel out. <A> Headphones aren't particularly susceptible to induced noise, when compared with audio amplifier inputs. <S> An audio amplifier may have an input impedance of many kilohms. <S> Headphones are typically around 32 ohms. <S> If a nearby signal induces a small imbalanced current into the signal cable, then that can produce a significant voltage across the input of a high impedance amplifier (V = IR). <S> Whereas to a 32 ohm headphone speaker, the current is too trivial to be heard. <A> However some signal pairs have differential drivers such as car audio power amplifiers and RS485. <S> But don't confuse that with the fact that what is received is a differential voltage, even if the return reference is 0v or ground. <S> However, differences in earth grounds can cause noise when using earth ground as a common reference. <S> Many audio sources use a floating ground unlike PC towers earth-grounded for EMI reasons. <S> When an additional voltage or current is coupled to both wires equally by say mutual inductance or capacitance then a Common Mode voltage is applied according to the CM impedance* CM current. <S> This becomes a Differential voltage if the impedance of each line is significantly different. <S> For a speaker the driver is <<1 Ohm just like the ground return, so the difference is not significant. <S> But for a 600 ohm microphone with a ground return the difference is significant so these are used in balanced mode. <S> For a long digital signal, the line inductance is significant so a balanced differential low impedance RS-485 works better and <S> over longer distances with a balanced receiver. <S> A coaxial cable is unbalanced but the inner wire is shielded so that any ingress voltage is common mode, so it behaves like a balanced signal. <S> However poor grounding along the distributed cable can cause stray CM signals such as from a nearby train to get into CATV coax. <S> Whenever the stray magnetic or electric field interferes with the unbalanced impedance of a pair of wires to create a "significant" differential signal , then a BALUN choke must be added or high quality STP with balanced driver and receiver must be used.
| Any time a voltage is applied thru a pair of wires to a load, it is differential but not necessarily balanced.
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Can I Build My Own USB Hub? Would it be possible to make a USB Hub from scratch? Like, designing a PCB, and hand soldering components on to it? What microchip would I have to use as the controller? Is there a simple controller chip I could use for, say, a 4 port hub? I know I'm getting a bit hopeful here, but are there USB hub ICs in the DIP package? <Q> You do need to pay careful attention to the layout of the traces associated with the crystal and the USB data pairs (impedance-controlled). <S> And some chips have a thermal pad that's difficult to solder without hot-air tools. <A> With some usb hub ICs it's as simple as the ic, a crystal, and a few passive components. <S> Easy to design if you feel like taking the time. <S> A single sided PCB is possible too. <S> Here's a simple (two-sided) example <S> PCB: <A> I have done this using TUSB4041I but had to use a professional assembly house to populate the board, it is incredibly fiddly.
| I use chips from Cypress, but they're certainly not the only game in town. Sure, I do it all the time — custom hubs for embedded applications.
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How to Attach a Nichrome Wire to a Regular Copper Wire? I am trying to make a room heater. It comprises of a few coils of nichrome wire connected in parallel with each other and in series with a regulator to control the voltage. I am also having a step-down transformer connected in parallel with this setup powering a small dc-motor to spin a rotor and push the air out. I was wondering how I could connect the nichrome wire coils to my regular copper wires which I am using to connect them to rest of the circuit without the copper melting at the junction of the two wires where the hot nichrome will be of a temperature higher than the melting point of copper. How do I join them without this happening? I haven't tried this but I think this is what is going to happen if I do. Any ideas anyone? <Q> Attach both wires to opposite ends of larger strip of metal using screws or welding. <S> The attachment strip can be stainless steel or something that will not corrode or otherwise be damaged by the higher temperature at the nichrome attachment point. <S> The attachment strip needs to be large enough so that it does not get too hot from the current and so that temperature at the copper attachment point is acceptable for the copper wire. <S> The copper wire may need a high temperature insulation near the attachment point. <A> That kind of depends how hot your plan on driving the ni-chrome. <S> If you are well under 1,085C, the melting point of copper, a good crimp will do with appropriate insulation on the copper. <S> If you plan in going closer to the ni-chrome max, 1,400C, you will need an appropriately sized intermediate block of material like stainless steel. <S> The block has to be sized so the TD from the ni-chrome side to the copper side is enough to bring the temperature well below 1,000C. <S> Here is a general temperature colour chart. <S> If your wire is at the yellow/white end, you need some intermediate material, if it's down at the dark red end, a direct crimp may be sufficient. <A> Choose a sleeve that slides over solid copper wire, which in this case is ~1mm OD. <S> Rip the plastic off the sleeves, slip the plain ends over the copper about ½ way, solder in place and also fill the belled end of the sleeves with solder. <S> An advantage is that ordinary soft (electrical) solder can be used and no special flux required. <S> In my case ~80mm of the copper wire is exposed to ambient air. <S> Nichrome coil <S> ~32 <S> awg and 3mm OD is attached by straightening a section ~10mm long at each end, reheating the soldered sleeves and dipping the nichrome into molten solder. <S> Strictly, it is not the same as a soldered joint because the nichrome will not be wetted but it is electrically sound (conducts well) and mechanically robust, since the comparatively heavy gauge copper sucks the heat away from the connection far faster than the nichrome can put it back in. <S> Of course if the joint is somehow allowed to get hot the solder will melt and quickly fall out; then the show is over. <S> One can also crimp the sleeves to the copper with nichrome coil in between (ie concentrically) but it is not as elegant and this also introduces a lot more heat to the copper wire and sleeves. <S> Plain old resin core solder works well <S> , just keep it cool. ! <S> Sleeves with plastic funnels still attached
| A simple method that seems reliable running a nichrome coil up to ~900°C is to use copper wire as a heat sink and attach to the nichrome with cable sleeves - those thin wall tinned copper tubes with a plastic "funnel" and belled entry at one end, normally used to make decent quality crimped terminations with finely stranded wire.
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ESP32: how to keep a pin high during deep sleep (RTC GPIO pull-ups are too weak)? I'm using an ESP32 in a battery-powered device. The ESP spends most of the time in deep-sleep to save power. The device also has some 5V-powered sensors which I'm feeding from a MCP1640 step-up converter, which is on during the brief active part, and powered off via its "ENABLE" pin during deep sleep. The relevant part of the schematic is: simulate this circuit – Schematic created using CircuitLab So, I need to hold pin #4 high during sleep.Due to sloppy testing of this part, I didn't know that the ESP32 shuts down its GPIOs during deep sleep , so pin #4 isn't kept high, and the sensors remain active. This drains the battery at a much faster-than-expected rate. I'm wondering whether it's possible to workaround this blunder by a software patch (of course, it's easy to just add an external pull-up resistor to pin 4 - but I have a few devices on the field, which I would hate to have to travel a few hundred kilometres just to solder a resistor to! And the people around aren't tech-savvy to do this themselves; on the contrary, remote software patching is easy and well-tested). For example, I tried the RTC's pull-up resistors: gpio_num_t pin = (gpio_num_t) PIN_DISABLE_5V;rtc_gpio_set_direction(pin, RTC_GPIO_MODE_INPUT_OUTUT);rtc_gpio_pulldown_dis(pin);rtc_gpio_pullup_en(pin); // set the pin as pull-upesp_sleep_pd_config(ESP_PD_DOMAIN_RTC_PERIPH, ESP_PD_OPTION_ON); // keep the RTC IO domain powered Executing this just before entering deep sleep almost worked, but it turns out the weak pull-up is too weak : probably ~38k if it is a resistance, or 90µA if it's a current source. This is insufficient to drive the ULN2003's pin. I also tried RTC_GPIO_MODE_OUTPUT_ONLY + rtc_gpio_set_level() too, but this didn't seem to increase the drive capability. Another possibility is light-sleep, which I'd avoid as it was buggy some time ago and I don't want to get my devices bricked. So my question really is: have I exhausted all software options (since the hardware ones also require burning a tankful of gasoline)? <Q> It's not clear that making the resistor change you propose will be effective at solving your problem. <S> The MCP160 requires that the enable input is taken below 20% of Vin to effectively switch the device off. <S> If Vin is (say) <S> 5 volts then 20% is 1.0 volts and, because the ULN2003 is a darlington, it may not reliably switch this low. <S> However assuming it does switch lower than 1 volt then there is still the possibility that my guess about the value of Vbatt of 5 volts doesn't cover the low end of the range. <S> For instance if Vbatt is expected to work down to say 2 volts then you can only switch the MCP160 off if the enable pin is taken to less than 0.4 volts. <S> This sounds all to close for my liking and I would recommend you think about the problem a bit more. <A> Well, sorry to revert my acceptance to the answer by Harry Svensson! <S> And I shouldn't really be adding to a question that is 14 months old by now. <S> However I just found the perfect software solution, which would have saved the tankful of gasoline I eventually burned (sorry, nature!). <S> Instead of setting the pin to be pulled-up during deep sleep, you can use the pin hold feature to keep the pin to whatever state it was before sleep: <S> #include "soc/rtc_cntl_reg.h"#include "soc/rtc.h"#include "driver/rtc_io.h"... <S> digitalWrite(pin, HIGH);gpio_hold_en(pin);... <S> That's all you need! <S> The pin will be kept as a CMOS output, and strongly driven to HIGH or LOW, whatever you need. <S> I verified it with the following schematic: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> While running, the current measured is approx 110µA (as expected). <S> During deep sleep, with the old code (in the question, that sets up a weak pull-up), the current falls to ~50µA <S> (so the pull-up resistor value is in the 30k range). <S> With the suggested code here, it stays 110µA. <A> Here's two other links of people who have tried. <S> fail , probably fail . <S> So it doesn't look promising with solving it in software, as you have tried. <S> So right now there's 3 failures, you and the 2 links. <S> Proposed solutions: <S> A) Burn a tankful of gasoline <S> Pro : Every device you fix you'll know be functional correctly. <S> Con : <S> A lot of your time and money will be wasted on gas and driving. <S> B) <S> Fix 1 device nearby and make a tiny tutorial for those who got your devices, pay them to solder a pull up resistor, e.g. follow the tutorial. <S> Yes, pay them because they will be doing some work, work that you failed to do during the testing stage of your product. <S> Pro : This will be cheaper, and whoever that has the devices might think less of you professionally, but higher of you in terms of trust. <S> Con : <S> Some people... just fail with simple things, including soldering. <S> So there's a chance that someone will just ruin their device and have to buy another one off you, which you will give away for free, which might be a huge loss. <S> Pro : <S> Whoever that has the devices will feel that you're professional. <S> Con : <S> Whatever random person that solves your problem now knows how to properly mess with your instrument. <S> D) Do the same thing as C , but use friends (who might live nearby) or some colleagues instead of some randoms. <S> Pro : <S> Whoever that has the devices will feel that you're professional. <S> Con : <S> Your friends will think you're unprofessional. <S> E) <S> Do what the user DoxyLover proposes, " <S> Fix some units locally and ship them to your users, along with pre-paid return shipping labels so they can ship the originals back to you. " <S> If I were you <S> I'd go backwards, start with E . <S> If that is not an option then continue with D , if you can't then go with C , if you can't then go with B and lastly go with A . <A> The problem that everyone is missing is that the input current needed to keep the the ULN2003 active sort of defeats any attempt at saving power by putting the microcontroller into deep sleep. <S> The proper solution is to replace that section of ULN2003 with something that doesn't consume any current such as a MOSFET. <S> However, I'm wondering why the ULN2003 is in the circuit in that location in the first place. <S> Why can't the microcontroller drive the EN pin on the DC-DC converter directly? <A> It seems that the lib has evolved from @anrieff's answer at Mar.25. <S> I have implemented this function rather using the RTC APIs of ESP32.Of course this functionality is still intended for low power usages, so I'm using a MOSFET & ESP32 to control another power source. <S> The official examples are quite useful: https://github.com/espressif/esp-iot-solution/tree/master/examples Code & comments: #include <driver/rtc_io.h>#define <S> SWITCH_PIN <S> GPIO_NUM_4#define <S> uS_TO_S_FACTOR 1000000ULL <S> /* Conversion factor for micro seconds to seconds <S> */#define <S> TIME_TO_SLEEP <S> 5 <S> / <S> * Time ESP32 will go to sleep (in seconds) <S> */RTC_DATA_ATTR <S> int pinState = <S> 0;void setup() { Serial.begin(115200); <S> pinState = (pinState+1)%2; //flipping between on & off rtc_gpio_init(SWITCH_PIN); //initialize <S> the RTC GPIO port <S> rtc_gpio_set_direction(SWITCH_PIN, <S> RTC_GPIO_MODE_OUTPUT_ONLY); //set the port to output only mode rtc_gpio_hold_dis(SWITCH_PIN) <S> ; //disable hold before setting the level rtc_gpio_set_level(SWITCH_PIN, pinState); //set high/low esp_sleep_enable_timer_wakeup(TIME_TO_SLEEP * uS_TO_S_FACTOR); //gpio_deep_sleep_hold_en <S> (); rtc_gpio_hold_en(SWITCH_PIN); // enable hold for the RTC GPIO port <S> Serial.println("Going to sleep now"); Serial.flush(); <S> esp_deep_sleep_start(); //sleep}
| C) Do the same thing as B , but instead of telling whoever that got your devices to do the job, hire some random people who live nearby and can do it for you. The most that you may have to add is a high-value pull-down resistor from EN to ground.
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Powering three separate shaded pole axial fan motors with one 3-phase VFD I want to power and control the RPM of three 115v .47amp shaded pole axial fans each powered by a separate leg of the 3-phase coming from a VFD. Can it be done? I'd like them to run at full speed (60hz) down to possibly half of that or less if that's an option. <Q> Most VFDs have a mode of control that is not senseless vector. <S> That mode is often called the "V/Hz" mode. <S> If you select that mode, it should be possible to power three shaded pole motors driving fan loads. <S> Some VFDs may have a "load loss" or "load imbalance" protection feature, but I don't think that either of those features is common. <S> VFDs usually have sufficient configuration adjustment ranges to allow setting up for 115V output at 60 Hz. <A> Given any reasonable VFD has a current regulation, no. <S> Reactive currents in the individual legs will differ depending on speed. <S> Active currents in the individual legs will differ depending on the load situation. <S> This will make the VFD go into failure mode, expecting the single motor it was designed for to be broken. <A> I have been working on a related question, so although I do not know for sure (please check on the bench before deploying) <S> Let's look at some factors that can push the answer in one direction or another. <S> Are your motors and loads identical? <S> the more unbalanced your current loads on the vfd are unbalanced the more likely the control loops in the vfd will be confused. <S> How coupled are your loads? <S> The good news is that unless all your fans share a shaft, your load can tolerate unbalance (synchronization of air flow is probably not necessary). <S> The bad news is that depending on layout the fan layout air from one fan could affect load on another (either reducing or increasing), and other airflows may affect different fans disproportionately. <S> this may significant affect the vfd current enough to cause problems. <S> Where is your neutral? <S> I have never seen a vfd with a four wire output (not counting ground), which means that either you need to use line neutral (which probably will start a fire), find a neutral inside the vfd (which means you need to modify the vfd, which I do not recommend), or you need to wire the fans in either a delta (which does not need a neutral) or a wye (which can float it's neutral). <S> remember that the output voltage of your vfd is a phase to phase voltage so if you are wiring a delta you use the name plate voltage, but if you are using a wye you need to multiply the nameplate voltage by the square root of three. <S> Also not that wye configuration is probably more vulnerable to a lost phase and neutral imbalance. <S> If you got this to work, I would love to hear about it. <S> ( I would also be interested in any problems you have, including HCF events. )
| I believe the answer is maybe, depending on how the vfd is designed and other factors.
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Isolating audio ground between devices I'm planning a build for a guitar amplifier with 2 stages, the first being a tube pre-amp and the second stage being a solid state amplifier. Both amplifiers will be powered from the same source, a 31v DC power brick. The pre-amp will be powered with a DC-DC boost converter from the 31v DC power brick to obtain the high voltage required; the boost converter ground output is tied to the input ground. The audio signal ground will be tied to the same ground. The solid state amplifier requires a dual supply which I will create with a voltage divider, giving -15.5v, virtual ground and +15.5v. As the audio signal ground will be relative to the ground of the 31v power brick (equivalent to the -15.5v of the voltage divider) then this signal would cause an issue connecting from the output of the pre-amp to the input of the solid state amp, effectively shorting -15.5v and the virtual ground. It would seem that I need to isolate the audio signal so it can be provided as a floating AC voltage. Can the signal and ground be isolated by adding a capacitor in series with each component like so: Or would the only solution be a 1:1 audio transformer? <Q> I am going to suggest that you want a differential connection from the preamp to the power amp. <S> This can be any of three techniques: 1) <S> audio coupling transformer 2) <S> differential line receiver such as THAT 12xx series <S> THAT differential line receivers 3) <S> if it is available, a differential input on your power amp. <S> The audio coupling transformer has the advantage that it is easy to implement. <S> Jensen makes nice transformers but they are expensive. <S> My company makes small PC boards with a semi-custom transformer on it that sells for considerably less than the Jensen transformers. <S> Performance is not as good as the Jensen but is exceedingly good nonetheless. <S> Trinity Electronics transformer card <S> There are many different differential line receivers available. <S> My current favorites come from THAT Corporation but TI, Analog Devices, many others make pin-compatible parts. <S> It is important to match the input coupling capacitors and use precision resistors in the input stage. <S> Any mismatch causes poor common-mode rejection. <S> Some power amplifiers have inherent differential inputs, even though the amp may normally be used with only a single-ended input. <S> Again - it is important that the input network (resistors, coupling capacitors, HF rejection capacitors) are all matched so as to achieve the best common-mode rejection. <S> The reason for using a differential input to the power amplifier is to eliminate any hum or ripple on the negative supply rail. <S> Using a simple singled-ended connection as suggested by others can lead to excessive hum or noise in the amplifier. <A> This stops the input from being ground referenced, you can then reference it to something else by adding DC (through a pull-up/down or other means). <S> Just make sure that the capacitor can handle the voltage between the virtual ground and the actual ground, 15.5 V, with the additional swing of the audio level. <S> You would use another capacitor again at the output when you are switching back to being referenced to the supply ground as opposed to the virtual ground. <S> You do not need to connect the grounds, the audio can be referenced from either. <S> Capacitors block the DC offset from either supply from transferring between stages. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The main thing to consider from an audio stand-point would be to choose a large capacitor value. <S> The capacitor will form a high-pass filter with any resistances to ground and so will limit the low-frequency content in your signal. <S> Series capacitance tend to add noise at low frequencies as well <S> so it is good to choose high-quality capacitors. <S> Also make sure that you are not trying to drive any DC from the circuit using the virtual ground as it cannot handle very much. <A> You can use a series coupling cap, but bear in mind that that cap (if the DC on either side is appreciably different) has to charge when the circuit is powered, which might cause an appreciable "thump". <S> You might want to disconnect the output for a while, until the circuit has stabilised.
| In this case you would want to use a capacitor at the audio input.
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33 LEDs in parallel over-heating unprotected Li-Ion Battery I'm flummoxed. I have 33 LEDs (17 "lagoon blue" and 16 white) wired in parallel on 22 AWG stranded stereo wire. Blue LEDs have 20 ohm resistor in series, white have 27 ohm. The circuit has a slide switch in-line and the whole thing is powered by a 3.7V Li-Ion battery. I did this once before, no slide switch and only 21 white LEDs w/ resistors. My on/off was to unhook the battery. Worked like a charm. In my current project the big changes are 1) 12 LEDs added 2) two different kinds of LEDs used and 3) slide switch added. Now, when I put a protected Li-Ion battery in the circuit the LEDs only light if I insert the battery while the switch is in the ON position. Then, if I switch the circuit off, everything turns off...and when I switch it back on, nothing. I have to pull out the battery and put the switch in its on position, then reassemble for the LEDs to light. In addition, if I leave it on, two of the cyan LEDs start to turn deep blue. If I put in the unprotected battery it will turn back on but the LEDs fade on slowly to full brightness, and the battery gets hot. What is going on? UPDATEOne battery is protected AW Li-Ion 14500 3.7V 750mAhOther is unprotected UltraFire Li-Ion 14500 3.7V 1200mAh UPDATE 1/16/18FYI this is an LED hoop. I disconnected the switch and ran straight from AW battery. LEDs still change color from cyan to blue, but sounds like that may be a separate issue with my resistor choice. Hard to fix or test that now. I wired another switch and the same issue occurred. Here is a sketch not to scale. <Q> Ahem... <S> In theory you can add as many LED's as you want, but you MUST add or increase the resistor in series with each LED. <S> 27 ohms sounds good for starters. <S> Beg, borrow or buy a DVM so you can see the current drain from the battery vs. its amp/hour rating. <S> The total current should not exceed 10% of the amp/hour rating, and that lets the LEDs run for ten hours. <S> LEDs are very efficient so do not be afraid to double the resistor value in series with the LED. <S> They are non-linear devices so that should dim them only slightly, yet save much current drain on the battery. <S> Be sensible about the total number of LEDs used. <S> The more you add the dimmer each of them will become as you are forced to increase the resistor values. <S> Red LEDs have the lowest turn-on voltage <S> so use a 33 ohm resistor as a minimum. <S> If the battery gets hot, measure your current and increase resistor values as needed. <S> There is a finite limit . <S> Thanks to @TonyStewart for doing the basic math, which shows you are already reaching the batteries limit. <S> Double the resistor values, then check the battery to see if it cools off. <S> A very hot Li-Ion battery is like a small bomb, so go easy on it. <A> simulate this circuit – Schematic created using CircuitLab <S> Estimated current (depends on LED specs) <S> BG <S> 14mA*17=238mA <S> (Blue Green or Cyan) <S> W 18.5mA*16=296mA <S> (White) Total = <S> 534mA <S> If the LiPo is getting hot , I imagine some of the LED's are getting hotter due to missing series R's or wrong values. <A> When the switch is in one position, it will short out all the LEDs and the battery, essentially sucking all the power out of the battery and away from the LEDs. <S> In the other position it will allow the LEDs to light. <S> This is not the correct way to wire a switch and you are lucky your battery hasn't exploded or something. <S> With this configuration I would expect the battery to get very hot whenever the switch is "off"; does this match your observations? <S> The switch should be wired in series , in between the battery and the LEDs.
| You may be close to the maximum number of LEDs you can use already , so double the resistor values to see if that cools the battery down, or limit the number of extra LEDs. I personally would remove some of the extra LEDs, or at least NOT add any more. From this diagram: it looks like your switch is wired incorrectly.
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why resonance is not considered in RC and RL Circuit Why resonance is considered in RLC circuit and not in RC or RL Circuit <Q> For resonance to occur, energy needs to flow from one component to another and back. <S> Resistors (R) can only absorb electrical energy and convert that into heat . <S> Then the electrical energy is "lost" and it cannot be retrieved through the resistor. <S> Capacitors (C) and Inductors (L) <S> however can store and release electrical energy. <S> So with an RC and RL circuit the energy "escapes" (as heat) through the resistor. <A> why resonance is not considered in RC and RL Circuit <S> There are times when resonance is considered in what appears to be a simple RL circuit. <S> Inductors are notorious at having parasitic capacitance and therefore some circuits (that on the face of it are "simple") must consider it as a problem. <S> Even a simple resistor has ESL (effective series inductance) and parasitic capacitance and, at a high enough frequency will become a resonant tuned circuit. <S> Capacitors have ESL and can become very resonant to the point that they become ineffective as decoupling capacitors on circuit boards. <S> If you are just considering theoretically perfect components connected together and the following equation cannot be used in some form or another: - \$f_{res} = <S> \dfrac{1}{\sqrt{LC}}\$ <S> Then that is because L or C are not present and the circuit is not resonant. <S> However, if you are considering the shape of the complex s-plane then any first order filter has an infinite point of "resonance" called a "pole". <S> It's unclear in your question as to whether you are referring to the complex s-plane <S> but I suspect you aren't. <A> General, for a system to be an oscillator, it needs at least two independent enegery storages. <S> Only oscillatably systems can have a resonant frequency. <S> On electronics, this can be achieved by a capcitor and an inductor, which are independent by nature. <S> As a side note: The fact of at least two independent energy storages is not special to electronics. <S> It is a general rule of phsyics and math. <S> In analogy zu mechanics, a spring mass system behaves in the same manner.
| In an LC or RLC circuit, the energy goes back and forth between the Capacitor and Inductor making resonance possible.
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Blocking cell phone interference on LM386 amp I built a small amplifier using an LM386 on a protoboard. The problem I have however is that it makes those typical cell phone sounds if my mobile phone is close to it. Since I want to 3d print an enclosure for it I was wondering what the most efficient way of getting rid of this interference is? Installing it inside a box with aluminum foil on the inside? Should connect the foil to ground? Will the holes that I need to make in the box for the buttons and leds not cause any leakage? This is what I built: http://www.instructables.com/id/Tales-From-the-Chip-LM386-Audio-Amplifier/ <Q> This will get a lot better once you lay out the circuit on a real PC board. <S> Even with a two layer board and the bottom layer being mostly a ground plane, there is a good chance the interference problem will be below the level you care about. <S> I wouldn't try to go for a shielded enclosure before determining it <S> is really necessary. <S> Use good layout, proper bypassing caps, and small caps to ground on all signals that go off the board. <S> Include a chip inductor in series between the external connection and the cap on any line that can tolerate the extra few 100 mΩ in series. <S> For the caps to ground, look at capacitor datasheets carefully to make sure the capacitors have the low impedance at the RF frequency of interest that you expect. <S> You'll probably end up with 100 pF or so ceramic, and then choose a particular model from a particular manufacturer. <S> At these frequencies, there can be significant differences between 100 pF, 50 V, 0805 ceramic caps, for example, between models and manufacturer. <A> Perhaps this app note from Maxim will help you. <S> Minimizing RF Susceptibility in Cell-Phone Headphone Amplifiers <S> Of course it pitches their audio part MAX9724 that claims low RF susceptibility. <S> TLDR quote from article <S> : When using external headphone amplifiers, two methods can ensure that RF noise does not become audible (Solutions 2 and 3 above):Shielding and shortening the input-signal traces to minimize the amount of RF energy that the amplifier sees;Choosing an RF-immune amplifier that internally rejects RF energy, minimizing the amount of noise coupled to the outputs. <S> Hard to offer specific advice without the pcb, but keeping traces short, using good gnd plane and decoupling. <S> Maybe adding a can could all help you. <S> You still have to watch coupling on the inputs from whatever your source is. <S> You can get shields you don't have to solder with shield clips and bendable shield material. <A> Since I want to 3d print an enclosure for it <S> I was wondering what the most efficient way of getting rid of this interference is ? <S> Installing it inside a box with aluminum foil on the inside? <S> Should connect the foil to ground? <S> Will the holes that I need to make in the box for the buttons and leds not cause any leakage? <S> You are on the right track, one solution for you would be to make a Faraday cage . <S> This is what you are thinking about, I think. <S> Loosely speaking: It's essentially a conductive shield enclosing your apparatus. <S> If you do decide that you want some holes through your shield, then that's fine because the wavelength of the RF waves that the cell phones are using are in the MHz range (I believe), which have wavelength's of several centimeters. <S> This means that for them to pass through the hole, the holes need to be at least a quarter of the wavelength in diameter to pass it through. <S> I am not 100% sure about the quarter wavelength part though... <S> so someone will probably correct me <S> And yes, connecting it to ground will allow any currents that comes from noise to dissipate safely through ground, instead of your signal traces. <S> TLDR; <S> The leakage will most likely not be noticeable due to the wavelength's of the RF waves that the cell phones are using. <A> Electrical shielding becomes gradually less efficient as frequency decreases. <S> For audible frequencies, the skin depth for aluminum is of order of several millimeters, so half a millimeter of foil will not achieve anything. <S> A much greater effect can be achieved by eliminating the loops in your circuit which pick up EMI: <S> Make all wires as short as possible. <S> Reduce the area of the contours carrying signal. <S> Run signal-carrying wires very close to the ground, or twist them with ground wires. <S> Building a proper PCB also helps a lot, as it improves both.
| Make some small holes for wires to pass through, the smaller the holes the better. Install it inside a box with aluminum foil around it that is connected to ground.
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Explain why the NodeMCU auto-program circuit would be useful I'm looking at the NodeMCU "auto-program" circuit but could not understand why it is there at all. From my point of view, this circuit enables open-collector outputs, but makes programming dependent on the time between switching the DTR and RTS pins being smaller than the ESP goes from reset to sampling the GPIO0 level. Another reason I could imagine was to avoid invalid startup/connection resets, but I can't see why proper pulling couldn't fix this. Is there any reasoning I couldn't see from my analysis? Why isn't DTR/CTS connected directly with one transistor on each line? Why controlling each line directly (and thus allowing to switch GPIO0 low before resetting without glitches) wouldn't be better? EDIT: while the question is about the same structure shown in Is this a flip-flop? , I understand it's function and know it's not a flip-flop (no need to redraw it). What I'd like to do is know why was it chosen instead of a much simpler open collector configuration like the one below, specially since this one doesn't generate a glitch in the transition between the two programming states (from reset to/from GPIO0=0). <Q> Again, yet another cute way of drawing the schematic. <S> (See: Is this a flip-flop? for another cute way.) <S> And once again, a less-cute drawing would look like: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The RTS line needs to pull down (be LO) on its emitter in order to have an impact on nRST. <S> The DTR line needs to pull down (be LO) on its emitter in order to have an impact on GPIO0. <S> Either way, the nRST and GPIO0 results are open-collector outputs, so they actively pull down (if active at all) but will need some kind of passive pull-up (often found inside the MCU) in order to have a definite output voltage in all cases. <S> If both DTR and RTS are pulled LO (or HI) then neither GPIO0 or nRST are actively pulled down. <S> So DTR and RTS must be oppositely engaged in order for either (one or the other, but not both) output to be actively pulled down. <S> An advantage here is that the external controlling device can choose to set both DTR and RTS to LO in order to decouple itself from the target device's <S> I/O lines. <S> (Setting them both to HI could lead to a problem where there is conduction via the base-collector diode.) <S> This frees up the I/O lines on the target device for other circuitry to be added for target device purposes not related to being reset. <S> Note added on 8/14/2020: <S> The above was entirely written before the OP's EDIT was added. <S> None of it applies to that addition. <A> This question doesn't seem to be fully answered, so I'll take a stab at it. <S> I can't say what was on WeMos's mind when they designed it, but the difference between their circuit and your redrawn alternative is that the original one only asserts NRST or BOOT0 <S> (or EN/GPIO2) when both RTS and DTR are being controlled in opposite of each other. <S> Some cursory searching seems to imply that there is no standard "default" state for DTR/RTS to be set. <S> Or at the very least, it's not consistently adhered to by manufacturers. <S> With the existing WeMos circuit, one has to be active and intentionally controlling those lines to have an effect, so your device won't just reset when you plug into a host that is allowing RTS/DTR to reset at an improper state. <A> I would like to answer this part of the question: <S> What I'd like to do is know why was it chosen instead of a muchsimpler open collector configuration like the one below, speciallysince this one doesn't generate a glitch in the transition between thetwo programming states (from reset to/from GPIO0=0). <S> NodeMCU was designed to be compatible with the Arduino IDE. <S> The Arduino IDE has a built-in programmer. <S> ESP8266 and ESP32 chips are a bit different. <S> You can read more about that here: https://cdn.sparkfun.com/datasheets/IoT/esp32_datasheet_en.pdf in section 2.4 "strapping pins". <S> Most importantly, if you want to flash the device, then you cannot keep both lines high at the same time for a long time. <S> Some programming tools that are specific to ESP chips work out of the box, without the "cross bjt" circuit. <S> For example, look at the reset code of esptool: <S> https://github.com/espressif/esptool/blob/master/esptool.py#L463 <S> There you can see various line state changes and delays that are used to properly reset the chip. <S> If you use such tools, then you can connect GPIO0 to DTR and RST to RTS (well, only if you have a 3.3V USB to serial converter!) <S> NodeMCU <S> however, was created with compatibility in mind. <S> You can use the Arduino IDE and Arduino (basically, the C language) <S> the write programs, and flash them to the device from within the Arduino IDE, and you can also program it with esptool. <S> It works either way. <S> If you are designing a custom PCB then you can save some costs by leaving out this "cross bjt" schematic - but then you must use the right tools to do the flashing.
| That programmer was originally made for Arduino MCUs, and they have a specific wiring that requires both DTR and RTS lines pulled high/low at the same time to put the device into flashing/programming mode.
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Type of Header for Motherboard PWR Switch I am trying to make my own 2-position connector cable to connect a Rasberry to the PWR_SW 's PWR and GND pins on the motherboard system panel headers, as shown in the diagram below. Can you identify the type of connector? Also attached a photo of how it looks like. If it's simply known as a connector, how should I filter down to this type of connector on Digikey? <Q> This should be a standard 0.1" (2.54 mm) pitched header. <S> A name of that thing goes along the lines of SIL (single in line) female crimp housing (for the part which goes on the cable) with fitting crimp contacts for the cable. <S> Something like <S> this and this . <S> There are certainly some already confectioned cables out there (like in the picture you posted). <S> I'm not so good at finding cables, but I'd go with the keywords: 0.1" 2 pin cable, and see what I'm getting. <S> Probably because jumpers are just the same 0.1" pitched header. <A> Even when not made by DuPont. <S> Universal connectors for standard 0.1" headers. <S> Otherwise they can be found as female to male or female jumper wire . <S> These will often be 1x1 single connectors. <S> Which are used as often as dual connector for what you want. <S> Preassembled to various lengths, instead of needing to crimp (which requires the right size wire, the parts, and a good crimping tool). <A> The problem is that damn near every connector manufacturer makes the things but every connector manufacturer has a different name for them. <S> You will find the term "Dupont connector" is used by the direct from china sellers on the likes of Ebay but you won't find it in a serious supplier catalogue. <S> What I tend to find works best is to go into the "wire to board connectors" section of the suppliers site, select the pin count and pin pitch (2.54mm), sort by price and then look at the pictures until I find what I want.
| Looking around a bit, a good name for a cable seems to be: 2 pin jumper cable A popular trade name or search terms got these are Dupont connectors.
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What is this glue? <noob-disclaimer></noob-disclaimer> For context, I need to solder a wire onto a pcb. However, the pad was destroyed during demounting the original component (picture 1). The plan B is to solder the wire onto the copper trace next to the pad. I am however worried that A) soldering alone won't be sufficient for a durable bond. B) that the piece of exposed wire/trace might interfere with something. At the same time I have this guy lying around (picture 2) and noticed that it also has wires soldered onto the board, with some glue(?) being used for, as I understand, similar reasons. What is this substance, and would it be suitable to use in my case? Picture 1 Picture 2 I appreciate you tolerating my knowledge of electronics. Thanks. <Q> Carefully scrape the soldermask (the green stuff) off of the PCB trace that routes to the damaged through-hole. <S> You can solder a wire directly onto that trace to form a good electrical connection, but it won't be a very durable mechanical connection. <S> I typically use hot glue in this application because it's easy to work with and remove if needed. <S> The substance in the second image might be flux residue left from when the wires were soldered to the pads, but it's hard to tell. <S> If you can wipe it off, it's probably flux residue. <S> I would try to heat it up and fill it with solder. <S> If the plated through-hole is thoroughly damaged, more serious repairs may be necessary. <S> Through-hole repair kits are commercially available. <A> It is hot glue from a glue gun. <S> There is a stacke exchange question dealing with the issue here <A> For permanently stabilising repairs/modifications two part epoxy (araldite or similar) works well. <S> Just make sure your repair is electrically correct first.
| You need something, like glue or tape, to hold the wires in place and decouple mechanical stress from the solder joint. As others have noted in the comments on your post, it appears there may be damage to the through hole.
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What is the color code for CAN Bus? I would like to know if there is any color code specified for CAN Bus communication wiring, maybe in ISO-11898-1? (I have no free access to the file) I looked for any official documents without any luck. When I asked to my colleagues about it, they all agree on the colors (Yellow and Green) but they do not agree if it is: Yellow for CAN-H Green for CAN-L or instead: Green for CAN-H Yellow for CAN-L <Q> CAN does not have a formalized physical-layer specification for conductor colors, or things like connector type or pin-out. <S> There are common practices (like using a 9-pin D-sub connector) but no official standard. <S> Vehicles these days also tend to have multiple CAN buses, so colors will, of course, vary to keep the different buses straight. <S> I have seen some buses adopt a solid color for CAN-L and a different color striped with the corresponding CAN-L color for CAN-H, to give a visual impression of belonging together. <A> I found a link for J1939 cables (not sure if this is generic): see link Excerpt (see last two lines): CBL-CAN-01: <S> CAN Cable for DB9 Male Connector <S> This is a 4-wire color coded cable. <S> One end is DB9 female connector, it is designed to mate with Au J1939 products at bus side, such as: J1939 Message Center System, J1939 Data Center System, J1939 Simulators, etc. <S> The other side of the cable are 4 pig-tail wires. <S> Red wire: Power supply, +12V Black wire: Ground White (or Yellow) wire: CAN High Green wire: CAN Low From 'John Deere' : Link Red is Power - 12 VBlack is <S> Ground - 0 VYellow is <S> CAN High - 2.5 VGreen is CAN Low - 2.5 V <A> The ISO 11898 standards do not mention practical things such as cables and connectors. <S> The second-most authoritative source has therefore become the CANopen standards, where DS303-1 specifies things such as standard connectors, pin-outs, cable lengths, baud rates etc. <S> Unfortunately, this document does not mention color-coding either. <S> However, just because these colors are commonly used, it does not make them more correct. <S> I'm guessing that the yellow and green comes from the universally standardized color-coding (same as we use on through-hole resistors). <S> That is: 1=brown, ... 4 <S> =yellow, 5=green. <S> For the "mini style" and "micro style" (round M12-like) connectors standardized by DS303-1, CAN High happens to be on pin 4 and CAN Low on pin 5. <S> From DS303-1 7.2: <S> This enumeration is however not at all consistent with other common, standardized connectors such as d-sub, RJ45 and terminal socket. <A> In every implementation I've seen, if you can imagine that one color represents "Sky" and the other represents "Earth", those happen to be Hi and Lo, respectively. <S> Yellow sun, green grass. <S> White clouds, green grass. <S> White sky, blue ocean. <S> Et cetera. <S> These mnemonics seem to fit <S> so well I have to imagine they're intentional. <A> CAN_H = yellow, CAN_L = green per SAE J1939-11. <S> This would only apply to J1939-compliant CAN installations.
| Overall, as long as you keep track of your conductor colors and pinouts, you can use any color scheme that suits your preference without violating any standards. Yellow and green seem to be commonly used, though I have seen yellow mean either CAN High or Low.
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Problems with Conversion of 1960s era transistor radio to a guitar amp I have an early 1960 Silvertone transistor radio that I'm trying to convert to a guitar amplifier for a cigar box guitar, aka a CBG. (CBG's are typically built with primitive equipment and the inherent distortions and defects are considered part of their charm.) The old radio works fine, I put 6 D-cells into it and I can pick up AM stations just fine. Instructions for conversion are all over the web but basically consist of this: Ground one wire from the guitar cord to ground on the radio. Find the volume pot. Attach the other lead from the guitar random posts on the volume pot and strum until you hear your guitar through the speaker. We're not talking rocket science here. Now my radio has 2 concentric pots, one for volume and one for tone. I've been able to identify the volume pot by hooking up my Ohm-meter and turning the volume knob. Now the problem is, when I attach the leads from the guitar I don't hear the string from the radio's speaker. Here's the back of the volume pot: Tap 1 has 9v when the volume control is clicked on, 0v when off. Tap 2 has 9v all the time. So it's clearly the power switch. Underneath we find 3 more posts: Taps A, B and C also have 0v when the volume is off, and 9v when the volume is on. I pull the batteries and measure resistance (and yeah, I know, the pot is in the circuit so it isn't accurate but I'm just trying to find the "input" lead. A always reads 4k Ohms, B varies from 2k to 4k, C is at 2k no matter what I do with the volume knob. 1 in always at 2k Ohms, 2 is 0 Ohms when off, 2k when on (which makes sense since it is connected directly to the battery) So it seems to me that I should be hooking into tap B. So when I hook the rod-piezo to B and to ground I still hear the static of the radio and when I pluck the string, I do not hear the note out the speakers. One member of the cigar box forums said that the rod piezo pickup might not put out enough power, so I tried a wound magnetic pick and got the same results. And here's the 'schematic' of the radio: (ok, so it is just a parts list...) All the online tutorials make this look easy, but I'm baffled at this point. And go easy on me, my degree does have an EE in it, but it also has a CS in it and I do CS stuff for a living, that EE was really a long time ago... Bonus question: I'd love to know why this works, it seems to me that I'm mixing the guitar signal with the radio signal but all the web instructions just say that you don't hear the radio. Bonus question #2: Just how would I find and cut out the radio section? I'm guess that once I find the right place to tap into I just cut there to remove the radio from the circuit, is it really just that easy? Update I have something which may be the real schematic! The radio is almost certainly a Silvertone 700. See https://www.radiomuseum.org/r/sears_roeb_silvertone_700_1217.html Hopefully this helps! Note: It turns out this is not the schematic since it is 6v, not a 9v circuit. Thanks to glen_geek for pointing this out. Update #2 Here's some photos showing the connections. First, the battery has a green and black wire coming off it. With the VOM's black on black, and the VOM's red on green, I get 9VDC. Here's the connection to the pickup: And my connection of the positive ground (green) wire. (The green wire goes directly to the battery pack's green wire, it's actually the same wire.) And the connection to the center tap of the volume pot, B, in the above diagram, which should be the + side of C10. And just for grins I put the ground to the other wire off the battery. <Q> A rod piezo (or any piezo) has very high impedance: it puts out a good voltage signal but at very tiny currents. <S> It will not be able to drive the common-emitter input stage of that radio's audio amplifier. <S> You would need to switch to a magnetic pickup, or add a buffer stage so this amp does not load down the pickup. <S> A common simple circuit is to use a JFET preamp. <S> [ This was from http://www.till.com/articles/GuitarPreamp/ <S> This is a very common circuit and used for things like buffering a condenser (aka electret) <S> mic capsule to drive a standard audio input. <S> Essentially no current is taken from the signal input since the junction is back-biased. <S> The JFET converts the signal voltage into a modulated current through R3. <A> The two terminals you marked as "1" and "2" are the off/on switch. <S> The terminals you marked "A","B","C" are for the inner of the two concentric knobs. <S> " <S> A" should be the clockwise terminal, "B" is the wiper (moving contact), and "C" is the counter-clockwise terminal. <S> There are another set of three terminals, closer to the front panel, shown in the first picture above the pot (not connected directly to the circuit board), for the pot operated by the outer knob. <S> You should connect your guitar signal to the clockwise terminal of the volume pot. <A> Why this works: In your schematic, diode D1 is a rectifier that turns the amplitude modulated signal to a varying DC voltage in the low audio range (< 5 KHz). <S> You would put your signal on the "+" side of capacitor C10; Q4, Q5 and Q6 are the audio section. <S> Your guitar only has to overdrive the diode and capacitor signal after it has been attenuated by the pot. <S> Put the pot about half scale to give the highest impedance. <S> Your signal must be with respect to ground, which is the positive battery terminal on your schematic. <S> Tie the ground of the guitar to the positive terminal of the battery. <A> Fortunately you have a schematic. <S> That makes this pretty straight forward. <S> The important part of the schematic is: One important thing to notice is that this is a positive ground system. <S> The positive end of the battery is actually ground. <S> When the power is turned on, it is negative relative to ground. <S> Everything to the right of where it says " <S> VOL. <S> 5K" is just a audio amplifier. <S> You can use it as-is. <S> You have to disconnect it from the radio receiver part, then connect it to this guitar thing. <S> It seems you have already found the wiper of the "VOL 5K" pot. <S> The end that is permanently connected to the + side of the battery is ground. <S> The other end is where you want to connect your audio signal to. <S> Before you do that, remove whatever else is connected to that end of the pot. <S> With nothing connected to the audio input end of the pot, you should be able to crank the volume to maximum, then hear some hum when you touch that point with your finger. <S> That's a good test that everything else is basically working. <S> This audio input is already AC-coupled due to C10. <S> You can therefore just bring out two wires from this radio, which are the two fixed ends of the pot. <S> Any audio you put on those two wires will come out the speaker, subject to the setting of the volume control. <A> The audio amplifier section seems to be having a low impedance which is not matched with high impedance output of the guitar. <S> I can suggest you to do following, Change C10 value 5uF to 1uF or 0.1uF . <S> This may increase the input impedance. <S> Try removing the connection to the volume controller from C10 and directly connect to guitar. <S> Whoever drew this circuit has changed the polarities upside down. <S> (+) in bottom and Ground in top. <S> Pay special attention to this. <S> Connect your guitar to C10 and the upper wire of the circuit. <S> Hope this will fix your problem :-)
| The problem you are facing seems to be 'impedance matching'.
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Digital counter with analog output I have a process that produces a digital pulse train from a rotary encoder. I need to be able to stop the process once a stepper motor has moved a given distance. The trick is, I need to manually set the length of the process, probably using a potentiometer. So what I'm looking for is something that takes the digital pulse train as an input and produces an analog voltage as an output, the voltage increasing with the number of pulses received. I can then compare this output voltage with the voltage at the potentiometer and stop the process when they match. This stopping is secondary to the main purpose of the stepper motor, and is not a precise requirement; the exact number of pulses received is not important. I want to stop the stepper motor if it overruns its intended distance, so a potentiometer control is adequate. I've been searching the web for articles that give a solution to this, but haven't found any. I've looked at digital-to-analog converters but can't find one that does what I need. Basically, it's a digital counter that produces an analog voltage output which increases with the input count. <Q> Typically this would be done by making the potentiometer an input to the processor (via an ADC) and having that make a decision. <S> Especially as it can apply somewhat more sophisticated filtering and decisioning than you can easily accomplish in the analog realm. <S> One crude, but sometimes suitable form of DAC is to use a hardware timer on a processor to generate pulses of varying width (PWM), and then smooth this with a filter made from at least a resistor and capacitor, or perhaps a more sophisticated one. <S> A key question here will be how smooth the output needs to be, vs. how agile you require it to be in responding to changes of value. <S> If it only needs to change slowly, it can be made very smooth. <S> But making an output that is smooth at rest but rapidly changes when required places more challenging demands on the filter, and may be a reason to use a "real" DAC rather than smoothed PWM. <S> Your question is rather short on details, so a more specific answer will not be possible. <S> And even if there were more application details, seeking recommendations of specific parts isn't really a permitted goal of questions here. <A> An analogue integrator appears to do what you want: - A single positive pulse will cause the op-amp output to ramp down at a specific rate <S> and, when the pulse is finished, the output level will be at a new level and remain at that level until another pulse arrives. <S> If the polarity of the input pulse changes then the output will ramp in the opposite direction so care has to be exercised to make sure your input is "reliable". <S> You also have to choose op-amp and capacitor carefully to avoid the output drooping over time in the absense of a pulse. <S> If you are "collecting" say a hundred pulses in ten seconds and relying on the output as an approximate measure of the number of pulses you should be OK. <S> If it is much longer you won't be OK and I would recommend a digital technique. <S> If the pulses have a lot of variability in their width then this will also be problematic because the circuit will "weight" pulses of longer duration more highly than thinner pulses. <S> If you can overcome these problems it should work. <S> Bear also in mind that you need a reset circuit to discharge the capacitor each time you start a new "run". <A> I am afraid you just need to use the straightforward approach to solve your problem, just as you formulated it: "digital counter with analog output". <S> Take some minimal-size MCU, whichever you feel comfortable with (and has a full development support), and count the pulses coming form the encoder. <S> There are plenty of ready-to-go development/reference designs, some are really inexpensive, which will serve the purpose without much soldering. <S> It could be just an interrupt input. <S> One pulse - one call, counter++. <S> In the same call you then simply output the current value of that counter into a DAC. <S> It could be an internal DAC in the MCU, or external DAC controlled by I2C interface. <S> You will need to get the DAC with sufficient resolution (number of bits) to cover your range of pulses. <S> An analog integrator will likely fail due to unpredictable leakage, so the output will drift.
| Alternately, more in keeping with your stated idea you can use a Digital to Analog converter (DAC) to turn a value in software into an analog voltage.
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Is there any formula to get number of windings for a 50W voice coil of a speaker? Is there any formula to calculate number of windings to get a 50 Watts voice coil of a speaker? let's say, I want to wind a 20mm diameter voice coil for a speaker. And my coil is made out of copper wire which is 1 mm thick (diameter) and has a resistance of 13 ohms per 1000 meters. And I want the voice coil to use 50 Watts. How do I calculate the number of windings required? P.S: I'm developing a new speaker design, so, I'll experiment with overheating issues. Just ignore all other parameters and let me understand the relationships between given parameters to initially wind the voice coil. I'm stuck in finding out the number of windings to experiment with, so if some formula gives 50 turns, I'll initially wind it and experiment with all other parameters one by one (eg: overheating). <Q> Calculate the length of the wire required first by its resistance. <S> If 1000 m per 13 ohm, that simply means 500 m per 6.5 ohm or 76.92 m per ohm. <S> Now if you are wiring a coil for 4 ohms, you need 307 m wire on the coil. <S> Now calculate the length of one turn of wire on the coil. <S> If one turn is 2 inches, for example, then you need 2000 turns for 4 ohms impedance. <S> The coil is designed on ohms not watts. <S> Watts indicate the capacity or load a coil can handle, which depends on the gauge of wire. <S> And before wiring, you can check gauge and watt chart for adequate watt calculation. <A> No, you cannot calculate that with only these parameters. <S> Think about the relations between the parameters of the voice coil you list: <S> coil diameter wire diameter wire resistance number of turns <S> And how that relates to the maximum power handling capability. <S> Suppose a certain voice coil can handle 50 W, what happens when I put 100 W into it? <S> I think it will get hotter compared to the 50 W situation. <S> But how hot is OK? <S> That isn't listed anywhere in the parameters for the voice coil. <S> The coil itself is OK until the copper starts to melt. <S> But I'm sure that at much lower temperatures other things had heat issues already like the material we're winding the coil onto. <S> Some voice coils are wound on an aluminium tube which will help in dissipating the heat. <S> That would increase the power handling capability of voice coil. <S> But hey, you didn't list which material is used for winding the voice coil onto. <S> Also, if the voice coil is allowed some movement (like in long-throw speakers) that moves the air around the speaker which also helps in cooling. <S> If you're using the speaker in a free air, a closed enclosure or a bass reflex enclosure that will also have a significant impact on voice coil movement and therefore cooling and therefore power handling. <S> See how I got from just the voice coil to the complete design of a speaker? <S> That is <S> because it is not only the voice coil which determines the maximum power handling capability, it is the complete design. <S> If you look up datasheets of loudspeaker drivers (so the speaker themselves, not a finished box with a speaker in it) then in a proper datasheet you should find the maximum power handling capability listed with a note saying how that's measured like what volume of enclosed box the driver was mounted in. <A> The wattage question is moot because your wire can handle not only 50 Watts, but <S> around 5000 Watts if you keep the Amperage around 9 Amps continuous.. <S> And it can handle about 15,000 Watt spikes... <S> But that's not really what you want to know probably. <S> What you need to know, is you need thinner speaker wire. <S> Your ~17 Gauge wire, to get it down to the correct ohmage would be about - 1000m / 13 <S> Ohms = <S> x/8 <S> Ohms (if it is an 8 ohm speaker system ( <S> generally home systems)- or 4 <S> if it is a 4 Ohm (generally, a car speaker system)). <S> Thus, you'd need 615m (8 ohm) or 307m (4 Ohm).. <S> Extremely large, It's pointless to calculate because the number of turns in the thousands and with the thickness of the wire, you won't be able to fit it in your magnet. <S> You can save your time winding, and money, buy using thinner copper. <S> each halving of diameter will increase the ohmage by 4 times. <S> So if you use 0.25mm solid wire, you'll be happy to use 16 times less length of wire. <S> Or about 17m for 4 ohms, or 34m for 8 Ohms. <S> And this is about what most woofer speakers use - 28 Gauge sold copper wire, about 120 feet of it, give or take. <S> Once you know the length of wire, and I'm just giving approximations here, you can calculate the humber of turns (and layers) required. <S> You have to know things like, how long is the throw if your woofer (as you will need to make sure your VC length is as short or shorter than this. <S> Also, how many turns do you need on the VC to make it fit perfectly in your magnet? <S> So calculating the number of turns is the least of all the problems here. <S> Making a speaker can be fun, but if you are doing experiments - you don't need to reinvent the wheel - there are many people who have done this - and far more - and compiled these into beginner guides. <S> Q factors, capacitance, inductance, etc. will all be necessary to understanding making a speaker <S> and it's enclosure.
| Well, your exact question can't be answered as it stands - or rather, the answer is, you won't be able to fit that much wire on a voice coil that needs the correct impedance.
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Step up 3.3V to 5V I'm new to more advanced electronics (I've tinkered with some arduino stuff before) and I'm wondering how I'd go about getting a 3.3V supply to a 5V supply. I've read around and saw some things about using opamps, but I have no idea how I'd use them either, and everything else with resistors seems to just be stepping down the voltage. So how do I step up 3.3V to 5V in a simple manner? <Q> I am guessing from your question <S> you need the first option. <S> They're not the simplest of circuits, but you can find prebuilt and tested modules available. <A> There are two possible answers here depending very much on the situation. <S> You have a 3.3V power rail and need a 5V one at some current. <S> In this case you need a boost converter, examples already given. <S> You have both 3,3V and 5V supplies and just need to move a signal from the 3.3V to the 5V domain, say you have a processor outputting 3.3V logic, but need 5V logic for some part powered by 5V. <S> In this case you need a Level shifter, something broadly like a SN74AVCH4T245 gets this done (There are variants with different numbers of data lines). <S> these is also the case that you have only 3.3V as a supply, AND 3.3V logic and need to both power and supply data to a 5V part, in which case you do both of the above. <S> I would note that for a 3.3V part when you need to drive an input from the 5V domain, series resistors and clamp diodes or a voltage divider can work, as long as the speed is low enough. <A> To step-up a supply voltage of 3.3 V to 5 V you should use a step up converter or also called boost converter . <S> Here's an example of a ready-made module from ebay : <S> This kind of circuit is normally used in a power bank to step-up the battery voltage (between 3.6 V and 4.2 V) to 5 V for USB. <S> You can also buy a separate chip and design a PCB yourself <S> but that's some work. <S> Also PCB layout and component choice are critical, you really must follow the recommendations in the datasheet (assuming there are any). <S> In practice it is much easier to just get a ready-made module. <S> Often that's much cheaper as well especially for hobbyists who aren't going to need these in large volumes. <S> Note that if you draw for example 100 mA from the 5 V side, that's 0.5 Watt, at the 3.3 V input you need to supply at least 0.5 W as well <S> so that's at least 0.5W <S> / 3.3 V <S> = 151 mA. <S> That excludes the power lost in the step up converter so in practice up to about 170 mA is needed. <A> As this site is about designs I would recommend the circuit below for a small, 200 mA, 5 volt output from an input of 3.3 volts: - If you want more power out then there are plenty of boost converters that can deliver amps at the output. <S> All the usual vendors (like TI etc.) can provide similar offerings and both Linear technology and TI have very good search engines that allow you to enter input parameters and get recommendations of suitable devices. <A> A very simple way to do it would be something like this: <S> http://www.ti.com/product/ptn04050c <S> I say something like, because it depends on exactly how much current at 5V you want and <S> whether you want a more expensive off the shelf solution or to build something yourself. <S> If this module doesn't provide you with enough power, there are many modules you'll find in places like RS, Farnell, Mouser etc with a bit of research. <S> If you look into boost regulators/converters, you'll find a plethora of information on-line about how they work. <S> TI (among others) have some great tools for putting input and output requirements, suggesting parts and giving you schematics if you want to do it yourself. <S> Along with the datasheets giving layout advice, these tools are really good and practical.
| If you want to step up 3.3 V to 5 V you need a boost converter , if you already have a higher voltage supply available and you need to increase a signal from 3.3 V to 5 V you need an amplifier .
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Would an unconnected coaxial cable act as a wifi antenna? I am building an underwater drone and I connect to it via wifi when it is at the surface. The wifi signal often gets blocked by water waves though. I don't really want to have another penetration of the pressure housing and wondered what would happen if I took 50cm of RG58 coaxial cable, stripped 5cm of shielding from either end, attached one end to the housing close to the wifi transmitter and had the other end poking up out of the water. The shielding of the housing cable end would be exposed to the water for grounding. Would there be any useful wifi transmission/receiving signal at the above water cable end? The connection runs Remote Desktop so I don't need a particularly large bandwidth, just something to make the wifi connection slightly more reliable. <Q> A better idea would be to get a real WIFI antenna, small enough so it can be sealed, and use that. <S> Something like this: <A> took 50cm of RG58 coaxial cable, stripped 5cm of shielding from either end, attached one end to the housing close to the wifi transmitter and had the other end poking up out of the water. <S> That would allow water to get into the cable, rendering it useless. <S> RG58 performance at 2.4GHz is not great even when dry. <S> And even when you could magically keep the water away from either cable end, this construction does not work because only a tiny amount of power actually couples into the cable - barely enough for a sensitive receiver but not enough to re-radiate <S> it witout some ampilifier. <A> I have experimented with crude mono-pole antennas made out of coaxial cable by stripping off the shielding at both ends and letting lambda/4 piece of cable exposed and it worked quite well. <S> This was over the air, I am not sure about underwater.
| There's no chance that a stripped piece of wire will perform better than an antenna specifically tuned to 2.4 GHz.
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Safe use of Coiled Cable So we have this machine at work, a strapping machine, and it has a long power lead which is left loose on the ground, and it's getting gradually destroyed - accident waiting to happen. I know that there are fire hazard risks associated with coiling wire, but I don't know how to work out how much you can coil with virtually no risk. I would prefer that we don't permanently shorten the cable, because it may come in handy to have it long in the future, and we'd have to raise engineering tickets etc... major hassle for what I suspect is a non-issue under these conditions. Please help! Let me know what other info might be pertinent. Rated Power: 5A Volt/Hz: 230/50 Cable length: Around 3m <Q> The hazard with coiled power cables is they can't dissipate heat as effectively because it's exposed to less air. <S> It's especially an issue for long cables since some layers of the spool are exposed to no air at all, instead being surrounded by more hot wire. <S> Fortunately 3 meters is not very long, and you will not need a very tight coil to take up that length. <S> So I'd say a safe guess would be: Coil it loosely, so there is only one layer of coiling. <S> Derate the the maximum current by 50%. <S> This will be a 4x reduction in heat generated, since power is proportional to the square of current. <S> So be sure the attached equipment will not draw more than 2.5 amps. <S> To really be sure of that, add a circuit breaker or fuse. <S> Be sure there is ample airflow around the cable. <S> Don't coil it inside a small cabinet or in a wall, for example. <S> Of course I can't take responsibility if this starts a fire anyway. <A> Is 3m a typo? <S> I can't believe it is real or the question is not really an issue if you are not coiling the wire REALLY tight and already approaching the maximum rated current for the wire. <S> A 30m cable for a 3m span would be a different story. <S> There are ways to alleviate that issue, but since all of them reduce the safety of the system, the only real answer to this is to properly wire it up in the first place. <S> Have an appropriate outlet installed where the power is needed, or buy the required length of suitable cable and the appropriate end connectors so the cable can be routed safely to where it needs to go. <S> Besides, minimizing the cable length will improve the performance of whatever you are powering. <S> "Accident waiting to happen" should be all you need to know to justify the expense. <A> Cable is current-rated assuming it's laid straight out, by itself. <S> Coiling the cable means that it has other heat producing things next to it, so will get hotter. <S> Heat softens insulation, and a lot of heat could cause it to burn. <S> The heat produced by cable is proportional to the current squared. <S> If a cable is carrying half its rated current, it will be generating 25% of the heat it's been designed to dissipate. <S> If you coil the cable 4 times, then the coil will be generating the same heat per unit length as a single cable at full current. <S> While it will run cooler than a single cable, as the diameter of a bundle of 4 cables is larger than a single one, it's not safe to estimate how many extra times the cable could be coiled. <S> You need to know the actual current consumption of the machine, and the rated current for the cable. <S> You can coil the cable that ratio squared. <S> It's not clear from your question what the 5A refers to. <S> If you don't know the rating of the cable, it tends to be 10A/mm2 for cables up to about 1.5mm2.
| The cable will get warm before it catches fire, so if after running the equipment for a while you notice the cable getting noticeably warm, it's probably time for a thicker cable, or a shorter one that doesn't need to be coiled. It will be safe to work from the rated current or the fuse rating of the machine, as these tend to be more than the actual consumption, an actual measurement may well be more favourable.
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Spike on graph generated I had draw a DAC by using an adder circuit.Can i know why there is spikes on the graph generated? <Q> Before posting such question its actually a good idea to isolate the problem. <S> For that you should throw away all but the resistor ladder, and you can see the same problem. <S> If you now zoom in at the point of one of the spikes, and show all voltages of your voltage sources, you will notice that the source that ramps down only starts doing so once the others are ramped up. <S> As Trevor says, this is more or less an inherent property/problem of resistor ladder based DAC designs and will happen to a degree on real implementations too, as the timing of the switches will never be perfect. <A> The issue is the simultaneous change of the bits in the presented DAC digital value. <S> During that change, the resultant presented value can be any combination of the bits that change during the propagation delay. <S> Theoretically a simulator should be synchronous to the set data parameters, but as Andy mentions it really depends on how granular the simulator takes it's samples. <S> Simulators like SPICE tend to sample in a reactionary sample rate based on what is changing so can present rather more spurious results as you have indicated. <S> How much the simulator demonstrates the effect is dependent on the granularity of the simulation rate. <S> By messing with the simulator granularity you can mask the effect, but that is actually a bad thing. <S> Since your real world DAC design will have propagation delay differences bit-to-bit, those spikes will be real and will be propagated through to the op-amp, and beyond if the op-amp is fast enough. <S> Your circuit should be designed to limit the response time/slew rate to block that. <S> For a DAC design that is low frequency, such as setting some reference value, you can simply filter out the switching frequency. <S> For something that is expected to output some waveform closer to the sample rate you need to synchronise the output to the data rate using a sample and hold circuit on the output. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> This is a feature of most SPICE simulators especially when a value for maximum time duration is not specified. <S> I see ".tran 10" <S> so you haven't specified a maximum timestep. <S> Try making this maybe 1 us and see what happens. <S> Sure it will be slower in the transient analysis but it should improve. <S> It wouldn't hurt to place a 1nf capacitor from inverting input to ground too - if you were making a real circuit with really fast edges on the inputs you might do this. <S> Alternatively slow down the rise and fall times of the generators. <S> You are also using an old op-amp that requires (in some applications) <S> the addition of a compensation capacitor.
| What you are seeing is a common and irritating feature of DACs in general and is prevalent in simulators. This might be the problem but certainly timestep issues and SPICE nuances quite commonly produce spikes like these.
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Parallel led segment with (1 led) and (2 led in serie) without resistor? I try to understand this datasheet: How it's can be possible to have in parallel (1 led) and (2 led in serie) without resistor for the led alone? [Edit] Here the only "datasheet" that I found: https://datasheet.lcsc.com/szlcsc/1-2inch1bit-Red-digital-LED_C53679.pdf https://lcsc.com/product-detail/Led-Segment-Display_1-2inch1bit-Red-digital-LED_C53679.html [Edit] Here what is inside: <Q> Its possible, but why they did this is hard to say. <S> (if in fact this is the actual diagram, it is a one color LED so this could be correct). <S> If the diagram is drawn correctly, at a lower voltage only the red LED will conduct and the two in series will be off. <S> At a higher voltage all three will be on but the LED that is not in series will be getting more current. <S> Since nothing else is known about the LED's (no voltage, color, or IV curve) nothing can be said of the actual operation. <S> This may make more sense if the LED's were two different colors. <S> If you use this you'll need to provide current limiting with a circuit or resistor. <S> You could find the IV curve but increasing the current in steps and measuring the voltage,and that would tell you more about this terribly undocumented product. <A> Looks to me like they made the PCB (and the datasheet) <S> so they could include either one LED die or two series LED dice per segment. <S> Since the color is stated as red (红) the forward voltage should be around 1.8-2V for one LED and around double that for two in series. <S> Since the display is relatively large (1.2 inch 英寸) it would be more even and brighter with two LEDs, but the forward voltage may be too high for some applications, particularly if they used the same PCB for green, blue or white- <S> in which case even 5V is not enough. <S> So you can easily figure out which of the two it is by measuring the forward voltage at a reasonable current (5 or 10mA). <S> It's a bit hard for me to see from your photo <S> but it does look to me <S> like two are fitted in series and the third is not populated. <A> Normally you would expect an datasheet to indicate wavelength, Vf, <S> If, Luminous Intensity (mcd) etc etc. <S> Also you would expect the Decimal Point, DP to be small compared to the segment, but not expect 2 diodes shunted by 1. <S> My guess is <S> the supplier has 2 sources , one with 1 single diode per segment and another with 2 and <S> since Vf is not given, you have to guess. <S> It is normal for digits larger than 1" to have 2, 3, or 4 LEDs in series for the segments and 1 or 2 for the DP and 1 or 2 in series for < 1" in RED 7seg + DP displays. <S> This drawing is obviously WRONG. <S> My Guess, all series chips. <S> Here all they tell you is the pinout and mechanical dimensions of each segment and list it as "1.2inch 1bit Red digital LED " <S> Conclusion: <S> Distributor with poor drawing skills has misrepresented the part. <S> How desperate are you to buy poorly documented parts? <A> Judging by the photo of the internals you supplied the datasheet is right! <S> (Your schematic diagram is not correct.) <S> Figure 1. <S> The LED arrangement highlighted for one segment. <S> LED 3 and 1 are in series between the anode pin (orange) and the cathode (blue). <S> LED 2 is directly between the anode and cathode. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Schematic of one segment. <S> Maybe 1 and 3 are a fallback for when 2 burns out! <S> It certainly is strange.
| To me it looks like someone with primitive engineering skills made a drawing with 3 LEDs per segment and made it your guess for multiple source.
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fast or slow fuse in transformer I have a 110VAC input, 24 VAC output transformer that includes an Elsen WA128 fuse on the 110 v input side. I accidentally momentarily shorted the 24v. This blew the fuse. I want to replace the fuse with an equivalent, but I need to know if this is a slow or fast acting fuse. This transformer supplies power to a skylight window controller. The window is opened and closed with a DC motor. The description of where slow and fast acting fuses are used is confusing: fast for electronics, slow for motors. But this application does both. Thank you for your responses. I admit that I was a bit sloppy - I said the transformer was 110 v, it is actually 120 v, I apologize. For further clarification, I have had this transformer in use for over 10 years without problems. It is rated at 55 VA. I accidentally touched the 24v lines together momentarily. A day or so later I realized that the skylight no longer worked. I found the problem to be the transformer. A VOM test showed that the secondary had continuity, but the primary winding was open. I drilled out the rivets and unwrapped some tape around the primary winding. I then found the Elsen fuse, connected in series with the primary winding. A VOM test shows continuity through the primary winding is good, but the fuse is open. Although I am not certain about this, from the comments above, and other comments on the web, it appears that the Elsen fuse is both a thermal fuse, AND a current fuse. I am not sure about this, but it appears that the Elsen part is no longer made. I was able to find a possible replacement on eBay, but that is when I started thinking that a suitable replacement might be a better approach. To my surprise, the momentary short circuit of the transformer also caused the skylight electronic circuit board to fail. This is a different problem, and also one that I have not yet solved, but a different subject. The transformer is mounted in an open area in the basement, and the duty cycle is really really trivial. The skylight had not been used for several days before my momentary short circuit mistake. I seriously doubt that the fuse blew due to transformer heat over 128 deg C. I think it was due to a few milliseconds of secondary short circuit. The part referenced above, thermal fuse NTE8070-8242, will pass up to 15 amps. I have not been able to find the current rating of the Elsen WA128. My guess is that this transformer was NOT made specifically for this skylight, and therefore the fuse in the transformer was intended to protect the transformer PRIMARILY from over temp, and only secondarily for over current. But I seem to have blown the fuse by over current. A 55 VA transformer at 24 v translates to a load of about 2.25 amps (ok, 2.29). I think the NTE part would not protect against a transformer current overload. So, my questions are: have I analyzed the problem correctly? Should I pursue the NTE (15A) thermal fuse, or try to find the obsolete Elsen part? <Q> A Current fuse opens above a protected current level defined by some I*t curve. <S> It can also protect appliances that are supposed to get hot like coffee pot heating pads. <S> When the current can be 800% during a startup, it is not harmful to the transformer until the temperature rises above it's protected limit. <S> This would apply to low energy transformers not kW , MW sized devices where faster protection is needed. <S> Consider these, available in any appliance spare part dealer. <S> http://www.weisd.com/test/WEISD_TBL_view.php?editid1=NTE8125 <A> That is a thermal fuse, not a current fuse. <S> It opens when the transformer gets above about 128 Celsius, hence the part number. <S> You should replace it with a thermal fuse of equal or lower temperature rating and reattach it to the transformer with solid, permanent thermal contact. <S> If it were a current fuse it would be a slow blow type. <S> The transformer can handle overloads for a while before it gets too hot. <S> Since overheating is a primary and dangerous failure mode of transformers, protecting them with a thermal fuse is appropriate. <S> Fast blow fuses are used where faults cannot be tolerated for any duration, such as in sensitive microelectronic circuits, and must be cleared as quickly as possible. <A> 110 transformer? <S> you need to get a new 120V transformer if you are on standard US power (120V) <S> otherwise its going to run hot. <S> Usually you don't bypass a thermal fuse on a transformer. <S> There are several safety checks that a repair guy should do before they do that, but usually its better to replace the transformer than to be opened to get sued if the appliance catches on fire. <S> But to let you know, a thermal fuse is a special type of slow blow fuse.
| A Thermal fuse opens above a protect temperature limit which is appropriate for this.
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Why do Rogowski coils work for measuring current? Rogowski coils measure current by completing a loop of coils around the conductor seen below. This makes sense, but question 1: This is a current transformer, and transformers couple, so in this system when the Rogoski coil couples with the measured wire, does this not affect/damage the measured wires current (we are measuring) as we are drawing/collapsing its magnetic field in order the induce a voltage in the coil? Question 2: Why can the circuit diagram below not do the same job? The Rogowski coil only works when you form a loop around the conductor. Why does the Rogowski coil only work when its a closed loop? Why can a coil no just be in parallel with a wire and measure current? simulate this circuit – Schematic created using CircuitLab The right hand rules says that in the coils are wound in the correct direction, in the circuit diagram would couple the magnetic field the current is inducing. Image source: http://www.electrotechnik.net/2009/09/what-is-rogowskis-coil.html <Q> A Rogowski coil is not a current transformer. <S> The current in the main conductor generates a magnetic field round itself. <S> The Rogowski coil samples the field, generating a voltage as the field changes. <S> It is weakly coupled to the main conductor. <S> As little, ideally no, current flows in the Rogowski coil, there is no effect back on the main conductor. <S> Any loops couple. <S> The sampling coil would generate a voltage whether it formed a closed loop around the main conductor or not. <S> However, that's not terribly useful, as the coupling, and hence the generated voltage, would change if the conductors changed position. <S> The sampling coil would also respond to any current changes anywhere in the vicinity. <S> There is zero coupling to any current that does not pass through the hole. <S> An incomplete circle, or a non-uniform winding, would not have these two desirable properties. <A> A Rogowski coil senses AC current. <S> This is a current transformer, and transformers couple... <S> This is certainly a transformer (one magnetic circuit, with two or morewindings), but <S> 'current transformer' implies a high coupling ( <S> nearly 1),and a low impedance on the secondary, both of which do not apply here. <S> Why can a coil no just be in parallel with a wire and measure current? <S> [as in, a helical coil with the helix axis parallel to the primary wire] <S> The coupling in a Rogowski coil comes from the inner semicircle partsbeing close to the sensed wire, while the outer semicircle of the wireis farther away. <S> Both semicircles are roughly parallel to the sensed wire, so they couple, but the inner half couples more strongly than theouter half. <S> Thus, the positive coupling on the inside and negativecoupling on the outside do not quite cancel entirely. <S> The 'return' wire is not parallel to the primary, and its path inside the R-coil isinsensitive to the current being measured. <S> The coupling of the R-coil isdetermined by that coil geometry without regard to the return wiresgoing to the integrator. <S> The 'helix-in-parallel' would have couplingdue to the wires from the endsof the solenoid as they are routed to the integrator. <S> Coiling (otherthan as it moves the wire path closer/farther from the primary) of the solenoid is not productive. <S> The R-coil does NOT have full flux coupling, so an open circuiton the Rogowski coil only loads the primary winding with thesame series inductance (maybe a microhenry per meter) as any wire inspace; a closed circuit on the Rogowski coil would only changethat by a few parts per million. <S> The EMF in the Rogowski coil is produced as a side effect ofthe self-inductance of the primary wire. <S> Using the side-effectmeans one can measure with no significant energy cost. <A> 1: The load impedance is high <S> so there is no burden to the source. <S> 2: L1 is not shown as a transformer. <S> The key to the circuit is the Mutual inductance , L which capture the magnetic flux that only flows thru the primary wire. <S> For peak of a square wave, V=L*di <S> /dt then integrated to become Vout = <S> I <S> * k for k = calibration const. <S> For a sine wave V = <S> ωLI the same units of peak, or RMS as desired. <A> more simple based on energy laws: there is no substantial difference (in theory) between a transformer and this. <S> Both used induction, BUT a transformer works with the scope to transfer energy, so typically efficiency is 90%. <S> this circuit is thought to measure, so follows the basic rule of measurement: do NOT modify what You want to measure, so do NOT sink energy. <S> And above circuit has so little induction that energy sunk is very small, and current sunk must be conveyed to an opamp, usually (so µA or nA are involved)
| The reason the Rogowski coil takes the form of a uniform toroidal winding is because due to the symmetric situation, it couples to all the current flowing within the toroid, regardless of the position of the conductor.
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Can I use electric tape instead of soldering for Speakers? I want to get a 3W 4Ohms Speaker for a bluetooth speaker project that I'm doing, but the speakers require me to solder wires onto them, while I don't have any experience in soldering. Am I able to use Electric Tape as a substitute, or are there any better substitutes, if any, that I can use? <Q> Get a soldering iron already! <S> Or if this is really the only time you would ever use it, have someone with a soldering iron solder the wires for you as a favor. <S> With the right equipment, it should only take a minute or two. <S> You are also confusing means of making a connection with the means of holding that connection in place. <S> Solder does both these things. <S> Electrical tape only does the latter. <S> Therefore, no, electrical tape is not a substitute for soldering. <S> Not recommended <S> You can strip maybe ½ to ¾ inch of insulation from each wire end. <S> Then run each wire thru the eyelet hole in the speaker lug, then twist it tightly around itself. <S> The twisting will tend to keep the wire in place, but it still won't make a great contact. <S> After you've done the best you can to electrically connect the wires with the speaker eyelets and to keep them there, you can wrap each connection in electrical tape. <S> That will protect the bare area from unintended electrical contact, and will help stabilize the kludgy joint mechanically. <S> While the above will largely work if done right, when the speaker is powered you can probably hear little crackling sounds as you pull on the wires or flex them around near the speaker. <S> Again, not recommended. <A> Yes wrap and tape wire connections are old as dirt. <S> The key though is to ensure the wire makes a good mechanical connection. <S> If the connection tab on the speaker has a wire-hole in it, strip enough of the end of the wire to let you poke it through then bend it back on itself so you can twist the wire back on itself. <S> The goal is to make that mechanical connection strong enough so the wire does not move on the tab when you wiggle it. <S> (And in this case, obviously, when the speaker vibrates!) <S> When you are happy with it, tape it up. <S> Note <S> : The connection will never withstand much pulling on the wire. <S> Tying the cable down somewhere with a tie-wrap to prevent pulling on those joins is a good idea. <S> Add: <S> If you have or can buy some heat shrink tubing that will work better than tape. <S> The shrink action adds compression to the join making it stronger. <S> If you use an inch or so it will also add considerable strain relief to the joint. <A> You can probably fit a screw terminal strip to the speaker lugs. <S> https://www.amazon.com/dp/B01MFEOTSW <S> these strips can be separated into individual wire connectors using a knife, so chop of two pieces and connect the wire using one screw and the speaker using the other. <S> ypu may need to trim the plastic back a bit at the speaker end of the connector. <A> Frankly, soldering wires together is pretty easy. <S> I don't really see any alternative connection technology than soldering here.
| No, electrical tape as far as I know it is an insulator , the opposite of what you want to have when you connect two cables.
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Replacement for 747 opamp (dual 741) I'm trying to find a "same or better" equivalent replacement for the 747 opamp. The footprint / pinout can be different, I don't mind. Using the parametric search on Arrow, while reading the datasheet for the 747 I found that an LT1355 might do as a replacement. However, they are quite expensive; $16.89 for one single IC (DIP package). Searching for LT1355 on Octopart shows that Arrow are the cheapest for this one, but the price is too steep for me. I'm fine around $8/pc. What other dual opamps can serve as a replacement for 747? <Q> Usually asking for products/parts is off-topic but since this can benefit anyone who wants something better than a 741, I'm making an exception. <S> The parts you list are waaaay <S> too expensive. <S> They likely have some special properties only few designs would need. <S> You should not be looking for a 741 replacement per sé <S> but more a "generic opamp". <S> There are very usable opamps to be had for much less money. <S> Like LM358, 10 for 1 Euro (shipping included!) <S> on Ebay other examples: MCP602, CA3140, TLC271 <S> Most of these have versions with 1, 2 or 4 opamps in one housing. <S> You will have to compare the datasheets to see what suits your needs. <S> Do pay attention to the supply voltage for these opamps as those can vary. <A> If your looking for a very quite op-amp for audio <S> then I would choose the TL071/TL072/TL074 series. <S> They have the same pin-outs as the 741/RC4558, with the TL074 being a quad package. <S> They cost about $1 USD each, less in quantities. <S> They are good enough for 16 bit audio, but not 24 bit or 32 bit audio. <S> Those type of op-amps are expensive. <S> The TL071 <S> /TL072/TL074 series are JFET op-amps with high-impedance inputs <S> so they do not load down their source, and you can use gain and offset resistors in the meg-ohms if you have to. <S> The outputs can source or sink about 5 mA. <A> The RC4558 is essentially a dual 741 ( <S> sans Vos adjust pins) in a DIP-8, SOIC-8 or TSSOP-8 package and was (and still is in some circles, at least the 'JRC' version) popular for audio work. <S> It does not have huge crossover distortion of the cheaper LM358. <S> There is a bit of crosstalk between the two halves that you may have to consider in some applications. <S> About 10 cents in 1K, Digikey lists one version of them at 35 cents in singles. <S> For many 741 applications you can use the LM324 (quad) / LM358 (dual). <S> They are very cheap, and the input common mode range includes the negative rail, and the output swings near the negative rail. <S> A pullup to +V can help mitigate the crossover distortion. <S> Generally in the same class as the 741. <S> For precision the rather old LT1013 is pin compatible to the LM358/4558, handles similar voltages and has precision but about 10-15x the cost. <S> There are op-amps with better (sometimes spectacularly better) specs than the above <S> however there are often a gotcha or two such as limited supply voltage (some op-amps are limited to around 5V or lower maximum total supply voltage) or high price. <S> For example, the TLV276x is recommended maximum 3.6V supply voltage ( <S> +/- <S> 1.8V if you need bipolar supplies). <A> I can only assume that you're using a 741 or similar because you need the offset null. <S> If you need zero input offset voltage <S> you can use a chopper op-amp like the OPA2180 . <S> Chopper opamps have an input offset of some microvolts.
| There are literally hundreds of suitable 741 replacement opamps.
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Proper name for "a switch-over time-delay relay" I have a joystick wired to a relay that spins a motor either one way or the other way. The problem is that I need to enforce a small "dead-time" when the user pushes the joystick quickly from one side to the other to avoid stressing the gear to much. What is the name for the component I am looking for? Its not strictly a time-delay relay since there should not always be a delay when the joystick is pushed, only when it is pushed from one activation state to the other. Is it a standard component or will I need to design my own? <Q> You can not get what you want in a single relay. <S> You will need two on-delay relays. <S> It should be easy to find appropriate relays and design a relay circuit that will function in the way you have described. <A> As you already suspected: I am not aware of a standard component that will do it. <S> I am afraid you have to make your own. <S> Of course a micro controller can do it, in discrete logic you mostly need to detect the change of direction. <S> From that you can trigger a one shot which keeps the signal to zero for a while. <S> Detecting the change of direction depends on what your joysticks outputs as signal. <S> For the one shot look up mono stable multi vibrator, NE555 or 74xx121. <A> I don't know what you building, but in the older electric crane/wench controllers they just use two relays that the coil for the other relay goes through the normally closed when not energized contacts. <S> This direction interlock circuit ensures protects fron short circuit conditions when swapping the volage around on a dc motor. <S> you would put these after the controller relays (with direction interlock) and before the motor contractors. <S> I would also be aware of arcing at the motor contacts, and arc suppression should be looked at. <S> Mostly, this is just at the main motor contacts where you would get a special relay that has a arc suppressor magnet in, or in some cases, gluing a small neodymium magnet on the case of the relay is sufficient enough to bend out the arc when it happens (if you can't get one small enough that has this protection) . <S> Now these days its better to use solid state as the longevity of these old relay based systems were never good. <A> It's called an anti-plug(ging) or "AP" relay but you may not find something suitable for a small motor. <S> A micro-PLC is probably your best bet, just write a few lines of ladder logic to make it happen.
| Internally they have dynamic braking, but in your case, where you not current braking the motor, you could use an extra pair of relays they call delay on relays.
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Why high impedance source in ADC input causes error? Recently, I had a problem with two sensors (LDR and LM35), when I tried to read them in Arduino ADC, both measures were totally wrong. Searching in the internet, the answers are related with high impedance of voltage divider provides to analog pin of ADC. So, I used a op-amp to provides a low impedance and result was great, but the doubt is still in my head.The sample and hold of arduino have a tiny leakage current (0.1 uA), i'd like to know how the high impedance source affects the ADC. <Q> A simple model to show how an ADC behaves is this: simulate this circuit – Schematic created using CircuitLab <S> There, you have a source (\$V_{\text{Analog}}\$) and its impedance (\$R_{\text{Source}}\$). <S> As well as the internal model for an ADC with a switch which is open and close every sampling period, and the ADC's impedance which is a combination of \$R_{\text{ADC}}\$, <S> \$C_{\text{Hold}}\$ and the leakage current. <S> The ADC also has a successive approximation register (SAR), which does a binary search to map the analog voltage value to a binary number. <S> Back to your question. <S> When the switch is closed, the capacitor, \$C_{\text{Hold}}\$, starts to charge and the time its going to take depends on the impedances and the capacitance value. <S> If we neglect the leakage for a second and resort to the well-known charging equation for a capacitor, you get: $$ V_{C_{\text{HOLD}}}=V_{\text{Analog}}\bigg(1-e^{-\dfrac{t}{\tau}}\bigg)$$ <S> Where \$\tau=(R_{\text{Analog}}+R_{\text{ADC}})C_{\text{HOLD}}\$ <S> The time it takes to fully charge the capacitor is about 5\$\tau\$. With that, the problem arises when the sampling time is not enough to allow the capacitor to fully charge. <S> For example, say your ADC samples (\$T_s\$) <S> every 1ms, and consider your source impedance \$R_{\text{Analog}}\$ to be very small compared to \$R_{\text{ADC}}\$. <S> In that case, it could make sense to approximate \$\tau\$ to be: $$ \tau\approx (R_{\text{ADC}})C_{\text{HOLD}} <S> $$ And just to make up an number say, after plugging in values for \$R_{\text{ADC}}\$ and \$C_{\text{HOLD}}\$, \$\tau\$ turns out to be: $$ \tau\approx 0.15\text{ms} <S> $$ <S> Since the sample time is greater than 5\$\tau\$, that is plenty of time for the capacitor to fully charge, and your ADC should capture the correct value. <S> At sampling period, the switch is opened. <S> Now, as your source impedance starts to increase, so does \$\tau\$. Say your source impedance <S> , \$R_{\text{Analog}}\$ increases so that now you can't neglect it and say this leads to \$\tau=1\text{ms}\$. Under this condition (\$\tau=T_s\$), every time you sample, the capacitor value will be at approximately 63% of the actual value of the analog source. <S> That would lead to an incorrect measurement. <S> It's about giving it enough time to charge up the holding capacitor and the source impedance, as it increases, prevents you from doing so. <A> - lack of shielded twisted pair cable <A> When a "cheap" ADC measures a signal voltage, an impulse of current is taken that can produce an error. <S> This happens when the signal source has a relatively high impedance (usually several kohm). <S> Take a look at the data sheet for the ADC you are using and forget about static leakage currents and concentrate on what the DS tells you about the current "surge" it can take when sampling a signal voltage. <S> Putting an op-amp between signal and ADC resolves this issue greatly because an op-amp can present an input impedance that is tens of Mohm whilst being able to drive tens of mA into the ADC when it samples thus, the sample and hold capacitor (inside the ADC) is fully charged to the signal voltage.
| Active Sensors sometimes give rise to errors from ; - EMI induced on unbalanced impedance of 2 long wires - differential voltage with 1 side as a shared ground shifted by other sources - stray noise coupled unevenly to the differential input or - lack of noise suppression capacitance.
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Difference Between Waveforms When Using An X10 Attenuator I noticed a thing that I cannot explain on my GDS-1052-U scope: while the probe and the oscilloscope are set on 1x the noise from the LM317 power supply is about 5-6mV pp and while the probe and the oscilloscope are set on 10x, the noise from the same power supply (a LM317 power supply with PNP pass transistor) is about 50-60mVpp (see the images below, in the first image the probe and scope are set on 1x and in the second image the probe and the scope are set on 10x). Is that normal ?I tried to use another probe but it does not change the amplitude of the noise. 1x probe: https://imgur.com/tV0kUqD 10x probe: https://imgur.com/nyeucoi <Q> This means the noise is the oscilloscope input noise. <S> The noise voltage is the same at the oscilloscope inputs in both cases, but the displayed value is multiplied by 10x when set up to use a 10x probe. <A> It must be scope noise and/or cable pickup. <S> I have a Tektronix TDS2024 just back from calibration <S> and I see the same thing with the tip grounded to its ground clip - 1.6mv p-p on 1x and 16 mv p-p on 10x. <S> This would make sense since 10x probes only attenuate the signal, and the scope just changes the scale accordingly. <S> A millivolt or two of noise is pretty good, and you can use 1x to measure small signals for this reason. <S> So, the answer to your question must be: yes, this is normal. <A> Usually probes set to x1 have much lower bandwidth than when set to x10. <S> More or less it goes like this: $$V_{noise(rms)}= (noise~voltage~spectral~density) <S> \times BW_{probe}$$ assuming a flat spectral density and a flat bandwidth (otherwise you'd need an integral). <S> Hence the more the bandwidth of the probe, the higher the RMS value of the noise voltage, hence the higher the peaks of the noise. <S> This is an excerpt from GWinstek GDS1052-U manual <S> (p.126 - emphasis in yellow is mine): <S> So, assuming you are using one of those probes, in x1 position you have a BW which is more than ten times smaller. <A> This is a result of the attenuation ratio you are using. <S> At a 1:1 probe ratio, you only get 1x the oscilloscope's front end noise. <S> At 10:1, you get 10x <S> the oscilloscope's front end noise. <S> The 10:1 probe attenuates the signal by 10x when it comes to the scope, and the scope then has to add in a x10 multiplier to get an accurate reading. <S> It gives you better isolation than the 1:1 probe, but more noise.
| It is possible that you are getting more peak-to-peak noise at oscilloscope input because the probe set to x10 has a greater bandwidth, hence it lets more noise reach the oscilloscope input.
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Will two capacitors with different voltage ratings and identical capacitance work the same? Suppose I have a 200 V 200 µF capacitor and a 16 V 200 µF capacitor, and suppose I have a circuit that requires a 16 V 200 µF capacitor. Can the 200 V capacitor be used instead of the 16 V one and will they work the same? Apart from the obvious overkill are there any disadvantages (speed, power dissipation... certain limitations)? <Q> They both would work if they are within the voltage range. <S> However, they do not work the same. <S> For ceramic capacitors you have to consider the bias voltage (VDC). <S> That is, when increasing the voltage the capacitance decreases. <S> You can see the difference between two 100 µF capacitors, one rated at 25 V and the other to 6.3 V. Both would support 5 V, but the capacitance of the 6.3 V capacitor would be lowered (around 50%). <S> The bias voltage is actually dependent on size rather than the capacitor voltage level, but it is true that higher voltage levels usually means higher volume. <S> Aluminium electrolytic capacitors are not as vulnerable to the bias voltage problem as ceramic ones. <S> But capacitors with a lower voltage rating usually have higher ESR . <S> That is something that might be considered (especially if there are no ceramic capacitors that have a very low ESR around). <S> Also, tolerance values are usually worse for a higher voltage level. <S> So in general they would work the same for most applications. <S> But there are some issues to consider when choosing the voltage level. <S> The safest choice would most likely be to have just the voltage level needed for your application with some margin. <S> But going with higher voltage levels, like in your case, is not usually a problem. <S> It is more critical when going for lower voltage levels. <S> Here you have a small recap of different capacitor types. <S> Sources: <S> Graph: Murata Electronics. <S> Table: NIC Components. <A> As long as the working voltages are within range nothing will explode. <S> But be aware that for certain capacitor types (especially high voltage aluminum electrolytic), the tolerance tends to be worse at high values of voltage and capacitance. <S> HV caps can also exhibit more "memory effect" than their low voltage counterparts. <S> For the values you stated, all of the above effects will be minimal. <S> But as others have noted, the most noticeable disadvantage is size. <A> You will find almost every Alum Electrolytic cap improves on Dissipation Factor with rising breakdown voltage rating. <S> This means lower D.F. = <S> lower tan <S> δ = lower ESR ( all equivalent yet different units related to max ripple current rating) <S> However you can also find a wide range in e-caps spanning a 100:1 range in ESR*C=T products. <S> This means one could find a 16V cap that can handle more ripple current than the worst 200V cap at the same C rating. <S> Will it work the same? <S> Almost. <S> But that depends on what "same" means. <A> 200 V 200uF capacitor means, it can tolerate upto 200 V which is its dielectric breakdown voltage. <S> So it can be used for the mentioned application.
| Of course, besides dimension differences, there are different parameters that change when having different voltages and also depending on the capacitor type.
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AC induction motor controlled by TWO TRIACS. They blow down I am getting crazy with a board intended to drive an AC induction motor, which has two symmetrical windings (for the two run direction), a center tap, and requires a run capacitor. The motor is rated about 900 watts, it can draw maximum 5 amperes, but untilnow I used it with no load, so 1-2 amperes are consumed. The idea is to drive a single TRIAC at a time, in the usual way, modulatingthe pulses in order to control the speed (a rotary encoder provides feedback). Just to be clear, speed control/braking/reverse work well, my problem is that the board is not reliable, the TRIACs melt quickly even without big solicitations, but when they are sane, the results are good. The schematic is the following: By driving one or the other TRIAC it is possible to command the running direction, and also decelerate (when pulsing the TRIAC opposite to therunning direction of the motor). The circuit works as intended, but is very fragile. If the system is powered with low voltage (30 volts AC), everything is smooth; if the system is powered with the intended voltage (mains 230 VAC), it survives tens of seconds and then fuses blow and TRIACs burn. If I mount only one TRIAC everything works well (of course, in one direction only): it does not burn. Mounting the second TRIAC, the opposite tap of the motor is no more free but it is connected to the circuit. Apparently the opposite tap develops high voltage and spikes which need to be canceled. I used an oscilloscope to analyze what was happening: I was looking forcross excitation of the TRIACs, extra voltage spikes and so on. Everythingseems normal, but the TRIACs keep to blow up. They are rated 800 volts. I then tried to protect them using two 750V varistors in parallel to the anodes. The varistors heat a lot, and this suggests me that there are high voltages running around, even if I don't see them with the scope. The next move has been to use 1.2KV rated TRIACs (don't havethe part number at hand now). Things go slightly better, but when the TRIACangle, from "low" power (near to right right end of the semi-cycle) is increasedto more power (near the middle of the semi-cycle), the TRIACs blow again. Fuses blow also, but no other components are affected. My thought is that first aTRIAC fails (or two TRIACs fail), then the short-circuit blows the fuse. When the board fails, it seems that both the TRIACs melt, but I amnot very sure of this - I mean, until now I've never seen a single TRIACblown, always two. I think I am missing something: to blow a 1200 volt rated TRIAC, a 1200 voltvoltage is necessary! I understand that windings can develop high voltage,but I can not see it. I tried to power that part of the board with 21 VAC (rms) from a transformer. The resulting sinusoidal wave has 67,2 volts peak to peak. The motor runs in the expected way (of course with very little torque). In no way I can see any voltage over 88 volts between any 2 points of the circuit. A pretty high voltage (peak to peak) is found across the motor capacitor: it can be as low as 44 volts with "low power" TRIAC angles, up to 85 volts (peak to peak) when exciting a TRIAC near the middle (top) of the semi cycle. Calculating some proportion: if instead of 21 VAC, I use 230 VAC, then I should have around about 800/900 volts peak to peak. They would not be sufficient to burn the 1.2KV TRIACs, but it happens! I don't know what to do. Somebody can help me please? Thanks, (many) thanks in advance. <Q> Can you add more details about the PWM driving? <S> min /max duty cycle and frequency?Also:-disclaimer <S> no experience with triacs but many experience with power mosfets-Did you measure the triacs after failure? <S> With mosfets, overvoltage failure usually shorts gate to source, while thermal failure only shorts drain-source. <S> Possible overvoltage causes are caused by inductance as you imagined, thermal problems can be caused by ringing. <S> The advantage with fets is that you can play with gate resistance (hence dV/dt of output) which you can't do with triacs. <A> When switching off your motor is acting as a generator. <S> The triacs try to block current, which leads to voltage spikes. <S> A very safe solution is to add a braking circuit. <S> I've tried that with motors blowing up VFDs used with long conveyor belts (lots of inertia) - the motors were pumping supply above input voltage and above max voltage of the VFDs. <S> When you switch off the triac you have to connect a resistor [bank] across your winding to safely dissipate the power. <S> This will prevent energy blowing up your triac. <S> You can try different resistor values for different braking speeds. <S> Of course you can try a "bigger", more expensive triac, but braking may be more cost effective depending on your exact machinery. <A> If it's an AC induction motor, which it sounds like from your description, you can't expect to reliably control the speed this way (there are exceptions for tiny motors < 100W, in ceiling fans or HVAC pumps). <S> You need to control the AC frequency : the controller you need is a Variable Frequency Drive. <S> (Most of these are designed to control 3 phase motors; you need one that can drive a single phase (or split phase, that's what the capacitor is for) motor. <A> You may be using 'snubberless' triacs <S> but you still need to include a snubber to keep the voltage within limits when driving an inductive load. <S> see section 1.2 of this app note: http://www.st.com/content/ccc/resource/technical/document/application_note/38/88/44/b8/2c/26/44/b8/CD00004096.pdf/files/CD00004096.pdf/jcr:content/translations/en.CD00004096.pdf <A> At 230V and standstill of the motor, you have a gate current of 230mA to the triac. <S> Peak gate current is 4A for the BTA16, but it's better not to overstrain this. <S> 100mA is sufficient in all cases. <A> You don't need to wait for the load to coast to a stop, but you should allow time for the motor's magnetic field to diminish. <S> My guess is that 150 to 500 milliseconds would be sufficient. <S> Reducing the voltage as a means of speed control is actually reducing the torque capacity of the motor. <S> Look at my answer to this question for additional information. <S> There you will see why this method of speed control is usually used only for fan and centrifugal pump loads. <S> The motor will tend to stall if the speed is reduced too much. <S> Since speed reduction is achieved by reducing the motor's torque capacity thus allowing the load to make the motor slow down, there will be little or no speed reduction at no load or with a very light load. <A> Not sure if you were able to resolve this problem. <S> I had a similar problem and was a able to solve it by adding a 4.7mH inductor in series with the Triac connection to the motor. <S> I don't think you need such high ampere triacs.
| It can be used for constant torque loads over a very limited speed range, but there will be a risk of overheating the motor. But you could try snubbers to see if it makes a difference. In addition to adding a snubber circuit as suggested by @TEK, you should prevent switching directions without some off time in between.
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7 segment binary to hex I'm trying to display a 4-bit binary number on a 7 segment LED as a hexadecimal number (0-F). I have an assortment of 7400 series ICs including the 7447. But that one only works for BCD (0-9). The 7400 series doesn't seem to have a hex to 7-segment decoder and I don't have one on hand. So I figure I'd have to build my own. The datasheet for a 7447 comes with the internal circuitry but I couldn't find a similar circuit for hex to 7-segment. I did K-diagrams for each LED segment by hand but the terms are rather large, much larger than for BCD. Does anyone have a finished circuit for this that I can check my work against? Does anyone have an optimized circuit for this that takes advantage of common subterms? There aren't many duplicate terms (like A0 & ~A1 & A2) but maybe using (A0 & A2) & ~A1 would allow sharing the (A0 & A2) subterm and overall reduce the gate count. Or maybe some tricks to use NAND, NOR or XOR gates for some parts? I don't care about different path length or races in the circuit as it's only going to drive LEDs and should be far too quick to see any of that. <Q> Find what museum is missing their 74xx logic ICs and return them. <S> Then get a small microcontroller and do all this in a single chip. <S> As a bonus, you'll be ready for other advanced projects from the late 1980s and beyond. <A> /drive the stuff. <S> :) <S> Here's a crude schematic of the whole contraption; If you happen to use some highish efficiency LED displays, you may even be able to drive a common anode one directly. <S> The '! <S> EN' pin can be either of the gate pins of the '138. <S> You can also connect both of the other gate pins together, letting you turn the display output on or off as you please. <S> This also has the benefit over the suggested ROM solution that it's inexpensive to build/prototype if you have the parts on hand, or can't be particularly bothered to program ROMs. <S> Besides, one could even call this fun to make :) <A> If you want to fiddle with gates, another possibility is to use a small FPGA or CPLD. <S> You can get started with a flash-based FPGA for a few tens of dollars. <S> Reprogram it as many times as necessary to get it right, and you generally get a lot of I/O pins per dollar. <S> Example <S> VHDL code here <S> For example, (shaky and not quite a complete loop, but you get the idea): <A> But that one only works for BCD (0-9). <S> Actually 7447 physically can accept hex values of A-F, but they display wrong images for them, see here page 3. <S> In my opinion, making hex-to-seven-segment decoder might be a good for education and visualization of how digital electronics work, <S> but it is really bad idea from practical point of view <S> - it is unnecessary complex and consumes much power in comparison to other solutions. <S> I recall having a PC diagnostic board back in ~1995 which was displaying error/status codes in hex (something like this ), and this board used GAL chips - predecessors of CPLD. <S> If you have searched over internet for the solution, you might have seen a lot of similar projects using programmable logic or MCU-based devices like Arduino . <S> @dirac16 suggested another way - using parallel ROM chip, a kind of lookup table with bytes in it representing segments being on/off for respective address (input value). <A> MC14495, DM8880/9368, V40511, D345, D346, 4311, 4368 or 74C915 <S> All of these are hard to get your hands on nowadays. <A> I think I have managed to find the optimal solution (one of them) now. <S> It uses all 24 gates of 6 ICs: 2x 7486 (quad XOR) <S> 1x 7408 <S> (quad AND) <S> 2x 7402 <S> (quad NOR) <S> 1x 7432 <S> (quad OR) <S> I've managed to find different solutions using only 23 gates but more ICs. <S> I have a feeling there isn't a solution with just 5 ICs. <S> Note: The circuit outputs LOW when the LED should be on since my 7 segment LED has a common VCC. <S> simulate this circuit – <S> Schematic created using CircuitLab D3-D0 are the inputs with D3 being the MSB. <S> A-G are the outputs with A being the top LED, then going clockwise around and last G being the center LED. <A> If you want to design the functions by hand an check them for correctness before (or instead of) implementing them in hardware, you can use an online calculator like this one . <S> There are also tools which can minimize multiple logic functions sharing common inputs. <S> One is expresso , which has free implementations in Windows and Linux. <A> It's been said you can try the ICM7218 or some other chips. <S> You may also try to build your own combination circuit. <S> This is a great video tutorial on how to do it and its mechanics. <S> If you want to optimize it, there are some datasheets ( 1 , 2 ) with the BCD (0-9) part. <S> As a follow-up, this guy also made a follow-up video on how to do the same with an EEPROM. <S> Granted, you still have to program it, but it may be worth it. <S> And talking about programming, you may also do it all with a microcontroller. <S> It is usually easier to program than a CPLD ( which you may also use ), and there are both code and tabular implementations. <S> I'd use this option, since I'm familiar with microcontrollers (you're not limited to the Arduino) and for just the chip <S> it might even be your cheapest option. <A> Use Google to find "hex 7-segment" items, including http://en.wikipedia.org/wiki/Seven-segment_display which has a hex truth table (in hex!). <S> Other items include diagrams and video demonstrations. <S> I was actually looking for four-digit, seven-segment displays that do the hex decoding for you, but they are considered to be "ancient" (sniff) and modern ones use i2c rather than parallel interfaces. <A> I've been using the DM9368N for that purpose for years now, and it's reasonably easy to get (15$ for a 10-pack on ebay or Aliexpress).
| Maybe, if you feel particularly adventurous, you could even use a diode ROM, with perhaps two 74'138 3-to-8 line decoders to decode
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Attaching a resistor to a oscilliscope probe I'm doing some ultra low power electronics, and I need a probe with an impedance higher than 10Mohm. Is it alright to just attach a 90Mohm resistor to the end of my 10x probe to turn it into a 100x probe? The signals are all under 1Mhz. Are there any problems that could arise? <Q> As an alternative to doing this, you might want to consider making an active voltage follower using a low Ib op-amp. <S> Even an inexpensive TL071 will have something like 1T ohm input resistance and input capacitance of if you stay inside the input voltage range <S> (eg. give it something like +/-10V supplies and stay within +/-8V. <S> Bias current is under 100pA typically at room temperature so 10M ohm source resistance will contribute negligible error (in comparison to the offset voltage). <S> There are much lower Ib op-amps available (down to fA). <A> New resistor and old capacitance make a lowpass filter. <S> The pulses get rounded due hf attenuation. <S> You should add also a small capacitor in parallel with the resistor. <S> Learn how normal probe compensation works. <S> You can still use oscilloscope's own square wave output for probe adjustments and checking the new scaling. <S> You also lose possible automatic probe type detection and automatic scaling. <S> No obvious other problems. <S> Add the resistor as near as possible to the measured point. <S> Do not try to insert it at the input of the oscilloscope. <S> Otherwise the inserted resistor doesn't reduce the reactive loading of the measured signal. <A> If you want to modify a standard 10:1 passive 10 MOhm probe, you will run into bandwidth issue. <S> Best (Keysight, Tektronix, etc) probes would have about 10 pF input impedance, 10^ <S> -11 F. <S> With a 100M resistor (10^ <S> 8 Ohms) <S> this will form a RC filter at about 1ms (10^8 <S> * 10 <S> ^-11). <S> Which is about 170 Hz at -3dB level. <S> Keysight/Agilent has a 66 MOhm/3pF/500MHz probe, model 10076C . <S> It is however 100:1, so signals have to be strong.
| A 1MHz probe with 100 MOhm impedance needs to be carefully designed, there is some literature on this.
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Is a diode's forward voltage stable enough to be used as a reference voltage? Assuming a resistor is in series with a diode (say, a 1N4001) so that the diode has around 1mA of current, will its forward voltage be usable as a voltage reference to within 1% (relative to itself over time and with ambient temperature changing by 10F max)? The voltage doesn't matter, as long as it doesn't change to a noticeable degree. <Q> Is a diode's forward voltage stable enough to be used as a reference voltage? <S> No, it will change with temperature: - Picture taken from here At 25 degC and a bias of 5.1 mA (for example) <S> the voltage will typically (that's typical and not always) be about 627 mV. <S> At 30 degC <S> the voltage will have dropped to about 618 mV - that's a change of 1.4% for a minor difference in temperature. <S> In voltage reference terms that's a stability of 2878 ppm/degC. Compare this to just fairly cheap voltage references having a drift of 50 ppm/degC. <S> Assuming a constant, preferably low current (1mA) is applied to a diode <S> If you have a constant current source then apply it to a resistor and you'll get pretty much stable results compared to the diode. <A> I have seen some instrument schematics, that use LED diode as power indicator and voltage source. <S> Could the LED be a better alternative? <S> Search for current, voltage source schematics examples, perhaps you will the answer. <S> P.S: The "instrument" is the LeCroy scope differential probe, not a toy. <S> Source : http://www.diyaudio.com/forums/equipment-and-tools/248505-differential-probe-reverese-engineered.html <A> Some opamp circuit is needed to extract the CTAT. <S> You combine the PTAT and a proper amount of the CTAT, and extract a very stable voltage reference. <S> If you do this, please honor Paul Brokaw for the invention.
| You can design BandGap voltage references, using multiple diodes running at different currents (thus different current densities and thus difference temperature coefficients.
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Combining two exact USB battery packs in parallel to increase current in a simple fashion? I am wondering how I could put two USB battery packs in parallel together in order to increase their amperage/current. I do not want to tamper with the Battery packs, as breaking them apart and soldering a new connector is out of the picture for the scope of the project, so I was wondering if I could do the following to accomplish the task: Attach USB cables to each of the two exactly same battery packs, cut off the positive and negative ends of the USB exposed cable and solder positive to positive, negative to negative. I would then solder the input red/black wires of the item I want to power to the now negative-negative positive-positive ends of their respective USB wires. Would this work? Can a quality USB cable be able to transfer 4.8 amps? <Q> No, you cannot just parallel two USB outputs and expect to double the current capability. <S> The USB battery packs work using a DC-DC invertor to provide the 5V/2A output, and if you measure the 5 V produced you will find differences between the units ( <S> just from the variations in components used).Since one unit will have a higher voltage than the other, it will supply more current. <S> How bad the mismatch will be is unknown. <S> Since both units will have a negative output voltage slope .... <S> as load increases the 5 V will drop, it is possible that they could balance at a point. <S> That does give you more current than a single unit could supply, but there is no way to know when any given USB pack may shutdown, so you are unlikely to be able to produce consistent results. <A> Would this work? <S> Can a quality USB cable be able to transfer 4.8 amps? <S> Yes, you'll get a voltage drop and heat up the cable. <S> USB wires are typically 28AWG that is 213mOhm/m <S> Consider a 5V source connected to a shorted 1m usb cable. <S> The resistance will be <S> 0.213*2Ω (you have to return through the ground, the resistance is double). <S> With that 0.426 of cable resistance connected to a 5V source, you will get 11A of current (and <S> ~50W of power dissipated in the cable, which will probably ignite it, or at least melt insulation). <S> This is your best case current transfer scenario. <S> But don't try it with 28AWG. <S> At 4.8Amps there would be 4.8^ <S> 2/0.426=10W dissipated in the cable, which is still quite high. <S> Lets say you find a USB cable with 24AWG wire (I don't know if these exist) in it which has 84.22mOhms/m of resistance. <S> At 4.8Amps there would be 4.8^ <S> 2/0.16=3.68W dissipated in the cable and at 5V <S> They do actually make 20AWG usb cables, which only would be 4.8^2/0.06=1.38W dissipated in the cable at 4.8A. <S> Another way to avoid cable losses is to up the voltage, but if someone were to accidentally plug in a device to a USB port with a higher non standard voltage, there would be smoke. <S> So choose a bigger cable, there will be less power loss and more for the peltier. <S> You will also get voltage drops from the resistance in the cable, but you are probably more interested in power transfer. <S> One problem with USB is you can't force people to use bigger cables, someone could accidentally plug in a smaller cable. <A>
| This is a no, as I have read onward that a single usb wire will not be able to accomplish the task.
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Reduce time to stable value in PID I am working with PID controlled temperature controller ( SELEC TC544 ) . I have set P = 5 / I = 0 / D = 3. System get long time to stable at my set point. How to reduce that time? <Q> Before you mess with the system make sure it's safe before you start playing with gains. <S> If it overshoots it could have disastrous consequences if the system isn't bounded (either physically or with the controller output). <S> If you need to, then increase the integral term. <S> Rarely do you need to use integral, but a small amount can help reduce noise in a first or second order system. <S> I'll bet that your system could benefit from some integral. <S> Below shows a graph of a system that shows an example of what different gains do. <S> There are also a few methods like Ziegler-Nichols that can be used to efficiently tune PIDs. <S> You may also be limited by control authority, meaning if your system hits the limits or rails (like if the thermal system is warming up and your PID is output 100%, or if your heater or peltier is at max power) then you will need to increase the control authority (or heater power). <A> Fast answer, you need some integral action. <S> The I in the PID usually gives a faster response but less stable. <S> I have seen in the data sheet of your controller that they provide some default values for the PID, where P is 10 <S> , I is 120 and D is 30. <S> Have you tried those values? <S> If it safe to try they may be suitable. <S> Note that I and D correspond to time values. <S> So, when you say you set P=5, I=0 and D=3, are I and D representing time as shown in the data sheet? <S> Since it is not the same Ki and Ti. <S> Either way, I recommend some integral action and compensate it with the derivative, since most likely it will not be a high order system. <S> For the record, the correct way would be to get the transfer function and apply some easy method like Ziegler-Nichols like other colleagues suggested. <S> But if you do not know how to get the transfer function or the response graph of your system. <S> You can just first tried as I said. <A> You should really understand how a PID system works, if you don't know. <S> Anyway, that I=0 (integral component) is probably wrong. <S> I took a look at the manual of your controller, but I can't understand what is the effect of having I=0. <S> Consider that without integral component, a system can not reach the setpoint (the target). <S> If the I component is too aggressive, the system gets unstable. <S> You should at minimum tell us whether your system is unstable (too much integral part / too quick) or if it always heats too little (in this case, more I is needed). <S> Giving that in your controller <S> the I parameter is given in seconds (time), and heating systems are fairly slow, I would suggest to set something like 120-240 seconds, or even more. <S> The P parameter can be fairly high in this case (heating system): the manual suggests 10, it is a starting point, but you can try to augment it. <S> The D component is used to stabilize the system - leave it alone or set it pretty high (long times): only if you see excessive response from the system you can diminish the time, giving the D component more importance. <S> But usually it is easier to <S> not use it, at least at the beginning.
| Try turning up your gain, realize that if you do so you will sacrifice stability.
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Linux on board without 3rd party modules. Where to start? I would like to be able to design my own Linux board the way I design my own micro-controller boards. The complexity is obviously going to be higher from PCB design and assembly point of view, but I can manage that. What I don't understand is the software part. What all does it take to get a get Linux running on a CPU. Are there some choices that make your life easy?. Such as choosing a processors where manufacturer provides XYZ software components. Actually, to begin with I don't even wan't to be able to go full-custom with processor choice. I am happy to stick to one processor (or its variants) as long as I can put it on a custom board I design. <Q> I understand that you want to know how to make a processor work with Linux without having so much trouble. <S> I will give you an answer based on my experience, but that has always worked so far, since I believe it could be useful for you. <S> Therefore, it is quite specific for one particular case, that you can copy or take as a reference for a different processor. <S> If it is not what you look for, it is no problem :-) <S> Boot (like U-Boot) <S> for the system and the file system . <S> I will refer to the whole thing just as kernel from now on. <S> First of all, I recommend to use a tool that would automatically give you all the kernel to make it easy. <S> The following image shows a full kernel mapped into a Flash, so you get an idea. <S> Instead of having to get each of the layers one by one, these tools give the complete stack at once for the chosen architecture. <S> If you can choose your microprocessor (of course it has to support Linux). <S> I recommend to go with the AT91 family . <S> It includes all the SAMA5Dx boards as well as AT91SAM9x5. <S> They are some of the simplest processors that can run Linux. <S> In this link you can see step by step how to get your full kernel with Buildroot. <S> Remember that you can decide to boot the kernel from a Flash or from a SD card (or micro SD). <S> Both methods are explained in the guide. <S> Furthermore, these boards have an evaluation kit where you can check that your kernel works <S> and you learn how to implement it in advance. <S> You will have the schematics of the evaluation kit and PCB design guide to make it easier. <S> I do not know if the answer would be useful for you, but I thought it might be. <S> You can always look for similar guides from other manufacturers or try to go with Buildroot for a specific processor from the beginning. <S> Best of luck! <S> Sources: <S> AT91 Linux4SAM <S> Buildroot <S> Yocto/Poky <A> The CPU isn't going to be the hard part – you just set up a cross-compiling toolchain for it <S> , that's it –, it's the peripherals built into the chip which give you a headache. <S> You need a proper device tree for that <S> and this is going to be hard, manual work. <S> I recommend to make this device tree as simple as possible. <S> Some RAM and Flash definitions, a means for a serial console, nothing more. <S> So you are able to see kernel messages as it boots. <S> Another hard obstacle is having a boot loader supporting the SoC you use. <S> It's basically the same thing again but less flexible than the Linux kernel. <S> Again, you have to tweak an existing boot loader <S> so it has the right settings for RAM, Flash and a serial console. <S> If you want to make your life simple, choose an SoC for which a Linux development environment already exists. <S> Look up the OpenWRT project, it has dozens of SoC to choose from. <S> Some may be freely available. <S> As soon your kernel boots, you can continue with the next, tedious work. <S> Compile libc and all the software you want to use for the target CPU. <S> While its compiling, think about additional device drivers you need. <A> You need: A kernel built to support the target architecture, including endianness <S> A means of actually booting the kernel on the target platform (e.g. U-Boot ) <S> Drivers included in the boot kernel or kernel initrd for required hardware (e.g. serial console; see also "device tree") <S> The hardware itself needs to have sufficient RAM to run Linux. <S> It's not a particularly simple project. <S> It might be best to start from someone else's design, e.g. https://www.olimex.com/Products/OLinuXino/open-source-hardware
| When it comes to the software part, as some colleagues suggested, you would need a Linux Kernel that supports the chosen microprocessor or architecture, a way to I recommend Buildroot or Yocto/Poky . Finally, as some others said, you would need enough RAM, but that is also shown in the schematics.
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How to ground a 2 prong plug electronic? I am installing a strip of 110v led strip outdoors and placing them on an aluminum channel. I will be cutting the strip, waterproofing them and placing different sections of the strip throughout the channel. What I am worried about the most is eventually the weathering may/will cause a short or ground fault, and it will become dangerous to touch the aluminum channel. From what I see the led strip is powered via mains to a rectified 110v DC, and only comes with a 2 prong plug. Is there anyway to wire/rewire the led strip such that it can detect a ground fault from a 3 prong outlet. <Q> This will protect anyone who might be subject to electric shock. <S> [Edit] <S> A GFCI works by monitoring the current difference between the Line and Neutral lines. <S> If a leakage path develops from the Line lead to Ground, the GFCI detects that difference and trips. <A> Since this is outdoors you probably should not mess with the wiring. <S> The driver itself needs either to be meant to be used outdoors or enclosed in a hermetically sealed electrical box. <S> Since it does not have <S> it's own ground <S> you should take steps to add one. <S> Tap and wire a separate ground line either back to the outlet or direct to another grounding point (gas meter, metal water pipe). <S> Wherever your channel is broken, say to go round corners, you need a ground strap to electrically join the parts back together. <S> Ultimately, you would be better to use low voltage LED strips that are intended to be used outdoors. <A> You can use a three-conductor cable or the two-conductor cable you have with an additional separate earth conductor. <S> Route live and neutral to your LED strip. <S> Terminate the earth conductor to a crimped, not soldered, ring terminal. <S> Drill a hole in the channel. <S> Clean the hole, surrounding area, and terminal with fine abrasives. <S> Assemble the ground stud onto the aluminum and seal the area with outdoor silicone. <S> Be sure to use zinc-plated hardware to avoid galvanic issues caused by dissimilar metal contact in the outdoor environment. <S> Use a GFCI as well. <S> Is this up to code? <S> I don't know. <S> Is it safe? <S> Yes. <S> The intended function of the earth connection is to ensure that large current flows when a short circuit occurs, causing protective devices like circuit breakers to activate. <S> The intended function of circuit breakers is to prevent fires caused by overheated wires, but they have the added bonus of ensuring that grounded equipment cannot become energized and pose a hazard to humans. <S> A GFCI trips if current flow to earth is detected, such as may happen while someone is receiving a shock, to protect human beings.
| The easiest way to protect against leakage to the aluminum channel is to install a Ground Fault Circuit Interrupter (GFCI) - also known as a Residual Current Detector (RCD) in the AC power supply to the LED strip. Clean the whole bond area and all hardware with alcohol.
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EEPROM with high endurance I am currently working on one embedded project in which I have one counter which will be active all the time. If the power goes down then also I have to store last counter status and load it back in next startup. So that I was planning to use EEPROM in which I will continuously write my counter value.But I have read somewhere that EEPROM has read/write endurance at about 100,000 and I will be updating that counter probably 120,000 per 24 hours.So I am finding alternatives to accomplish this task.please give me your suggestion for doing the same. <Q> FRAM doesn't suffer from the same limitations on write cycles as EEPROM. <S> Some of the MSP430 products from TI are available with FRAM, here's a link to an application similar to what you describe: <S> MSP430 with FRAM save state on power failure <S> Here's the Wikipedia article on FRAM: FRAM <A> I have this issue in a current project. <S> The way I'm dealing with it is to keep the live value of the counter in RAM. <S> I added a little bit of hardware so that the microcontroller can detect that the raw input power voltage is low. <S> If so, it stops what it's doing, saves the live counter value in EEPROM, then waits watching the raw power voltage. <S> If it goes back up, with some hysteresis, then the micro essentially restarts. <S> Otherwise, if power continues to go down, the micro will eventually get stopped. <S> On the next restart, the counter value is loaded from EEPROM, then used live in RAM again until the next power-down. <S> It doesn't take long to write a small value to EEPROM. <S> Most likely your existing power supply system has enough energy storage that you can detect the voltage going low, and still have enough guaranteed run time before power to the micro <S> goes below the operating or EEPROM write threshold. <S> In my case, the only additional hardware was a Schottky diode to prevent the DC power supply from sucking charge from the local reservoir on the way down, and two resistors as a voltage divider so that the micro can read the raw input voltage. <S> The rest is firmware. <S> It's important to note that you should be watching the voltage on the input to whatever final supply powers the micro, not the micro's power voltage directly. <S> By the time the latter goes low, it may be too late. <S> Hopefully there is a voltage range that is below the worst case when everything is operating correctly, and above what the micro's power supply needs to guarantee regulated voltage to the micro. <S> In my case, the micro's supply was a buck regulator fed from 48 V, so there is a large range that is below normal but where the micro can still operate reliably. <A> Old old old solution, cmos counter + lithium battery or Ram + lithium battery. <S> The power supply for the storage element comes from the normal power supply when it is available or the battery when it is not. <S> A lot of modern micros in sleep will maintain their state with a very low current supply. <S> So you can use this technique with power down detect to go to sleep then use a battery to maintain state during the sleep period while the main supply is off. <A> Microchip has a series of I 2 C "EERAM" parts which will allow data to be stored in SRAM, then write it to EEPROM (using energy stored in a capacitor) <S> when power is lost, to be loaded when power returns. <S> This seems like it would be perfect for your application. <S> A representative example of these parts is the 47L04 . <A> Another solution. <S> Detect power down and use a supercap or non super cap to keep power up for a few milliseconds. <S> Use this time to write your counter value to EPROM. <S> Only write to EPROM at power down. <S> Number of EPROM cycles = <S> no of power down cycles. <A> Use a FRAM chip such as the FM24C04B.They have very high write endurance and are non-volatile. <S> https://www.mouser.com/ds/2/100/001-84446_FM24C04B_4_KBIT_512_X_8_SERIAL_I2C_F-RAM-477782.pdf <S> You could also use a battery backed SRAM (NVRAM) module. <S> For example M48Z02-150PC1 <S> https://www.mouser.com/ds/2/389/m48z02-955115.pdf <A> I decided to go with "ds1307 RTC". <S> Because it has 54byte of power-backed SRAM. <S> which allows infinite read/write cycle. <A> If your embedded project has NIC on it send your counter to remote computer/server. <S> It seems 120,000 iterations in 24 hours is about one iteration in 0.72 seconds, should be OK for network traffic. <S> Server will always have last value of the counter stored. <S> No counter value corruption on power loss because valid packet needs to be issued to update value on the server; however requires constant connectivity, or special time-out protocol must be designed. <S> In addition, as a bonus, you will be able to control your device from remote if it is needed. <A> A) Use a 100 µF capacitor (or larger) to power up the counter during the off time. <S> Or whatever logic that is required to hold the counter value. <S> B) <S> Use magnetic-core memories , they might be a bit iffy to set up. <S> C) Make a motor controlled potentiometer (like a servo), at some point your counter will overflow, right? <S> Map that to 360 degrees. <S> Then make a feedback loop so you can set the value of the potentiometer digitally and read it digitally. <S> D <S> ) Send your counter value once every minute to some server, or servers, and let them remember the value for you during the off time. <S> Then once power comes back up retrieve the counter value back.
| Another solution could be to use a microcontroller with non-volatile FRAM.
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3.3v regulator problem when sharing power source with LED strip I have this circuit implemented on a prototype board. The 12v power source is a 4A switching power supply. This powers the 12v LED strip, the 7805 regulator for the Arduino, and the IRU1015-33 3.3v regulator for the ESP8266. The led strip is a 5 meters 12v 30leds/meter RGB, controlled by PWM from the arduino. The problem: power on the 3.3v regulator output fluctuates depending on whether the LED strip is ON or OFF. When LEDs are full OFF (pwm 0% duty cycle) I get around 3.29v. When LEDs are ON (pwm 100% duty cycle) I get around 3.5v/3.6v, reaching the max input voltage for the ESP8266 module, wich is unsafe. Wich ever duty cycle I put in between (say %50), I get almost the same signal variation on the 3.3v regulator output. If I unplug the LED strip, the issue disappears and the regultar output stays around 3.3v. If I unplug the ESP module, the problem stays the same. (In that case R8 would be the only load for the regulator) Below are some oscilloscope captures I did.CH1 is the output from the 3.3v regulator.CH2 is signal between R3 and OK3, just for trigger and visualization purpouse. PWM 0% duty cycle. PWM 100% duty cycle. PWM 40% duty cycle. Strobe effect from 0% to 100% duty cycle. This one shows the voltage difference better. <Q> The IRU1015-33 has absolute maximum input voltage of 7 V. <S> You are operating the IC way outside its design parameters. <S> Change the regulator to whichever can handle 12V input. <S> Or switch the input of IRU1015 to +5V rail. <A> As @Ali has swiftly pointed out, you neglected the ABS.MAX.SPECS. <S> in the datasheet, and by a huge margin. <S> You were also concerned why there might be so much ripple in the 12V from a 4A supply from only one 5m reel of stripLEDs. <S> Load regulation and conduction losses are the main reasons for this. <S> So what is the dynamic load for reel? <S> ~ <S> 1 Ω Speculation: <S> Assuming typ. <S> RGB 0.6 <S> Amps <S> /m ... <S> *5m = 3A... <S> * 12V=36W. <S> If it was a linear R load , <S> R= 12V/3A=4Ω <S> but it's not. <S> If the current rises linearly from 9 to 12V then it is 3V/3A <S> = 1Ω due to all the series shunt current limiting resistors. <A> Not knowing how this circuit is wired up, or how much current the LED strip draws... <S> One thing you may want to check is the integrity of the GND net. <S> Does the voltage at the ground terminal of the regulator stay at zero, or does it increase with LED strip current due to resistive voltage drop along the current return path? <S> I'm amazed the IRU1015 survives and regulates pretty well at nearly double the abs max input voltage!
| Anything can happen, and 12V on your communication module is a real possibility.
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Is hand soldering a 48-VFQFN possible? Is hand soldering a 48-VFQFN possible? I could not find any videos on this and I guess it must be quite hard. <Q> With reflow, the part will align itself and solder will 'suck' towards the pins and pads. <A> Yes, I have done it. <S> I use a lot of flux for delicate solder jobs and to check the pad alignment I have a small USB microscope <S> so I can check all sides without having to move the PCB. <S> I put a tiny bit of solder on one pad. <S> Then a bit more flux and last the chip on top. <S> Check with the USB camera and push the chip so all pads are aligned. <S> Then I only have to heat the solder and not mess with a soldering iron AND solder wire. <A> It's quite possible to reflow QFNs (even those with a big thermal pad) without any special tools or materials: Just a soldering iron, solder wire, flux, and a kitchen stove. <S> Tin the pads with a layer of solder of consistent thickness. <S> Apply flux and align the chip <S> Place the assembly on an old frying pan or directly on a hot plate, and crank up the heat until the solder melts and the surface tension pulls the chip into position. <S> Remove the board and let cool to room temp. <S> Obviously this only works with boards that only have components on one side. <S> I wouldn't use this for any production boards, but IMO the results are still better than a hand soldering attempt. <A> Hand soldering QFN chips is difficult, but can be done with fine solder and a small tipped iron. <S> The standard footprint (intended for paste) needs the pads to be extended.
| It 's possible by hand when you're very experienced with fine hand soldering, but for lesser souls it's also possible with solder paste and manual reflow with a hot air (rework) station.
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What does it mean to "gate the clock"? In my lecture notes I keep reading "do not gate the clock". I tried searching on the Internet, but I'm unable to find the exact meaning of this phrase. <Q> simulate this circuit – <S> Schematic created using CircuitLab <S> The diagrams above show <S> and AND and OR used to gate the clock. <S> One forces the clock low the other high. <S> To prevent clock pulses which are 'too short' either high or low ("runt pulses"), we must make sure that: The control signal for the AND gate should change only when the clock is low . <S> The control signal for the OR gate should change only when the clock is high . <S> Gated clocks are very useful for reducing power in CMOS as the logic stays 'quiet' while the clock is stopped. <S> You will find that modern synthesis tools have special option to insert clock gating automatically. <S> simulate this circuit Above <S> are two circuits which safely generate a gated clock. <S> The circuits rely on the fact that there is as a small delay (clock to Q) for the control signal to come out of the register. <S> Thus the control signal changes at the gate when the clock has a known polarity. <A> It means: Do not use an AND or OR gate (or any more complex combinatory term) to derive a clock signal from another clock signal. <S> The reason for that rule is that race conditions among the multiple inputs of the combinatory term may cause multiple clock edges (glitches) where you expect only one clock edge. <A> Gating, in this context, means to pass a signal through a logic gate to control it. <S> Passing it through one input of a 2-input AND gate allows a control bit on the other input to force the AND gate output low or to let the signal pass through and out. <S> A similar function can be done by an OR gate, with the signal being forced high or let through. <S> So gating a clock means forcing it low/high or letting it pass through. <S> Not gating clocks is good advice. <S> It can be done, with care and thorough understanding of the possible consequences. <S> These include metastability when taking clocked signals into the gated clock domain and worse results from timing-driven synthesis/layout. <S> But there are nearly always other ways to achieve the same control over a circuit as gating the clock, without all such risks and penalties. <A> For high speed it implies do not add logic gates which add propagation delay to the clock as it may cause race conditions with working with data using original clock.
| To "gate the clock" means put a logic gate in the clock line to switch it on or off.
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Resistance in Ohm's law Hello everybody I have a question about resistance in Ohm's law, according to the formula "I = U/R" the less resistance you have, the more current you got and accordingly the more power you drain. So I don't get it, is this meaning the more load you got the less power you drain? That's a little counterintuitive, so I need a little advice from you guys to make this clear.Thank you in advance. Just to make things clear. An electrical load is an electrical component or portion of a circuit that consumes (active) electricpower. E.g. solenoid, motor, lightbulb etc. The stronger load (solenoid), the more wire it has, the more resistance it has. The higher resistance in circuit, the lower current flow, and the power. <Q> ... <S> according to the formula "I = U/R" <S> the less resistance you have, the more current you got and accordingly the more power you drain. <S> Correct. <S> ... <S> is this meaning the more load you got the less power you drain? <S> That's a little counterintuitive, ... <S> The smaller the resistance the more current will flow. <S> (That should be fairly obvious - think of the water analogy: a bigger pipe has lower resistance and will have a larger flow for a given head of water.) <S> Figure 1. <S> Water "current" and flow analogy. <S> Note that the pressure (voltage) is the same in both cases. <S> Source: Sparkfun . <S> The "load" is the load on the source, the energy supply or, in the case of the water analogy, the water supply or reservoir. <S> The lower the resistance of the load the more current will flow. <S> Update after question updated: An electrical load is an electrical component or portion of a circuit that consumes (active) electricpower. <S> E.g. solenoid, motor, lightbulb etc. <S> So far, so good. <S> The stronger load (solenoid), the more wire it has, the more resistance it has. <S> Not necessarily. <S> It's the ampere-turns that will determine the strength of the solenoid. <S> If you double the number of turns but keep the wire gauge and voltage the same then the resistance will double and the current will halve. <S> Double turns x half the current gives the same ampere-turns so no improvement. <S> To increase the ampere-turns you could: Increase the wire size. <S> (Lower resistance, more current -> higher ampere-turns.) <S> Increase the turns with the same wire size and increase the voltage in proportion to maintain the current. <S> (Same current, more turns -> <S> higher ampere-turns.) <S> Any combination of the above. <S> Correct. <A> It is a bit counter intuitive. <S> "More load" means a lower load resistance. <S> More current flows into a smaller resistor. <S> The device driving the load has to push harder than it does for a "small load" i.e., higher load resistance. <S> We can measure loading in units of Amperes. <S> Placing a smaller resistor across the terminals of a power supply causes more Amperes to flow. <S> This is an "increase in loading." <A> the less resistance you have, the more current you got and accordingly the more power you drain <S> Well, not exactly: that's more power IF I*U goes up, but there'snot necessarily any more current unless you think the voltageis a constant. <S> Voltage and current are both variables, andonly by examining the source character <S> do we know how they vary. <S> the more load you got the less power you drain? <S> The more resistance, the less power from a voltage source. <S> An ideal battery and light bulb and switch, with the switch 'off', i.e. high resistance, doesn't drain the battery. <S> Current can be the constant, too. <S> A flashlight lamp filament (high resistance) <S> heatsgreatly when the switch is closed, but the switchand wiring stay cool with the same current. <S> Ohm's law gives one equation in two unknowns, I and U. Power source characterdetermines a second equation, and you can then solve two equations forthe two unknowns, and compute the power I*U that istransferred.
| The higher resistance in circuit, the lower current flow, and the power.
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Why is a caseless PC not an EMI problem? As I understand it, the case of a PC is metal so that it can form an EMI enclosure around the electronic parts. My view is reinforced by there being conductive fingerstrips around the mating edges. This is a home built PC without any form of case:- I suspect that the black backboard with the four arms is scratch built too. It's probably plywood. This is 100s of Watts (motherboard + 2 high end GPUs) of totally exposed electronics running at ~4GHz spread out over at least a square foot. That's got to be an ariel somehow. There are many other examples if you Google "wall mounted PC" . And there are of course all of the fashionable single board computers like Raspberry Pis running inside shoe boxes at 1.2GHz. Or glass PC cases. Since EMI regulations get stricter all of the time, I thought that EMI is a problem. I read that it can be a major headache for equipment designers. Is it not really the problem the authorities make out? Or will this PC's owner never receive a mobile signal, be unable to watch clear TV or be sterilised within hours? I've read EMI/RFI emissions and computer cases , Non-metal cases and emi standards/best practices for PC cases . The general tone of answers to these questions speaks to enclosing the PC in a shield of some sort, be it metal or conductive paint/ tape. Why bother? <Q> Here's why: This digital circuitry contains a bunch of highly integrated circuits providing many functions on its interfaces. <S> It will generate a wideband noise rather than emissions of single frequencies. <S> Wideband noise is much less likely to produce problems on other appliances/electronics A tight enclosure of a PC does not emit nothing but a reduced wattage. <S> Hence an open case will only make more noise of the same kind, and most electronics are already designed to cope with that type of emissions. <S> Even if there's an emission capable of let's say interfering in the ISM band, it will probably only reduce the usable bandwidth on a certain transmission by lowering the SNR. <S> Even if there are serious interferences suppressing a specific application e.g. in your neighbourhood it is still unlikely that they will find out, that you are the culprit. <S> My assumption is, that the designs of these "custom made PCs" did not undergo proper EMI-testing, because it is so unlikely that they will be ever made liable for violating CISPR regulations or similar code. <S> Nevertheless, there's a question of morals. <S> Is it ok to break rules, just because you'll never be hold liable for it? <S> Probably not. <S> This is a typical problem of collective behaviour. <S> To keep things working, the majority of people (and their computers) have to adhere to regulatory code. <S> Otherwise the emissions of too many bad designs would sum up in a way preventing the use of correctly designed hardware. <A> There is a big difference from a consumer or commercial product and an OEM module designed to be integrated into something else. <S> If it is a complete product intended for wide distribution it must, or is supposed to be, tested to meet the requirements of regulatory body of the region where it is to be sold. <S> Products designed to be integrated into other products do not need to meet those requirements. <S> The company integrating the module is instead responsible for ensuring the final product complies with the requirements. <S> That does not mean all such products are without certification, most at least pass safety rules. <S> That may sound like a loop-hole <S> but it really is not. <S> The fact is even if the thing was tested and complied, there is no guarantee that after you stick it in a box with connectors and other parts that it will still pass testing. <S> As such you need to certify the whole assembly anyway. <S> Further, everything is made of parts, made by other companies, and some of those parts are in turn made by someone else. <S> Each in their own right is a potential source of EMI when you connect them together no matter how good the individual certification is. <S> If you are building your own stuff, and decide to pack someone else's module in a plastic box, you do not need to get it tested, however you do run the risk of having the local radio emissions authority come knock on your door because your neighbor complains when he cant watch his favorite TV show. <A> The hard drives, for instance, are in metal cases, and the wiringgoing TO the hard drives is hidden (perhaps in a metal channel?). <S> The power supply (upper left) is in a steel case. <S> The RAM modulesappear to have their own metal shields (a pair of plates affixedat the factory). <S> The (big, fancy) dual video cards have some kindof housing as well, which might be both shielding and a ductworkfor cooling air. <S> With short wires, kept close to the wiring ground plane, electromagneticemissions are minimal and (if the circuit board designer cares to doso) can be made zero by printing shielding layers into a wiring board. <S> Really, an overall grounded case is only one last layer of a defense-in-depth strategy of keeping all signals from interfering witheach other. <S> The remedies for electromagnetic interference are well-understood and don't always require an overall metal box. <A> It is often an EMI problem. <S> GPU miners runs into this problem often. <S> It is not typically a bad enough problem to warrant FCC action though, and most of them would not fail FCC class A.
| I'm pretty sure, that this device will cause a considerable amount of EM emissions, but it is rather unlikely that this will lead to a problem for the owner. There's electronics in that PC, but there's shielding, too. That looks like a computer module, not a consumer product, in which case the rules are different.
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BJT Transistors - On Characteristics Concerning the technical documentation of the BC337 BJT NPN Transistor. https://www.onsemi.com/pub/Collateral/BC337-D.PDF I'm just learning the jargon. I am under the impression that in the 'on specifications' Vce = 1.0Vdc means the voltage from the collector to emitter pin cannot exceed 1.0V? Is that correct? Thanks <Q> It means (100mA case) that if you drive the transistor such that the Vce is 1.0V and the IC is 100mA, the required base current will reflect the stated hFE range. <S> Imagine an 11.0V supply voltage, a 10 ohm resistor to the collector, emitter grounded. <S> You apply increasing base current until you see Vce drop to exactly 1V. <S> Then hFE is Ic/Ib = 0.1A/Ib. <A> No, If you look on the first page of that datasheet there will be a "Maximum Ratings" table. <S> This has the maximum Vce voltage. <S> The Vce = 1V specifies the conditions when the relevant parameters were measured. <A> It's showing the test conditions under which the measurement was made. <S> The datasheet states that Vbe can be up to a maximum of 1.2 V. <S> It says that the test conditions which pushed Vbe up to its maximum drop were conducting 300 mA of collector current, a current which caused a 1.0 V drop between the collector and emitter. <S> Hence, conditions: Ic = <S> 300 mA, <S> Vce = <S> 1.0 <S> V.
| The characteristic you're referring to in that datasheet is for Vbe.
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What are my prototyping options when working with smt devices? I have been working on a project which will allow me to learn how to charge Li-ion batteries. I recently purchased some simple PMIC's made by Texas Instruments. I received them the other day and realized I made a mistake in my purchase, the PMIC's were sot-23 smt packages. I don't have much soldering experience, and most of my experience with electronics is with through-hole bread-boarding and through hole soldering. I hear a lot that smt is the way to go if your serious about electronics, so I don't know if I should consider returning them and sticking to what I know, or taking the leap into smt now. Here are some things that I have to consider: I already have a breadboard power supply, which I intended to use during the prototyping phase for charging. I don't know enough about SMT to know if I can still use it in such applications. I can return the smt PMIC's, but the through-hole PMIC options I have found so far cost three times as much as the smt alternatives. And after reviewing some datasheets, I have found them to be way more complicated than whats actually necessary for my needs, I would be going from 6 pins to 16, with a bunch of added features I didn't need to deal with just to learn how to charge my first lithium battery. I have reviewed the datasheet for the smt's I purchased about three times now, and it requires less than a half-dozen supportive components, most I had already or have purchased, such as NTC resistors. Based on all that, I would at least like to know what are my prototyping options when dealing with smt devices, so that way I can make an intelligent decision on whether keeping them is even practical at this moment. I'm assuming the four breadboards I have laying around are useless, so what can I use to setup and test circuits with smt components? <Q> Use adaptors to get you going... <S> like this commercially available one, or make your own.. <A> Don't be afraid of SMT soldering! <S> It's easy! <S> The 2.54 separation between holes on prototype board matches most of SMT component pin spacing. <S> When it doesn't look for "PCB board adapter" (add the package code to your search like "SSOP8") and connect them with pins (see Trevor_G's reply) or with wires to the main board. <S> You don't need them for <S> SOT-23 and the likes: just solder them between the holes. <S> Except when you want the component to be connectable individualy. <S> You will also save a LOT OF space on your board compared with through-hole resistors and so on. <S> IMO breadboards are irreliable when you have more than 5 or 6 components or are used more than once. <S> They are just fine to test parts of your circuit. <A> I ordered for this exact reason convertor PCBs on Amazon. <S> These were for bigger ICs, mind you, but I bet they exist for the TOS-23 package as well!
| What you need is a good adjustable solder station, a magnifying glass with light and a feet, some prototype boards (through hole) and maybe an hot air gun (if you become serious with very small SMT packages).
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Why is MATLAB a worthwhile investment in electrical design? Is MATLAB a worthwhile investment when developing semi-standard circuit designs? such as fitlering, DSP, power supplies, usual things we all encounter. From what I have read and used, it provides great filter design, easy interface for mathematical calculations and some block level design using Simulink and some ultra-specialist stuff. <Q> I am slightly bias'ed on my views of Matlab as I have been using it for decades. <S> I am also part of their advisory board. <S> The immediate benefits of Matlab is job opportunities. <S> Being able to put onto your CV that you have <S> x number of years leveraging Simulink(tm) in a multidomain model to optimise the time to market of a new program speaks volume. <S> There are aspects of matlab-script that appear really poor design semantecs but make sense from a mathematical perspective. <S> Personally python+numpy+scipy+matplotlib <S> > Matlab. <S> If you add Simulink... I can describe a 30kVA BLAC motor-drive (3phase in, inverter, control, motor, loading response) in python and provide results comparable to the real system and a simulink model. <S> That doesn't mean others would easily be able to take my code to determine system responses. <S> Now Simulink... <S> Simulink is a very very powerful suite of tools to simulate systems. <S> If you are a circuit designer you may question the usefulness of it over say... <S> SIMextrix <S> but thats because SIMextrix is a spice simulator and Simulink is a systems simulator. <S> Simulink can simulate electrical circuits, hydraulic, mechanical, fluid etc <S> but it does not do this as well as dedicated applications. <S> However those dedicated applications do not describe other physical domains. <S> So the moment you need a cross-domain simulation what are you going to do? <S> you need a suite that can cater for this to then provide domain-specific information to the more detailed design suites. <S> Second to this is the concept of "model base design" where a simulink model captures the intent of a system model for requirement management to finally facilitate autocode generation. <S> When subsystems/blocks are linked to requirements (say in DOORS or <S> ... excel if you must) a design team can gain coverage and provide an audit-able trail if the business environment you operate in has some sort of agency oversight <A> Directly, nothing, aside from familiarity as it is taught at a number of engineering schools. <S> However, there is a lot of legacy code and toolboxes that are built on Matlab (either by mathworks or by third parties) that can be very powerful. <S> Well, I suppose Matlab can also be used to generate code for signal processing for a variety of systems (including generating C and Verilog), which could be a good time saver in certain circumstances. <S> I'm not sure what the maintainability of that generated code is like, though. <S> As an aside, the Matlab 'programming' language is absolutely horrid. <S> It's an excellent example of a language that has evolved organically over many, many years while maintaining backwards compatibility. <S> As a result, the semantics of the language itself have serious shortcomings. <S> I had to put together a small GUI in Matlab several years ago, and that required a huge number of what are essentially incredibly ugly hacks to work around basic language mechanics such as variable scope and the inability to pass around anything by reference, among other things. <S> Doing simple things like bit manipulation is like pulling teeth. <S> I saw a python script a while ago that was doing bit manipulation essentially by converting everything to strings and doing string operations... <S> apparently it was the first script the guy wrote in python after only ever writing code in Matlab. <S> Another researcher I know has worked exclusively in Matlab for quite some time, and now that he has to work with normal programming languages (mostly C and C++), he basically has to unlearn most of the Matlab techniques that he's familiar with (cue Yoda, "you must unlearn what you have learned"). <A> MATLAB isn't really about designing circuits. <S> It's a programming language and IDE (Integrated Development Environment) that's geared specifically towards doing mathematical stuff. <S> It can be easier and more efficient to do heavy number crunching in MATLAB than in many mainstream programming languages. <S> If you're willing to pay extra, there are all sorts of libraries that may be useful to you. <S> And, as Brian Drummond has suggested, if you just want to have a play, then Octave isn't as polished as MATLAB, but is highly compatible with it. <S> For what it's worth, my experience with MATLAB has mainly been in digital image processing - nothing to do with circuit design. <A> It isn't a good investment. <S> There are several free alternatives that fulfill the same requirements for algorithm development and high-level modelling that are nearly 100% code-compatible, which means that your skills will transfer directly to Matlab if/when that ever becomes appropriate for you. <S> I happen to use GNU Octave to develop signal and image processing algorithms, and regularly exchange code with a colleague who uses Matlab, with only a very few minor compatibility issues that are easily worked around. <A> MATLAB is supplied to Universities and to students under very concessionary terms. <S> So when you hire a new graduate she will probably know MATLAB and you can avoid the cost of them learning some other package and taking it with them when they go to the next job. <S> If you have a bunch of toolboxes and floating licenses it's pretty expensive. <S> I had a guy working with it to develop some signal processing algorithms which were moved to FPGA. <S> Worked okay for that project. <S> The Simulink software is good for testing control algorithms. <S> Consider the free open-source alternatives.
| Matlab at its core is a programming language written by mathematicians for mathematicians.
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Simplest, cheapest, fast and minimum footprint current limiting circuit with low resistance in normal state I have a digital output, driven by the high-side driver with nominal voltage of 24V DC. The load current normally is below 100 mA. The output is monitored, so I can switch it off quickly if I detect a short circuit at the load side. The problem is that the driver itself is not protected and short-circuit makes it generating a lot of smoke. So what I need is a simple circuit at the output of the driver which: has low resistance of below 10 Ω if output current is under 100 mA rapidly increases its resistance to limit the driver current at 500 mA level or lower withstand capability at short-circuit current shall be at least 20 ms for short circuit to be detected and driver switched off has a working voltage of 50V or higher has minimum components and cheap (0,20$ per channel max) is not single-source supplier I tried PTC resettable polyfuses, but they are too slow. Microchip's FP0100 should be good but it's expensive (I need at least 60 channels on my PCB). Bourns TBU series are also OK, but also expensive. Any other options? UPD1. My current output circuit is MIC2981/82 driven by 74HC594 shift register. On each output I have Littelfuse 1206L012 PTC. On my board I need 64 channels like this, and this is small series board so total price per channel and footprint are important. <Q> Your typical double transistor current limiter may be your best bet. <S> Shown below is the top-side and bottom side versions. <S> simulate this circuit – <S> Schematic created using CircuitLab Note there is a penalty of about a volt drop with this circuit. <S> Buy dual transistors in a single 6 pin package. <S> The small resistor will cause the current to fold back when it reaches Vbe. <S> The other resistor sets the base current and needs to be calculated to produce sufficient collector current taking into account Hfe. <S> HOWEVER <S> : Be aware that transistor needs to handle a few watts for the duration of the short since it only limits the current to your threshold value. <A> Take a look at the ProFET high-side driver ICs. <S> These devices give you a switchable high-side drive with protection from all sorts of things, including output over-current. <S> Have a look at the BSP752T, which is cheap, small and can be driven directly from 3.3 V or 5 V logic. <A> To build upon Trevor's excellent answer : <S> There's semiconductor devices that are constant current sources (or sinks); many of these will internally look exactly like Trevor's circuit (maybe adding a few temperature-compensating elements). <S> One very simplistic device (constant current sink with exactly two pins, designed for voltages <= <S> 50 V and a max/constant current of 350 mA) is the NSI50350AD . <S> I don't know what it internally does, but the datasheet calls it "self-biased transistor", so chances are it might be a combination of some bipolar transistors, a JFET and a couple resistors internally. <S> Now, your 50 V limit really hurts – it's hard to find integrated current sources that will work at that voltage. <S> For smaller currents, a self-biased JFET might work, but at 100 mA that's going to be expensive. <S> So, I'd really roll with Trevor's solution, albeit I might recommend a few things: <S> Check whether you can't simply increase the speed of your fault detection. <S> That would solve the problem. <S> Because (as far as I know – correct me if I'm wrong) it's hard to come by transistor arrays (which you'd prefer if you need reduce effort and board space), you might want to spend a little more on the component than just a NPN for Q4, but save on pick&place cost by using a device with multiple comparators in one case. <S> Luckily, 4x comparators and 4x opamps cost around 13 ct when bought in hundreds, so that's ca 3ct in opamp per channel; use the opamp/comparator to compare the voltage over R2 to a constant reference voltage (here, a simple zener might do) and to drive Q3. <S> Notice that you don't need an R3 for every single channel anymore. <S> (the same applies for the high-side approach with Q5/Q6) <S> Use resistor arrays instead of individual resistors, thermal design permitting. <S> Another relatively crazy approach would be using a high-side 8.2Ω resistor before your load. <S> Design that series resistance so that for 100 mA \$I_\text{Load}\$, the transistor is in saturation, but for 500 mA, you significantly pinch of. <S> Put the C-E of the optocoupler output in low-side series with your load: simulate this circuit – Schematic created using CircuitLab <A> This one works out to $0.2/port x16 <S> https://ca.mouser.com/ProductDetail/NXP-Freescale/MCZ33996EKR2?qs=sGAEpiMZZMuCmTIBzycWfKe9ppy40BrEybgj5eCsa3I%3d <A> Here is the basic idea for the SCR circuit. <S> The total resistance in parallel with the base emitter junction of Q1 will set the trip current. <S> Once Q1 starts to conduct, the SCR will fire, and then the load will be protected until the PTC trips. <S> Q1 can be a SOT-23. R3 and R4 are just guesses. <S> They are just there to prevent over-current damge to Q1. <S> Most SCR's are kind of large. <S> I will let you look to see if you can find one small enough to suit your needs. <S> Note <S> : Once the SCR fires, you will probably have to de-energize <S> the power supply before it will stop pulling down the rail. <S> simulate this circuit – <S> Schematic created using CircuitLab
| After that, insert a current divider between your load and the LED side of a transistor-out optocoupler, with an appropriate series resistance. Might have to add a resistor in series with PTC1 to get the right value of resistance. You can find and select ProFETs readily enough from distributors. A cheap candidate for the optocoupler would be Lite-On CNY17 .
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Wiring diagram of aller retour switch with lamp I need to connect a lamp with 2 aller retour switches. In the wiring diagram how many cables will i have between the switches if the lamp is between them? <Q> There are three ways to hook them up. <S> (Thanks to Janka for the third) simulate this circuit – Schematic created using CircuitLab <S> simulate this circuit <S> The first or third methods are the preferred ones. <S> As Dave points out, the second leaves the light itself hot half the time and is prone to make before break switch shorting. <A> With SPDT ("three-way" in English) switches, you can't put the load (the lamp) between the switches, at least not electrically. <S> This does not affect the number of wires "between the switches". <S> Normal arrangement: simulate this circuit – Schematic created using CircuitLab <S> Folded arrangement: simulate this circuit Funky German arrangement: <S> simulate this circuit <S> Note that the last one requires 4 wires in at least one of the cables, more than either of the other arrangements. <S> It also runs the lamp current through the largest number of wires. <S> I fail to see any advantage to this, other than the fact that in the region between the two switches, you can connect both switched and unswitched loads. <A> Actually if you´re using a three-way you´ll use 1 cable as the hot line, coming from the circuit breaker. <S> This cable gets to the central screw of one of the switches. <S> Done that, 2 wires will leave that switch and meet the other one. <S> Finally, this other switch takes the last wire to the lamp. <S> So counting on the neutral cable, you´d have 1+2+1+1 = 5 wires. <S> If you want to count the earth one it would be 6.
| If you want to "fold" the wiring so that the lamp is physically between the switches, you'll have to run extra wires from one of the switches to the load.
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A problem regarding my Slayer Exciter circuit? I have made a Slayer Exciter circuit for my school project. I'm going to be assessed on it. Now, it works just fine. But if I have to light a tubelight or something, I have to touch the terminal to the wire sticking out of my secondary coil. Then I can move the tubelight about, close to the coil and it won't go off. But why do I have to touch my light to the wire initially? Because in most Slayer Exciter circuits that I have seen on the internet, the light only has to be brought close and not actually touched. Can someone please give me a technical explanation? The examiner may ask me this and more than that, I want to know. <Q> But why do I have to touch my light to the wire initially? <S> You need to create a situation where the electric field across the tube is big enough to start ionization. <S> Given that your body is actually part of the circuit, this feeble connection is the weak link in my opinion. <S> I would connect the negative terminal of your battery to a piece of tin-foil (aluminum foil) that forms a 12" ground-square under your base board. <S> Now you have a significantly increased ground capacitance and a less-feeble AC connection to your body. <S> This will improve things. <S> However, there is no guarantee that your circuit is producing enough voltage so although the ground-square may improve the situation, you may still need to adjust the output coil. <A> To start discharging in the gas needs stronger field than your system generates without touching the wire. <S> It continues more easily if there's some ionization left. <S> Fix: get higher output voltage or more easily ionizing tube. <A> It takes more energy to ionize the gases in the first place than it does to keep it ionized. <S> Therefore in your low-power setup you need that initial direct high voltage connection in order to transfer enough energy to the tube to "ignite" the gases. <S> Once "ignited" it will take significantly less energy to keep the tube lit, which is why you can move it around the coil without touching it. <S> You'd need to make sure your circuit can handle the new supply voltage though.
| If you wanted to improve the output of your coil so that the tube lights without even touching it, you could increase the supply voltage. Your circuit is battery powered and appears to have very little ground capacitance hence your body makes a very feeble connection to the "current return path" of your circuit.
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In digital systems do we discretize both time and magnitude or only time? In studies of Digital Circuits or Digital Electronics or Digital Systems, the very first step is to convert an Analog signal / Analog system into a Digital Signal / System. In this process of converting an Analog function to a Digital Function, do we need to discretize both time (X-axis) and magnitude/functional value (Y-axis) or only time (X-axis)? Can anyone give me the correct definition of this (Discretization / Analog to Digital Conversion), as I found at some places it says both time and value (both X-axis and Y-axis) i.e take (X,Y) ∈ ℕ ,whereas at other places it says only time/X-axis values i.e only X ∈ ℕ and Y can have any values i.e Y ∈ ℝ . <Q> An analog signal that is quantized is a discrete-valued signal. <S> A signal that is discrete in time and amplitude is called a digital signal. <S>  <A> The answer depends on what you are doing. <S> A sample and hold circuit will sample the analog signal at discrete time points, hence converting to digital in only the time (X) axis. <S> An analog-to-digital converter will sample the magnitude at discrete time points, hence converting to digital in both time (X) and magnitude (Y). <S> This is the most useful form of a digital signal, which you can process in a digital signal processor or FPGA. <A> Both. <S> An analogue-to-digital converter usually emits a series of quantized values, each of which represents the value of the input at a particular time. <S> Usually they are sampled regularly at a particular sampling rate. <S> To complicate matters slightly, there are types of ADC which ouput discrete digital values but are continuous in the time domain: http://www.electronicdesign.com/analog/what-s-difference-between-continuous-time-and-discrete-time-delta-sigma-adcs <A> A discrete time signal can be called as a digital signal when its amplitude is made discrete too. <S> i.e., when the amplitude is quantised. <S> In ADCs, it does both sampling and quantisation. <S> Means, the analog signal is made discrete in both amplitude as well as time axes.
| An analog signal that is sampled is called a discrete-time signal.
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Is there any reason that any speaker connected to any computer couldn't be used for listening/surveillance? I understand that it's possible to use the headphone port of a laptop as a microphone, but what are the physical limitations to using any speaker in the same way? I'm curious about what makes amplification a one way street and if it is necessarily always the case. <Q> I take your question to mean "can software use the speakers in computers at the moment as microphones". <S> And not to mean "can we change computer electronics to use their speakers as microphones". <S> Currently, computer speakers are generally (almost always) driven by an amplifier, with no electronic circuitry to read that speaker signal. <S> So there is no mechanism there to 'listen' through the speaker using the existing electronics. <S> As you suspected, amplification itself is a one-way street. <S> It is possible to design more elaborate circuits that aren't one-way. <S> But that's not what your desktop or laptop PC has driving the speaker in it. <A> Yes it is possible to hook up a circuit to a speaker to turn it into a crude microphone... <S> for example.. <S> However, your laptop speakers will not have such circuitry built in, and as such, you can not use them that way. <S> The sound quality from said microphones is also very limited, and the output and frequency range is highly dependent on the type and sixe of the speaker. <S> Generally, this method is not used. <S> It is so much simpler to add a small microphone which gives much better performance. <A> If active speakers (i.e. speakers that require power) are connect to a computer then the amplification circuitry is a one-way street and it would be impossible for microphonic signals to be received by the computer. <S> If passive speakers or headphones (these are essentially the same) are connected to the computer <S> then there is a more complicated answer. <S> Modern computer audio codecs often support multi-purpose audio jacks. <S> This allows each jack to be software configured to function as: headphone output, microphone input, line out output, etc. <S> It is therefore possible for headphones connected to what the user presumes to be a headphone jack, to function as a microphone. <S> Such a hack has already been demonstrated: http://vrzone.com/articles/new-software-lets-hackers-turn-speakers-microphone/117315.html <S> This hack is only possible if the headphones are not being used to output sound. <S> However, any microphonic current induced will be absorbed by the output driver so that the microphonic signal is never manifested as a measurable voltage. <S> Even if the audio output was 'silence', the output driver would still be driving a zero voltage signal and absorbing microphonic current. <S> In summary: Active speakers cannot be used for listening/surveillance. <S> Passive speakers or headphones can be used for listening/surveillance but only if they are not being used to output audio.
| The speaker can not be driven as speakers at the same time as they "listen" as the transmitted sound will swamp the receiver. It is plusable that an audio codec could be hacked to have a jack simultaneously function as a microphone input and audio output.
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Costs of Conductive Silicone/Rubber Keypad vs. Mechanical Switches? I am toying around with some new mechanical design ideas. One thing I've never looked at it is the rubber keypads with the conductive "pucks" underneath. I've always used mechanical switches, or overmolded silicone buttons onto the finished enclosure for waterproofing. Example of the conductive rubber keypads: http://www.rubber-keypad.com/Conductive-Keypad-pd6229345.html I take it the manufacturer gives you a footprint that matches the conductive "puck", and the footprint goes to GND so when the puck hits, your logic line goes low. Does anyone have any experience implementing these? Are there any gotchas or things to watch out for? Any experience on the cost side? <Q> Does anyone have any experience implementing these? <S> Are there any gotchas or things to watch out for? <S> I have tested a lot of tactile/membrane keypads in the past and the biggest issue I found <S> was <S> bounce time was very variable between one manufacturer's product and another. <S> This can be annoying if interfacing with a spcific chip (like a DTMF encoder in a telephone) because you quite often got double digits when you thought you'd only pressed once. <S> Same when releasing the switch - it can bounce then. <S> Some keypads I tested that had a seemingly beautiful tactle click <S> didn't actually make contact until you pressed a little harder. <S> Now I'm sure the inducstry has moved-on from those days back in the late 1980s but caution should still be your watchword. <S> Any experience on the cost side? <S> They are cheaper on production costs for low to medium volume but don't ignore the time and effort into guaranteeing a good design. <S> The main reason for using them is of course that they can be designed to have buttons in irregular positions i.e. they are easily customizable. <S> Not wishing to counter anything said by anyone else (@Jonk) <S> but a good technology should give you over ten million operations. <S> We (back then) modified a motorized hack saw like this: - It produced a repetitive forward and backwards stroke <S> and we used a spring/cushion to set the impact force onto the target keypad. <S> We easily got ten million operations from quite a few but very few could meet the debounce times at end of life. <A> So the rubber keypads are less expensive unless you're buying a hand full for a hobby project. <S> They make silicon keypads with compression molding. <S> Similar to injection molding where you make a mold but instead of injecting hot plastic <S> I think it's a powder. <S> There's not a whole lot of labor involved in cranking out a bunch of these. <S> A recent small soft tooling mold cost me $1000. <S> You do design the button in a program like solidworks, or your favorite 3D tool. <S> And you have your choice of pills, carbon, metal, little click devices. <S> Getting the right feel of your button can be a little tricky though if you don't have experience. <S> Things like shape, wall thickness, material hardness etc all lead to different feeling buttons. <S> Anyway it's pretty easy to make a button and then have marketing unhappy with the "feel" and have to do it again a few more times till everyone is happy. <S> Also like anything else molded you have to beat the cost of the mold and make enough that you really get your price down. <S> Making 10 at a time will cost you more. <A> There are numerous methods for creating keypads, from expensive individual switches down to cheaper membrane and or puck type keypads. <S> The decision of which type is best for a particular application can be a difficult one which needs to factor in life expectancy and wear of the switches and of course cost to produce. <S> The main issue with the puck type, other than they wear out over time, is there is zero tactile feedback to the user. <S> As such issues with "not actually pressed" or "multiple pressed" is rampant with this type of keypad. <S> A more rugged approach is to use tactile dome switches... <S> These are mounted on a film that adheres to the PCB which has similar layout to the puck design, and are pushed through a similar rubber and silkscreened keypad which lays over the top. <S> The domes provide good tactile feedback, click, and are reasonably inexpensive. <S> In the end it really depends on the application and how often the buttons are expected to be pressed and how accurately those presses need to be received. <S> By the way, there are companies that specialize in and will manufacture the customized rubber part for you. <S> You will find the quality is much higher than something you can do on your own.
| If you have a high volume product or a product that gets a lot of keypad use I would seriously consider a lot of mechanical testing of different supplier's products if they do not have a technology that guarantees a closed resistance within a certain time period of the button being pressed.
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How is "signal strength" characterized on a coaxial cable? I've come across the Shannon-Hartley theorem, so I assume that a "signal" must have some notion of strength or "power" that can be referenced to some noise level (and whose power can be varied at the signal source). But what form does that power take on a coaxial cable? E.g., is it a maximum voltage that can be observed between the conductor and the shield? Or something else? And what is a typical value for this power measure or limit – say, as applied to RG6 cable for residential data transmission? <Q> The SNR used in the Shannon-Hartley theorem is the ratio of the signal power to the noise power . <S> So you want to use the power in watts or milliwatts to plug in to the formula. <S> Of course, if you have the power in dBm, you can convert that to milliwatts. <S> Or if you have the voltage amplitude (and the characteristic impedance of the system) you can also use that to find the signal power. <S> Power in dBm Voltage amplitude <S> Voltage peak-to-peak Voltage rms <S> You should be prepared to convert any of these to the others to be able to compare specifications provided by different vendors or customers. <A> The Shannon-Hartley theorem couples Bandwidth and Signal to Noise ratio to how much information can be transferred over a medium. <S> Let's just assume that there is a certain Bandwidth available, it doesn't really matter what it is. <S> Then the Signal to Noise ratio becomes the only important factor. <S> Note how it is a ratio where Signal and Noise both have to be expressed as power . <S> A coaxial cable is a from of a transmission line where both ends of the line (cable) <S> have to be terminated properly. <S> Most Coaxial cables you will encounter in the real world have a characteristic impedance of 50 ohms. <S> That 50 ohms determines the ratio of voltage and current. <S> Also, if you know the power of a signal going through that cable you can calculate the voltage and current. <S> So the signal (or noise, <S> noise is also just an unpredictable signal) going through the Coaxial cable is just voltage and current, same as with any other cable. <S> The cable is also just a "medium" for transport of the signal. <S> Ideally the cable doesn't care about the voltage, current or power level. <S> In practice there will of course be limits to what a cable can handle <S> but usually we can ignore these limitations. <A> For cable modems, Arris gives example signal levels of -2 dBmV and SNR of 29 dB. <S> To give practical values for loss of RG-6 coax (more correctly referred to as Series 6 coax since RG-6 refers to an obsolete military specification), you can consult the Society of Cable Telecommunications Engineers' standard for drop cable, ANSI/SCTE 74 . <S> At 55 MHz, the insertion loss of 100 meters of Series 6 coax is 5.25 dB. <S> So a signal level of 3.25 dBmV (1.45 mV) travelling 100 meters becomes -2.00 dBmV (0.794 mV) at the cable modem. <A> as you noticed, the theorem is about ratio of powers . <S> A power is, for a linear medium, proportional to the square of voltage . <S> However, in cases where you deal with coax cables, it's rather unusual to calculate in voltages, anyway; most things are defined as powers, as this makes handling easier (and also, the proportionality is given by considering the characteristic impedance).
| In practice, I've seen all of the following used to describe signal powers: Power in watts Power in milliwatts
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How can the contact points of jumper cables handle the hundreds of amps required to start a car? When jumpstarting a car, you connect those big alligator clamps to the terminals of the batteries. The cables themselves can handle the current just fine, but the contact points between the alligator clamps and the battery is extremely small. I'd imagine a maximum of 4 teeth are touching each terminal. How do those contact points not simply explode under the hundreds of amps? <Q> Hint: resistance is not just a matter of cross section, but also of length. <S> Would you expect the jumper cables to work if they were say, 10 meters long? <S> What about 100 meters? <S> You got my point. <S> Even if the contact surface is very small, the effective length of the contact is very short. <S> One could argue that having flat, rounded surfaces would lead to a better contact anyway, but this is not true. <S> The battery contacts are dirty and oxidized, the pointy alligator clips, together with the strong spring, help the stranded driver to perforate said oxide and guarantee a good enough contact. <A> Have you noticed how the clamps get <S> hot after a jump-start? <S> That means power is lost, it ends up in those clamps. <S> You're right <S> the connection isn't ideal so power is lost but not so much that anything melts or explode. <S> What is there to explode anyway? <S> Nothing :-) <S> You wouldn't want to make that huge starting current flow for more than a minute as the clamps will heat up severely and the plastic insulation would melt. <S> But since we just want to start the other car, which usually takes much less than a minute, it is OK. <A> You seem to overestimate the effects of current on conductors. <S> Let's take a regular AWG17 wire for example, which is a mere 1mm² of copper. <S> Guess how much current it can take for 1 second before blowing up? <S> Now check if you guessed right. <S> AWG 17 (1.04mm²) fusing current <S> : 10s @ <S> 99 A, 1s @ <S> 316 <S> A, 32ms @ <S> 1.8 kA <S> Jump start clamps have a contact area of about 5mm², and the teeth are cooled by the thermal mass of the rest of the clamp, so handling a few hundreds of amps for a few seconds is not a problem. <S> AWG 10 (5.26mm²) fusing current <S> : 10s @ <S> 333 A, 1s @ <S> 1.6 kA, 32ms @ <S> 8.9 kA <S> ... <S> AWG 11 (4.17mm²) fusing current <S> : 10s @ <S> 280 A, 1s @ <S> 1.3 kA, 32ms @ <S> 7.1 <S> kA <A> After starting your car, examine the lead battery terminals. <S> The heat melting the lead is taken away from what is needed for the thermonuclear ignition that you fear. <A> Consider that copper is an excellent heat conductor. <S> It is so excellent that, for example, you would have really hard time trying to solder the clamps to anything with even moderately powerful soldering iron. <S> Also, when you connect clamps there is typically a bit of arcing. <S> The clamp actually gets welded a bit increasing the cross section of the actual contact with the battery leads.
| The clamps can handle that huge current for a short time . Sometimes when the contact is bad you will find a tiny pit where the lead has melted.
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Best way to connect two adjacent PCBs by their outermost hole rows? I have two separate circuit boards that each have the same 8 hole row near their edges and I need to make a bridge from one PCB to another.Placed like that, the distance from the first PCB’s first hole to the second PCB’s mirrored one is about 4mm - 4,5mm. What is the best way to permanently connect these to one another in a neat and compact way? Some sort of staple shaped pins that I don’t know the exact name of yet because I’m new to this? Usually I can easily find the answers by searching online but somehow the key words in this are involved in far too many rather general topics, so I'm afraid that a mere search engine won't understand the relations of the terms involved here; better ask a human. <Q> I would consider connecting them with 90° angled male and female header pins, while on each end of those add 2 "spacers" and use small (such as M3) screws to hold in place, as shown on photo below. <S> Different shade of green shows on 2 different PCBs, while on one side there are female and on the lighter green side they are male pin headers. <S> I've drawn straight instead of 90° angled, but you get the point. <S> On each side, you have those small grey "boxes", which are aluminum parts with 2 drilled holes for your screws (as I've said, such as M2, M2.5 or M3, depends on size of your PCB) and the red are the screw heads. <S> Needles to say, use washers and nuts on the bottom of your PCBs. <S> I think this will be quiet a good, permanent connection between them. <S> EDIT: 3 new options came to my mind;Both ones start with setting one PCB on top of another, so the holes match. <S> I wouldn't recomend this option very much, as other 2 are more stable. <S> Second one is to stack holes again, one on top of another and push through copper wire, as thick as possible, that goes through the holes, then solder it on both edges (on top of top PCB and bottom of bottom PCB). <S> The third one is very simmilar to the second one, but use pins for this application, such as these: <A> It depends how many units you intend to produce, if this is a one off, bent zinc-plated paper-clips are cheap, easy to solder and very strong. <S> just make staples out of paper-clip wire and solder them in.alternate top and bottom for extra rigidity. <A> For staple-shaped connectors you are either looking at: U-Shaped Headers. <S> For example, Farnell sells them as Board-to-Board connectors, such as these here . <S> Solder anchors. <S> Digikey sells these, they are single U-shaped components used on motherboards to hold heatsinks in (typically). <S> Digikey link is here .
| One option is to fully cast both holes together with the solder.
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Can a Li-ion battery be discharged in CV mode? I have a load which can also be set to "Battery test". I can easily discharge a Li-ion battery using the CC (contant-current) mode, which is also a condition recommended by the manufacturers. For example, a typical recommendation would be: Discharge CC (18A) to 2.0V @ 20degC I have worked with CC before and it works as expected. But what about CV (contant-voltage)? One can never see that on the specs/recommendations. If I set one knob to CV and use the "Battery Test" setting, it says that I can't use it. So I assumed that the load doesn't really like the idea of CV and I was wondering why? I could switch to another type of load (not "Battery Testing"), where I can actually set the CV, but I'm afraid that this would destroy the battery. Can anyone help me to figure out whether or not I could force CV into the battery and why is there no CV into the battery-testing option? Thank you <Q> I think you misread or mis-interpretted the datasheet. <S> If it read... Discharge CC (18k) <S> to 2.0V @ <S> 20degC <S> .... <S> then I would recognize this as a standard (one time) <S> capacity test for 3V Lithium coin cells which start at 3V and never ( ever !) <S> for LiPo as this causes permanent damage. <S> But may ok for max current for Li-Fe. <S> Li-Fe example <S> 70Ah (ESR 2mΩ) Charge 3.6VDischarge 2.0V <S> Max Charge <= <S> 3CA <S> Max Discharge <4CA or <12CA <S> pulse Recommended Charge 18A(CC), 3.6V (CV) , 10%A cutoff Since the model for a battery is similar to a capacitor Ic= <S> ΔV/ESR+CΔV/dt <S> For ΔV= <S> Vinit-Vload = 1.6V and applying a CV load of 2.0V to a cell at 3.6V ESR = 2mΩ, the 1st part becomes 1.6V/2mΩ ~ <S> 800 Amps or Pd=I²ESR= 800²*2mΩ = 1280 Watts <S> ( Pfft or kaboom) <A> That, being small, means the initial current is very high and uncontrolled. <S> This will result in overheating and potential fire/explosion. <S> The whole point of CC discharge is to limit that heating effect. <A> The datasheet has specified a CC discharge at 19A to 2V. <S> This suggests something like a LiFePO4 battery which has max 3.65V and anywhere from 1.8 to 2.5V low voltage cutoff. <S> Let's say capacity is 20Ah, so you're discharging at the C-rate. <S> Typical internal impedance is somewhere between 0.5 mohm and 5mohm, let's call it high at 5mohm. <S> A step change of just 5mV difference to the terminal voltage would produce an instantaneous current of 1A, and 500mV would cause an instantaneous current of 100A to flow. <S> The very low source impedance produces large variations in current for small variations in voltage - it is better to control the current until the voltages have stabilised.
| If you use constant voltage less than the battery voltage the only thing limiting the current is the battery resistance.
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Electric wiring color codes: BROWN, BLUE and BLACK 31.01.2018 16:40 start /Thank you all for the answers, because this is not a standard wiring plug, i need assistance wiing the wifi adapter in the middle of cable which connect the energy master expert lcd display to the energy master expert plug. Is this correct? WIFI ADAPTER LIVE IN = Energy master expert PLUG BROWN live wire WIFI ADAPTER LIVE OUT = Energy master expert LCD DISPLAY BROWN live wire WIFI ADAPTER NEUTRAL IN = Energy master expert PLUG BLUE neutral wire WIFI ADAPTER NEUTRAL OUT = Energy master expert LCD DISPLAY BLUE neutral wire WIFI ADAPTER EARTH IN = Energy master expert PLUG BLACK wire WIFI ADAPTER EARTH OUT = Energy master expert LCD DISPLAY BLACK wire \ end 31.01.2018 16:40 31.01.2018 16:36 start / Update from customer service:**The Ground Line is bridged in the connectorThe meter does not require a Ground Line in the cable.Blue is the Null Line and black and brown are the Live Line Should you have further questions, please feel free to contact us via email. Yours sincerely ELV Elektronik AG Technical customer service department** \ end 31.01.2018 16:36 Which colors are Ground Line, Null Line and Live Line? The standard says the Ground line is always green-yellow.None of these wires is green-yellow. Thanks. updateIf black is Earth, then why the on off switch button is using the black wire? Because the cable is too short , less than 2 meters, i want to extend it. I also want to join the cables with a wifi adapter SONOFF® POW 16A 3500W DIY WIFI Wireless Long Distance APP Remote Control Switch Socket Power Monitor Current Tester For Smart Home 80-160MHz AC 90-250V Support 2G/3G/4G Network. That is why i need to know exactly which colors are Live, Null and Earth. <Q> The ground prongs are connected internally, no wire needed. <S> You have to connect the blue wire to the (also internally connected) neutral screw. <S> The brown and black wires go to the other plug/socket screws, it's live in/live out. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If this cable could be considered as "internal to the equipment", normal electrical code colours may not apply. <S> If it is a power monitor, the monitor circuit has to measure current somehow. <S> I suspect that R4 is the current sense resistor, and power comes in from the plug on the brown wire, and returns to the socket on the black wire. <A> <A> You will have the german standard . <S> But your application is not a standard plug. <S> Brown will be live. <S> Blue will be neutral. <S> Green/Yellow must be earth. <S> Or something else, black is the trivial color. <S> In your measurement plug, only one line is passed trough the measurement circuit. <S> Brown -> <S> measurement -> <S> Black. <S> Neutral is passed to the measurement as well to provide voltage measurement and supply. <S> Earth is not required, and thus not wired, since thee enclosure of the measurement device is plastic.
| This is a combined plug/socket. If the instrument box has no exposed metal parts, it would have no need for a Safety Ground connection. It depends specifically where you live, but usually you find Ground Line as Green-Yellow, Null Line as Blue and Live Line as Red, Gray, Black or Brown. Black can be another phase, or a switched live.
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Implementing a voltage sensor with voltage dividing resistors I am implementing a power storage system. The voltage of the storage device must be sensed in order to direct current to and from the storage device. The storage device voltage will vary between 0 and 60 volts, and the system's current will be up to 60 A. The micro-controller operates at 3.3 V. To map 0 to 60 volts onto 0 to 3.3 volts:60 V x 10 kΩ / (10 kΩ + 170 kΩ) = 3.33 volts The current through the resistors is 0.333 mA. (Corrected from original 3.33) Would this work as expected. Is there something I am neglecting? <Q> Your system will work but there are a few of things you need to be mindful of. <S> You are dividing the 60V to the max that the micro can handle. <S> This means if the storage device is ever over 60V you will be presenting too high a voltage to the micro and your ADC will not be able to detect it. <S> You would be better to use something like 10/190 <S> so you present 3V at 60V <S> so you can allow 10% overshoot on the 60V. <S> The numbers also are easier to work with. <S> As WhatRoughBeast pointed out, the divider will always drain some current from the source, be that only about 333uA. <S> You may want to consider adding some form of switching circuit so you only attach the divider to the 60V when required. <S> If you are using 1% resistors, the possible measurement error through the divider is +-2%. <S> That may or may not be a problem for your application, however it is a problem if you use 10/170 since +2% will make the presented voltage over the magic 3.3V. <S> Most micros these days have the option to use a stable internal voltage reference for the ADC instead of the rail voltage. <S> If yours does (you did not specify the micro) <S> you should scale the divider to use that level instead, again with that 10% overhead. <S> That will remove errors caused by whatever the 3.3V supply is doing and get you more temperature stability. <S> Adding some small capacitance to the division point will also help make the system less sensitive to noise, both ambient, and on the 60V rail. <S> There is a balance here though, do not make it too large or your sample time needs to extend to cope with the slower step response time. <S> You should also consider splitting the top resistor to make it two in series. <S> This will provide you with a little extra isolation from the 60V, split the power dissipated, and remove the potential of a single fault short in the resistor blowing up the micro. <S> A little extra protection would not hurt either. <S> Finally, this design is high impedance. <S> That means it is affected by the input impedance of the ADC. <S> Using a voltage follower buffer between the divider and the ADC is prudent. <A> One thing you ought to be aware of is the possibility that the MCU will not power down correctly if on its own 3.3 volt supply whilst the potential divider is still producing a peak voltage of (say) 3 volts - current may be passed through the IO pin to Vcc pin and keep the MCU powered. <S> It can happen and the way around this is to choose an op-amp buffer on the same rail as the MCU. <S> You have to choose a rail-to-rail op-amp that can have an input voltage a few volts greater than its rail voltage (3.3 volts). <S> With the resistors mentioned you might fall-foul of not providing a low enough impedance to correctly and accurately drive the ADC input. <S> This is usually solved with a capacitor from ADC input to 0 volts and this capacitor might also prevent spikes damaging the ADC. <A> Define "works as expected". <S> Yes, assuming the micro-controller has a very high input impedance, 60 volts will map to 3.3 volts using a 10k/170k voltage divider. <S> The current through the divider will be 0.333 mA, not 3.33. <S> Of course, the voltage divider will continue to draw current even when there is no energy available to replenish the battery. <S> 0.333 <S> mA may not seem like much, but it will drain the storage device 24/7.
| You should also consider that if your IO pin connects to an ADC it will have offset and gain errors that might mean you cannot detect the low-end or the high-end voltages in the range hence, you should slightly bias the potential divider junction to avoid 0 volts and choose an attenuation factor that avoids running to close to the 3.3 volt nominal upper limit for ADC measurements.
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How does an ARM processor in thumb state execute 32-bit values? What I understand is, the ARM mode can execute 32-bit of instructions and Thumb mode can execute 16-bit of instructions. For instance, Here is the ARM instructions set: And Thumb instructions set: From these both instruction set tables, please see ADC mnemonic that describes add two 32-bit values and carry . So, 32-bit is common in both the modes. What I didn't understand is, how a thumb mode which can execute 16-bit is able to execute 32-bit value? I referred the other books (which I could) and in those books also same description is given. Please explain me this concept/correct me which I misunderstood. <Q> The data bus width of the processor has nothing to do with the length of the instructions. <S> The ARM processor can manipulate 32 bit values because it is a 32-bit processor, whatever mode it is running in (Thumb or ARM). <S> It just means its registers are 32 bits wide. <S> And the registers don't change when you switch mode. <S> Now, it doesn't have any implication on the length of the instructions . <S> The instructions could be encoded in any length. <S> The x86, for example, uses 8-bit instructions but is also able to work on 32 bit values. <S> For ARM, this is what changes when you switch to/from ARM and thumb modes. <S> For example, the instruction MOV R0, R1 (copy the contents of the 32-bit R1 register to the R0 register) is encoded in the following way: <S> E1A00001 for ARM <S> (32 bit) 4608 for Thumb (16-bit) But the processor, in the end, will perform exactly the same operation, and it will do it on 32-bit wide data, whatever the mode. <S> This ability to switch modes simply allows you to decide on the compromise between code density and flexibility. <S> You can pack more instructions in a kB of code with 16-bit instructions, but the 32 bit instructions are more flexible (they offer more features and you can do more with a single instruction). <A> The canonical reference for this information is the ARM Architecture Reference Manual. <S> The situation is complicated by there being many versions of this document, and many architectures. <S> Generally, the most recent versions will be more complete in their formalisation (since there can now exist 16, 32 and 64 'flavours' which could be conflated). <S> For example, ARM DDI 0487C.a ARMv8, for ARMv8-A (edited) The Execution states are: AArch64 <S> The 64-bit Execution state. <S> This Execution state: <S> • Provides 31 64-bit general-purpose registers, of which X30 is used as the procedure link register. <S> • Provides a 64-bit program counter (PC), stack pointers (SPs), and exception link registers (ELRs). <S> • Provides a single instruction set, A64. <S> AArch32 <S> The 32-bit Execution state. <S> This Execution state: <S> • Provides 13 32-bit general-purpose registers, and a 32-bit PC, SP, and link register (LR). <S> The LR is used as both an ELR and a procedure link register. <S> Some of these registers have multiple banked instances for use in different PE modes. <S> • Provides two instruction sets, A32 and T32. <S> In ARMv8 the possible instruction sets depend on the Execution state: AArch64 <S> AArch64 state supports only a single instruction set, called A64. <S> This is a fixed-length instruction set that uses 32-bit instruction encodings. <S> AArch32 AArch32 state supports the following instruction sets: <S> A32 <S> This is a fixed-length instruction set that uses 32-bit instruction encodings. <S> T32 <S> This is a variable-length instruction set that uses both 16-bit and 32-bit instruction encodings. <S> In previous documentation, these instruction sets were called the ARM and Thumb instruction sets. <S> Critically, the A32 and T32 instruction sets are used within AArch32 where the general purpose registers are 32-bit . <S> The T bit only affects the instruction decode, selecting between the 32 bit instruction set, or the mixed 16/32 bit instruction set. <S> It's also worth noting that any documentation which refers to Thumb being a 16 bit instruction set <S> is very old. <S> Assuming you are studying now, rather than working with an old design, the most restricted subset which is relevant is the ARMv6-M architecture, and the instruction set implemented by the Cortex-M0 processor. <A> Thumb mode doesn't turn the ARM into a 16-bit CPU. <S> The registers, ALU and memory buses are still 32-bit. <A> Roughly speaking, a CPU instruction is a particular sequence of bits which, when presented to the CPU, means "Add two 32-bit values and carry". <S> The exact value of bits in this sequence has nothing to do with values being added. <S> In 32-bit instruction set, this sequence is composed of 32 bits. <S> In thumb, it only has 16. <S> Obviously, thumb has less instructions (because there's less bits to encode them), but instructions which are implemented do exactly the same thing in both instruction sets.
| In Thumb mode, the CPU uses 16-bit instruction op-codes instead of 32-bit ones, to reduce program memory usage.
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If a datasheet doesn't state I2C address pins have pull-up/down resistors, does that mean I must connect them? I've got a handful of DS75S+ temperature sensors. They're SO-8 and I'm going to solder them onto prototyping boards for use with a Raspberry Pi. I want to set the addresses sequentially so I have the option of using as many as I like on one Pi (up to 8, but I only have 5). The datasheet says nothing about internal pull-up/down resistors on the address inputs, so does this mean they're floating? Tying them high is trivial, tying them low might be a little messy. But I don't want to set it floating only to find later that I've caused a problem (especially as, once tested, I'll probably pot them in epoxy). The "Detailed pin description" says "7 A0 Address input pin." etc., there's no typical circuit and the block diagram has nothing of any help. I've seen how do you typically tie the address pins and WP pin of I2C device? Do these lines need a pull-up/down resistor or just tied directly (closed) which has a helpful answer that unfortunately starts "Completely dependent on the specific IC you are using, and would be listed in the data sheet. " which isn't true here <Q> I had a quick check of the datasheet (No I did not read the complete sheet but looked at every instance of the text 'address'). <A> The datasheet says nothing about internal pull-up/down resistors on the address inputs, so does this mean they're floating? <S> As explained in other answers, if the datasheet doesn't mention any defaults, then you should specifically set the address inputs as you require them. <S> Even if there were (weak) internal pull-ups / pull-downs, if they aren't documented in the datasheet, then you can't rely on them. <S> In this case, the address pins appear to be normal CMOS inputs with all the pros and cons that implies. <S> Some I2C devices have much more complicated address inputs than the DS75. <S> For example, some devices can allow address inputs to be floating <S> (as in that case they are essentially analog <S> inputs and have internal bias resistors) to set a different I2C address from when it is being pulled high or low. <S> However that behaviour would definitely be explained in such a device's datasheet. <S> the block diagram has nothing of any help One source of information for the DS75 which has not been mentioned in other answers, and gives you specific confirmation in your case, is the original Dallas Semiconductor datasheet, before Maxim took over the company. <S> Not for the first time after a takeover, the change to a "new company" datasheet format actually lost some of the original information :-( <S> Look at the difference in the block diagrams shown in the new and old datasheet versions. <S> Here is figure 1 in the Maxim DS75 datasheet: <S> But figure 1 in the old Dallas Semiconductor version of the DS75 datasheet showed this - see the part which I've highlighted in red: <S> That confirms that the DS75 address pins should be connected to 0V or VDD to set the specific address you require. <S> Whilst that can happen (e.g. when a Fab belonging to the "new company", starts to be used to manufacture a device, which was previously manufactured at a different Fab belonging to the "old company") <S> my experience is that big functional changes are avoided, if it would cause a lack of compatibility with the original devices. <S> The lack of any documentation to the contrary in your case, means that I'm confident the part highlighted in red above still applies to Maxim-branded DS75 devices. <A> There is a clue here: - Input current each <S> I/ <S> O pin <S> 0.4 < VI/ <S> O< 0.9 VDD <S> -10 <S> +10 <S> μA <S> This tells me that there will be either a leakage into or out of the pin of up to 10 uA <S> and therefore if you leave the pin open it might go low or <S> it might go high (when trying to leak out 10 uA). <A> For completeness, having inadvertently put it to the test with my own heavy-handedness when using the heat from my hand to test my code: The address lines in this case are internally, and very weakly, pulled high as discovered by breaking the connection to ground and scanning the bus. <S> A floating input picks up noise and causes the address to change, leading to comms errors.
| The data sheet does not mention anything like pull-ups/downs or defaults thus you must assume the worst case: you have to tie each pin either high or low. Just for completeness, there is a tiny possibility of a new company making changes to a device after a takeover, such that old datasheets don't fully apply.
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Is a decoupling capacitor theoretically required across the supply pins of a Hall sensor? I am using the Hall-effect sensor OPTEK 3075S to ensure every step commanded to a stepper motor is executed. In the datasheet, they recommend connecting a 100nF capacitor across he input pins for "stable operation". I have been successfully using this sensor without cap with short supply wires for months, and now that I am planning on using this sensor after a few tens of meters I wonder if this is required. I know it is easy enough to add the cap anyway (although I would need to build another PCB), but I do not like to add components without understanding their added value: how would the current source of the Hall element (which apparently is a voltage regulator across a constant resistance) be less stable without cap, given that there is nearly no load, and that the power supply rejection is not so important (since this is a digital sensor)? Certainly the opamp is the one which may become unstable? Is it possible that stability would only result in unstable transients but the output would not be affected after the step has been made? Is there a way to quantify/estimate the gain in stability margins by adding the cap in (with respect to what? Supply variations? Hall element output?)? Note: the sensor state transitions are located dead-on between two steps (45° per step) for stable output for every motor positions, and the motor step rate is 5 steps per second. <Q> There are several reasons to use the 100 nF capacitor: <S> - The data sheet recommends it <S> The internal chip current changes by 2 mA when it switches <S> The external load (attached to the open collector output) might switch an extra 20 mA <S> The supply voltage may not be a perfect voltage source i.e. it might have ESR and ESL components that would cause a ripple voltage on the supply when the device switches <S> The cable supplying power to the chip might be long and have more ESR and ESL <S> Switching times are around 100 ns i.e. it switches at a fairly high speed. <S> I don't see any good competant reason NOT to use it. <S> how would the current source of the Hall element (which apparently is a voltage regulator across a constant resistance) be less stable without cap, given that there is nearly no load, and that the power supply rejection is not so important (since this is a digital sensor)? <S> The first part is impossible to say without detailed knowledge of the device. <S> There is an internal load and that load changes by 2 mA each time the device switches and, it switches in about 100 ns. <S> There is a graph in the DS that shows this. <S> Power supply rejection being unimportant is not a healthy place to start making a counter argument. <A> You probably know that most voltage regulators require decoupling capacitors and their input and output. <S> Is there a way to quantify/estimate the gain in stability margins by adding the cap in? <S> If you had the complete design (including transistor models) of the chip you could in theory simulate it. <S> In the real works only the people who designed this sensor can do this. <S> A workaround could be to just try it with the long cables and use an oscilloscope to see how much ringing there is at the slopes of the signals and on the supply voltage of the sensor. <S> You could hand-solder a capacitor where it needs to be <S> , repeat the measurements and look for any differences. <S> I guess that the manufacturer simply suggests to use a decoupling capacitor as that would eliminate a lot of issues with their customers beforehand. <A> I've diagnosed oscillations in Brown_out_Detectors, where the designers were ignoring the onchip gain-of-10,000,000 (yes 10 Million), interacting with the onchip bandgap voltage reference, and the lead inductance. <S> Put a cap right by the sensor. <S> So-----theoretically?
| I think the capacitor is mainly needed for the voltage regulator. Yes, the cap is required.
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difference in reverse voltage of diodes? I have a 9V circuit to operate latching relays. Because there are instances where two independent DC power supplies meet briefly, I decided to implement schottky diodes in the circuit to prevent them from affecting one another. Initially, I used IN4004 blocking diodes ( https://docs-apac.rs-online.com/webdocs/14f5/0900766b814f5b51.pdf ), but took them out because I realised they could only carry 1A of current. I subsituted them with SBYV27-200-E3 switching diodes ( https://docs-apac.rs-online.com/webdocs/14ae/0900766b814ae442.pdf ). I understand that switching diodes are used in instances where there are rapid and frequent switches in the power supply in both directions i.e. AC, however I have used them since they were already available and could carry up to 2A of current (vs 1A of IN4004). When I used my relay to disconnect a part of the circuit (9v measured before disconnection), I found it unusual when I measured around 2V in potential difference (with reference to GND) upstream to the diode at the disconnected part of my circuit. I was puzzled by why there was 2V when that part should have been disconnected by the relay, and any reverse current blocked by the diode. I then took those switching diodes out and switched back to the IN4004 blocking diode. With these, I measured 0V when the relay perform the same disconnection, showing that the circuit was effectively cut off by the relay and no reverse current was allowed through the blocking diode. Thus, my question is why are my IN4004 diodes able to effectively block all current, while my SBYV27 diodes seem to allow some reverse current through? I want to clarify it this is indeed the case and whether a switching diode is similar (or not) to a blocking diode in blocking reverse current. simulate this circuit – Schematic created using CircuitLab I added a schematic. This schematic only has one DC source, but I tested the voltage measurement on this as well. Circuit maker does not have the exact relay model I am using so I combined two of them here. When one the coil is activated, one circuit closes and the other opens, and vice versa when the reset coil is activated. My circuit might appear unusual but I've designed it to work such that the coils are operated by the very same current that runs through the switching circuit it controls. <Q> Schottky diodes tend to have lower forward voltage at the same current than similar conventional diodes, but also tend to have considerably higher reverse leakage. <S> One follows from the other if you look at the Shockley diode equation. <S> Usually a 1N400x or even a 1N4148 <S> (200mA rated diode) is fine for the flyback diode on a small relay coil. <S> It only sees the coil current and only briefly. <A> The SBYV27 diode has a reverse leakage current of 0.4 uA (at 10% of its maximum reverse voltage and at 25 degC). <S> See figure 4 in its data sheet . <S> So it could be leaking maybe 0.4 uA through D3 (then via the coil) and through D1 to your open circuit measurement node. <S> If you are using a multimeter with 1 Mohm input impedance you could measure 0.4 volts due to 1 uA flowing. <S> If you look at figure 5 in the 1N400x data sheet from Vishay <S> you will see that at about 5 volts (reverse) <S> the leakage current is about 20 nA at 25 degC. <S> That current into a multimeter of 1 Mohm input impedance would produce a voltage of 20 mV. <A> What's all this stuff about reverse diode leakage? <S> Note above diode reverse voltage leakage is 0.5 uA typ at 25'C and curve is fairly flat (i.e. its a really unstable constant current source). <S> In general, reverse leakage current is fairly flat (until zener voltage) but temperature dependent (rise with T) and is higher for diode rating and higher with faster switching silicon diodes and much higher in lower voltage Schottky diodes. <S> When relay coils release, the voltage reverses due to reverse EMF and the diode clamp current is initially the same as the coil as it decays down with somewhat a linear L/R=t time constant. <S> Since the question is about latching relays, we know they take much more power initially to switch from a magnetic resting position than a non-latching but can save energy over time. <S> We know everything has thermal mass and thermal resistance and a time constant to reach critical temperatures and we need to keep safe margins and not choose parts operated at their Absolute Max limits for reliability considerations. <S> Check your assumptions <S> 1N400x is rated for 30 A for 1 cycle 60 Hz and not 1 AMP. <S> The old workhorse 1N400x is rated for 1 A continuous and 30 <S> A surge for 1 cycle. <S> (The x suffix indicates peak inverse voltage (PIV) in units of 100 hundred volts. <S> Like all semiconductor switches can handle much higher peak currents than the sustained current. <S> This is often described in several ways; I vs t with an inverse characteristics with graphs (trending towards deleting these curves in datasheets for common legacy parts <S> Maximum non-repetitive current for 1 cycle (based on either 10 ms sine or 16.7 ms (60 Hz) sine) <S> The other part of this question is about a fast recovery diode which has more leakage (0.5 uA) than a 10 Mohm DMM <S> so it reads a pullup voltage <S> when reverse biased (no big issue with power drain here, I expect). simulate this circuit – Schematic created using CircuitLab Conclusion <S> 1N400x is rated for 30A max 1 cycle 60 Hz and is likely to work perfectly for your miniature latching relay.
| At high junction temperature you can have VERY significant reverse current flowing in a large low-voltage Schottky diode (tens or even hundreds of mA).
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What is the difference between narrowband and wideband? Im a bit confused between the difference of Narrowband signals and Wideband signals.Is there a specific value of bandwidth which classifies a signal into Narrowband or Wideband?What values of bandwidth would be considered as Narrow? <Q> These are just rough labels that depend on context. <S> This is no different than "big" and "small" in everyday language. <S> Saying something is big or small without context is meaningless. <S> A small house, for example, is still a lot bigger than a big coffee cup. <A> My experience is that "narrowband" usually means "whatever new thing we're switching to that uses less bandwidth", and "wideband" usually means the opposite. <S> Of course the specifics are contextual; but for example, in the context of commercial / amateur FM radio communication, wideband right now means 5kHz deviation / 25 kHz bandwidth, and narrowband means 2.5kHz deviation / 12.5 kHz bandwidth. <S> But in a more historical context, the same terms once referred to 15kHz deviation and 5kHz deviation -- <S> as technology marches on, what was once 'narrow' is now 'wide'! <S> In the physics context, there are more specific definitions: "Frequency modulation can be classified as narrowband if the change in the carrier frequency is about the same as the signal frequency, or as wideband if the change in the carrier frequency is much higher (modulation index > 1) than the signal frequency." <S> ( https://en.wikipedia.org/wiki/Frequency_modulation ) <S> But which signals exactly will be 'narrow' or 'wide' <S> depends on how you quantify "much higher". <S> Modern FM communications are definitely narrowband; commercial FM radio is definitely wideband. <S> Stuff in between is fuzzy. <A> I've heard the terms used for many things mo to me over the years. <S> A narrow band signal is either a single tone or a carrier with on a single modulation applied. <S> So it's frequency spectrum is narrow. <S> Like a clock or a some AM signal. <S> A wide band signal may be modulated at a higher rate (hence a broader spectrum) or have multiple modulation signals applied such as OFDM. <S> Or the sidebands aren't suppressed well. <A> I think, the background of the definition for these terms (narrowbadn vs. wideband) is the ratio "occupied signal bandwidth - to - available channel or system bandwidth". <S> For example, to demonstrate the existence of a negative group delay we can send a amplitude-modulated signal through a band-stop network. <S> This experiment works only under the following restrictions: <S> The bandwidth of the modulated signal must be small enough not to exceed the quasi-linear region of the filters phase response. <S> Hence, it must be a "narrow-band signal" which, certainly, is much smaller than the filters bandwidth- <A> Examine your dataeye, or your Bit Error Rate, as you change the input bandwidth for a receiver. <S> Many of the sidebands are not needed, if you accept sloppy dataeyes. <S> Thus a signal transmitted wideband, with lots of sin(x)/x sideband (USB & LSB) energy,may be used in a narrowband receiver.
| Obviously, a narrowband signal has a narrower bandwidth than a wideband signal in the same context, but that's about all you can tell from that.
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Powering 12V device from 5/9/12V powerbank I recently bought the Litionite Falcon powerbank that has one of its outs designed for 5/9/12V. The 12V is at 1.5A which I thought would be enough for my 12V/0.8A camera. However, connecting the camera to the charger with a USB to 0.7mm jack doesn't work. Indeed, I can measure only some 5.3V coming through the USB cable which explains why the camera doesn't recognise the power. How do I get the 12V from the powerbank to charge my camera? Do I need a special USB cable? Do I need one with a step up? Or is it enough to wire the USB differently? <Q> Your Litionite Falcon power bank uses the Qualicom <S> proprietary Quick Charge 3.0 technology. <S> To enable 9 or 12V, your device have to provide certain sequencing of voltage levels on D+ and D- data lines, aka "handshake". <S> The details of QC protocols are not publicly disclosed, and only occasional information is available on how to conduct the handshake. <S> The most comprehensive details were eventually published in US patent Application US2014122909 . <S> If you can read the awkward patent language, you can start there. <S> Alternatively there are certain ICs that support the protocol, namely AP4370 by Diodes, NCP4371 by ONSemi, and CHY103 by Power Integrations. <S> So some bits of information about actual protocol have been leaked. <S> For example, Texas Instruments PMP9773 Reference Guide describes the protocol as follows: <S> According the description in the CHY100 datasheet, the processes to enter <S> QC2.0 are: <S> − Apply a voltage between 0.325 V and 2 V to D+ for at least 1.25 seconds <S> − <S> Discharge the D- voltage below 0.325 V for at least 1ms while keep the D+ voltage above 0.325 <S> V − Apply the voltage levels in Table 3 to set the output voltage. <S> (must keep the D+ voltage above 0.325 V) <S> The table of DC voltages that you need to set on on D+/D- wires looks like this: To get an idea how the QC protocol has evolved to version 3.0, the following presentation can help. <S> Version 3.0 introduces pulsing protocol, each pulse can decrement or increment VBUS by 200 mV. <S> So this is up to you which way to experiment with. <S> Given your 12V@0.8A camera requirement, you probably be better off with a 5-to-12V booster from eBay. <A> Found the following at this site: https://hackaday.com/2017/03/04/unlocking-12v-quick-charge-on-a-usb-power-bank/ <S> no idea <S> if it actually works <S> Sam Mallicoat says (November 13, 2017 at 4:03 pm) : <S> It’s easy to set a QC3 supply to 12V with just two resistors and a toggle or push button switch. <S> Here’s how: Take a 10K Ohm and a 2.2K Ohm and solder in series across the Vbus (red) to ground(black wire). <S> The tap between the two resistors will measure about a Volt. <S> Solder D+ (green to this tap. <S> Then wire the D- (white) through a N.O. switch to the same tap. <S> Apply adapter or power pack supply and wait 1.5 seconds to push the button. <S> Presto, 12V @1.5! <S> No need to hold the button, the supply stays at 12. <S> the hackaday article also links to this: http://blog.rnix.de/12v-from-a-usb-powerbank/ <A> Your power bank uses Quick charge technology. <S> Since your device doesn't do that, you need to use a middle man.
| You can make them yourself, or you can buy a quick charge trigger board on the usual eBay-like websites. To enable 9 or 12V, you have to handshake between the device and power bank. Since your camera doesn't do that, you need to make a device between your camera and the powerbank that generates the D+/D- signals.
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With parallel termination of a transmission line, why does the receiver see the full voltage? With parallel termination of a transmission line, the line characteristic impedance \$Z0\$ is matched with the pull-down resistance \$Rt\$. But, in that case, shouldn't the receiver see half of the transmitted voltage? In my understanding, with no reflection, the whole situation can be brought back to a voltage divider with equal resistances. Obviously, I made an error somewhere. So, how can the receiver see the full voltage with parallel termination? <Q> Line impedance is not the same as a series resistance which would cause a voltage drop. <S> Line impedance, altough it happens to be measured in ohms, tells how the electric and magnetic fields are related outside the wires (=in the space where the wave actually travels, it's not in the metal). <S> You can use the line impedance in calculations of what happens to a wave in line joints and terminations, but it has no use in ohms law. <S> In ordinary 2 wire cables, but not in waveguides, you can make most of the wave calculations by using the voltages and currents which they cause to the wires, but remember, that the wave and thus also the actual energy flow is outside the metal, it's only guided by the wires. <S> In your case the wave comes from the source, which obviously has very low internal series resistance. <S> The wave meets the matched load, no reflection is caused. <S> If the signal is DC, the rest can be calculated with ohms law. <A> The characteristic impedance of a cable is the value that should be taken into consideration when terminating to avoid reflections; it doesn't mean that it acts as that impedance in a potential divider sort of way. <S> Consider a really short 50 ohm coax terminated in 50 ohms driven by a zero ohm source at a low frequency - the full driving voltage will appear at the load both mathematically and intuitively (if you think about it). <A> I think you are confusing, case A: simulate this circuit – <S> Schematic created using CircuitLab <S> With, case B: <S> simulate this circuit <S> The difference between A and B is only at the source. <S> The voltages I indicated is the amplitude of the signal at that point. <S> Indeed you're right that there is a 2:1 voltage division between the voltage at Vsource and the output voltage. <S> That's because Rsource and Rload make a voltage divider. <S> However the voltage at Vsource isn't really the input voltage of the transmission line. <S> That voltage is only present at the right side of Rsource. <S> In case B this is more obvious as I have replaces the voltage source and series resistor with their Thevenin equivalent, a current source and parallel resistor. <S> Note how the characteristic impedance of the transmission line is not actually present in the circuit. <S> The characteristic impedance is the impedance with which you should feed in and extract power from the transmission line. <S> If you don't do that you get signal reflections.
| You get full voltage if the wire hasn't remarkable DC resistance.
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