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Longest, thinnest trace length to decoupling capacitor before capacitor is effectively useless I was told time and time again to always put decoupling capacitors as close to the IC pin's VCC as possible. Sometimes when I make complex boards, I try my best to put the capacitor lead so close to the VCC pin, that the distance between the inside of the capacitor (where the pin runs into) and the VCC pin is less than 3mm. There are times where routing the single-sided board gets so complex that I have to use 10 or 12 mil wide traces and make a track length of about 1/2 inch to even 1 inch in rare instances between VCC and the capacitor. Whats the longest length I can go with traces between IC VCC and nearest capacitor before the decoupling effects of the capacitor become useless if the width is 10mils wide? <Q> The answer is unhelpfully "as short as possible", and/or "it depends". <S> Whether it needs to be 1mm, 1cm, or even 1m will depend on factors such as switching frequency, signalling standards, whether sensitive analogue components are involved, how much EMI you can get away with etc. <S> The purpose of the capacitor is to provide a nice low impedance path between supply rails. <S> If you have a signal that toggles very slowly, and/or can cope with noise generated on the signal due to poor decoupling, then you can get away with longer narrower traces. <S> If however you have a signal running at hundreds of MHz, or into the GHz range, even a few millimetres of trace might be too high of an inductance. <S> At this point even a via might add a lot of unwanted inductance. <S> In circuits with analogue electronics, things also get critical because supply noise and ground noise have a nasty habit of finding their way into analogue front ends. <S> The simplest way to find out is to build a prototype, and test it. <S> If things work as expected and performance is within parameters, then your traces are short enough. <S> If you find issues with noise or signal problems, poor decoupling may be a culprit. <A> Here are some resonant-capacitor-inductor-Rdampen screen shots. <S> Rdampen is 1 milliOhm in the first screen-shot (the R in each of the 4 LRC is 1 milliOhm). <S> The following screen shot has all the dampening Resistors be 10 milliOHms. <S> Notice the higher frequency peaks are NOT AFFECTED. <S> You'll need more than10 milliOhms (either lumped or ESR inside the cap or trace) for 10nH and 100pF. Notice that resonance at 150MHz brings up the energy from -25dBc to +5dBc <S> Your Question was "What is the longest thinnest trace"? <S> Depends on the inductance you can tolerate. <S> Traces over air (no underlying plane) have about 1nH/millimeter or 25 nH/inch. <S> There is mild dependency upon width. <S> [its a log(1 + length/width) ] <S> Traces over planes have about 0.1nH/millimeter, depending on distance from trace to plane. <S> Thus an underlying plane (GND or VDD or otherwise) will raise the resonant frequency by sqrt(10) or 3.16X. <S> Be aware of that. <S> What can you get away with? <S> Clearly, as you are already aware, you have at least 3 degrees of Freedom: capacitor value, trace inductance, and losses due toESR or trace. <S> A mere 10 squares of trace (0.08" long, 0.008" wide) has, at 500uOhm per square, an Rdampen value of 5 milliOhms. <S> Thus many crimes of this nature are hidden under the Rdampen of the traces and Vias. <S> How to pick Rdampen? <S> Use the (tiny) formula Rdampen = sqrt( L / C ) <S> .Thus <S> 100uF and 10nH, which is the left-most LRC shunt circuit in each screen-shot, needs Rdampen of sqrt( 10nH / 100uF) = <S> sqrt(1e-8 / 1e-4) = <S> sqrt(1e-4) <S> and for that left-most LRC, use Rdampen of 1e-2 = 0.01 ohm = 10 milliohm. <S> Does this value of 10milliOhm suffice? <S> At 1milliOhm, the dip at 160Khz is -93dB. <S> At 10 milliohm, that dip has risen (been dampened) to -74dB and is quite broad. <A> It is dependent on the capacitor, the capacitor packaging, trace geometry, and PCB stackup. <S> Your best bet is to do an FEM simulation with the s-parameters of the capacitor, if you have access to the tools. <S> The effect of the trace length can be quite dramatic.
| The longer and thinner the trace, the larger the inductance of said trace, and the higher the resulting impedance (remember inductors are short at DC, and increase in impedance with frequency).
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Single Suppy Opamp for inverting amplifier produce negative voltage signal i made an inverting amplifier circuit using CA3130, the supply just the positive voltage and GND for V- supply. the circuit is used for full wave rectifier because the signal is around 1 MHz and 500 mV and i don't find any good diode for that frequency. the problem is, i expected only the positive voltage signal is produced but then the negative signal was appeared too in oscilloscope, how is this even possible? simulate this circuit – Schematic created using CircuitLab Thank you <Q> 5V/100 ohms= 50mA. <S> The output current of a CA3130 with a 10V supply is only about 4mA which reduces the output swing to 3V peak. <S> Try 10k or 20k for the resistor values. <A> Instead of discovering wheel again, how about using Ideal Diode configuration? <S> The voltage drop VF across the forward biased diode in the circuit of <S> a passive rectifier is undesired. <S> In this active version, the problem is solved by connecting the diode in the negative feedback loop. <S> The op-amp compares the output voltage across the load with the input voltage and increases its own output voltage with the value of VF. <S> As a result, the voltage drop VF is compensated and the circuit behaves very nearly as an ideal (super) diode with VF = 0 <S> V. <S> Source: <S> https://en.wikipedia.org/wiki/Operational_amplifier_applications#Precision_rectifier <S> This way you will stay in safe side in means of absolute maximum rating for input which is Vss-0.5v (Ref: https://www.intersil.com/content/dam/intersil/documents/ca31/ca3130-a.pdf ) <S> Edit: <S> I didn't state my reasoning clearly. <S> As far as i understand, the problem arises when 0.5V input voltage spec is taken into account. <S> It's hard to find low Vf and high speed diode but this changes when low Vf spec is relaxed; since 1N4148 alike components become possible to use. <S> So consider this an alternative solution. <A> I don't see any source of negative voltage other than the input signal. <S> I suspect your op amp circuit isn't working at all, and you're seeing the input signal appear at the output. <S> Connect a probe to the input and see if the output is getting inverted as expected.
| Your 100 ohm resistor values are way too low and the input signal goes directly through them and to the output.
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How to find out whether serial standard is RS232 or TTL when stated as "serial(RS232/TTL)" I am recently moving from C programming into automation engineering. I have learned about RS232 and TTL standards, and I understand that they are based on the same concept of serial communication, however TTL uses logic voltages (3.3 V or 5.0 V) to be compatible with microcontrollers, while RS232 uses higher voltages for historic reasons (signal-to-noise-enhancement). To start connecting to the "real world", I planned on using a simple printer to start practicing sending bits via the COM-Port. Here is a link to an example: https://de.aliexpress.com/item/JP-QR203-58mm-Micro-Receipt-Thermal-Printer-RS232-TTL-USB-Panel-Compatible-with-EML203/32693670343.html This device seems suitable for my needs, however, some questions remain, as I am very eager to learn, but still a newbie in electrical engineering: The interface is stated as "Serial (RS-232,TTL)" - Is there any way to find out which standard is implemented on the PCB exactly, RS232 or TTL? Another suitable device seems to be this one here: https://www.amazon.de/WELQUIC-Thermodrucker-Bondrucker-Bluetooth-Standard-Art-1/dp/B075GG7VJT/ref=sr_1_2?ie=UTF8&qid=1533543476&sr=8-2&keywords=welquic+printer Again, how can I figure out whether RS232 or TTL is implemented? This device seems to feature a Mini-USB port for USB-communication and another Mini-USB port for RS-232/TTL communication... Which cables would be needed? Is there something like a serial/Mini-USB adapter cable? As I understand, USB standard has to be converted to RS232 or TTL - So my second question basically is, how can there be a port in Mini-USB-format for RS232/TTL? <Q> The problem lies in the fact that the data stream as produced and consumed by a UART does not have an official name. <S> In the 'good old days' the UART data was always send through line drivers <S> * before they came out of the equipment. <S> The connector and the signal levels outside the equipment where formalized in the "RS232" standard. <S> Nowadays we use a lot of micro controllers en single board computers and the signals out of the "UART" interface are used directly. <S> Which raises problem: how do you call that type of interface? <S> With which the mean the RS232 data format but with TTL signal levels. <S> Therefore there is a high probability that the equipment has TTL level signals. <S> But you can only find out for certain after you bought one or talked to the vendor. <S> To connect such an interface to a standard PC there are USB-RS232 converters/adapters. <S> Yes, they are called as such although they are not conform to the RS232 standard in neither the signal levels nor the connector pin-out. <S> Which proves the point I made above. <S> * <S> The line drivers also inverted the UART signals which gave raise to more confusion: is the idle state high or low? <A> Main difference between industrial RS-232 and most of simple USB to COM adaptors is the voltage levels. ' <S> True' RS-232 use positive and negative voltage values. <S> USB/COM adaptors are mostly working in TTL levels (which is enough for most of project when you want to communicate between PC and your microprocessor). <S> To ensure about full industrial RS-232 you would need an oscilloscope and confirm negative/positive voltage swing. <S> (You can expect signal levels +/- <S> 15V. Check here section Voltage levels ) <S> Your first printer is well documented, for example here . <S> It uses TTL levels and you can safely connect it to your microprocessor (check if you using 5V as Vcc or 3V3: <S> in case of 3V3 you may need a level shifter). <S> To confirm second printer, you need a full datasheet, because information on the webpage is confusing. <S> Also, using this connector as COM port is not bringing any clarity. <S> Ask the manufacturer/seller for datasheet. <A> it's unclear what the device does <S> it may support both UART and RS232 voltage levels with a dip-switch or jumper setting to configure them. <S> or it may support a form of RS232 with a reduced voltage swing (0 to +5 instead of -12 to +12 ) Despite 0V being in the "indeterminate" no-mans land <S> most RS232 reveivers will see anything below about 1V as being low.
| You will find that the interface is often referred to as "R232-TTL".
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AND Logic on 24 Volt Signals What kind of chip and/or plug and play device could handle multi-input boolean logic with 24 volt signals? I have three sets of 20 reed Hall effect, three-wire switches that each output a 24 Volt signal, and I need to combine them down to simply 3 24 volt signals using AND logic (where the signal is high only if all reed switches in a given set are also high). I am currently using a PLC, but I think that is way more functionality and price than I need. I thought about powering each one of the reed switches using the signal wire from the previous one, but I sincerely doubt these reed switches can handle the amperage that would need. Edit: @Transistor You are right, these are three wire Hall magnetic field sensors, not true reed switches. A = AND(A1, A2, A3, A4...) or rather A = AND( AND(A1, A2), AND(A3, A4)) etc.The AND logic is the A is high only if all A# are high. The same is true for B, and C. <Q> For each input of the AND gate connect the anode of a diode to that input. <S> Connect all of the cathodes to together, and connect a pullup resistor from the diode cathodes to 24V. If any input is low the output will be pulled down to about 0.7V. <S> If all of the inputs are at 24V the output will be 24V. Diodes that can handle 24V are quite inexpensive. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Standard relay logic. <S> Diode D1 protects the reed switches against arcing when the relay is switched off. <S> In the old days this sort of problem was all solved by relay logic. <S> In this example RLY1 is energised when all the required switches are closed. <S> Depending on the logic this may be enough but if you are using single-contact reed switches then additional relays may or may not be required to give additional contacts. <S> Diodes can also be used so that contacts can be used on several circuits but this depends on your logic. <S> Edit: @Transistor <S> You are right, these are three wire Hall magnetic field sensors, not true reed switches. <S> Right-ho. <S> They're not reed switches at all. <S> Start again. <S> simulate this circuit Figure 2. <S> The problem. <S> You can't daisy-chain too many of these switches before voltage drop becomes an issue and following switches won't work. <S> Some of these sensors have a wide operating voltage and low quiescent current. <S> I suggest the following to see what your options are: Get the specification (and share it with us). <S> Find out The operating voltage range. <S> The queiescent current (when no magnet near). <S> The no-load on current (magnet near) which will probably be mostly due to the indicator LED, if there is one. <S> The voltage drop between the input and output when on and load present. <S> With this information you should be able to work out how many you can connect in series. <S> simulate this circuit Figure 3. <S> If, for example, the switches will work down to 12 V and the voltage drop across each was less than 1.2 V <S> then ten could be connected in series to drive a 12 V relay. <S> Two chains of these would drive two relays and series connection of their contacts would give an output when all 20 switches are on. <S> Note that a downside of this arrangement is that all sensor indicator lamps (if they are fitted) will be off after the first open switch. <S> You will always have to fault find from left to right. <S> Report back. <A> Something simple like a 3 * 8 input AND would seem to work. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Obviously the diode input element can be expanded to suit any grouping you like. <A> Use a 7400 series AND and/or NAND gates, cascaded if necessary, and to make your 24V fit its 3V to 5V input range (depending on what family (HC, HCT, LS…)) <S> you use, add voltage dividers to your two 24 V outputs. <S> On the output, use a (weak) pull up resistor against 24V and an NPN transistor to pull it down. <S> It's an inverter, hence the NAND recommendation. <S> In a two-input AND function, this would look like simulate this circuit <S> – Schematic created using CircuitLab
| A 24V AND gate is easy to do with diode logic.
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Can I use a 3.3V linear regulator to level shift a 5V square wave down to a 3.3V square wave? I have an output of a comparator that is a 5V square wave. Can I feed that square wave into a 3.3V regulator to get a 3.3V square wave? The 3.3V regulator I have is the LP2950. <Q> What is the comparator part number? <S> Does it have an open collector output? <S> – <S> Transistor <S> It's an LM393, and it has an open collector output. <S> – Shock-o-lot simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) Comparator with 5 V pull-up. <S> (b) Comparator with 3.3 V pull-up. <S> Figure 2. <S> LM393 internals showing the open-collector output, Q8. <S> The open-collector output can be used as a level shifter for this application. <S> It also makes a very simple way of ANDing the outputs of two or more comparators as, for example, in a window comparator circuit. <A> No, this is not an appropriate use of a linear regulator. <S> Your comparator may not be able to drive enough current to start up the regulator. <S> Most linear regulators require capacitors on the input and output to guarantee stability. <S> These capacitors will take some time to charge when the output of the comparator goes high, so the "response" from the regulated comparator will be much slower. <S> If the output of the comparator changes frequently enough, the regulator may never turn fully on. <S> Linear regulators generally do not make strong guarantees about their behavior when their input is cut off. <S> If there is no current drawn from the output of the regulator, its output may take some time to go to zero. <S> Use a resistive voltage divider, or a level shifter IC. <A> Unless the square wave is extremely slow (<10Hz), I definitely would not use a linear regulator due to the input and output capacitance required for it to function. <S> Since, as you have said, the voltage may not be "exactly 5V", I would use a 3.3V Zener Diode for this application. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The Zener will have some parasitic capacitance, so give it a go <S> and hopefully your application isn't too high-speed. <S> Hope this helps. <A> You can use this https://www.sparkfun.com/products/12009
| Linear regulators have a minimum input current. Voltage regulator weren't meant to be use that way. If your input (whatever the 3.3V square wave is going into) is high-impedance, you can choose a decent size resistor (1k+) and reduce wasted power (which Zener regulators are notorious for).
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Best way to flatten PCB again after cutting with utility knife I cut my presensitized PCBs down to size with rulers and a utility knife when I need a custom PCB. I noticed every time I cut one, the cut is always clean, but the copper on the board at the same edge as the cut I made is uneven. In fact it curls up and that makes PCB development difficult because the board isn't 100% flat to the glass when exposing it. Here's a picture illustrating what I am talking about: Currently, I use sandpaper to try to sand the excess copper so the board is as flat as possible but it wears the sandpaper fast. Sometimes I use the utility knife at a 45 degree angle on the edge. While this often removes the curl, it also removes at least a good 30mils from the copper. What's the best way to remove the curl without wasting a lot of PCB material or wasting any other materials? I don't think sandpaper is up to the task. <Q> You should use a "scoring knife" (carbide-tipped), and make pre-cuts on both sides of PCB, and you won't have this issue. <A> I use a large metal file for that. <S> I put the file on a flat surface and move the PCB against it. <S> It removes copper burs and finishes the PCB material. <S> I'll even do that to industrialized PCBs that were separated using V-scores and to remove mouse bites (tabs that keep individual PCBs connected to the panel during routing/milling). <A> They can afford to waste boards as the customer picks up the bill over time, but for simple stuff Ali Chen might have a better solution.
| Board houses put 'dummy' boards on top and below the boards they are machining to prevent burrs at holes, slot cuts and edges.
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Logic: Output a 1 when a = 1 and b = 0 I'm a hobbyist just discovering logic gates and would like to know, is there a logic gate that will only output a 1 when a is a 1 and b is a 0? Truth Table a | b | y---------0 | 0 | 00 | 1 | 01 | 0 | 11 | 1 | 0 Here is a schematic I made, that I think should work in theory, but would like it in one package. Are chips like this readily available or is everything made up of individual gates? OFF TOPIC EDIT:IMGUR is currently not working with SE for me or for people in the comments so the way I got around that was to go imgur.com and uploaded the picture manually. Then right click the picture > Copy image address and then paste it here like so:  <Q> I'm pretty sure half a 7402 is all you need: use one gate to invert A and the other to NOR it with B : <S> simulate this circuit – <S> Schematic created using CircuitLab <S> A <S> A' B <S> | <S> Y------------+---- 0 <S> 1 <S> 0 <S> | 0 0 <S> 1 <S> 1 <S> | 0 1 0 0 <S> | <S> 1 1 0 1 <S> | 0 <A> This can be implemented with an AND gate with one inverted input, or with a NOR gate with one inverted input. <S> Devices that implement this logic are the (SN)74xx1G58 , (SN)74xx1G97 , and (SN)74xx1G98 . <A> Something like an SN74LV1T125 <S> might work. <S> Hook A to A, B to OE, Y is your output. <S> Since the output tri-states, you will need a resistor to ground on the output. <A> You could implement the logic with a single 7437 logic chip using only 3 of the 4 gates in the package. <A> There are many chips which have both enable Hi and Lo for other features like One-Shots but not specifically this. <S> You can also use INV + NOR gate and swap inputs. <A> If all you want is to get the function in one package, you can do it with a multiplexer: half of a 74x153, for example. <S> Connect A and B to the address pins and make all but one of the inputs low, while the AB' pin is high.
| You could simply use a buffer chip with an active low enable.
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Is there a lower limit on operating frequency of CMOS SRAM? I study computer science, and so I am in the deep end here.I have set upon designing a machine which requires RAM. I found a listing for Toshiba's 128KB (8 bit per word) SRAM ( TC551001BPL ). I intend to run this at a very slow clock, maybe 1-10kHz or so (maybe even single stepping it with a button). Due to this I figure that I never have to worry about any sort of transition, propogation, etc times , as a clock cycle of 100 microseconds is orders of magnitude higher than the timings of any chips today, which is usually in 10's of nanoseconds (bus transcievers, D-Flip Flops, binary counters, RAM, etc). However, this made me wonder, is there a minimum frequency at which I must operate the RAM? I could not find anything like that in the datasheet, but I'm still not sure. Could the fact that this RAM is asynchronous cause this issue? This is slightly off topic, but I only need 1024 word, 8 bit word of RAM or so. Any IC recommendations that better suit my application would be extremely helpful. <Q> No, there is no minimum frequency because it's static RAM. <S> Any digital circuit that's described as "static" doesn't have a minimum frequency - it can operate with all signals held static for an extended period of time, hence the name. <S> "Dynamic" circuits have a minimum frequency. <S> Just because your circuit operates at a low frequency doesn't mean you can ignore all timings, however. <S> You still have to meet the required setup and hold times, as well as the maximum rise and fall times. <S> The latter usually aren't a problem if you stick to a single logic family and don't use 4000 CMOS. <S> (74HC/HCT/LS will be just fine.) <A> I checked a 1997 data sheet for the TC551001BPL. <S> Cycle times for read and write cycles are given on page 4. <S> Note that there is no maximum time specified, so you can go as slow as you like. <S> If you step with a button, debounce the pulse carefully! <S> Availability of your Toshiba SRAM seem limited. <S> A quick parametric search on DigiKey threw up the AS6C62256-55PCN from Alliance Memory. <S> This is a 32k x 8 SRAM, is low cost, and comes in a 28 pin DIP package. <A>
| It is a static RAM, so no there is no minimum clock cycle.
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How to measure tilt on a dynamic device? I'm trying to instrument a vehicle component. I want to measure a component's angle during a drive cycle. The component has a single degree of freedom, it has a fixed axis. I'm trying to figure out an efficient way to measure this. I'm hoping to use Internet of Things (Raspberry Pi / Arduino) technology as appropriate. Without going into details, I must power this thing via battery. I'm assuming the testing will be done during daytime, on asphalt or concrete pavement. Sensors that seem to offer possibilities: Accelerometer (X,Y,Z) Electrolytic Tilt Sensor Passive InfraRed (PIR) sensor Radar sensor Sonic sensor Laser Time of Flight My original thought was to use an accelerometer, attach it to the bottom of the device and analyse the direction of gravity to deduce component's angle. I know this would work on a fixed vehicle, but my situation is anything but fixed. I won't be able to deduce vehicle travel direction. Additionally, my vehicle's speed is not constant. I will be monitoring acceleration (but subtracting out gravity) while I'm looking for other characteristics... Because of my huge variance in accelerations, I don't think a fluid based Electrolytic Tilt Sensor would work either. My next thought is to accurately measure the distance from the component to ground. I'd fix the sensor to the movable end of my component, then calibrate distance observed to component angle, while vehicle is at a complete stop. I can monitor distance in real time to deduce component angle relative to the ground. Scratch passive infrared. I'm pretty sure sunlight will mess that up bad. Radar is a pretty nice technology, but seems to be oriented to metal, water or humans , and has a very slow response time. Scratch that. And that leads me to either a sonic distance sensor or Laser Time of Flight sensing. At first glance both methods are cost effective and seem to work in the ranges I'm interested in. But I have no experience here. What I don't know is if on road testing in daylight will affect these devices. Will ambient noise mess things up? What about reflected sunlight? Does anyone have any experience here with those last two technologies (Sonic Distance or Laser Time of Flight)? Are there any technologies that offer an alternative method for measuring a component's angle (relative to the ground) in a dynamic, moving vehicle, perhaps something I've missed? Note: I've done a lot of work using magnetometers and accelerometers (Ugh, remember the Vector Cross Product?) but I've never touched a gyro device. Are there opportunities there? <Q> My Ipod 6th gen, and other mobile devices have a Gyro controller (Tilt sensor) which with a suitable app can display X,Y tilt with a 0.1deg resolution fast enough > 10 sps. <S> My iPod has a 2D bubble level application in a Multi-function App. <A> I want to measure a component's angle... <S> The component has a single degree of freedom, it has a fixed axis. <S> I'm trying to figure out an efficient way to measure this. <S> " <S> This" being what? <S> An angle in vehicle coordinate system or an angle with a horizon? <S> If your component has an axis then a simple potentiometer or encoder can give you an angle between component and vehicle. <S> My original thought was to use an accelerometer ... and analyse the direction of gravity to deduce component's angle. <S> OK, let's assume you want an angle with a horizon. <S> Yes, that is what accelerometers often used for. <S> And this is probably your best option. <S> I won't be able to deduce vehicle travel direction. <S> Umm... <S> so, what is it <S> you are trying to measure with accelerometer, angle or travel direction? <S> These two are completely different. <S> If you mean that you won't be able to interpret accelerometer's readings because of vehicle movement then you haven't done your research. <A> My first thought, if I understand your application, is an integrated Orientation Sensor. <S> The de-facto is the Bosch Sensortec BNO055. <S> You can find a breakout board on Adafruit, or put it on your own PCB. <S> The sensor has a 3-axis accelerometer, a 3 axis gyroscope, and a 3 axis magnetometer (not super useful in this case). <S> Most importantly, the chip has its own CPU, loaded with a proprietary Sensor Fusion algorithm. <S> In addition to the raw sensor data, the chip outputs Euler Vector and Quaternion orientation data. <S> The chip supports i2c, so you should be able to connect it to most IoT platforms. <S> The limitations of this approach: <S> The Sensor must be mounted on the thing you want to measure. <S> I'm not totally clear on your application, but I'll assume that's possible. <S> The Sensor Fusion can only handle so much angular velocity. <S> Again I don't know the details, but if you are trying to measure the angle of a crankshaft or something that's really moving, you will have trouble. <S> Feel free to add more application details if I've misunderstood.
| By using a combination of 3-d accelerometers, 3-d gyros and sensor fusion algorithm the sensor orientation can be easily calculated with good precision.
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What is the purpose of the resistors in this schematic? I’m looking into using a CD4021B shift register with an Arduino Uno, and found a schematic online detailing how to use it. However, I am confused as to what the point of the 10kohm resistors is, would the current not just pass straight through from the 5V to the shift register when the switches are closed? To me it seems like the resistors are pointless, however I am rather new to this all so if someone could enlighten me as to their purpose it would be much appreciated. Schematic: <Q> These are called "pull-down resistors". <S> They prevent the inputs to the IC from floating, and thus being in an indeterminate state. <S> The point is to put the inputs in a known state (logic 0) when the switches are open. <S> When the switches are closed the resistors are effectively "overridden" and a logic 1 is applied to the inputs. <S> If you have an IC with active-low inputs, then you would want pull- up resistors, which pull the inputs up to Vcc until the switch connects the pin to ground. <S> See this question or this question for more details. <A> It's a set of drop down resistors to pull the input low when the button isn't pressed. <S> You can think of the switch and resistor as forming a voltage divider where the switch is either \$0 \Omega\$ or \$\infty\Omega\$ depending on whether it's pressed. <S> so the voltage at the input is either 0V when not pressed or 5V when pressed. <A> The CD4021 is a CMOS device. <S> These have very high input impedance and if left floating (disconnected) could assume either a high or low input condition. <S> Closing the switch will, obviously pull the input to +5 V which will be read as logic '1'.
| Addition of a pull-down resistor ensures that when the associated switch is opened that the input is pulled to zero rather than leave it in the undefined state.
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Creating required voltage across a device using basic electrical principles Let me explain what I mean by an imaginary case. Let's imagine a 10 V battery in series with two resistors equal in resistance value. The voltage in between those resistors is 5V according to the voltage division rule. I have a component (does not matter what it is) which needs to be powered up by 5 volts. When I try to use that 5V voltage value in between those resistors to power it up, the voltage division rule breaks and I get less than 5 volts across that component. I cannot figure out how to provide it 5 volts to power it up by using voltage division or basic circuit analysis techniques. What is the way to achieve any voltage value across any component we want? <Q> The answer by Oliver is indeed correct. <S> However, I will just add why you are seeing the results you are seeing when trying to power something through your divider. <S> You are correct in saying that for a power supply of 10V DC, dividing it down via 2 equal resistors will give you 5V. <S> we see that here: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is because the current flowing through this would be 10/20k = <S> 500uA. <S> the voltage drop across a 10k resistor is: 500uA * 10k <S> = 5V. <S> So, what would happen if you fitted a load on this? <S> Let us see: simulate this circuit <S> You have now added extra resistance on to the bottom resistor. <S> Assume that the load resistance is 10k. <S> This would make the bottom resistor 5k (resistors in parallel). <S> inputting this new figure into our previous equations will give is a middle point voltage of 3.3V. <S> So you can see that the voltage division rule does not break, you have just adjusted the values. <S> Adding a voltage regulator (for 5V the classic 7805) will be able to ensure you have a stable output voltage no matter what the load. <S> This is usually done with an internal compensation circuit such as this: <S> simulate this circuit <A> A voltage regulator. <S> A voltage regulator is a fairly fundamental concept in electronics. <S> The simplest voltage regulator is the Linear (or sometimes LDO - Low DropOut). <S> The regulator is essentially using two resistors to split the voltage to create your target output voltage (potential divider). <S> However, the regulator also monitor the output voltage, and when it changes (because the load changes), the regulator alter your resistors to compensate. <S> In a typical LDO device, a transistor is used as a variable resistor to adjust the potential divider and achieve the desired output voltage. <S> There are other types of regulators that use different methods, but the common feature is the device regulates the output voltage to remain as constant as possible regardless of the load presented, or the input voltage. <S> The wiki page has more information than anyone would ever want <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Various voltage regulation circuits. <S> As you have discovered, the simple voltage divider of Figure 1a is not stable. <S> Once a load is connected the voltage drops below the divider-only value and, worse, it varies depending on the load. <S> A simple voltage regulator can be formed using a Zener diode. <S> These do not conduct in the reverse biased mode (as shone in Figure 1b) until the voltage rises above the breakdown voltage, 5.1 V in this case. <S> The trick here is to set R4's value low enough to provide 5 V on the load at maximum load current. <S> If the load current reduces then the voltage would tend to rise but D1 will break down and hold the voltage at 5.1 V. You will find plenty of reference material and tutorials online. <S> The capacitors are required to prevent oscillations so don't be tempted to leave them out. <S> Note that all of these arrangements waste half the supplied energy as heat. <S> Switching, rather than linear, regulators are much more efficient. <A> We can't build a circuit that will supply a perfect 5V to any possible load. <S> What we can do is build a circuit that will provide a sufficiently good approximation of 5V to a load whose current consumption is within a defined range of values. <S> The simplest soloution would be to reduce the values of the resistors in our voltage divider, the smaller the resistors <S> the less sensitive the divider is to load. <S> However this also increases the power wasted in the voltage divider. <S> So it's not a good soloution. <S> The next step up is to replace the bottom resistor with a non-linear component, for example a reverse biased zener diode. <S> The voltage across these components varies much less with current than the voltage across a resistor. <S> So the current through the top resistor need only be slightly larger than the maximum current drawn by our load. <S> The next step up is to use an "emitter follower" to buffer the output of our resistor/zener combination. <S> So the current drawn from the resistor/zener combination is much smaller than the current delivered to the load. <S> This is essentially the design of the most basic linear regulator, real regulators may have additional components but they are likely to have the same basic elements. <S> "Low drop out <S> " regulators are a subset of linear regulators that are designed to work even when the input and output voltages are similar. <S> A basic linear regulator design needs at least a couple of volts difference between input and output, low dropout designs can get that much lower. <S> The final option is a switching converter. <S> A switching element is rapidly switched on and off <S> and then an inductor/capacitor circuit is used to filter that into a stable voltage. <S> This can be very efficient, but it adds substantial complexity and noise considerations.
| The regulator circuit adjusts its "resistance" to maintain the output at its rated voltage, 5 V in the case of the 7805 regulator. A three-terminal linear voltage regulator, Figure 1c, provides a very simple and cheap solution.
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Can I connect Inductive Proximity sensor to two different PLC Some background, I have a tractor that I want to make semi-autonomous. The tractor has it own PLC and an NPN Proximity sensor to measure speed. I have industrial PLC by Unitronics (V130) with I/O that can act as NPN or PNP.I wish to connect my PLC to the same sensor and save me a hassle to install another one next to it. I wonder if it safe (no harm to a sensor or either of the PLC) to split the sensor wires and if it possible to sense the sensor in both PLC? The sensor model is NBN8-12GM50-E1-V1 by Pepperl-FuchsDatasheet: https://files.pepperl-fuchs.com/webcat/navi/productInfo/edb/293454-0057_eng.pdf?v=20180423144546 EDIT:The two PLCs don't share the same power source and ground Thank you <Q> If the two systems share a common negative then <S> yes, if the PLC has an NPN (sourcing) input but it could cause problems. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Opto-isolated feed to PLC. <S> With the addition of a resistor and opto-isolator you can ensure trouble-free operation due to the complete isolation between the two systems. <S> Calculate R1 to give about 10 mA through D2. <S> D2, being an infrared LED will drop about 1.4 V at this current. <S> Q2 can be connected to the PLC input and common but observe correct polarity. <S> One of the beauties of this arrangement is that Q2 can drive a PNP (sinking) or NPN (sourcing) input. <A> The output is an NPN transistor with a maximum load specified to support 200mA to ground. <S> You PLC inputs will have either an onboard pullup resistor or you get to add one externally. <S> It's extremely unlikely that you have anywhere near the maximum load in your current configuration. <S> If you want to test it, measure your output with an oscilloscope and then attach your second PLC load and ensure you still meet the <2V low signal. <S> If the input resistor (pullup) is configurable in you PLC, then you only need 1 resistor configured for the two inputs. <A> I wonder if it safe (no harm to a sensor or either of the PLC) to split the sensor wires and if it possible to sense the sensor in both PLC? <S> ... <S> EDIT: <S> The two PLCs don't share the same power source and ground <S> But read below... <S> Aside from electrical compatibility, what about functional compability? <S> Is it disirable that two PLCs, for different functionalities share a sensor? <S> What if a mechanic for system A replaces the sensor, being unaware that system B relies on it? <S> I would place a secondary sensor. <S> All problems avoided.
| If the PLC's do not share the same ground, it is electrically not possible to share this sensor. It looks extremely likely you can connect two loads to the unit: Depending on the frequency of the signal this could be solved by using isolators or optocouplers.
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Circuit design with multiple voltage levels I'm working on an art piece based around an Arduino Mega that essentially requires power for three types of devices: Solenoids (4): 24 V, ~100 mA a piece (400 mA, 9.6 W total) NeoPixel LEDs (1280): 5 V, ~60 mA a piece (77 A, 384 W total) Base stepper motor (1): 5 V, not really an issue because it can be powered with an H-bridge driver from the output of the Arduino Since the LEDs are the primary current consumer here, we're thinking we'll use this 600 W, 5 V DC power supply (SE-600-5): http://www.meanwellusa.com/webapp/product/search.aspx?prod=SE-600 This power supply has three pairs of output terminals, and is rated for enough current to power the LEDs. The concern here is how to step up that voltage from 5 V to 24 V to power the solenoids (with enough current). We are considering voltage regulators, boost converters, as well as a high voltage amplifier like the PB64 from Apex Technology (shown in Figure 4 here: https://www.digikey.com/en/articles/techzone/2018/apr/how-to-combine-high-and-low-voltages-in-a-single-design ) Does anyone have any experience with issues like this? How would you recommend managing two voltage levels? Thanks so much! <Q> IMO it would make much more sense to use an SE-600-24 and use buck convertors to create a local +5V supply for your LEDS. <S> You have the potential to need conductors that will support many 10's of amps (if you have 4-5 major wiring runs) if you use a 5V supply and it is best to get this done at the highest voltage and lowest current possible. <S> Running 5V any distance you will end up with significant voltage lose, and may end up with signal ground problems for your LED data. <S> There are plenty of 2-5A Buck convertors like this , this <S> (I've used a bunch of these <S> and I like the input capacitors) or this available at low cost that would support groupings of up to 50+ of your LEDs allowing much smaller wires to be used in your installation. <A> There are a few ways to do this: 1) Use a separate 5V and 24V supply, the advantage is you don't need any DC to DC controllers in the design, this would likely be the cheapest and least complex method. <S> 2) Use 24V and use voltage regulators or DC DC buck converters to step down the voltage to 5V. <S> The advantage to this is only one 5V buck DC to DC converter would be needed. <S> The disadvangtage would be that 5V buck converter would be large (and expensive) you would also have losses of roughly 80% to 90% in the DC to DC converter. <S> Off the shelf DC to DC converters mostly range from 0.5$ to 1% a watt. <S> 3) Use 5V as the main rail and step up the 24V with a boost converter. <S> The advantage is the boost converter would only need to handle 10W <A> You can use a boost converter to get the 24V from the 5V. <S> For example, the ITX0524SA , can boost a voltage in the 4.5V-9V to 24V @250mA <S> (for your application may need two, but this is just an example to give you an idea). <S> You can find one with more power capability. <S> This is more like a standalone solution. <S> You may use an IC such as the LM27313 (for example). <S> This one can You can use a boost converter to get the 24V from the 5V. <S> For example, the LM27313, can boost a voltage in the 2.7V-14V to voltages up to 28V @800mA. <S> This should be able to power the 4 solenoids (400mA @24V). <S> Efficiency may not be great at that power for this particular IC <S> but this is just a guideline. <S> This requires more work since you'd have to add the external components but could be cheaper.
| Some AC to DC supply's have swapable modules that would allow you to generate both voltages in one supply.
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Generating a Negative Voltage In general, there are many ways for DC-DC conversion: SMPS, charge pump, linear regulators, zenner diodes, even resistive dividers. But lets say I want to create 2 voltage rails, +Vcc and -Vcc. I can use an explicit negative voltage converter like an inverting charge pump or a buck-boost inverting topology converter, or even build a dual PSU that produces 2 rails centered around earth. Can I achieve the same by just have 2 DC-DC converters, calling the middle node GND, and calling the low voltage -Vcc and the high voltage +Vcc? Are there any downsides to this method rather than using an already established GND and pulling a rail down with reference to that, aside from potential shorting risks to earth? <Q> At least one of the converters must be galvanically isolated (input to output). <S> There cannot be a common between input and output. <S> Using two isolated converters can help you deal with noise and ground loops in an analog design. <S> There is a potential issue with most single output converters not expecting a load that goes beyond their negative or positive rail. <S> You can prevent any issues by putting a reverse-biased Schottky diode across each output. <S> There is an advantage over using an inverting buck converter for the negative rail in that the startup surge of an inverting buck can be very high, enough to potentially pull down the input source and prevent proper starting. <A> In general, there are many ways for DC-DC conversion: SMPS, charge pump, linear regulators, zenner diodes, even resistive dividers. <S> But lets say I want to create 2 voltage rails, +Vcc and -Vcc. <S> Most of us would exclude linear regulators, Zener diodes and resistive dividers from a list of DC-DC converters and call them regulators as they burn-off excess voltage rather than convert it. <S> I can use an explicit negative voltage converter like an inverting charge pump or a buck-boost inverting topology converter, or even build a dual PSU that produces 2 rails centered around earth. <S> You don't have to use earth. <S> You can call any point on your circuit GROUND and use that as your reference for everything else. <S> A battery-powered, hand-held device, for example may need a dual supply but can't have an earth connection. <S> Can I achieve the same by just have 2 DC-DC converters, calling the middle node GND, and calling the low voltage -Vcc and the high voltage +Vcc? <S> Yes. <S> You might be able to avoid one of them if your battery voltage or primary power-supply suits, say, the positive rail. <S> Then you just need one DC-DC converter to generate the negative rail. <S> Are there any downsides to this method rather than using an already established GND and pulling a rail down with reference to that, aside from potential shorting risks to earth? <S> No, if the outputs are isolated from each other you are free to connect any one point on one to one point on the other. <S> That's standard practice and good engineering. <A> This WILL NOT work with a non-isolated DC-DC converter!!
| If you use an isolated DC-DC converter, you can just connect its positive output terminal to Ground, and call the negative output terminal "-xV".
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Looking for a PCB flag/indicator component I am designing a test jig (within a 6U rack) that mates with 6 removable trays. Each tray can hold 16 sensors which can also be removed. The system will be intended to function in the following way: Technician will populate all trays with sensors, Trays are entered into rack and all sensors are tested, Trays are removed from rack and sensors that failed are removed first and binned. The rest are passed and bagged. My issue is production have requested for indicators next to each of the sensors on every tray that indicate a pass or fail. When the boards are disconnected from the rack they lose power and although LEDs and external battery would be the first solution....they have asked for alternative solutions.Is there such a component as an indicator relay or something along those lines? If anyone has a good solution for this issue then please let me know. Software GUI is currently ruled out also as they believe looking at a screen and then at a PCB could lead to human error... <Q> There are non-volatile electromagnetic indicators <S> that can be used. <S> They are a little oddball though. <S> They were originally developed by the Canadian unit of Ferranti-Packard , but the patent has long since run out. <S> They work like a bistable relay (electromagnetic) and retain their state with power off. <S> In the old days they were used in quantity for airline status displays in airports. <S> Personally I would consider fitting the trays with batteries (or applying power later) and using LEDs. <S> You could store the information in a non-volatile memory such as a small microcontroller with on-board EEPROM or FRAM if the power has to be interrupted. <A> That's a completely non-volatile display, that can be directly driven by 7 pins of a 5V microcontroller, no additional circuitry required. <S> (I'm not sure if 3.3V would be enough to get usable contrast. <S> These displays want 15V for full performance, but the main issue with using only 5V is that it takes about a second to change state.) <A> I was running on a similar train of thought as Spehro. <S> Figure 1. <S> Magnetic latching indicators - flip-disk and flip-dot displays. <S> Source: eldisrl.com . <S> These are similar to the elements that make up seven-segment displays on some petrol pumps. <S> Figure 2. <S> These paraticular units have a two-wire coil and flip state is determined by polarity reversal. <S> Source: Datasheet . <S> Figure 3. <S> A manually reset and electrically unlatched indicator was used on the old manual switchboards. <S> Source: Telephone switchboard , Wikipedia. <S> They were also used for call-bells <S> but I can't find an image of anything modern. <S> Would you 3-D print your own with a little solenoid release mechanism? <A> I managed to find these Miniature bite indicators that come in a variety of sizes, colours and voltage ranges, so thought I'd share these with everyone who is interested. <S> Thanks for the help in pointing me in the right direction! <S> http://www.flamecorp.com/PDF/L3-Electrodynamics/L3%20Ball%20&%20Flag%20Indicators.pdf <A> Producing a battery + LED setup or even adding a permanent flipdot to each sensor may not be suitable for your needs <S> but I would recommend adding a couple pins to attach a test indicator that could be placed quickly on each board before testing and then removed and reused before the next batch is inserted for testing. <S> In your case this could either be a flipdot + driving circuitry or a battery with some LEDs and indicator logic to keep your sensor footprint to a minumum. <S> We have some interface cards with a setup like shown below: <A> You might want to have a look at electronic paper (e-paper) displays. <S> The displays come in all kinds of sizes and shapes, and they retain the set images for atleast weeks some even months when power off. <S> We use them in a system where we have limited power, so we set the image and then cut the power to the display. <S> The image is still visable, and when we need to update the display we power it on again. <S> More info can be found on https://en.wikipedia.org/wiki/Electronic_paper <S> biggest drawback is you can not see them in the dark, and the pixels are only two toned, (whith many different combinations)
| The flip dots mentioned in other answers would probably be more suitable if you can still find them, but another possibility is an E-Ink display: they're available in forms with individual dots, rather than the usual dot-matrix or 7-segment form.
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How fast can we measure using IR sensors Supposing I have an repeatedly object passing through IR sensors and I am looking at the output waveform. What is the maximum frequency the output waveform can have? I think this maybe related to the frequency of IR rays used for detection but I am unable to come up with a concrete logic as to why it may depend on that. PS: ignore other limitations such as the speed of the object to give rise to such high frequency and measuring instrument limitations. <Q> From a fundamental point of view there is practically no speed limit for IR detection (i.e., convert light intensity to an electrical signal). <S> The fastest photodetectors for the IR that I know of, have cutoff frequencies in the 100 GHz range . <S> This corresponds to rise times in the order of picoseconds. <S> Well this is an extreme example. <S> Practically you have to take into account other constraints, e.g., the size of the detector (smaller is faster) <S> which has an impact on the field of view with optics applied, the cutoff wavelength (different detector materials have different "speeds"), the detector noise level (there are detector types that are very fast but not low noise), the desired amplification gain (the gain bandwidth product is usually somehow limited) or, the detector price. <S> Often the amplifier imposes the bandwidth limitation, and for the high speeds, an integrated module (i.e. detector and amplifier) is required. <A> In addition to the other answer, the practical limitations include current levels, diode junction capacitance and load resistance as the breakpoint. <S> Thus some have a rise time >>10x decay time. <S> A typical 5mm PD like LTR-323DB has a junction capacitance of 50pF and useable up to 3MHz. <S> By converting the PD current to voltage the trans impedance amplifier, (TIA) which is an OpAmp Vin-/Rf and making Rf low (xxx Ohms) you can cascade more gain to get a more useable voltage to detect. <S> If you have a narrow optical beam restricted by a PD aperture of 5mm and a recessed PD and narrow emitter Beamwidth of <10 deg. <S> , it is possible to define a small path in between , you can now compute speed. <S> But since emitter power is the biggest variable you need AGC. <S> This poses a problem with detection but is easily solved by having 2 equal emitters , pulsed alternately , one blocked and the other not so that AGC always has a stable input to compare the dropout . <S> Then you need a simple sync pattern to discern 1 source from the other such as time duration. <S> Now you “can” detect the object relative transmission loss not only in short pulses ~1us <S> but at high velocity in a very small aperture such as 1mm in the path between using a recessed 5mm 10 deg PD and detecting a 50% drop in transmission loss. <S> 1mm <S> /us is 1km/s <S> which is pretty fast assuming high SNR with daylight blocking lens on PD. <S> Anecdotal <S> Now that does seem too fast to me <S> but I have only done this with 1m path and detected a 1mm wire passing fast across the middle using 100us pulses to verify the optical design. <S> It worked. <S> With multiple detectors I detected direction. <A> Here are some facts about silicon. <S> The thermal timeconstant of a cubic meter of silicon is 11,400 seconds. <S> The thermal timeconstant of a cubic centimeter of silicon is 100*100 faster <S> , or11,400 / 10,000 = 1.14 seconds <S> The thermal timeconstant of a cubic millimeter of silicon is yet 10*10 <S> faster, or 1.14 seconds / <S> 100 = 11.4 milliSeconds. <S> The thermal timeconstant of a cubic micron of silicon (perhaps the junction depth of some detector) is 11.4 nanoSeconds.
| The frequency of the light is so high (in the 100 THz range) that there is practically no limit from that, given typical light modulation frequencies that can be electrically handled.
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Connecting KSZ8863RLL to a MAC using RMII (Edited to ask further question about the MDIO/MDC pins.) I'm adding a Lantronix xPico 250 to an embedded deivce to provide WiFi access. The device has an ARM based SOM which includes a PHY and there is also an external RJ45. The three are connected using a Microchip KSZ8663RLL 3-port switch. The SOM's PHY is capacitively coupled to one of the PHYs on the switch, and the external RJ45 is connected to the other. These can communicate with each other no problem. The Lantronix xPico 250 supports an RMII interface so this is connected to the RMII port on the switch.The RMII pins on the switch and the Lantronix are direcly mapped to each other by pin name. i.e. TX -> TX, RX -> RX. It doesn't work though. I can access the xPico 250 wirelessly and accessing its internal webpages can see that the status of its ethernet link (eth0) is down. I don't have much else to go an at the moment. I've checked the clock. The Lantronix xPico 250 expects a reference clock as an input which the switch is providing. I've done some Googling and am aware that RMII isn't designed to work between two MACs. The datasheet for the Microchip switch refers to the RMII interface supporting both MAC and PHY modes but doesn't explain how to configure this. Every example reference design I've seen on the internet has it set up as a MAC. How can I be sure the switch is set up as an RMII PHY? If it is actually a MAC and I'm connecting MAC to MAC, do I need to swap the signals round (ie. RX -> TX, TX -> RX etc) Here's a schematic: Additional queestion about MDIO/MDC..I have left the MDIO and MDC pins not connected/floating on the Lantronix. On reference designs the Lantronix has been connected to a PHY and they've been used. Is it possible that this is causing a problem? Should they generally be pulled up/down if not used? <Q> Ok. <S> I've finally got it working. <S> There were two issues: <S> Firstly, the MDIO and MDC lines do need connecting from the Lantronix to the switch. <S> I added a 1.5K pullup to the MDIO line and also pulled up SPISN (pin 39) to disable SPIQ. <S> Secondly, the Lantronix by default uses a PHY address of 0 when sending the read register commands over the MDIO/MDC. <S> The two PHYs on the switch use addresses 1 and 2 so <S> the switch was igoring them. <S> The firmware in the factory-shipped Lantronix doesn't have scope for changing the PHY address but the latest firmware on their website (xPico200_1.9.0.1R4.signed.rom) does. <S> I reconfigured the Lantronix to use PHY addr 1 and everything seems to work now. <S> eth0 link up, can ping etc.. <A> Once this is fully understood the option of whether you connect pins named TX<->TX and RX<->RX OR if you need to swap as TX<->RX and RX<->TX will become clear. <A> (Edited to correct references to 50MHz crystal) <S> Register 53 (Port 3 Control 5) bit 7 should be 0 for port 3 to be in PHY mode (this is the default). <S> You have connected REFCLKO_3 back to REFCLKI_3 and also to the xPico 250. <S> This looks correct with respect to table 3-7 <S> (MAC to MAC connection). <S> For this clocking setup, you need to set register 198 bit 3 to 0 (default), to float pin 17 (internal pullup), and to float pin 18 (internal pullup). <S> Probe REFCLKO_3, make sure you see 50MHz. <S> If you see the output clock but still have problems, try disconnecting REFCLKI_3 (from REFCLKO_3) and pulling it down, then set register 198 bit 3 to 1 (internal reference clock). <S> The pin description for the REFCLKI_3 pin says "Output in PHY MII mode". <S> It does not mention "PHY RMII mode". <S> Confusing... might be worth trying to isolate REFCLKO_3 and drive a clock to the xPico 250 from REFCLKI_3 instead. <S> Dump all port 3 control and status registers and inspect them. <S> Check your layout to make sure any path length/matching constraints for the RMII have been met. <S> If your traces are longer than an inch or so, you may need series termination resistors. <S> Ensure your crystal is correctly specified. <S> It's probably fine if the KSZ8863RL PHY ports are working OK.
| You need to read your data sheets very carefully to ascertain which lines of the RMII on each end transmit and receive data (including the control lines).
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What can happen with the PCB foil while SMD resistor destructive overheat? I need to protect the gate driving circuit by putting the resistor between the gate and the control circuit: During the testing of the mockup MOSFETs can go be destroyed which leaded to high voltage on the gate circuit. I'm going to place 0805 SMD resistor (R19) which will be destroyed in milliseconds as well (as it will be exposed to 50 W continuous overheating). However it will be much simplier to replace just this resistor insted of looking up for the whole circuit for destroyed elements which was a real pain. My concerns are: If R19 will be overheated and destroyed - will it break the circuit? Or it is possible that it will go to short circuit (as FETs usually do)? If the resistor destroyed - how likely that the PCB will be destroyed too? I'm using 35 um foil. <Q> It is likely that R19 will fail open circuit. <S> It is rare but not impossible for a film resistor failure to result in a short. <S> I've never seen it happen. <S> Consider using an appropriately rated resettable fuse in place of R19. <S> Predicting the damage to the PCB foil is hard to do. <S> I've seen 35um (1oz) foil pads apparently survive thermal destruction of a resistor, but then fall apart during (very careful) rework. <A> I agree with @amb with using a resetable fuse, or even a normal fuse. <S> A resistor that should break exactly what a fuse does, but a fuse is designed to do this well. <S> So why not use a fuse? <S> Also, while the resistor heats up (before it burns up), all that current is still running through the rest of the circuit. <S> So other components can still be damaged. <S> I'm sure a fuse will burn out much quicker and more consistently. <A> The Zener will also see some substantial dissipation, especially if the voltage is positive. <S> If it was +240VDC, the Zener would dissipate more than 5W and the resistor ~50W. <S> A TVS may be better for this purpose than a Zener. <S> As far as your question goes, carbon film through-hole (eg. 1K) resistors can arc and drop in resistance with 220V on them (becoming a glowing ceramic rod), but usually thin film SMD resistors, even cheap ones, use something like ruthenium oxide, which probably has less tendency to arc. <S> You can certainly test a dozen resistors or so and see what happens (just solder them onto a test board and try it- <S> a scope with slow sweep will let you measure the duration of conduction). <S> Maybe try something like an 0805 size. <S> I expect it will work. <S> As an aside, this circuit looks like it would switch very, very slowly (especially in the 'on' direction - several milliseconds most likely), which can be very hard on the MOSFETs if they are carrying substantial current. <S> This could be what is causing your trouble. <S> I assume you are trying to switch AC mains voltage and are driving the gates with an isolated voltage source.
| In my experience a metal film resistor (through hole, such as 1K 1%) will probably work. I would also try to find a fuse that trips as quick as possible.
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STM32F103 - Single pin shorted to VDD? I have a custom STM32F103C8T6 board. Long story short, the MCU works fine, I can connect to it, program it, read sensors, toggle GPIO etc. The problem is that, one of the crucial pins seems to be shorted to VDD. The pin is PA9, which I wanted use for USART_TX1. I have no control over this pin. I can not digitally toggle it, can not use uart, nothing. I checked the pin via a multimeter ,with the board powered on and off, and both times I noticed a short between VDD and the pin. The length of the track out of this pin is about 7 inches. I visually inspected the route to see if short was on the board but I didn't notice anything, route is neat and isolated. Thus, I believe the pin is internally tied to VDD somehow? I also performed a full chip erase, still tied to VDD. I also think it is worth noting that sometimes there exists an amount of resistance that the multimeter measures when checking for shorts. I read values ranging from 2 to 190 ohms. What is going on here? Edit: schematic and layout of the relevant parts. Top layer being ground, and the bottom being VDD. Let me know if more info is needed. <Q> If the trace is still high impedance, it's probably the MCU and the MCU will need to be replaced. <S> Odds are the port is blown out, especially if this has been connected to a non TTL uart by mistake, or ESD damage. <S> While the PIN is up you can check the I/ <S> O on the PIN independent of the PCB <S> I've had a few through hole vias that shorted planes together. <S> The nice thing about this trace is it's only tied to one pin, which makes it easy to troubleshoot. <S> I've had problems like this where there are 5 pins, and you have to go down the line and cut traces and remove parts until you've found the offending part or trace. <A> I have suffered from exactly the same symptom, in my case it was due to tin whiskers . <S> These were forming in real time - within minutes of the rework that I was doing to clear them - and were invisible to the naked eye (and we didn't have optical instruments to hand <S> so I never saw them). <S> They caused shorts ranging from units to 10s of ohms. <S> Absolutely amazing. <S> The root cause was a problem with a cleaning process at the PCB fabricator. <S> They proactively contacted me to advise that other customers were seeing the same issue with boards made that day. <A> I soldered a new chip to the same board, and it seems to work fine for now.
| If the trace is shorted (still reads a value lower than 100kΩ (although should be more like few MΩ)), it's the PCB (and all PCB's are not perfect). Unsolder and carefully lift the pin of the part then measure the trace with a DMM and check for shorts to ground.
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Causes for a burned up flyback diode on a relay? I'm repairing a control board for a premium washing machine (model #s at bottom for anyone that's interested). The machine was displaying an error code for relay failure. I took the board out of circuit, injected voltage on the DC bus, and detected two shorted 1N4148 "T4" diodes, each in parallel to a 12v SPST-NO relay (marked with arrows below). I replaced them. Each relay switches one side of the water heating element (+ or -). I have not tested the repair yet, but the machine does work without issue with a substitute board. Trying to figure out what may have caused the diodes to short? Note: After repair Machine: MHW6000X Relays: Omron 12V G2RL-1A-E Control Board: W10406607 / W10406604 (alternates, W10342325 / W10342327, W10388205, W10427972, W10384506, W10354088, W10406635) I found these short diodes using isopropyl alcohol (99%) and a microscope (loupe or optivisor works for less $$). Here's a video from my microscope camera showing the alcohol https://youtu.be/w-QVvYimghc This is one of the larger "neighboring" diodes, MURS160T3G "U1J" in SMB package. Same relay, so perhaps a strange bit of design here. <Q> Figure 1. <S> Relay datasheet extract. <S> Checking the relay datasheet we find that the coil current should be about 33 mA. <S> At the instant of switch-off the coil inductance will maintain that 33 mA through the flyback diodes. <S> Figure 2. <S> Extract from the Vishay 1N4148 datasheet. <S> Checking the Vishay (chosen at random) datasheet we see that 33 mA is well within specification. <S> Update for "T4" diodes: The Diodes Incorporated datasheet rates their diode at 250 mA continuous. <S> That isn't the problem either. <A> Speculation: — Improper hand solder temperature and duration. <S> Perhaps they were hand soldered from fallout during wave solder with Relays due to proximity or installed later for some other reason. <S> Although excess solder is not normally a bad thing , excess heat and duration >3seconds this close to the epoxy seal can be a cause for moisture ingress in the field. <S> This is a Maytag washing machine probably next to a dryer <S> so humidity in this area will be high and improperly soldered plastic parts <S> will stress epoxy seals to fail their moisture barrier more rapidly. <S> It can cause epoxy LEDs in the field to fail prematurely too as LED clear epoxy is easier for damaging the moisture seal when soldered less than 5mm from the base for >3s , otherwise <5s. <S> Heat velocity is rapid <1s in these parts, but only 1mm/s on LED leadframes, so the epoxy can move(microscopic) when exposed to rapid thermal stress. <S> I suggest you follow any hand-solder document for this plastic package or use 600’F irons and ensure soldering <S> is <3s per lead with cool down after 1st lead. <S> You can see the right epoxy package has “popped a corn” or epoxy on the corner. <S> (Or last time...) <A> Look at the other two relays. <S> There are much bigger diodes. <S> So one reason could be that failed diodes are not suitable to it's task (too low current handling capability or too low reverse polarity voltage) <A> For what it's worth, there is anecdotal evidence about failures of glass 1N4148 diodes in this service (flyback diodes that fail shorted). <S> Several of us have reported these failures over the years, with no real plausible explanation other than bad batches of parts. <S> Since the failure is to short physical damage (eg. <S> broken package) is less likely. <S> For example, an exchange I (and others) had with Tony Williams (RIP) about 11 years ago on sci.electronics.design, which can be easily searched. <S> The use of 1N4005 or M5 SMT diodes is probably not a bad idea.
| I would conclude that the designers did their job OK but that the purchasing department may have found some bargain diodes. This could be from a kernel of moisture and metal diffusion into the silicon causing the excess heat in the first place.
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Why can't high current substitute low voltage I know this look stupid, but why high current can't substitute low voltage. I read many articles about the difference between the two, and asked my professors, but still have some doubts and blind spots to differentiate between them. let's say a diode, I need 0.7V to make it operate, why if I decrease the voltage below that and significantly increase the current it still doesn't operate? I tried to visualize the current as an army (quantity) and the voltage as the power of the that army. And thought even if the power is low if I increase the number of my army I would substitute the low power. I know this is not true, but why? <Q> Consider using the water in pluming metaphor to explain voltage and current. <S> Using this metaphor, the voltage is the pressure in the pluming and the current is the amount of water making it through the pluming. <S> Batteries become water pumps. <S> Resistors become constrictions in the pipe. <S> Diodes become check-valves which need a little pressure to open : <S> A diode is equivalent to a one-way check valve with a slightly leaky valve seat. <S> As with a diode, a small pressure difference is needed before the valve opens. <S> And like a diode, too much reverse bias can damage or destroy the valve assembly. <A> Seemingly your professors couldn't see or believe how big hole you had in the basics. <S> You cannot alter voltage and current independently. <S> The operation of the electronic components IS how they make the current dependent on the voltage. <S> The operation of the component ISN't something that you can start by applying a good voltage and a good current, which you select both as you like. <S> You apply a voltage and the component sets the current. <S> The setting by the component happens as exprcted = <S> the component works. <S> A silicon PN diode for example WORKS when it lets only a small current go when the applied voltage is much under 0,7V, say 0,2V or less High current and low voltage is a substitute of high voltage and low current in one special case: What the power calculation equation P= <S> U*I gives when you insert numerical values to U and I. <A> The diode is a good example. <S> The quantum physics of a silicon PN junction determine the forward voltage of the diode. <S> You can't change that voltage by being pushy with current. <A> I explained why you need 0.7 V in my answer to your question of yesterday, Why do we care about voltage and current - not power only . <S> " ... <S> why if I decrease the voltage below that [0.7 V] and significantly increase the current it still doesn't operate? " <S> You can't get significant current to flow until you exceed 0.7 V. <S> I tried to visualize the current as an army (quantity) and the voltage as the power of the that army. <S> No, the voltage would be like how many soldiers are in each line (file) pushing one behind the other. <S> The current would be like the width of the rank (how many side by side). <S> It's not a great analogy. <A> The controls on your lab power supply are labeled as "Voltage" and "Current", and so you might think that they're setting those physical parameters directly, but they're really just limits . <S> The relationship between V and I is determined by the circuit, and the supply simply stops pushing when it hits one or the other limit. <S> If you want to "push more current", then the only way to do that is to increase the voltage because that's precisely by definition how you push current. <S> I agree with user287001's assessment of your professors , and it's a bit unsettling to me that they would miss that. <S> What other fundamentals (or fundamental misunderstandings) would they miss? <S> Or perhaps they did see and failed to address? <S> That's also unsettling. <S> Or maybe they did explain it all correctly in their way of thinking <S> and you just don't think that way. <S> Welcome to university! <S> A big part of it is not the subject matter itself, but learning to think differently than you're used to. <S> Build up a library of different thought patterns and see what works. <S> Hint: <S> Asperger's and Autism are often associated with technical understanding. <S> You might want to read up on how they think and mimic that. <A> why high current can't substitute low voltage. <S> There are situations where high current DOES substitute forlow voltage, like in a transformer. <S> But, all real-world applicationshave current limits, voltage limits, and thresholds. <S> A wire thatcarries 15A of current is thick; hard to bend, expensive to build from copper, and requires good low-resistance terminations to connect itto anything. <S> A wire that carries 30 kV to fire a spark plugis always fat with insulation, so that it doesn't make St. Elmo's fire(corona discharge) <S> when active. <S> I have a 6V 100W light bulb: it takes about 15A to run. <S> And, a 120V 100Wlight bulb takes 0.8A. <S> They are very different designs, and extendingto 0.6V or 1200V would require very different switches, wires, sockets. <S> let's say a diode, I need 0.7V to make it operate, why if I decrease the voltage below that and significantly increase the current it still doesn't operate? <S> A diode that does useful rectification of electrical power, made ofsilicon, has negligible power output with less bias voltage than about0.5V. <S> There are some RF rectifiers (point-contact diodes) that canwork at much lower voltages, but burn up at high current. <S> So,the lower-voltage/higher-current options are impractical. <S> Thisis <S> because there is a material property (the silicon bandgap) thatdetermines the useful range for the (easy to produce) silicon rectifier. <S> Folk who DO redesign circuitry often find that theirgoals are inconsistent with low currents (because cosmic rays inducenoise) or with low voltages (because transistor switch thresholdshave as much variability as the total power voltage), or withhigh currents (because wiring gets too fat) or high voltages (becauseinsulation cannot be assured). <S> A typical CPU chip needstwo or three precise voltages in order to function, and thosevoltages are NOT arbitrary.
| The only way you could reduce voltage AND increase current would require a redesign of the diode itself, rather than using off-the-shelf components.
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Flooring for home lab I am setting up a home electronics lab for projects with my kids, currently ages 4 through 10. What type of flooring would work best? We will be doing art projects in the same room, so we need something easy to clean (no carpet). I would prefer something relatively inexpensive as well. I have found lots of information about flooring for a pro setting, but I have not found recommendations for home use. <Q> ESD is a matter of how much intermittency or damage you can afford. <S> In a medial device, or aerospace application or other industry <S> the costs are very very high into the hundreds of millions (or a human life) for a failure. <S> In these cases every available option to mitigate ESD risk is taken, which include floor mats, wrist straps, humidity control (60% if I remember right), making sure no materials that promote ESD (like paper and plastic) come into contact with electronics and even monitoring the local electric field. <S> The cost in time and money is also high. <S> Even in a regular production environment, the costs can be high (for scraping a few hundred dollars worth of components). <S> In your case, since your costs of failure are quite low (if you ESD'ed a MOSFET, that's ~1$ and tens of minutes to troubleshoot whereas a mat is 10's of dollars or more) and the costs of ESD mitigation would be higher than a few components, it may not be necessary to go to such great lengths for ESD lab coats and ESD flooring. <S> It might be better to just use wrist straps and an ESD mat for educations sakes and call it good. <S> I'm not aware of any home materials that would satisfy the conductive requirement. <A> What is most important is a static dissipative bench e.g unpainted wood is ideal. <S> I dislike pvc antistatic mats that melt under soldering heat, and find wood a better work surface. <S> If it has a metal frame then connect that to mains ground via a 1M resistor. <S> Train your kids to work with bare arms resting on the wood, and to contact the bench first <S> A bare wire or metal angle/strip/foil connected to mains ground by a high value resistor (1M) stapled along the corner or front edge where you will touch <S> it is another common idea. <S> Chair covers can sometimes be static generators - staple cotton fabric on if this happens. <S> Modern chairs can be a bit of a nightmare of insulated plastic bits. <S> Bare feet not flip-flops-on-vinyl are better. <S> Make sure to have an earthed soldering iron. <S> Low voltage soldering stations are much better. <S> If it is chinese you must test it is actually grounded. <S> (I have seen two that weren't grounded, and one of them, supposed to be a low voltage iron, but actually had the tip wired live to mains). <S> Get a plug-in RCD or change the wall socket for an RCD/GFCI for the whole workbench. <S> (saves your ears from swmbo complaining its too dangerous for a 4yo to be rewiring the toaster). <S> Arrange to have a single visible switch that can turn the entire work bench off at one point. <S> Your kids love to leave soldering irons, glue guns, flame throwers etc running overnight. <S> (I never do this, no siree). <S> Fires while we sleep are my #1 home worry. <S> Asking at an industrial flooring supplier is likely to get you an offcut of vinyl a good size to go under and around the bench as far as your chair rolls. <S> (industrial vinyl is also solid not paper with a film of vinyl) <A> My earliest shop used 1/4" Masonite over top of the existing (very) short pile commercial-space carpet that was installed when we rented the space. <S> The substrate beneath the carpet was concrete and we simply used Tapcon screws to hold the Masonite in place. <S> The local ESD specialists told us that as long as we maintained reasonable humidity levels, the resistivity of the Masonite would be in the Dissipative range. <S> This was born out by regular testing. <S> When we eventually purchased our shop space, the old Masonite and carpet was replaced with standard 12" x 12" tile right on top of the concrete. <S> We didn't have enough money for ESD tiles, so now we are stuck with stripping and waxing with ESD wax on a far too-frequent basis. <A> It would be much simpler and less expensive to cover the bench with an anti-static mat and install a wrist strap earthing point.
| You can put an antistatic mat if you want, but in a fairly humid environment wood is fine. For a proper ESD floor, the floor needs to be conductive and connected to ground (with 1MΩ of resistance between the user and the floor), proper footwear that is conductive is worn with a foot strap. Static disippative vinyl floor is common in many manufacturing industries not just electronics. For a simple shop and provided you can maintain a decent humidity level, I would seriously consider a Masonite floor.
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Is there such a thing as a resistance oscillator? I think I understand how a Oscillator works (VCO). One uses an Oscillator when one wants for example a square wave oscillating between sy, \$0\$ volts and \$5\$ volts. My question is: Is there such a thing as a resistance oscillator? That is, how can one create a circuit where it is the resistance that changes as a function of a time? Can this be done as a square wave? Sine Wave? <Q> The design of the circuit and its power supply voltage determine its high output voltage and its low output voltage. <S> If you want to change its output level then you need a voltage controlled amplifier (VCA). <S> Do you know that a variable resistor is commonly used as a volume Control? <S> It is a variable voltage divider. <S> But instead of audio as the input you could use a DC voltage that is changed when you turn the variable resistor. <S> Connect +5V to one end and 0V to the other end. <S> The center pin is the output which goes from 0V to +5V and it can control the frequency of a VCO. <S> It is easy to change a voltage as a function of time with a resistor connected to a fixed voltage that charges a capacitor. <S> The voltage rising in the capacitor can control the resistance of a transistor or a Mosfet. <A> There are several different types of voltage controlled resistor. <S> One is the "Vactrol" whish <S> is a photoconductive element illuminated with a controllable light-source - often an LED shining on a light-dependant resistor. <S> Another is the "Carbon Button Amplifier" which uses a voice-coil to vary the pressure on a pressure-sensitive carbon contact (which varied the resistance of the contact). <S> Long ago these were used to amplify telephone signals on long-distance wires. <S> Another is to use a field effect transistor. <S> FETs behave somewhat like resistors at small signal levels. <S> These methods are all becoming <S> less common, and electrically controllable variables resistors are simulated by switching fixed resistors in and out of circuit - this technique is used in digital potentiometers etc. <S> If you take a time-varying signal from an oscillator and feed it to a vactrol you'll get a time-varying resistance. <A> One can gradually collect voltage to a capacitor with a current or electric current to an inductor with a voltage. <S> Having one quantity A which grows or decreases <S> gradually due <S> another quantity B is essential for oscillations. <S> In addition we need a control circuit which reverses the polarity of B to force A in turns to increase and decrease. <S> The control circuit must directly read A to decide how B should be changed. <S> We have several mechanisms where resistance increases or decreases due some quantity X. <S> For example a voltage over a resistor makes a resistor warmer due the dissipation. <S> The resistance of an adjacent NTC resistor decreases gradually. <S> As well a voltage can run a DC motor which turns a potentiometer. <S> These can cause oscillations if we allow some test signal to be applied to measure the resistance. <S> Resistance cannot be seen directly, a voltage is needed to see what current is caused or a known test current must be made and the voltage is measured. <S> A simple resistance oscillator with an electric sensing circuit can be a bridge, where a motor turns a pot and that pot supplies a voltage to motor from a bipolar DC source. <S> Without any inertia the motor stops to the position where it gets 0V, but the rotating mass easily continues the movement and the balance is never found. <S> If we need a system where the resistance swings and at the same time just that resistance must not be used for any electric signal in the oscillator, but it still must be the seen quantity of the circuit which controls the resistance, we unfortunately are out of the luck. <A> normally you should always have some imaginary part for an oscillation. <S> Something ideal, that is only resistive will never oscillate. <S> Maybe this can help. <S> So it is necessary to have a resonance for an oscillation. <S> This is only possible with introducing an inductor and a capacitor as well. <A> A 1mm cube of silicon has a thermal time constant of ...... 11.4 milliSeconds. <S> You could thermally-bond a resistor to be heated, and a thermistor that senses the heat and moves the output of a voltage-divider above or below some threshold. <S> Why use silicon? <S> Some resistors have cores of clay, or ceramic, and thus will propagate heat similar to silicon.
| A VCO is an oscillator that uses a variable DC voltage to change its frequency.
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I want to draw 20 amps from my home power outlet is there any way to do that safely? I'm wanting connect up a bunch of Raspberry Pis (micro computers) to one power outlet in my home. At about .5 amps per computer * 40 computer means I will need to draw 20 amps. Now, I know that I'm only supposed to draw a max of 12-15amps from my standard 120volt North American outlet, but I'm wondering if there is any way to amplify the amperage and then use it to power all the computers, or alternatively, whether there is another way to somehow achieve this that I'm unaware of. If there is no possible way to do this, what other method for drawing 20amps of power would you suggest? EDIT: Computers run at 5V DC, I'm using a 5V, 20amp power adapter which I plug into the wall to get that but I don't want to draw the full 20amps because the wall socket can't handle it. <Q> Not a problem. <S> You have 40 devices drawing 0.5 A each at 5 V. <S> 40 x 0.5 <S> A x 5 V = 100 <S> W <S> So if your AC to DC converter is 100% efficient then you need to draw 100 W from the mains and in the US <S> that is I= <S> 100/120V = 840 mA which is well within the limits. <A> 5V 20A means 0.8A at 120V. <S> you'll be OK <A> Think about conservation of power. <S> Power in must equal power out, just as you can't mix 100 g of chemicals and end up with 200 g of result. <S> So the first question to ask is: how much power do the raspberry pies need? <S> (volts * amps * count.) <S> The next question it how much power your outlet can supply. <S> (120 volts * 13 amps, or 10 amps if you have old wiring.) <S> Since you have more power available than you need, all you must ask is how to do the necessary conversions. <S> The common answer is a transformer for high voltage or a switch mode power supply for lower voltage or when lighter weight is needed. <S> Once you account for inefficiency (some output power is waste heat and waste EMF rather than electricity, a safe assumption being at least 75% efficiency), a transformer's output equals its input. <S> Hence 0.75 * 120 volts * 10 amps == <S> 5 volts * 180 amps . <S> You won't easily find a transformer that meets these specs, but no current will flow except the current required by your load. <S> Your load will absolutely not draw 180 amps. <S> You can either use a power supply that can supply enough current, or use more than one without exceeding the current rating of any individual power supply. <A> If you draw more than 15 amps for an extended period of time, the circuit breaker will trip. <S> Kitchen counter and bathroom outlets are wired for 20 amps, but it would be quite unusual to find any other circuit connected to a 20 amp breaker.
| You can not change the circuit breaker because the wire size will not be sufficient to carry more than 15 amps safely. The most likely possibility would be to find that there is space for another breaker in the box and access to wire from the box to a new outlet in a convenient location.
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What is the use of two power supplies in an opamp? I am learning about opamps. Now I want to experiment with them but the problem is with -Vee (that negative power supply). I felt like I couldn't design it so opted for ground instead of negative supply, as I heard about single supply. Now anyone please elaborate what is the use of two power supplies also the consequences of making the -ve supply as ground. My college course prefers IC 741 OPAMP. <Q> The uA741 does NOT require dual supplies. <S> It requires the ability to bias the operating point of the internal amplifiers between the power supply voltage and the easiest way to achieve that is with a dual supply. <S> Look at the datasheet in section 7.2. <S> Although the supply is shown as VCC+ and VCC- there is no Ground shown. <S> The decision about where the ground point might be is complex and depends on both input bias and output load configuration. <S> For example you could have a load that goes from output to VCC+ which would change the output current characteristics you would expect. <S> The biggest problem that new users have is that the 741 is actually a very high voltage op amp, so needs a 10 V (+/-5V) supply to meet the recommended operating conditions. <S> Notice in section 6.4 that the output swing is guaranteed to be 20 V peak to peak within a 30 V supply (+/-15V). <S> The output amplitude relative to the supply is very restricted, especially as you put a greater load on the output. <S> The 741 is used in course material because it is easily analyzed: here <S> is a typical DC analysis, but there are many, both AC and DC. <A> See this: The ground line isn't normally connected to supply a 741. <S> There's no pin for it. <S> Thus the 741 cannot have a slightest idea what happens to be your ground potential compared to VEE line, it's happy if there's high enough voltage between VCC and VEE lines. <S> Let's assume your 741 does something useful and does it as we expect an opamp to work in its linear region. <S> Then internal transistors Q3 and Q5 need proper operating voltage. <S> It comes from Q1. <S> Let's assume UE=2V <S> is enough, more is allowed but less than 2V isn't enough. <S> As you surely know from the theory of the emitter follower, then U1 must be about 2,7V or more. <S> Let's assume your application needs that 741 must be fully operational even in case <S> the +input is connected to ground or it gets a signal voltage which happens now and then be zero voltage above your ground potential. <S> To keep 741 operational there's no other possiblity than to connect VEE line to negative voltage and it should be at least 2,7V below zero. <S> If you must put negative signal voltages in the +input, then even more room is needed. <S> -3V <S> to +input forces to use VEE=-5,7V <S> Be informed that you can connect VEE to the ground, but then all input voltages must be far enough from zero. <S> If the 2V minimum UE is true, then U1 must be at least 2,7V <S> The 2V limit is my coarse simplification to keep the numbers short, you can read from the datasheet how much lower VEE must be than the input. <S> ADD: <S> You can easily find simple opamps (say 3704) that are advertised to work from a single positive supply. <S> If you look at their interior schematics, you see that they have just mirror PNP input stages. <S> There the voltage reserve is a must between the input and VCC line. <S> Dual supply is still needed, if you want to have both + and - input or output voltages. <S> Rail to rail opamps use input circuits that do not need that voltage room between rail and input. <S> They have fet outputs which can pull the output to both rails if needed. <A> The picture below shows what you need to think about when choosing the power rails for your op-amp: - <S> Most common op-amps <S> (741 included) can't: - Work with input voltages too close to the positive rail Work with input voltages too close to the negative rail Produce output voltages that are too close the the power rails <S> So, if your circuit application does not need to directly measure and produce voltages close to the power rails then the -Vs supply can be ground or 0 volts. <S> If your application is a DC amplifier whose input signal can be 0 volts to a reasonable positive value <S> then you should either look for an op-amp whose common-mode-input-voltage-range includes the negative rail or provide a negative rail of 3 volts or more. <S> If the output has to also fall to close to the negative rail then you might consider a "rail-to-rail" output op-amp although there are still serious limitations of getting to within 10 mV of the negative rail. <S> Again, this is solved by providing a small negative rail to the op-amp to give more headroom. <S> If your application is a DC amplifier whose input signal is bipolar <S> around 0 volts <S> then you would naturally choose a design that has a negative rail power supply. <S> I'm sure there are some exceptions to this rule with some more specialist op-amps but the 741 is definitely not one of them. <A> The design of a 741 opamp is 50 years old (!) <S> and its performance is very bad compared to newer opamps. <S> Kiss it goodbye and bury it. <S> Maybe your college professor is 80 years old and has never seen a newer opamp's spec's?
| If you want to get negative voltages or even zero from the output, you also need negative VEE.
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Why is there a system controller and a microcontroller in the same FPGA? I am starting to dig into the FPGA world. While exploring the FPGA SmartFusion2 architecture I found out that there is a microcontroller (ARM Cortex-M3) and at the same time a system controller . Why is there a need for both? Can't one of them replace the other? <Q> It may or may not have a CPU, but in any case, it is not user-programmable. <S> In contrast, the primary function of the ARM processor subsystem is to execute application code. <A> That's not an FPGA architecture, that's a System-on-Chip that happens to contain a microcontroller and an FPGA. <S> Having a hardware MCU in there makes a lot of sense: in reality, systems containing an FPGA often need an MCU to load the configuration into the FPGA, and to do system maintenance tasks, which, would you not have the MCU, you would do by implementing a small CPU inside the FPGA and letting that run software. <S> Also, dedicated silicon implementing the exact same functionality as implemented in an FPGA usually is faster and uses less energy. <S> So, if you just need your microcontroller to e.g. receive command packets via SPI, then extract the command from them, and set off the right calculations to be done inside the FPGA, you can save a lot of energy because much more components of your system can be put to sleep, and the ones that always wake up use less power. <A> The system controller is responsible for booting both the FPGA and the ARM CPU, among other things. <S> Standalone ARM CPUs also have similar logic to configure a minimal memory map and conservative clock PLL setting before starting the CPU core, which will then do the remainder of the setup. <S> In a combined FPGA/MCU fabric, this is usually extended to loading the entire application program from configuration flash, since we already need to do this for the FPGA tables, so loading a bit more RAM doesn't add much complexity. <S> Keep in mind that FPGAs need to boot really fast, for PCIe and USB applications where standby mode doesn't have the power budget to keep the memory contents, but the device is expected to return to full operation within a few milliseconds. <S> Booting the MCU and then having it shovel data from flash into the FPGA tables would simply take too long — this would start the FPGA PLLs after the CPU PLL rather than in parallel.
| The system controller is a group of dedicated hardware functions that manage the internal operation of the overall chip.
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Which type of capacitor to use for matching the data sheet? I want to choose decoupling capacitors for some microcontroller peripherals, which in their data sheets only specify some capacitans. Since you have quite a choice of passive components and the data sheets not always specify which type to use (ceramic, some electrolytic, aluminum or tantalum) I feel insecure. The data sheet says to use a 1uF and a 0.1uF cap. I feel sure that the 0.1uF one could be a ceramic. But could the 1uF be one to? I want to operate at 3.3V will a 1uF ceramic cap lose to much capacity to make the recommended 1uF. And is a 0402 cap as good as a 0603 for the same capacity? Thank you for any help! <Q> Decoupling capacitors are usually of the ceramic kind, sometimes combined with an electrolytic capacitor. <S> In most cases ceramic capacitors will fit your need just fine <S> I usually use the biggest footprint I can fit on to the board (0805-1206) unless I don't have the space. <S> Capacitors with bigger footprints usually have a higher max voltage rating without a lot of added costs. <S> Also they tend to have improved temperature stability. <S> So the 0402 cap will probably have a reduced max rating, and slightly reduced temperature stability in comparison with the 0603. <S> But if you are not planning on using the product in extreme environments and the Vmax of the 0402 fits your needs (including some extra tolerance as buffer) it should be fine. <S> Another thing to watch out for is the type of capacitor X5R, X7R and C0G/NP0 are less susceptible to a broader temperature range as Y5V. <S> This is <S> I think pretty much the basics, other things that can be taken into account are the ESR/Q needed for example. <S> Which can be of importance when designing filters of switching power supplies. <S> ADDITIONAL EDIT: <S> I just briefly want to add that if you are working with high frequency, where the performance is determined mostly on the inductance. <S> That a 1206 SMT capacitor will have approximately twice the inductance of a 0603 capacitor. <S> Therefor for high frequencies it is recommended to use the smallest practical size for the application. <S> And find the largest capacitor value available in this size. <S> (A smaller value will have no improvement in high frequency performance but will degrade low frequency effectiveness) <A> I think you're over thinking this. <S> The datasheet says that you need to use 1 uF in parallel with 0.1 uF and that is because the combination of those will give you proper decoupling over a wide (enough) frequency range. <S> See this video from the EEVBlog . <S> Since the 0.1 uF will (should) indeed be a ceramic type the type of the 1 uF does not matter that much. <S> You can use the type which you prefer. <S> My guess is that the 1 uF capacitor will also be a ceramic type as there are quite common. <S> I want to operate at 3.3V <S> will a 1uF ceramic cap lose to much capacity to make the recommended 1uF <S> Not sure I understand you here, <S> since you want to use the decoupling at 3.3 V you need to make sure the capacitors can handle that. <S> Any decent 1 uF capacitor will have a value of 1 uF at 3.3 V. <S> Even if the value of the capacitor deviates somewhat over DC voltage that is not a problem for supply decoupling. <S> The actual value of the capacitors does not matter that much. <S> The size of the capacitors 0402 or 0603 does not matter (much) either, use whatever you prefer. <S> Either size will decouple the supply lines equally well. <S> You might also want to watch: <S> Are bypass capacitors really needed? <A> Ceramic capacitors are fine in bypass applications, but you're right to worry about the voltage coefficient. <S> Along with temperature and aging, it can affect the actual capacitance--in fact, larger capacitors are often at the low end of their target value even at nominal test conditions. <S> It's an issue for all class II ceramics, and as package sizes grow smaller, it becomes more prevalent. <S> App notes for bypass caps are typically fairly generic, so you don't need to be concerned about matching the value exactly, but for the larger caps (>1uf) you'll get better results from a larger package. <S> This is a reference I find indisposable. <S> I don't work for MuRata, but it's a great tool, and most caps from other manufacturers (given the same size and specs) will behave similarly.
| Ceramic capacitors are widely used for most applications, the electrolytic ones are useful if you need higher capacitance values in a relatively small package.
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Looking for a dc 2pin connector with latch I'm looking for a pair of connectors with following features. about 10V rating (no high voltage) 5A 2 pins latch which will prevent from accidential unplug panel mount I want to mount the socket on alluminium scheet. I dig through internet, but the only connectors which I found (which comply with above requirements) is BNC. Can you help me and propose something else? I'm not sure if i can use BNC as about 4Amps connectors, it's kinda unproffesional. <Q> Aluminium sheet, or aluminium pcb. <S> Not exactly sure how you want to mount it. <S> But maybe these through hole molex connectors <S> Part Number: 26-01-3127 <S> Mini-Fit Jr. Header, Dual Row, Right-Angle, with Snap-in Plastic Peg PCB Lock, 2 Circuits, PA Polyamide Nylon 6/6 94V-0, <S> 0.38µm Gold (Au) Plating <A> One way to deal with this is to find a suitable connector family that has more pins and the wire up multiple pins in parallel through the connector. <S> This lowers the current rating needed per pin whilst opening up the available selection of connectors. <S> So for a 4-pin connector use two pins for PWR and two for GND with a current rating of 2.5A per pin. <S> Using this technique can sometimes be an advantage when it comes to building the mating cable. <S> Instead of one larger conductor in the cable you can run multiple smaller conductors which can end up with a more flexible cable and not having to bridge multiple connector pins to a single conductor in the connector backshell. <A> If it's a hobby application, try what the Chinese call "aviation connectors". <S> Two pin are rated at 5A. <S> They're circular connectors that have a nut that holds them together. <S> Very inexpensive and fairly sturdy.
| A 6-pin connector would offer three pins each for PWR and GND and get by with a rating of less than 2A per pin.
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How to turn on/off a converter using OpAmps I am designing a controller board for turning a DC/DC Converter on when the voltage level is below 24, and turn it off when it's above 29. With this circuitry I'll be turning it off above 29V. I can design similiar comparator to turn it on below 24V. What would be the best way of combining these circuitries? And how do I add hysteresis? Is it a feedback resistor to the OpAmp? <Q> The window comparator circuit is the best for selecting a range of voltages, which requires another comparator, you may have to have to watch the impedance of the transistor stage, or provide a buffer as this circuit is more for a high impedance input as it has a pull up. <S> Source: <S> https://www.electronics-tutorials.ws/opamp/op-amp-comparator.html <S> Another way to do this would be to use diodes between the window comparator and the transistor. <S> Source: <S> http://www.electronics-tutorial.net/analog-integrated-circuits/op-amp-comparators/window-comparator/index.html <S> With hysteresis the diode design might be better, however with a feedback resistor and a high impedance reference voltage <S> this may prove difficult. <S> Make sure you size the resistors of the feedback network to be higher impdeance than the voltage dividers so the reference voltage doesn't sag. <A> You only need a single hysteretic comparator that covers the entire range, for a total of two thresholds. <S> In your case, those thresholds would be 5V apart. <S> When the input exceeds 29V, the circuit switches and starts looking for 24V. <S> When the input drops below 24V, the circuit switches and starts looking for 29V. <S> This circuit should get you started: simulate this circuit – <S> Schematic created using CircuitLab <S> The VR's are the adjustment points. <S> The two other resistors around each one are to limit the range if you need, or you can calculate the exact values that you need and eliminate some adjustment. <S> s <S> is the scale factor that the input is reduced by. <S> Set this first so that the input is always in the opamp's range. <S> c is the center of the hysteretic range. <S> r is the hysteretic range itself. <S> The feedback to the positive input is what creates the hysteresis. <S> In this variation, you might think of it as the threshold staying fixed and the apparent input "jumping" up or down following the output. <S> An alternate version would swap the V+ and Input ladders so that the threshold seems to jump instead. <S> Either way works, and you can make it inverting or non-inverting simply by the way you connect it. <A> simulate this circuit – Schematic created using CircuitLab <S> Using a 555 timer in schmitt trigger mode. <S> When your Vin is lower than 24V, the TRIGGER voltage is below 5V when it passes thru a 19V zener diode; this will cause the OUTPUT to go high. <S> When the Vin exceeds 29V, TRIGGER voltage will >10V. <S> This will cause the OUTPUT to go low. <S> The IC cost less than 10 cents if you buy in bulk.
| You don't really need separate hysteresis at each end of your range, for a total of 4 thresholds.
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Low power way to switch on MOSFET with ~100mV input singal? Is there a low power (single digit uA's quiescent) technique to switch on a MOSFET when an input signal rises above ~100mV (too low for direct gate drive)? You can buy ICs like the MCP6548 that functionally accomplish this, but I am looking to do it with discreets. I am thinking there might be a way to bias a MOSFET gate so that the 100mV falls right in the transition region, and triggers positive feedback loop to throw it in to full active region, but I can't figure out how to do this. <Q> Consider this circuit simulate this circuit – <S> Schematic created using CircuitLab <S> You can drop the +9 volts to +6v or maybe <S> +4 or +3v. <S> The slewing of Q11 collector, from fully one direction to the other, may set the propagation delay. <S> Headroom on Q2 and Q3 is marginal; you may need triple Darlingtons to create more headroom, to drive the NPN diode loads. <A> Here is an interesting solution I found that uses an optocoupler as a constant current source to boost the input signal up the gate threshold voltage... <S> https://www.electronicdesign.com/power/simple-circuit-overcomes-mosfet-gate-threshold-voltage-challenge <S> While this is clever, it is not low power since the optcoupler's LED is always on. <S> It is also non-optimal since you must manually adjust the booster so it does not automatically adjust to changing temperature. <A> No there is not a way of doing what you think. <S> You can indeed bias the input up, but the gm (change in drain current / change in gate voltage) is so low that switching does not happen, just a small change in current from on to a bit more on. <S> Then with a fet, the voltage where this happens varies enormously from fet to fet. <S> With a bipolar it works a bit better, but still gm is very low, so multiple transistors are needed. <S> As you try to make the threshold closer to 0, and further from 0.5V, it become more sensitive to supply voltage and temperature etc. <S> This circuit gives you your 100mV at 2.5uA, and is a hard switching schmitt trigger. <S> In theory. <S> Its more like a useful circuit for a 300-400mV level. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> That said, this just runs on the 100mV... <S> simulate this circuit
| There has to be a lower power way to boost that input signal that uses feedback to automatically adjust to exactly the bottom of the MOSFET's linear region.
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How to test for USB dropouts on STM32F? I've got a devices that is really strange, it drops USB periodically. By periodically I mean on a weekly (or biweekly) basis. By dropping out I mean windows loses the handle to the driver. Dropouts a not tolerated well by those using the devices. The weird thing is I use a similar design on most of the products that we have (more on that later), for some reason this one doesn't want to play nicely, some of the other devices run for months to years (with the same design but different layout) with zero issues. It's only on some of the devices so that probably rules out firmware/software. The USB is 2.0 FS, running to an STM32, with an ESD diode chip in the middle. Schematic (D+ and D- and OSC_IN and OSC_OUT (oscillator design was inherited) go to their corresponding pins) A better question would be, how do I test for these dropouts? Is there some method that could monitor a device for very long periods of time with millions of packets going by and find the source of the error? <Q> Is there some method that could monitor a device for very long periods of time with millions of packets going by and find the source of the error? <S> Yes. <S> The device is called "USB protocol analyzer". <S> If you monitor only the host software side, the maximum you can see is that there was some "transaction error", and the port can or can't recover after dropping. <S> USB protocol has hardware-assisted means to re-try failing transactions, and software doesn't have any visibility into "error count". <S> So you need to identify the root cause of the error at physical level, on D+/D- wires. <S> There are affordable USB analyzers, especially for USB 1.1 (FS 12 Mbps) rate. <S> A good analyzer can be set for a sophisticated trigger while monitoring traffic in a long loop, or even recording the entire traffic up to capacity of your hard drive. <S> I would recommend a small Teledyne/Lecroy model Mercury T2 , but other guys like Ellisys and Totalphase <S> Beagle are getting better and better. <S> However you need to be careful, since the analyzers are somewhat invasive, and their connectors/internals do have some effect on signal integrity. <S> In case of flaky connection and rare error rate the analyzer can either improve the signal (and you might never see the problem) or can kill the link functionality (which will helpful to pinpoint the problem). <S> So in short you need to identify who is at fault when the device drop happens. <S> It could be <S> (a) device makes wrong responses to a valid USB protocol, (b) channel signal integrity problem, or (c) <S> host hardware has a bug in handling some peculiarities of USB protocol. <S> I would start with (b) and check if all signals on the bus meet basic USB signal specifications: pattern frequency within 2000 ppm, jitter within the norm, signal edges are monotonic, and signal eye meet the diagram mask, all over your specific cables, devices, and hosts. <S> There are standard procedures described in USB-IF website how to perform the electrical tests within USB compliance program . <S> If the signals meet basic FS signal specifications, the Protocol analyzer would be the next thing to deploy. <S> If might be challenging to set up proper trigger and have correct interpretation of bus events leading to error. <S> If you don't have experience with USB analyzers, you might need to take some training or get a consultant. <A> I had a similar problem with something I was working on. <S> It seems the signal levels were rather marginal and the whole thing became sensitive to the quality or length of the USB cable. <S> You could start to look at signal levels, and maybe try a shorter cable and see if that improves things. <A> We had similar problems in past. <S> USB serial device would on some PCs fine for months and on some we would lose communication every week. <S> We started with running three devices on two identical windows PCs (combination that would have problems sometimes) and one Ubuntu box with simple log to disk software. <S> After some time, one windows box lost it - port just disappeared. <S> Windows USB driver stack (this was Win7) is bunch of legacy stuff that apparently can't handle quirks of FTDI chips. <S> So my suggestion is run test on two identical configurations + third different one. <S> If it stops responding, you have problem with device + usb controller + driver combination. <S> If not, its probably signal issue - USB monitor could help you.
| All USB cables are not equal - some have thicker conductors for the data lines than others.
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Standard ways to make a variable low pass filter I tried out a basic low pass filter on one of my analog synthesizers: simulate this circuit – Schematic created using CircuitLab and this seems to work. I was wondering how I can create a variable low pass filter. I first thought that I would replace the capacitor with a variable capacitor, but I have a hard time finding these at the range of about 470 nF. Then I did replace the resistor with a variable resistor and this seems to work somewhat. But I am wondering if it is just that simple. My question is therefore: What is the/a standard way of making a variable low pass filter? (used for synthesizers.) <Q> You can use a digi-pot to make the resistor variable. <S> Then select multiple capacitors that have values that define the filter ranges that work with the resistance range of the digi-pot. <S> Deploy a microcontroller to control the pot and mux. <S> The MCU can also support the user interface of your choice. <A> One method I have seen used in bass guitar add-on circuits is a PWM modulated R using a CMOS transmission gate. <S> If you choose the frequency such that it is not audible and vary duty cycle,d then the effect Reff=R/d. <S> This results in a reduction of f(-3dB)=1/{2piReffC} , but also a rise in output impedance so it must be buffered. <S> A simple vbl. <S> PWM can be made by offsetting the DC bias input towards Vcc/ or Vcc*2/3 of a CMOS Schmitt Relaxation Osc. <A> Your varipot approach works "somewhat" because the simple filter you built (one-pole) doesn't have a very steep drop-off. <S> For the fairly narrow band of the audio signal, what you're trying to do will yield mediocre results. <S> You're better off making a few higher order filters, and using a multi-position switch to switch between them. <S> Alternatively, you can buy reasonably-priced switched-capacitor filters (sort of what @TonyEERocketScientist was talking about). <S> Such devices use the frequency of a square wave clock to set the cutoff. <S> You would then need to generate a clock signal, maybe with a microcontroller or 555 timer. <S> For an example part, there's the MAX7490
| Use an analogue mux chip to select which capacitor is switched into the filter circuit at any given time.
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Series inductor to limit motor current? Short question : Would it be possible to use a series inductor to limit the inrush current when switching on (or briefly loading) a DC motor? Longer question : I have a battery-powered motorized toy which for convenience I'd like to run from a power socket instead (to avoid having to recharge and replace batteries). Currently it uses 4 1.5v batteries and if I measure it in steady state it draws about 0.8A at about 6V. The way that the toy works is that the motor is usually running continuously with very low load, and occasionally it briefly gets some load on the motor. I don't know exactly what kind of motor is used, but assume it's cheap and simple. I bought a (cheap) 6V power supply which is allegedly capable of delivering 1.2A, but its current protection cuts in when switching on the motor. As I understand it (please forgive my very basic knowledge here!) the motor presents an extremely low resistance when it's starting up, and so tries to draw a lot more than its usual 0.8A. When running from batteries, I guess they have a maximum current that they're able to provide, so everything gets automatically limited. When the motor is up to speed, its "effective" resistance rises due to its motion, and it reaches its steady state of drawing 0.8A. I expect that when the motor experiences some mechanical load, then the drawn current also temporarily rises, but again the batteries can only provide what they can provide. When it's connected to the mains power supply, the motor tries to draw much more than 0.8A (I'm not sure exactly how much more), trips the power supply's protection and the power supply cuts out. The motor slows down again, the power supply recovers and switches back on, but then the motor tries to draw too much again and it cuts out again. I've read about "motor inrush current" and "current limiting" but for this simple toy I'm hoping it won't be necessary to build a custom driving circuit with transistors or mosfets and I can't believe that a temperature-based thermistor solution is ideal (partly because of wasted heat, partly because my intermittent load is not temperature-related). From what (very) little I know about inductors, it seems that one of those in series would "fight against" the sudden change in current and develop a voltage across itself at startup. So I'm hoping this would temporarily limit the current to say 1A. In the steady state this voltage across the inductor would drop to zero and it would then act as a pure low resistance, which is exactly what I think I want. It doesn't need to be super-precise or to have a perfectly-flat 1A current limit, it would be fine to curve around 0.9A or 1.1A, but I don't know of an easy way to measure what's going on with a simple DC amp-meter. Could this be the right way to go, or is it too simplistic? And how would I calculate what kind of inductor would be suitable? <Q> The time is too long for a usefully small inductor. <S> A big electrolytic capacitor across the power supply can supply the start surge. <S> Try 4700uF or 10000uF. <S> You can also make a soft-starter using a power fet. <S> Adjust R and C to get right rate. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> eg: <S> https://www.aliexpress.com/store/product/DC-DC-CC-CV/1326062_32803647489.html <S> The three brass screws control the output parameters, one sets voltage, one sets output current, and the other sets input current, set the voltage one for 6v then set the input current one so that the powersupply's over-current current limit does not activate. <S> You'll have to feed this thing with more than 6V though, the output is always less than the input, 9V or 12V etc would work fine. <S> you're going to need a voltmeter so you can set the voltage. <A> You could fit a simple inrush current limiter (ICL) in series with the power supply. <S> These are basically negative temperature coefficient resistors - when you power the device, a t=0 the ICL has a high resistance which limits the current. <S> As the small amount of current slowly heats up the ICL, it's resistance decreases. <S> This eventually leads to the resistance being near zero when the ICL stabilises at a certain temperature. <S> The effect is it squashes inrush currents and stretches the pulse out, then once stable has almost no effect on the circuit. <S> Added bonuses, its one small cheap 2 pin device. <S> Although its possible to model the behaviour of ICL devices, and select an appropriate through simulation alone - its incredibly difficult. <S> There are a lot of factors in play (heatsinking of ICL to board, ambient temperature, shape of current pulse, repetition etc). <S> In my experience its best to just buy a selection of values and give them a try. <S> In the lab I test them with a scope and current probe, but just works/not works testing is also fine.
| I'd go with an off-the-shelf DC-DC converter with built-in current-limit and feed that from a plug-pack with a higher voltage
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How feasible is it to localize a robot based on EM beacon? I am designing a little tabletop robot with a very simple goal of keeping track of where it is on the table as he traverses it. This, as it turns out, is a lot more difficult than I imagined. All the obvious solutions like using an accelero-gyro combo and using distance sensors seem to be flawed as they either drift very much or are noisy and unreliable. So I came up with the idea to install some kind of EM beacon on the vertices of the table. I don't have a detailed theoretical understanding of EM waves but here is what I have gathered so far: I can generate a simple EM wave by exciting an antenna with a sinusoidal wave (with an R-2R DAC) An EM receiver is an antenna output hooked to an op-amp input with its output going to a DAC. I can use a microcontroller to get the peaks of this signal and collect data on how distance affects amplitudes. If I plot this data on the computer, I should see a relationship between distance and amplitude of waves. Then I can use this relationship to measure the robots distance from that particular beacon. If I use different frequencies for the 4 different beacons, The robot should be able to calculate its position on the table. I'm a programmer first so I am sure I can handle the digital portion of this setup. I am, however, concerned if EM waves work the way I expect. Am I being too naive with this? Basically what I want to know is this: Is it reasonable to expect that this setup will work reliably, assuming that the digital parts of the system work fine? Has this been done before, and reliably? What are the sources of error that I have not been able to gather? Any suggestions/tips/resources are welcome. Edit : While the answers are interesting and have provided me with lots simple ideas to accomplish what I'm trying to do with my convoluted setup, none of them really answer my question. As such, I think that unless I conduct the experiment myself I am not going to find an answer. <Q> Basically what I want to know is this: Is it reasonable to expect that this setup will work reliably, assuming that the digital parts of the system work fine? <S> Has this been done before, and reliably? <S> What are the sources of error that I have not been able to gather? <S> First off, you can minimize your sources of error with a Kalman filter. <S> The drawback is they are difficult to implement and understand. <S> Or you can buy an Inertial Measurement Unit (IMU) that has already done the work and corrects the data. <S> Secondly there are already off the shelf systems that can help localize robots. <S> One is RTLS systems which work to 10cm (like Pozyx ): <S> Precision RTLSs have emerged as an effective method for determining the location or tracking of people or mobile assets in office complexes, warehouses, manufacturing plants, and assembly lines. <S> In this approach, a mobile object (tag) exchanges information with fixed-position devices (anchors) using standard formats and UWB technologies specified in IEEE 802.15.4-2011 for low-rate wireless personal area networks (LR-WPANs). <S> By determining the distance between the tag and several anchors, applications can determine the tag’s relative position to those known anchors – and thus, the tag’s absolute position. <S> Source: <S> Digikey <S> The second is RTK GPS, which works better outdoors but has 1cm accuracy: <S> Real-time kinematic (RTK) positioning is a satellite navigation technique used to enhance the precision of position data derived from satellite-based positioning systems (global navigation satellite systems, GNSS) such as GPS, GLONASS, Galileo, and BeiDou. <S> These systems can be had for roughly 500$ <A> I've looked at magnetic field sensors. <S> Perhaps at 10MHz or so, because you have a very short working region. <S> Previously I have 10 meter ranges at below 1MHz carriers, with signal recovery still needing narrow-banding before feeding to ADC. <S> With 100KHz bandwidth, my SNR was about 0dB. <S> Had I taken the project forward, a 100 Hertz bandwidth would have produced 30dB SNR. <S> And 100Hz allows quite rapid switching between signal sources, thus perhaps only 1 second of signal processing is needed. <S> You'll need I & Q coils in the robot. <S> Extract a vector to each beacon, and use the " resection " math. <A> Mark a 2 dimensional non repeating (pseudo random) pattern on the table and have a camera on the robot looking down. <S> Print it with same coloured ink on coloured paper that is visible in IR or UV <S> light only and no one will be the wiser. <S> Mouse camera can give you speed of movement in X and Y and a Web cam can calculate your position and angle based on the decoded pattern. <S> Resolution to sub pixel of your camera is achievable. <S> Resolution to your pattern cell size should be possible in a micro controller with rudimentary camera library. <S> EDIT:
| There are a few projects that have hacked mouse cameras to supply the raw diagnostic video that could be used to establish location with a single cheap camera.
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Class D power supply filtering for conducted emissions I have recently designed an audio player that uses the TI TPA3112D1 power amplifier, and mostly based it on the reference design and schematic . I find however, that the design is failing the conducted emissions test due to noise feeding back into the 24VDC supply. This noise starts at 340khz with harmonics present proceeding up the freqency range. I assume this to be from the amplifier, as the noise is starts as soon as the amp is unmuted. The noise is present regardless of audio level output of the amplifier. I have tried adding different types and variations of bypass caps, which seems to have really no effect. I also tried a 12uh inductor (AISC-1210HS-120K-T2) I had on hand on the positive supply input, which also made no noticeable difference. The test house also tried a Steward LFB310190-000 , which made little difference. The PCB is a 6 layer board, the pictures below don't include layers 3 and 4, as there are no traces or planes close to the power and power amp circuit. The power circuit uses a common mode choke , which probably doesn't target these low freqencies at all. Googling this is largely ineffective, as most app notes are geared towards emi on the output side of a Class D amp. It may be I need low freqency filtering of some sort, but it isn't clear to me. Power amp datasheet My own lab results Pre and Post 12 uH inductor, prior to common mode choke 24V + and -, with A-B, at output, through common mode choke(It would almost appear to amplify these signals!?) Attempted fix 1 In this fix, I have cut the traces between the via and the sm caps, routing the supply through the cap leads. This cap is a 470uf 35v electrolytic. I have also double stacked 1nf and .1uf on each side. L4 and L5 are part HI0805R800R-10 Original C12 and C64 100uf 50v are part UCD1H101MNL1GS Changed to 470uf 35v EEU-FR1V471L C29 and C13 1nf 50v are part C0402C102K5RACTU C30 and C31 0.1uf 50v are part C0603C104K5RACTU Results are no better, probably slightly worse. The most noticeable difference is caused by touching output leads or speaker body, which causes these spikes/harmonics to approximately double in amplitude. This would seem to indicate that output filtering (L4 and L5) play a part in how this works. Probe setup <Q> Do re-read the datasheet: <S> The TPA3112D1 is a high-performance CMOS audio amplifier that requires adequate power supply decoupling to ensure that the output total <S> harmonic distortion (THD) is as low as possible. <S> Power supply decoupling also prevents oscillations for long lead lengths between the amplifier and the speaker. <S> Optimum decoupling is achieved by using a network of capacitors of different types that target specific types of noise on the power supply leads. <S> For higher frequency transients due to parasitic circuit elements such as bond wire and copper trace inductances as well as lead frame capacitance, <S> a good-quality, low equivalent-seriesresistance (ESR) ceramic capacitor from 220 pF to 1000 pF works well. <S> This capacitor must be placed as close to the device PVCC pins and system ground (either PGND pins or PowerPAD) as possible. <S> For mid-frequency noise due to filter resonances or PWM <S> switching transients as well as digital hash on the line, another goodquality capacitor typically 0.1 µF to 1 µF placed as close as possible to the device PVCC leads works best. <S> The 220-µF capacitor also serves as a local storage capacitor for supplying current during large signal transients on the amplifier outputs. <S> The PVCC pins provide the power to the output transistors, so a 220-µF or larger capacitor must be placed on each PVCC pin. <S> A 10-µF capacitor on the AVCC pin is adequate. <S> Also, a small decoupling resistor between AVCC and PVCC can be used to keep high frequency, Class-D noise from entering the linear input amplifiers. <S> Get at least 220uF low impedance cap. <S> Also here are a few pointers to your layout. <S> 3 circles in red, traces to thin to deliver higher power. <S> 2 arrows in orange, shift the via holes to a new location; punch at least 3 vias per side. <S> Layout examples in page 23 is better to follow for first prototype. <A> the design is failing the conducted emissions test due to noise feeding back into the 24VDC supply <S> This requires some source impedance in order to achieve any attenuation with C12,C29,C31 for 100uF, 100nF , <S> 1nF. <S> If the source impedance of the supply is lower then noise is not attenuated as much. <S> A low ESR ceramic 1 <S> uF may be better than 100nF <S> > <S> 1MHz <S> Here is a solution using 10uH. <A> Nothing like having skin in the game, so I am posting this as none of the other answers really targeted the core problem. <S> This is after all a high performance chip, so that means fast switching to a load is going to reflect right back to the PVCC, and no amount of power input filtering is really going to mute this level of noise. <S> This answer on the TI forums points to that assumption. <S> So in the vein of that answer I put 15uH and 0.47uf reconstruction filters on the output rather than the ferrite bead filter. <S> Original part scan: <S> Modified part scan:
| For filtering lower frequency noise signals, a larger aluminum electrolytic capacitor of 220 µF or greater placed near the audio power amplifier works well.
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How to find the total resistance of this circuit Without giving resistor values, what are the steps to breaking down this circuit to find total resistance? Edit: Like this? <Q> First, identify the nodes that are have the same voltage on one side and the same drop voltage on the other end. <S> Take the parallel of the resistors and substitute that value for the R-equivalent. <S> If you redraw the circuit, you will see that voltage drop is the same for R6 and R1 path <S> R5 and R1 path R4 and R2 path <S> Let's call the low side node, attached to the voltage source, X; and,the upper side, where R5, R6 and R4 are connected, Y. <S> The voltage drop across the aforementioned resistors are Vy-Vx, corresponding to the nodes at location high side Y and low side X. Focus on this portion of the circuit and simplify: <S> Here we see R5 and R6 share the same node with R1. <S> Call this node Z. <S> These two are in Parallel, because R5 and R6 have the same voltage = <S> Vy-Vz. <S> Simplify these two resistors, and this becomes R56 equivalent. <S> Now R56 is in series with R1; Simplify and substitute this with R561 and attach it between node Y and node X. R561 is parallel with R4 (and R2 along that path), b/c <S> it is between nodes X and Y. Simplify this one and make the substitution. <S> The process is identifying the parallel combination of resistors and then resolving any series resistors attached. <S> Make sure that when two or more resistors are in parallel, we can identify the total voltage across to be the same. <S> In your example, it is easy to identify R5 and R6 in parallel, but to identify (R5||R6 + R1) <S> || (R4+R2) is not so easily seen. <S> Redraw the circuit each time you simplify with an equivalent resistor. <S> The rest is left as an exercise. :) <A> <A> Here is ... <S> (+) -- (R3 || R7) -- ( (R5 || R6) -- <S> R1) <S> || (R4 - R2) ) -- <S> (-) <A> Figure 1. <S> OP's answer. <S> Yes, you appear to have it. <S> You could improve the notation a bit using '+' for series and '||' for parallel. <S> Your end result would be Upper: (R1 + (R5 || R6)) <S> || (R2 + R4)Lower: <S> R3 || <S> R7Total: <S> (R1 + (R5 || R6)) <S> || (R2 + R4) + (R3 || R7) <S> That way it's clear how you reached your conclusion and you can generate the mathematical expression to calculate from the series and parallel resistor formulas.
| Try to break down the circuit into simpler series/parallel resistors and get the equivalent resistor, then repeat until you get only one resistor.
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IC that would pull output to the ground, perhaps NOT gate? I have a board with four relays on it. It has a standard Vcc, Gnd, and In1-In4 pins. I want to control it with Arduino or in my case an ESP32. However, I noticed that when output pin on my controller is pulled HIGH nothing happens when it's pulled LOW the relay will engage. This is a problem when the board is reset because of all the relay switches will engage until and this isn't something I want. After some digging, I made this: This works as desired, there's always 5V on the relay and the relay is disengaged all the time. When I output a signal to the base of the Q1 it will pull the output to the ground and the relay will engage. Right now I have four of these relays and pretty soon I'll have more. So, I'd like to solve this with an IC. In particular, I was looking at the IC7404 which is a bunch of NOT gates, however, I am not sure this will solve my problem. So, is there an IC that would pull the output to the ground when the signal is present on the input? Will 7404 work and I'm just not seeing something? :) Edit: This is the relay module I am using . I wasn't able to find any useful data sheets on it. <Q> The fact that relays are actuating at reset when pins should be in a high impedance mode is worrisome, and something that you need to fully understand in order to validate your system and fix it. <S> Edit: now that it is clear that this is a software bug in your code, to solve the inadvertent drive low during starting, configure the pin's output data register bit to high before you set the pin to be an output. <S> That said, a literal answer to your original question would be a chip containing several open collector inverters . <S> The 74xx05 is an example. <S> The 74xx06 is similar but allows applying acceptable voltage to the output even when this exceeds the supply voltage, for example you can operate the part on 3.3v but pull down outputs from 5v. <S> But these are not the solution to your actual problem. <A> The simplest solution is to swap the connections of the NO and NC pins of the relay. <S> You can add a few constants #define RELAY_ON LOW#define RELAY_OFF HIGH <S> To bias the relays into a certain state during reset (while the output pin is high impedance) you can add a pull-up or pull-down resistor as needed. <S> The IO pin will easily overpower the current leaked through the resistor and still properly switch the relay. <S> Using those 2 techniques you can pick the default state of the relay to be the unpowered state. <S> Either because it's the one used most of the time or because it's the fail-safe option. <A> Here is a "User manual" for your relay. <S> You should really never design a system without having all datasheets first. <S> From the schematic supplied there it can be clearly seen you can just connect your controller directly to the InX pins. <S> That is, none of the schematic you devised is needed. <S> Since the input of the optocouplers is able to operate in open-collector mode I would suggest to you to just keep the corresponding pins of your uC as input/ <S> High-Z when you want to keep the relays at NC. <S> Since I'm not familiar with your controller, here is some pseudo-code to help you: Initialization <S> , for each pin controlling a relay: Set direction to inputSet level to "0" <S> (do NOT change direction to output) <S> Command relay to NO: <S> Set direction to output Command relay back to NC: <S> Set direction to input Usually a uC will wake up with its pins already as input, so the first line of the initialization is redundant, but it's always safer to perform it to make sure the system is at a known state. <S> Make sure you do not have some code that does a general init of all pins to some values as part of some boiler-plate code, beginners are known to make this mistake. <S> P.S.For the time when you actually will need some inverting current driver just use something like a ULN2003A !
| The next option is to invert the programming logic for that pin.
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Detect 220 VAC with a Raspberry Pi I want to build a simple device to check if the light is on or off. I found some schematics but I need to be sure that it is safe. Source : How to detect 220 VAC voltage using an opto-isolator My problem is that I do not understand this very well. I found that PS2501 have a chance of working up to 80V so how can I safely connect it to 220 VAC? Which resistors make a limit for the voltage here? Why do we need to add a capacitor and diode (D1,C1)? I'm a beginner in this field so please be patient. <Q> If safety is a primary concern, rather than build that circuit up, you could consider purchasing a standard AC-input module for about $10-$15. <S> They are UL, CSA, CE, and TÜV safety certified (it's still possible to go wrong and create a dangerous situation, especially if the wiring is sloppy, but less likely). <S> Best to have someone knowledgeable look it over before you power it up in any case. <S> The circuit inside the housing is somewhat similar to the one you show, however it will respond much faster. <S> With regard to the circuit you show the PS2501 has a short-term isolation voltage rating of 5000V RMS. <S> That is adequate to safely withstand the 240VAC mains and most transients that might appear on it. <S> For safety you need to keep the creepage (surface leakage) distance between input and output leads (of the opto) to at least 8mm and make sure it can never get wet or otherwise contaminated with conductive materials. <S> R1 and R4 may or may not be acceptable depending on the type. <S> They can certainly burn up under some conditions. <S> Without an earth connection on the isolated side you are depending on a few mils of plastic inside that opto for safety. <S> As far as your other questions, the 90V rating is only the output transistor- in operation, it sees only 5V and the 240V input is reduced to the 1.2V the LED needs through the components to the left of the isolator. <S> All the latter parts are electrically "hot" (including that side of the opto) and need to be well protected against accidental contact. <S> The capacitor C1 MUST be an X2 type which is a safety certification for cross-mains use. <S> R1 and R2 must be capable of withstanding mains voltage and transients. <S> Vishay VR25, 35 etc. <S> series is appropriately rated. <S> C1 is what really does the work of dropping the 240VAC down to 1.2V. <S> Most of the mains voltage appears across it. <S> On positive half-cycles the current flows through the LED in the optocoupler, on negative half-cycles the current flows through the 1N4007. <S> The resistors are there mainly to limit the current if powered up when the voltage is not zero. <A> <A> I strongly suggest to reconsider using your circuit in favor of detection without direct connection to mains voltage. <S> Some examples are: Photodiode to detect light Coil-based current transformer to detect current Current transformer with Hall effect sensor to detect current <S> All of the above have the benefit that they do not trigger if your light bulb is burnt out (if that is what you want). <S> Non-contact voltage sensor (NCV) circuit to detect AC voltage <S> The latter can trigger when the switch is turned ON even if the bulb is burnt out. <S> Not only the options above are much safer, their circuits are usually simpler than the monstrosity in your question. <A> That circuit detects AC current from your wall; it doesn't directly detect light. <S> It could work if you plug it into the same connections as your light, but it's not really a great (or safe) solution. <S> To answer your questions: I found that PS2501 have possiblity to work up to 80V <S> so how can be work if we want to connect it to 220 VAC ? <S> Take a look at the datasheet. <S> The 80V rating is for the collector-emitter voltage for the transistor; this is only applicable to the right side of your circuit, which is only exposed to 5VDC. <S> Why we need to add a capacitor and diode (D1,C1)? <S> D1 clamps the nodes in front of and behind R3 and the optocoupler when that side has a positive voltage. <S> This is to make sure that the diode in the optocoupler doesn't break since it can only withstand 6V in reverse. <S> Clamping is a common way of protecting circuits from unwanted voltage. <S> The capacitor in series with the resistor was added to smooth out the AC voltage waveform to avoid false positives.
| The circuit on the low voltage side should be earthed and fuses or other current limiting used so a failure of the opto cannot cause a hazardous condition. The cheapest and simplest solution is just to buy a tiny mains-powered USB charger and wire up the USB +5 and ground lines to detect when the charger is powered by AC.
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Do an Op-amps supply pins need to have the exact same nominal voltage? By power supply unit should ideally output +12 and -12V, however the readings are closer to 12.2V and -11.8V. Is this likely to cause a problem, even though I only need an output range of 10V to -10V? <Q> There is no requirement for balanced voltages. <S> You can use -0.1V and +20V if that works as indicated above (and, of course, if the op-amp can handle a 20.1V supply). <A> The PSRR specs include DC meaning , other than saturation near the rails, if bipolar outputs , there is neglible influence. <A> If you operate within specifications your choice of +ve and -ve supply rail magnitudes is not that relevant. <S> Often we see a few volts negative to handle ground references inputs and outputs and the positive supply common with other needs. <S> Supplies of +12V and -5V would be common for reliable 0-5V operation. <S> Your choice of OP amp or other components will dictate the power rails. <S> If you need to operate at the mid point of the output swing and want to amplify ground referenced signals then you need to keep them equal magnitude. <S> Some new OP amps can operate very close to the rails on input and/or output. <S> These developments are the result of the drive for ever lower operating voltages. <S> You cannot afford to loose 1.2V of headroom if your circuit is working with a 3V supply that you want to run from a single cell that is partially depleted. <S> Most OP amps with single and dual rail operating specifications are inherently designed to handle uneven supplies. <S> One of them is by design specified to be zero. <S> Bypass, compensation and other circuitry has to be referenced to the correct point, often the negative rail <S> so it works in both configurations.
| If your op-amp can swing within 2.2V of the positive rail and within 1.8V of the negative rail under all possible load conditions , and if the inputs likewise are within the allowable common mode range then there is no problem. Usually systems are designed with same voltage but this is not a hard and fast rule.
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Create small variations in voltage for opamp reference pin Hello and thanks beforehand. I'm designing a circuit to measure 4 wheatstone bridges using 4 Analog Devices AD8422 in-amps. Now, i'm feeding the AD8422 output to a 3.3V ADC, so i'm offsetting the output 1.65V in order to be able to measure the full scale output of the bridge. However i'm aware that tiny differences in the bridges, opamps and resistors will require slightly different offsets for every bridge (i'm expecting something like +-0.05V would do). The AD8422 requires the reference to be fed from a low impedance source Now, to create the 1.65V reference, i'm using a 3.3V source, a resistive voltage divider, and a Texas instrument's OPA333 as a voltage follower. However, to individually adjust every AD8422, I would need 4 of those circuits with potentiometers in the voltage divider, each one feeding into one AD8422. The boards that have the AD8422 are very small and integrating another opamp (even a SOT23 one like the OPA333) is pretty difficult (and expensive if we want to expand the system), so, here comes the question: How would you apply small variations in a reference voltage without altering the low-impedance nature of the output? I'm aware that the potentiometer aproach would not work, as you would need precise knowledge of the intensity going into the REF pin and pretty high resistances (2k+) which would be like a voltage divider. Thanks for your help! simulate this circuit – Schematic created using CircuitLab <Q> However i'm aware that tiny differences in the bridges, opamps and resistors will require slightly different offsets for every bridge <S> (i'm expecting something like <S> +-0.05V would do) <S> I wouldn't bother trying to do this in hardware; I'd just compensate the ADC reading digitally to overcome the slight discrepancies in offsets. <S> I'd produce one solid 1.65 volt reference voltage and feed this to all AD8422 InAmps. <S> After all, if there are resistor variations that produce voltage offsets, those variations will also produce gain variations that are likely to need compensating in software. <A> If you take a look at the datasheet there's figure 55: showing us how the REF input is used. <S> If the REF input would have an additional impedance in series (like a reference voltage coming straight from a resistive divider, without a buffer, then this impedance adds to the 10 kOhm resistor present on the chip. <S> Then accuracy suffers and offsets might be introduced. <S> This sort of defeats the purpose of using this Precision amplifier . <S> What I think is the right way to go forward <S> is to determine how much inaccuracy you can tolerate. <S> From that you should be able to determine how much series resistance can be allowed in your design. <A> Something like this: simulate this circuit – Schematic created using CircuitLab <S> You will need one op-amp per ADC to get the best accuracy. <S> The voltages are different, you want them to be ~zero ohms impedance compared to 10K. <S> Depending on your accuracy requirements, it might be better to have one divider and pot per op-amp <S> or maybe you can put several pots in parallel and share the three resistors, which will result in worse tempco typically and a slight interaction. <S> R1 and R2 (in parallel with the pot element(s)) and R3 set up the limits of your Vref adjustment. <S> R2 is used because the tempco and tolerance of pot elements is typically vastly inferior to that of a precision resistor. <S> So you'd have R2 << R4. <S> And you'll pick an op-amp with low enough Ib that the the source impedance does not cause excessive error or drift. <S> Calculating the values, worst case range and drift I will leave to you. <S> Having a single voltage and b*ggering with it is less accurate but might be "good enough" if you're looking for less accuracy. <S> This involves 3 parts per ADC plus two resistors and an op-amp to generate the 1.65V. simulate this circuit <S> The added 50 ohms or <S> so will affect the gain accuracy by less than 1%. <S> If you are adjusting each one then you could trim that out, however the two trims will interact slightly so a couple iterations might be required. <S> Either way you're looking at 9-20 parts or so. <S> Accuracy and power consumption will vary with choice of resistances and op-amps. <S> For example, the circuit #1 with paralleled pots, dual op amps would be 9 parts. <S> With individual dividers and dual op amps 18 parts, Circuit #2 would be 15 parts.
| You can indeed compromise and tune the offsets as you propose but performance (accuracy) will suffer.
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why do most sensors have an electrical output, regardless of the physical nature of the variable being measured I have been trying to figure out a explanation to this question, can someone explain this ? <Q> That's easy- spirit of the time. <S> Current technology is all about electronics. <S> A hundred years ago it was different, Litmus for example hasn't electrical output :) <A> Sensors could press on or retract a pushrod. <S> With air-bearings, the wear factor would be excellent (long duration sensing). <S> But imaging the task of instrumenting a nuclear-test-site, with lots of crucial measurements having impulse-responses of nanoseconds or microseconds. <S> The pushrod inertial serves as a low-pass-filter; the exploration of uranium at Los Alamos might still be underway, and USA might now be speaking German. <A> Why do most sensors have an electrical output, regardless of the physical nature of the variable being measured? <S> Electrical control systems have the following advantages over alternatives such as mechanical, pneumatic, hydraulic, etc. <S> Speed . <S> Electrical control systems are fast. <S> Size . <S> The control systems can be miniaturised. <S> Power . <S> Electricity is readily available. <S> No compressors or pumps are required. <S> Reliability . <S> Electro-mechanical (relay) systems can have high reliability despite the moving parts. <S> Solid-state (transistor, triac, etc.) can have far higher reliability as there are no moving parts. <S> Distance . <S> Cost . <S> Mechanical control systems require precision components which can be more expensive to manufacture and calibrate. <S> Versatility . <S> Electrical systems are very versatile. <S> Transducers are available to convert light, sound, pressure, force, strain, radioactivity, viscosity, temperature, velocity, etc., into electrical signals. <S> The last one is probably the most important. <S> While, for example, pneumatic logic and control systems exist and have their benefits (explosive atmospheres, for example) it is difficult to imagine how a light to air pressure transducer is going to beat a light-dependent resistor. <S> In other applications such as temperature control mechanical solutions - car coolant thermostat, for example - are more than adequate.
| Electrical signals can be transmitted long distances without significant loss.
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Identify this rechargeable battery pack I want to identify a rechargeable battery pack from inside an old portable hard-drive. It is unlabelled. (I ripped open the blue covering - it isn't labelled on the internal metal cover either.) Battery is ~55m x ~33m x ~5mm - so relatively flat. The plug is 3-4 mm wide. Device takes +5V DC, so I imagine that is the charging voltage.One part of the circuit takes 3.3V DC, so I imagine that's the output voltage. How is this battery specified? <Q> It looks like one of a standard Li-Ion or Li-Po pouch cells, size 5.0 x 33 x 55 mm, more like 50 mm, or 503350 for short notations, could be 503250. <S> An example from Adafruit of a 503035 battery: <S> The example has h=5mm, w=30mm, and L=35 <S> mm <S> To see the actual label and determine the size, you need to carefully remove the blue wrap and exclude battery protection part from measuring. <S> CORRECTION: <S> the battery protection part is included into length measurement. <A> Aside from physical dimensions, such rechargeable cells are specified by: Chemistry (LCO, LTO, LFP, etc.) <S> LCO is the most common for older cells. <S> Min and max operating voltage ( <S> those largely depend on the chemistry) <S> Capacity (in A*h) <S> Max charging and discharging currents. <S> Those are usually proportional to the capacity for a given battery type, so they are often specified in relative units called C rate. <S> A battery capable of 1C charge/discharge can be fully charged/discharged in 1 hour. <S> 2C means 30 minutes, etc. <S> Other parameters like temperature range which are less relevant in this case. <S> AFAIK there is no reasonable way to determine the parameters above for an unlabelled battery. <A> It could be Li/Ion, Li/Poly or Li/FePO4. <S> If you replace it with "some other battery that is also square-shaped", you might end up damaging the cell or in worst case blow it up. <S> Notably there are only 2 wires, not a 3rd one for a NTC thermistor, so you can't just use any charger either.
| Given a blue chunk of plastic, there is no way to tell what the underlying battery might be. It could have one of several different charge voltages and charge currents.
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Can I replace a battery with a source of the same voltage but different capacity (mAh)? I have a device that has a bad battery and I am trying to find a suitable replacement. There are no replacement batteries for the device so I am trying to put something together myself. The original battery has written on it 3.7 V, 1.41 W. I have found a battery (CR123A) that is 3.7 volts and 700 mAh. Can I safely use it? Also, the original battery has 3 wires - red, white, and black. How would I connect the new regular battery? ** It's for my Nest thermostat. <Q> The mAh rating of a battery is two-fold generally. <S> First it tells you something about its capacity (at full state of charge). <S> The product of Current and Time is definitively Charge (as in Coulombs). <S> Second, for rechargeable batteries, it tells you something about the safe rate at which the battery can be charged, often stated in C's. <S> So a 1400 mAh battery might ideally be charged at 1.4A maximum. <S> This is just a rule of thumb type of thing, certainly refer to your battery manufacturer's data sheet for definitive information on this. <S> But if you are replacing a 1400mAh battery with a 700mAh battery, I would be really careful about using the same charging system with a lower mAh battery pack. <A> The new battery (700 mAh) comes up with 3.7 * 0.7 = 2.6 <S> Wh, so it should be capable to power your device. <S> Your original battery (1.41 Wh) comes down to 1.41/3.7 = 380 mAh, or it is smaller than CR123A. <S> It should be expected that your built-in charger uses no more than 200 - 380 mA charging current, which should be fine for CR123, it will just take a longer time to re-charge. <S> More challenging would be the handling of "white" wire. <S> I assume it is a plain thermistor, although it could be more complicated. <S> You can measure the resistance between thw white wire to ground (black wire). <S> If it comes up as, say, 10k +- 25%, then it is a thermistor. <S> You would need to fool the white wire with the same value, plus-minus. <S> Without the termistor the charger would think that the battery is overheated (or broken), and would refuse to charge it. <S> The negative side is that the thermistor was there for a reason of extra protection, and your new setup will be lacking it. <S> However, the Nest thermostat is a fairly popular product, so the replacement batteries should be freely available. <S> It is advisable to get a direct replacement. <A> An Amazon ad says <S> Duracell CR123A batteries but <S> their photo shows CR123 that is very different. <S> Somebody else's ad says Energizer CR123A <S> but they don't make it, theirs is also a very different CR123. <S> If you try to charge a CR123 then it might explode.
| Using a lower rated mAh battery pack will most likely just mean its battery life is shorter.
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removing wire from through-holes I'm currently attempting to mod my snes jr for rgb output. This requires me to solder some ribbon wire into 4 through holes. I did this once but I (stupidly) made the wire too short and had to remove it using solder wick. I must have done this incorrectly because now the through holes have a few tiny bits of wire in them and I am unable to insert a new wire (I apologize if the picture is low quality it was taken with a cell phone). What soldering/desoldering techniques would one use to clear out these through holes? I have most common soldering tools outside of a desoldering station (I just have a self-heating pump). Worst case scenario, is there anywhere I could send this to be finished off by someone who's better at soldering? Thanks. <Q> Heat with a soldering iron and use a stainless steel pick or needle with pliers to push through the remains out of the way. <S> Or add more solder and use a desoldering pump. <S> The extra solder will help bring the wire strands with it. <A> Apply liquid flux (rosin or no-clean) to the area around the pads and on the pads themselves. <S> Now apply more solder. <S> Hold the soldering iron on long enough to ensure that the solder is molten all the way through the circuit board. <S> Now use tweezers to pull the broken wires out of the holes. <S> Be very gentle - both with the soldering iron and the tweezers. <S> Flood the solder joints with solder again, one at a time. <S> While the solder is molten, sharply rap the circuit board flat against your table. <S> The solder will fly out of the holes and land on the table. <S> Inspect the now clean holes. <S> Repeat the entire procedure for any holes still having wires inside if necessary. <A> Use a solder pump not solder sucker. <S> Solder pumps are bigger and have a more powerful suction. <S> Add more solder then very quickly remove the soldering iron and place the solder pump over the via and suck it all up. <S> Make sure you do not overheat the vias because you can damege them.
| You can also try adding solder and using pliers or twizers pull the pieces out.
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How to avoid a electrolytic capacitor on a (audio) signal path? I've built a compressor basing on the all-in-one chip THAT4301. It works well. But I'm still not happy because of the electrolytic capacitor I used in the signal path. My design was based on the reference design of THAT corporation: The capacitor above is a polarized type. The signal is not biased before. As far as I know, this is a bad use. I also verified my concerns with a simulation. If the signal really containts a negative DC (which this cap tries to avoid) the cap would/should fail or blow. So I want to replace it with a polarized type. However there is not really a good/small capacitor with such a high value, so I also would like to decrease its value. My circuit works with 10u and I cannot detect any cutoffs in the LF-Response. Still a big cap in the signal path. I cannot calculate properly the value of this AC-coupling cap because I cannot find the input impedance of VCA (Voltage controlled amp) in the datasheet. http://www.thatcorp.com/datashts/THAT_4301_Datasheet.pdf 47u would be necessary to pass a 10Hz-AC-Signal if the load had 330 Ohm input impedance! Isn't this freaking low? Can anybody read the input impedance out of the other values in the datasheet? Why are they recommend a polarized capacitor for a non-biased input? I've seen this also in wikipedia (also frequently mentioned here). What kind of possibilities do I have to avoid such big AC-coupling capacitors in the signal path? <Q> The input is virtual earth <S> (It is basically the inverting input of an opamp buffer, so the time constant is defined by R1 or R6 depending on the channel. <S> I would possibly use a non polar part there, but really the trick with electrolytic caps as DC blocks is to make them large, a cap forming a pole way below the audio band has by definition <S> little signal voltage dropped across it and thus contributes negligible distortion. <S> Obviously a cap forming a pole in a filter network (Something like a passive crossover) has issues, but for small signal DC block a non polar elco is generally fine. <S> There is a LOT of crap talked about capacitors, and it IS possible to use the wrong ones in the wrong applications, but a non polar elco there will do just fine in practice. <S> C. Bateman did the classic work on this stuff back in the day in a series of articles in E&WW, well worth tracking down. <A> It's a virtual ground. <S> The datasheet says: As mentioned in the section on theory, the VCA input pin is a virtual ground with negative feedback provided internally. <S> An input resistor (R1 , 20k\$\Omega\$) is required to convert the ac input voltage to a current within the linear range of the 4301. <S> A small amount of reverse voltage is just fine on a regular polarized wet electrolytic capacitor ; it won't "blow" with a few hundred mV of reverse voltage. <S> There is an anodic film on the opposing electrode, it just isn't as thick. <S> You can replace it with a film capacitor of equivalent capacitance if you really want- <S> it will be very bulky and will cost a lot of money. <S> For example Kemet R60DR54705050K (photo from Digikey): <S> Or you could use a multilayer ceramic capacitor with a reasonable dielectric (eg. <S> X7R or X5R, you can't get NP0 in that size realistically), eg. <S> TDK C5750X5R1C476M230KA which would be relatively small and somewhat cheaper <S> but there might be concerns about introducing a microphonic part in the signal path. <S> Alternatively you could (say) use an input voltage 10x higher and increase R1 to 200K, and reduce C1 to 4.7uF, which is a relatively reasonable size for a film cap. <S> You probably want something that behaves reasonably linearly- while the capacitor won't have any effect on audible signals when the cutoff frequency is chosen properly, if there are sub-audible frequencies present near the cutoff the harmonics could be audible. <S> That would tend to point against ceramic caps, which have a large voltage coefficient. <S> Film and electrolytic caps are fine. <A> Agree with earlier response that I personally never use electrolytics of any kind; I feel (perhaps no longer true) that they evolve over time since they are liquid-based. <S> Also, the polarized nature implies that its behavior is inherently asymmetric for negative versus positive voltage drops across it, which to me implies some small amount of distortion and generation of even-order harmonics (though perhaps very very low in amplitude if the capacitance is high enough). <S> I also wonder if the ESR is similarly asymmetric for positive versus negative voltage drops. <S> So I would probably use as many Ceramic caps as I needed to (in parallel and series) to get the voltage and capacitance value I needed. <S> Also, a trick--if you DID want to use polarized caps like electrolytics or tantalum, you could use two 100uF caps in series (or four 47uF caps, two parallel pairs in series), but orient the series devices in opposite directions, so their + terminals are touching each other. <S> Then, apply a 1 megaohm resistor to this center terminal, giving the other end of the resistor a positive DC bias equal to half the voltage rating of the capacitors. <S> That way, you're using them right at the center of their voltage range, and giving maximum excursion in either voltage direction. <S> And also, any asymmetric I-V characteristics would largely cancel and you'd be left with a symmetric behavior, less likely to generate even-order harmonics. <S> Two series capacitors of 94uF will give a net capacitance of 47uF. <S> I would also add a 100k from the input to ground, to keep leakages through the cap from letting its DC point drift up in cases where there is another DC block on the incoming signal.
| But I would suggest just using the inexpensive and perfectly suitable 47uF polarized electrolytic cap the manufacturer recommends.
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Is a capacitor bad when it looks like this? Hello I am new to the electronics field. I am looking inside of one of my synthesizers and I see some capacitors (I think they are Mylar, correct me if I'm wrong) that look like this. I know it is a bad sign if an electrolytic capacitor is bulging, but is it the same on these? Is there anything wrong with these capacitors? The issue I have is that the brass sound disappears over time after you turn the synth on. It starts with high volume but it slowly goes down in volume and sometimes it disappears completely. The other problem is that there is bleed between the higher and the lower part of the keyboard. The keyboard is split in the middle. The sounds come from 5 different presets that you can turn up and down for the higher and the lower part of the keyboard. But if you turn for example the strings up in the higher it also plays a bit in the lower and vice versa. I have studied the schematics and diagrams over the past week and only now did I discover that a significant amount of the electrolytic capacitors are not the same value as what it should be according to the schematic. But it is always a higher capacitance and higher voltage. But I don't know if the schematic is wrong or if there are wrong valued capacitors on the board. <Q> Since this a heat related problem, as everyone has stated it is probably an electrolytic capacitor which is more than likely true. <S> The best way to test for the heat related capacitors is with a can of circuit cooler. <S> Canned Air Dusters will work very well if turned upside down. <S> Allow equipment to warm up and volume has gone down. <S> Then spray some capacitors in one area of the board, allow a few seconds for the coolant to take effect. <S> If you have hit the right one the volume will return. <S> If not select another area to cool four or five capacitors. <S> When finding the right area allow it to warm up again, then selectively spray one at a time each capacitor, allowing time for it to take effect. <S> When you hit the right one, replace it. <S> Some cases may have several capacitors bad. <A> The film caps look fine, but barring something really destructive (unlikely with such a part in this application) they usually do. <S> The electrolytic caps dry out over time, power supply caps first generally because they are hotter (every 10'C halves the life). <S> You usually can't tell by looking at them either (newer applications they sometimes bulge). <S> Soviet-era parts are old enough to be faulty. <S> To test you need an ESR meter, and a chart of approximate expected values. <S> It's usually possible to check in-circuit with a good such meter. <A> I'm talkikng about the k50-12 capacitors. <S> I would advise you the following rule: Download the documentation and schematic diagrams: <S> http://www.ruskeys.net/base/1501tom.php <S> Start by replacing the electrolytic capacitors in the power supply (if you have an oscilloscope, the indicator of degradation is the presence of 50 Hz on the power buses) <S> Replacement of electrolytic capacitors in the synthesizer modules: <S> The operating voltage of the new capacitors should not be less than the equivalent voltage indicated on old capacitors <S> If the capacitor shunts the supply circuits (see schematic diagrams), the capacitance may be larger than the capacity of the original capacitor. <S> If the electrolytic capacitor is in frequency-setting circuits, it is desirable to install a new capacitor with the same capacitance <S> I think no other substitutions are needed. <S> P.P.S. One from my instruments after replace all of electrolytic capacitors - works fine
| The degradation of these capacitors can lead to a strong distortion of all modes of operation of the electronic circuit in direct current, which can cause a total inoperability.
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Calculating the ESR of Capacitor I try to calculate the ESR of a capacitor. I checked this datasheet: http://www.farnell.com/datasheets/2245214.pdf?_ga=2.167304311.817060034.1534770853-676261198.1531404202 I can see from the datasheet for the Aluminum Polymer Capacitors that they specify the max impedance ( Z ) and the tan( δ ) factors I would like to calculate the ESR for some specific capacitors ( for example 220uF) Can I just multiply these values in order to get the ESRmax value at 100kHz = tan(δ)* Z ? Is that correct? Thanks a lot for your help. <Q> This occurs at SRF (self resonant frequency). <S> The data sheet you provided only gives impedance at 100 KHz which is unlikely to be the SRF but is as close as you are going to get without measuring it. <S> 100 kHz is a common frequency used for specifying the impedance of electrolytic capacitors as they are commonly used for low frequency filtering. <A> In theory, you can find the ESR at a certain frequency by multiplying the dissipation factor at that frequency, also knows as tan(δ), by the capacitor's reactance at that frequency. <S> In practice, capacitors aren't completely linear. <S> You can make a decent guess at the ESR, which lets you engineer for the range the ESR might be, but you won't know the exact value. <S> See this graph from Murata for a visual example: To estimate the ESR for your capacitors, I would look up the tan(δ) value, finding 0.12 at 100 Hz, and I would calculate the reactance at 100 Hz, ignoring parasitic inductance and ignoring the fact that it's nonlinear. <S> Sticking that into Wolfram Alpha gives me an ESR of about 0.87 Ω: <S> As an aside, in case you're confused as to why datasheets list dissipation factor under "tan(δ)", the graph below, which I found here , illustrates why the dissipation factor is equal to tan(δ): <A> I computed Xc(f) after your component specs. <S> \$Zc(f)= R(f) - jXc(f) <S> ~~~~,~~Xc(f)= <S> 1/\omega <S> C\$ <S> I use R(f) <S> since ESR is frequency sensitive and is mainly a percentage of dielectric impedance well below SRF and called "dielectric loss" then above SRF, it is the loss of the inductive electrodes with skin effects. <S> The lowest impedance is at the series resonant frequency and the shape of the "notch" like shape is due to the Q factor which is somewhat inverse to tan δ. <S> The important thing to learn is that DF and tan(δ) <S> does not predict the ESR performance at 100kHz. <S> So if you see caps with only DF or tan(δ) <S> then you cannot guess the ESR at high frequency and most likely, not a low ESR type. <S> ** <S> Volt <S> uF <S> tan(δ) <S> ESR (100kHz) <S> Xc(100Hz) Xc(100kHz) <S> Zc <S> 16V 220μF <S> 0.12 <S> 0.022 723 0.723 <S> 0.723 16V 270μF <S> 0.15 <S> 0.012 <S> 589 <S> 0.589 <S> 0.590 <S> If you compare aspect ratios, you will find e-caps have lower ESR with taller cylindrical shapes and higher voltage ratings when volume and C(uF) in the same type.
| ESR of a capacitor occurs when the impedance of its capacitance and inductance are equal and cancel out leaving only resistance.
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Electrical lug for button switch So I bought a few push buttons that came with lugs but I'm not sure what tool I am supposed to use crimp them to wire. Advice is appreciated. <Q> Something like this. <S> The loop in the wire keeps it from pulling out. <S> You can even fold the wire more than once. <S> Use the end of a screwdriver blade to fold the ears over the wire. <S> Crimp down using pliers. <S> Note: solid wire will not withstand continued stress from being moved repeatedly. <A> Crimping tools like this one are used for this. <S> They are usually quite expensive, though. <S> The exact model of crimping tool depends on the dimensions of your contacts. <S> Note that most contacts like the one on the photo are designed for stranded wire. <S> If this is one-off project and you want to use solid wire you can try carefully soldering it taking care the solder does not flow into contacts. <S> Then use pliers to squeeze larger "ears" around the insulation. <S> This would be sub-optimal solution, though. <A> - this is OK (but far from optimal) with stranded wire, but I wouldn't do it with solid wire unless I also soldered the joint.
| Unless you expect to "unplug" the switch from the wires frequently, I would just solder to the switch terminals. I have occasionally crimped similar terminals with small-nosed pliers, bending in the sides one at a time
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Using shielded audio cable for potentiometer sensor I'm trying to measure a potentiometer from an Arduino's ADC. The potentiometer has a maximum value of 100k ohms and is located about 1 meter from the Arduino. The potentiometer's wiper is wired to the ADC pin and the pin to ground. I've enabled the 20k internal pullup on the ADC pin, so the potentiometer creates a logical high and low at the extremes. I have the potentiometer wired using two stranded 24 awg unshielded wires, twisted together. To filter out noise, I'm using an exponential moving average on the Arduino. However, even with this, I'm still seeing a +/- 10 unit change several times a second, even when the potentiometer is completely still. The two ways I've read about reducing noise in wires carrying analog signals are: Using a shielded cable Would it make any difference if I instead used a two-conductor shielded audio cable? I have one audio cable that has a red and black wire, as well as metal shielding. If I attach my potentiometer's pin to the two wires, should I also wire the shielding to the ground or leave it unattached? I see a lot of conflicting opinions over how to leave the shielding, with commenters in this thread suggesting to leave it ungrounded, or grounded at the sensor end, or grounded at the Arduino end, but definitely not grounded at both ends. Which is best for my case? Using a capacitor How much of a difference would it make if I placed a 0.1uF capacitor across the wires, as a low pass filter? Should I place it on the sensor side or the Arduino side? Again, I've seen a lot of conflicting advice about this. <Q> Why don't you connect potentiometer as people normally do, between GND and VCC ? <S> If you use 5-10K potentiometer you'll see much less noise on the input. <S> The reason you see so much noise is that your source impedance is too high, you simply don't allow enough current through to charge sampling capacitor in ADC. <S> You cannot use less then 22k if you use internal pull-up, while ADC in MCUs usually optimized for about 10k source impedance. <S> There are three methods we use on electrical scooters with about the same distance to multiple resistive sensors. <S> add 0.1 uF capacitor between the wiper wire and GND right before MCU pin. <S> The capacitor performs two functions. <S> First, it filters out high frequency noise. <S> Second, it further reduces source impedance by acting as local low-impedance source during sampling. <S> add ferrite core to high current switching power lines. <S> Again, this might not be the case for you. <S> In short: 1) use 5-10K potentiometer 2) connect it between VCC and GND 3) disable internal pull-up 4) <S> add 0.1 uF capacitor near ADC pin <A> Your question does not define your PSU conducted or radiated noise or method of grounding. <S> Thus radiated noise, conducted ground noise and common mode local noise from SMPS is unknown. <S> To isolate your problem you can place a cap directly across the source output and measure, then move to receive input and measure. <S> Zc(0.1uF)=30kOhm @50Hz <S> If the noise problem is fixed, with 0.1uF <S> then you know the noise is not 50 Hz but much higher than this. <S> If not fixed then you may suspect 50 Hz and try 100uF or more. <S> If this makes an improvement, then you can suspect a common mode noise creating a differential noise signal due to the layout of cables and supply. <S> Then consider a 0.1uF cap from Earth ground to 0V DC on the Arduino. <S> retest and compare results. <S> Earth shielded twisted pairs may be a good solution as well. <S> By grabbing both DC power cable and signal cable and check noise results you can infer that if it increases, that CM noise coupling is an issue with unbalanced input impedance for the ADC, which causes CM stray noise current to be converted into a differential voltage. <S> Cable proximity and 0V ref. <S> and decoupling may also be improved. <A> Yes, to 1. <S> A good piece of coax will almost certainly solve your noise issues completely. <S> No cap necessary. <S> It can be either foil or copper wrap <S> but if the cable is moved at all or even if there's a small vibration, a foil shield can generate noise so a wrap shield is better. <S> If the pot terminals are 1 for CW, 2 for wiper and 3 for CCW - connect the coax signal wire to both 1 and 2 and then the shield to 3. <S> Try to make the shield get as close to pin 2 as possible. <S> At the ADC end, obviously the conductor goes to the ADC pin and shield to ground of uC. Use internal pull as stated. <S> If that doesn't resolve the issue to your statisfaction, you can improve things further by using a 2 conductor w/ shield cable (like the kind used in wiring audio patchbays like Redco TGS-02 <S> with the jacket removed) and use one conductor for pot terminal 1, the other for the wiper and pin 3 to the shield. <S> Then supply the logic high voltage to terminal 1 through a 100R resistor and use a 10k pot. <S> Dot NOT use internal pullup in this case. <S> This will reduce the impedance of the whole thing drastically and reduce noise accordingly. <S> Neither of these scenarios leave the shield unconnected at either end. <S> You would only consider that if you had more circuitry at the far end that is connected to a different ground.
| In your case this is not an option, but you might see some noise reduction if you connect shield to GND. use shielded cable with shield connected to system ground near MCU. But it has to be a full-coverage shield.
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Low power switch input with MCU wake-from-sleep interrupt? Im trying to read the falling edge of digital logic input to wake up a microcontroller. Below is the circuit I have used. There is an RC filter to provide a ~5ms debounce time, and a 100K pullup so the circuit doesn't use lots of power while the switch is in its on-state. The switch may be in its on-state for roughly 10% of the circuit operation time. simulate this circuit – Schematic created using CircuitLab The problem with my circuit may seem obvious to some but I seem to have overlooked it; The input is pulled low quickly (discharges) just how i wanted it, but it is pulled high (charges) very very slowly due to the 100K pullup. I don't want the circuit to take a long time to charge up (ideally <20ms) but I also don't want it to draw lots of current when the switch is in its on-state. Are there any engineering tricks i can employ to achieve these two things together? I assume its a classic problem but I cannot find any solutions. <Q> Why are you using falling edge for interrupt at all? <S> However small, you still have some current permanently flowing through the pull-up, which defeats the purpose of sleep mode. <S> Plus the pull-up makes you RC filter lopsided. <S> Anyway, you have several options. <S> First, you can use SPDT switch which will guarantee same timing for charge and discharge. <S> See first schematics below (BTW, it works for either edge). <A> You could, for example, insert a resistor in parallel with R1 only during the charging process. <S> This would reduce the charging time. <S> See my simple idea sketched in the figure below. <S> In normal operation, when the switch is open, the P-channel MOSFET is cut off. <S> When the switch is closed and the voltage in the capacitor reaches a low value, the voltage \$Vgs\$ will be negative and if it becomes smaller than \$Vgs_{TH}\$ , it will cause the mosfet to conduct and insert the additional resistor in parallel with R1. <S> I leave you with the proper dimensioning of the parts and the addition any other. <S> EDIT: <S> After you answered: "... It could indeed be a high side switch ... <S> ", I suggest the modification below considering the low power application and the processor must wake up in occurrence of interrupts. <S> The advantages: <S> The upper resistor will not consume power when the switch is open (so, can use a lower value as 10k for example). <S> Using a Schottky diode (with a lower forward voltage) will bring a higher voltage to the MCU interrupt pin - a good thing when the battery drains or the supply voltage is 3V, for example. <S> Warning: <S> The resistive divider should be calculated taking into consideration the hysteresys levels of the GPIO pin for the specific MCU (to avoid false interruptions or to lose one). <S> Also, disable the internal pull-up resistor. <A> There is an RC filter to provide a ~5ms debounce time Get rid of it. <S> The MCU can debounce in software. <S> This allow you to lower C1 to something like 100nF , yielding a much shorter charging time. <S> Another optimization could be the internal Pullup/pulldown resistors inside the MCU - which can be turned off an back on after some time. <S> While this would delay a "button up" event (because you can only detect it when pullup is on), there would be no current flow in pullup OFF (pulldown on) state.
| If you want to keep using rising edge and SPST switch you can use two identical resistors, but shunt one of them with diode, resulting in almost the same rising and falling times (tweak one of the resistors to counter voltage drop on a diode, if you wish), and recalculate your RC for high resistance and much smaller capacitance, to keep quiescent current low for the same time constant, e.g. 50k resistors and 0.1 µF capacitor.
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Is everything OK with this circuit? Coming from this question, I decided to have some changes. So my case is I have an LED with this description: Dimensions (diameter x height): 20 x 8 mm Operating voltage: 3,4-3,6 V Rated current: 700 mA Power Dissipation: 3 W Lighting angle: 120° Luminous flux: 160 lm Color temperature: 6500-7000 K Mounting: Installed on an additional radiator Glow Color: White Then I have AA Li-ion 18650 3.7V, 3000 mAh rechargable battery, 21*15*10MM TO-220 aluminium radiator, TP4056 Li-ion battery charger and 2 A, 1 Ohm resistor. simulate this circuit – Schematic created using CircuitLab If I'll construct it this way, is everything is gonna be alright? I don't have every detail shipped yet and I'm out of time UPD : updated my circuit by adding 1S PCB Bms protection board . Will this flashlight work properly for couple of months? <Q> Is everything OK with this circuit? <S> The LED will light when the switch is closed. <S> For a lantern or flashlight it is not okay. <S> Mediocre at best. <S> White LEDs are available with forward voltages (V f ) below 3V. An A 18650 battery discharges to 3V (see red line in discharge graph). <S> With your LED if the V f is 3.5V you will have insufficient voltage when the battery is discharged to about 50%. <S> With a 1Ω resistor and full charge on the battery, the current will start at 700 mA (@4.2V) <S> then quickly drop 400 mA (@3.9V) and not too long after that drop to 100 mA (@3.6V). <S> This is why I mentioned in your previous post a battery powered lantern should a current regulator and not use a current limiting resistor. <S> If you use an LED with a max V f of 3V or less it will operate down to the end of the discharge curve. <S> Source: <S> Panasonic Li-ion <S> NCR18650F Datasheet <S> If you do not have SMT capabilities then use a Cree XP-G3 (2.7V-3V) <S> mounted to a star board . <A> You would need a boost converter in current mode. <S> This way you will get constant LED current, whatever is the battery voltage. <A> is everything is gonna be alright? <S> That depends (on what is "all right") <S> When the switch is closed, the LED will discharge the battery. <S> But when the battery is nearly empty, to what voltage will it be discharged? <S> Ideally the discharging should stop when Vbat < 3 V or so. <S> It depends on how you want to treat the battery, do you want a long lasting battery? <S> Then cut discharging below 3 V, if you don't care you can discharge more to 2 V. R1 will dissipate some power especially when the battery is full because then it will be 4.2 V, not 3.7 V! <S> Did you account for that in the dimensioning of R1? <S> 2 <S> A, 1 Ohm resistor for resistors <S> I want to see a value, 1 ohm, and a power rating like 1 W. <S> If the resistor has a current rating maybe you have a fusable resistor? <S> You could also just use a normal (SMD) fuse. <S> Note that the TP4056 needs not much more than 5 V to charge the battery. <S> At higher voltages power the TP4056 will get too hot while charging (it will lower the current to prevent overheating though). <S> Charging can take a long time (hours) as the TP4056 cannot charge with a very high current.
| Your LED "operating voltage" is too high for a Li-ion battery powered circuit.
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Modify my old (1978) car's existing dash light dimmer (10 ohm Rheostat) to control LED replacement globes Firstly my electronics knowledge is limited. I am trying to work out a way to use the existing Rheostat (10 ohm) that is used to dim the instrument lights in my 1978 car. I have changed the globes to 12V dimmable LEDs, but the voltage drop on the dimmer is not sufficient to dim the LEDs. I have tested the LEDs and they are turned off at about 4.5V and are almost at full brightness at 10V. I could use a PWM device with a different trimpot, but I would like to use the existing dimmer switch. I thought maybe an Op amp with a 800mA transistor on the output might work, but I am struggling with that concept. Any ideas or available modification circuitry available? Thanks. <Q> The 10 ohm Rheostat will be in series with the light, that would work with light bulbs. <S> Suppose the light bulbs are 10 Ohms in total (that's 10 ohms for all of them in parallel) <S> then with Rheostat = 0 ohm you get full power and with Rheostat = <S> 10 ohms you would get about 1/4 of the full power. <S> (1/4th since the voltage is halved but then the current halves as well). <S> The LEDs consume much less power <S> so less current flows so less voltage is dropped. <S> Suppose the LED us 1/10th of the original light bulb <S> current <S> then the voltage drop over the Rheostat is also 1/10th. <S> So 6 V (12 V / 2 for bulbs) becomes 0.6 V. <S> That's useless for dimming ! <S> So if we could increase the current back to it's old value then dimming would be much better. <S> Those give light and might break. <S> Instead of bulbs you could use resistors, these behave similarly to bulbs but give no light and should not blow either (assuming you're using them correctly). <S> So what value resistor should you use? <S> Suppose there used to be 4 light bulbs of 2 W each that is 8 Watt in total <S> 8 Watt / 12 <S> V = <S> 0.67 <S> A then an equivalent resistor (to "fake" <S> the load of the 4 2 W bulbs) is 12 V / 0.67 <S> A = 18 ohms <S> Since the power is <S> 8 w you need a 10 W resistor of 18 ohms. <S> If you do not have one of those lying around, try a 12 V 8 W (or 10 W) bulb, connect it in parallel with the LEDs and see how that works. <S> Note that this isn't a energy efficient solution ! <S> Normally we use LEDs to save power. <S> My proposal is an "easy fix", it is not intended as an energy efficient solution. <S> Then a car isn't energy efficient anyway. <S> Also the 8 W for dashboard illumination pales to the 2 x 50 W of your car's headlights ( <S> assuming they're still bulbs). <A> Here's a circuit that should work <S> but you may need the help of someone with a little electronics experience to put it together. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> A variable current source. <S> How it works: <S> Q1 is turned on by R4 drawing current from its base. <S> Current now flows through Rmin + R1 (your dimmer), Q1 and out to the LED lamps. <S> As the current increases through Rmin + R1 <S> the voltage across them increases. <S> This voltage is applied to Q2's base. <S> When Q2's base voltage drops about 0.7 V below its emitter voltage it starts to turn on, reducing the emitter-base voltage on Q1 and preventing Q1 turning on any further. <S> The circuit settles with about 0.7 V across R1. <S> You mention 800 mA in your question <S> so let's say we want to adjust between 80 and 800 <S> mA using as much of the range of the dimmer as we can. <S> At 10 Ω we will get 700 mV at \$ <S> I = \frac { <S> V}{R} = \frac {0.7}{10} <S> = 70 \ \text {mA} \$ which is very close to what we want. <S> To get 800 mA we need a resistance of \$ R = {V}{I} = \frac {0.7}{0.8} = 0.875 <S> \ <S> \Omega \$. <S> So setting Rmin to something between 0.5 to 1 Ω should work. <S> If you want to be picky then recalculate the 10 Ω with 10 + <S> Rmin <S> but I'd be surprised if you could see the difference. <S> A couple of notes: Calculate the power rating of Rmin from \$ P <S> = I^2R \$ and buy one with twice the rating so that it runs cool. <S> Q1 should be rated at twice the max current you expect to draw. <S> Q1 should be mounted on an isolated heatsink. <A> You could use a cheap microcontroller (e.g. ATTiny85 or similar) to measure the voltage output of the existing circuit using its analog to digital converter (using a voltage divider to reduce the input voltage to a range the microcontroller will tolerate) and use that to set the duty cycle of a PWM (which the microcontroller also has). <A> Well this is what you asked for: an opamp circuit. <S> R7 is setting what the voltage will be when R1 is a max R.R4/6 set the gain U1 is acting as a voltage follower somewhat like using an NPN transistor here. <S> I use an LM317 as it has current limiting and thermal shutdown, and is kind of an easy way to get these. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is a constant current circuit. <S> It is a bit inflexible since R1 has a fixed value and Vref is fixed, so you don't have too much range to adjust. <S> However it might be in the correct range for you. <S> Note that this is really the same circuit as Transistors , but using LM317 with a 1.2V reference instead of Q2 with a 0,6V reference. <S> So current will be 2x. <S> Again, using lm317 gives thermal protection. <S> If you add D1, then current will approximately halve. <S> simulate this circuit <S> You can run the simulator ("DC solve") to look at the voltages, and see what changing R does. <S> You need to adjust R4 and R6 to set the dim and bright points. <S> When R7=R1, Vout is ~250mV. <S> Op amp output should be 1.2V below VOUT. <S> Opamp outputs can't go to V+, nor can LM317, <S> so max Vout will be ~10V.
| What you could do is connect light bulbs in parallel with the LEDs. Then just use a transistor or mosfet to drive your LEDs from the PWM output. That depends on how much current was flowing originally, we do not want to exceed that current as that will damage the Rheostat.
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Fiber-optic transceiver ICs for applications other than fiber-optic Ethernet? When we talk about fiber-optic transmission, we almost always talk about copper to fiber-optic Ethernet conversion using fiber-optic transceivers. But since fiber-optic communication is just transmission of any electrical signal in the form of light, I was wondering if there are other transceivers available for fiber-optic transmission, I mean for interfaces/protocols other than Ethernet. In my search I could not come across any such transceivers. Whenever and wherever I see the word fiber-optic, it is related to Ethernet/networking only. Any comments are appreciated. <Q> SONET is kind of popular in the telecomms world, very different to ethernet, and kind of cool in its way. <S> In reality many non ethernet uses exist, but we tend to try to use rates that are close enough to something used by either the phone company or the datacenter because economies of scale make optical modules for those line rates all kinds of cheap. <S> MADI for example <S> (multichannel digital audio used in pro environments) often goes over 155Mb/s SFP modules marketed for SONET use by the telecom crowd, IIRC it actually runs the line at 125Mb/s, but close enough. <S> SDI digital video is another example, either at 1.55Gb/s, 3Gb/s or 12Gb/s, again SFP modules that are close enough usually exist. <S> Particularly with SFP interfaces they usually don't care about the protocol as long as DC balance is not too horrible and whatever you are sending can survive being highpassed at a few MHz <S> (The ALC will stuff up the levels at the the slicer if you go slower then that). <S> Apart from that, generate a PECL version of whatever you like and stuff it into a SFP+ socket, odds are it will work if you pick a reasonable module for the speed you need. <A> As already mentioned there are many possible communication transport layers possible over fiber optic links. <S> Here is one useful application that uses a fiber optic link to extend a USB interface out to far beyond what is possible with standard USB cabling. <S> (Source Vendor Page) <S> This example shows a USB3.0 link of 50 feet <S> but there are many product options in this space. <S> If you search you can find USB2, HDMI, Ethernet, Audio etc. <A> How about S/P DIF , Fiber Channel , Thunderbolt optical cables , to name a few ? <A> I am surprised no one mentioned good ol' TOSLINK. <S> The simple board-mounted transmitters and receivers for it are available from digikey and the cables are cheap. <S> TOSLINK technology has evolved over the years. <S> There are general purpose devices for up to 8Mbps and full-duplex SMI transceivers.
| There are many other link-layers transported on Fiber-optic medium, apart from Ethernet.
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How does the AXI-Interconnect know where to route the data? Im interested in where exactly the Addresses (BASE_ADDR) set in the "Address Editor" of a Vivado Block Design come into play in the FPGA-Part. I have multiple Blocks with AXI-Lite connected to a Zynq via an AXI-Interconnect, which works fine. But i want to know how: To my current understanding, the routing must be somewhere in the AXI-Interconnect-IP-Block, however i can not see how it knows where to route the bus as "Customize IP" shows nothing related to the Addresses.It just seems to magically connect everything correctly... So how does it work? Edit: For clarification - this is how the part of my system looks like: ______________ | |---------- Block 1 (AXI-Lite) | AXI |---------- Block 2 (AXI-Lite)Zynq (AXI Bus) -------| Interconnect |---------- Block 3 (AXI-Lite) | |---------- Block 4 (AXI-Lite) |______________|---------- Block 5 (AXI-Lite) The Zynq itself is not relevant in this picture, there is just a full AXI-Bus on the left and multiple AXI-Lite-Blocks on the right, which can have Address-Ranges set in the "Address Editor" of Vivado. Further clarification: I know how the lower bits of the address-lines are used to address single registers in AXI-Lite, but the Xilinx-Template doesn't seem to care about the higher bits, therefore my assumption that the addressing has to be done outside of the AXI-Lite block. <Q> The bus is physically routed to every block that needs access to it. <S> Each block is responsible for decoding its own addresses, which is why each one has a "base address" setting. <S> It's up to you to make sure that the address ranges don't overlap. <A> Your diagram helps. <S> The AXI infrastructure is a 'star' network, not a 'bus'. <S> The reason is that AXI uses a handshake protocol ( valid and ready ) for each transfer. <S> These are paired with every valid having one ready. <S> (You can't give out many valids and wait for any ready to come back). <S> Thus if you have one AXI bus going of to many destination you need a to make a dedicated AXI stream with it's own valid and ready. <S> Thus is done inside the interconnect. <S> The address is used to decide to which output port the transaction should go. <S> Therefore the address decoder you are looking for is inside the AXI interconnect. <A> The address decoder is inside the interconnect. <S> Maybe they are not exposed at the top level as I think the top level file is generated, but they are in the heirarchy somewhere.
| There are parameters at some level inside the IP block that set the addresses.
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How to crimp a ferrules(end sleeve)without a crimper? Hi. I know that the best way and the safe method to crimp those uninsulated ferrules is to use the appropriate crimper.But,since i need to crimp only 3-4 wires,buying a special crimper only for that is not economic.Which other method can i use in order that the crimp of those uninsulated end sleeve will be good enough? Thanks. <Q> It will be better than trying to use pliers, which doesn't really make a crimp because it squashes them flat <S> and they can easily open up from that. <S> Flattened ferrules are also very wide and don't fit in screw terminals that would otherwise hold that gauge of wire. <A> If you are using these ferrules for a cable that needs to be robust and last, I'd go for the crimper or borrow one. <S> The crimpers have exact dimensions to crimp the cable properly. <S> That being said, if you only have a few crimps, maybe a terminal crimper might be a close second. <S> I have bought decent crimpers on ebay for a fraction of the cost of the new one, and I don't need a new one. <A> You might consider an ordinary crimping tool. <S> Most everybody has one of those.
| Based on experience trying myself: Just get the cheapest ferrule crimp tool you can find.
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Eliminate noise from IR sensor I have a 5v infrared beam reflection circuit for capturing heart beat from fingertip. I am using LM324 low pass filter to filter out the noise and i have isolated the ir sensors by covering the sides with a black tape but the noise is not reducing.My output is a blinking LED and the LED is not supposed to light until a finger is placed on the ir pair but the LED lights up whenever i power the circuit.Below is my schematic <Q> The ultimate problem/mistake here which makes noise such as issue is the use of an always-on IR LED as the source. <S> If you want to build something like this which actually works , then you should modulate the LED at a frequency many times faster than a hearbeat, and use a demodulating detection circuit which measures only the desired signal. <S> The most robust form of this would be a synchronous detector or Lock-In Amplifier . <S> Essentially what you do is have your MCU turn on the IR LED, measure the received level, then turn off the IR LED, measure the received level again, and subtract. <S> Repeat <S> this many times and filter (or at least average) the difference to produce your output, though be sure not to average so much that you filter out the desired heartbeat. <S> Since there's no meaningful delay to want to measure here <S> and there are unlikely to be in-band interference sources, it's also possible to use a band-pass filter to pass only a stable IR LED modulating frequency, amplify that, and then measure the amplitude - with care to circuit design this may give greater dynamic range. <S> Typically IR remote controls work this way (often with a frequency around 38 KHz), but the output is interpreted as simply on/off, while IR distance sensors may give an analog output which varies with the strength of the synchronous reflection. <S> To detect the presence of a finger (vs some inanimate near reflection), you may have to first measure a heartbeat frequency and then determine if it is stable enough to be a biological source. <S> Or you may more simply be able to build something that decides the modulation-frequency reflection is "strong enough" to probably <S> indicate that a finger (or at least something ) is close, and turn on the crude presence indicator LED. <A> Your noise density comes from the 68Kohm resistor PLUS opamp internal noise (which will vary from brand to brand, even if the part number is the same) PLUS Power-supply noise PLUS magnetic and electric fields PLUS photonic noise. <S> Assume your bandwidth is 20 radians per second (set by 470K and 0.1uF), or 3Hz. <S> Perhaps a bit low for best heart-beat amplification. <S> The noise density of 1Kohm is 4 nanoVoltsrms/square-root-of-one-hertz, <S> thus 2Hz provides 1.414 * 4nV, <S> 9Hz provides 3 * 4 = 12 <S> nanovoltsRMS total over that bandwidth. <S> The noise density of 100Kohm is 40 nanoVoltsrms/rootHz. <S> There is a factor of pi/2 for noise integrated DC---infinity <S> , even tho 3Hz is your "bandwidth". <S> Computation of total input-referred noise: 40nanoVolts <S> * sqrt(3) <S> * pi/2 = <S> 40nVrms <S> * 1.7 <S> * 1.6 <S> ~~~ <S> 40 <S> * 3 = 120 nanoVoltsrms, at input. <S> Scale this up by 2,500, to 300 microVolts output. <S> The low frequency noise of opamps, caused by silicon traps emptying and refilling, can easily boost the noise (and 3Hz is low frequency) by 100X if the corner frequency is 10,000Hz for a cheaply-made lotta-surface-contamination foundry. <S> This noise has you at 30 milliVolts output. <S> But you also could have oscillation. <S> Show photos of the "noise". <A> You need to consider the different sources of noise in your circuit. <S> Most of them come in the form of Johnson-Nyquist noise, aka White Noise, across your resistors, its generally modeled as a voltage source in series with a resistor. <S> it's equal to $$V_{Noise} = <S> \sqrt{4K_{b}TRW}$$ <S> T - temperature (Kelvin) R - resistor (Ohms) W - frequency (Hz) <S> Its due to the current fluctuations as they jump across a junction. <S> If you think of a PN junction, aka diode, as the depletion layer between them get smaller and smaller electrons jump the depletion layer creating noise in its variance. <S> $$V_{Flicker} = <S> \sqrt{2qI_{DC}W}$$ q - charge of electron (Coulombs) Idc - DC current (amps) <S> W - Frequency in Hz <S> Many (if not all?) <S> OpAmp spec sheets have a line breaking down their noise source. <S> Another thought could be, do you need a different OpAmp configuration? <S> You might benefit from having two IR sensors and using an Instrumentation Amplifier to look at the difference between them to generate your pulse rate.
| Another major noise source is "Shot Noise" as is caused by nearly all semiconductors. Normally the goal of something like this is to actually measure the heartbeat frequency (or even measure blood properties). Or you can do this in the circuitry rather than software domain. You've got a gain of 2,500X.
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PWM dimmer for off-the-shelf 12v LED bulbs I am building a PWM dimmer for loads of up to 10 off-the-shelf 12v LED bulbs. Here is my current circuit. I'll describe a fundamental problem and I ask if I'm missing something here. The driver (everything outside the dotted box) contains a basic astable 555 configuration for generating a square-wave of adjustable duty cycle.** . The right-most component is a potentiometer for adjusting the duty cycle; this one is voltage-controlled for simulation convenience, but in real life it will be a digital pot chip. The square-wave (the 555's OUT pin) gates a power nMOSFET (a Darlington NPN pair would work too). The bulb (dotted box) contains a full-wave rectifier with low-pass filtering (for use with 12vac), and a simple LM317 current source (yet these bulbs are $2 each). The driver needs to be able to work with up to 10 bulbs, connected in parallel, though I've only simulated one here (multiple bulbs will not affect the simulation significantly, I believe, as long as the FET can handle the current). The problem is that if the pot is too large (say 100K) then the frequency of the PWM waveform is low enough (about 100Hz) that flickering becomes evident under some conditons, but if it's too small (say 10K) then the frequency is high enough (about 1KHz) that the bulb completely filters it out and almost no dimming occurs (see waveforms). The too-fast case is actually a little more complicated: when OUT is high the bulb turns on quickly, but when OUT goes low, the bulb's capacitor has to discharge via the 57ma current through the LEDs, and recH/recL drops low enough for the current source to collapse little, if any. The only solution is to simply make the best compromise between flickering and dimmability (a 200Hz frequency might be the ticket, but apparently disturbing visual effect can still occur with saccades). Or to use LED bulbs designed for DC operation, which probably don't have such a honking cap. PLEASE NOTE: I am well aware that better way to dim LEDs is with an adjustable constant-current driver. But I need for this driver to work with THIS bulb (in as much as it's typical of off-the-shelf 12v LED bulbs). ** One finds a variety of 555-based PWM generators on the web. They config the duty-cycle setting resistors/pots in different ways among pins 2/6, 3, and 7 (and some don't even use pin 3); and some even feed the charge/discharge ramp into an opamp comparator and adjust the comparison voltage. Possible advantages of these other circuits are unclear to me, and this one seems simplest. In any event, none of the alternatives is likely to help with the current problem. <Q> if you add an inductor in series with each lamp assembly that will stop the inrush current from immediately charging the capacitor and make them respond to PWM control <S> you could alternatively use resistors instead at a lower efficiency simulate this circuit – <S> Schematic created using CircuitLab <S> This means you'll need to modify each light socket. <A> That capacitor left of the LM317 C3 is indeed the problem. <A> Consider a simple PWM to disable a current limiter based on NPN Vbe driving an Nch MOSFET <S> gives reasonable low dropout of < 0.6V <S> is all you need. <S> You may use any PWM method you prefer.
| Variable current drive will work with these lamps, when the current goes below the limit set using the LM317, the LM317 becomes like a constant voltage drop of about 1.2V, but the current setting setting where this takes effect will depend on the design current of the lamp. The PWM should not be driving a capacitive load.
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What keeps the kickback voltage from reaching an infinite voltage? We know that the voltage over an inductor is defined by the formula: \$V = L * \frac {di}{dt} \$ So in the case where the current flow is suddenly interrupted (like when a mechanical contact is opened), voltage spikes occur in real life. However, this is not always the case: we don't see arcs happen in small inductive loads. (By small inductive loads I mean a toy car motor, for example.) However, the formula says that the \$ \frac{di}{dt} \$ term should approach infinity when mechanical contacts are opened, therefore the \$L\$ term (which should be small in small inductive loads) shouldn't have a significant effect. Simply, we should be able to see sparks any time we open any inductive load - independent of the inductance. What are the practical factors that stop the voltage from reaching infinity? Does the current flow actually decrease slower, or is the formula perhaps insufficient for such a "discontinuity"? <Q> A real inductor looks like this (shown below is an inductor with 4 coils) there is a small amount (usually in the pF-fF range) of capacitance between each coil. <S> Each piece of wire also has some resistance associate with it. <S> Because each coil in an inductor has resistance (or each section of wire if you consider one coil) <S> this impedes the current and reduces the voltage. <S> These all soak up energy that prevents the Electro Motive Force (EMF) that has been stored around an inductor from generating an infinite voltage. <S> An inductor can actually be simplified into a circuit such as the one to the left below. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> A superconducting coil would be able to generate much more massive voltages because of much lower losses due to parasitics. <A> Any energy storage system (an inductor) has non-zero size. <S> Anything of non-zero size has non-zero electric fields, or capacitance. <S> Device junctions are usually a large source of parasitic capacitance. <S> Flyback systems use a diode to transfer energy into a load capacitor. <S> At peak voltage excursion, all the inductive energy has(1) been dissipated as heat(2) been radiated as EM field(3) been stored in the electric-field of the intentional and the parasitic capacitances. <A> The series resistance matters a great deal with the "kickback" voltage due to the series capacitance of the "switch" when opened. <S> This forms a classic series RLC resonant circuit which has properties of voltage gain by impedance ratio of \$Q=\dfrac{|X_C|}{R} = <S> \dfrac{|X_L|}{R}=\dfrac{\omega <S> _0 <S> L}{R}\$ at resonant frequency <S> \$\omega _0= \dfrac{1}{\sqrt{LC}}\$ <S> For the situation of kickback voltage peak, it can be proven that \$|V_p| = Q <S> * V_{dc}\$ for Quality Factor, Q (above) and loop supply voltage Vdc at some resonant frequency. <S> When de-energizing a circuit with a contact switch as t goes to 0, V/L=dI/dt, <S> V does not go to infinity due to this parasitic capacitance. <S> Example simulate this circuit – <S> Schematic created using CircuitLab <S> e.g. Consider a series circuit, Vdc=1V, L=1uH, <S> R=1 Ohms , Idc= 1A . <S> What is the switch voltage kickback, when just opened, if Csw = 1pF ? <S> 1V , 100V, 1kV, 1e6 V or infinite? <S> Now consider the same for a FET switch with 1nF output capacitance with RdsOn << 1% of R=1. <S> What is <S> dV? <S> p.s. <S> if you learnt something, then comment your answer. <S> The intuitive answer is that the switch goes from a conductor to a tiny stray capacitor which limits the slew rate of the voltage and as does the inductor limit the slew rate of the current and at their resonant frequency the voltage gain, Q at ω0 is inversely proportional to R, so bigger series R dampens the voltage. <S> Answer \$ <S> V_p= I_{dc} <S> \sqrt{\dfrac{L}{C}} \$ <S> = 1A <S> * √(1uH/1pF)= 1kV <S> Misc <S> It can be proven the open circuit impedance like a transmission line "characteristic impedance" \$ <S> Zo= \sqrt{\dfrac{L}{C}} \$ <S> We see the voltage kickback looks like Ohm's Law. <S> \$ <S> V_p = <S> I_{dc}*Z_0\$ The peak voltage Vp, generated from interrupting an inductive current, \$I_{dc}\$. <A> Just consider a simple example of 100 uH and 1 amp flowing. <S> When the contact in series with the inductor opens, there may be 5 pF of parasitic capacitance left across the inductor and that 1 amp will create a high kick-back voltage but how much? <S> $$I = C\dfrac{dV}{dt}$$ <S> So potentially (no pun intended) <S> the voltage across the 5 pF capacitor could rise at a rate of 200 kV/microsecond. <S> Given that its starting voltage is potentially neglible in comparison, within a few micro seconds a pretty big voltage could develop. <S> However this is mitigated by the lack of energy stored in the inductor: - $$W = <S> \dfrac{L\cdot <S> I^2}{2}$$ Or 5 micro joules. <S> All this energy will cyclically transfer to the capacitor and we can equate the capacitor energy formula to 5 uJ to give us the maximum voltage: - $$W = <S> \dfrac{C\cdot <S> V^2}{2}$$ <S> This produces a peak capacitor voltage of 1414 volts.
| The small amount of capacitance will also store some of the voltage and prevent an instantaneous change in voltage.
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Why can we hear the full range of square waves from 1 Hz-20 kHz, but not sine waves of the same frequencies? I connected the output (50Ω) of a function generator to a 35 watt 5Ω speaker. When I sent square wave signals of 1 Hz-20 kHz to the speaker, my friends and I were able to hear them. But when I sent sine waves across the same range, no one was able to hear anything below 100 Hz. <Q> Most people can't hear sinewaves below about 20 Hz. <S> The fact that you couldn't hear them until they got to 100 Hz is probably due to poor low-frequency response in your amplifier/speaker. <S> When you use a square waveform, you are adding a rich series of harmonics to the fundamental frequency — these are what you're hearing, and your brain extrapolates the fundamental frequency for you. <S> Down to a point, that is — at some point you just start hearing individual clicks. <A> The basic difference is that a sine wave is only a single frequency, but a square wave is actually made up of the fundamental frequency plus odd harmonics. <S> So for example a 50Hz square wave isn't just 50Hz, but also 150Hz (3 x 50HZ) <S> , 250Hz (5 x 50Hz), 350Hz (7 x 50Hz) <S> etc. <S> Amplitude decreases as the harmonic number increases, but may still be significant out to the 10th harmonic and beyond. <S> The other factor involved is the response of the human ear. <S> The ear is less sensitive to lower frequencies, and this gets worse as the sound level decreases. <S> Below the threshold level you won't hear any sound. <S> At ~25 <S> dB SPL you may be able to hear the harmonics of a square wave whose fundamental frequency is below 100Hz, when a sine wave of the same frequency and amplitude is inaudible. <S> I connected output (50Ω) of function generator to 35 watt 5Ω speaker. <S> If you connected the generator directly to the speaker (without an amplifier in between) then the sound output will be very weak because:- <S> a) <S> function generators are designed to produce a low power signal (typically 0dBm or 1mW) not drive a high power speaker. <S> b) there is a large mismatch between 50Ω and 5Ω, which greatly reduces the amount of power transferred. <S> c) <S> the speaker also has a frequency response which drops off below its resonant frequency, and its impedance varies with frequency <S> so when driven by a high impedance the frequency response is less flat. <S> All this means is that to get a good sound level you need to put an audio power amplifier between the generator and speaker. <A> All non-sine wave signals can be broken down into a collection of sine waves interfering with one another. <S> This is done using a Fourier Transform. <S> Each of the component sine waves are different frquencies, and when they come together in non-ideal scenarios, such as a real finction generator, they dont quite create a perfect imitation of the signal they are trying to create. <S> So the square wave you're sending through the speaker is not a perfect square wave. <S> Thus you are probably hearing mostly noise and some of the component fequencies as well. <A> Perhaps you've connected a bare loudspeaker with no baffle (box) around it to your signal source? <S> At low frequencies, air pushed or pulled out the speaker's front simply whips around to its back, and is poorly radiated to your ear. <S> At a high-enough frequency, this process is foiled by speed-of-sound and you hear a louder sound. <S> A speaker mounted in a baffle-board or stiff wall increases the front-to-back path length, which extends the low-frequency audible response. <S> And as others have mentioned, a pure single-frequency sine wave ensures that what you hear is a pure tone at the frequency displayed by your function generator. <S> A square wave includes 3X, 5X, 7X, 9X frequencies in addition to the 1X frequency displayed by your function generator. <S> The square wave harmonics (those other than 1X) are more clearly audible than the 1X frequency, for function generator frequencies below about 100 Hz.
| With a square wave, the human ear's increased sensitivity at higher frequency more than makes up for the reduced amplitude of the harmonics.
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Should I wear a grounded ESD strap when probing/debugging my raw boards? Manufacturing and handling the unpowered boards, the assembly workers often use strict ESD control: dissipative mats, smocks, and grounded wrist bands. As an embedded engineer, I am always powering and probing up my raw boards (without the plastics enclosures) at my cubicle... JTAG debugging them without any sort of ESD safety. What is the recommended guidance on this? Should I be worried about ESD? Do I need to be wearing any ESD gloves/strap? note: the humidity in my office is around 40-50% <Q> My entire office is carpeted, and we constantly debug/test raw PCBAs at our desks. <S> I take care to ground myself on metal objects around me, and I have an ESD mat on my desk. <S> When I have to transport boards, I put them in ESD bags. <S> I also take care to ground myself after getting up from my chair, or upon sitting down. <S> When passing off boards to others, we always do the "EE's handshake", where we touch hands before exchanging the PCBAs. <S> This has worked well for us, and I have yet to see a PCBA damaged due to unintentional ESD. <A> Technically, yes. <S> However, in my experience most boards seem to be much less ESD sensitive after they are assembled. <S> Some logic families can still be very sensitive; we used to work with FCT (Fast CMOS Technology) logic from IDT and that definitely needed wrist straps all the time. <A> It is generally a good idea to wear a wrist strap, keep in mind there should be 1MΩ of resistance between the strap and whatever ground your connecting it to, which reduces the chance of you becoming part of a circuit in the event of a short and possible injury. <S> I have seen ESD discharges ruin flash on micros before and the board needed to be reflashed. <S> Most ports are protected these days with ESD diodes\diodes, but intermittency caused by ESD can cost you hours of debugging to find the problem is the hardware and not the software.
| Any form of ESD mitigation is recommended, in other words, wear the wrist strap.
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Should I add bypass capacitor for every IC and Microcontroller? I am expecting a noise from a Power Source caused by motors . However my setup already includes a buck-converter , and have the input and output capacitors. Should I add bypass capacitors for the powers on the IC's and Microcontroller like the Particle Photon. What will be its effect since it would add up capacitance values since they are all parallel. <Q> The "Particle Photon" device already has bypass capacitors on it. <S> In general, for any design you would want to decouple every IC. <S> At low frequencies you are correct, the capacitance adds up. <S> But at higher frequencies, like harmonics generated from a buck converter, the bypass capacitors will act in a distributed way. <S> An IC that has no bypass capacitor near it will have a higher (read worse) impedance to ground at higher frequencies. <S> The "capacitor on every IC" rule of thumb makes it so that each individual IC sees their own nearby bypass capacitor first thing, so there will be a lower impedance to ground and better power-integrity for higher frequencies. <S> tl;dr: <S> yes. <A> Integrated Circuits will have 2--10nanoHenry inductance in each pin. <S> Onchip capacitance from well-to-substrate reversed-bias junction is where the charge is supplied during those 50 picosecond logic-level transitions. <S> If you have 100pF onchip and 10nH pin inductance, that LC will ring with 150MHz frequency. <S> That is 6 nanosecond sinusoid period, and 3nS to the first energy rebound zero crossing. <S> Thus until 3nS (3,000 picoSeconds) passes, only the onchip silicon capacitances can provide charge. <S> Having a discrete Surface Mount cap right at the pin is the fastest way to re-supply the just-consumed charge. <A> Just to add to this question and @user55924 answer, because you mentioned parallel capacitance. <S> A capacitor can be modelled as follows: where Rp is the lumped parasitic resistance and Lp is the lumped parasitic inductance. <S> every capacitor has a special frequency point—a natural frequency: the impedance of the circuit is dominated by the capacitor at low frequencies, and is dominated by the inductor at high frequencies. <S> The self-resonance frequency of a capacitor determines the frequency at which the capacitor switches from being mostly capacitive to mostly inductive effects. <S> It is important however to be vigilant of what capacitance values you are using. <S> As you can see in the figure above from "from Electromagnetic Compatibility Engineering, by Henry W. Ott, section 11.4.4. <S> " Introducing multiple capacitance values in parallel, also allows for multiple anti-resonance peaks. <S> And while at this peaks the impedance is lower, the dashed line consisting of four exactly the same parallel capacitors provides an overall better impedance. <S> Therefor when putting capacitors in parallel, try giving it a moment of thought if you want to put them in parallel to reduce the impedance, and if you will then use several capacitor of the same value, or capacitors of different values. <S> Also read: Antiresonance of multiple parallel decoupling capacitors: use same value or multiple values?
| Usually the capacitor is meant to have a low impedance to ground, putting a lot of capacitors in parallel can help reduce the impedance.
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High Voltage ADC Measurements using a PIC and OpAmp I need to make a device to measure voltage of a series stack of capacitors which will reach a maximum of 100V (DC only). On the schematic I have omitted several from the stack, balancing circuitry and the PIC support devices (xtal, decoupling), for simplicity of getting my point across. Each cap will have 5V max across it (in house made supercapacitors if you are interested!). As the dynamics are quite fast, my usual Arduino solution just isn't quick enough so I am back to the old days of PICs, which I haven't touched for many years! 1-10ms for all 20 channels in total, ideally. In order to measure the high voltages I assume a simple resistive divider would suffice, however this would need to be as high an impedance as possible so as to not skew the readings and require high power resistors. This puts the impedance well over the max 10k for the PIC I am using (PIC18F25K83). The PIC is running at 5V. Current thinking, using this as inspiration, is to use an opamp to buffer the output. Not used opamps before. Is there anything glaringly obvious that would make the magic smoke appear, or is there a better solution? EDIT: Image cropped as requested. Quantified how fast. <Q> Take care while routing, sometimes through hole parts can be advantageous to avoid arcing and to provide separation as creepage and clearance for 100V is 0.71mm. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The AD8067 is largely unsuited for this application. <S> When running from a single 5 volt supply, the input common mode voltage range is 0 to 2 volts and this means you would need to radically rethink your resistor values if you are sold on that particular chip. <S> The Ad8067 is also quite hungry power wise. <S> Current consumption is 6.4 mA per op-amp <S> so it all adds up. <S> This is because it is a fairly fast device and this is wasted in your application. <A> I'd probably be lazy and avoid building (and, potentially, calibrating) 20 voltage follower ADC buffers. <S> TI sells a 10-channel MUX + PGA IC which is meant pretty much for this kind of application. <S> PGA116 or PGA117. <S> It takes at most 5 nA of input current even for the active input that is muxed to the programmable amplifier. <S> So, let's say you have 0.5 MΩ of series resistance, that'd be a voltage drop of <S> \$5 \cdot 10^ <S> 5\,\Omega <S> \,\cdot\, 5 \cdot <S> 10^{-9}\,\text <S> A= <S> 25\cdot 10^{-4}V <S> = <S> 2.5\, \text{mV}\$. Over a range of let's say 2 V, that's an error of \$\frac{2.5\cdot10^{-3}}{2}=0.125\%\$; your PIC has a 10 bit ADC, so its ADC quantization allows for only slightly better resolution, anyway. <S> If that's too much, decreasing the voltage divider impedance to 300 kΩ would already put the error below the resolution. <S> So, you'd be replacing 20 Opamps + 2 PICs by 2 PGA116 / 117 + 1 PIC. <S> You can control their switching using SPI from that one PIC easily! <S> To reduce the chance of high-speed oscillations / RFI stuff, Nick correctly recommended small (1 to 4.7 nF or something) ceramic capacitors in parallel to the lower side of your voltage dividers. <S> Cheap! <S> Use each channel as little as possible – that keeps these capacitors charged and further reduces the impact of biasing current. <S> Use arrays of TVS diodes. <S> They can be had cheaply (and with low parasitic effects) <S> e.g. as high-speed / super-speed USB protection diodes.
| You may want to put a protection diode on the input of the op amp, the high voltage could easily burn up the op amp if it somehow crossed the barrier of the resistor.
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Replace a one shot thermal fuse with a resetting one My mother owns a device for drying fruits and stuff. Now every year or so, the thermal fuse breaks. I'm always checking if the fan is damages or something else, but the device is always fine. It is just the thermal fuse which breaks for no known reason. So instead of replacing it with the same type of thermal fuse every time. I was wondering if I could use a resetting thermal fuse. The current thermal fuse is labeled DYE ZRJG DF110S which seems to be a DF type thermal fuse made by Limitor with a function temperature of 110 °C. The tolerance is listed as +0/-5 K and the maximum rated current is 16 A. The device has a rated power of 600 W, so the fuse needs to handle at least 2.6 A. It is a resistive heater, so that should be about it. As I'm not really familiar with these temperature cutoff switches (is that the right product name?) is there a reason why I should not replace the fuse with a cutoff switch? I could imagine, that if there finally is a failure in the device and it gets turned on and off by the thermal switch, it might cause more damage. What I want to achieve is, that the device will run for more than a year without having to replace the thermal fuse (but it should still be protected, so a wire is not a solution). Sorry - should have done this the right way the first time: simulate this circuit – Schematic created using CircuitLab After the request by @jsotola for better pictures I took the device apart again. I also removed the control unit cover to understand how the temperature control is implemented. What I found was: a micro switch is turning on and off the whole device. A (probably bimetallic) bending strip is the temperature control, the knob controls how much pressure is on the strip and thus controls the tripping temperature of the strip. Close to the thermal fuse (on the other side of the picture I took first) there seems to be another thermal switch, which should probably act as a first safety measure. <Q> You could try a "Bimetal Cutout". <S> (It's a bimetal switch, which performs a thermal cut-out.) <S> Image from cantherm.com <S> These act like you might expect: <S> They are normally closed, and open when the temperature rises above a certain threshold. <S> They close again after the temperature drops down (including some hysteresis). <S> You can find them for 250V+, 10A+, and in the 110C range. <S> Note that you may have a difficult time fitting them in the same tight space. <A> This is a safety device, so tampering with it may compromise the safety of the product. <S> It's the last watchman that prevents a potential fire (say if the inlet air flow is blocked). <S> The thermal fuses are very simple and reliable and (most important) are very unlikely to stick 'on'. <S> That's why they are used in this application. <S> I have some sympathy with your position, we have a very nice Krups sandwich press which similarly has failures (particularly if it is accidentally left on for a long period of time). <S> I changed the position of the thermal cutoff slightly internally and it works for longer now. <S> You could consider going slightly higher ( <S> eg. <S> 114 <S> °C or 115°C) in cutoff temperature, but of course that does not comply with the original designer's intent (nor the testing that was performed when the safety agency allowed its mark to be put on the device). <S> So, I'm not recommending that. <S> But you definitely should make sure the way your mom is using the dehydrator is not contributing to the failures. <S> I know you put the device on a carpet to photograph it, but if it was actually on something like that it could obstruct the air flow and lead to higher temperatures inside, and the fuse would do its job and shut down the appliance. <A> You need to do some measurements, and establish likely causes. <S> I would imagine that normal operation of a dryer only involves air at 40-50C, so thermal fuse failure implies a major fail. <S> Either <S> she is overfilling it up, and thus restricting the airflow (or do so some other way) <S> the fan is failing to start or stopping (bearing lube failure, or something sticks down and catched the blades) <S> These two seem less likely: the design simply runs too hot and should have the heater power reduced. <S> normal operation involves a high heating wire temperature vs air temperature, and that is making the temp switch much hotter than the airflow, by direct radiation <S> A PTC protection/heating thermistor, or the thermal switch Bort shows , may be able to regulate the temperature of the airflow, to something far below 110C <S> (e.g 70-80C). <S> Since the likely root cause is failure of the airflow, have a buzzer to alert her that it has gone off. <S> It might be able to be across the switch, or certainly it would be easy to add to a w1209 type controller (some have a place for the buzzer to be fitted) <A> After my further analysis (thanks @jsotola) <S> I came to the conclusion, that the device should already have a self resetting thermal switch. <S> I hence took the idea from @spehropefhany and bent the thermal fuse a bit inside the plastic carrier in order to gain some distance to the heating winding. <S> So maybe that fixes it, but if not I end up replacing the fuse every year. <S> This question is a good example why repair questions are not the best fit for EESE, as my edit would probably change the answers quite drastically.
| You would leave the thermal switch in as a failsafe. Another alternative is to get a cheap little temperature controller from CN like W1209 and again set a lower control point.
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What if I cut off an NC pin from an IC? I am designing a circuit using an IC, LTC7004 , which is a high-side MOSFET driver. There is one pin (10th) marked as NC. NC (Pin 10): No Connect. This pin should be floated. In my PCB design, I'm pulling out a track from the 8th pin which is the pin to connect the source of a FET as in the figure below. If I can pass the bootstrap (9th) track over the 10th pin, I can have a wider track for the source connector than the one I currently have. Since the 10th pin is a no connect pin, will it make a mess if I bend and break away the 10th pin from the IC? (Then I can design a custom foot print to have only 9 pins with the MSOP10 package dimensions). Should NC pins need to be soldered onto a PCB with isolated pads by any means? <Q> Sometimes "No Connect" means there is no connection between the package pin/pad and the IC die. <S> In that case you can route whatever you want through the corresponding PCB land. <S> Sometimes the datasheet makes it clear which of these applies, sometimes not - from the text you've quoted it's not 100% evident but points in the direction of the latter, which is always a safe assumption to make. <S> In this application the trace impedance is going to be trivial compared to the gate drive IC's 1+ ohm output resistance, so you really have nothing to gain by changing the layout from your present design. <A> Ripping it off or soldering it to a (small) pad makes no difference. <S> From a mechanical point of view, having it soldered or not can make a difference though. <S> I am fairly sure that the manufacturer recommends a certain footprint in the datasheet, and that that footprint includes the pad for pin 10. <S> An even better option, from a manufacturing point of view, would be to just remove the pad for pin 10 from your PCB, and route the track from pin 9 right below it. <S> The solder mask will guarantee electrical isolation, and there is no need to add a manual step to the manufacturing process. <S> As Tom Carpenter as commented, there is a drawback: the solder mask thickness can prevent the chip to seat flush on the copper, thus leading to soldering issues in some cases. <S> Whether you trust your solder mask enough for this is a decision you should make. <A> The major reasons are already covered in other answers/comments, but also: <S> You risk damaging other pins, either physically or through electrostatic discharge. <S> In some cases, heat is dissipated through the pins. <S> Modifying a pin can hinder this process. <S> It looks really unprofessional, and your customers/boss might care about that.
| If you are confident that the other pins give enough mechanical stability, you can just cut it away as per your plans. From an electrical point of view, NC pins need not to be soldered. Sometimes "No Connect" means "thou shalt not connect anything to this pin", because it is connected to something on the die but reserved for factory test or some other purpose.
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Auto LED dimly lit when off I bought an LED bulb, W5W, to replace the position lights, number plate lights and maybe the interior reading lamps. However they stay dimly lit when they're off. I am thinking that the current is already flowing through the wires when the lights are off and the key is out. So the circuit is closed, but with an incandescent lamp, not enough current exists to heat the coil, but with LEDs there is enough of it to light it. Some people advise to put a resistor before the bulb, to fix this. Will these LEDs drain the battery if they stay partially lit (i dont mind them staying like this)? If I add a resistor, won't that one drain the battery too? <Q> Cars often pass a little bit of current through "off" bulbs in order to determine if the (previously incandescent) bulb was working normally. <S> Depending on the replacement LED bulb circuitry, this can either cause intermittent flashing (when the LED driver circuitry builds up enough charge to light up), or constantly light an LED dimly (if it's a simple resistor in series with the LED). <S> Yours sounds like the latter. <S> Olin is right with the suggested workaround - adding a bleed resistor in parallel with your LED bulb. <S> The parallel resistor can provide an alternate path for the small "off" state current, meaning the voltage across the LEDs never gets high enough to cause them to light up. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Please note that all values here are placeholders. <S> You'll need to do some experimentation to find a bleed resistor value that works. <S> 10k is probably a good starting point. <S> The replacement LED bulb will almost certainly be wasting less power than the original incandescent when "off". <S> You could confirm that by checking the voltage across a working incandescent and your new LED bulb, and calculating the current through the bulb check resistor with V <S> =I*R . <S> Lower voltage across the bulb means higher voltage across the bulb check resistor, which means higher current wasted. <A> First, it doesn't sound right that there is a little current thru the lights when they are supposed to be off. <S> You really should check into that. <S> However, if you are truly OK with that current, then put a resistor across the LED. <S> Find what the current is, then size the resistor to produce about 1 V with that current. <S> That won't be enough to light the LED, but should draw so little extra current when the LED is lit to be irrelevant. <A> It is not unusual to have leakage currents through relay coils or transistors that are supposed to be fully OFF. <S> The driving source may leak a few milli-amps of current, not enough to turn on the relay or a series indicator lamp. <S> 12 volt relay coils are about 400 to 1,000 ohms of DC resistance, enough to turn off a conventional lamp but not an LED. <S> To fix the problem add a small value resistor across the LED such that the LED voltage drops to <= <S> 2 volts down to 1 volt. <S> That is below the 'ON' threshold of most LED's. <S> The leak is already there so a resistor to bypass the LED is not going to increase the leakage current. <S> Remember there was a bulb there at one time, not impeding the flow of leakage current at all. <S> Measure the leak current if you can. <S> A 1 K resistor creates a 1 volt drop across it per each mA of current flowing, so a 1 K 1/4 W resistor is a good starting point. <S> It may have to be as low as 200 ohms if several mA of leakage current is present, but you said "dimly lit", which implies only 1 or 2 mA of leakage current. <S> NOTE: <S> 'OFF' is not always an ideal OFF, as this case proves. <S> I had to use diodes often to block leakage currents from electronic modules used to control overhead lights, and sometimes used to trigger a third-party alarm. <S> Often it is a transistor that is not fully OFF simply because of a cheap design. <S> It worked at the factory with standard light bulbs, not expecting LED replacements.
| In "on" mode, the parallel resistor will just waste a tiny bit of power - it make the bulb visibly dimmer.
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I'm planning to put lots of phone chargers on one circuit, how to calculate this This would be primarily for samsung S7 devices. Would like to be able to charge, say, 256 of them. They will work in groups of 16, trying to design the power flow of this. Was imagining that each group of 16 would plug into a USB power strip of some kind, and then each group of 16 would link up to another power strip. So, 16 groups of 16. But I think that this will not work on a single 120V/15A power source. I would like to understand the math involved with how to scale this. It will probably be more like 100 devices at a time, but trying to understand how to calculate this and know what the limits are. <Q> 0.5A in. <S> 30 of those and your general purpose outlet is at capacityso <S> you'd need nine 15A circuits to run your collection of chargers. <S> Now, 0.5A at 110V is 55W which suggests the charger has a power factor of only about 0.3 <S> If you need 3KW of charging, start with a power-factor corrected 5V supplies. <S> and connect a bunch of USB sockets to that, you then you only need two 15A circuits. <A> If you are using a quick charger, this will be higher. <S> 256 devices at 10 Watts each will draw 2,560 Watts, plus a bit more for power supply inefficiencies, so say 3,000 Watts to get a round number. <S> A US 15 amp circuit a handle 1,440 Watts, though, to be safe, you should derate it to 80% or 1,152 Watts. <S> Summary, you’ll need to split them up onto separate circuits, either 3 x 15 amp or 2 x 20 amp. <A> I'd really avoid this. <S> Having several 10s amperes around will give heavy and unpredictable voltage drop on the 5V lines. <S> Heavy <S> because you'll find that even huge cable gauges will not keep power within specs, not mention that USB connection to your groups of 16 will probably melt down. <S> Unpredictable because voltage drops will depend on actually sunk current, varying with number of phones and their batteries status. <S> Something like the above <S> , it can be installed like a conventional AC socket but it can supply a couple of phones
| In turn I believe the only viable option is a bunch of individual 230V to 5V USB power supplies, one for each phone. Each device would draw up to 10 Watts (assuming 2 amps at 5 volts). A 20 amp circuit will handle 2,400 or 1,920 Watts at 80%.
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Why plug is polarized in the US? What are the benefits? I could not get a good hold of it, I searched online quite a bit. This must have to do with safety but then non-polarized switches are used through out the world with no issues are risks, so my question why use polarized socket where one blade (neutral) is longer than the other (hot). The best explanation I came across is , it has to do to with the lamps used in United States where you have to reach out to turn the lamp one and off and may accidentally touch the bulb part which in case of non-polarized socket could either be hot or neutral and can give a shock, but is that the only reason? Somebody could do the wrong wiring and that means you will still get a shock with polarized socket? I could find other reasons like capacitor between ground and hot wires or making the chases connected to ground etc but I don't think they are the main reasons because again non-polarized switches are used through the world for appliances. So my question is why use polarized switch in the US? If electric lamp is the main reason, why not re-design the lamp and make it safer? <Q> It's more than that -- polarized plug make cheap single-pole switches much safer. <S> For example, look at the extension strip: <S> one just makes sure it interrupts the "live" contact With a non-polarized plug, one either has to put a more expensive double-pole switch, or to accept that half of the time, powered-off power strip would present a shock hazard. <S> Neither of these options is very good. <S> It gets even worse with the fuses -- they are single-pole by definition. <S> So if you have something protected by a fuse, and the fuse blows, it better be the fuse in the live wire! <S> And in the non-polarized-plug land, a blown fuse might leave device non-working, but still energized -- which is pretty dangerous. <A> The reason is historical. <S> My dad grew up in the 30s and <S> it <S> he said it was very common for washing machines and drills and light fixtures to have their metal cases connected to the neutral wire. <S> Since there was always a few volts on the neutral due to wire resistance it was common to get a tingle if you grabbed a water pipe while touching a running appliance. <S> I guess copper was expensive back then and an extra ground wire was a considered a luxury. <S> So it was essential back then to polarize the plugs or people <S> would die from full voltage appearing on the case. <S> You can imagine that the transition to requiring a separate ground wire couldn't happen overnight because you couldn't require people to throw out existing appliances. <S> The fact that we still have polarized plugs means we are still in that transition. <A> Of course, Hubbell also invented the polarized outlet in the 1910s. <S> Knapp at Hubbell also invented an grounded outlet that was incompatible in the 1910s (used in China and Australia today). <S> At the time, both US and most of Europe used 110V to 125V with a neutral. <S> When Europe switched to 220V, it was often off a 127/220V three phase feed, which means that the 220V had no neutral (still common until recently for example in Belgium and Norway). <S> Fortunately most of Europe already had grounded outlets by the 1930s. <S> In the US, it took until 1948 before NEMA 5-15 was standardized with the round ground prong we use today added to the original Hubbell outlet. <A> If you search hard enough, you will probably find that "it seemed like a good idea at the time. <S> " It provides some increase in safety to have the outer part of a bulb socket connected to ground. <S> The same is true of certain internal parts of some products. <S> Some parts are more at risk of contacting external metal parts than others. <S> Sometimes design detail decisions and electrical code rules are made for reasons that are rather weak. <S> Even though we have double insulation and ground-fault interruption devices today, there is no particular reason to do away with polarized plugs and receptacles. <A> Polarized plugs ensured AM radios worked on DC power, and that blenders rotated in the correct direction with DC power.
| With a polarized plug, it is perfectly safe to have a single-pole switch there-- WAAY back in the old days the neutral was used as a ground. In the US, the current two-prong power outlet was invented by Hubbell back in the 1910s.
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Heat problems when using 5 buck/switching converters in a double isolated enclosure? I want to use 5 step down converters (220 VAC -> 5V DC) inside an enclosure inside an enclosure (double isolation). The step down converters are similar to these. There will be 5 of them, where one is connected to an RS485 transceiver (MAX1485CPA), and the other 4 have each an optocoupler (6N137) and RS485 transceiver (MAX1485CPA). The inner enclosure is completely closed, plastic, and just big enough to contain them with about 1 cm / 0.4" space between each converter. In the outer enclosure (also plastic) will only be XLR (DMX) connectors and max. 10 small LED lights. Also I will use a fuse (1A?) for safety. Questions: Will heat be an issue? If so, should I make 'holes' in both enclosures for air flow? As you can see I cannot screw the devices in an enclosure (no holes), so I want to glue them (upside down with the yellow flat area glued for a maximum and a flat glue area). Besides that, I'm intending to use wooden popsicle sticks to 'physically' make sure they never can touch each other (by glueing them somehow between the converters. Is there a problem or better solution (without having to buy new converters)? Update After a discussion it is best to use a switching supply with enclosure already around it (being an amateur): 5V 3W mini power supply module Hi-Link HLK-PM01 <Q> Any enclosure that has power being dissipated inside of it will create a situation where the temperature will increase. <S> The more that you cram into a smaller space <S> the worse the problem can become. <S> You should evaluate just how much heat each of the modules creates when in the proposed enclosure before you commit to a final design/build. <S> Remember that excess heat is the enemy of electronics. <S> The flat yellow surface is just some tape wrapped around the transformer core. <S> It is probably not the best gluing surface and the tape could easily come loose. <S> I would suggest that after attaching your input and output wires that plopping each adapter into a good pool of hot melt glue may be a better mounting scheme. <S> Just make sure that the hot melt glue adheres to the inside of your enclosure before committing to use it. <S> I have to ask why you think it is necessary to have five separate power adapters in one enclosure when the load on each one is so small. <S> Maybe you could explain that with a schematic. <A> Small transformers are very inefficient, it's just the way that the parameters of iron and copper turn out. <S> You will have much smaller heat problems if you use a single power supply of 5x the current output. <S> This is definitely a case where one bigger is better than many smaller. <A> Based on your previous question, this looks like an X-Y problem. <S> I can't give a model suggestion, but even eBay gives plenty of suitable listings. <A> Will heat be an issue? <S> If so, should I make 'holes' in both enclosures for air flow? <S> Yes, you either need to ventilate or derate enough that they do not dissipate much energy. <S> Is there a problem or better solution (without having to buy new converters)? <S> Gluing PCB's is bit of an amateur solution. <S> Potting is more common if that's what you need. <S> Buy better converters, or put each in their own enclosure. <S> Remember that mains is a fire hazard.
| It could be that you may even have to add a fan to force air through both enclosures to keep things cool. You can actually get your task (powering isolated DMX lines) done with a single mains-powered AC-DC converter and 5 isolated DC-DC converters - they exist and they aren't that expensive, and it certainly will be safer in terms of operating voltages. Sometimes it's worth paralleling smaller devices, and sometimes it isn't.
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Why does power dip during a lightning storm? If lightning were to strike power lines, wouldn't you expect it to increase the power, and hence your room lights get brighter? Why do they go dim for 1 second at a time? <Q> Janka comes close, but there are are several more details. <S> (Note, recalling from EE classes about 45 years ago.) <S> On many high voltage lines there are arc electrodes at various points. <S> This helps to dissipate the voltage of the lightning strike. <S> But the spacing of the electrodes is such that, once the arc has formed, the normal voltage on the line is sufficient to keep it going. <S> So the electrodes are formed in the classical V shape, close together at the bottom and wider at the top. <S> Heat causes the arc to rise (see "Jacob's ladder") <S> and so it gets longer. <S> Eventually, when it gets to the top, the voltage will (hopefully) no longer sustain the arc. <S> If the arc does not extinguish itself, eventually a nearby over-current detector (fuse) will trip and cut off the power. <S> But to save the lineman a trip to reset it, the over-current detector is often designed with a timer so that it resets itself after a few seconds. <S> But usually there's a limit to how many times it will reset ( <S> in case the over-current condition is due to, eg, a downed power line). <A> Lightning brownout procedure: <S> When lightning hits an overhead power line, there is overvoltage at first, for about 100 milliseconds. <S> This overvoltage creates an arc at a nearby pole. <S> The arc works as a short circuit, so current from both sides of the overhead lines flows to the arcing pole. <S> The voltage at other places of the grid dips because of the huge current flowing to the arc. <S> The arc eventually extinguishes. <A> There's another explanation that does not require a direct lightning strike. <S> Storms strong enough to generate lightning also typically have strong winds. <S> A strong gust may bend a tree branch or other vegetation so that it touches a power line and creates a temporary short circuit to ground. <S> This draws a large current, which drops the voltage and causes your lights to dim. <S> Once the tree bends back, the power is restored to normal, either immediately or after the operation of a recloser, as described in this answer to a related question. <S> Branches weighed down by snow or ice may be closer to the power lines than in their unweighted position, so it might take less motion of the tree to contact the power line in this condition.
| When lightning strikes the line, the increased voltage causes an arc to form across the electrodes.
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Measuring a binary number with maximum electrical certainty I want to be sure, with arbitrary precision (let say, for example, with 10^-100 certainty), that I can check the following statement (in binary): x = 49 That is, I want to check that a number that is (correctly) coded in a binary electrical signal is the same as a second number (also correctly coded in binary). I'm a novice at electrical circuits, but I've worked out that I can perform this operation using 6 exNors and 5 And gates. I'm assuming there is some small uncertainty in these logic gates, and that there's a super small chance that these gates can misfire (maybe even so small it's associated with the quantum electrical noise). If I wanted to make this system have arbitary precision, what could I do to prevent these errors from /ever/ occuring? One option I thought of is to perform the same operation multiple times and AND-gate the results together. (I am perfectly fine losing "efficiency," that is, I don't mind if the circuit has "false negatives" where the number is correct but since there was a misfire in one of the circuits it does not return a 1.) Is there a way that I can analyze the probabilities of the circuit being correct? And what types of electronics would be ideal for this type of system (my guess would be that higher voltages would have less fundamental noise, so they would be the best to construct the system.) It seems as though there is some work on simulating the propagation of these errors using some advance techniques like monte-carlo simulations and baysian analysis. But is it not possible to get a crude, order-of-magnitude estimate for the amount of uncertainty/error per gate? *Moved information in comments to the question where it belongs* The purpose of this device I think is outside the scope of electrical engineering, but it's sort of related to metrology and quantum mechanics. I agree it's a very unconventional thing to be concerned with. I am first trying to get some kind of grounding on the topic because of how unusual it is. Any implementation (or discussion of errors, no matter how small) would be helpful (not concerned with hardware, frequencies, as long as it's digital). Maybe rephrasing the question. If I performed this comparison operation on a microcontroller over and over, how much time (or how many repetitions) would it take until it misfired and gave an incorrect comparison? (Minutes? Days? Millennia?) And does this error loosely scale linearly/polynomially with the number of gates? So, for example, if I have 600exNors now instead of 6, does my operation misfire 100 times faster? (implying the errors scale linearly) <Q> I'm a novice at electrical circuits, but I've worked out that I can perform this operation using 6 exNors and 5 And gates. <S> Yes you can (if 6 is the number of bits you want). <S> I'm assuming there is some small uncertainty in these logic gates, and that there's a super small chance that these gates can misfire <S> (maybe even so small it's associated with the quantum electrical noise). <S> The chance is negligible, by which I mean, it doesn't happen. <S> You may as well ask whether there's some small uncertainty in a tennis ball's position, so that there's a super small chance that it will teleport across the room (maybe even so small <S> it's associated with quantum noise). <S> Or the probability that every now and then, your car will go backwards for a moment when you press on the accelerator, because of quantum noise. <S> If I wanted to make this system have arbitary precision, what could I do to prevent these errors from /ever/ occuring? <A> Bit Error Rate BER can be predicted for different types that affects margin to error. <S> Errors can injected by a measured amplitude of energy required to cross the binary logic threshold . <S> A driver has a defined source impedance and connections have a characteristic impedance and bandwidth as well as the input has a bandwidth limit. <S> It can also be controlled by design of supply voltages and thresholds as well as filtered in time , when it is synchronously latched into memory. <S> Since 10e100 is a really big number of operations, to measure this many cycles is very impractical. <S> A test can be designed with confidence level of achieving a certain error rate with a combination of stress tests for low supply voltage, high temperature, vibration with radiated and conducted noise including ESD and power-line transients. <S> Noise effects from all sources including gamma rays can be reduced by redundancy, but the cost of verifying extreme levels of low error rates must be supported with a healthy budget, great design specs including; the expected operating environment and diligent design verification test DVT methods. <A> Assuming this is not about Quantum Mechanics.. or tennis balls and walls even... and, simply about dealing with "stuck bits" or other hard failures on the IC and how to protect against computing failures if that occurs.... <S> You could always use Triple-Mode Redundancy (TMR) for the design. <S> This is a technique that is used in "high-upset" environments such as space (radiation effects) to reduce probability of a failure when an upset occurs. <S> It votes 3 outputs to determine the correct value. <S> You can do this by design, or let (some) tools do it for you at synthesis. <S> If concerned about radiation, there are Rad-Hard processes Honeywell offers at their foundry.
| You can't, there's always the possibility that your entire apparatus will quantum-tunnel into the middle of the Milky Way never to be seen again.
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Do I need to fuse batteries in parallel? I have two deep cycle batteries in my Airstream, in the panel there is a fused bus bar (that appears to be faulty) so the solar panels only charge the one battery. Can I just run a heavy gauge wire between the two positive and two negative terminals? Or do I have to put a fuse across them? ... if so do I fuse the positive or fuse the negative? What's the worst that can happen if I don’t fuse them? <Q> Remember that in a DC power system like you have, fuses prevent electrical fires by protecting WIRES , not the devices connected by the wires. <S> This is not to say that blowing a fuse never prevents damage to an attached device, but that the "correct" way to think about circuit protection is to examine the circuit topology and identify faults which could result in a "wire" being subjected to current (and hence temperature) sufficient to start a fire, either in the insulation or in a device connected to the wire in question. <S> This includes all fault modes identifiable. <S> In particular, this includes at least damage to insulation from abrasion, thermal stress, water absorption, and chemicals (oil, gasoline, solvents), also parts and tools being dropped into a panel or battery box. <S> One other case to consider is that a fuse performing its act of protection can lead directly to a fault as well. <S> If you have never seen one of the big T-series 400amp fuses go bang! <S> you have likely saved pair of underwear. <S> Fuses blow by melting metal, producing liquid metal which can be ejected to land on some other connections causing another fuse to blow. <S> This is why fuse enclosures are worth paying to get quality products. <A> The worst case scenario is one of the cells is shorted and before it explodes/catches fire itself it also shorts the other cell. <S> Sure, you "can" bypass the fuse with heavy gauge wire, but you will take the risk. <A> It is very common to have two or more lead-acid batteries in parallel, with no fuses between the batteries - but you MUST have a fuse close to the batteries, between them and other wiring in the boat/vehicle. <S> For marine use, ABYC says the fuse must be within 7 inches of the battery. <S> I have had a couple of cases where one cell in one of the paralleled batteries shorted - <S> there was no fire,explosion or other excitement - I just couldn't charge the battery bank to its proper voltage.
| The fuses are there for a reason (and there is a reason the fuse went open).
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relay coil voltage: can I have trigger with both 110v and 240v? I'm searching for a relay that can both trigger and switch on either 110v or 240v. To clarify - both the coil voltage and the switched power would be the same voltage - and as low as 110v or as high as 240v. (think of an appliance that could ship in the US, or internationally). The relays I've found so far trigger at ~80% of max coil voltage, which isn't enough latitude to run in both scenarios. Am I likely to find a relay like this, or should I devise another solution? Per request: here's the relay I was looking at with a datasheet for LY2F-AC220/240 <Q> No. <S> "70%V or less ofnominal voltage" means it won't be more than 70% <S> but it also means it may be unreliable or slow or not activate at 50% Vac rating and 200% would exceed power rating of coil. <A> You may need to use two relays - one 240 V and one <S> 110 V. simulate this circuit – Schematic created using <S> CircuitLab <S> How it works: <S> On 240 V RLY1 is energised, disconnects the coil of RLY2 and feeds 240 V AC to OUT. <S> On 110 V RLY1 is de-energised, supply is connected to the coil of RLY2 which is energised and feeds 110 V to OUT. <S> When supply is low (maybe < 40 V AC) then both RLY1 and RLY2 are off and there is no connection between L and OUT. <S> Relay hold-on voltage is generally quite low relative to the pick-up voltage and this can cause problems in some applications. <S> In this case, however, both relays will have dropped out by the time you get from one continent to the other <S> so I don't think there will be a problem. <A> It's possible but unlikely you could find something like that- <S> the relay would have to be designed for that purpose meaning it would be larger (more expensive) and less efficient than a regular relay. <S> You might want to consider a universal input DC power supply (or wall wart) and a 12V relay. <S> That would be particularly convenient if you actually needed the 12V elsewhere.
| There are many alternatives, but appliances usually have a low Vdc for power relays.
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What equipment do I need to test an eye diagram for USB? I'd like to test USB full speed with a goal of testing high speed (480 Mbit/s), I have a tek scope that does 300 MHz (which I could upgrade to 500 MHz) and I'm looking at a 500 MHz differential probe. As I understand it I also need a breakout board (which I'm not quite sure is the best thing to get), but I'm looking at this board from tek and one listed here . What are the minimum requirements for a test like this? Is this equipment list sufficient to preform an eye diagram test for full speed USB? Is this equipment list sufficient to preform an eye diagram test for full speed USB if I have a 500 MHz scope? <Q> What are the minimum requirements for a test like this? <S> Minimum requirements for test equipment to use for USB 2.0 signal quality evaluation is listed at USB.org in the following place . <S> There are links that describe electrical test procedures and tool requirement for Rohde&Schwartz, Tektronix, Agilent, LeCroy, and Yokogawa oscilloscopes. <S> Typically the eye evaluation software tools are offered on scopes with no less than 2GHz bandwidth. <S> For Tektronix, the eligible scope series are TDS7254/B, <S> TDS7704/B, <S> CSA7404/B, TDS6604/B, <S> TDS6804/B, TDS6404, DPO7254, DPO7354, and DPO/DSA70000. <S> The smallest eligible oscilloscope for USB 2.0 testing is MSO/DPO5204 . <S> For FS eveluation you don't need differential probes, the scope does it mathematically using single-ended probes. <S> However, the software package can't be installed on smaller bandwidth scopes, so, even if 500 MHz bandwidth is OK for FS eyes, it is unlikely that you can use this scope. <A> If your signal has a fundamental at 500MHz, and you're trying to measure its characteristics, then you're not going to do well with a 500MHz scope because the scope won't capture any of the harmonics. <S> You will need a higher bandwidth scope and probe if you want to make an accurate measurement of what's happening. <A> The horizontal position of point 3 is at 37.5% UI, and point 6 is at 62.5% <S> UI <S> , so you are trying to measure a rise-time of about 75% of 2.08 ns, or about 1.5 ns. <S> A 500 MHz scope will measure a minimum risetime of about \$0.75/500\ {\rm MHz}\$, or 1.5 ns. <S> You can estimate the risetime you measure will be about $$\tau_{meas}\approx\sqrt{\tau_{scope}^2+\tau_{sig}^2},$$ <S> so a 500 MHz scope is not going to cut it. <S> I'd look for at least 1 GHz, and 2 GHz will help if your product doesn't have much margin. <A> I ended up using a single ended measurement and using matlab to poll the textronix scope and using the instrument control toolbox: <S> myScope = oscilloscopedrivers(myScope)availableResources = getResources(myScope)availableResources{5}myScope. <S> Resource = ' <S> TCPIP::xxx.xxx.xxx.xxx::INSTR'connect(myScope)get(myScope)waveformArray = getWaveform(myScope, 'acquisition', true);figure;plot(waveformArray)enableChannel(myScope,'CH3');enableChannel(myScope,'CH4');Time = <S> linspace(0,myScope. <S> AcquisitionTime,myScope. <S> WaveformLength);aqs = <S> 1000;for <S> i = 1 <S> : <S> aqs[w3, w4] = <S> readWaveform(myScope, 'acquisition', true);pause(0.05);w3mat{i} = w3;w4mat{i} = <S> w4;endfigure;subplot(2,1,1),hold on,legend('D+','D-'),subplot(2,1,2),hold onfor i = 1 <S> :aqsw3 = w3mat{i};w4 = w4mat{i};subplot(2,1,1)plot(Time,w3,'b'),plot(Time,w4,'r')subplot(2,1,2)plot(Time,w3-w4,'r')endsubplot(2,1,1),legend('D+','D-'),ylabel('Volts');subplot(2,1,2),legend('D+ subtracted from D-'),xlabel('Time'),ylabel('Volts');subplot(2,1,1)x = <S> [8.2 21.15 69.93 75.13 <S> 69.93 21.15 8.2 ] <S> *1e-9/2;%These come from the USB IF website specsy = <S> [1.65 2.5 <S> 2.5 <S> 1.65 <S> 0.8 <S> 0.8 <S> 1.65];fill(x+1e-7,y,'r')title('USB <S> Full speed <S> EYE test')line([1e-7 <S> 1.4e-7 ],[-0.9 -0.9])line([1e-7 1.4e-7 ],[4.4 4.4])
| To test a USB high speed transmitter, you have to test the TP3 mask:
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Is a motor-generator set a reasonable way to get the purest waveform electricity? It looks like audiophiles go as far as installing separate powerlines to power their high quality audio systems. The motivation is that the amplifier and other hardware would run cleaner when the powersource provides cleaner waveform and with a separate powerline the amplifier will not be affected by various electrical devices run by other households. Let's not assess whether any audible difference can exist because that's highly subjective. I'm interested in the purest waveform of the powersource part. That separate powerline can be hit by lightning is also parallel to many other wires around it and so can be affected by processes in those wires is connected to a large grid which has tons of transient processes happening all the times so it doesn't really sound like a reliable best purest waveform source. Would a motor-generator set run on the original powerline be better? It looks like it doesn't care much about all those lightnings, parallel wires and the grid. How good is a motor-generator set for generating the purest waveform electricity? <Q> The "purest waveform" you want for noise-sensitive equipment is flat DC . <S> The both simplest and best way to get it is a battery. <S> Why? <S> Any kind of AC needs rectification of some kind. <S> Even a motor-generator with a standard DC generator produces a rectified sinusoidal current. <S> But even then, you have a second concern, and that's AC currents on ground. <S> In your motor-generator the two machines are connected with a steel rod. <S> It's not possible not to have at least some AC noise on the DC ground without replacing that one by glass fibre or something like that. <S> Use a battery. <S> Charge it if the device is not used. <S> Medical equipment is sometimes designed that way. <S> (It makes also the compliance testing simpler.) <A> Would a motor-generator set run on the original powerline be better? <S> The smaller the generator, the more distorted the waveform will be. <S> This is simple physics since smaller generators will have less and less iron, and will produce more triangle-like waveforms. <S> Especially when loaded. <S> You'd need at least a few dozen kVA, preferably even higher, to get near a "clean" cosine. <S> The grid is (for households) "infinite" kVA due to the low impedance, thus in theory more ideal than any generator. <S> If your goal is to purify your input to filter anything beyond 50 Hz, transformers do typically perform a really good job at this. <S> Combine these with line reactors, and you'll have a pretty decent filter. <S> But high energy loss. <S> However, when a large customer in your area generates a very high energy, low frequency harmonics (up to a few kilohertz). <S> This will be audible in some systems, and very hard to filter out. <S> A dual conversion UPS or generator might be necessary. <S> Or batteries, that's always the best. <A> If you want to sell clean power solutions to audiophiles, you might consider two products. <S> One product should be a dual battery DC supply with a charger and automatic micro-computer controlled battery alternation system. <S> The second product should be a motor-generator system with two stators in the same housing and two rotors on the same shaft. <S> That would eliminate any coupling backlash due to torque variation caused by power level variations. <S> Design the motor and generator as short, large diameter machines. <S> That would allow space for a shaped interior-magnet rotor design and optimized distributed windings on the stators. <S> A proprietary design of that type would eliminate competition from cheaper, two standard machines bolted to a base, products. <A> Consider a 100 amp peak waveform with 10uS turnon, in the rectifier diodes of the Audio Power Amplifier. <S> The dI/dT is 10^7 amps/second or 10,000,000 amps/second. <S> Assume <S> the HOT lead of the 117VAC power is 10cm from a twisted pair running between the phono cartridge and the phone preamp/de-emphasis circuits. <S> We'll first assume the twisting of the phono wires does nothing to attenuate the power line interference, then we'll come back and modify various assumptions. <S> Assume the twisted-pair wires are 1mm apart and of length <S> 1meter. <S> The voltage induced into the wiring from the phono cartridge is the standard Vinduce = <S> 2e-7 <S> * Area/Distance <S> * dI/dT <S> Vinduce = <S> 2e-7Henry/meter * (1mm <S> * 1meter/ 0.1 meter) <S> * 10^+7 amp/second Vinduce = <S> 2e-7 <S> * 0.001/0.1 <S> * 1e+7 <S> Vinduce = <S> 2e-7 <S> * <S> 1e-2 * 1e+7 = <S> 2e-2 = 20 milliVolts <S> Given the strongest output voltage from a MM moving-magnet cartridge is 5milliVolts??? <S> ? <S> , the SNR is -12dB, with the interference an evil-sounding singing. <S> And if the signal source is MC Moving Coil with 500 microvolt output (or even smaller), the preamp is hugely overloading and clipping. <S> What to do? <S> (1) use DC power(2) place the 117VAC power blocks several meters away(3) <S> be glad the usual phono twisted pair should give at least 20dBcancellation, depending on the quality/consistency of the twisting(4) <S> be glad the standard power line wiring has the HOT wire adjacent to the RTN wire <S> , thus another 20dB attenuation is yours for free.(5) pay $1,000 for a special woven HOT/RTN/Safety power cable(6) use steel chassis on the power block(7) <S> use steel chassis on the RIAA preamp(8) <S> keep the power cable far away from the phono-cartridge cable(9) <S> install heavy power-line filters in the Audio Power Amplifier rectifier circuits, to avoid conducted emission of the rectifier fast turnon
| You could use an unipolar DC generator, as it was done for ultra-low-voltage ultra-high-current DC applications as galvanic plating in the past.
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What kind of breadboard and prototyping board do I need for this right-angle DB25 connector? What kind of breadboard and prototyping board do I need to fit this right-angle DB25 connector? I'm new to soldering and don't know what the standards are. Additional information on this subject would be helpful as well. <Q> Example <S> They also exist in a smaller (double row) format like on this website <A> That type of staggered pins are more or less standard because they must also fit straight-through DB-25 footprints which are standardized, but you need a protoboard that is designed to hold them. <S> This image was the first I found on google, but you can find them in all shapes and forms. <S> This is for DB-9. <S> Note the odd staggered part to the right: Since the normal protoboard spacing is 2.54 mm while the DB-25 pin spacing is 2.77 mm, you need to search for a dedicated DB-25 protoboard. <S> It could be easier to buy an adapter board which includes the correct connector, but have pins or screw terminals at a more convenient spacing. <A> Another possibility is that you could find an old ISA prototyping breadboard with 1 or 2 DB-25 connector pads. <S> The advantage of this breadboard is the size of the breadboard available for you circuitry. <S> Thought is was designed for PCB compatible ISA slots the connect is easy to find and you could wire it for your own use. <S> Radio Shack used to sell them <S> but I believe I have seen them around on sites that sell surplus electronics from time to time.
| What you are searching for is a DB25 breadboard adapter, looking like this: Website (example):
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Why do we use a DB9 connector for serial com? We only use 4 wires (pwr, common, and two data), so why does rs232 use such a big connector? <Q> why does rs232 use such a big connector? <S> Assuming you mean the standard 25-pin connector, then there is a simple answer: <S> The RS-232 specification states the defined function for every one of the 25 pins bar three. <S> There are lots of features of RS-232 which are little used these days! <S> This diagram is a brief overview of the pinout of the 25-pin connector: <S> (Figure above courtesy of Dallas Semiconductor Application Note 83, Fundamentals of RS–232 Serial Communications ) <S> Why do we use a DB9 connector for serial com? <S> It's a similar explanation for the 9-pin connector made popular by its use on the original IBM PC/AT (the earlier PC and PC/XT used the full 25-pin RS-232 connector). <S> All 9 pins have a defined function, whether you use that function or not. <S> (Strictly, it's a D E -9 connector, not a D B -9. <S> The first letter, D, refers to the D-shaped connector's metal shell. <S> The 2nd letter is the size of the connector shell, and this smaller connector is size "E", whereas the larger 25-pin connector is size "B"; other sizes exist too. <S> See the Wikipedia article on the D-subminiature connectors for more details.) <S> (Figure above also courtesy of Dallas Semiconductor Application Note 83, Fundamentals of RS–232 Serial Communications ) <S> We only use 4 wires (pwr, common, and two data) <S> In fact there is no power pin in the RS-232 specification. <S> Of course people can adapt an RS-232 connector to include a power connection, but then it doesn't fully meet the RS-232 specification. <A> As it turns out, there were nine signals specified within the DB9 serial standard, including a ground. <S> The list of signals included handshaking signals such as DTR and DSR, flow control signals (RTS and CTS), an out-of-band ring indicator that was useful for modems, and (of course) the actual data signals themselves, as well as two grounds. <S> They mapped nicely to the nine pins of a DB9 connector when one of the grounds was removed (RTR and RTS shared the same pin and only one was in use at a time). <S> The DB9 connector and serial cable did not supply power (but as pericynthion mentioned in the comments, some devices tried to steal power from the RTS pin). <S> The fact that only TXD+RXD+Power+Ground are used today is likely related to the availability of more capable hardware with larger transmit/receive buffers and higher clock speeds, which meant that data could be sent with less handshaking requirements. <S> Modems have also fallen out of fashion, so we also don't see many of the issues with flow control that they originally required. <A> We only use 4 wires (pwr, common, and two data) <S> As has been pointed out, this was not always the case, although the extent of signal pruning has been significantly understated. <S> so why does rs232 use such a big connector? <S> What do you mean, "such a big connector"? <S> 9 pins? <S> This only started on the IBM AT. <S> Back in the dark ages (say, the 1980s) RS232 connectors were normally 25 pin connectors, and a common part of PC construction was a 25 pin to 9 pin adapter cable for running from the input connector on the chassis to the AT-compatible header on the serial board or motherboard. <S> The original RS232 spec (ca 1969) called for 22 signals. <S> See, for instance, <S> https://www.camiresearch.com/Data_Com_Basics/RS232_standard.html#anchor1155404 for a discussion of all the pins.
| RS232 specifies a number of out of band control signals as well, which were highly useful when devices of varying capabilities, often with limited memory, processing power, or other resources, needed to reliably interoperate with flow control and the need to avoid dropped characters.
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What is PVDD and AVDD? I know VDD,VCC,VEE and VSS and what they stand for. But when I came across this circuit: I saw this. Those PVDD and AVDD terms?What to do they convey or stand for? This circuit is amplifier circuit of PAM8403 DS. <Q> The first thing you should do when trying to figure this out is to find and thoroughly read the datasheet . <S> You will find the answer on page 2: PVDD (Power Vdd) is the pin that the current to drive the speakers comes from (a relatively high current). <S> They are kept separate so they can be bypassed separately to keep PVDD from affecting AVDD. <A> The answer is right in the datasheet for PAM8403: <S> So P stands for "power" and A stands for "analog". <S> Chip vendors and board designers are essentially free to name pins or nets however they like. <S> The "AVDD" designation is pretty common for the pin powering the analog part of a chip. <S> I've never seen "PVDD" before, but I don't work in audio, so I wouldn't know if it's common or not in that field. <A> This datasheet explains it all, but basically Vdd is for the analog pre-amp portion of the IC, while PVdd is for the digital/PWM portion and audio outputs, so can draw much more current than the Vdd pin. <S> It is common sense to separate the 2 supplies even if they run at the same voltage. <S> If not then even more reason to separate them. <S> This way you can use external RCL filters for each power input so digital noise does not get back into the analog section. <S> For a class D amplifier that could cause severe distortion. <S> Be sure to have 100 nF capacitors <S> right at each power pin and connected to a broad ground plane. <S> Follow manufacture guidelines for IC board layout pattern to get the best performance.
| VDD/AVDD is analog Vdd which is the lower current that is used to power the rest of the chip.
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Three Different Voltage Lines, Same Common Ground? I am trying to send PWM, clock, latch and blank signals from Arduino through a series (16) of TLC5940 Breakout boards and on to a grid of 512 individual controlled LEDs six feet away (RJ45). I am using op-amps to help get the serial signals pushed through the large number of boards. Power and Ground are in parallel. Of course, I am having problems. The set-up, as pictured, works well until the 9th or 10th board, then things get wonky. They are getting power but not individual control -just flickering. I am trying to troubleshoot. The whole thing is powered by 12 V, 30 A switching power, buck converters for three different voltages (10 V for Arduino, 4.7 V for op-amps, 6V for boards and LEDs). I am wondering if I should tie all grounds to a single common line back to the power source (V-) or do the individual voltages (10 V, 4.7 V, 6 V) need their own ground line back to power source? Example: the five op-amps - 4.7V+ and 12 V- single line common ground? - or should they have their own ground line all wired back to the 4.7 V converter and its 4.7 V-? <Q> There's nothing inherently wrong with using power line buses. <S> Star wiring might not solve the problem. <S> Ideally, you'd use an oscilloscope to see how stable the power lines are. <S> But I'll assume you don't have one. <S> Consider centre-feeding the power lines, rather than feeding the power in at one end. <S> That reduces the cable runs. <S> Consider using fatter wires. <S> The volt drop in a thin wire could be too high. <S> Digital devices often require sudden spikes of current, so the power lines could be dipping even if the average voltage measured by a voltmeter looks good. <S> Consider adding capacitors - electrolytic and/or ceramic - between the power lines and ground near each board (you already seem to have a few ceramic ones on some boards only). <S> This could end up a bit trial-and-error. <S> Too little capacitance won't achieve what you want, too much will overload the power supplies on power-up causing them to shut down. <A> Just a word of warning - having multiple parallel ground lines designed for different currents is a recipe for disaster. <S> If high-current power ground (e.g. LED) is accidentally disconnected all that current will go through (usually) thin wires for control supply. <S> I've seen spectacular meltdowns of control boards connected by CAT5 for data and logic power and separate 10 AWG power lines. <S> In this case it would be much safer to tie all supplies with just one ground wire to common bus. <A> Your DC-DC Modules have almost certainly an internal common Ground <S> (Go ahead an measure resistivity). <S> So you have them connected anyways. <S> If you have problems the Power "Bus", as your diagram suggests, you should deploy a star configuration for your power, esp. <S> ground. <A> Some points taken from old rules (about 1970 ;) ) of powering the large analogue/digital systems, containing large number of modules: Add 10...100uF capacitors (16 or 25V will be enough) on each PCB between it's each power pin and ground. <S> You can remove existing 104 (=100nF) <S> ceramic capacitors, shown on wiring diagram; Add 0.1uF <S> (=104) <S> ceramic capacitors between each power pins of each IC (as close to IC as possible) on each PCB; Use fatter wires - especially for ground and for powering LEDs (6V line). <S> Wires taken from CAT5 network cable may be too thin for ground. <S> Fatter wire is always better than thin ;) <S> So, these 10...100uF capacitors count should (has to) be equal to count of separated PCBs, and 0.1uF capacitors count should (has to) be equal to global count of ICs on all PCBs together. <S> Should help. <S> These rules are something like "best practices of powering large systems". <S> I'm using these rules on every system I made (regardless to the system scale: small or large), and I can confirm that they are always effective - it they are applied consistently. <A> Warning: these cheapo "LM2596" are probably counterfeit . <S> Last time I tested one, it definitely wasn't a LM2596 <S> (the frequency was wrong), inductor was saturating, output ripple voltage was terrible, and the input/output caps were extremely hot due to being "optimized" for cost rather than performance (ie, ESR way too high). <S> These LM2596 modules are not going to last, the cheap "general purpose" caps are not designed for this kind of abuse, they are going to die <S> and then the DC-DC may lose regulation and possibly destroy the rest of your setup. <S> But before that happens, output ripple will become so bad that it will stop working anyway. <S> So if you need high current 6V <S> you should really use a 6V power supply , these are not expensive. <S> Considering the time you put into this it would really be a bummer if it failed due to counterfeit DC-DCs. <S> You can power the arduino from 6V using a LDO to output 5V, or use a cheap 5V USB phone charger. <S> You can also power your 74HC7014 from a 5V LDO or the same 5V supply as the arduino. <S> I am using op-amps to help get the serial signals pushed through the large number of boards. <S> Power and Ground are in parallel. <S> Note <S> : they are not opamps, they are logic gates, specifically buffers. <S> Since each package contains 6 buffers, make sure you connect the input of the unused ones to ground or they will pick up noise and switch randomly. <S> Note <S> I didn't answer your question about grounding, the others did and I'd like to see a photo too. <S> But you really need to get rid of the fake LM2596.
| From the picture in your question it seems you are using ubiquitous non-isolated Chinese DC-DC, which (again, usually) have common ground passing through.
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Converting an AC switch signal into a reliable Digital ON/OFF signal I am working on a project to add some digital processing, using Arduino, to an old Electro-Mechanical Pinball Machine. Specifically, I want to detect the score reel triggering on the pinball machine and make a digital version of the score. When a score reel is triggered, a switch closes, putting power into the coil of the score reel mechanism. This then advances the reel one position, and the power is switched off to the coil. The pinball uses 25V AC through the switches and the coils. I have connected up and opto-isolator circuit across the switch, and so I can detect when it is powered. If I sample it constantly, of course with it being AC, I get a lot of ON/OFF's per second (UK 50Hz). I have programmed in a delay, which is partially working, but I either get too many or too few digital readings, compared to the mechanical movement of the score reel itself. Part of the problem is that within the pinball machine, lots of different targets, rollover switches, etc. can trigger a point score. These work at different speeds and switch the coil on for different time periods. Also, say, if the pinball rolls over a rollover switch slowly, that switch is on for a long time, compared to a fast moving ball that triggers the switch for much less time. So, I cannot rely on fixed timing. I need to treat one instance of the switch being truly on as one digital read (or point score). I have tried a full bridge rectifier, to turn the AC signal into DC, but this doesn't help fully either, although has improved. So I'm looking for any other ideas please. <Q> Good question. <S> Then I would basically try to model the score reel using software. <S> It will probably not be perfect, but it should be able to get pretty close. <S> I would assume that the wheel needs to be energized for X peaks before it "latches". <S> I would also assume that there's a certain reset rate, where if there are no peaks then the solenoid retracts. <S> This assumes the wheel is spring loaded to return to an integer position if it gets a partial trigger. <S> To model that, I would decrement the peak counter at a specific rate over time. <S> Once you register a trigger, just wait for the peak count to go back down to zero and then reset the whole thing. <S> Another option would be to attach some kind of encoder to the first disk of the score reel. <S> I think that's more likely to get a perfect count over a long period of time. <A> I've worked on pinball machines for 15 years, Electro Mechanicals produce all kinds of stray noise when coils energize and De-enegized. <S> You will have to deal with the noise. <S> Below is a simple schematic, the output should be from across the coil not the switch. <S> The size of the capacitor and resistor will have to be determined based on: time on <> verses ripple from the 60 Cycle. <S> I spent many weeks trying to make an electro mechanical slot machine work. <S> It was built by another company but was plagued with errors from switching noise. <S> Got it close but not prefect. <S> Electronic Pinball Machines have noise diodes, capacitors and resistors. <S> The schematic editor keeps sending the lead from (C1 to LED) to the rear. <S> It looks fine in editor but not here. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Relay isolation. <S> I would first attempt to find a quick-acting 24 V AC relay, wire that in parallel with the reel solenoid and use a debounced contact of the relay to drive the micro-controller GPIO with internal pull-up enabled. <S> Figure 2. <S> A selection of 24VAC+relays found in an image search. <S> If your counter reel can react to the shortest pulse then your mechanical relay should be able to as well. <S> I would be inclined to mount the relay close to the micro and run a twisted pair of wires to the coil to minimise any coupling between the AC wiring and the logic signals.
| Here's what I would do: First I would build some kind of AC peak detector, and maybe even give it a threshold voltage so that it doesn't detect a "peak" unless it's >15v or something.
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Efficient way to selectively unpower USB ports (Edited and expanded from original question for clarity) USB interfaces as a means of controlling external hardware from a desktop or laptop computer are ubiquitous; many thousands of hardware models are available, ranging in price from 10 -1 to 10 6 USD. It is reasonable to require that the +5V DC voltage line be able to be switched off from the host computer, but this is by no means guaranteed across every mainstream PC/laptop running every widespread OS. I am looking to have a true cross-platform solution to be able to selectively power USB ports down and up from software, for totally arbitrary lengths of time. This is not because I am prototyping or breadboarding, and nor is this my first foray into electronics making an LED blink on and off; I require a long-term reliable solution. I am looking for an efficient solution which minimizes: Cost Implementation time Other general hassles, such as volume or amount of trailing wires Based upon comments, I have thusfar ranked my options as follows: Option 1 Buy a dedicated USB splitter, with fully cross-platform proprietary software, such as the Yepkit YKUSH hub : $42 Negligble software compilation time + some time to "box up" Compact and neat Option 2 Buy a programmable USB hub : $300 Small amount of time to implement command-line software control Professionally boxed At the moment, I would be tempted to get a 4-way USB hub, splice its power lines through a separate arduino-controlled relay (as I have an arduino that need never be unpowered). Option 3 Buy a fairly off-the shelf hub compatible with uhubctl . Then either run Linux or program a way to make it work in Windows. A VM would also be a headache, because programs in Windows will have trouble interfacing with Linux programs inside the VM. ~ $20 Long time to implement cross-platform software Professional USB hub Option 4 Buy a USB hub and a Normally Closed 5V relay, controlled by e.g. an arduino to cut power to the USB hub as required: ~$25 Long time to solder and tidy connections + short time to implement switching in arduino code Fairy ugly due to splicing wires (Surprisingly not suggested in responses given that this is an Electrical Engineering site) <Q> It's certainly possible in some cases, but it depends on the hardware you're running. <S> Some USB chipsets allow "per-port power control" (the name of this feature seems to change from one manufacturer to another). <S> Your mileage may vary because this feature appears to be an optional part of the USB3 spec. <S> There is an unofficial and probably incomplete list of chips that support this feature here: <S> https://github.com/mvp/uhubctl/blob/master/README.md <S> Works for individual ports on PCIe-to-USB <S> adaptor cards too. <S> It can be run from the commandline or a shell script and you just nominate which port you want to turn on or off. <S> If your hardware contains one of the supported chips (and the manufacturer chose to implement it on your motherboard / PCIe card) <S> then that is one option that works for me natively under Linux. <A> First, not all desktop PC have individual port power control, actually very few have it. <S> You will have a better luck with external hubs, although it will be also a challenge. <S> Second, USB port power, connect, reset, enable etc. <S> port function are integrated steps of USB protocol, and are implemented at kernel level. <S> Enabling port power is the first standard function for any root (or normal) hub port, and turning it off is not in the standard USB protocol. <S> Therefore you would need the usb driver source code to implement new custom calls and functions. <S> If this is theoretically available for Linux/Android, it is highly problematic under Windows. <S> However, if this is for testing of individual basic USB functions under manual control and no "normal" USB class functionality is needed/required, there is a stand-alone software package called "USB3CV" - command verifier , available from USB.ORG, which replaces the standard Windows USB stack and allows to execute individual port commands via graphic interface. <S> This again assumes that the hardware (USB hub) has individual port power controllers, which is hard to get in first place. <S> If you need a normal USB stack functionality but want to exercise systematic power-off (VBUS off/disconnect) of connected USB devices, your best option is to get a hub with individual port power control, and hack the control wires to your Arduino controller. <S> Again, it all depends what is the purpose of your exercise, there might be some caveats in the disconnect process. <S> ADDITION: In classic USB environment with Type-A ports <S> the individual port power switching is OPTIONAL and should't be generally expected from on-the-shelf products. <S> However, for all devices with Type-C downstream connectors the individual port power control is MANDATORY by USB 3.2 specifications. <S> Therefore, Type-C is the way to go (into future). <A> On Windows, you can simulate disconnect event using utility devcon provided by Microsoft. <S> Use it as outlined in this answer . <S> However, it doesn't necessarily mean that power will be turned off. <S> On other platforms (Linux, Mac) you can use my utility uhubctl . <S> It will work and actually turn power off per each port selectively for any USB hub that supports per port power control (check compatibility list for tested devices). <S> Good news is that uhubctl supports USB 3 and many USB 3 hubs properly support per port power control.
| Uhubctl is a Linux based program that lets you turn on and off USB ports on compatible chipsets.
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What PCB Plating do I need to use Exposed Copper Pads with Pogo Pins? I am designing a PCB that will have some Mill-Max pogo pins make direct contact with exposed copper pads on the PCB. This will have many connects and disconnects, think hundreds to thousands. I know that it needs to be at least gold coated in order for this to be resistant to corrosion, but does that mean ENIG plating is enough? Or do I need selective hard gold plating? <Q> I work in the semiconductor test industry. <S> We've designed large boards, 40 layers, sometimes more. <S> And typical pcbs too. <S> We make a connection from our lager pcb to a probe card using pogo pins. <S> These connection like yours are in the thousands.. <S> Even 10k <S> .. <S> Because of the forces involved and the number of pogo pins making the connection. <S> With very good results over the years. <S> Hope that helps.. <A> I have been putting solder on my pogo-pin pads. <S> I apply the solder by hand on my prototype boards and have openings in the paste mask for those boards that have solder-paste applied with a stencil. <S> I find that the solder bumps have been far more reliable than bare pads for repeated use. <S> For those situations where the pogo-pin pads are used only occasionally, I find that whatever plating the board house uses is reliable. <S> For me, this is both HASL and ENIG. <S> But: the pads will not stand up to repeated use. <S> The cure is simple - just add a tiny bit of solder to each pad. <S> A rounded bump is all that you need. <A> That's a lot of connection cycles! <S> In that case, in addition to the plating, it is important to choose the correct pogo pins. <S> A simple sharp point will eventually wear through the plating. <S> The rounded, Mill-Max tips will last longer. <S> But there are more available options. <S> For example, these are a few of the different options from IDI <S> (now Smiths <S> Interconnect ): <S> (web link to full document) <S> The H-style tip, especially, should always have a good point of contact regardless of wear. <S> Not only will the probe find some high point on the contact, but it will rotate slightly over time and so find entirely new contact points. <S> If you couple that with @DwayneReid 's recommendation of putting solder over ENIG (or HASL) pads, it should last a very long time. <S> This combination is particularly effective because the "soft" solder will provide a good landing, even as it is deformed by multiple connection cycles. <A> Is this just for occasional (testing), or one time (initial programming) use? <S> Hard gold would be great, but overkill. <S> ENIG would be good, even the Hot Air thing (drawing a blank on the name) that iteadstudio uses as their base option would be good enough. <S> I have iteadboards that I've had quite a while <S> that look as good as when I received them, some are months old, some even older.
| We use a selective hard gold plating, exactly: Minimum 30 micro inches gold over 150 micro inches nickel.
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Power meter current transformer: possible to sum mutiple phases I am currently looking at this inexpensive power meter modules with TTL interface. It seems to use internally the same ASIC as many chinese wall-wart power meters: It comes with a current transformer for measuring the current through a live wire and has connections for sampling the voltage on the wire. My question now is this:Assuming I have a 3-phase AC setup, can I thread all three live phases (i.e. shifted 120° relative to each other) through the current transformer to measure the sum of the POWER in all three phases? My hunch is that this will not work. My primary concern is that the voltage on only a single phase will be sampled while the current transformer sees the current on three phases which will make calculating the power difficult. Also I believe on AC the direction of the current in the three phases will generally not be aligned leading to cancellations. Is this correct? Edit 1:Clarification. I am not attempting to measure a 3-phase load but multiple single-phase loads attached to the three phases. The return current goes through neutral. <Q> The current sum through all phases is zero , given you don't have neutral current. <S> If you have, things are even more complicated. <S> By multiplying with the phase voltage, the phase difference vanishes. <A> can I thread all three live phases (i.e. shifted 120° relative to each other) through the current transformer to measure the sum of the POWER in all three phases? <S> If you mean directly through the aperture of the CT well, it won't cause a safety issue but the magnetic fields will sum to zero on a balanced load/supply and you won't be able to use it to measure power. <S> Information: <S> You can run two phases through the CT aperture to get a current reference that is in phase with the voltage reference. <S> For example, if your three phases are red, blue and yellow you can route red forwards then blue in the reverse direction to produce a net current phasor that is in phase with your red-blue line voltage. <S> Maybe this is an alternative idea? <S> I've seen 3 ph power meters use it <S> but I can't vouch for its accuracy and it will need scaling appropriately. <A> The power meter interface is designed for single phase. <S> It cannot have a slightest idea what is going on, if you put three wires to its current transformer, its continues to assume all output is caused by single current. <S> Unfortunately that output from the transformer doesn't reflect at all the total 3 phase power. <S> Besides you have no place for 2 of the phase voltages. <S> In theory you can get some statistical estimate of the total power if you have a multiplexer that connects one phase (voltage and current) at a time to the system and changes the phase under measurement ater every few AC cycles. <S> But that kind of multiplexer surely costs 500% more than having one interface for each phase.
| You need one current transformer per phase and have to multiply it with the phase voltage before summing.
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Identifying FR-4 and G-10 I am evaluating a couple products -- lighting ballasts -- and noticed one of them seems to use G-10 as the PCB laminate. I assume that the beige, solder-mask-free top layer is the clue. Top: Bottom: Given that FR-4 is F lame R etardant, whereas G-10 is not, I prefer to stay away from the latter laminate. But I cannot say for sure if the second board uses FR-4 or not. Top: Bottom: It appears that the second board has more of the FR-4 look to it (that pale green), but that could just be a top-layer solder mask. Am I on the right track here? Is it even possible to tell the difference between G-10 and FR-4 like this? <Q> The top one looks like a cheap paper-based phenolic board eg. <S> CEM-1/CEM-3. <S> The outline and holes appear to be punched, which is very common for low end consumer products and power supplies. <S> It punches fairly nicely under the right conditions, is cheaper and wears the tools less than epoxy-glass laminates. <S> Eg. <S> Nan Ya Plastics CEM-1-97 <S> The bottom one looks more like epoxy-glass such as FR4. <S> You can test for UL94V0 with a Bunsen burner following the procedure . <S> Basically it's not supposed to keep burning (much) after the flame is removed. <A> Am I on the right track here? <S> Is it even possible to tell the difference between G-10 and FR-4 like this? <S> The color difference is in the solder mask. <S> Many manufacturers list datasheets with FR4 and G10 as the same product. <S> You could tell chemically if you could get an analysis done to check for bromine, as FR4 has a bromine additive to give it <S> it's flame retardant capabilities, G10 does not. <A> You could apply a flame to the corner and see if the flame self-extinguishes in 30 seconds under a fume hood. <S> There is a logo ® partially exposed in silkscreen for the ratings. <S> CEM-3 is very similar to FR4. <S> Instead of woven glass fabric CEM-3 has a milky white color and is very smooth. <S> It is very popular now in Japan.
| You won't be able to tell the difference between G10 and FR4 by look or feel, they are essentially the same from a materials perspective.
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Dual power supply with two boost converters I currently have a dual power supply, that supplies +/-15V. One -output of the PS is connected to the other channel's +output (GND). Now, the problem is that I need +/-50V. I bought two step-up boost converters (cheap ones from Amazon), but now I realise that I may have the wrong idea. The converters have +/- input and +/- output, and I connected the +15V from the PS into the +input of one, and -15V into the -input of the other. The unconnected inputs are grounded. Now, my idea was that one of them would give +50V (which it does), and the other -50V. Note that my +converter has its -input and -output grounded, my -converter has its +input and +output grounded. However, the -50V is not working, and the PS keeps getting current locked. Am I connecting things wrong, or is my idea wrong? EDIT:Sorry for the confusing explanation. Here is a rough sketch of what I'm trying to do: ( Source ) I am using this to power a high voltage op-amp, specifically OPA454. I'm not sure what "isolated" boost converter, but I have two of these from Amazon: DROK Micro DC Converter Power Transformer 120W 12V/24V/48V 10-32V to 35-60V Volt Regulator Module Boost Step-up Inverter Board DIY for Car Auto Vehicle Motor Thanks for all the help. <Q> Your circuit cannot ever work. <S> The modules provide a regulated voltage between the output terminal and -ve lead of the power convertor. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> One other way to do this is to get a Boost convertor that supplies 100V output and use an electronic ground (rail splitter circuit). <S> I've used a bunch of <S> these to provide 75V for stepper motors ... <S> but they extend to 120V. <S> You could use 30V input to reduce the input current requirement. <A> I disagree. <S> I think your circuit can work with one minor modification...remove the connection between the +/-15V supply. <S> You don't mention where you're getting +/-15 <S> from, but if it's a bench supply, the outputs are usually isolated unless you connect or ground them externally. <S> Your -50V supply is probably trying to generate 50V over a common ground between its negative input and its negative output. <S> If you're using a dual supply with a common ground, you may need an additional one--and they have to be isolated from each other, which means isolated from any common inputs as well. <S> Compared to your output ground, the -15 supply will be providing -50V on its negative terminal, and -35V on the positive. <A> Note that after clicking on Product Details <S> you get a bullet list of features. <S> Near the top, it clearly says "non-isolated". <S> That means the negative of the power input is internally connected to the negative of the power output. <S> For your circuit to work, the modules would have to be isolated . <S> That means there is no direct connection between input and output. <S> The output can float (within some range, usually a few 100 V at least) with respect to the input. <S> With such modules, you could power two of them from the same 15 V supply, but connect their outputs to make a ±50 V supply with ground. <S> In your case, you say that your ±15 V supplies are really separate (apparently isolated) supplies, and you connected them together to make a ±15 V supply with ground. <S> Don't do that. <S> Leave them unconnected, and have each power one converter. <S> It seems silly to use 15 V supplies only to power 50 V supplies. <S> Make 50 V directly, if that's what you ultimately want.
| However, the real answer is probably to get two isolated 50 V supplies that can run directly off of whatever is powering the 15 V supplies. You would need to have the lower DC-DC convertor be an isolated module (no connection from input to output) to make this work.
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How to know when to tune up a digital multimeter? I have bought the multimeter with temperature measurement from the following website: http://futurlec.com/Multimeters.shtml Currently I use the voltage, resistance, and short circuit detection functions on it. The short circuit detection function works fine. but now when I measure high resistance values over 100K, the results are inaccurate. For example, the 560K resistor was measured at 408 instead of 560 on the meter (I turned the knob to the 2000K setting). When I measured a 1K resistor, the meter reported 0.8 instead of 1.0 on the 200K setting. All resistors measured were carbon film 1/4 watt with 5% tolerance. One time when I opened up a multimeter to replace the battery, I noticed it had a fuse and a potentiometer as part of its circuit. Would adjusting the potentiometer fix this issue? I mean the low battery symbol never appeared on the meter so I don't know if battery is an issue. Could I hack the meter somehow and measure something that could give me an indication if the battery is dying? <Q> For the 560k resistor reading, if you had your fingers touching both probes, you body resistance would be in parallel with the resistor, so would lower the meter reading. <S> This effect would be insignificant with a 1k resistor, however. <A> When I measured a 1K resistor, the meter reported 0.8 instead of 1.0 on the 200K setting. <S> That's an error of \$ \frac {0.2}{200} = 0.1\% \$ of full scale on that range. <S> That's much better than might be expected on such an instrument. <S> If unsure what to expect then start on the highest range and work down so that you minimise the chance of applying an over-voltage to the electronics. <S> (This isn't important on the resistance measurement ranges as the meter itself is the voltage source.) <S> You don't know what the potentiometer is for so don't mess with it. <S> You don't have any reference to measure from anyway. <A> If all of the other ranges are reading correctly, you have a bad resistor which sets the gain for that particular range. <S> Without a schematic, there is no way to tell which it is, and I'm assuming it's a SMD part anyways. <S> So, no, the pot won't fix the problem - or if it does it will cause all the OTHER ranges to be wrong. <S> The problem is probably fixable, but not without both knowing what you're doing and having access to documentation. <S> Generally, unless you're just in it for the experience, you're better off either just not using that scale or trashing the meter - DMMs are dirt cheap these days, so why put up with the hassle?
| For accuracy you should always make your measurements on the lowest range that will accommodate the value being measured.
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Switching a divided voltage with transistor or MOSFET I have a problem that I cannot solve and need help. I have an analogue input on a microcontroller and I want to have multiple switches which the micro can see and act on depending on which switch was pressed.I have 5V that the circuit works on. So I am trying to get different voltages from 0-5 into the analog input of the micro. The catch however is that I want to control it with a p type transistor or MOSFET so that the user can just short something to ground to activate the divided voltage. I've sat here for days trying to figure this out using MOSFETs and transistors but nothing I do works! The divided values are never correct! I've been playing with BC327's and BS250 MOSFETs to try and make something work, but everything I try fails. I realize I need help, so I'm throwing this out there on the chance someone can help me! <Q> Below are two methods. <S> The simulated voltage for the BJT circuit is 2.497V, for the MOSFET circuit 2.500V. <S> With 3 base/gate resistors increased from 10K to 100K (R7 to 10K), the values are 2.495 and 2.500V. simulate this circuit – <S> Schematic created using CircuitLab <A> Your circuit won't work when the transistor base is sinking current to ground. <S> Re-position your resistor and try again. simulate this circuit – Schematic created using CircuitLab <A> There is no reason at all to use a transistor or MOSFET in a switch detection circuit like this. <S> You can simply connect as follows: <S> This takes way less components and results in easy switch detection via the ADC reading. <S> No switch pressed reads a nearly 5V value. <S> S1 produces a voltage less than 2.5V, S2 produces a voltage of about 2.5V and S3 produces a voltage above 2.5V. <S> Of course you are free to change the resistor values to suit your application. <S> I just recommend keeping them in the Kohm range so that some input impedance at the ADC does not change the switched voltage dividers very much. <S> Also keep in mind that when you detect each switch you need to check a range of the A/D readings for each to accommodate resistor tolerance and noise. <S> Edit: <S> (in response to OPs query about long wires) <S> This should work fine with wiring to tact switches mounted on a circuit board. <S> It can also work if the switches are wired remotely with wires but consider several things. <S> First there can be noise on longer wires so make sure to select the resistors such that the voltage difference from switch to switch is much larger than the noise voltage. <S> Also keep the impedance lower in Kohm range like I suggested. <S> If you used Mohm range resistors the lines will be more susceptible to noise pickup. <S> Instead each switch should be dealt with in a digital manner. <S> Digital treatment makes it more practical to: Provide ESD clamping on the signals at edge of circuit board. <S> Buffer switch signals if necessary to protect sensitive MCU pins. <S> Debounce switch signals with 1's and <S> 0's instead of varying analogue values. <S> Organize larger switch counts into a matrix to save GPIO pins. <A> Originally you said you wanted to use a PNP transistor and a few different buttons that short something to ground in order to produce an output of 0-5V to send into a single input of your microcontroller. <S> Here's one way to do that. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> [the values shown are just a rough example, you should play around with them and see what works best for your application] If you push each button one at a time, the output will change according to the resistance connected. <S> Adjusting your resistor values adjusts the output. <S> (This won't always work if you're planning on hitting two buttons simultaneously though since that would put the resistors in parallel.) <S> Alternatively, you could put Ro on the collector and take your output there, but you won't be able to use as wide a range of resistor values at the base (not sure how many different values you need at the output), and in that case R1/R2/R3's value will vary inversely to the output. <S> With the configuration shown, the output will increase with their resistance and the change will be slower (you can do a DC sweep to observe this). <S> Of course you could skip the transistor and Ro altogether and take your output directly from the resistive divider.
| With remotely mounted switches where wire may be very long or if there is concern with ESD then it would be advisable to not use this analog sorting for switch detection.
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Can I get away with that poorly designed board? I am designing a board but space is not a luxury that I have at the moment, can I get away with that design or I must use a SMD resonator? <Q> Load caps should be between the MCU and resonator, otherwise they act as an antenna and might pick up extra noise, which, depending on the operational conditions, might mess with your timings. <S> I would suggest leaving resonator "horizontally" (space-wise seems to fit) and move load caps to the side of the MCU, between resonator and MCU itself. <S> In theory some older PTH resonators could be more thermally stable than SMD due to larger die, albeit suffer from some frequency deviations from nominal due to manufacturing tolerances. <A> A good clock design keeps the traces as short as possible, and the capacitors and traces symmetrical to keep the parasitics equal. <S> A copper pour can also be a good idea on the top layer around the components to increase capacitance to ground, and shunt high frequency signals to ground. <S> The biggest problem with crystals is radiated emissions, and depending on the country your in this could be a problem. <S> It's best to use good clock design practices to avoid problems. <A> It depends on what you mean by getting away with it. <S> There are a few considerations: 1) <S> Your Xin and Xout traces are of different lengths. <S> It looks like the difference is small enough not to be an issue, but without knowing more about your application, I don't know. <S> 2) <S> If it's an HC49-U, it will really loom over your chip and make it hard to probe. <S> If HC49-S, it won't get in the way quite so much. <S> 3) <S> If you decide to switch to SMD, keep in mind that they have very different -R and pullability than HC49 (they will be worse than HC49-S, which in turn is worse than HC49-S).
| One thing you might be able to do is mount the crystal on the other side of the board to decrease spacing. You could "Get away" with a design like this.
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When are fiducials needed? On a single-layer board with no through-hole and no machine placed components, is there any need for fiducials (for example to route the board outline)? Board house has asked for permission to add fiducials to gerbers. Of course I will say yes, but I am surprised. I definitely don't need the fiducials for assembly. But the vendor seems to be saying that the fiducials will assist in cutting the board outline. <Q> Fiducials are required when you use any kind of computer vision. <S> You need them for vision assisted drilling, pick and place, but also Automated Optical Inspection (AOI). <S> Basically any machine that automatically recognizes the board position and angle. <A> According to this application note from norcott, three global fiducials are used to determine the correct orientation of the board (or panel). <S> Relevant quote: <S> It is important that only three fiducials are used. <S> This ensures that if a panel is accidently inserted into the placement equipment rotated through 180°, the equipment can detect it and halt assembly. <S> And for a single board without a panel, those fiducials are placed on the board, and will show up in the end: (images taken from the application note) <S> There is no specific mention of routing the board outline in that document, but I can imagine that they use this also in the routing machine to make sure that the board has the right orientation. <A> You don't know their process, and just how many things they have automated, requiring machine vision, for processing and maintaining the quality of their boards. <S> Perhaps the most important thing is that they want to process your boards through their standard flow. <S> If your board is a 'special', so it doesn't need this station, and it can't be put through that station, then it has to be handled differently, and it will cost them 10x more. <S> If their outline router uses fiducials, they will want to add them. <A> Fiducials are also used to line up stencils for applying solder paste.
| The fiducials are the calibration point for the vision system and the board coordinates.
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Why does this DC motor have such a wide range of power? This DC motor has a power range of 9.0W-300W, while its voltage range is from 6 to 20 volts. Does this mean that its internal resistance is somehow dependent on the load? Or does the user determine the actual power with using a resistor in the circuit? Performance Table MODEL ---- VOLTAGE ---- -- NO LOAD --- ---- AT MAXIMUM EFFICIENCY ----- ----- STALL ------ OPERATING NOMINAL SPEED CURRENT SPEED CURRENT TORQUE OUTPUT TORQUE CURRENTRZ-735VA RANGE V r/min A r/min A mN·m g·cm W mN·m g·cm A9517 6 - 20 18 20400 2.8 17990 20.9 149 1523 281 1265 12895 156 <Q> This is true of all motors. <S> When a motor is lightly loaded, it generates a high level of back EMF, which opposes the flow of current. <S> You can think of this as a high effective resistance, although the actual resistance of the coils doesn't change. <S> What is actually changing is the net voltage (= source voltage - back EMF) across that resistance. <S> When a motor is heavily loaded, the back EMF is reduced, allowing more current to flow. <S> This decreases the effective resistance by increasing the net voltage across the internal resistance of the motor. <S> 1 <S> The wide range of power ratings for this motor indicate that it is both efficient (low power consumption at low loads) and robust (can handle the current associated with high loads). <S> The methods for controlling a motor depend on how you're using it. <S> If you're primarily interested in controlling its speed, you regulate the voltage, allowing the current to vary (within limits) with the load. <S> If you're primarily interested in controlling its torque, you regulate the current, and the voltage varies with load. <S> A resistor is NOT a particularly useful way to accomplish either of these. <S> 1 <S> With AC (e.g., induction) motors, the situation is a bit more complicated. <S> The magnitude of the back EMF doesn't change as much as its phase relationship to the source voltage. <S> This still has the effect of increasing the net internal voltage. <A> Think of the motor as of a device that is meant to convert electrical energy into mechanical energy. <S> If your mechanical load needs a lot of energy, the motor needs a lot of electrical energy to generate that. <S> That's why it doesn't have a fixed power rating. <S> Does this mean that its internal resistance is somehow dependent on the load? <S> Yes. <S> You can't really use a resistor model for a motor, because its resistance varies due to a lot of factors : mechanical load, input voltage, etc. <S> Or does the user determine the actual power with using a resistor in the circuit? <S> No. <S> Using a resistor in any power electronics application is generally a bad idea. <S> Resistors turn electrical energy into heat. <S> That's a massive waste of energy unless you are making a heater. <S> However, you can use a motor speed controller that basically varies the voltage given to the motor which in turn changes the power delivered to the mechanical load. <S> Now some extra info about the datasheet that you provided, just to make things clear: <S> In the performance table, you have a current of 2.8 A under no load. <S> That's what the motor draws at the nominal voltage of 18 V if there is nothing attached to it. <S> There is a certain operating point (dependent on the load) for which the motor gives the maximum efficiency, which is also shown in the table : 149 mNm of torque for which the motor draws 20.9 A. And finally, if the motor is stalled, meaning that it cannot spin at all, it draws 156 A. <A> A typical DC brush motor will behave like an ideal motor-generator in series with a resistor and an inductor. <S> Ideal motor-generators may be characterized by either of two parameters (either of whose value would be implied by the other): the amount of voltage required to produce a certain rotational speed, or the amount of current required to produce a certain amount of torque. <S> At any moment in time, the voltage on an ideal motor, scaled by the appropriate parameter, will match its rotational speed, and the current, scaled by the appropriate parameter, will match the applied torque. <S> These relationships hold in both directions. <S> Any time the voltage of a practical (as opposed to ideal) motor doesn't match what the rotational speed would imply, there will be some voltage drop across the modeled series resistance and/or inductance. <S> Voltage drop across the series resistance will convert electrical energy into heat, but that and mechanical friction on the bearings are generally the only two aspects of the motor that would cause the amount of electrical energy flowing in (or out) to differ from the amount of mechanical energy <S> being put out (or fed in).
| Yes, of course the effective resistance of the motor depends on the load.
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Need help choosing a power supply for a breadboard Hey i want to (in the long term) start experimenting with remaking modules from old analog synthesizers. For example remake an oscillator from a Minimoog or a chorus from a Roland Juno-60 or something for my own pleasure. And i figured that i would just start out by experimenting on a breadboard. But i have never used a breadboard so i don't know that much about them. I know how it works when there is power put into it. But i don't know how to power it. I can only really find power supplies that supply 3,3/5V and that is too small a voltage for what i want to do. I think i need something that is variable and that can at least output 30V. It is important that it can send out different voltage and that it can be used with a MB102 breadboard. What could be a good solution for me? If you have any tips for using a breadboard for something like this or if you just have some tips in general then it would be greatly appreciated. Thanks in forward <Q> They come with cables that have a banana plug on one end and an alligator clip on the other. <S> You can strip a short piece of wire to go between the hole in your breadboard and <S> the alligator clip to provide power easily. <S> Bigger breadboards have binding posts which is a round connector you can use for power, that also accepts a banana plug. <S> That way you can use a cable with banana plugs on both ends to conveniently route power from your power supply to the large bread board. <S> Here's an example of one: https://www.adafruit.com/product/443 <S> For your application you may want to pick a power supply that can provide the maximum voltage you need (say 30V) and then add voltage regulators on the breadboard to provide lower voltages as needed, such as 5V or 12V that are derived from the 30V source. <A> What could be a good solution for me? <S> Choose a circuit that you wish to build. <S> Examine the schematic to find the voltages required to power the circuit. <S> Estimate the maximum current drawn by the circuit. <S> Find one or more mains powered power supplies with the required output voltages and adequate current rating. <S> This might be a split rail ±15 V supply for a synthesizer module or a 9 V supply for a guitar effects unit. <S> Variable power supplies are very useful for the experimenter but more expensive - especially the split-rail type. <S> The breadboards have power rails along the edges - usually two on each side. <S> Normal practice would be to have a GND (ground or 0 V) rail on each side and positive on one side and negative on the other. <A> I would build my own power supply for this because it would be a lot cheaper than a lab power supply and it would meet my custom requirements. <S> I would start with a wall adapter that provides two isolated voltages (from a transformer with two secondaries) that provides a few extra volts more than what I need to compensate for the drop on the regulators. <S> The two secondaries are necesary for a symmetrical power supply, which is probably what you want so that you don't have nasty offset resistors and stuff like that. <S> If I needed fixed regulated voltages, I would use LM7812 and LM7912 for +-12 V. <S> For and adjustable power supply <S> I would use LM317 and LM337. <S> I would build the whole thing on a perf board so that it plugs right into the power rails of the breadboard using pin headers to power the rails and screw terminals for the inputs from the wall adapter. <S> I would also avoid switching mode power supplies because they might add high frequency ripple. <S> Linear regulators would be a good choice for such a project. <S> Anyways, the project idea sounds really nice to me <S> and I've played with similar things myself. <S> Good luck.
| You will want to invest in a "lab power supply" or "benchtop power supply" which is typically a single or dual channel adjustable supply, such as this one: https://www.sra-solder.com/korad-ka3005d-precision-variable-adjustable-30v-5a-dc-linear-power-supply-digital-regulated-lab-grade
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How to fix a PCB inside this enclosure? This is a related question to How to fix this in an enclosure without glueing . However, for a protoboard/PCB, how could I attach it to the enclosure? The bottom of the enclosure has holes for attaching, but it doesn't have screw thread. How should these holes be used and how can I attach a PCB? Glueing is not what I want, in case I want to add/change components on the protoboard later. <Q> You buy the complementary inner panel/base plate to mount your products on. <S> It will fit matching the mounting holes, and it will arrive with the correct set of fasteners. <S> Or you could just make a PCB matching the mounting holes. <S> images from Hammond MFG. <A> The small standoffs molded into the bottom of that plastic box can be used with the proper sized metal machine screws. <S> The screws will cut their own thread in the soft plastic. <S> There are several things to consider: Make sure to select the proper diameter machine screw. <S> Make sure to select a machine screw that is not too long. <S> If too long it can bottom out in the hole and strip (see next item) or even make the plastic in the other side have a bump out. <S> Use care to not over torque the machine screws. <S> It is relatively easy to strip out the holes. <S> When re-mounting an item and re-using a mounting hole that was previously threaded use care to try to get the screw thread to reengage the existing threads instead of trying to form new ones. <A> Figure 1. <S> Relative size of box-top screws and PCB mounting holes. <S> If those are M3 screws for the lid then it looks like the mounting posts are about 3.5 mm outside diameter and maybe 1.5 mm inside diameter. <S> You're looking for a short self-tapping screw with a 2 mm thread. <S> Figure 2. <S> A selection of 2 mm self-tapping screws. <S> If you have a set of drill bits you can use these as a hole gauge. <S> I use a set with 0.5 mm size increments. <S> The screw thread should be probably about 25 to 33% larger than the hole. <A> I would like to offer a generic solution that I have found highly useful because it does not depend on the type of project box and does not require that the PCB have mounting holes in specific locations. <S> This approach allows you to mount any PCB in any project box with a minimum of fuss. <S> 1. <S> Get (or make) standoffs tall enough to position your PCB at whatever level you prefer. <S> My local hardware store has hollow standoffs in nylon and aluminum, but I've used wooden dowels and even drilled-out sawed-off lengths of pencil on occasion. <S> 2. <S> Any screws that will grab the standoffs are fine. <S> Self-tapping sheet-metal screws are great, but sometimes an existing PCB doesn't have room for the heads on those screws, so I often use smaller machine screws. <S> If you buy pre-threaded standoffs, obviously you need to match the screws to the threads. <S> 3. <S> Put your PCB (with standoffs attached) into the project box and move it around until you are happy with the position. <S> Mark the position (or just eyeball it). <S> 4. <S> If you want the ability to pull it apart and re-position things fairly easily, you can use a glue gun. <S> If you want things to be really secure, use 5-minute epoxy or a 2-part polyurethane glue. <S> If you want an instant set, use superglue with a spray accelerator. <S> Or just use whatever you have on hand. <S> 5. <S> Put the glue on the bottoms of the standoffs, and put the PCB into the location you have chosen. <S> Leave it until the glue sets. <S> DONE! <S> Any PCB in any position in any enclosure. <S> I have a 3D printer and I print my own enclosures, but I don't bother to print standoffs any more because it's not worth the hassle to get the holes aligned perfectly with PCB holes <S> that may not be evenly spaced. <S> It's just too fast and easy to mount standoffs on the PCB and then glue them in place. <S> Hope <S> this helps somebody!
| Get screws that will grab the standoffs you've chosen, and screw them through the PCB mounting holes into the standoffs. Get some fast-setting glue.
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Do I need a heatsink for a MOSFET I'm currently working on a project where I need to be able to switch 7m of 9W/m LED strip (connected from both sides) using an Arduino with PWM . So the power of the led strip is about 63W at 12V. I have a plenty of IRF540N mosfets lying around but I need to calculate if I need a heatsink, or not. The intended circuit looks like this: Note that I don't want the high current to flow on the tiny PCB traces, that's the purpose of the screw terminal. I know that a logic level MOSFET would be better, but if I understand the datasheet correctly, IRF540N should be able to switch even much higher currents with 5V at the gate than I need. Or would using e.g. IRL540N make any difference? Do I need a heatsink? If so, how to choose one properly? Is there anything else I should be worried of considering my circuit?Thanks in advance! <Q> The simple answer is that you really do need a heat sink. <S> Take a look at figures 1 and 2 of the data sheet . <S> At a 4.5 volt gate voltage, a typical Vds at 5 amps is about 0.6 volts for a case temperature of 25 C. 0.6 volts times 5 amps is 3 watts. <S> As Linkyyy pointed out, the nominal maximum thermal resistance is 63 degrees/watt, so this suggests a 190 degree rise, for a nominal temperature of 215 degrees C. <S> If you now check figure 2, you'll see that at 175 degrees, for the same Vgs, Vds is now about 0.8 volts, and power is now 4 watts. <S> This is not a good trend. <S> For that matter, figure 9 establishes that the absolute maximum junction temperature is 175 C, so <S> you know you're looking at trouble. <S> And all of this is predicated on using typical values. <S> Trust me on this, if there is one thing you should learn it is never, ever, to design a circuit using typical values for critical functions. <S> Always, always, always use worst-case. <S> Murphy's Law reigns supreme here. <S> Granted, since this circuit isn't failing by an order of magnitude, or anything like that, you probably don't need all that much of a heat sink to stay safe. <S> But you do need one. <A> The IRF540 can have a gate threshold voltage of up to 4V according to the datasheet, so it would probably not be able to supply enough current for the LEDs to run close to 12V. Actually, taking a closer look at the datasheet for the IRF540 , figure 3, it seems that it would be more than adequate of driving a 5A load at 5V gate drive. <S> I would try this if i were you. <S> The heatsink is still necessary though! <S> The IRL540 on the other hand is more suitable as it is a logic level MOSFET. <S> According to the datasheet, the RDS(on) is around 77mΩ. 63W at 12V equals 5.25A, lets just say 5A as there will be a small voltage drop across the MOSFET, and that will lower the current in the LEDs. <S> The dissipated power in the MOSFET will then be: \$ P_{dissipated} =RDS_{on <S> } <S> *I^2\ <S> = 77mΩ <S> * 5 <S> ^ <S> 2 = \textbf{1.9W}\$ <S> According to the datasheet, the junction to ambient thermal resistance is 62 °C/W <S> This means the MOSFETs temperature will increase with <S> \$1.9W * 62 <S> °C/W = <S> \textbf{117°C} \$ from ambient temperature. <S> Assuming 25 <S> °C degrees ambient, that will result in 142°C . <S> That is still in spec assuming ideal condition.. <S> But you know, the world is not ideal.. <S> EDIT : I forgot to take into account that the on-resistance has an significant increase with temperature, so you should definitely put a heatsink on it! <A> The IRF540 switches on well when the gate-source is 10V. Some of them switch with less voltage <S> but others simply get very hot and cause dim LEDs. <S> The datasheet of a IRL540 show it heating with a maximum of 2.2W when driving your LEDs slowly so a little heatsink will be fine.
| A Mosfet has a high gate capacitance and an Arduino cannot charge and discharge it quickly which causes heat in the Mosfet if you have it switching at a high frequency.
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Very precise position motor recommendation I've been searching for a suitable motor for my solar panel and I need two motors to rotate to two angle that is solar azimuth and altitude. Since it involved a very precise angle calculation and so I hope I can find some motor that suit the project. My tracker is much more smaller compare to the others. Any motor that can recommend to me? I hope the price can be a lot cheaper. <Q> Since it involved a very precise angle calculation <S> Exactly what do you mean by "precise"? <S> Keep in mind that the power from a solar cell is proportional to the cosine of the error angle - that is, maximum power is produced when the panel is perfectly perpendicular to the incoming light. <S> So how does error affect you? <S> The cosine of 1 degree is 0.99984. <S> A one degree alignment error will reduce panel power by .00016, or about .016%. <S> As a matter of fact, an 8 degree error will only drop your power by 1%. <S> Is this really cause for "a very precise angle calculation" ? <S> Another thing to consider is that, in principle, you need rather high torques available to deal with balance/wind loads. <S> Both torque and resolution are multiplied by the gear ratio. <S> For that matter, unless you can justify angle precisions much less than 8 degrees, you can couple a rather coarse encoder to the panel rather than the motor, and not worry about losing track of rotation counts. <A> The "best" well, simplest and most effective for the cost that I saw was the use of two small solar panels (2" * 3") mounted at 90 degrees to each other with a jam jar filled with water acting as a lens. <S> The panels were wired in series but the two negatives connected together and the two positives drove a small motor which was geared to move the panels. <S> Whichever panel had the greatest output caused the motor to rotate, as this assembly was mounted with the support system it would find the "balance" point. <S> Small diagram: If you check out the Homepower magasine - they have had a lot of different solutions to this with some very good or at least interesting ones. <A> If you use direct drive, you need a BLDC with a high resolution encoder, no doubt about it. <S> Although you should also admit that the panel precision is not like you are pointing to a star. <S> If you use gears, i would still go to BLDC, maybe even with encoder after the gear. <S> But it's a very common practice to use 200 steps stepper motors. <A> See this answer regarding positioning precision. <S> Also, you have to use some of the generated power for your tracking system. <S> If your panel is small it might be comparable or even bigger than the increase you get by tracking. <S> Finally, MPPT provides comparable efficiency increase while requiring much less power to operate. <S> This would be your best option. <S> With all that said, if you still want to use solar tracking, then choose motors by their efficiency, not precision. <S> You will get your precision with high gear ratio, as @LongPham commented earlier. <S> Also some brushless motor controllers give you an access to encoder feedback as internal counter. <S> You can use that for precise positioning (since it is read before the gearbox). <S> I also recommend taking a look at Roboteq controllers. <S> Not only they support encoders and have built-in closed loop position mode, they also allow you to write your own script, so you might be able to implement your entire system without any additional parts, just motor controller and two brushless motors.
| Since you don't need to do high-speed slewing (worst-case is 15 degrees per hour), the obvious approach is a rather small motor with a cheap, low-resolution encoder coupled to a rather high ratio gear box.
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Not enough pins, looking for a "demultiplexer" that keeps state I want to control a simple display with a microcontroller. The display has 8 input pins, the microcontroller only 6. Is there an IC that I can interpose that lets me set the state of each of its output pins and keeps it that way requires the least amount of input pins necessary Something with a serial bus maybe? I know that typically display controllers are used, but are there general purpose devices? <Q> What you are looking for is a GPIO "expander" chip. <S> The Microchip MCP23017(I2C)/MCP23S17(SPI) <S> is just one example that I have used in the past. <S> There are many others to choose from. <A> A typical shift register is 74HC595. <S> What you pass is a byte containing 8 bits, thus 8 signals. <S> You only have to store that byte (8 pins) in memory which only cost one byte. <S> For this you only need 3 pins. <S> Except for GND/VCC you only need 3 pins. <S> You can even daisy chain 4 of them, controlling 32 pins (which cost 4 bytes to store) and still use only 3 pins (except GND/VCC). <A> You most likely want either a "Shift Register" or a "Latch". <S> Shift registers allow you to load data serially, and are commonly used with MCUs - 74HC595 is a highly common one. <S> Latches allow you to use a control signal to hold a value. <S> These are used a lot in memory access where you want to latch an address - for example if you want to load a 16-bit address from an 8-bit bus, you can load the upper half, latch it, and then load the lower half. <S> You could accomplish an 8-bit bus using two 4-bit latches. <S> Each latch could have its own control signal using an additional 2 pins. <S> That gives 6 in total. <S> You could do it also in 5 pins by using an inverter such that one latch is holding its value while the other is transparent.
| There are many available, typically using I2C or SPI to connect to the microcontroller.
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Voltage drop on diode (0.5V ---- 100R resistor --- diode --- gnd) How do you figure out voltage drop on this diode? (Lets assume forward voltage of this diode is 0.7V) <Q> You can measure diode current right down to zero volts. <S> However, below the magic figure of 0.7v (for silicon), it tends to be very small. <S> It depends on the type of diode what the current is. <S> Here is a picture of some measurements I did a while ago to answer a related question. <S> Do click on it if you want to read it. <S> What surprised me is how bad the 3v zener was at stabilising a voltage. <S> So how do you figure out the diode voltage? <S> For a first order approximation, we know the diode current is going to be 'very small' (as the voltage is below 0.7v), so we can neglect the resistor voltage drop, and say the diode voltage is about 0.5v. <S> For a second order approximation, you can see from the figures above the current for my 1N4148 at around 0.5v at whatever ambient temperature it happened to be was about 100uA. <S> That dropped over 100 ohms is 10mV. <S> So with more accuracy we can estimate the diode voltage as 490mV. <S> Interestingly, the figure for a much bigger 1N4006 turns out to be very similar. <S> There's absolutely no point going to a higher order approximation without specifying the diode, and the temperature. <A> Diodes have a current-voltage relationship. <S> (see image) <S> To make a long story short, a diode needs an amount of positive voltage applied to "turn on" the diode and let current flow through it. <S> In your example the current flow is I= <S> U/R = <S> > <S> 0.5/100 = <S> 5mA.The supplied voltage is 0.5V. <S> There is always some voltage drop across the resistor (but that is negligible) so to answer your question <S> , in this case the voltage drop is 0.5V. <S> How do I come to this answer? <S> It's by looking at the i-v characteristics: <S> (Image source: https://learn.sparkfun.com/tutorials/diodes/real-diode-characteristics ) <S> This characteristic is saying the following: In order to let a current flow through a diode, a certain voltage has to be applied to the diode which is in the rule around 0.7V. <S> So you have to look at the voltage and following from that, you can read from the graph the amount of current that can flow through it. <S> Since in your case only 0.5V is applied, it is all used by the diode and only a small amount of current can flow through it. <S> If the supplied voltage would for example be 5V, then there would be a voltage drop of 0.7V and you would measure 4.3V after the diode. <S> There are equations to calculate how much current exactly can flow on certain voltages, but since this is an exponential signal it is hard and not advised to use in real life applications. <S> The reason for this is that if you supply 5V, the voltage is never always 5V; it moves up and down because of noise and of course can be stabilized by capacitors and other circuits, but making it perfect for real life applications is hard and expensive. <S> I hope I could be of some help for you :) <S> To help you in the future, LTSpice is a great program for these question to figure out yourself. <S> You should take a look at this program. <S> It is free to download and massively used by engineers <A> To answer your question, what is the diode voltage drop with Vin=0.5V with 100Ω in series? <S> My answer 430mV with 70mV drop on the resistor from 700uA for an ideality factor of 1. <S> You cannot figure it out without some assumptions on the Ideality Factor or knowing the current in the diode. <S> more <S> (@analogsystemsrf previously illustrated if Ideality Factor=1, then the diode voltage rises <S> 58mV/decade in current <S> @ <S> 25'C. <S> The important thing to remember is that it is continuously logarithmic over at about 4+ decades of current. <S> For simple applications, this log. <S> characteristic may be limited by noise at the lower limit and linear behavior from internal resistance, Ri at the rated current , although not often stated in datasheets. <S> From the datasheet curves, the log <S> Vf <S> / <S> If slope and Ri may be computed near rated DC current. <S> After repetition, you remember for future use he values for 2 common diodes; 1N4448 <S> rated for 0.1 <S> Adc @ 1.0V ... <S> Vf/If= 113 mV/ decade <S> (If), Ri= <S> 600 <S> mΩ <S> 1N4005 <S> rated for 1.0 <S> Adc @ 1.0V ... <S> Vf/If= 140 mV/ decade <S> (If), Ri = 66 <S> mΩ <S> Notice that the current rating and Ri product are similar. <S> This is not coincidence. <S> (meaning high the current rating , lower the Ri internal resistance. <S> The higher current diode here also has a higher Vf/ <S> If slope. <S> Now if you put a sawtooth current source into a diode to plot voltage by using a large series resistor and voltage, you get this. <S> So in short, we use 0.7V for diode drop for convenience. <S> But it is hardly a hard switch , but still very effective. <A> Here is a simulation of diode current from 50mV to 500mV. <S> As you can see, it's a continuous function. <S> I've used a logarithmic Y-axis scale to make it a bit more clear. <S> Generally the algorithms used in SPICE require the functions to be continuous and differentiable in order to converge. <S> Things like step functions are typically approximated with functions using tanh(), for example.
| So in summary, the amount of voltage drop that is possible across the diode, determines how much current can flow through it. In addition to other fine answers, looking at common Silicon diodes with Ideality Factors other than 1.
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Replace 3.5 mm AUX with USB This might be a touchy subject, but bear with me. I have a 2.1 speaker set which has typical 3.5 mm AUX input and output. The problem is that sometimes the left and/or right audio channels cut out and I have to jiggle the cables to get them to work again. I've tried duct taping the cables in certain positions, but the problem always reoccurs eventually. I tried buying a better quality aux cable, but the problem seems to be with the female connectors on the subwoofer. It's a massive annoyance. It is for this reason that I am a shameless supporter of the death of the 3.5 mm connector. I know, I know. Hate me if you must. Call me whatever names come to mind. All I know is that I am currently listening to Zeppelin's 'Ramble On' with one audio channel and it's torture. I think I may have a solution, but I'm not sure if it's viable. On the inside of the sub is a PCB with a couple of female 3.5 mm jacks soldered on that look like this: I am wondering if I can remove them and replace them with something similar to this: Is it just a matter of soldering certain connections together? Or would I need a DAC and/or other components? What gets soldered to what? It's a pretty cheap speaker system so if something breaks I'm not too worried. I have very little electronics experience, but I would like to make a small project out of this anyway. I appreciate any tolerance of my unpopular views on tech. Go easy on me. <Q> Assuming this is a line-level signal, sure, you could replace the 3.5mm jack with a USB connector. <S> But why? <S> It would likely be a lot simpler to simply repair or replace the existing jack. <S> It my simply be a bad solder joint to the board. <S> If the jack itself is bad, you can probably find an exact replacement with a bit of looking. <S> Even if you can't find one that fits the existing footprint, it's still probably going to be easier to fit a new 3.5mm jack even if you have to do a bit of bodging than to try to fit a USB connector there. <S> The 3.5mm TRS connector has some drawbacks, but it's been the standard for this sort of thing for a long time with little trouble. <S> A bad experience with one device doesn't really warrant such an extreme response. <S> If you do have repeated problems with cables failing, you might want to look for a cable with a right-angle connector at the speaker end. <S> This may reduce the strain on the connection and improve the lifespan of both the cable and the speaker connection. <A> Using a standard usb connector or cable or a non usb system is asking for trouble. <S> Yes you could use the connector but just don't. <S> Last thing you want is somehow plugging it into a live usb jack and something breaking. <S> You could add a usb audio card internally, connecting <S> it's analog out to the speakers audio input, and connect it to your computer using a standard usb cable. <S> This would bypass the need for the 3.5mm connector and cable. <S> Or replace it with a Bluetooth audio receiver. <S> Same exact change. <A> If you're determined to replace the connector, change it for something more durable, like a binding post ( https://en.m.wikipedia.org/wiki/Binding_post ) <S> If you're looking for something compact a USB could work, I don't think you'll cause any problems if you one day plug your memory stick into it :) <S> - the process would essentially be to work out which connectors on the board are L R and GND, connect them with wires to the connectors on the USB (the USB module's connectors are in a different place to the jack <S> so you'll need wires for reasons of flexibility) and ensure when making up your USB-to-jack cable that you wire up to the same pin. <S> You arbitrarily decide this order when soldering the USB socket in. <S> Just bear in mind that the connectors of a USB plug are upside down relative to a USB socket when connected (if you were to dismantle both, hold them as if you were about to connect them but then turn one of them over so you can see the gold connectors on both plug and socket, the rightmost connector on the socket contacts <S> the leftmost connector on the plug - this is different to a jack connector where "furthest in connects to furthest in" etc <S> For my money I think I'd try use an Ethernet style RJ45 first because making the patch lead up would be easier/require less soldering and heat shrinking (but then I already have a crimp tool)
| The other option is replacing the audio jack connector.
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is a voltage regulator required for atmega328p when powering from boost converter at a set voltage? I know that linear voltage regulators are more accurate at holding the required voltage but the boost converter i am using seems to keep a pretty steady one. the voltage will of course be within the chips requirements. what would be the main drawbacks of not using one? <Q> is a voltage regulator required for atmega328p when powering from boost converter at a set voltage? <S> No <S> it is not. <S> The ATMega is "just" some logic circuits in a chip. <S> So is the CPU in your PC and in your smartphone. <S> Both of these are powered directly from switched regulators (usually buck converters but that doesn't matter that much) and they work fine! <S> As long as you keep the supply voltage of the ATMega chip within the specified limits (see the manufacturer's datasheet, from my memory it is 1.8 V to 5.5 V) <S> then it will work as long as there is not too much variation (ripple) on the supply. <S> To keep the ripple in check, any boost, buck and linear regulator has at least one decoupling capacitor across its output. <S> The ATMega chip also needs it's own decoupling capacitor. <S> It is good practice to place at least one near each supply pin. <S> The accuracy of the supply voltage is not that relevant . <S> The logic will work just as well at 3.456 V as it will on 3.3000 or 5.0000 Volts. <S> The only exception can be if you use the build-in ADC of the ATMega. <S> This ADC can use the supply voltage as its reference voltage. <S> If the supply is not accurate then the ADC's output data will not be accurate either. <S> There is a build-in more accurate reference for the ADC that can improve this. <S> You can also use an external reference voltage if you like. <S> Also if you would use the PWM output function, filter the PWM signal and want it to be accurate then also it will be as (in)accurate as the supply voltage. <S> But in general, if you need this then you would know and take precautions. <S> Oh <S> and there is no reason to assume that switched regulator (boost or buck) are in any way less accurate than linear regulators. <S> It depends on the model (chip, circuit) <S> you're using and if you calibrate it or not (measure the voltage with a known accurate meter and then adjust to the right voltage). <S> Both linear and switched can be as bad or good as the other. <A> If you just have an all digital circuit, there are little drawbacks (some ICs do not react very well to voltage ripple, but those can be fixed with a RC combination on the supply usually). <S> You should make sure it starts fast enough for all the chips. <S> This can be improved by the linear regulator, but there are also other ways to accomplish that (voltage monitor). <S> If you are planning on using the ADC (mixed analog and digital design), you might want to evaluate if the readings are within your required range of noise. <S> Even then you can use the boost converter to power the digital part and the linear regulator for the analog one. <A> One thing which might be an issue is also the ramp up of the voltage for the chips. <S> You should make sure it starts fast enough for all the chips. <S> This can be improved by the linear regulator, but there are also other ways to accomplish that (voltage monitor). <S> I would also enable the Brown Out Detector in the microcontroller. <S> That will keep the chip in reset state until the power supply voltage gets high enough. <S> There are a couple of voltage levels that can be selected: Off, 1.8V, 2.7V, and 4.3V typical, +/- <S> 0.1 to 0.2V.
| One thing which might be an issue is also the ramp up of the voltage for the chips.
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Can you use a respirator for soldering? I know that breathing in solder fumes is not really good for you. I happen to have some respirators lying around. Could I use this instead of a fume extractor? It might save me some money if I could just use the respirator? <Q> Sure, especially the ones with charcoal built in. <S> However, in all the years I've been in this racket <S> I've never seen a fume extractor or a respirator actually used. <A> If the air filter is not expensive, sure you can use it with suitable air flow. <S> Solder fumes are entirely vaporized flux (evaporating volatiles). <S> Rosin core is distilled from acidic tree sap which is basically the nutrient and immune system of trees but its properties are suitable since it melts with solder heat, reduces solder surface tension and the coating over copper reduces oxidation. <S> * <S> *Do not accept <S> ** that not all people react negatively to these fumes, but limited exposure may not be harmful. <S> Although it may smell nice of pine smoke, there may be trace amounts of nasty formaldehyde and toluene. <S> Rosin is a subset of Resin acids that are made from different mixes including organic acids. <S> Inorganic strong acid flux (HCl) is used for non-electrical solder work. <A> I simply take a deep breath before starting to solder, and then exhale a steady slow stream across the work, which blows the stuff away from my eyes, and reminds me not to breathe in. <A> I don't think solder flux fumes are particularly bad for you, unless you are allergic to them or something. <S> I'm 50+ and have been soldering with tin/lead solder for years, with no noticeable health affects.
| The solder itself only melts, it is not vaporized, so you're not breathing in any harmful metal vapors. I find flux fumes pretty unpleasant when they're hot and concentrated straight off the iron, but less of a problem once they're cold and diluted in the air.
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How can I tell if a MOSFET is enhancement-mode or depletion-mode? Today, from ignorance I have fallen head-first into the world of MOSFET transistors. In my scramble to find some information on the MOSFET I will be using as a switch (HEXFET actually), I learned that MOSFETs in general come in two modes, enhancement mode, or depletion mode. When I tried to find out which mode the IRF3710 was, from the datasheet , I found that it does not say (or maybe I need glasses). At this point I started searching to find how to tell the difference between the two modes. After some time I gathered that the schematic symbols differ: Enhancement-mode MOSFET: Depletion-mode MOSFET: The difference being the highlighted part below. Three separate lines means enhancement-mode (left) and one solid line means depletion mode (right). So, my question: Is this the only way to tell which is which, or is there a quicker way to tell (by markings on the device maybe?). Also, are there symbols out there which use a different method to differentiate between them? I am asking here for my own learning, but also for other people who might have the same experience as me. I did not find that much helpful info in my searching. <Q> Two things I want to add to the answers already given: <S> Don't trust the schematic symbol. <S> You'll see the depletion-mode symbol used pretty often for an enhancement-mode part because it's easier to draw. <S> (The symbols suggested on the manufacturer datasheets won't make this error, but some random application circuit schematic from the web is not trustworthy at all) How to tell from the datasheet whether the part is enhancement mode or depletion mode. <S> For an n-channel FET, if the \$V_{gs({\rm th})}\$ is greater than 0, then it's an enhancement mode device. <S> If \$V_{gs({\rm th})} < 0\$ it's a depletion mode device. <S> For p-channel, it's the opposite: \$V_{gs({\rm th})} < 0\$ means enhancement mode, \$V_{gs({\rm th})} > 0\$ means depletion mode. <A> Rather than look at symbols you need to examine the device datasheet. <S> An Enhancement mode FET such as your IRF3710 will have a couple of VGS = 0 characteristics with a Vds current that is very low, usually this is the Vds breakdown voltage or Vds leakage current specification. <S> There will also be a VGs(threshold) specification, which is the beginning of conduction (starting to turn the FET on): <S> For a depletion mode FET such as the DN2625 <S> the same leakage current specifications will show a non-zero value for VGS (the device must be turned off to measure the breakdown or leakage currents).The inverse of the VGS(threshold) to turn on the enhancement mode FET is the VGS(threshold) required to turn off the depletion mode device <S> (be careful here, in that the value is that required to reduce ID to the leakage current, not the threshold to begin lowering the current): <A> You can tell by looking in the datasheet at the numbers. <S> Usually it will say up top, because depletion mode is relatively rare, but that's not always true. <S> For example, small VHF Mosfets are often depletion mode and that's sometimes not mentioned. <S> For a depletion-mode MOSFET the Idss will be relatively large, and the cutoff current will be relatively small and specified with a negative voltage for N-channel and positive for P-channel. <S> Below is a small N-channel device (as far as I can tell, discrete P-channel depletion mode parts are not manufactured). <S> For the much more common enhancement mode MOSFET, the Rds(on) and Id(on) will be specified with Vgs positive for N-channel and negative for P-channel, and Idss will just be the leakage current, and relatively small. <A> Here is an example from Digi-Key : <S> AFAIK, there is no common convention on how the depletion-mode transistors are designated, all is vendor-specific. <A> Is there another way to tell which is which based on the schematic symbol alone? <S> No, not really. <S> You can also deduce it from the context: <S> If there's a part number, look up the datsheet. <S> Can the circuit even work with a depletion mode device? <S> Depletion mode MOSFETs are on with zero gate-source voltage, which is unlike enhancement mode MOSFETs (be they N or P channel). <S> Depletion mode MOSFETs are really rare. <S> At the time of writing, digikey lists 242 depletion mode parts out of 47915 discrete MOSFETs. <S> Is it likely that some engineer included one in the design? <A> It's possible to tell a depletion device from an enhancement one using a single, simple rule: <S> is the Vgs(th) <S> the opposite sign of what you'd expect it to be from the device polarity (PMOS vs NMOS)? <S> If so (negative Vgs(th) on a NMOS, positive Vgs(th) on a PMOS), then you're looking at a depletion mode device. <S> If not (positive Vgs(th) on a NMOS, negative Vgs(th) on a PMOS), then you're looking at an enhancement mode device.
| In most supplier's search engines (and in manufacturing selection webpages) the enhanced-mode MOSFETS are considered as "normal" or "standard" and may not have any clear label as "enhanced", while the depletion-mode transistors are explicitly designated as "depletion mode". Look at the Vgs(th) on the datasheet
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GPIO state of MCU (AVR) with no supply voltage In general:what state do the GPIOs of a MCU (an AVR in my case) have when there is no supply voltage and the MCU is "off". Am I right that the pins are floating as there isn't any "reference" for the logic level at all. In special:I have a bilateral switch (BS) shifting a variable voltage (provided by a DAC) for an analog device (AD). Both the BS and the DAC are controlled by a MCU. As the BS's output will be floating when its input is not set high by the MCU, I need a pulldown at the output to define the low-state. I've attached a simplified circuitry illustrating the scenario. If the MCU is running, everything is clear: the BS will supply the AD with an analog signal as long as gpio1 is high. But if (for a reason) the 3.3V power supply is not there, I still need to have a defined voltage at the AD's Ain pin. So if my supposition, that the gpios of an inactive MCU are floating, is correct, the following circuitry should meet my requirements: MCU active: gpio2 is configured as low output -> strong pulldown (1k & 10k in parallel equals 0k9) outweighs 100k which leads to a negligible voltage divider. Ain is pulled to low voltage (not ground exactly, but close to it - 0.2V) MCU inactive: gpio2 is floating -> weaker pulldown (10k) and not negligible voltage divider (100:10) leading to a defined voltage of 0.9*24V = 2.2V. I'm quite sure that the circuitry as such makes sense, but if it works or not completely depends on the behaviour of gpios when there is no supply voltage for the MCU.Can this damage the MCU? (I guess not, as long as R1 is high enough) <Q> No, the inputs are not floating, and <S> yes, applying a voltage to them when power is off can lead to damage. <S> This is true for just about any modern IC, where the data sheet does not explicitly say otherwise. <S> In the case of say an ATmega328p, we can see it in the "Absolute Maximum Ratings" section Voltage on any Pin except RESET with respect to Ground . <S> -0.5V <S> to VCC+0.5V <S> These limits are typically due to the presence of ESD protection diodes which divert currents on the I <S> /O pins to the supply rails. <S> When the supply rail is at ground potentially, the allowed input range is just less than a diode drop higher or lower. <S> Applying a higher voltage (at least without large series resistance) would cause the protection diode to conduct, and could easily provide enough current to damage it. <S> Additionally, it would attempt to raise the entire board's supply net. <S> Even if this did not cause damage it can lead to putting various circuitry into a state from which power on reset would not cleanly work (for example, <S> while it's not a problem the AVR seems to have, I've seen other systems which would not boot if a USB-TTL adapter's data lines were connected before the system was powered, even if that adapter was connected via substantial series resistance) <S> If you need a defined level, that level should be ground via a pulldown resistor. <S> Even better, also powerdown whatever the MCU is connected to. <S> Remember also that even if the pin is not floating when power is off, it will be floating between reset and when it is configured as an output. <S> Especially if something like a serial bootloader is involved, that could potentially be a substantial period of time. <A> The simplest way to imagine the MCU pins with no power is like this: <S> What's going to happen in your circuit, is that once the voltage increases past the diode threshold (0.6v), current is going to start flowing through that 1k, and the voltage isn't going to get much higher than that. <S> Without doing the math, i'd say like.. 0.8v <S> I'm not sure what exactly the purpose of the gpio2 pulldown is, but you could hook it up like this instead: simulate this circuit – <S> Schematic created using CircuitLab <A> One typical problem with unpowered MCU while its GPIO have some path to a voltage rail (like 2.2V from 100k : 10 k divider via R3) is that the MCU power rail will "leak-up" via ESD protection diodes (typically built in GPIO pads). <S> The Vcc voltage rail will be charged up to the input voltage minus 0.5 bandgap diode voltage, minus some leakage via other resistors, hard to predict. <S> Since Vcc will charge up, the "absolute maximum ratings" will unlikely to be exceeded. <S> Some part of problem is that usual LDO regulators don't have anything to keep their output at zero when Vin is not there. <S> To keep the LDO output down, there are special models with "output discharge", or you need to add explicit bleeding resistors on Vcc rails. <S> I didn't analyze the exact path in your design, but I hope this answer will give you an idea where to look. <S> ADDENDUM1: <S> There is a classical example when an un-powered CMOS IC (like CD4017 counter) would even function and count states if just a clock is applied, say a 5-V clock. <S> The Vdd rail will charg up to 4 - 4.3 V level via upper ESD diode, and the counter will happily toggle. <S> ADDENDUM2 <S> : Similar effects frequently occur if an unpowered MCU is connected to a powered interface (say, I2C pull-ups, or USB). <S> The MCU rails might be charged up partially, to some intermediate (1 - 2 V) level. <S> When powered back, the ICs internal circuitry might fail to receive valid internal reset and fail to function properly. <S> This is the major reason why self-powered USB devices are prohibited from any active "back driving" when the host is in unpowered state.
| Basically I would say that the input is floating, but if you apply any voltage to that pin, then it will flow through the upper protection diode and begin parasitically powering the circuit.
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