source stringlengths 620 29.3k | target stringlengths 12 1.24k |
|---|---|
How to define a microcontroller pin as open collector or open drain? Source I have understood the fundamental theory of open collector and open drain configuration. But the problem is how to define/make a micro controller's pin as open drain? Can any pin of a micro controller can be used in open drain configuration?? If I define a GPIO pin as an input pin, can I use this pin in open drain configuration? Is it mandatory to define a pin as output for using it in open drain configuration? <Q> Whether an output is an open drain or open collector or something else is a question of hardware design. <S> An open collector output has to be built so that the output pin is connected to the collector of a transistor of the chip - and the collector is connected to nothing else. <S> If a chip isn't designed with an open collector output, the no amount of configuration will get you one. <S> That said, many microcontrollers are designed to have switchable GPIO pins. <S> You can configure a pin to be input or output, and you can configure what type of output. <S> What combinations are possible depends on what the hardware manufacturer decided to build into the hardware. <A> If I define a GPIO pin as an input pin, can I use this pin in open drain configuration? <S> Maybe. <S> if the pin sees a voltage higher than the supply voltage of the microcrontroller bad things will happen.if it never does you can fake an open drain. <S> Is it mandatory to define a pin as output for using it in open drain configuration? <S> If you're required to define the direction of the pin at compile time you can't use it as open drain. <S> ` <S> The trick to faking open drain is to set the output bit to low, and send the signals by flipping the data-direction bit.on and off. <S> with the direction set to input the pin looks to the external world much like an inactive open drain, with it set to output <S> it looks like an active open drain. <A> How to define/make a micro controller's pin as open drain? <S> Does any pin of a micro controller can be used in open drain configuration? <S> If so, you can configure the behaviour through registers. <S> Most MCU manufacturers offer macros or functions to set the inputs and outputs correctly. <S> If i define a GPIO pin as an input pin, Can I use this pin in open drain configuration? <S> The open drain configuration itself does not influence the input. <S> Furthermore, you can emulate an open drain (open collector) output: Set the (push-pull) output to low for OUTPUT LOW. <S> and switch the data direction from output to input direction for an emulated OUTPUT HIGH. <S> Does it mandatory to define a pin as output for using it in open drain configuration? <S> You may define a GPIO to open-drain, e.g. use it as normal input, and switch to open-drain when changing the data direction.
| In the datasheet of a MCU, you can see if an GPIO pin is capable to be push-pull, open drain output.
|
How to keep the state of relay In schematic below when current exceeds the set point of PR1 it toggles the relay and as soon as current drops the relay goes back to normal state: How can i keep the relay state as triggered even when the current drops and only switch the relay back to normal state when i pressed a push-button? <Q> You can realize the required function using a diode. <S> What needs to be done is to pull one of the inputs of the opamp to such a voltage that the opamp's output stays high (and the relay powered). <S> Add a diode (1N4148) with the anode to the opamp's output and the kathode to the + input of the opamp. <S> Then when the opamp's output is high (and the relay is powered) <S> the + input of the opamp will be pulled to a higher voltage than R3 and R7 supply (which is about 6 V). <S> The - input of the opamp will not be able to reach that voltage (it will always be at a lower voltage) <S> So the output of the opamp remains high. <S> To add the reset function, add a 1 kohm resistor in series with the diode and add a switch to ground from the + input to force it low so that the output of the opamp becomes low as well: simulate this circuit – <S> Schematic created using CircuitLab <A> You can add some positive feedback, with a N.C. pushbutton in series. <S> For example, a diode from the op-amp output to non-inverting input. <S> With said switch in series. <S> Edit: This is similar to Bimpelrekkie's solution except it requires a N.C. pushbutton, which is less common, and does not force the relay state when the button is pushed. <S> The diode can be any common silicon signal diode such as 1N4148. <S> Note: If you require a particular guaranteed state at power-up you might have to add some circuitry. <A> I'm adding a latching relay below. <S> BTW I couldn't check current and voltage ratings of your circuit. <S> This component is just an example. <S> https://www.digikey.com/product-detail/en/kemet/EE2-3SNUH-L/399-11008-1-ND/4506460 <S> there is 2 control input in this relay. <S> once you triggered first input and the other one is logic0, relay switches inputs. <S> Until you apply trigger to the second input this will stay same.
| If you have a chance to change your relay, you may try to use latching relays.
|
How do I terminate a directional coupler? I'm using a directional coupler for a radar application using frequencies of 5.3-5.9GHz. The directional coupler linked above is matched to a 50ohm characteristic impedance. How do I terminate the isolation port to prevent reflections? Various sources online just say that I should terminate with the characteristic impedance, but is it sufficient to place a short microstrip at the isolated port and then via to a common ground plane? Is it bad to use the common ground plane? Do I need to worry about RF vias here, or is a regular via sufficient? I've seen a design that uses a 0402 49.9ohm resistor between the isolated port and a via to ground (connected by short microstrips). Does that provide any additional protection? <Q> How do I terminate the isolation port to prevent reflections? <S> Various sources online just say that I should terminate with the characteristic impedance <S> The ground should be the same ground as the shield of the signal output. <S> Use any other ground and the signal path will be longer which adds inductance to the 50 ohms which will prevent a proper termination. <S> What if you do not properly terminate the signal? <S> Then the output signal at that port will reflect back into that same port. <S> Terminate it properly and all signal will be converted to heat by the resistor. <S> Then (almost) no signal will reflect back. <S> but is it sufficient to place a short microstrip at the isolated port and then via to a common ground plane? <S> A microstrip is just a path for the signal to a different location. <S> It is not a proper termination. <S> At the end of the (50 ohm) microstrip you would still need a 50 ohm resistor. <S> Is it bad to use the common ground plane? <S> That would increase noise levels. <S> If your RF signals are quite large it might not matter. <S> Do I need to worry about RF vias here, or is a regular via sufficient? <S> My guess is that an RF via has less inductance. <S> Less inductance is almost always a good thing in RF design, so if you have the option: yes, use RF vias. <S> I've seen a design that uses a 0402 49.9ohm resistor between the isolated port and a via to ground (connected by short microstrips). <S> Does that provide any additional protection? <S> Protection for / against what? <S> This is the proper way to terminate a 50 ohm (49.9 ohms is close enough to 50 ohms <S> ) port. <S> Just do the same! <A> The matched termination that you provide on the board is vital for the specification of the coupler. <S> If you leave it reflective, it will reflect the energy coupled from the backwards direction through the coupler and send it out through the coupled port. <S> You will then have lost all your directivity, the reason you are using a coupler instead of the simpler and cheaper voltage pickoff (which has no directivity). <S> Take a microstrip far enough away from the package so you can handle the footprint easily, then put a resistor to ground. <S> Usually at this frequency, I'd use two 100ohm resistors in parallel, which tends to provide a better load than a single 50ohm at these frequencies and component sizes. <S> However, the specifications of this coupler are so poor that there's little to be gained by that refinement. <A> You need to connect it to something that both: 1) matches the impedance (ie doesn't provide a reflection at the connection) and 2) absorbs any signals that come out. <S> A microstrip terminated in a via to ground does 1), but reflects the signals straight back in, so doesn't do 2). <S> A microstrip terminated in a resistor will (if the resistor functions like a resistorat the frequencies you're operating at) will do both.
| And that would be correct , any output with a characteristic impedance can only be terminated properly by terminating it with a resistor to ground with the same value as the characteristic impedance, so 50 ohms in this case. It depends on your design, if you are making a sensitive (easily disturbed) RF circuit then you don't want to share the same ground (plane) with some digital circuits.
|
USB current draw and CE marking We have developed a docking station product which charges multiple devices simultaneously, and currently ships with a 5V, 3A barrel jack adaptor. We are looking to move this over to USB power, and have recorded maximum current draw with a full complement of devices around 1.8A (quickly returning to 1.5A, which I understand is the minimum required for a USB Dedicated Charging Port). The factory are telling me that if exceeding 1A it will not pass CE certification, but I can find no reference to this in any of the documentation I have read, so my question is: Is the factory correct? If so, could anyone refer me to a standard whichstates this? Given the prevalence of USB supplies with 2A or greater rating,would we be violating any rules by simply stating in the manualthat it must be connected to a supply which can provide 2A or greater? Thanks in advance. <Q> CE marking and USB standard are completely different to each other. <S> CE is for safety, USB is functionality. <S> So this is at least two different questions. <S> No idea what the factory is quoting the 1A limit for CE. <S> Ask they for what standard this breaks, they should know, that's what you're paying them for. <S> CE is a lot more worried about EMC, EMI and voltage limits rather than a current value. <S> But be aware, many users don't know about current out of USB, so will claim your device is broken when it doesn't work. <S> This is why the Amazon Echo comes with a dedicated USB supply, as they clearly state that you have to use their USB supply due to the higher current capability of it. <A> I think the factory is confusing CE marking with USB specification. <S> CE marking is purely a safety certification, meaning that any test made on the product are there to certify that the product is safe to use. <S> The only place where you will find the USB specification is on the USB consortium website , and here you will find the specified limits for current consumption. <S> Complying to the USB specification is just the first step, after that, you need to comply to the CE marking standards, which will look at the electrical safety of the product, e.g. It won't catch fire when you start to draw 2A. <S> What I think the factory means, is that if you are doing a design change on a CE certified product, you will have to re-certify it, <S> so make sure you are still within the EMC rules and electrical safety defined by the standard. <S> The amount of test you will have to put the new hardware through, depends on how big are the changes compared to the previous version. <A> If your device doesn't have a captive lead, (at least in some jurisdictions) you will need to have some mechanism to handle the case where someone uses a different, USB certified cable to connect your PSU to your device. <S> If you're pulling 1.5A through a cable designed for 0.5A you'll be looking at a fire. <S> The usual tricks are to check the droop along the cable using the shield for reference (but be careful of cables which lack one), or just shutting down if the supplied voltage droops.
| As for the current question about USB, no rules are broken if you don't claim to be USB certified. Assuming you've done a good design, EMC and EMI should be fine (and they'll give you a report of emissions), and USB is 5V (unless you're doing some of the higher power chargers).
|
Why would you need an op amp for reference voltage when the voltage divider does the trick? Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? Can the AD8031 op amp be removed? Also, to save power, can the voltage divider be replaced with a buck converter? <Q> Is there a reason this schematic uses an op amp (AD8031) for the reference voltage of the op amp (AD8544) filter when the voltage divider already brings down the voltage anyways? <S> Can the AD8031 <S> op amp be removed? <S> In this case, since the AD8544 has only 4 pA input bias current, I'd expect the AD8031 can be removed without much change in performance. <S> Another issue to watch for, since this reference is connected to two different signals, is whether removing the buffer could allow the two signals to crosstalk with each other. <S> Given the high resistor values connecting the two op-amp inputs to the reference, it's unlikely this would be a real issue, but to be sure you could simply make two different dividers and use one for each of the filter stages. <S> Also, to save power, can the voltage divider be replaced with a buck converter? <S> Any buck converter will have some output ripple. <S> If you used it here, that ripple would be coupled directly into your filtered signal. <S> I wouldn't do it just to save something like 150 uA. <S> (You'd also need to find a buck converter design with less than 150 uA quiescent current to make this a positve trade) <S> If those 150 uA are really important to your application, you might rather find an op-amp with very low quiescent current <S> (the AD8031 has 800 uA <S> , you'd be looking for 10's of uA), replace the AD8031 with that, and increase the resistor values in the divider to 100 kohm or more. <S> Aside <S> The AD8031(A) is only rated to drive capacitive loads up to 15 pF and maintain stability. <S> C2 and C4 in your schematic are probably causing the op-amp to generate noise (it may even be oscillating) rather than reducing noise. <S> I'd remove them. <A> Is there a reason ... <S> Yes. <S> The two 10k reisistors give the voltage reference an impedance of 5k. <S> This means that if the current drawn from the reference changes by 0.1 mA that the voltage of the reference would change by 0.1m x 5k = <S> 0.5 V. <S> This would be a very unstable reference. <S> The op-amp buffer fixes this. <S> The output impedance of the buffer is close to zero in comparison. <S> This is a stable reference. <S> Can the AD8031 <S> op amp be removed? <S> Maybe, but probably not a good idea. <S> Also, to save power, can the voltage divider be replaced with a buck converter? <S> The voltage divider consumes <S> \$ <S> I = \frac { <S> V}{R} = \frac {3.3}{20k} = <S> 165 \mu \text A \$ . <S> A buck converter is designed for power supplies rather than a voltage reference. <S> The converter would likely consume more than 165 μA <S> so there would be no advantage. <A> That's a horrible circuit, I wonder where you got it from. <S> The AD8031 is very intolerance of capacitive loads, see Figure 46 in the datasheet , so most likely that op-amp will be oscillating at high frequency, which will, at a minimum, cause increased power consumption. <S> You can use a TLE2426 , which will consume only 170uA typically at 5V. <S> Below is a way to connect a conventional op-amp in a stable manner (from a TI ADC datasheet): <S> That's a low-noise high speed amplifier, for yours you might try increasing the resistor values by an order of magnitude.
| The usual reason to use an op-amp to buffer a divider like this is to ensure the reference voltage doesn't change if whatever it's connected to sinks or sources current.
|
Choosing the right microcontroller How do you choose a micro-controller for your project? My hardware requirements are : I2c SPI ADC channels SMD package 3.3V operating voltage A few GPIOs. There are so many microcontrollers available from various companies which satisfy these hardware requirements(almost all I know of satisfy 1,2,3,4,6)? So then should I choose the microcontroller based on my familiarity with the controller and resourses/documentation available about it or is there any other method to choose one? <Q> I think it depends on many things, like: <S> You mentioned already: <S> Capabilities/features Familiarity (yourself) <S> Others can be: <S> Cost Familiarity in your company Expected support from the manufacturer Swappability (how easy it is to later convert to different models when hardware requirements change) <S> Software/tools/IDE support (thanks to Peter Smith) <S> This list can probably be much longer. <A> Almost all microcontrollers have / can support the features you list. <S> So in my opinion you will not find what you need by looking at the features in your list. <S> The Arduino IDE is easy to use (I think) and free. <S> Also you will find loads of examples for that platform. <S> But then you're "stuck" with the uCs that are supported by that platform. <S> So that's mostly the ATMega uCs. <S> Is that an issue? <S> For me it is not as I use uCs only for hobby projects. <S> I simply buy an Arduino-NanoPro clone on ebay for less than $2 and use that. <S> If you need a uC for some gadget and it needs to be mass produced and as cheap as possible then there are uCs that cost only 3 cents each. <S> But these need a special development platform and can only be programmed once . <S> But to learn about uCs for hobby projects: just use an Arduino. <S> With the knowledge you gain from doing that you will be able to make better use of any uC you will be using in the future. <A> Nano won't do 3.3V that I know of. <S> But a 3.3V/8MHz Promini will, plug on an FTDI Basic or equivalent clone to program/debug it. <S> Same Atmega328P uC as the Nano and Uno.
| A very important feature (for me) is how easy it is to use and develop using a microController.
|
Circuit to convert 24VAC digital signal to 5V DC digital signal I am designing a circuit to convert 24VAC signal to 5V DC signal. The 24VAC signal is designed where 24VAC is on and 0VAC is off. The signal needs to be fed to an MCU which is 5V DC. The circuit designed has to be very cheap so more expensive ICs and full wave rectifiers are not options. There are two designs that I came up with The first design outputs a 5V DC signal for high and about 0VDC for LOW. The second design eliminates the capacitor but outputs a 5VDC for low and 5VDC square wave for high which will be accounted for in code. I would like to know if there is an advantage for one over the other considering that the circuit needs to be as cheap and safe as possible since it will be used in an industrial environment and will be heavily affected by dust, debris, and humidity. <Q> Figure 1. <S> R4 is redundant as R1, R3, Q1 and R2 shunt it. <S> Similarly in the second scheme R1 is redundant. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Solution with component count of 2. <S> If you use a bi-directional opto-isolator and series resistor you can cut the component count to two. <S> The opto-isolator output transistor will pull low every half-cycle if you use the MCU internal pull-up resistor. <S> Your software will need a timer to check that AC has been lost for 10 ms or so before determining that AC is off. <S> I have written more on opto-isolators here . <A> Industrial environments can be harsh from an electrical point of view. <S> If the 24VAC isn't close to the input, the signal can pick up noise and transients, also possible differences in ground potentials, all of which which can cause bad readings or even damage the input. <S> For these reasons, galvanic isolation is often used. <S> Having said that, for lowest cost, look into using a basic resistor divider with a couple of diodes to clamp to +5V and ground. <S> I would add a small capacitor to reduce noise and help out with transients and ESD. <S> Functionally, there isn't much difference between the two circuits, except that the first would be better at sourcing current, the second better at sinking current. <S> But since there's a 10k resistor in series with the output, this probably isn't a concern <A> Assuming a \$5\:\text{V}\$ DC supply rail and assuming that a broad approach is satisfactory, then something like this might be okay: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It's mostly just a AC high pass filter and resistive divider ( \$C_1\$ , \$R_1\$ , and \$R_2\$ ), followed by \$D_1\$ to protect \$Q_1\$ 's base-emitter junction from reverse voltages, <S> followed by a 2-BJT shaper ( \$Q_1\$ , \$Q_2\$ , \$R_3\$ , and \$R_4\$ ), followed by a simple RC lowpass filter/integrator ( \$R_5\$ and \$C_2\$ ), followed by a simple saturated switch that normally is OFF unless there is a sufficient number of low-going pulses at the collector of <S> \$Q_1\$ to pull down the voltage on <S> \$C_2\$ . <S> When \$Q_3\$ is active, it pulls the output high. <S> Otherwise, \$R_8\$ pulls the output low. <S> I've no idea what your source impedance is, but the values of \$R_1\$ and \$C_1\$ can be adjusted per need there. <S> The design is more "slapped together" than it is carefully crafted, because I have no idea if this achieves any of your goals. <S> If the broad strokes are okay, then fine... more could be done to craft it better. <S> If not, then at least this provides enough for you to clearly say so. <A> So, I finally got around to drawing up a simple circuit using a bridge rectifier. <S> I can order a bridge rectifier that costs less than a 2N3904 from the same supplier. <S> The blue section represents whatever is generating the 24VAC. <S> The red section represents the microprocessor input, with clamp diode to 5V and 100 kOhm as the input impedance. <S> The AC-detector is really just the three resistors, the rectifier bridge, and the capacitor. <S> The three resistors reduce the 24VAC to something that will be a little over 5V when rectified. <S> They also limit the current that can reach your processor in case you get more than you expected on your 24VAC. <S> This is the simulation: <S> You can't see it, but the current through the clamp is less than 1mA (red line across the bottom.) <S> So, probably safe enough to let the clamp diodes on the processor's input handle it. <S> If not, a schottky diode is pretty cheap. <S> The voltage builds up a little slowly. <S> That's about 75 milliseconds before it hits 5V. Ought to be fast enough, though. <S> The decay when the 24VAC disappears ought to be similar. <S> That's a cheap, dead simple AC detector that offers at least some protection against over voltage. <S> Probably as cheap as if not cheaper than the transistor based circuits.
| As mentioned in the comments, there are SMD bridge rectifiers for low current available for about the price of a single transistor.
|
PCB design for outdoor use without coating I am making a PCB for a home project: a small weather station that I will place outdoors to collect data. There is now a question of environmental resistance. I can mill out the PCB for free on a relatively old machine. This means that I will not get any protection against the elements out of the box. My question then is: how quickly will the PCB degrade when left outdoors (in a housing) with only tin covering the copper and are there any relatively cheap methods of protecting the PCB against the elements? <Q> It only takes a few degrees of difference between moist air temperature and your board to get moisture condensing on the board. <S> Soldered/tinned connections degrade surprisingly quickly. <S> If you want to test this out get a moisture detector PCB from Sparkfun and see how long it takes to corrode. <S> Even in a non direct wet environment outside it's a problem after just a month. <S> Potting or deep conformal coating is the only way to get real longevity (many years). <S> Unless you are in a gland protected moisture proof box the ends of your cables (copper) will degrade over time. <S> It's unlikely that you have significant current involved, but even at lower currents you cannot solder the ends of wires without long-term corrosion problems. <S> It is best to use solid copper cables and IDC or crimp the connections rather than use soldered or screw terminals (perhaps start reading here ). <S> there are plenty of options for IDC connections and these are particularly effective. <S> AVX have a broad range including SMT parts , and my personal favorite is Phoenix PCB IDC which does not require a pushdown tool and can handle both solid and multistrand wire. <S> For conformal coatings you can't beat MG 422 IMO, but there may be other good ones out there. <S> MG 422 is easily peeled back for repairs, which is why I like it. <A> Buy a cheap spray can of acrylic protective lacquer for PCBs <S> (eg <S> https://www.electrolube.com/products/conformal-coatings/apl/acrylic/ ) , and spray both sides of the board. <S> If you protect the coating from UV and mechanical stress, it should last a very long time. <S> You are presumably going to place this board in a mostly water tight enclosure? <S> Other possibilities are potting your circuit, either in the traditional epoxy, or one of the gel based products like raytech's magic gel ( http://www.raytech.it/product/low-voltage/magic-gel ) <A> Then let it thoroughly dry and maybe do a second or third coat if needed. <S> I made a pin stand for this where a took straight pins (the T-type) and ran them through some thin foam. <S> Then I set the coated board directly on the pin points to dry (i.e. flat which tends to give it the best uniform coating). <S> Others use a small wire attached usually via a hole and hang the board from it, but I found that causes to much of the polyurethane to drip off before it solidifies reducing the total amount of coverage. <S> Although sunlight will yellow the polyurethane over time, I've had boards that look great even after years of being outside with no visible signs of corrosion. <S> And if you have attached wires that can't be completely coated with the polyurethane, just put a few dabs of hot glue on them and usually your good to go on that as well. <S> What I also like about this method is that if/when you need to do maintenance or modifications, aside from the nasty burning of the polyurethane (which you should be using an exhaust fan anyway), you generally can fix the problem and then just recoat the board with polyurethane again. <S> With many (not all) conformal coatings this isn't possible to do (especially if you pot the board). <S> One small can ( <S> 1/2 pint) of polyurethane will generally run you around $7 at WalMart and will last for 100's of small sized boards (now that's cheap!). <S> Of course aside from this, an enclosed box is always helpful to prevent the sunshine issue of oxidizing the polyurethane along with additional protection from really harsh weather conditions (i.e. like high winds causing debris to nick your board).
| On the cheap...buy a small can of polyurethane and after assembly, dip the whole board in it. If your project is exposed to large variations in temperature and humidity, you may begin to see serious degradation in just a single season.
|
Building AC to DC circuit that can not be run on a breadboard First I have an AC power source. Can I solder the power source directly to a rectifier to make it DC? I plan on using a protoboard using the DC power but I need the board to be able to handle 2Amps and from what I found regular breadboards can't be used so I've been looking at veroboards. What are trace lines on veroboards? I found this online calculator for max amps of a board. Lastly would it be safe if I solder circuit together directly instead of using a protoboard? The circuit would require 12 volts at 2 amps. <Q> One can buy a 12VDC, 2A supply, able to run from 100-240VAC at 50/60 Hz, for $7. <S> Making one is not worth the effort anymore. <S> https://www.mpja.com/12-Volt-Plug-Power-Adapter-2A/productinfo/34142+PS/ <A> First of all it would be more helpful if you could define a few things. <S> What type of AC source, is it mains? <S> lab bench supply? <S> What voltage is your AC source and what is its current rating? <S> Without a smoothing / reservoir capacitor your output will not be a clean DC signal. <S> A bridge rectifier by itself will provide a waveform like this: <S> Taken from: About rectifiers Further to this, your question about running 2A on protoboard entirely depends on what board you use. <S> The traces are the strips of copper running down the board. <S> Go to a website like RS / mouser / digikey and search for protoboard / veroboard, when you find one it will have a datasheet giving you the parameters to put into your online calculator. <S> To answer your final point, maybe. <S> Again this would depend on what this 'ac power supply' is, the quality / size of the contact joint and physical safety. <S> As opposed to soldering directly you could consider using lengths of wire. <S> 12AWG wire should be more than suitable for 2A. Good luck. <A> Can I solder the power source directly to a rectifier to make it DC? <S> Yes, you can solder components directly together. <S> There's no law of physics that says you have to have a circuit board at all. <S> That would be a short circuit. <S> I plan on using a protoboard using the DC power <S> but I need the board to be able to handle 2Amps and from what I found regular breadboards can't be used <S> so I've been looking at veroboards. <S> What are trace lines on veroboards? <S> This question has already been asked - the latest answer on this question estimates you should be able to run at least 5 amps through a veroboard trace. <S> Lastly would it be safe if I solder circuit together directly instead of using a protoboard? <S> The main thing to worry about, that you don't worry about with a protoboard or breadboard, is accidental short circuits. <S> After building the circuit you'll find out how rigid it is. <S> You could use something like electrical tape, heatshrink or even hot glue to keep wires separated. <S> The electrocution risk would be the same as it would be with a protoboard or breadboard. <S> The usual advice applies: don't touch the high voltage bits, or better yet, <S> if you aren't confident then let someone else do the high voltage bits.
| If you do it, make sure all your wires can't touch each other when they're not supposed to.
|
Multiple input to single output converter and vice versa First question so please forgive any mistakes. I'm trying to design a system by which multiple LEDs are controlled by multiple switches, but without using as many wires as there are switch/LED pairs. I'm planning to have roughly 20 momentary switches connected to the same number of set-reset latches which control LEDs. I'd like to know if there are any chips or circuits I can use in order to take these 20 inputs, compress them down into as few wires as possible so that I don't have to cable manage 20 wires (which, for my application, will be unfeasible), then provide the same number of outputs at the other end. The outputs will only need to provide a brief pulse along the correct line, due to the set-reset latches I'm using, and no two inputs will be able to be activated simultaneously . Thanks all. <Q> You are looking for a serializer/deserializer. <S> These devices are intended for just this kind of thing. <S> They take a large number of parallel inputs, convert them to a high speed serial communication, and then convert back to parallel at the receiving end. <S> A similar problem occurs with the displays on laptop computers... <S> all of the information for the display has to travel through a thin, flexible cable through the hinge. <A> No need for any ICs or microcontrollers. <S> You could do the wiring with only 5 wires. <S> The only things you have to consider is to use 2 channel momentary switches and some diodes after them. <S> Greetings <A> Are you looking to condense the input signals, the output signals or both? <S> Even if you have an IC which handles the inputs, you're still going to have to wire the buttons to that IC. <S> If your micro can handle it, you can read in the 20 inputs and then use something like a TLC5972 which is a 16 channel LED driver with serial inputs and outputs. <S> It is super easy to use, just requiring 1 resistor to set the current. <S> This way you'd only need 3 or 4 wires between your uC and the LEDs. <S> If you need to condense both ends you can essentially make a grid of buttons. <S> A grid of n^2 buttons only requires 2n pins. <S> For 20 buttons you can get away with 10 wires. <S> I'm not going to draw all 20 buttons, but this 4 button example can be expanded. <S> simulate this circuit – <S> Schematic created using CircuitLab Node 1 and 2 are outputs, <S> Node 3 and 4 are inputs. <S> You pull Node 1 high, then check if Node 3 or 4 are high. <S> If they are, then that button has been pressed. <S> You then pull Node 1 low again, and repeat the process with Node 2 high. <S> If you have 10 free GPIO pins on your uC <S> you won't need any external ICs to read the buttons and can get away with half the wires. <S> Of course, if 10 wires is still too many for the inputs you can do something similar with an 8 to 3 multiplexer and 3 GPIOs by replacing Node 3 and 4 with the inputs to the mux. <S> Let me know if any of these ideas appeal to you and I can elaborate further if need be.
| with the given parameters (20 Switches, 20 LEDs and no simultaneously pushed buttons) Charlieplexing could be a good way to go for you.
|
How do I figure which LED I need? I need to replace two LED diodes on a TP4056 board (bottom right corner on the picture). The board is going to be in a box and I want to move diodes to the small holes in the box. There are two resistors with marks 102, and that is 1 kilohm. I measured voltage on diodes and my multi-meter said 2.7 V. I want to use something like this: Those diodes are 3 V 20 mAh. Can they replace exiting diodes? update: Thank you guys. You helped me a lot. Btw, current LEDs are blue and red <Q> Figure 1. <S> Typical IV curves for various colours of LEDs. <S> Image source: LEDnique . <S> A little bit of background theory may help. <S> LEDs have a non-linear relationship between applied voltage and current. <S> The forward voltage also depends on the colour as shown in Figure 1. <S> Your measured 3 V LED voltage suggests that you've got either white or blue LEDs on the board. <S> With the 1 kΩ resistor in series the current will be limited to a safe value even if you change to one with a lower forward voltage. <A> Based on the schematic in the datasheet , they're not critical, any LED will do. <A> Yes, you can most likely replace the SMD LEDs by any "jellybean" LED. <S> You can easily try it. <S> Do observe the correct LED polarity.
| Replacement of any of those small indicator LEDs with 3 or 5 mm LEDs should be fine.
|
How to figure out SD Card optimal write size I am writing the data logging portion for a real time embedded application. I have an SD card I communicate via SPI at 33Mhz. I need to be able to log at least 1024 bytes of data ever 1 milli-second. I have tried sending the pre-erase command with 2 blocks erased and then sending the multi-write command with 2 blocks but this takes longer than 1 milli-second. My SD card is rated at 20MB/s so it should be good for 20KB/ms at optimal speeds. I have tried reading the SD card specification to see how to calculate the optimal write size or sector boundaries and can't tell exactly what to do. Should I be writing in multiples of the sector size at blocks that are multiples of the sector size? <Q> I have tried reading the SD card specification <S> The document will not give you much information about performance tuning - including block size effect on the performance. <S> My SD card is rated at 20MB <S> /s <S> Rated for which operations - <S> or is it maximal <S> SPI communication speed? <S> Let's do simple math <S> : you drive card at 33 MHz, without considering the control overhead each clock cycle is one bit, thus to transfer a byte in SPI mode (single wire) <S> you will need 8 clocks, thus 33/8 = 4,125 million bytes per second. <S> Source , page A-2. <S> Thus ideally SPI allows you transferring 1024 bytes in 248 us, which seem to be well below 1 <S> ms. <S> However card's performance depends on some other things than SPI bus speed. <S> It has its internal controller, which can be a bottleneck (sometimes - e.g. when CRC error or any other exception occurs, or always because it is slow), memory array with its timing can be a bottleneck. <S> Example: <S> Altera serial flash device I used recently (EPCS16 <S> ) - same marking and same component source - earlier device erases sector within 1 second, later device erases it within 200 ms. <S> This is dramatic performance improvement. <S> The same for SD-cards - their timing may vary a lot. <S> I also have heard that performance depends on the size of the card - bigger cards are slower, but I can not comment on it - just keep this possibility in mind. <A> I've used SD cards in a few projects as a maker, and one thing I've learned is that the SPI access mechanism - in contrast to the more modern 4 bit bus - is generally going to perform poorly compared to the card's specifications. <S> I learned this through multiple iterations of my Orthrus project, the last of which used an ATSAMS17 chip with its HSMCI interface. <S> Orthrus was doing crypto, and only single block <S> I <S> /O, but even then the results I get are better than SPI even taking into account the 4 bit versus 1 bit per clock difference. <S> To get the specs they promise, you need to negotiate very high clock speeds and reduced (1.8v) power, none of which are available via SPI. <A> fat16lib in the Arduino forum has done lots with SD cards, and has found 512 bytes to be optimal.
| The actual speed of the cards can differ dramatically, thus you must dig into the specifications of the card deeper to find out read and write speeds, or better consult with manufacturer on what card may suit you and your pattern of access . For compatibility reasons (and for some other reasons) use 512 byte block size.
|
Power determination for resistive load I am a mechanical engineering student and i need to measure electrical output. I have some voltage and current measuring instruments. I used them to measure the voltage & current of a blower(resistance heater). The results were 3 ampere and 155 volts. But the appliance is rated at 2kW. So according to the formula, P=VI, the load of the appliance should be 155*3=465W. Why does this deficit exist between the rated power and the one i measured. Do i need to introduce power factory in the P=VI equation?Someone told me it involves the concepts of apparent power reactive power, real power. If so can someone please explain those terms to me in a simplified manner, my knowledge of electrical engineering and circuits is poor. <Q> From your readings you can calculate the effective resistance of the load at \$ R <S> = \frac {V}{I} = \frac {155}{3} = 52 \ <S> \Omega \$ . <S> If a resistance of this value was connected to a 230 V supply then the power would be given by \$ P = <S> \frac {V^2}{R} = \frac {230^2}{52} <S> = 1 \ \text {kW} \$ . <S> You would need to check: <S> Loss of <S> one or more heater elements. <S> My bathroom heater has 2 x 1 kW elements in parallel and one is switched off by thermostat once the air reaches a certain temperature. <S> Power selector switch. <S> Any kind of dimmer power control. <S> I have assumed that the fan motor is a small load - maybe 50 W - in comparison with the heaters. <A> 2 kW is the rating of the Blower for a standard voltage (say 220 V or 110 V for AC Systems).for 220 <S> V max ampere is around 9A. for 110 V around 18 A. <S> I am not sure of working of Blower, I think the blower has a resistance inside. <S> you can adjust this resistance to get max ampere of blower. <S> check for the standard voltage of blower and try adjusting the resistance(change the settings-hot/cold) and calculate results. <S> You would probably get a value near the rated value. <S> Since you have mentioned 230 V is rated voltage, you can apply that and adjust the setting and find value closer to rated one. <A> There are multiple types of "power ratings" which can produce very different results. <S> Power in product documentation and advertising. <S> This is typically determined by what marketing thinks "sells best" and doesn't involve outright lying (although sometimes it does) <S> What the device actually does: which can be measured but may vary a lot with the specific use case and/or environmental conditions <S> For example let's look at https://www.crownaudio.com/en/products/ma-12000i <S> The name implies that it's a 12000W amplifier, the data sheet lists the highest use case as 9000W, the power on the label is 1950W <S> and I'd expect it to draw maybe 1000W or less in a real use case.
| Power on the label: This is determined by local legal procedures which may or may not represent actual use cases. This ampere is dependent on the load on the system (here the blower).
|
Making a parallel connection = increasing the total current? Considering the scheme below simulate this circuit – Schematic created using CircuitLab Am, I understand correct, in real life, if I will step-by-step add the parallel resistors, the total curent will grow, and even there only 1V, the value of curent can be over 10 50 100 or more A? And hence, even an 1V voltage source can melt a wire, or create strong magnetic field? <Q> <A> That is correct until you exceed the ability of the 1 V source to maintain its output voltage. <S> Batteries have some internal resistance which as you increase the current draw causes the terminal voltage to drop. <S> This is known as the "equivalent series resistance". <S> You can see this yourself if you measure the open-circuit voltage of a battery and then measure again with a significant load. <S> Mains-powered unregulated power supplies may have a similar voltage droop with increasing current. <S> Regulated power supplies will have some active circuitry to maintain a constant output voltage up to a specified maximum current. <S> Beyond that the power supply goes into current limit and the voltage reduces to that value which will maintain the maximum current through the load resistance. <S> In your example a 1 V supply with a 1 A current limit will work fine until the 101st resistor is added. <S> Then the voltage will decrease to \$ <S> V = <S> IR = <S> 1 <S> \times \frac {100}{101} = <S> 0.99\ \text V \$ and will decrease further for each additional resistor added in parallel. <A> if I will step-by-step add the parallel resistors, the total current will grow? <S> Yes, by adding resistances will parallel the effective resistance is going to decrease. <S> the effective resistance of a parallel circuit is always less than the least resistance in that network. <S> even there only 1V, the value of current can be over 10 50 <S> 100 or more A? <S> Yes, if the power supply is capable enough and the resistors also to be able to dissipate that power. <S> And hence, even an 1V voltage source can melt a wire? <S> Yes, current is the one responsible for melting and voltage <S> is the driving force for current. or create strong magnetic field? <S> yes it will.
| Yes, if the various resistances are low enough.
|
FIR Filter on Cortex M0 Is it possible to implement basic DSP (such as an FIR Filter) on a Cortex M0 or is the architecture too limited? <Q> Of course you can. <S> Practically it will depend on the size of your filter (ie. <S> RAM usage) and required speed (if it is real-time). <S> Of course the implementation will have to be fixed-point, as M0 does not have a floating-point unit (so floating point operations are slow). <S> I recommend using CMSIS <S> DSP which is a DSP library provided by ARM and optimized specially for their cores. <A> Yes you can. <S> An FIR filter is essentially a cascade of multiplication, addition and registers y[n] = <S> b[0 <S> ] * x[n] + b[1] <S> * x[n-1] + b[2] <S> * x[n-2] <S> + ... <S> + b[numTaps-1] <S> * x[n-numTaps+1] <S> What makes Digital Signal Processors (DPS's ) so attractive for this type of operation is a lot of optimisation effort went into providing a platform best suited for Digital Signal Processing. <S> So can such algorithms be implemented on non-DSP platforms? <S> of course. <S> You can implement them in Excel if you so wish <S> and you have a need to post-process data. <S> The time to execute such filtering algorithms may present a bottleneck in your uP designs and this is where tradeoffs in implementation is required https://www.keil.com/pack/doc/CMSIS/DSP/html/group__FIR.html http://wseas.us/e-library/conferences/2012/Prague/ECC/ECC-73.pdf <A> I once implemented an adaptive least-mean-squared filter algorithm on a 16-bit processor running at 3 MHz, so <S> yes of course you can. <S> Now, if you had given us information about the filter size and sample rate we might have been able to give a better answer.
| You can implement any kind of DSP on a Cortex M0.
|
How do electrons carry thermal energy in Peltier coolers? I've read that when electrons enter from metal to semiconductor type N, they gain thermal energy and make that side cooler and the reverse happens when they leave semiconductor to the conductor.Why does this happen? How do electrons gain, hold, transfer and lose this thermal energy in this process? <Q> What happens is that the electrons need to gain energy to jump the energy gap at the metal-semiconductor junction. <S> The valence band is shown in the shaded area. <S> In conductors the conductance band and the valence band overlap. <S> In semiconductors there is an energy gap between the conductance and valance bands, as shown in the figure. <S> This energy is the thermal energy, hence heating or cooling. <S> A similar argument applies for p-type semiconductors. <A> This appears to simply be an observable effect. <S> The best explanations are probably in mathematical equations, but the simplest phrasing I could find was that: <S> "The Peltier effect is caused by the fact that an electric current is accompanied by a heat current in a homogeneous conductor even at constant temperature. <S> The magnitude of this heat current is given by \$\Pi <S> \cdot <S> I\$ " <S> (Pi times current), where \$\Pi\$ is the peltier/seebeck coefficient of that material. <S> This coefficient can be positive or negative and as a result materials can be paired and alternated in a zig zag to produce a favorable series electrical arrangement along with a parallel thermal arrangement. <S> I've been unable to find anything regarding the precise mechanism (it makes sense to me that electrons would carry thermal energy in the direction they are flowing, but I'd like to know why thermal energy would flow against the actual flow of material). <S> It could be an effect at the barrier regions of the materials as some of the other papers seem to suggest , with the electrons accepting a transfer of thermal energy in order to cross the barrier between different materials. <S> This would appear to better account for negative coefficients. <S> I think you should consider migrating this question to the physics stack exchange as they may be able to provide a more in depth model. <A> Better question is, how do electrons carry thermal energy in any metal? <S> On the other hand, we could crudely answer OP's question in terms of the semiconductor physics behind LEDs and PV cells. <S> In LEDs, if a wandering electron falls into a hole, both carriers cancel, and this releases energy, either EM waves (photons) or lattice vibrations (phonons.) <S> The opposite is also true, where either lattice vibrations or EM waves can be absorbed, elevating a carrier to a higher-energy band and producing an electron-hole pair. <S> The hot-junction of a Peltier device is like an "LED for phonons," emitting energy as the carriers cross a junction and fall down an energy-hill (the built-in junction-potential,) falling to a lower energy-band. <S> The cold junction is opposite, where carriers can cross the energy-hill but only if they randomly absorb ambient energy and rise to a higher energy-band. <S> Don't forget that Peltier arrays involve Nonrectifying Junctions , where no depletion zone is blocking reverse currents. <S> Won't ordinary diodes exhibit these effects? <S> Yep. <S> But any thermal emission at the PN junction is exactly countered by thermal absorption at the metal-N and P-metal connections, where the diode's metal leads connect to semiconductor. <S> If you made yourself an inch-long germanium diode with a pn junction in the center, you should notice that when the junction becomes warm, the metal contacts both become cold. <S> And no diode is needed: your junction could have been (++p)-metal-(--n).
| A conducting electron must gain energy as it transitions from the conductance band in the conductor to the conductance band in the semiconductor and must lose energy vice versa.
|
Paralleling output of multiple BLDC motors used as generators First of all, I apologize for anything stupid I say as I am not yet well versed in the proper technical terms and schematics. I wish to use several identical BLDC motors that will be spun at the same speed (chain driven off same source) and used as generators rectified to DC power.I am curious if there is a simple way to safely parallel their output (combine their output current). Since the motors are identical and spun at the same speed, I may potentially get away by using identical three phase bridge rectifiers for each motor and then paralleling the output of the rectifiers. However, I know that is not a reliable solution as manufacturing tolerances and other variables will still mean that the output from each rectifier will not be 100% equal, thus possibly causing a catastrophic situation. If anyone has any advice on how to do this properly, I will greatly appreciate it.Or is it even possible? Thank you. <Q> You should use the same model motor (with the same turns per winding/phase), mixing different motors will not work well. <S> There will be no catastrophic results from this paralleling, simply a mismatch in current provided by each generator. <S> Even with the same model motors, you will have potential differences from each generator, so the current/voltage from each will not be equally balanced. <S> However since the motors have winding resistance this will help to balance the current from each motor/generator under load. <S> Providing the voltage output curves of the coupled motors are within a volt of each other you could take active control of the balancing. <S> You could (if you can tolerate the losses) put a single high side diode on the output of each 3 phase rectifier and use a P-FET to partially short it out to even up the current flows. <S> This could be driven by an analog signal or a PWM signal. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You could also replace the three low side diodes with a FET and control the balance using those signals, but it's more complex. <A> Ideally you want the torque to be constant with rotation and not pulsed by pole positions , so the more motors the better with 360 phase split evenly between gear timing belt position like a V-8 with 8 motors uses the camshaft split the cycle or crankshaft position and a flywheel dampens the vibrations of torque. <S> Alignment of phases and sizing of a flywheel depends on your motive power source and inertia you need relative to the torque curve. <S> You will still need a MPT tracking system to prevent stalling, allow buildup of speed to develop torque on your power source and maintain adequate efficient RPM as power = <S> Torque & RPM. <S> This would be done by PWM ganging of motor FET rectified output. <S> Each pair of motor wires will need a FET Half bridge for low performance losses. <S> You would also have to evaluate the profile of torque vs RPM of your power source which may not be linear. <S> With 3 motors you can simulate 3 phase power with 6 diodes or 6 MOSFETs using current and RPM sensing to either maximum Power or limit the torque load as you see fit. <S> I have taken my stepper bridge driver system in full-step mode and turned off the power then moved it only to see the 12V fan start to move. <S> You may also find that moving the motors by hand can generate the DC voltage back to your battery source. <S> But I cannot guarantee your system will be as reliable or safe to do so for CMOS. <S> I left USB power on and disabled the 12V motor power. <S> No load voltage may be sine but <S> current depends on magnetic pole current commutation method used and tends to be distorted when loaded. <S> The battery ESR, charge C ratio, gear ratios of cogged belts ( like a timing belt in a car) bridge commutation deadtime and other system variables requires a lot of design to optimize. <A> A low resistance (like an 1-2 ohms, or maybe even less) at the output of each rectifier just before the node where the currents combine would balance out the currents due to any voltage differences (such as those due to differences between the kV of individual motors). <S> Of course, it would affect efficiency. <S> Each motor has its own winding resistance though which might be able to take the place of a discrete current balancing resistor <S> but I am not sure if they are sufficiently high enough. <A> I think y’all are overthinking this. <S> Tie the coils together and sync up the shafts (they’ll tend to do that anyway when they’re coupled.) <S> Then tie the coils to the 6- diode set. <S> Or wire them to separate diode sets. <S> Ideally, phase the shafts for best interleaving / minimum ripple.
| Providing you use a 3 phase diode bridge for each generator, you can parallel the outputs with impunity.
|
Selecting coaxial cable and connector for PCB referring to required frequency I want to transmit my 24GHz signal from PCB through coaxial to antenna. I wonder how to chose best match for it, also the power is -5dBm, 50 ohm. It's based on ADF5901 2ch PA. I have found two good site but still I think I didn't found my way: https://www.rosenberger.com/us_en/test-measurement/precision_connectors/ http://precisionconnector.com/ https://www.te.com Most of them are for cable and not for connecting to PCB, specially I want it to be surface mount. The cable will be too small, the antenna is placed on other board and have few Centimeter distance from transmitter, I want it's cable be flexible and thin. It is my first project on RF board I have zero experience, please help to chose the best cable and connector. <Q> EDIT: <S> Subsequent to posting this answer, the OP has clarified his original question. <S> To get your match, you need your PCB tracks to have the same impedance as the coax and the coax connector. <S> You'll need to calculate this based on your chosen track topology (ie. <S> microstrip/stripline/cpwg) and your stack-up (specifically, the height and the permittivity (er) of the substrate). <S> There are calculators for these all over the place, such as on Microwaves101: https://www.microwaves101.com/calculators <S> Other users will probably give you a more detailed answer, but I do suggest you read up on some RF design guides/tutorials, such as this one on the Maxim website: <S> https://www.maximintegrated.com/en/app-notes/index.mvp/id/5100 <A> See here <S> I would choose SMA as it is a widely available part and the cable assembly tools are easy to find. <S> When it comes to cables, look at the pre-made cables, as those are tested to work at those frequencies. <S> If you still want to make your own, look at the datasheet for these cables to see what connectors/cables the use. <S> Please note that not all connectors or cables are made the same, so make sure you check the datasheet for the maximum frequency supported, and the impedance. <S> As others have mentioned, ensure that you choose a substrate that can take these frequencies, such as Rogers or other exotic materials. <A> At 24GHz any flexible cable/connector system is going to have horrible losses, I would be thinking about handi-form if not hardline for this, probably with an isolator on each end to handle the inevitable mismatch. <S> I would not be at all surprised to see 6dB of losses between the connectors, the isolators and the line loss! <S> Actually, I would just put the damn chip on the antenna board, less losses that way and overall a much simpler design. <S> Connectors for 24Ghz are NOT trivial, so go to some lengths to avoid them, far better to move the IF and reference frequency between boards and only have the quick stuff on the antenna board. <S> A reference and IF at a few MHz to a few GHz are much easier to deal with then 24GHz across connectors. <S> To give you a flavour of just how tough this stuff is, my Gore test port extension cables cost well over $1,000 for a foot of cable with two 2.92mm precision connectors (And they are NOT specified to 24GHz!). <S> You do have access to a VNA good for 24GHz? <S> And the cal kit to make it useful? <S> Otherwise you have basically no chance of pulling this off. <A>
| You need to select the best design for dielectric properties, controlled impedance, microvias, layout, connectors sources and have a VNA full calibration experience using semi-rigid copper cables to make this happen. There are plenty of connectors that can support 24GHz, I usually use Hirose, as their range is quite comprehensive, as well as they are usually available from most vendors.
|
Make an LED (20 LEDs) circuit battery efficient I want to wire up a miniature wargaming figurine to have 20 LEDs. As I've poked around options my concern is the battery life and how often the battery would need to be replaced. Since each gaming session lasts 4-5 hours, it'd be great if I could find a battery that would last multiple gaming sessions (though I think this is impossible...) I'm VERY NEW to all of this so I would appreciate explanations/solutions that understand I have little knowledge of the terms or techniques (I still don't understand what current is...) EDIT: Here are more details on what I'm trying to do I am trying to light up this figure (Knight Castellan). Locations for the lights would be (and their specs): 8 blue for right arm (3.4V, 20mA) 2 Red-Blue-Green for the eyes (3.4V, 20mA) 10 red/orange lights for left arm and engine (2.2V, 20mA) So altogether that's 20 LEDs, all at 20mA and none above 3.4V. The battery would go inside the torso -- I can't find an exact image but the dimensions are roughly 2" wide, 2" tall, 3" deep. My concern is battery life, as I'd like the batteries to last at least a few game sessions (so let's say 10-15 hours). I don't mind replacing or recharging a battery, but there's no point if I have to do it halfway through a game. I should add that I don't know how bright these LEDs are (the specs say 8-10,000 mcd for the blue ones), and that I don't know how to dim them, but would prefer them bright since they are usually in a well-lit room. <Q> The capacity of this battery is not 1000mAh, it is about 750mAh at low currents. <S> Note that the linked battery only provides 9V for a short period of time. <S> The rated capacity assumes that your device still works if the voltage drops to 5.4V. <S> You will actually get about 700mAh and the voltage will be 7V to 8V. <S> If the LEDs are not in series and you need 400mA you should expect no more than 1.75 hours of operation. <S> It will be difficult to use 2 LEDs in series if the voltage drops close to 7V, and the brightness will change greatly as the battery discharges. <S> Even if that was OK, you will get about 3.5 hours of operation in this case. <A> At 20mA your LEDs will be incredibly bright for such a small figure. <S> There are several solutions to the problems you have: <S> You could use the current sources I showed in the other answer and provide a small PWM to lower the brightness. <S> This would reduce the power consumption considerably. <S> Using a small simple MCU could give you PWM control of perhaps 3-5 strings each at a preset brightness for the individual LED strings. <S> It may even be possible to reduce the number of LEDs required by making a Fiber optic connection to a small number of RGB LEDS. <S> Use a small hole in some plastic to position multiple fibers above one of the LEDs. <S> The WS2812b comes already in LED strips form from many sources , and might be a good starting point for you. <S> This could be a very efficient means of providing your lighting and getting the runtime out to many hours. <S> The controller for setting brightness and color would not actually need to be within your model, you could send the serial data when you turn on the model, and then remove the controller. <S> The same controller could be used with multiple models (there are endless opportunities and variations for creating the control system). <S> The WS2812 works from 3.5-5.3V so a single LiPo battery without any regulation <S> is all that is needed; this would significantly simplify your wiring. <A> You don't want to run all of these at 20mA. 20 leds in a 6" cube of a model like that knight (It took a while to find the size of these things) at full brightness are not just visible in daylight, they will absolutely illuminate a room. <S> You can power them at a lower current, and they will still be visible in daylight and quite nice in the dark. <S> Since you power them at a lower current, they will have a lower forward voltage so you can use a smaller battery for longer. <S> 20 leds at 1 or 2 mA will run for days off a pair of AAA batteries. <S> Now you could power them all individually with a resistor each, but that's a bit wasteful and take up space. <S> They can be powered in parallel with a single resistor (Ignore those that scream that's a sin!). <S> The only issue with that is that you want red/orange, so you would have one string for the blue and one for the red. <S> The last thing is the RGB you want for his eyes. <S> You haven't said how you want those to work. <S> A fix color (4 pin RGB led that you combine multiple resistors to get a rough color)? <S> Or flashing (2 pin led with a built in micro controller that runs through a pattern, just add power). <S> The flashing led would work just the same, normally expecting 20mA at 3.3 volts, so power it at less <S> and it's good. <S> simulate this circuit – <S> Schematic created using CircuitLab Example schematic. <S> 3x AAA batteries for size. <S> About 13mA for both the blue and red strings. <S> 10 for the two color changers. <S> You could use NiMH 1.2V batteries but would probably want to use 75Ω resistors instead. <S> You could use 75Ω resistors if you want them brighter but <S> obviously that changes how long they last by a bit.
| About 22 hours with good Alkaline batteries (plus more due to the way the batteries discharging lead to dimmer leds which pull less current, but that depends on how dim you mind them going). You should be looking for another approach, possibly one that uses a high-efficiency switching regulator as a constant current source. You could use the WS2812 integrated RGB LED to get any color or intensity you want. Plastic Optical Fiber (POF) is dirt cheap ( see here ) and readily available.
|
Why do we use a transformer on some heating element projects when the wattage is the same on both sides of the trasformer? On some projects involving heating elements there's a transformer being used.and the explanation given is "we are using a transformer because need more current" but since the watts on both sides of the Transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the Transformer ? <Q> Power in a heater is given by the equation \$ <S> P = <S> VI \$ <S> where P is power (watts), V the applied voltage (volts) and I the current (amps). <S> You can see that there are multiple ways to achieve a given power - high voltage / low current - low voltage / high current. <S> For a transformer the relationship between input and output is given by \$ P_{in} = <S> P_{out} \$ <S> (ignoring the few percent losses in the transformer). <S> From the previous formula we can write \$ V_{in}I_{in} = <S> V_{out}I_{out} \$ . <S> This allows us to step the voltage up or down to meet the requirements of the load. <S> ... <S> but since the watts on both sides of the transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the transformer? <S> Yes, if the resistance of the wire is the same. <S> A high voltage heater will use a very thin wire with high resistance. <S> If this proves too fragile or too thin for the application then a thicker wire can be used but the current will have to increase to get the same current density in the wire. <S> Since the wire is thicker <S> it has a lower resistance so a lower voltage can be used too. <S> An example is the butcher's bag sealing machine which uses an 'impulse' element nichrome (or similar) wire to seal the bag. <S> Here a certain thickness of seal may be required and so a lower voltage, higher current is required. <S> The transformer 'transforms' the voltage and current to match the heater and provides electrical isolation from the mains eliminating a shock hazard. <A> If you have a 1000 Watt kettle designed for use in North America, where we have 120V power in the kitchen, then take it to Europe where they have 240V, the kettle will consume 4000 watts (resistance unchanged, but double voltage, power is voltage squared over resistance). <S> To use the kettle safely in Europe, you will need a step-down transformer to convert the 240V supply to the 120V that the kettle expects. <S> With the step-down transformer the kettle will consume 1000 watts, and the transformer will draw 1000 watts from the 240V source. <A> A heater is essentially a resistor with some resistance \$R\$ <S> given in Ohms ( \$Ω\$ ). <S> Suppose we have (for example) <S> a \$36 <S> W\$ <S> heater designed for \$12 V\$ . <S> This heater would have a resistance of \$4 Ω\$ . <S> If \$12 V\$ is connected to the heater, a current of \$I = V/R = 12V/4Ω <S> = 3 A\$ would flow ( Ohm's law ). <S> This would dissipate a power of <S> \$P = V\times <S> I = <S> 12V\times3A <S> = 36 W\$ . <S> but since the watts on both sides of the Transformer are supposed to be identical, wouldn't the heating element dissipate the same energy with and without the Transformer? <S> If we connect the (nominally \$36W\$ ) heater directly to the ( \$220 V\$ ) mains (without a \$220V\$ to \$12V\$ transformer), then a current of \$220V/4 Ω=55A\$ would flow, causing a power dissipation of \$220V\times55A=12100W\$ . <S> This will destroy the heater and/or blow the fuse/breaker. <S> A \$36W\$ <S> heater designed for \$220V\$ would need to have a resistance of \$1344Ω\$ . <S> If we were to connect this heater to \$12V\$ <S> it would only dissipate \$0.1W\$ . <S> The power dissipated by a heater depends on the voltage applied to it . <S> If you connect a heater to a too high voltage it will heat up too much, if you connect a heater to a too low voltage it won't heat up enough. <S> A transformer increases the current while decreasing the voltage (or vice-versa). <S> In the scenario where the \$36W\$ heater is connected to a \$220V\$ to \$12V\$ transformer, the heater draws \$3A\$ from the \$12V\$ supply (provided by the transformer), but the transformer, in turn, draws only \$0.16A\$ from the \$220V\$ mains (ignoring losses in the transformer). <S> So the transformer "increases" the current, but this is only part of the story.
| The transformer is needed to get the correct voltage.
|
Why do LCR meter manufacturers give the details of measurement frequency to us? When you want to purchase a DMM, more probably you don't see at what frequency the DMM you considered would work, but in the situation you're considering to purchase an LCR meter, manufacturers give you big details on the frequency of the LCR meter. Why do they give us the details on the frequency of their LCR meter? It seems that they gives us the measurement frequency for all of three components. Resistors, inductors, and capacitors. Okay, but how the best frequency be chosen for each one of these components? Here is several links to some LCR manufacturers: IET/QuadTech 7600 Plus Precision LCR Meter Lutron LCR-9184 HIOKI IM3536 CEM DT-9935 MS5308 <Q> They give you the details, because for real-world components there isn't one concrete value; they are all dependent on the operating frequency to a greater or lesser extent. <S> The best frequency will depend on your application. <S> Ideally it would be the same as your target operating frequency. <S> For example, if you are measuring an SMPS inductor then a high frequency will be more useful ( <S> as your circuit operating frequency is likely to be high), an electrolytic capacitor would be best measured at a lower frequency (as they don't have, and you wouldn't design expecting them to have, good high-frequency performance) and a line-frequency choke would be best measured even lower. <A> If we consider a real inductor there are several factors to take into consideration. <S> The windings have resistance <S> There is capacitance between turns The permeability of the core is frequency dependent. <S> As such, trying to produce a single figure for inductance is difficult and the result will vary depending on the measurement frequency. <S> Similar arguments exist for why a capacitor or resistor is frequency dependent too. <S> Since the answer varies with frequency the manufacturer has to tell you the test frequency. <A> Any good DMM will specify very clearly what is the operating frequency range, the measured signal characteristics, and the DMM parasitics. <S> Some DMMs even have 'fast displays' that operate at a different bandwidth than the slower digits, these will be clearly specified in the marketing material itself. <S> Some markings will only be understood by the experienced user, e.g.: "True RMS". <S> So it is not true that DMMs don't specify these characteristics, look at the low-end general purpose Fluke 114 marketing pamphlet . <S> It is clearly seen that it is designed for signals below 1kHz. <S> Although they do not specify loading impedance, you can go to the user's manual to find that. <S> (BTW: the models just above this one, 115 & 117, measure capacitance by applying a known charge and measuring voltage <S> , these do not use a frequency sweep). <S> But this is a general-purpose low-cost (for a Fluke) automotive and household multimeter, it is not really intended for the electronics professional. <S> If you look further up their extensive and expensive DMM product line you will find more detailed and specialized specifications, geared towards one market or another. <S> When it comes to an inductor or capacitor these are far from ideal components, their actual value will depend on many factors. <S> Temperature, DC bias, applied signal magnitude, and yes, test frequency. <S> Depending on the application some of these factors will be important while others will not. <S> Test instruments are designed with these factors in mind. <S> That is, the instrument specifications will strongly depend on the intended market. <S> The market for most LCR meters is not your common household as it is the case for the cheaper DMMs. <S> You cannot really compare a $10 disposable multimeter with a $6000 professional LCR meter. <A> When you look at manufacturer's specifications for inductance you will sometimes see odd-ball test frequencies. <S> e.g. 7.7 MHz, 25 MHz, 2.5 MHz. <S> These are an artifact of the old Boonton/Hewlett Packard Q-meter. <S> In the old Q-meter you measured inductance by determining the resonance point of the test inductance with a variable capacitor. <S> The dial on the variable capacitor was calibrated in picofarads. <S> But in addition, the dial was calibrated in inductance (mH, uH, nH). <S> The calibration for inductance was only good at specific frequencies. <S> So at 2.5 MHz the dial would show 100pF as well as 40.6uH. <S> These would be another set of inductance numbers on the dial for 7.7 MHz (4.3uH).
| It is a matter of the expected user knowledge and the intended market.
|
Are there any traditional colours to use for the wiring in a 4 mA - 20 mA current loop? I am designing a cable that carries a 4 - 20 mA current loop. Considerations: The positive lead is connected to VCC. The negative lead is connected to a 120 Ω resistor (other end of the resistor is grounded). <Q> Traditional Traditional standard colours for DC are red for positive and black for negative. <S> In this case it could cause confusion and result in the black being grounded. <S> To avoid this I would try red and white or red and clear, if available. <S> Current practice Industrial sensor leads are now standardised as brown for positive, blue for negative and black for signal. <S> That would leave you with a brown-black pair possibly leading to maximum confusion. <S> If you supplied all three cores and leave the blue unconnected at the sensor end then that confusion would disappear but a new one would be introduced: where does the blue connect to? <S> Figure 1. <S> Wiring for the ifm <S> TAD981 sensor. <S> Two-wire colour coding shown on left. <S> Selecting a random 4 - 20 mA sensor from the ifm range shows brown and white. <A> I'll bite. <S> Traditional: <S> Red = Positive = <S> (+ <S> ) = Vcc Black = Negative = <S> (-) = Resistor. <S> Not necessarily twisted pair. <S> Perhaps due to how current-meters and/or voltmeters were connected in prehistoric times. <S> However, any color combination will will work , as long as you document it. <S> There are dangers of choosing unwisely - like your poor users constantly hooking things up backwards. <S> I bet dollars-to-donuts that some physicist is responsible for that. <S> In essence, the "hotter" color tends to be connected to (+). <S> Yes, very technical ;-) <A> white & red <S> (rare but seen in the wild). <S> The convention that is common in my locale for signal cables is "White Hot". <S> Even the local electricians respect this convention even though it is contrary to the Canadian Electric Code for power conductors (White is always Neutral). <S> But you aren't wiring power here. <S> In the above pairings: black is low-side for the first two pairs, white is low-side for the white & red pairing. <S> Again this is what is considered to be common practice in my locale. <S> This may be different where you are.
| There are three common pair colors in these single-pair cables: black & white (most common) black & red (not as common) For an example of a non-traditional color-code (from another device type), see the color code for a J-type thermocouple ( https://en.wikipedia.org/wiki/Thermocouple#Type_J ). The easiest jacketed cable that I can get in my locale is 22 AWG twisted-pair in PVC or polyurethane jacket.
|
Alternative to alternating relay? or help? I'm using a littelfuse relay shown in the link to switch motor directions. I'm sending a pulse to the relay switch of 0.02s in order to switch between the L1 and L2 contacts. The pulse seems to be too short and the relay picks up on the pulse sometimes, not always, which is a major problem for me. If i increase the time of the pulse duration, the relay works perfect but the accuracy of my project reduces, so the shorter the time the relay picks up on the pulse, the better. What are some alternatives to this relay, or how can i improve my design to make sure the signal is picked up by my relay? Thicker wire, shorter wire, caps? I dont know. I'm only knowledgeable in the basics plus just a tiny bit extra and i can't figure this one out. Thanks in advance for your time and help! edit: I've included a wiring diagram with part numbers available. The counter is counting up to a pre-set number XX. Once it gets to XX, counter sends pulse to Relay #2 and reverses motor through the speed control. The pulse duration needs to be AT MOST 0.02s in order for my accuracy to be sufficient. I was thinking if there were some kind of solid state alternating relays? The motor is a 90V DC from Dayton. I hope that's a little more information. I didnt want to add too much so that you all didn't have to do all the work, but any and all suggestions are welcomed. LittelFuse Relay Diagram: <Q> The datasheet for the relays says that these are intended for operation from an AC supply. <S> This is either 50 or 60 Hz. <S> You state that your pulse is only 20ms long and results in unreliable operation. <S> Increasing the pulse width increases the reliability of the relay but degrades the accuracy of your process. <S> You may now see where I'm going with this. <S> Assume that the signal detect network for the control input to the relay <S> uses a half-wave rectifier. <S> The control signal timing is asynchronous relative to the AC mains that the relay operates from. <S> Depending upon the phase of the AC mains supply when the control signal is initiated, the relay may not see an entire half-wave of the control signal. <S> It is entirely possible that the relay sees only two pulses: the end of one half-cycle and the beginning of the next half-cycle. <S> The datasheet does not mention a minimum control pulse width <S> but I suspect that a minimum of two or three cycles of the AC mains is required for reliable operation. <A> from https://www.grainger.com/product/DAYTON-DC-Gearmotor-90VDC-1LPW5 that torque and RPM computes to 0.02 HP output power or equiv to 16W or 180mA full load and 1.8Ap start and 3.6Ap change direction <S> so a 15~20A relay is what I suggest. <S> with an RC snubber for more contact life and EMI. <S> Required <= <S> 20ms <S> Operate Time 13ms <S> Release Time 10ms <S> 12V or other options avail in datasheet https://www.digikey.com/product-detail/en/te-connectivity-potter-brumfield-relays/K10P-11D15-12/PB324-ND/254541 <A> Your wiring diagram was useful. <S> I can tell you what I would try <S> but you may not be able or willing to do this. <S> I see two approaches that may allow you to keep using your existing Littlefuse relay. <S> Both involve increasing the pulse width of the trigger signal without affecting the normally-closed contact on the counter module. <S> Problem is that this added relay may have the same limitation as the Littlefuse relay. <S> But it is something that you can look at. <S> 2) Use a diode & capacitor to stretch the pulse into the Littlefuse relay. <S> This requires some experimentation on your part because we don't know the polarity of the trigger signal. <S> Nor do we know the input impedance of the control signal pin. <S> But this is the approach that I would look at. <S> Start by adding a diode (1N4005 or similar) in series with the control input right at the Littlefuse relay (pin 5). <S> Test to see if the system still works as before. <S> If it doesn't work, reverse the polarity of the diode and test again. <S> If it still doesn't work, abandon this approach. <S> Now add a 1uF 400V capacitor between pins 3 & 5 of the Littlefuse relay. <S> Try the system again. <S> There is a really good chance that 1uF is way too large <S> but it's a starting point. <S> Now just try different values of capacitor until you get satisfactory operation. <S> I would iterate until I saw a small change between capacitor being present or not, then try a capacitor value perhaps 3 to 5 times larger than that minimum value. <S> Something like this is extremely easy for me to do - I'm an old-timer <S> and I still have my old resistor and capacitor substitution boxes hanging around. <S> These are very old-school and intended for working with tube-type equipment. <S> The capacitors in the capacitor substitution box are rated at 600 Vdc.
| 1) Use another fast-acting timer to stretch the pulse coming out of the counter module that feeds the Littlefuse relay. In other words, the relay that you have chosen is most likely NOT suitable for your requirements.
|
Opto switch for detecting translucent polypropylene I am building a liquid handler (a CNC machine that pipettes small quantities of liquids). The machine switches polypropylene "tips" between pipetting different fluids. Positional accuracy within 0.5mm of the pipette tips is key. However, I have sourced tips from several different manufactures and in each batch there are some tips that are slightly bent/warped. I would like to find the position of the bottom of the pipette tip to calculate any offset into the g-code if the pipette tip is bent. (Note: The diameter of the bottom of the pipette tip is 0.75mm.) These tips are very fragile, so I would prefer not to bump them into mechanical switches. I wanted to use optical switches, but polypropylene has a high transmission of IR wavelengths, and thus do not trigger optical switches. So my question: is there a way to optically detect polypropylene? Polypropylene has a low transmission at 3250 nm. This is my first post to StackExchange let me know if detail/formatting is lacking. Thank you for your time. <Q> For an industrial application the normal approach would be to submit a sample of your product to an industrial sensor vendor and let them propose a solution. <S> A through-beam slot sensor should be able to give very good resolution. <S> You would program the machine to move the part across the sensor, stop when the beam is broken and then make your move relative to that position. <S> Figure 1. <S> An opto slot sensor has an infrared LED on one side and photo-transistor on the other. <S> You might be able to find one with a logic output which would simplify your setup rather than have to amplify the photo-transistor output current. <S> Figure 2. <S> Using polarized light to detect translucent part. <S> Image by LEDnique . <S> Another approach would be to use a vision system to report the position. <S> Illuminate the part from the rear using an LED backlight covered with a polarizing filter. <S> Add a polarizing filter to the camera and rotate to 90° so that the backlight appears black. <S> Now when a part is positioned in between the filters it will randomize the polarization of the light passing through it and the part will show up as a light image on a dark background. <A> Droplets of Poly. <S> will diffuse light easily and may be detected dropping past a vertical camera with a diffused back reflector such that the attenuation loss from the reflection coefficient and light spreading is easily detected with 1/60th second captures over a 1cm area or a droplet speed of 60 cm/s. <S> If it is faster, then define it. <S> It is possible to locate an optical sensor 10mm effective aperture camera array to accept a calibration drop and measure the offset from the optical datum home switch to measure any change from the working position home defined as the xy axis droplet centre position. <S> This can be used to create a new home work position from which a stepper gantry can then locate to the system design accuracy be it full steps of say 0.5mm or 1/16 steps of < 0.05mm accuracy. <S> The g-codes for doing the correction calibration is easy. <S> The challenge is to choose an optical reflective array sensor like a 9600 dpi laser mouse chip to detect the peak centre of the droplet and convert this into a geometric work 0,0 position which is just near the datum 0,0 optical home positions in a corner or single linear axis. <S> Cleaning the array sensor after calibration might be a manual operation. <S> Here the SNR of the optical array signals only have to report the error in the geometric xy mean of the optical array. <S> This can confirm as a go-no go or be used in error correction for position error in the servo by some method. <S> TBD <A> I doubt that you can easily find LEDs/Photodiodes/photo transistors in the 3250nm band, but you should be able to target the much narrower ~1450nm set of bands. <S> This LED operates at 1450nm with a bandwidth of 90nm. <S> It might have enough sensitivity around that wavelength peak to detect both of the polypropylene absorption peaks in that region. <S> You can use the same LED as a photodiode, although these are not specified for that purpose, its absorption spectra will be in the same region and should work well for your application. <S> You might probably have to mask them to achieve the mechanical tolerance you require.
| Since your part is translucent you could use polarisation of light to your advantage.
|
Protecting against a SSR failing closed In a project I'd like to control a heating circuit with an SSR, and I'd need a contingency plan for when the SSR fails closed, and the microcontroller cannot turn off the heating element, which would be unsafe. The easiest option I can see is to have two SSRs in series, and do a check on startup to check that turning off either will interrupt the current, and display an error as well as refuse to operate. What would be considered best practice here? <Q> The best solution for this situation is to use the single SSR to electronically control the heater. <S> But then add a couple more items to increase the safety aspect of the product. <S> Add an AC voltage detector circuit with an optocoupler that feeds into your microcontroller. <S> When ever you have the SSR on have the software look to see that the AC voltage detector shows voltage present. <S> Likewise when the SSR is off have the software check to see that the AC voltage is off. <S> It is a good idea to have the software do the AC voltage checks periodically to detect when the SSR is not maintaining proper operation. <S> Install a circuit breaker in the AC line in front if the SSR. <S> The best circuit breaker to use for this is one that has a built in solenoid trip circuit that will open the breaker under command of the microcontroller software. <S> You would so this under conditions that the SSR has shown a failed condition according to the monitoring of item (1) above. <S> An alternate method to provide a somewhat lower cost solution than the remote activated circuit breaker (which can be spendy) is to install a fuse instead of the circuit breaker in the AC line in front of the SSR. <S> You then install an additional circuit that intentionally places a load on the AC line that exceeds the rating of the fuse so as to intentionally blow the fuse where there is a detected fault condition with the SSR. <S> I do not recommend the idea of placing two of the same SSR in series. <S> Similar components have a risk that they could both fail due to some aberrant condition. <A> I would add a normally open mechanical relay in series with the SSR. <S> Let the microcontroller close the relay on boot, and make sure the pin's state is such that the relay is open when the controller boots and before any code runs (most controllers boot with tristated GPIO pins, so add a pullup/pulldown to your relay driver transistor. <S> Do not rely on the software pullup/pulldown to disengage the relay). <S> Enable the watchdog on the micro, so that a reset disengages the mechanical relay if the software crashes. <S> I assume that you have a temperature sensor for measuring the process parameter. <S> Add a timeout and a threshold to detect a failed open SSR or a misplaced/disconnected temp sensor <S> : When the micro decides to engage the heater and the measured temp does not rise e.g. 2 degrees in 30 seconds, bail out and open the mechanical relay. <S> Also make the micro turn off the relay when the temp is unsafely high. <A> It really depends on whether you are trying to achieve the highest MTBF (which can be applied to each component such as an SSR) or the highest equipment reliability which involves component redundancy used in a design. <S> The easiest option I can see is to have two SSRs in series, and do a check on startup to check that turning off either will interrupt the current, and display an error as well as refuse to operate. <S> This is a great strategy overall for high reliability, but is dependent on the absolute stress for each series component. <S> In your case where a heater is involved you need to ensure you lower the stress on BOTH SSR's or else <S> you compromise the achievable reliability. <S> A heater typically has a significant power up stress profile, it draws much more current at turn-on and until it heats up (3-6 times rated current is not unusual). <S> Since your strategy is to have two in series to increases redundancy both carry the surge current and therefore are exposed to the same stress rating at startup. <S> To reduce surge current stress, you need to soft start the heater (phase control). <S> To reduce voltage stress you select SSRs with a higher voltage rating and put surge suppressors across the mains input. <S> To reduce thermal stress you need good cooling. <S> Hot components have much reduced MTBF rates. <S> A good place to start is here with Crydoms reliability report. <S> Since SSRs (Triac and SCR based) typically have quite large voltage drops they can become quite hot and as the graph below shows <S> even 65degC case temperature can significantly reduce the MTBF.
| Further, add a temp fuse or temp switch in series with the relay and the SSR with a temp a little bit above the software limit, to disengage the heater in case of fatal damage to the control hardware. It is better to use different technologies for your safety strategy due to overall different failure scenarios for each.
|
Interpreting rating label of DC power supply - supposed to be a constant voltage source or a constant current source? Link to product store page Hi all, bit of a beginner question here. According to the output section of the label, this is a 12V, 1000mA power supply. This is confusing me a bit. Is this device meant to output a constant voltage or a constant current? My gut feeling is that it's meant to be a constant voltage source, and that the 1000mA value is some kind of failure threshold. THEORETICALLY SPEAKING , would there be a danger if the (tampered with and exposed) terminals of the output came into contact with human skin (say, at the hand)? Surely 12V won't be enough to hurt anybody? Also theoretically speaking, can I use this power supply (after voltage division) to act as a cheap power supply for my ESP8266 circuit? Or will this deliver way too much current and fry my delicate microcontroller chip? Thanks for taking a look at my extremely beginner questions, I really appreciate it. I'm trying to get into a hobby here ;) <Q> 12 V is the value of the regulated, stable output voltage. <S> 1000 mA (or 1 A) is the maximum output current you can draw and still expect the output voltage to remain at 12 V. Note that the max output power is 12 W. <S> This is consistent because (combining Joule's Law and Ohm's Law), power equals voltage times current. <S> Another note: <S> most switching power supplies cannot maintain regulation all the way down to zero output current. <S> A typical minimum load is 10% of the rated max. <S> In your case, this means that for loads less than 100 mA, your supply's output voltage might be something other than 12 V. <S> You can test this by connecting resistors of 120 ohms or more to the output and measuring the output voltage. <S> Be sure to use resistors rated for sufficient power. <S> The power dissipated in a resistor can be calculated using Watt's Law: P = <S> E^2 / <S> R. <S> For safety, use a resistor rated for at least twice the expected power dissipation. <S> A simple resistor divider will not work as a step-down regulator for an external device because that device expects the supply voltage to remain constant asn the current load varies. <S> The 8266 draws more or less current depending on what the circuit is doing, and this will cause its supply voltage to vary. <S> Not good. <S> Best to add a simple voltage regulator circuit to step the 12 V down to what is needed and hold that output voltage value constant. <A> Just about every off-the-shelf DC power supply you can purchase will be a constant voltage source. <S> The current rating is approximately the maximum possible current that the supply can deliver to a load. <S> 12VDC is not considered dangerous to touch to outer skin, but reasonable caution should always be exercised when working with electricity. <S> For voltage division, be aware that you might be effectively changing a regulated power supply that maintains constant voltage into to an unregulated voltage supply at the point where you draw the divided voltage. <S> This will happen if the effective resistance of the overall load is changing with time. <S> When you do the math for your voltage divider, include your load as "resistor" in the circuit and see what happens when this load "resistor" changes value... <S> the calculated divided voltage will change. <A> 1000mA is the maximum it can supply. <S> Your question: Or will this deliver way too much current and fry my delicate microcontroller chip? <S> It will deliver as much current as your chip pulls. <S> Don't think of it as the power supply pushing current into your micro, your micro pulls current from the supply acting as a load. <S> Resistance of skin is somewhere between 1k and 100k ohm. <S> Taking the lower bound of this (1k), the maximum current draw would be 12mA which is in the 'danger zone' so to speak. <S> 1k would most likely be for moist skin. <S> Normal dry skin can be way into the 10k range, so you would barely feel it. <S> Taking caution with any electronics is always a good idea. <S> If you take the approach that any electronics can kill you, then you wont go far wrong. <S> You can use that supply to power your esp8266 but a voltage divider wouldn't be the right way to go about it, as the power dissipation will most likely cause the resistors to get very hot. <S> A good option is to use a cheap adjustable buck converter that will regulate the power supplies output down to the 3.3V needed for the esp8266. <S> Buck converter module Hope that helps.
| The supply is constant voltage.
|
Can’t reproduce oscillations using a simple inductor capacitor circuit? I’m trying to reproduce a simple oscillator conductor / inductor circuit as shown here. I replicated exactly the same scheme on a breadboard and check all the connections using a led However when I try to get a measure of voltage it seems to stay constant even though according to the simulation on here (with exactly the same parameters) I’m supposed to have an oscillation. Link to simulation Here’s my wiring where I connected my osciometer What am I doing wrong and how can I have voltage oscillations back and forth like in the simulation? <Q> A couple of points: that's not an Arduino board, it's a breadboard (although it may have come with your Arduino which is the PCB with the microcontroller on it). <S> You missed some details in your simulation. <S> Figure 1. <S> Screengrab from the OP's simulation. <S> The initial oscillation is caused by the simulator starting up. <S> In real life it will be caused by connecting the battery. <S> Note that the oscillation is decaying in very few cycles. <S> On the second cycle it's already below half the initial amplitude. <S> Note the time. <S> This is just 60 ms into your simulation. <S> The result is that the oscillation has decayed before you even have time to read it. <A> The oscillation decay time is OVERDAMPED <S> so there is none if you had R=10, L=100uH, <S> C= 100uF <S> so the voltage rise time or current decay time <S> T=RC here 10*100uf= 1ms with voltage at 64% of target or steady state. <S> I added realistic ESR of the cap , choke and voltage source which must always be considered in any analog design when considering Q values for Q=1/ζ with damping factor ζ. <S> Graphical Solution using RLC Nomograph expects Zo=1 Ohm and with series R=10 to have Q=0.1overdamped and RC= 1ms pulse decay time. <S> Now if you short the 10 Ohms <S> what decay you get of the oscillation envelope is your ESR total in the loop. <S> ( we call it DCR for chokes) <S> Read the graph see the resonant frequency from the intersection of 3 variables to get the 4th (resonant frequency) <S> You can choose any 3 of 4 variables or just 3 as in RC,f-3dB. Or add your expected ESR's and simulate But be aware that short circuiting the electrolytic or e-cap can result in oscillations that momentarily reverse the voltage more than it's 10% tolerance perhaps to short of time to cause permanent damage if done only a few times. <S> e.g. with ESR values reduced. <A> The RLC circuit continously oscilates after applying pulsed voltages, not DC. <S> The simulation runs because of power supply first shot. <S> Try with squarewave voltage power source. <A> Without an active device such as a BJT or FET acting as an amplifier with feedback the oscillations will die down after a few cycles, as in your simulation, and you will be unable to see them with a scope. <S> You need to construct a suitable oscillator circuit using a Hartley or Colpitts oscillator.
| You'll need to add some active electronics to make it a self-sustaining oscillator.
|
What is meant by 'local' in local oscillator? We know that a local oscillator (LO) is an electronic oscillator used with a mixer to change the frequency of a signal. But what do we mean by 'local' ? How is 'local oscillator' different from normal oscillator? Why can't a normal oscillator, like an LC-tank circuit or an opamp circuit, be used in a heterodyne receiver? <Q> You got that wrong:The "local" in oscillator doesn't describe the kind of oscillator used. <S> It could just be an LC-tank <S> , it could be a crystal-derived oscillator, it could be something synthesized from a reference clock or something recovered from the data stream received: The "local" in oscillator refers to the fact that it's what the mixer uses locally to mix down or up, as opposed to the oscillator at the other end of the communication, which simply isn't the same oscillator. <A> Think of frequencies. <S> The local oscillator generates a local frequency, that is only used by your mixer as the intermediate frequency. <S> Whereas the transmitter frequency is not staying on your local board but flying through the air as radio waves. <S> The transmitter and the receiver does not need to use the same intermediate frequency, but could use different local oscillators. <A> simulate this circuit – <S> Schematic created using CircuitLab <S> The LO could be fixed or a synth or a VCO with PLL to tune RF to the fixed IF filter. <S> It may be used as an up- or down-converter.
| LO to an RF Engineer means a "local" sine wave signal or "within the device" (or Radio) used for heterodyning or RF mixing.
|
Converting 220v 50Hz to 110v 60Hz I have a situation which I'm not sure how to handle. I've purchased an appliance from US. This appliance is powered by electric motor (175W) and then I realized I cannot use simple step-down converter because they usually convert 220v 50hz to 110v 50hz, but in electric motors the frequency matters. Of course, such voltage/frequency converters exist, but they usually in kW range industrial grade converters with price tag ten fold of the price of the appliance. So, I'm thinking about taking 220v 50Hz AC to 12v DC converter and connecting it to 12v DC to 110v 60Hz converter, like pure sine or something. Does it make sense? Would it work? is there any better solution which wouldnt break a bank? In case my solution is good to go, how to calculate the wattage of each of parts I will use? For example, 175W it is nominal power draw of the motor, but it has spikes, so I guess, 300W output will be ok, maybe 350W. Then what wattage AC to DC I have to use for the DC to AC converter to produce 350W output given the power efficiency is not 100%? EDIT001: The appliance is WorkSharp Ken Onion Edition Knife & Tool Sharpener, hope it helps <Q> It depends totally on your appliance motor. <S> If the motor is a universal motor (it has brushes), then the frequency is much less important. <S> In this case a simple 220/120 transformer will work just fine providing it can support 2A at 120V. If the motor is a 60Hz AC synchronous variety you will lower both the speed of the appliance motor and increase the losses in the motor (so it gets hotter). <S> Since you gave no idea of the appliance or motor type <S> it's hard to judge what might be the effects of a lower operating frequency. <S> Update: <S> Since you now have given more information, I would suggest that the frequency (50 or 60Hz) will have ABSOLUTELY NO impact on the tool. <S> The manual shows that this is a speed adjustable application. <S> These types of motors are almost always a 'universal' brushed motor with a phase controlled speed adjustment. <S> Today we start to see BLDC motors used in some tools, but this does not appear to be one of those (they have an SMPS power supply). <S> As I said, IMO all you need is a 220/120 voltage convertor and your tool will work fine. <S> I would recommend that you buy a voltage convertor that clearly supports your 175W load (though I didn't see this power rating in the manual). <S> Something like this is ideal. <S> No bells and whistles and clearly just a transformer (though this one is an autotransformer <S> it's ideal for double insulated tools). <A> Since excitation current depends on V/f ratio and inductive impedance, you must consider the 60/50 ratio of higher excitation current of running it at 50 Hz when trying to match the 120V 60Hz rating for no-load excitation current. <S> If you wanted the same RPM, then you need a 120V 60Hz inverter. <S> If you just wanted to use a transformer at same power capacity and slower RPM , then you can either use 50/60 * 120V = 100V for same excitation current and full rated power of the motor P= <S> V <S> * I = <S> Torque <S> * RPM <S> Since the current is the same for no load at lower RPM but same V/f ratio your more torque can increase but less RPM yet same "product" = <S> Power <S> If you use 110V or 120V @ <S> 50Hz then excitation current increases. <S> ( higher V/f) and may risk overheating if you applied the rated torque at a higher V/f ratio than expected. <S> If it is a vacuum motor then consider replacing the motor at a repair shop. <S> Might be $25 plus labour or DIY online purchase $15. <A> Chances are about 99.9% that this is a "Universal" motor, the same kind found in most portable power tools and small appliances. <S> It will care about voltage, but not frequency because internally the motor converts the AC to DC and is technically a DC motor.
| It might be cheaper to just change the motor and anything else voltage dependent or replace the appliance. Don't buy a 'travel adapter' as many are based on an autotransformer and can be very hit an miss in terms of reliability.
|
What is the purpose of capacitors parallel to crystal? In many schematics, we have two capacitors connected in parallel to a crystal?Why do we have capacitors parallel to crystal? Whats the purpose?How are the capacitor values determined ? <Q> The capacitors are effectively in series with each other and the combination is in parallel with the crystal. <S> The total "load capacitance" is thus (ignoring the oscillator circuit and stray capacitances) <S> \$C_P= <S> \frac{C_1 <S> C_2}{C_1 + C_2} \$ <S> In a typically microcontroller or similar circuit there is an amplifier input connected to one side and a current amplifier output connected to the other. <S> Both have a bit of additional capacitance that adds to the load. <S> It's called a series resonant circuit <S> The oscillator is called a Pierce Oscillator . <S> There are two things that help you determine the value. <S> First there may be suggested or limit on the maximum capacitance from the microcontroller or chip manufacturer. <S> Best stay within that or your oscillator may not start reliably. <S> Secondly, you would like, for maximum accuracy, to match the specified load capacitance on the crystal datasheet. <S> That means taking into account the input, output and stray capacitances. <S> The effective total load on the crystal is: <S> So you want to pick C1 (usually = C2) <S> such that CL is close to the number on the crystal datasheet. <S> In general, it's not that fussy usually and you can use a few pF <S> (like 5pF) for the values of Ci and Co and Cs. <S> Because the crystal itself is a very high "Q" resonator, slight mismatches in the capacitance won't have much effect. <S> If you're trying to keep time-of-day, you may wish to be more fussy and/or adjust the load capacitance, but usually 100ppm either way doesn't make so much difference. <S> There are enough variables that you may wish to use whatever the recommended crystals are from the chip maker. <S> In general there no guarantee that your oscillator will oscillate otherwise, <S> since chip makers very seldom specify parameters such as transconductance (gm) limits over temperature that could be used to guarantee operation. <S> There is thus ample opportunity for the chip maker to blame you and the crystal maker, and the crystal maker to blame you and the chip maker (the uncomfortable common factor being you). <A> The manufacture of the crystal makes them so they run a very tiny bit fast until you install the capacitors of the specified values. <S> They will then run closer to the 'ideal' specified frequency. <S> Also these need to be good quality low drift X7R or no drift NPO types. <S> Note that extreme changes in ambient temperature will cause the crystal to drift off the ideal frequency. <S> If very tight accuracy over time is required then crystal oscillators are mandatory. <S> For extreme long-term accuracy TCXO (Oven controlled oscillators) are required. <A> The capacitors implement a phase inversion simulate this circuit – <S> Schematic created using CircuitLab
| Such capacitor topology is used to fine tune the crystals resonate frequency closer to what is on the label.
|
Can you and/or why should you not replace optocouplers with transistors when dealing with slightly different voltage I stumbled upon this: Source. The MDB bus is communicating at a 34V level and the other side is at 5 or 3.3V. Why are they coupled with optocouplers rather than a voltage divider on the master tx and a normal transistor on master rx? What are some potential problems with such a design? I am a layman at electronics so low level explanations would be great. <Q> You have optocouplers for both transmit and receive. <S> The voltage difference by itself could be worked out with resistor dividers, or a simple transistor circuit if a boost in voltage is needed. <S> The reason optocouplers are widely used regardless of voltage differences is that they isolate grounds as well for galvanic isolation. <S> You may have heard of 'ground loops' where the signal and/or Earth ground of different pieces of equipment are not the same. <S> Some equipment has an isolated ground, especially if powered by a wall-wart. <S> Those with an isolated ground can have small non-hazardous leakage currents that may cause problems with data links, so using optocouplers isolates the "grounds" to prevent any ground loops. <S> Also equipment that is Earth grounded but at distant AC outlets may have enough current difference in their grounds to induce noise in direct coupled comm ports. <S> Some ask you to run a separate signal ground from device to device to avoid drifts in a common 'signal' ground which often has a medium value resistor tied to Earth ground. <S> RS-232 and RS-485 (multi-drop) ports have a signal ground shared by all connected ports, though it has a high impedance Earth ground by using a resistor high enough so it is only a 'static' reference. <S> I worked with surge generators for 15 years to simulate lightning strikes, and when it fired a 27,000 volt pulse at 20,000 amps our earth ground wire that went to an isolated ground rod had for a brief time (20uS) 1/2 of the surge voltage on it, or -13,500 volts on it. <S> The oscilloscope had differential inputs and we used probes rated for 75KV, but we had to use an opto-isolator on the GPIB bus to the computer, or we could fry the GPIB card in the PC. <S> We installed the isolator after 2 fried GPIB cards. <S> At times the optocouplers save equipment from real damage. <A> So signals can be transmitted without any electrical connection. <S> In this case it could be the possible since we don't what the circuitry (and especially the 0V line) at the 36V part is doing. <S> Another reason to use optocouplers could be to avoid EMC issuses. <S> LM2576 can switch up to 3A, these 3A are coming from the 36V part(since the LM2576 is a Buck Conveter) and so conduct through its ground line(0V). <S> These high currents can rise the ground potential between other parts of the circuit(conductive coupling) and distort the signal integrity. <S> This issue can be conquered by laying out seperate Power and Communication Grounds as its done here. <S> So this optocouplers could be to avoid a ground loop bringing further EMC issues. <A> Optoisolators are excellent for common-mode rejection. <S> These are bi-directional optoisolators which give excellent immunity from isolating common mode noise. <S> This galvanic isolation offers many other benefits for HV isolation and transient protection. <S> The only down-sides are the tolerance on LED emissions and the GBW product limitations of the detector and possibly the CTR or current transfer ratio choice tradeoffs. <A> Optocouplers are normally used for galvanic isolation. <S> The "connection" between the two sides of an optocoupler is just light. <S> The emitter and detector are physically separated inside the optocoupler, so no current can flow from one side to the other. <S> It would take a very high voltage to make current from one side "jump across" to the other side of the optocoupler. <S> The voltage required in the case of the 6N137 in the circuit is 8000V peak. <S> Apply 8000 volts to one side of the optocoupler, and the other side won't notice it. <S> The thing is, though, that the circuit referred to in the question isn't using the optocouplers for galvanic isoation. <S> Both sides of the optocouplers use the same ground. <S> That breaks the galvanic isolation. <S> If you zap one side of that circuit with 8000 volts, both sides will "feel" the effects. <S> In the given circuit, both optocouplers could be replaced with transistor circuits that would be cheaper, simpler, and electrically faster than the optocouplers.
| Normally optocouplers are used for galvanic isolation between circuits.
|
How do I identify this component? It just says 0 on it Can anyone help me identify this component? It just says 0 or D on it (I think it's a zero). I tested it for conductivity with a multimeter - doesn't seem to have any (tried both polarities in case it's a diode). Tested for resistance: nothing (i.e. infinite resistance), although I'm not 100% sure I was touching the contact points correctly, so don't take this as an ultimate measurement. Thanks in advance! <Q> This is a 0 Ohm resistor. <S> Probably in 0805 package. <S> If the multimeter shows infinity (open circuit), or any significant resistance > 0, when measuring the element's resistance it means that it is damaged. <A> That is an 0 Ohm Resistance. <S> they also called SMD jumper resistors. <S> They are used as wire links to connect the traces on Surface mount boards, which can be assembled using pick and place machines easily. <S> (same like jumper wires in through holes boards). <A> It is an SMD resistor . <S> As stated in the comments, the number indicates its resistance value; namely 0. <S> You mentioned you couldn't measure any resistance across it. <S> Are you sure it's not coated? <A> Be careful using a solder blob or too heavy a gauge of wire you might blow other components in the circuit. <S> An analogy: Your car has blown a 5A fuse and your radio stops working <S> , well you check and you don't have any 5A fuses on hand. <S> So.... you substitute a 10A fuse instead, you could cause damage to the radio or whatever shares that circuit because too much current flowing through the 10A fuse. <S> Better check to see if deices in that circuit could have caused to infinity reading in the 0 ohm resistor.
| The multimeter should show zero resistance when measuring such "resistor". On the PCB you can replace it with a solder blob or a piece of wire.
|
Controlling 12V Magnetic Latch Solenoid with RaspberryPi 3b As the title says I am trying to control a solenoid with a raspberry pi. I am currently using an npn transistor (will post link with parts at the bottom) however I am having a hard time getting it to work. The collector is at 12V and the gate is controlled by the raspberry pi gpio pin (runs at 3.3V and 16mA (this is the max amount I believe)). Attached is a picture of a circuit I used to verify my solenoid circuit. It is essentially the same except that it has LEDs and that it runs on 3V instead of the 12V for the solenoid. Unfortunately I do not have any pictures with the solenoid. It would look roughly the same except that the two positive legs on the solenoid replace the LEDs and the single ground wire is hooked up to the negative row on the breadboard. This verification circuit works perfectly, but the solenoid circuit does not. I have read online that MOSFETs can also be used as a switch, would this be better? In the case that it is, I would like to try to get the NPN transistor to work as I would rather not use more money. Any help is greatly appreciated. Thanks. Solenoid (15mm version): https://www.pneumadyne.com/way-normally-closed-latching-solenoid-valves-p-1448.html#1-YToxOntzOjQ6ImdyaWQiO2k6MDt9 Transistor: https://www.mouser.com/ProductDetail/Diodes-Incorporated/ZTX688B?qs=sGAEpiMZZMshyDBzk1%2fWizF0eVyYE44ZxbGt8e1SnMg%3d Images: https://imgur.com/a/gOnuPFT I have included a diagram of what my solenoid circuit looks like.For reference on the transistor labeling, as I am not sure what is common. B - base (gate)C - collectorE - emmitter <Q> Your circuit does not work for the solenoid or supply voltage that you have. <S> You have a 3 wire (2 coils) <S> self latching solenoid that specifies the common as the negative pin. <S> The data for the solenoid shows that you have a 4W drive requirement. <S> This means that at 12VDC you need about 340mA to activate each of the solenoid coils (latch and release). <S> The transistor you selected is completely unsuitable for your task. <S> It has enough Collector current capability but has a V(CBO and V(CEO) of only 12V. <S> You would never use a device at its absolute maximum rating. <S> A potential circuit would be as follows <S> and you would need two GPIO pins to drive the latching solenoid: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It would be up to you (in software) to ensure the A or B coil is activated for the appropriate length of time to latch and unlatch. <S> Note: <S> You must have the pulldown resistors on M3 and M4 to ensure that neither coil is activated as you turn on and boot the R'Pi. <S> The GPIO pins are all set as inputs on power-up, so the 10K Ohm resistors keep M3 and M4 off. <S> Update: <S> The documentation for the solenoid switch <S> you use specifies the polarity of activation <S> (I assume this is required since the latching is by a magnet, so pole polarity sensitive): <A> I would also add a diode to protect the transistor from the spikes caused by switching the inductive load. <S> BTW, the transistor value is just the default part that comes up when I drew the schematic. <S> Please look up the datasheet for the absolute maximum values. <S> I suggest that if you're are in doubt of how a circuit works (or doesn't) try simulating it in LTSpice; It is free, and fairly easy to use, and it will save you a lot of headaches. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As it has been pointed out that the solenoid is polarised, and that OP is a newbie, I shall present the easiest and probably cheapest solution possible.it also includes a protection from activating both coils at the same time. <S> simulate this circuit <S> The relay board can be found for just over a dollar on Aliexpress <A> When using transistors in switching applications, it is always a good idea that they are referenced to a voltage that will not move on you. <S> If you notice on @peter bennett ‘s suggested circuit, the NPN is driven by the GPIO of the RasberryPi referenced to GND (emitter is at GND, and the base is driven by the GPIO which supplies the base current via the base resistor as (Vgpio - 0.7)/Rbase), and the base of the PNP is driven referenced to the “fixed” battery voltage at its emitter. <S> Notice as @Jack Creasy pointed out, that you need a current limiting resistor in series with the collector of the NPN used to drive the PNP, or you will probably fry them both. <S> Otherwise, that should be a perfectly good circuit for driving a regular solenoid. <S> I need to review how the latching solenoids <S> work—-if <S> i remember, once they no lnger need to be driven anymore . <S> However, to disengage them, i think you need to drive the current through the solenoid in the opposite direction. <S> If so, @Peter Bennet’s circuit will not achieve that. <S> You could accomplish this by using an H-bridge.
| You are going to have to buy suitable devices to drive your solenoid. As mentioned by Peter Bennett, you should change the transistor so that it doesn't work at its limit, and wire it as shown here below.
|
RC Lowpass Filter between Amplifier and ADC input I have sensors (pyranometers that consist of thermopiles and measure sun irradiation) that ouput a low voltage signal so I need to amplify them using an instrumentation amplifier. I have chosen the AD8237 for this task: Datasheet I'm using a gain of 100 to amplify the initial low voltage signal (ranging 0-20mV) to 0-2V range. I'm then feeding the amplified signal to the ADC (MCP3422): Datasheet My sensor values change very slowly and I will read out the digitized ADC values only once every second, so speed is not important in my case. Now as pointed out in the accepted answer in this question I need a filter between the IN-Amp and the ADC to filter the noise.In many ADC-datasheets a simple passive RC-filter is suggested between the INA and ADC.I did quite some research and I still have some questions that confuse me and I hope you can help me with: I figured that a first order RC filter does not meet my requirements so I cascaded multiple RC stages: simulate this circuit – Schematic created using CircuitLab And this is the simulated filter response in LTSpice: Questions: Can I do that? What would be the disadvantages using the proposed filter? Capacitor values like 47µF or even 100µF give me a even better response (stronger attenuation,) would that have a negative impact on my signal or the ADC? I guess resistor values should not be further increased to prevent voltage drop on my signal. The filter response seems very promising: Signals at 10Hz already attenuated by ~50% and at 25Hz already by ~90%. As I only care about the DC signal I guess that response should be fine (also 50-60Hz range is covered strongly by the filter.) Resistors create voltage drops so how would these three cascaded resistors affect my amplified signal (thus my digitized value calculated by the ADC?) Ohms law should apply, but I do not know the current...Any clarification on this is highly appreciated. Regarding speed/time constants: As my data acquisition (readout ADC once per second) and change in sensor value is very slow, do I need to keep an eye on the speed/time constant of this filter? As many datasheets suggest an RC filter stage this approach should not be too far off. <Q> If you are reading the ADC just once per second then you need to eliminate frequencies above 0.5 <S> Hz to prevent aliasing. <S> If you think your system will have noise at, say 10 Hz, then that noise will contaminate your readings. <S> I recommend that you sample at a much higher rate, perhaps some multiple of the power mains frequency, and perform low-pass filtering in software. <S> Even a simple moving-average filter would work and wouldn't take much processing. <A> ADC inputs typically are quite high impedance. <S> I have often used 100K in series with an input with no dc loss. <S> (and a capacitor to ground for filtering) <S> If you are happy with the attenuation with that circuit then I would suggest scaling the resistors up and the capacitors down. <S> I would not use electrolytic capacitors as they tend to have more leakage compared to other types. <S> I would probably use a ceramic cap. <S> Edit: <S> I just looked at the data sheet for the part. <S> Go take a look at page 3, Input Impedance. <S> Loading certainly will not be an issue. <A> This 3-section RC should provide better rolloff at high frequencies. <S> The random noise is dominated by that 3,000,000 ohm resistor with the 5Hz bandwidth, less than 1uV RMS. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here is what Signal Chain Explorer (we used that to predict Gargoyles interferer levels)shows as the 3-pole rolloff. <S> With 2 volts PP input, the ENOB is 19.7 Notice we are NOT including ANY ADC noise contributions. <A> Check out the datasheet <S> : your amp has 4 mA <S> short current and response characteristics are measured with loads of 10-100 kOhm. <S> The equivalent resistance of your filter must be at least that big. <A> The ADC input will look like a SHORT, for 5 or 10 nanoSeconds, at beginning of the Sample time. <S> That "short" will disrupt any opamp directly connected to the ADC Vin or the ADC VREF. <S> To prevent this "disruption" (which shows up as ringing, and perhaps input-voltage-dependent quantization errors, we can place LARGE capacitors on the Vin and the VREF pins. <S> Assume <S> the ADC has 10pF capacitors on its Vin and VREF pins, and assume these capacitors have had their charge consumed during the just-prior ADC operation. <S> As the ADC once again grabs some charge, there will be surge currents demanded from the external voltage sources(Vin and/or VREF). <S> To minimize the voltage upset, use LARGE external capacitors: 100X or 1,000X or 10,000X larger than the ADC sample (10pF) capacitors. <S> In the 3-cascaded_RC filter I gave you, that final capacitor is 10nF (10,000 pF) and should work well. <S> Again, if the AVERAGE input current is 9nanoAmps (Vin of 3 volts, Cap is 3pF, Fsample being 1,000 per second), flowing thru 3,000,000 ohms <S> , there will be an error of 27 milliVolts. <S> This will show up as a linear gain error. <S> [ ERROR this had been 27 microVolts] <A> Forget the filter entirely. <S> It's a waste of time. <S> You have a uC at your disposal, so simply take multiple samples (10x, 100x.. as many as feasible) and average the result. <S> That'll get rid of any a/c <S> and/or noise. <A> I agree with Elliot - a little different approach could be to sample/hold, sort of a Nyquist filter where you pick the best over-sample frequency to get rid of the most prevalent noise. <S> I've done that with RTDs in a noisy aircraft and it gave me good results. <S> I was dealing with millivolt changes that had to be very accurate. <S> That gets rid of those big caps and the insertion loss from the resistors that you are concerned about. <S> I just threw this together in LTspice to give you the idea... <S> if you want the source I will send it. <S> I made the input the 2 volts which would include the amplifier in your design. <S> I added 50Hz and some random HF noise riding on top of that. <S> The filter uses active components except for the R/C tweak with filter 3. <S> Implementation is up to the designer but this can be done with tiny parts, 2x2 to 4x4mm for most active and 0402s for the rest. <S> I think that's smaller than the passive parts <S> but if RE is important an area study is necessary. <S> I just show a switch (S/H) for the concept. <S> Once implemented a value change or two will adjust the sample rate. <S> From a practical viewpoint, the input is almost DC with such slow fluctuations. <S> The noise is much faster and random with respects to the fixed sample rate so it averages out. <S> The assumption is that the noise excursions fluctuate around zero, typical of differential coupling. <S> I used this with RTDs which are slower and in an aircraft environment which is noisy (it qualified MIL-STD-461). <S> It seems like it would do a good job for this source as well, but will take some tinkering based on the real world. <S> I displayed the parameters on the schematic so you can lift them if you use LTspice.
| Your filter will effectively short out the amp if the signal has any significant AC component (even around mains frequency and first/second harmonic).
|
Is it correct to say that a smartphone cable is only a "1A" cable? A store owner said that some lower quality smartphone charging cables are only "1A cables", so if you want 2A, those cables won't do the job. Is it correct to say a cable is only 1A? Does the metal wire inside of a cable get so thin that it can only support 1 Ampere of current to pass through it? How thin does a cable have to get, before it limits the current that can pass through it to be 1A only? I was thinking the cable is bad only because it has higher resistance than usual (but should be still low, like 50 Ohm?). The current decreases, according to V = IR, because V stays the same but total R has increased due to the cable having additional resistance. So the current decreases, but it might still be 1.8A or 1.7A, not necessarily limited to 1A. <Q> What happens is the thinner cables, having a higher resistance, will drop V= <S> IR volts across the cable run. <S> Which in turn causes the phones charging regulator to drop down to another standard charging level when it sees the voltage at the phone end be much lower than it expects. <S> While USB is supposed to be 5V, phone chargers have been made to be 5.3V or higher on the charger side to make up for the voltage droop across the cable at higher amperage, and newer standards are using higher voltage at lower currents to sidestep these issues as well. <S> While he's painting the issue with a wide brush (it can only do x amps), in practice the actual current drawn will depend on the specific resistance vs length vs <S> current the specific phone will draw, so he is correct. <S> Cables with thicker wires will allow a higher charge. <A> The 2A is for faster charging. <S> The cable is not a really an issue for smartphones, the resistance is quite low ( ~ <1ohm)and <S> this is not the issue for smart phones. <S> Cables are rated for current for example in house wiring, and many other cases where there is a lot more power to worry about. <A> There might be some truth to it, depending on the exact gauge of the cables in question. <S> Assume that the cable is 1m long, copper, and <S> it has a gauge of AWG30 (very fine 0.254mm diameter). <S> This would mean that the total round-trip resistance would be ~0.67Ω, which results on a 0.67V drop at 1A and 1.34V at 2A (and would dissipate 2.7W). <S> This would likely be rather unacceptable, but could still work at lower currents or shorter distances. <S> The designer would have a tolerance threshold. <S> Let's say that the supply is 5V±0.25V, to satisfy this at 1A with 5V at the source you can tolerate a resistance of no more than 0.125Ω <S> /m which means ~AWG25. <S> But at 2A your resistance cannot exceed 0.0625Ω <S> /m which means ~AWG22. <S> Practically speaking those limits might be too tight, the power supply can provide a higher source voltage (e.g., 5.25V) or the device can tolerate a lower input voltage at the desired current level, both of which increase the tolerance to cable resistance. <S> For reference, short-distance Ethernet cable can be anything from AWG26 to AWG22.
| Other factors come into play such as what are the most common cable configurations in use, which translates into economies of scale. The current is really limited by the power supply (the small part that plugs into the wall) , they are usually 1A or 2A.
|
Understanding this Mean Well Power Supply I purchased, I believe, the wrong power supply for my purposes, but I'd like to understand a few aspects of its specs: https://www.meanwell.com/webapp/product/search.aspx?prod=ELG-200&mws=8519DC592A6AAA18 This unit appears to have no way to adjust the constant voltage aspect - so am I correct that this will never output above 24V, even if I were to hook up several LED strips in series (each one with an operating range of 23-24v) - so what is the purpose? I can hook up my LEDs in parallel, but then the constant current aspect seems wasted? In the INPUT section, it says lists 142 ~ 431VDC - what does this mean? This power supply can handle both AC and DC input? <Q> With LED's you want constant current <S> This keeps the LED's from exceeding their current rating and let's the designer avoid using a current limiting resistor or some other circuit to keep the current through the LED constant. <S> This supply does both, but it depends on the load placed on it. <S> If the supply is in constant current mode (with the dimmer option with a 0-10V signal or PWM to go from 0 to 100%) <S> then it will run from the voltage from 50% to Max. <S> If you use 100% PWM then it will output it's full voltage. <S> You can string LED's in parallel and still use constant current. <A> Some LED strips are designed to work with 12 V as they have build-in current limiting resistors for the LEDs so these can be powered directly from 12 V DC. <S> If you have two identical LED strips that are designed for 12 V then you can connect them in series so that each gets 12 V. You have to make sure that the LED strips are identical and designed for 12 V DC. <S> Do note that if some LEDs break on a strip the other LEDs might get more current and break sooner. <S> To avoid that use a 12 V supply and connect the strips in parallel (not series). <S> If your LEDs specify a current then use a LED driver with the rated current. <S> In the INPUT section, it says lists 142 <S> ~ 431VDC - what does this mean? <S> This power supply can handle both AC and DC input? <S> Yes it can. <S> That's because these type of supplies convert the input voltage to DC in order to work. <S> So AC will be rectified to DC. <S> DC will just remain DC. <S> The rest of the circuit is designed to work on high-voltage DC. <S> The electronics will adapt automatically to compensate for low (142 V DC) or high (431 V DC) input voltages. <S> It does that by measuring what voltage comes out at the low voltage side (here 24 V DC). <S> It is a common principle nearly all power supplies (most common example: a phone charger) use. <A> Since this 24Vdc max CC regulator works in both CC or CV 24V mode, it is safe to put 23~34V Strip LEDs in parallel but will not work in series. <S> For that you would need >=48V+ regulator. <S> But you still have up to 200W available LED power. <S> Since LEDstrips also have series R current limiting they can operate off 24V CV <S> and they dim at ~3/4 of rated voltage due the series R drop voltage on the strips, which can vary. <S> The % PWM of max current must be compared to your load current. <S> If your load current is greater than 8.3A then the supply starts to limit, so if your load was expected to be 10A in parallel or 240W <S> then you must adjust PWM to <=8.3/10=83% MAX to keep operating at 200W. <S> If the load on 1 string was only 50% of rated current when attached, then PWM from 50% to 100% would result in 24V or no change then full range regulation works from 0 to 50%. <S> added <S> Now that I read the datasheet from comments <S> The pertinent tolerance is @ <S> 1120mA P= 24.5 <S> to 27.1 W.This means round down 200W/24.5= 8 strings or 200/27.1 = 7 strings. <S> So you can use 8 strings but you must monitor current to stay within 100% which might be as much as 200/27W*8= 92%max PWM and 7 strings requires no testing but <S> full dim range might be from 0 to 24.5*7/200 = 87% PWM at full brightness and same above this at 100% current.
| It's a PWM driver, meaning it modulates the voltage to get the correct amount of current through the LED's.
|
How can I run my e-bike wheel on AC power? I have an electric wheel designed for an electric bicycle conversion. By my understanding, the wheel is essentially a 48V, 1000w rated brushless DC motor. It comes with a thumb throttle and kill switch built into a break lever. I am looking to set it up as part of a stationary rig to do some testing and want to run the wheel on AC power. The wheel has no regenerative breaking so only draws power from a battery (not included in the kit). Instead of wiring the positive and negative terminals of the motor to a battery, can I wire them to the appropriate terminals of an AC-DC transformer adapter such as this one and plug this directly into mains? If not, what extra components do I need to include? <Q> You can wire the + and - connections of the included controller to a power supply. <S> Bear in mind that the current, which will be limited by the controller can initially be quite a lot higher than the rated current. <S> A battery will tolerate that, but the switch mode supply you ink to may well limit or shut down when it sees that current, so you often need a large margin on rating the power supply. <S> Given that the battery's characteristics - it has a finite internal resistance, and the supply will droop under load - will affect the performance of the motor, so if you're using the rig to evaluate available power, you're better either using the intended battery, or a supply that has matching characteristics. <A> If you're using the same control electronics (thumb throttle and kill lever), then yes you can just hook it up to a Switch Mode Power Supply like the one you show. <S> You'll likely want some sort of very low ESR decoupling capacitor where you hand over power to the controller. <S> Make sure the power supply you choose is up to providing the instantaneous current required by your motor drive electronics, as that may be much larger than its average consumption. <S> (The decoupling capacitor will help you here) <A> Interesting, although I would be interested what kind of application you have in mind. <S> My suggestion would be to take a hybrid approach. <S> Use the 48V (nom.) <S> battery that is commonly used for the motor, which is able to provide the peak load currents at starting and accelleration. <S> In addition connect a mains-fed power supply in parallel to the battery that can provide roughly the average current needed for the setup (note: no need for precise balancing of the two current contributions!). <S> The voltage of that supply should be set such as not to overcharge the battery; as well the supply should have a properly adjusted current-limiting circuit in its output line, preventing shorting etc. <S> Note that the supply does not need to be capable of delivering the instanteous peak currents: the battery takes care! <S> In this way one also ensures that the voltage applied to the DUT is well stabilised by the (low Ri) battery. <S> Main concern is to keep a solid eye on all (!) <S> safety issues around Li-battery (dis)charging! <A> What you want to do sounds feasible, with one notable problem: the power supply that you have chosen may not handle startup current. <S> This isn't a problem if the wheel is not loaded but is definitely a problem if the load is at all similar to a rider on a bicycle. <S> All you can do is try it. <S> If the power supply collapses under startup conditions, consider adding a battery pack in parallel with your power supply via a suitable isolation diode. <S> You can keep this extra battery pack on a charger all the time. <S> Just be sure to set the output voltage of the power supply to be somewhat higher than the fully-charged voltage of the battery.
| The great advantage of using a large battery pack is that the battery pack can supply enormous amounts of current for the short time that it takes to get the wheel spinning.
|
For an inverting opamp, is inverting the same thing as an 180 degree phase shift? Many texts talk about 180 degree phase shift as a function of the inverting opamp. Is that correct? It seems to me inverting is happening almost immediately but phase is related to time and period and time delay between the input and output. <Q> An inverting op-amp inverts the signal; it does not phase change the signal at the output by 180 degrees although, if the input waveform were a sinewave, then it would look like 180 degrees of phase shift. <A> How an opamp behaves depends on how you configure it in a circuit. <S> But the opamp is actually irrelevant to your actual question. <S> I think that your actual question is: Is inverting a signal the same as phase shifting it by 180 degrees <S> What we mean by inverting <S> a signal is multiplying the signal by -1 , so +33mV becomes -33 <S> mVand -0.5 V becomes <S> +0.5 V. A 180 degree phase shift is indeed related to time but since phase is also coupled to frequency we only tend to use phase when talking about a single frequency. <S> The only signal that contains a single frequency is a sinewave . <S> Now for a sinewave inverting it (multiply it by -1) or phase shifting it 180 degrees will result in the same signal. <S> Also for periodic signals like square waves and sawtooth signals, which consist of a base frequency and harmonics, inverting and phase shifting with 180 degrees is the same thing. <S> Then the phase is only related to the base (lowest) frequency. <S> For non-periodic signals (these do not have a base-frequency) <S> this isn't the case. <A> Inverting a signal is the same as a phase shift of 180 degrees at all frequencies. <S> If you take the Fourier transform of a signal, and compare it to the Fourier transform of the inverted version of the signal, you will see that the latter has the phase shifted by 180 degrees at every frequency. <S> This means if you invert a sine wave you get a shifted sine wave, but also if you invert the sum of two sine waves, for example, you get the sum of two sine waves that are each shifted by 180 degrees. <S> You can describe this in either the time domain or the frequency domain. <S> Obviously it is more intuitive in the time domain ("it multiplies the signal by -1"). <S> You are correct that the frequency at a single point in time doesn't make sense to talk about - but we are not talking about inverting a single point in time, because the circuit will invert the signal at all points in time. <S> Or at least, all the points between turning it on and turning it off.
| So yes, for sinusoidal signals, inverting and phase shifting with 180 degrees is the same thing .
|
Connecting CMOS digital ICs that are powered from different power sources Imagine I have 2 CMOS devices with logic gates, etc. Both use the same nominal voltage - 5v but are powered from different sources. If I will use simple regulators like 7805 on both sides, they will give slightly different output. One may give 5.05v, some other 4.97v. The grounds of these devices will be connected. This means that sending logic zero in any direction will not cause a problem, but sending logic one to a device with lower power level may cause a problem. I can put a resistor to limit the current, but would that be adequate? Is there anything better than this resistor? What would be a good value there? Or maybe this is not a problem at all and I can simply connect the pins directly assuming that the difference of the power levels is small? What would be the difference when the problem will start to appear for our days CMOS logic gates and micro-controllers? <Q> A 100 mV difference in Vdd levels will not normally cause any problems. <S> No current limiting resistor would be required. <S> However, if it's possible for the upstream device to be powered up while the downstream device is not powered (for example during start-up or shut-down) and for the upstream device to produce a "1" output during this time, then you can get upredictable behavior. <S> If that isn't an option, in many cases a 5 kohm or higher series resistance can minimize problems, but even then certain circuits could still misbehave. <A> As long as the difference between the supply voltages is less than the forward voltage of a silicon diode <S> then you can connect the gates directly without a resistor. <S> If the voltage difference between the supply voltages is at any time more than the forward voltage of a silicon diode then you need to add a resistor. <S> As The Photon rightly comments, it might be that one of the supplies will be "up" later than the other so then you will have the 2nd situation and you will need the resistor. <S> Why is that so? <S> I'll explain this by drawing two CMOS inverters, each having their own power supply and adding the ESD protection diodes that are present in all ICs: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> (this circuit is somewhat simplified and does not show the full ESD protection to keep it more clear) <S> Suppose Ropt = 0 <S> ohms <S> (straight connection), then when VddA is 5 V but VddB is not present yet, VddB can be pulled up via the connection between the gates. <S> Assume that the input of the left inverter is "0" <S> then pmosB will be on, this will pull up the output of the left inverter, then current can flow through Ropt, through Desd3 and pull up VddB. <S> Then the left inverter will supply power to the inverter on the right! <S> This should not be allowed as it can give unexpected results. <S> To prevent any damage it would be enough to make Ropt large enough to prevent a damaging current from flowing. <S> As long as the signals you're using aren't very high frequency (the 4000 series logic you're using isn't that fast) a 10 kohm resistor would do the job. <S> This effect is also explained in this EEVBlog video <S> Dave uses a microcontroller to show this but the underlying principle is the same: power a circuit through a signal input because of the ESD protection. <A> Maybe there is more than one question... power supply and logic(?). <S> Anyway, Not sure if you are asking about this, but... <S> As far as voltage differences causing an issue for interpreting a logic "high", CMOS levels are 3.5v-5v for Logic "high" so that small difference won't be an issue <S> This page talks about it
| If you need to allow for this scenario, the best solution is to re-design to avoid it entirely (for example, use a tri-state buffer with the enable pin controlled by the power state of the downstream circuit).
|
As we go around the circuit, the total voltage drop must equal zero I am studying an advanced mathematics textbook. One of the examples is within the context of an electrical circuit (a series circuit, I think). The example makes the following claim: As we go around the circuit, the total voltage drop must equal zero. Can someone please explain what law/principle of physics this is, and elaborate on it a bit? My research indicates that it might be Ohm's law, but I'm too inexperienced with these concepts to link the formal description of Ohm's law to the above statement. Thank you. P.S. I'm not sure if my tags are appropriate, so feel free to edit with the appropriate tags. <Q> Can someone please explain what law/principle of physics this is, and elaborate on it a bit? <S> It's more common sense than anything else. <S> Kirchoff wrote it down first! <S> Let's use an analogy: <S> You are on the ground floor (Floor zero in Europe) of a large building (voltage = 0 V). <S> You go up to the third floor in the lift / elevator. <S> (Voltage = 3 V). <S> You walk across that floor to the stairs. <S> (Voltage is still 3 V.) <S> You come back down to ground floor. <S> (Voltage is zero again.) <S> No matter how you make the journey with any combination of stairs, lift or falling out a window <S> the sum of the height changes will be zero when you arrive back at the start. <S> My research indicates that it might be Ohm's law ... <S> No. <S> This isn't directly related to Ohm's law. <A> This is one of Kirchoff's Laws its the voltage law. <S> Consider a simple circuit simulate this circuit – <S> Schematic created using CircuitLab <S> This is a formal stating of what should be obvious common sense. <S> To understand what this means I will introduce the notation \$ V_{12} \$ as meaning the voltage measured between 'N1' and 'N2' treating 'N1' as the positive polarity. <S> \$ V_{23} \$ having similar meaning for 'N2' and 'N3' etc. <S> Now consider a closed path <S> \$ N1 \rightarrow N2 \rightarrow N3 \rightarrow N5 \rightarrow N1 \$ <S> All this is saying is \$ V_{12} + V{23} <S> + V{35} <S> + V{51} = 0\$ <S> Or by noticing that \$ V_{51} = <S> -V1 <S> \$ because it's the voltage across \$ V1 \$ but with the opposite polarity \$ V1 <S> = <S> V_{D1} + V_{R1} <S> + <S> V_{D2} \$ <S> This is true of any closed loop <S> so \$ V_{34} + V_{45} <S> + V_{53} = 0 \$ and \$ V_{12} <S> + V_{23} + V_{34} + V_{45} + V_{51} <S> = 0 \$ for example. <S> Kirchoff has another law for current that simply states: <S> The net current into any node is zero. <A> Well, it's one of Kirchhoff's laws and since you include a tag "electromagnetism" <S> it's worth pointing out that it's incomplete as you write it: <S> the sum of all voltages as you go round a circular path equals the rate of change of magnetic flux through that circle. <S> Basically, magnetic flux changes introduce "anonymous" voltage sources in such circles that are located between no two nodes in particular.
| Kirchoff's voltage law simply states that the sum voltages around any closed path is zero
|
Variance in ECG lead resistance ; its effect ; CMRR We are designing custom ECG electrode with flexible printed wires as to reduce wires clutter from different electrode. The manufacturer has made the patch but the resistance variation between all the three electrode varies, which is 20 ohm, 5 ohm and 30 ohm respectively. Is this variation acceptable or resistance of the electrode and the the lead as seen by the amplifier should be same. Considering the electrode-skin interface in which variation is in the order of 1000 of ohm, is variation of lead resistance by 15 to 50 ohm acceptable. If we consider common mode noise variation across leads with different resistance, will amplifier amplify common mode noise, thus amplifying noise. Will using the right leg drive can help to reduce the noise. <Q> Considering the load is > 1M and source is 1k min, adding a tolerance of 50 Ohms <S> has null effect considering 5% of source is good and 50 parts per million compared to load is effectively null. <S> I might consider < 100 ohms OK. <S> The bigger error source is contact pressure or motion producing galvanic skin voltage from the change in pressure and change in capacitance between electrode and skin. <S> Thus you want pads that maintain relatively constant adhesion pressure. <A> Any common-mode influences will be negligible in the pass band, but there may be subtle differences in cutoff frequency of an RF attenuation. <S> You shouldn't care about this, as there is tremendous space between your target signal and RF. <S> You can, and should, deal with any noise remaining after the conversion to a single-ended signal by filtering. <A> You need very high input resistance, then it all will not matter. <S> Actually, anyway skin variance is few orders of magnitude highet, so anyhow just forget it. <S> May your CMRR be 120dB!
| The difference you are seeing is far less than the differences in the skin-electrode interface at the different sites, and should not be your biggest source of noise by far.
|
What could be the reason for chip capacitor failure due to silicone potting? Is silicone potting of PCB board beneficial or risky? What could be the best material for encapsulating a power supply board? I used 2 part mixed silicone potting compound for encapsulating a microcontroller based (ASIC based, too) power supply, and one of the chip capacitors on the board failed within few hours of regression of device. Reason is still unknown, looking for some valuable suggestion. <Q> Well, any potting material will put mechanical stress on the board and the components. <S> Any glue or compound, when drying or becoming hard due to its chemical reaction, will do so with the side-effect of changing its volume, even if it happens only ever so slightly. <S> It may well be that one of your MLCCs failed because the compound subjected it to shear stress, maybe worsened by heat. <S> Also, any chemicals (read: dirt) on the board will remain just there under the potting compound, being trapped forever, and even a bit of moisture might make corrosion effects worse - compared to a board with no potting or coating. <S> This, however, is a long-term effect not likely to be the cause of the particular failure mode you observed on your board. <S> Try using potting as a last resort. <A> Ceramic chip capacitors are extremely susceptible to stress fracture, and they almost always fail shorted. <S> Any stress on the PCB can lead to MLCC chip capacitor failure. <S> There are lots of guidelines on chip capacitor placement (away from edges, orienting perpendicular to longest dimension of PCB etc.) <S> Stress fractures may accumulate over time, and lead to failure. <S> There is a lot of research underway on how to reduce ceramic chip capacitor failure, including coating them in elastic materials, orienting the pads on the longer edge etc. <S> It is possible that more expensive chip capacitors may be more robust, due to these additional protections. <A> Standard bathroom silicone is not inert. <S> It reacts long after drying. <S> So it may (or may not) have an effect chemically. <S> I would use only products certified for PCB.Any electrical insulation is also a thermal insulation. <S> Components may heat much faster than expected because there is no contact to air. <S> HTH
| Additionally, chip capacitors may even crack due to overheating (as in prolonged soldering).
|
What does n/o and n/c mean on a schematic diagram? Okay so I have two questions. I am looking at a schematic diagram of a remote control for a toy car and I can't seem to figure out what the "n/o" and "n/c" components stand for that are attached to the motor supply. Also not sure if they are the same component or they just have different looks. My other questions is what is the RL1 component with the swirls and the two lines? <Q> RL1 is a relay , a type of electrically-actuated mechanical switch. <S> The swirls are the relay's coil, and the two lines indicate that it's wrapped around a magnetic core. <S> The thing labelled <S> N/C likewise is "normally closed", which means that switch contact is closed, i.e. connected, when the relay coil is not energized. <S> When a current is flowing through the relay coil, the magnetic field it creates pulls the switch over, opening the NC contact and closing the NO one. <S> Note that in the context of integrated circuits you may see "NC" used to mean "not connected", but here with it being next to a relay's contacts <S> it's pretty unambiguous. <A> NO = normally open contact NC = <S> normally closed contact <S> Normally open (NO) contacts connect the circuit when the relay is activated; the circuit is disconnected when the relay is inactive. <S> Normally closed <S> (NC) contacts disconnect the circuit when the relay is activated; the circuit is connected when the relay is inactive. <S> All of the contact forms involve combinations of NO and NC connections. <S> Source <A> N.O. and N.C. stand for normally open and normally closed contacts. <S> These are connections that are part of a relay (or relay like components, ex. <S> Solid state relays or modules). <S> They are not independant parts. <S> As the name implies, normally open contacts will be open, or not connected to the common pin, when the relay is off. <S> RL1 is the relay which the NO and NC pins are part of. <S> The swirly lines is the inductor, the electromagnetic wire wound core that is used to make the relay work. <S> This diagram shows it more clearly as a single unit. <S> You can use a Single pole single Throw relay which will only have a normally open contact (normally closed SPST relays do exist too), or a Single pole double throw relay which will have both. <S> Other variations of relays exist with multiple combinations of NC and NO contacts.
| N/O and N/C is also part of the relay; N/O is "normally open", which means that switch contact is open, or disconnected, when the relay coil is not energized.
|
Protecting a 3.3V circuit from 5V power supply I'm designing a dev board for 3.3V parts. On the dev board, I will have a header for an "FTDI Friend" USB-to-Serial converter. The FTDI part can supply Vcc to the target board. However, FTDI parts are available both in 3.3V and in 5V versions. I want to protect my board against someone accidentally using a 5V FTDI and feeding that into Vcc. What's a cost-effective strategy for this? <Q> Here is the type over voltage protection circuit I would use if it was essential to protect against a greater than 3.3V input. <S> This circuit uses a low cost voltage comparator to compare a 2.5V reference voltage against a divided copy of the input voltage and then will turn off a P-channel MOSFET when the input voltage goes above 3.3V. <S> You can change the reference voltage part to some other similar part at another voltage if needed and the scale the voltage divider resistors accordingly. <S> One example of a low cost reference is a TLV431. <A> Ti has an over-voltage app-note utilizing the LMV431 (cheap voltage reference that can be used like a comparator) with a P-Channel MOSFET to block input when Vin is greater than your chosen threshold. <S> The part is available unencrypted, so you could use a free tool like LTspice or Tina to tweak values. <S> You could probably delete a few of the zeners and resistors that limit Vgs on a 5V implementation, compared to the automotive (12V bus with 60V excursions) type situations <S> the app-note was originally targeted at: TLV431 app note <S> It would probably cost 25-40 cents to do. <A> By combining the regulator with a voltage comparator, a P-mosfet and a voltage regulator, you can have both the protection and keep it functional at 5V or at 3V. <S> (For intermediary voltages data sheets for each part has to be examined carefully and additional or other components may be needed). <S> Note: I didn't test the present circuit yet. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> When the voltage is under the voltage comparator threshold the gate of the P-MOSFET M1 is at 0V (or GND level) and the MOSFET conducts. <S> The 3.3V regulator has no effect. <S> Current flows normally. <S> When the voltage is above the comparator threshold (e.g. 5V) <S> its output turns high and the P-MOSFET stops conducting. <S> Then the current goes through the voltage regulator and Vcc receives 3V. <S> The regulator must have a drop out voltage of less than 2V. R1 is necessary to bring current to the MOSFET gate with the BD48 serie because it's open drain. <S> There are other variants in the market. <S> This is only one non-exclusive example.
| A low voltage Zener diode could also be used but the precision of the circuit would be lost and the voltage divider would have to be tweaked to allow for greater margin above the 3.3V level.
|
How much do lithium polymer batteries expand in volume? I'm designing a device with a small lithium polymer battery (4x12x30 mm, 120 mA-h). Looks like this: I've heard that there is a rule of thumb that the space left for the battery in a case should be around 10% larger (I suppose primarily in thickness) than the nominal dimensions to allow for expansion. Extra 10% seems quite large. Where does this rule of thumb come from? Is there any official recommendation for how large a compartment to put lithium polymer cells in? How much do these batteries expand and shrink in normal use? For example during charging/discharging cycles, temperature cycles over normal temperature range (-20C to 60C), etc. What happens if the battery is in a rigid compartment in the case of malfunction? It's pretty common to have batteries "puff out" if they get shorted internally, but what happens if the battery is in a compartment that prevents expansion? (Assume the compartment is strong enough to withstand the pressure build up) Does the pressure/walls make the short worse, or better? <Q> This paper measured a cell, they reported a max of 0.5% expansion over a charge cycle: <S> Source: Expansion of Lithium Ion Pouch Cell Batteries:Observations from Neutron Imaging Figure 7 Over the lifetime of the battery <S> , it swelled more than 1.5% Source: Expansion of Lithium Ion Pouch Cell Batteries:Observations from Neutron Imaging Figure 9 <S> One could use these numbers for a baseline, but it's not that hard to make these measurements to a reasonable degree. <S> Measure the cell under maximum discharge, because the cells swell more with thermal expansion. <S> Give your self additional margin to account for differences in batteries and manufacturing tolerances. <S> Cells also bulge more in the middle if they are heated then the outside. <S> So make sure you measure the middle of the cell. <S> Source: https://www.researchgate.net/publication/283720424_A_novel_thermal_swelling_model_for_a_rechargeable_lithium-ion_battery_cell <A> There should be a distinction made between inevitable electrode expansion/contraction due to electrochemistry of electrodes itself at nanoscopic level (which was presented by lapto2d), and "battery swelling/puffing" of pouch-type cells (presented by K.Krull) due to electrolyte decomposition/outgassing, which is a sign of malfunction and/or poor manufacturing quality of a cell. <S> Regarding puffing, there are several theories about it, but it looks like the main cause is some electrolyte decomposition and metal build-up when the cells are left in nearly over-discharged state for long time. <S> The manufacturing issue is related to production process, where the cells are "formed" before being sealed , letting the electrolyte to outgas. <S> If the forming is done sloppy/too fast, the sell still has some gas build-up and will puff over time. <S> Obviously the overall cell expansion in practice is a combination of the two effects, and some studies of well-made cells show expansion up to 4% after 50 cycles, see this publication . <S> While the 0.5%-1% of electrode thickness expansion is natural and can/should be accommodated with some oversize of battery compartment, excessive irreversible puffing is a serious precursor to catastrophic failure. <S> At one point in dealing with customer's issues I came to realization that it would be very beneficial to have a pressure sensor to detect this state before the whole device case gets torn apart. <S> It appears that this idea is already patented, US8717186B2 . <S> I strongly recommend to put a pressure sensor inside your design. <A> Although I cannot compete with the wealth of detail given in previous answers, I think it might be helpful to give you an example why this much extra space is used for a good reason. <S> A Lipo does not only swell during normal operation with temperature and charging/discharging, but also when it ages. <S> Have a look for electrolyte decomposition to find more about this phenomenon, but ultimately it breaks down to the creation of gases (mostly oxygen) inside the LiPo. <S> Just to give you an idea about what you can expect: A frequently used 3.5 year old battery I measured during writing this expanded from 25mm (according to the reseller, couldn't find a datasheet) to almost 32mm, which is more than 25%! <S> I guess the datasheet values are not as optimistic as the reseller's ones, but still this is a substancial increase, which should be taken into account while designing your product. <S> See the comments on below your question, this has been described by K H in detail.
| If you really want to find out how much your battery is swelling, get a micrometer and measure it discharged, then measure it fully charged and see what the difference is. If the battery has no room to swell, it will become a possible risk of fire and - in the worst case - even explosion. Since batteries are made with different anode\cathode and electrolyte combinations that vary from manufacturer to manufacturer, it would be wise to consult the manufacturer on swelling or measure it.
|
How to connect these two resistors like on the schematic? I am trying to connect CAN to the LPC1768, like here Expect that I don't understand the picture or the meaning of "you need two 120 ohm terminating resistors at either end of the bus" . Does it mean I am suppose to connect a resistor of value 120 ohm between CANL and CANH (between pin 6 and pin 7) on both of the MCP2551s? For me this schematic shows two resistors connected together. <Q> If the bus is short enough you can probably get away with a single resistor of 56R or so <S> but because CAN is current driven the ideal layout <S> is to connect one resistor of 120R at each end of the bus. <S> In this case this would directly between and as close as possible to pins 6 & 7 on each of the two ICs. <S> This is to allow for the inductance of the wires and match impedance at either end. <A> Yes, you have to connect 120 Ohms between CANL and CANH on both of the MCP2551s. <S> In case you have many devices you don't have to put a resistor for every device. <S> Many devices: <A> The resistors are in parallel, but physically far apart at either end of the bus. <S> So yes, one resistor at each device. <A> You have, I suspect, not really understood CAN systems. <S> The important part is that all communication is conducted on a buss - that is, a single twisted pair which runs from one end of the system to the other. <S> All devices connected to the buss do so as stubs which should be as short as possible (0.1 meter is the standard max, and shorter is better). <S> No branches are allowed. <S> In general, the impedance of commercially available twisted pair is about 120 ohms. <S> These two resistors are what you have included in your schematic. <S> The two resistors on your schematic will not appear on a single pcb. <S> For each device, there will be one or zero, depending on where the device sits on the buss. <S> Another approach is not to include the terminating resistors on the device cards at all, but to install them at each end of the buss, then install transceivers where appropriate. <S> TLDR; The schematic is wrong. <S> Make the board with room for one resistor. <S> If you have more than two on the buss, install the resistor only on the two boards at each end of the twisted pair (and make sure the wires do not extend past the board). <S> Do not install the resistor on the intermediate units.
| If you have only two units on the buss, install a resistor on each. For proper operation, each physical end of the buss should have a terminating resistor of (nominally) 120 ohms.
|
Sensors to detect vehicle height from the side I am looking for a way to determine vehicle height measured from the side. The application is for a drive-through ATM, the sensors being mounted inside the ATM enclosure. Some ATMs have a roof as well, in those cases the solution would be straightforward - ultrasonic sensor (emitter and detector) mounted in the roof, measuring distance to vehicle roof, and with a known distance to ground, the height could be calculated. But I need a solution that works in all cases. I want to preferably avoid any detectors or emitters outside the ATM enclosure, such as an ultrasonic sensor on a pole sticking out the top of the ATM, or a detector on the far side. One solution is to have an array of ultrasonic sensors arranged vertically inside the ATM enclosure. This will require experimenting with the spacing between sensors, and knowing the overlap in each sensor's 'field of view'. The application is outdoors, with large temperature variations, which is why I looked at ultrasonic sensors first. I'd like to know if something else may work better in a cost-effective manner. If someone one wants to see what a typical drive through ATM looks like: <Q> Try this simulate this circuit – Schematic created using CircuitLab Offset the two sensors by 6". <S> This allows the Range^2 and (Range+6")^2 to give you some useful distance information. <S> The two diodes after the photodiodes will generate the LOG (beware, temperature sensitive) and greatly extend the dynamic range. <S> Have one of these every foot of vertical distance. <A> I would use only one or two ultrasonic sensors at different angles placed on the top of the ATM, two to get around with an open window on the driver side. <S> Use the fixed lower sensor to detect a vehicle presence , wait a bit to allow the vehicle to stop then move up the ATM if a proximity is detected by the top ultrasonic sensors or move down if not, until something changes then adjust the height according to your knowledge on vehicle height vs top roof height and stop. <S> You can make the machine learn from user input where to place the ATM for different top roof heights improving the automatic height adjustment in time combining the readings from the fixed sensor and top sensors and user input for similar previous values. <S> Don't do that without a driver available manual control, in case of a failure placing the ATM at an accessible level he will be in an funny impossible situation, he cannot open the door since is obstructed by the ATM. <S> Take care also about the side mirrors, the ATM might be at a lower height when the vehicle stops, maybe an obstacle before the ATM to prevent the vehicle to stop to close or some flexible roof extension on the top. <S> Of course , the ultrasonic sensors must be used one at the time, the lifting is slow so you can live with that. <A> Your best bet will probably be some sort of “visual” sensor, namely sensors that react to transmitted or reflected light. <S> These could be: <S> A camera (or two) with or without IR illumination, that feed into custom software to measure a well-defined area of the image. <S> An electro-optical assembly (e.g., LIDAR) that scans the area in front of the ATM. <S> A few linear optical arrays (basically 1-D tall cameras) that provide transmitted or reflected light information. <S> At least two strategically placed optical distance (parallax/trigonometric) sensors, to sense the roof from as high an angle as possible. <S> I believe the least expensive, and most reliable solution might be the camera. <S> As most of the cost would be in the software, but stereo vision is rather common in robotics, so there are a lot of projects out there <S> Here is a project that could get you started. <S> Parallax sensors can be built from a laser and a 1-D optical array. <S> But these are rather common in industrial applications.
| Low resolution 1-D optical arrays can be constructed from discrete devices and can be placed in the movable portion of the ATM to “scan” for the location of the roof. A must is a manual control available at any level (parallel UP and DOWN switches at many levels) to allow the driver to adjust the height at a comfortable level.
|
would like to verify a filter design I need to make for a backup camera in my truck I am installing a backup camera in my truck. I am getting some source of electrical "noise" in the electrical system in my truck and its not allowing the backup camera to operate. Let me explain a little further. (thanks for your patience) Installed the head unit and backup camera in the truck, Head unit is pretty standard, it has a 12 volt (+) input to trigger the head unit to "see" that the truck is in reverse. the camera is connected to the head unit via an RCA cable with an additional (+) and (-) line included. at the camera end of the RCA line there is (+) and (-) tag ends hanging out that you are to connect to the reverse light harness on the truck. Basically when you put the truck in reverse, the power is also tapped to the camera, when the camera is powered it sends a (+) signal to the head unit and it flips the screen over to the reverse camera automatically. When the truck is off but key in the on position and I put the truck in reverse the camera works not a problem. When I start the truck and put in reverse the screen simply goes to the yellow ! on the screen. According to the manual they are saying its because there is interference with another signal. Im assuming at this point its with the alternator or something with the engine running. In the installation manual they have a pic of a filtering circuit. I believe that in the diagram has the polarity on the capacitor is incorrect? Should not the negative lead of the capacitor be on the negative lead of the 12 volt circuit? In my local electronics store I could only get a 3300 uf cap but I believe this should be ok in this instance. also to verify please the band on the diode should be toward the backup camera correct? <Q> Seeing as neither the diode or the capacitor is actually shown with a standard symbol, it's difficult to argue if it is drawn 'wrong'. <S> However, for clarification, it should be like this: simulate this circuit – <S> Schematic created using CircuitLab <S> A 3300uF capacitor should be fine, but you could always get a smaller ceramic capacitor in parallel with it if you are unsure, they are cheap enough that it doesn't matter if you buy some and find they are not needed. <A> The diode is shown correct in the diagram. <S> Now the thing is that in most of electrical schematic diagrams positive lead of cap is shown by a marker or line. <S> Maybe they messed it up. <S> Either they (manufacturers) have shown the negative pin of cap with line or it was just a mistake, who knows. <S> The thing to remember is that positive pin of electrolytic capacitor goes to positive connection of supply and negative to negative of the supply. <S> By the way you could also add an inductor in series with the camera after diode to further smooth out the voltage. <S> Please let me know if adding filter solved the problem. <A> So after much frustration and many hours working with atoto found out the problem. <S> Bad rca cable from the stereo to the camera, specifically the end on the radio. <S> Was working on it and had the deck sitting on the dash with the cable a little tight, suprise the camera stayed on. <S> Started messing with the connections and got it to knock out by twisting the rca connection about 45 deg and 5 sec later out goes the camera feed. <S> Thank you to everone who viewed and commented. <S> Solution is stupid but still a solution.
| Talking of polarized electrolytic capacitors these should never be connected in reverse polarity with power supply or tgey will pop or explode depending on the conditions.
|
Does USB supply a "constant" 5V or "up-to" 5V? I am trying to reverse engineer a digitizer inside an old laptop. I am fairly certain it is a USB device because I noticed a twisted pair of cables next to a ground wire. Just to make sure it is USB I probed the pin to the other side of (what I suspect to be) D-. When I did this I saw different voltages flashing very quickly but nothing close to 5V. I believe the max I saw was 1.2V and min being somewhere around .3V. I don't want to wire this up to a USB cable out of fear of destroying the digitizer if it can't in fact handle 5V. This would make sense to be USB VCC if it does offer up to 5V but I couldn't find anything definitive that I could understand at least. <Q> TL;DR just because it's connected to the USB does not mean 5V or even that it is powered from USB. <S> For versions 1 and 2 of USB <S> The spec says 5V <S> +/- <S> 5%, so you should design to accept 4.75V to 5.25V. <S> USB Version 3.something introduces the availability of increased voltages (up to 20V). <S> but a device has to explicitly request increased voltage, else they just get the same Approximately 5V. <S> When I plug my phone into its charger via a power meter I see the voltage jump to 9V <S> My laptop's charger uses a USB C connector and can provide 5,9,15, or 20V <S> (according to the label) <S> Now, your laptop's internal device may be designed for some voltage other than 5V. <S> The USB data lines D+ and D- signal at only 3.3V. <S> So, 3.3V could instead be the supply to the (touch-screen) digitizer. <S> In general the USB host can't see the device's supply voltage, it only sees the data signals. <S> Some USB devices (eg: laser printers) do not use the USB VCC line for power. <A> USB Vcc is 5V - constant. <S> It should only drop if you draw too much current, in which case a PTC (automatically resetting fuse) will open and shut off the current. <S> In that case the voltage will drop to near zero. <S> The voltage on the data lines is a different story. <S> D+ and D- transmit data at high speed. <A> These are (and were) commonly used for internal data connections at fairly low data rates. <S> What did you probe the signal with? <S> If a voltmeter, the readings of 0.3 to 1.2V could be (or almost certainly are) misleading on a data line. <S> If an oscilloscope, can you post a picture of the signal? <S> It would also help if you post a picture (or details of markings) of any ICs on the digitizer, and any silkscreen markings on the PCB where the suspected USB line is attached. <A> USB will power 5v constant. <S> If you directly power it for experiment, it may blow some component or IC which maybe very difficult to find in the open market. <A> I recently bought two cheap measurent devices, one for Micro USB and another one for USB-C to verify the wall-charger/cable/QI-charger chain for different devices. <S> Nowadays USB seems to constantly haggle about voltage and wattage! <S> Some of my chargers would provide up to 12 V through USB (far off the original specification) but some devices would even consume it under varying circumstances. <S> To cut a short story shorter: You will not get around to measure your specific case. <S> The specifications are interpreted in enough ways to ake this necessary. <A> It sounds like you probed a data line. <S> However, signalling is at lower voltages. <S> USB 1.1 used 3.3V signalling, and USB 2.0 High Speed uses about 400mV. <S> That is if it was USB at all of course. <S> It could be PS2 or something else.
| If there isn't a clear sign that it is USB, consider that it could be SPI, I2C/SMBus or some other form of serial connection, particularly for an internal device. USB supplies 5V for power. Your multimeter can't make any sense of what is there - it changes way too fast for the meter. If your digitizer has unkonwn input voltage requirement, it is better to do more research or ask the manufacturer of the device or laptop. If you could see the line changing then it was likely a data line.
|
CAN Bus first steps In a few months, I will need to build a project that will let me connect from my PC to a power supply via CAN bus. I don't have this power supply here and have zero knowledge (even less) of how to do this. I want to prepare by arranging a demo, to put some device that I want to give commands to, from my PC, and see that it connects, and also be able to read something from that device. My guess is that I will need a device, a wire to connect to this device, and a converter from the USB or Ethernet connector of my PC to this type of wire. As you can see, I'm clueless. I will appreciate any help in these first steps, like what should I buy for this, nothing is too simple for me. Thanks in advance for any help. <Q> A good start is watching some tutorials. <S> The web is full of stuff, and even being totally clueless <S> it's easy enough to find something useful. <S> For instance, have a look at this one: https://www.youtube.com/watch?v=RRbrk3SdSKA <S> But actually, for your purpose you don't have to have really detailed knowledge on CAN. <S> What you have to know is how to control your PC software to send and receive messages, and of course you will finally need a specification of the power supply's CAN interface: - What messages does it expect? <S> - What messages is it able to send? <S> - <S> What's the meaning of the data bytes in the messages? <S> The PC software is usually bundled with a specific CAN hardware from the same supplier. <S> Kvaser was already mentioned, Vector is another (expensive) one, Intrepid has solutions matching your needs, ... <S> They all offer professional control software, but also have free libraries to allow interfacing their hardware with your own software solution or third party tools like LabView. <S> Let me add ETAS Busmaster as another very interesting tool: It can deal with the above mentioned and various other hardware interfaces, it's open source, and it's free of charge. <S> The Arduino solution mentioned in another answer may be the cheapest from hardware side, but IMHO not the easiest and quickest way to get things running - unless you are willing and able to deal with source code and building your own application more or less "from scratch". <A> First, I suggest doing some research on the physical layer of CAN -- Wikipedia is usually a good start. <S> The easiest and quickest way to get up and running is to purchase an Arduino or Teensy and a CAN breakout board. <S> These latter devices typically employ an MCP2515 <S> CAN transceiver that's controlled by a SPI interface. <S> Make sure the breakout board includes a 120 \$\Omega\$ termination resistor on the CAN signals. <S> Since you have a few months before your project begins, I suggest buying two of each of the above and making your own two-node network. <S> Look into the structure of CAN messages and try sending messages back and forth. <S> There are tutorials online <S> (search 'arduino can tutorial') <S> and you can come back here with specific questions if you get stuck. <S> EDIT: <S> As Lundin points out, these CAN controllers are not ideal -- especially if you plan on a permanent setup. <S> The next step up would be to use the Teensy, which includes a CAN peripheral (but no library support), and a transceiver. <S> The only transceiver <S> breakout <S> I found was here , and the company also provides a short tutorial on interfacing the board with a Teensy. <A> Well-known manufacturers are IXXAT, Vector, Kvaser. <S> You can buy 2 of these to "speak with yourself" if you have no other CAN hardware available <S> - CAN always requires at least 2 nodes to work. <S> CAN listener software on the PC. <S> You can buy this too from the above mentioned companies. <S> Typically they give you a limited "light version" of such a tool together with the USB-to-CAN adapter. <S> Better CAN listeners cost a lot of money, but they come with a PC programming API that you can use, to manage all communication directly from the PC. <S> You will always need a CAN listener no matter if you run CAN from PC or a microcontroller. <S> Forget about decoding messages manually with a scope. <S> Build some lab wire with 9 pin ribbon cable and as many IDC dsub connectors as you like. <S> Add an option to put 120 ohm terminating resistors at each end of the wire, between CAN_H and CAN_L. <S> The pinout is standardized:
| Obviously you will need to study CAN, but what you need in terms of hardware/software is this: An USB-to-CAN converter with USB in one end and a DB9M dsub connector in the other end.
|
Reduce EMI generated by old AC motor from an electric knife I have an old electric knife (220V 50Hz) which seems to produce a lot of EMI.Some clues on that were a "bzzzz" noise from PC speaker I noticed sometimes when someone use it. Recently I bought a device which has touch-buttons and I noticed when the knife is in use the device detect false touch. So at this point I think that knife produce a lot of EMI (maybe due to it is an old model?) So, I would ask: is there any way to reduce them (capacitor on AC motor,..)? Or can I make a sort of shield?(eg. whith alluminium sheets glued inside the plastic cover). Please don't suggest to replace it :) the question is just to know something more about EMI, shielding, etc.. <Q> Sounds like you have low frequency conducted EMI. <S> An X-capacitor straight over the motor would be the first order of business, but I've included the steps I would have taken down the line (litterally). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Couldn't find a symbol for you motor, nor lines to draw a correct CM choke. <S> More luck with LTspice: <S> All values are estimations, but at least not several orders of magnitude off. <S> To be compliant, your capacitor needs to be at least X2-rated. <S> The R in parallel would be required in the EU to bleed down to 60 V (thanks Martin Bonner!) <S> within one second or you will get nasty shocks from the plug if you toutch it once unplugged. <S> There are special X-rated capacitors with this resistor built-in. <S> This is called a bleeder resistor. <S> There is a tradeoff between time and power dissipation when plugged in so make sure you have sufficient voltage and power rating on it. <S> Your suggestion of aluminium foil wrapping won't do crap here due to too low frequency and conducted emission as opposed to radiated emission, where alumimium foil might have worked. <S> Probably measureable, but managable. <A> I've had success with a radio and blocking emissions from a PC with the mains cable from the radio wrapped around a ferrite toroid. <S> You can also get clamp on ferrites almost everywhere. <S> This will work if it is conducted emissions from the knife, <S> if it's radiated emissions (radio waves) then your best bet would be to shield the speakers. <S> It might be good to put a ferrite on the speakers AC mains cable too. <S> Source: <S> https://palomar-engineers.com/ferrite-products/ferrite-cores/ferrite-ring-toroid-combo-pack <S> This is what a clamp on ferrite looks like: Source: https://www.argentdata.com/catalog/product_info.php?products_id=44 <A> You can also tear down the machine and make sure the commutator and brushes are tip-top.
| One easy way would be to put a toroid or ferrite on the cable, they increase the inductance on the cable for high frequency signals and prevent conducted emissions. If you don't want to dissassemble your device, at low frequencies there is a fairly low penalty in placing the filer down the cable instead. The source of the EMI may not be the machine itself, but defects or wear in the machine.
|
What tools can I use for parametric circuit analysis? I have analog circuit designs that I want to analyze for accuracy, etc. according to the component variations based on tolerance, temperature, etc. To elaborate further, imagine there is a circuit designed with couple of resistors and an opamp, and based on their nominal values we have an expected output. How does this output change based on the possible changes in the circuit element parameters? (tolerance and temperature dependency of the resistors, offset voltage and bias current of the opamp, etc.) <Q> You want to do a Monte Carlo simulation. <S> In this kind of simulation you specify the tolerance for all of the parameters of interest. <S> You may also be able specify whether the expected variation of each parameter is a normal distribution or a gaussian distribution. <S> The simulator then selects random values for each parameter, based on the parameter's tolerance, and runs a simulation. <S> So, set up the Monte Carlo simulation and run 50 simulations. <S> Look at the results from these simulations to see how the output characteristics of the circuit have changed. <S> That gives you a reasonable approximation of how actual manufactured circuits will vary, if you have set up the component tolerances correctly. <A> I can't remember where I learned this technique. <S> You need two functions for this. <S> The first is a utility function to determine if a certain bit of a number (when represented in binary) is set. <S> The second is a function that gives the value of a component for a particular "run". <S> In LTSpice these functions look like this .function <S> bit_get(bits, index) <S> floor(bits/(2**index))-2*floor(bits/(2**(index+1))).function wc(nom, tol, id) <S> if(run==-1, nom, if(bit_get(run, id)==1,nom*(1+tol),nom*(1-tol))) <S> bit_get takes a number and the index of the bit to check and returns 1 if that bit is set and 0 otherwise. <S> wc takes the component's nominal value, its tolerance, and a unique (among all components that have a tolerance) "id". <S> Depending on the value of the global variable run , wc <S> returns either the nominal value, the maximum value, or the minimum value. <S> In order to determine operating points of the circuit for all combinations of nominal, minimum, and maximum value for each component you create a step directive of the form <S> .step <S> run <S> -1 max 1 where max is <S> 2^(num wc components)-1. <S> For example, if I want to analyze a circuit with 4 components then max is 15. <S> Then you create a directive to find the operating point for each step of run with .op <S> A trivial example of this all tied together: <S> Here the voltage source is nominally 5 VDC +- 5% and the resistor is 10K +- 1%. <S> The graph shows the current through R1 for each run . <S> Nominal values are when run is -1. <S> This method assumes only one symmetrical tolerance per part, but hopefully it is a helpful starting point for you. <A> Normally in production, the bottom line is yield to test specifications. <S> There is a statistical measure of margin to upper/lower specifications and a multiplier of standard deviation . <S> This is called Cpk. <S> Depending on the cost of scrap that cannot be repaired cheaply, and your cost margins , an analog statistical deviation may be 2 sigma to 6 sigma reaching the spec limits. <S> Like Diode <S> forward voltage , it may be asymmetrical tolerance on the high side from the mean. <S> I suggest you use Cpk to evaluate your design margin to specs using sigma and limits and use worst case tolerance analysis to compare. <S> then validate your component tolerance with datasheets and supplier qualification. <S> Specs may be guaranteed by source testing by supplier or stated as implied by design. <S> When there are too many variables, that are interactive with each other, we do a sensitivity analysis <S> Allen the Taguchi Method. <S> For simple designs however, one can analyze sensitivity by summing the tolerances with 2 sigma and differentiating the sensitivity curve to find the min, max points to improve the design.
| I use LT spice to find worst case DC operating points of circuits. In LTSpice one can specify the value of a component as a function of other parameters.
|
Can a battery be damaged by charging it with wrong amperage PSU? Can a battery (specifically a lithium-ion type that you would get in a small portable electronic device) be damaged by charging it with a 5V 1A PSU, where the manufacture specifies as 5V 2A PSU? Similarly can a charging a device that requires as 5V 1A PSU with a 5V 2A PSU damage the battery? If the battery was not damaged in either of the above examples, could other electronics in the device be? <Q> You clearly have some misconceptions regarding how Li-Ion battery charging works. <S> The 5 V input does not charge the battery directly <S> , there is <S> should always be a charging circuit in between. <S> If you would apply 5 V directly to a Li-Ion battery you damage it beyond repair and it might even catch fire. <S> There needs to be a charging circuit in between as Li-Ion cells must not be charged to more than about 4.2 V. So 5 V would be too much. <S> How much current is used to charge the battery <S> depends on the battery charging circuit. <S> The battery charging circuit decides how much current can be used. <S> It might decide that for example 1.5 A is needed. <S> When you use a 5 V 2A power adapter like is needed for your device, all will be fine. <S> But if 1.5 A is needed but you have connected the 1 A adapter issues can occur. <S> It can be that the 5 V voltage drops too much and charging is done at a lower current. <S> It can be that the 1 A adapter is overloaded and shuts off. <S> Note though that not many power adapters have an overload protection! <S> The 1 A adapter might be overloaded, become too hot and be damaged and maybe catch fire. <S> What will happen when more than 1 A is needed from a 1 <S> A adapter is anyone's guess as it depends on the design of the device you're charging (phone) and the design of the power adapter. <S> In general you should just use the recommended rating, so in your case 5V, 2A to be safe and not use an adapter with a lower rating. <A> First and most important, it is not safe to charge a battery with higher-than-specified current. <S> Doing so risks damaging the battery (at best) and causing a fire or explosion (at worst). <S> Fortunately, what you are asking about is not <S> the charging current, but the current available from the power supply that supports the charger. <S> The charger will draw only as much current as it needs from your PSU. <S> The PSU you select should provide the same voltage that the electronic device expects and at least as much current as it needs. <A> The charging device should always match the cell's specifications, even though that cell has a certain tolerance (the electronics that comes within). <S> Try to read here, there is a good resource: <S> https://batteryuniversity.com/learn/article/charging_lithium_ion_batteries Not properly used the Li ion cell could even explode.
| The actual battery charger is inside the "small portable electronic device" and the charger itself will limit the charging current to a safe level.
|
Connection for tungsten heater This is my first post, so hello all :) I have a tungsten wire thats wrapped around a crucible. I want to send ~20A through it in order to heat the crucible. Now the tricky part. Since this is part of a delicate setup, I don't want to use solder or glue in order to minimize contamination through evaporation (this is a physics experiment).At the same time the setup has to fit inside a 10cmx10cmx5cm box. Any ideas on components I could use to make these connections? Is using crocodile clips to connect the tungsten wires a completely stupid idea? :D <Q> A simple pair of copper busbars (0.5 * 0.125) with a high pressure contact would appear to be all you need for each terminal. <S> Separate your wires where they enter the busbar area and clamp each wire individually with a stainless steel screw and high pressure domed washer. <S> Separating the Tungsten wires may help in supporting your crucible. <A> That will involve only two materials- <S> the copper wire and the tungsten. <S> You can make the copper terminals large and with enough heat sinking that the vapor pressure of the copper is minimal. <S> Second choice would be to braze the joint, again if you keep it relatively cool the contamination might not be important. <S> The solder used ("silver solder") is typically mostly silver but is alloyed with a bunch of other metals. <S> Such as a ceramic insulated terminal block. <S> Image from here . <S> In ceramic kilns a simple arrangement with a pair of washers and is used to clamp the heater wire (typically Kanthal alloy). <S> Screw based methods need to be kept cool and they may eventually loosen up with cycling. <A> I do not know if tungsten wire is sufficiently strong, but I would use crimp connections. <S> You don't mention the tungsten wire diameter, nor do you tell us what the operating current will be <S> and thus we can't provide any detail. <S> In general, you can crimp pins or butt splices onto the resistance wire. <S> You generally need to provide some heat-sinking on the connections to keep the connection temperature low enough that the crimp doesn't fail. <S> Generally, just increasing the surface area of the connection is enough. <S> I am curious about your choice of tungsten wire - most common heating wire is Nichrome. <S> Kanthal is often used in vape pens.
| A third choice, if the tungsten wire is robust enough, might be to use screw terminals that are big enough and well heat sinked enough to keep them from heating much. If you are in a vacuum chamber you can connect your entry wire to one of the busbars, I doubt you need to do both at the low current you are using (20A). Probably your best bet is welding, such as spot welding.
|
ICL7660 - Cannot provide stable negative voltage for audio amplifier I built this audio amplifier mainly with UA741, ICL7660 etc. The VCC is 5 volt from USB. But I found that the -VCC is quite unstable and this caused noise from speaker, which can hear even when music was played. And here is what my scope picked when nothing is played on computer from output (pin 6) of the Op Amp on the left. And I picked the working frequency from pin 2 of ICL7660. They lined up so nicely that's why I think it is the ICL7660 IC that caused the audible noise. Also, when music is played I found the -VCC become less negative significantly while VCC stays really stable. So, can I improve this to get rid of that noise or must I make some other approach to fetch -VCC? <Q> As stated in the comments, ICL7660 and ICL7660A series switched capacitor voltage inverteres shouldn't be used for current requirements more than 20mA because of high output impedance (around 100 Ohms) . <S> You can check this from the datasheet . <S> You might know that these circuits draw only a few mA of current. <S> If required voltage swing is not higher than ±2V <S> then you can still use the single +5V supply with virtual grounding. <S> If you really need a negative supply with higher current then you can build a classical voltage inverter with MC34063 or LM2776 or equivalent: <A> It’s not the Op Amp but rather the -Vcc supply that cannot drive <100 <S> Ohms let alone 8 Ohms. <S> You also lose <S> ~4.5V <S> swing due to 741 output swing = <S> (Vcc+|Vee|)-3V and 2xVbe drops =1.5V . <S> So neither 741 nor charge pump work well here. <S> The 741 was intended for >+/- <S> 12V supplies and 20mA needs a current gain of >50 at 1A on emitter followers. <S> So read data sheets carefully and supply must deliver 1A regulated voltage. <S> The 741 has 77 dB PSRR and 70 dB CMRR . <A> The UA741 does not have good -Rail Power Supply Rejection. <S> Use a different opamp. <S> Notice location of the 25pF capacitor. <S> The right end is the large-voltage-gain node for the opamp. <S> The left end is two diode drops (Q16,Q17) above the negative rail. <S> At high frequencies (fast trash on the -VDD rail), the trash on the -rail is copied onto the VOUT (base of Q18) and then heavily buffered to the output pin.
| The most common usage for this type of inverters is providing a supply for opamp-based circuits and RS-232 circuits.
|
TTL IC logic high output doesn't reach high enough voltage I've been having a problem trying to build a breadboard circuit. The problem being, that some IC high outputs doesn't reach a high enough voltage, staying around ~2V. That is not sufficient for a clear high signal, and indeed the logic gates that receive such a signal end up not working correctly. Here is a pic of my wiring (in the pic there are some disconnected input pins, but the problem persists even when they are all connected): The IC in question is the 74LS173, a 4-bit tri-state register. Link for its datasheet. I feel like I must be missing something pretty simple, but I can't for the life of me find out what is it. In any case, thanks in advance for the patience and for taking the time to help! <Q> You appear to have some LEDs connected to your outputs. <S> These will limit the voltage to 2(ish) volts if driven direct from the outputs. <S> TTL outputs are not designed to drive LEDs to ground and have a valid logic high level. <S> Their weak drive current is however convenient for driving LEDs like this. <S> TTL is much better at sinking current than sourcing. <S> If you want to drive a LED, you should connect it to the +ve rail, in series with a current-limiting resistor. <S> Unfortunately, that means it will be 'on' for a low output. <S> This is a small price to pay for driving the LED properly, and delivering valid logic states as well. <A> You can't connect an LED to a typical TTL logic output and still get a valid logic signal on that output. <S> Typical LEDs draw more current than a TTL gate is designed to provide. <S> In particular, TTL gates cannot source much current at all for a logic '1' level. <S> You can add current limiting resistors in series with the LEDs <S> but you would be much better off using a dedicated driver for those LEDs. <S> The datasheet for you <S> LS173 states that you can draw no more than 5.2mA from a high logic output, and the output voltage will fall as low as 2.4V if you do. <S> If you have low current LEDs and add current limiting resistors you might make it work. <A> Without an schematic it’s hard to be certain <S> but I see two basic issues: <S> Its output driving current can be very small when driving a logic 1, as opposed to its sinking current when driving a logic 0. <S> More importantly, you seem to have directly connected red LEDs from the outputs of your IC to ground. <S> Take a look at the I/V curve of a typical red LED and you will see how much output current are you demanding from your IC. <S> No logic family would handle that.
| The TTL logic families (like the 74LSxx you are using) are not designed to generate a “strong” 1.
|
Variable Current Draw from motor for torque control I am building an electric go-kart. I would like to use regenerative braking. I would like to have a brake pedal that controls the amount of torque (slowing the motor and wheels down) that the back emf from the motor creates. In a hypothetical situation, the motor (90amp 36volts permanent magnet dc motor) will be spun up when the car accelerates. It will then spin freely once up to speed. This part of the circuit I have figured out this part of the circuit already. (in this image the important switch is the pedal microswitch which closes every time the throttle is engaged Also, the Main, Foward (F) and Reverse(R) notations refer to contractors used to provide power to and switch the polarity of the motors respectively) If we now treat the motor as a generator, from what I understand the amount of current that we draw from that generator the more torque will be applied to slowing the motor down. If we take the extreme and fully connect the motor leads together with zero resistance the motor will lock up. Where I am confused is how to draw variable current from the motor and control it with the brake pedal. For safety purposes, I plan on installing a mechanical hand brake as well. Any help on this would be appreciated, I am sure that I do not have a full understanding of this mechanism. Note this diagram is from a Curtis motor controller manual. Here is a picture of the cover. <Q> I don't think you realize how much is involved in building such a powerful motor controller. <S> I will explain some basic concepts to hopefully give you and idea of what you are trying to achieve, but I suggest you purchase a speed controller with appropriate ratings. <S> Building a safe and reliable speed controller for 90A is a big deal. <S> To a first approximation, motor current = motor torque. <S> If the current is flowing from motor into battery (regen) then the torque is acting in the "braking" direction. <S> If the current is flowing from battery into motor, then it is acting to push the vehicle in the desired direction. <S> To produce a regulated braking torque, you will need to produce regulated motor current flowing from motor into battery. <S> This can be done using PWM control in conjunction with a full bridge motor drive circuit. <S> Please note that regen must be limited to avoid charging the battery at too high of a current, and also to avoid overcharging a battery that is already fully charged. <S> Your solution must accommodate these constraints. <S> For example it could be a micro controller with features included to facilitate PWM control. <S> Here is an image of an H bridge which I found online here . <S> When Q4 is on, the duty cycle of Q1 can be adjusted to apply variable forward voltage to the motor. <S> When this voltage is greater than the back EMF, there will be forward current and the motor will push the vehicle forward. <S> When the applied voltage is less than the back EMF, this will produce regen current and braking. <S> By adding shunt current sense in the motor path, you can use a microprocessor to control PWM duty cycle to achieve the desired motor current. <A> V=IR + E. For a given speed, thus a given electromotive force (E) you will need to vary voltage in order to vary current. <S> You will need to include a DC converter which is able to absorb current from the motor. <S> Maybe a full bridge with optional filtering between it and the motor. <S> Then, you will need to design the control and connect it somehow to the pedals. <S> It can't be done just as it is, because you only have on-off speed control. <S> P.S. <S> The full bridge is indeed implemented with mechanical switches, thus your gokart would suddenly start and suddenly brake. <S> You need to switch faster than it can respond and thus vary the average value of V <S> and I you apply. <S> That is the purpose of the converter. <A> You can use field oriented control to control the torque of the motor. <S> The field oriented control does the torque control by regulating the current .first <S> you have to select a proper motor as per the torque requirements .You can select PMSM/Induction motor/Bldc motor for this purpose. <S> Then you have to design the motor controller for the motor. <S> If you are not familiar with circuit design you can buy ready made curtis/sevcon controller. <S> how to get regenerative breaking:Our aim is to activate the regenerative breaking as soon as the throttle/accelerator is released. <S> Suppose if the motor is running in clockwise direction with a positive torque to extract the energy from the motor back to battery you have to apply a negative torque to the motor from the controller. <S> The amount of negative torque applied will decide the stopping distance of the motor,also the energy dumped back to the battery. <S> if you wish to design your own motor controller hardware and software you can refer to texas instrument reference design. <S> Here is the link : http://www.ti.com/tool/TIDA-00364
| In order to control the motor current, you will need some kind of control system which can sense current and adjust PWM duty cycle to produce the desired current.
|
Can voltage and current be stepped up and connected together Ok if a boost convertor is used to step up voltage, then a seperate buck convertor is used to step down, increasing the current, can these two be configured to utilize the high voltage on one output, then utilize the high current on the other output and connect the both together <Q> Depends on what you mean by "utilize". <S> Beyond that whole second law of thermodynamics thing, each converter will be in the 75% - 90% efficient range. <S> If each is 80% efficient, then for every 100 W into the first converter you will get only 64 W out of the second converter. <S> Sketch <S> a basic wiring diagram of what you want to try, write in some input voltages and currents, and see what the outputs are. <A> No, absolutely not. <S> The output of the buck converter is at a lower voltage than the output of the boost converter. <S> Connecting a high voltage, low-current source to a low-voltage source will cause current to flow into the low voltage source, which is just the opposite of what you propose. <S> Energy must be conserved. <S> If you use a dc-dc converter to get a higher voltage then you must have a reduced current. <S> As a wise man once said, TNSTAAFL. <A> The Law of Conservation of Energy says (in one form) <S> that whatever you do, you can't get more energy out of a system than you put in, Laws sometimes get a bad rap, keep off the grass, don't drive faster than 70, but this one is not of that type. <S> It's a description of what happens. <S> Since scientists noticed this behaviour, they've done all sorts of careful experiments, and it's never been found to be wrong, ever. <S> It's rather nice to have this as a law, as a general principle, because if something appears to violate it, you don't need to flog through all the details, you just know that it's wrong. <S> If you said I have a black box, which takes a 10W input and delivers a 20W output continuously, then you just know it cannot happen. <S> No conceivable arrangement of converters can do this. <S> Energy can be stored. <S> The black box could have a battery, capacitor or inductor in it. <S> It could take 10W while charging, then discharge to deliver pulses of 20W, or yet shorter pulses of 100W. Increasing the the power output while reducing the output time is a perfectly respectable activity. <A> This question was asked of me, in high-school. <S> The person planned to use a vacuum tube with 2 cathodes and a single plate, to combine the high current and the high voltage. <S> I did not know how to explain the conflict, other than "energy is not free." <A> You can do this, but it won’t give you the result you’re asking. <S> You need to look at the wattage of the power supply, not the current or voltage. <S> They are always tied together so just voltage or just current doesn’t tell you enough. <S> For example, you can take a power supply, let’s say 100W, and boost it to 100V, but the current will be limited to 1A. <S> Or you can take that same 100W power supply and buck down to 1V and <S> the current limit will be 100A. <S> Now, let’s say you attach both a buck and boost converter. <S> However, the combined converters will only allow 100W/101V=0.99A. <S> If you tried to pull more current, the voltage would begin to drop because the power supply just can’t supply enough energy to sustain 101V at a current greater than 0.99A.
| If you use a dc-dc converter to increase the current then you must have a reduced voltage. Using some sort of isolation, like transformers, you could combine the two power supplies to get 100V+1V=101V.
|
Why does a car cabin fan need a big resistor? The blower motor resistor for the cabin fan in a Volvo 850/V70 is so big that it is not even part of the motor. It has its own mounting. Here is a picture: Why does the blower require such a large resistor? I have much larger fans in my house that have no such resistor. <Q> The resistor is used for voltage reduction to the DC motor in your car fan. <S> The power dissipated in the resistor can be calculated from \$ P = \frac {V^2}{R} \$ or \$ <S> P = <S> I^2 R \$ . <S> The fans in your house are AC <S> and so a capacitor can be used to reduce the voltage without any power being dissipated. <S> A capacitor has an impedance (the resistance to AC) given by \$ Z = \frac {1}{\omega C} = \frac {1}{2 \Pi f C}\$ . <S> The current through the capacitor is 90° out of phase with the voltage so the power dissipated is zero. <A> The V70 is 25+ year-old design so it uses fairly primitive methods. <S> A series resistor (switched for selection) is an acceptable way if controlling speed of a brushed DC motor driving a blower load. <S> The worst case dissipation of such a resistor is half the motor full power consumption, but the designers also have to consider the possibility of the motor stalling due to worn-out bearings or debris sucked into the fan (or possibly critters). <S> You will note the one-time thermal cutoff (fuse) mounted at one end to deal with over temperature. <S> The large physical size results from the required worst-case power dissipation and the requirement that the part should not run (say) red hot for safety and reliability reasons. <S> The resistances are relatively low and the voltage does not much exceed 12V. More modern cars may use a large MOSFET and pulse-width modulation to control the blower speed. <S> AC fans such as ceiling fans use different methods to control speed. <A> If that was only one effective resistor it would only need two wires. <S> It has several connections matched to the number of speeds available for the heater, 3 or possibly 4, which is why it is so large also due to the current it has to pass.
| The resistor needs to be able to radiate this heat away without getting so hot that it melts nearby plastic components or setting itself on fire - hence the size.
|
Male or female header on PCB? I'm designing PCB which has pin header connectors(0.1" pitch). So my question is - Is it the standard to put female headers onto PCB and solder male headers to wires or vice versa? My idea is next:Female pin headers is better to put on PCB. Why? They are protected with plastic around them, so you can't accidentally short connections and damage PCB(assuming that PCB is more expensive than female headers). PCB has 34 pins for Arduino and DS3231 module(those pins will be female pins), six pins are signal pins, one is VCC, one GND pin. <Q> My theory is that male headers are less likely to suffer damage by contamination since everything is "out there" and visible. <S> The females, when part of an IDC cable, are more easily replaced if they fail, rather than trying to replace a part on an expensive multilayer PCB. <S> You will find most products where there was a choice such as IDE disk drives and PCB motherboards will agree with this choice. <A> There is no standard. <S> These cables were equipped with 0.1" pitch female socket connectors which mated with male pin box headers. <S> Over time many applications eliminated the shrouded box header in favor of the open male pin header due to cost and space savings reasons. <S> These days there are so many choices it really becomes a case of finding the connection technology that best fits the packaging concept for your electronics gadget. <A> There's not a standard, but a good rule of thumb is to think about would happen if someone untrained took hold of the board. <S> For example, mains wiring is always a socket in the wall so it is impossible or difficult to touch a live wire accidentally. <S> If there is power being supplied by one side, I would generally make that a socket (female). <A> Contrary to your statement, male headers also come with shrouds. <S> If you are plugging in cables, then use male headers that include retaining clips. <S> I do not know if you can even get a female connector with retaining clips. <A> Ma Bell has the female parts <S> It applies to phone, electric, USB, and all manner of other signal circuits. <S> For safety, the electrically active parts of the circuit are protected; barring tampering, the matching plug is required to access the power or signal. <S> For usability, in general, the plug is more likely to be damaged by repeated or incorrect usage. <S> And a cable end is usually easier to replace than a mounted connector. <S> But there are exceptions, especially if the duty cycle for the connection is very low. <A> You do not provide many details, so I will be generic. <S> You should consider safety. <S> Safety for the device, safety for the cables, safety for the users (people). <S> Applied to electricity networks, inputs are male and outputs female. <S> In this way you make sure you cannot have a direct link to the electricity supply - which can be accidentally touched by people. <S> You cannot easily create a short-circut. <S> On the other hand, we (sometimes) use fuses not to protect the devices, but the cables - replacing cables can be a lot more expensive, because of dismantling, re-assembling... <S> So analyze your setup from these points of view. <S> Protect as much as possible. <S> Prioritize.
| Use the shrouded kind for more protection. Cost wise, male tends to be cheaper than female, so if you're doing big numbers that may also be a consideration for you. Male pin headers on PCBs got their start in the electronics world back when ribbon cables became popular. The standard, if there is one, may be based on the old rule of thumb that says
|
What is the meaning when we say power of circuit having a clock frequency 100Hz is 2W? When we say clock frequency is 100Hz, then there are 100 clock pulses in one second. So when we say power is 2W, is it 2W for 100 cycles or one cycle ? Or is it anything else? <Q> Power is energy per time. <S> It is up to you to define a time interval. <S> But usually we are talking about mean or effective power of a sine wave, which defines the time interval to be one or multiple full periods of the sine. <A> An average power of 2 watts means 2 joules of energy per second. <S> If you sub-divided this up into smaller time slots <S> (i.e. 10 ms thus corresponding to a 100 time slots per second) <S> the energy would be 0.02 joules per 10 ms. <S> This is still a power of 2 watts because 0.02 joules divided by 0.01 seconds = 2 watts. <A> Full wave Rectifiers are one way to produce 100 Hz from 50Hz <S> but this can be done in microwatts. <S> So power and frequency do not need to be relevant but might for some motor. <S> Power “may” be measured; in Zero Time as a sample or an Average , or a Peak or RMS value or in complex form X+jY or Real, and Reactive or Apparent power the hypotenuse of the above or in some trig <S> form A (cos wt)+B or estimated average power or true RMS power as DC power = <S> RMS AC power <S> [Wh] or watt-seconds = <S> Joules [J] <A> i recall using a 9 amp Power Driver for MOSFET gates. <S> The shoot-thru charge, during the 10 nanoseconds of internal slewing of control voltage, was rated in nanoCoulombs. <S> Assume this was 5amps (mid-rail) for 10 nanosecond, or 50 nanoCoulombs. <S> At 15 volts on the IC, the energy was 750 nanoJoules per transition, either going-positive or going-negative. <S> Thus a complete cycle was 1.5 microJoules. <S> We used 0.1uF bypass caps, with 1cm leads, on the breadboards, and tolerated the constant 0.5 volt rail sag and the scary ringing. <S> At 1,000,000 cycles per second <S> (Hertz) <S> the internal power dissipation was 1.5 watts. <S> You could buy these parts in DIP (and 1.5 watts is a LOT of power, for a DIP) or buy a TO-220 with a 1cm by 2cm copper tab you could BOLT to a heatsink. <S> I (and my team) quickly learned to only use the TO-220 in our prototypes, with a small fan pushing air onto the hot tab.
| Power has no Units of time , unless as additional information to compute Energy in watt-hours
|
Modifying land patterns generated with IPC-7351B wizard I generated a 4x4mm 0.45mm pitch QFN28 according to the Atmega328P datasheet : However, the pads on the corners (1+28, 21+22, 14+15, 7+8) are too close to each other, less than 0.2mm apart (when datasheet says the pins are 0.4mm apart because of the chamfer). The solution that comes to mind is to make the pads on the corner shorter, as I did with pads 1 and 28: Which gives about 0.3mm clearance: Is this the proper way to do it? If I just leave the pattern as generated I will most likely get DRC errors depending on the process I pick (0.15mm vs 0.2mm). I know the assistant is just a tool and not set in stone, but I'd just like some informed answers about the proposed solution. (Or alternatives) Update after answers: Thanks to DerStrom8 for the App note. Here follows a comparison of IPC generated and Atmel/Microchip recommended patterns: Left is Atmel/Microchip App note (~0.266mm), Right is IPC wizard from Altium (~1.77mm) <Q> Unless you're etching this yourself, a clearance of 0.2 mm probably won't be a problem for the manufacturer. <S> 0.127 <S> mm (5 mil) is not uncommon, and 0.152 mm (6 mil) is even more common. <S> I would check your board house's requirements and modify the DRC rules to match what they are expecting. <S> Then if there are any specific issues based on your manufacturer's rules, you can focus on them one-by-one. <S> If you want to bypass the IPC footprint altogether, you can go with the recommended land pattern from Microchip/Atmel themselves. <S> See page #12 in this app note: http://ww1.microchip.com/downloads/en/appnotes/atmel-8826-seeprom-pcb-mounting-guidelines-surface-mount-packages-applicationnote.pdf <A> The PCB pads will be slightly larger than the package pads to allow for slightly imprecise placement. <S> This will be corrected by surface tension when the solder is in the molten phase. <S> The PCB shape generated by the IPC wizard will be fine (as long as it meets your manufacturer's DRC minimums). <A> I created custom pad shapes (one cut on the right, one on the left), and then replaced the eight corner pins on the package. <S> I started with an IPC-7351B-compliant package created by Library Expert. <S> I calc'd the chamfer dimensions to end up with my required clearance between pins on the completed package. <S> Finally, I went to the symbol itself and replaced the 8 pins in question with my new pad shapes. <S> This was using Cadence OrCAD.
| For the new pads, I started with the dimensions of the IPC-7351b-recommended oval pad, and then manually modified them to create the chamfered corners. The drawing in the datasheet is the actual package itself, not the pad layout you have on the board.
|
Why aren't fully asynchronous circuits more prevalent? From my understanding, most modern consumer CPU's are based on synchronous logic. Some high-speed applications (signal processing, etc.) use ansync logic for its higher speed. However, in today's market, speed in consumer products is one of the main selling points (see AMD vs Intel.) Is the development of more complicated lithography faster than adoption of fully asynchronous logic? Or is ansync logic too complicated/impractical for VLSI applications? <Q> I spent some years in a startup commercialising async design technology, so I'm familiar with the reasons: <S> async isn't intrinsically faster. <S> The worst-case path delay remains the same. <S> It's just that sometimes you get to take advantage of a faster path executing. <S> async has overhead of completion detection too. <S> Design tools. <S> This is the really big one: there isn't really a full async "flow" of tools available to the same quality as synchronous design. <S> Training. <S> You'd effectively have to retrain all your designers on the new paradigm and tools. <S> Risk and conservativism. <S> So much of the industry is "produce something similar to the last one, but a bit different". <S> This has a very high chance of working. <S> Companies are much more reluctant to build something totally different since it has much more chance of being a total writeoff costing tens of millions of dollars. <A> It is very tempting to design asynchronous integrated circuits. <S> The other answers already cover many reasons to think twice before doing it. <S> Here is one more: <S> IC development is not finished with the design. <S> Verification and test are equally important. <S> Not only the design tools are very advanced for synchronous circuits, but it is the same with simulation tools and test equipment. <S> Verification <S> It is not sufficient to have the circuits working at lab conditions. <S> They need to be robust with respect to the operating voltage (V) range, the operating temperature (T) range, and the variation due to the manufacturing process (P). <S> For synchronous logic this can be guaranteed with the help of static timing analysis. <S> The circuit is broken down to all timing paths, from flip-flop to flip-flop. <S> Setup and hold times are checked for every single timing path, and for different combinations of P, T, and V. These PTV combinations are the so called simulation corners. <S> A similar verification could be done for asynchronous circuits, but it is much more difficult and much less supported by the design tools. <S> It also restricts the designer to asynchronous constructs that actually can be verified. <S> There is no reliable verification for arbitrary asynchronous circuits. <S> Test <S> Similar difficulties are there when it comes to testing the hardware. <S> Testing synchronous logic is fully supported by testing standards and equipment. <S> The circuit could fail due to race conditions at some PTV combination, that is not covered by the corners. <S> Summary IC designers have not given up the asynchronous paradigm, but asynchronous logic comes with heavy disadvantages during verification and validation. <S> In an industrial context, asynchronous IC design needs to be restricted to construct that can be proven to work over the whole parameter space of process variation, as well as the operating ranges for temperature and voltage. <S> The so called "Locally synchronous globally asynchronous" design is one way to get more benefits and less disadvantages of both timing paradigms. <A> Async binary counters are simpler because they only use 1 memory cell or T flip flop per divide by two. <S> Hence the old CD and 74HC4020 and 4040 offer many binary stages cheaply. <S> The prop delay in each stage means it cannot be used without race conditions or glitches with gate decoding of the binary addresses unless the prop delay is less than 1/2 input clock cycle using the trailing edge to latch the result. <S> The output latency is then multiplied by N stages. <S> Synchronous binary counters use an extra memory cell to D FF to delay but minimize the delay to 1 value for any length of counters so it consumes more area. <S> Hence all CPU’s use complementary clocks to optimize expected latency in address and memory read /writes to maximize thruput but not exceed prop delay, setup and hold times. <S> Memory now uses many phases such as DDR, 3DR, 4DR, 5DR especially for graphics memory but with CPU clocks going much faster than the single cycle RAM rate so that read and write address delays may be timed by single or multiple or half counts of the superclock ( eg 100MHz xN) designated by T fractional or integers counts for each parameter. <S> These prop delays increase with temp. <S> For CMOS and reduce with higher Vram voltage which if properly cooled might reduce latency or other increase Pd and temp rise and make it worse (slower). <S> So cooling, V, f, T all are critical for optimal latency whether it is used for Async or Sync operations.
| Testing asynchronous circuits not only is more complicated, but because of the lack of timing abstraction, it is not even sufficient to prove that the circuit will work for all PTV corners.
|
Why do we use alternating current and not direct current for mains power plugs within a house? I understand that for transporting electric energy it makes way more sense to have AC. But within my house, I don't think that argument holds any more. Devices I use (roughly ordered by power consumption): Stove / oven: Isn't connected to the usual power plugs anyway, but a "high power" one Washing Machine: max 900 W Micro Wave: 800 W Hand mixer: 450 W (I was quite surprised by that) Fridge+Freezer: max 110 W Laptop (usually 50 Watt, max 90 Watt) Smartphone: 15 W? electrical shaver: < 10 W LED lights: 3 W alarm clock: ? Charging USB 2/3 devices I've just seen that the car charger cable for my Notebook is WAY smaller than the one for the usual power plug. Looking at the power supply unit of my laptop, I see that it outputs direct current. Seeing this huge difference in the size of the charger unit (and the price as well), I wonder why the usual power socket does not provide DC, but AC. What are the advantages within a house to use AC? Why wouldn't it be a good idea to have DC power plugs (e.g. as in cars)? (I've also heard that solar pannels provide DC). <Q> Within your house, electricity is still transported . <S> The typical voltage loss on the average household cable of 1,5mm² cross section with an average length of 50m (forth and back) from the meter cabinet to your applicance is about 1V per 1A. <S> 1V isn't much for 230V. 1V is much for lower voltages. <S> For lower voltages, you need more current for the same power, so the voltage loss on the cable gets worse. <S> That's why you want to have 230V at your electrical outlets. <S> But your question was about DC. <S> Why not DC? <S> Because, until lately, it was highly impractical to turn one DC voltage into another. <S> You always need electronics to do that which have some sort of DC-AC-DC inverter inside. <S> AC on the other hand can be transformed to another voltage by a simple electrical component. <S> No electronics. <S> Copper on an iron core. <S> A transformer. <A> The power company uses AC in the distribution network so that it can easily use transformers to move power between high voltage segments (for long distance lines) and low voltage segments (for supplying individual customers). <S> To supply you with DC, they'd have to add additional equipment near your home to produce the DC from their existing AC network. <S> That would cost them money, so obviously they don't want to do that. <S> Alternately, you could take their AC and buy your own converter to produce DC for distribution around your home. <S> But then you'd be faced with a fairly complicated optimization problem. <S> What voltage should you use? <S> Large loads like motors would want higher voltages to reduce wire cost, smaller loads like smart phone chargers would want lower voltages. <S> How big a converter should you buy? <S> Do you know how many appliances you'll have in 20 years and how much current they'll need? <S> Should you use DC for all loads, making the wiring simpler? <S> Or should you stick to AC for (incandescent) lighting and heating loads, to reduce the size (and cost) of the AC-DC converter you need? <S> Currently we distribute AC around the house and use a separate converter for each appliance that wants DC. <S> The benefit of this is it lets us optimize the DC converter for each load that needs it. <A> Because converting to DC is not necessary and costly for your entire house. <S> Most of the devices you named can use AC without an issue. <S> Converting your whole house to DC would be expensive and inefficient. <S> You would have to buy a large and expensive converter to convert the whole house. <S> This converter would use energy as it was converting and would generate a good deal of heat. <S> Your electricity bill would go up 30% + <S> whatever extra your air conditioning would use to counteract the heat generated. <S> The converter would break every 10 years requiring expensive repairs or replacement, meanwhile rendering your house without power. <S> Buying small converters for devices the require DC is acceptable. <S> DC only devices are a small percentage of the overall usage of the house. <S> Small converters are cheap and easily replaceable. <A> If DC is used instead of AC the copper drop issues are still there so large loads would be best on a reasonably high voltage .100VDC <S> was normal before Tesla won the AC/DC battle over a century ago. <S> Arcing and its associated fire hazard is much worse with DC because there are no zero crossings for the arc to go out .Hence <A> I understand that for transporting electric energy it makes way more sense to have AC. <S> But within my house, I don't think that argument holds any more. <S> You'd have several issues. <S> Generation of the DC. <S> This would require a large AC-DC PSU in the building. <S> Choice of DC voltage: 12 V for car stuff, 19 V for your laptop, 5 V for you USB devices, 60 V for a particular LED light fitting? <S> Which one would you standarise on? <S> When you've done that everything else will require DC-DC converters. <S> Size of cable required to run around the house to keep voltage drop to an acceptable level for the higher currents drawn at low voltage. <S> DC circuit breakers. <S> I suspect that many off-grid residents trying to run on DC power banks are well aware of the advantages of a constant voltage AC supply.
| DC switchgear is much larger and more expensive than AC switchgear at high power.
|
Limiting the current into Raspberry Pi GPIO pin I have a dc dc regulator which can output .35A Current at 3.3V. I just need to check that the voltage it is producing is 3.3V by connecting it's output into one of the GPIO pins of Pi. So i was wondering what is the easiest way to do it, will a series resistor do the job. And what would be the value of it. <Q> just need to check that the voltage it is producing is 3.3V by connecting it's output into one of the GPIO pins of Pi. <S> No, that is NOT the way to check the voltage. <S> Beside the chance of damaging the Pi, even with resistor, the Pi can not tell you what the voltage is. <S> Buy a Volt meter. <S> (I bought five <S> ** good ones for $5 each, a few years back). <S> If you want to continue dabbling in electronics you are going to need one. <S> ** <S> I gave some to family <S> so when I visit and they ask me to 'look at my...' <S> I have a meter to do it. <S> I once managed to guide my parents by phone to use one and find the fault in a cable. <A> If you want your Pi to tell you the output voltage of your DC/DC regulator you'd need to use an external analogue-digital converter (ADC). <A> That sounds iffy. <S> I would use something with an inbuilt ADC such as an Arduino. <S> If you have something like a 1K resistor on hand then put that in series with it just in case. <S> Equally if you are just looking for a number (i.e. not constant monitoring) then buy yourself a multimeter online. <S> That being said it can be a good project to make your own at home!
| Nowadays you get a good universal meter for as little as $10. If you have some diodes they would be best, placing one with the negative side on the power supply you are measuring and the positive side to the power supply of the Arduino.
|
Difference between analog signal and battery voltage I want to find out if using an analogue sensor is feasible in my noisy environment by running some signal integrity tests before I buy it. Because I don`t have the sensor yet, I need to simulate its output. This can be assumed to be a constant DC voltage, since it will change very slowly. A reasonable value for this voltage is 2.25VDC. My approach to model my sensor is to take a 9V battery and load it with four 220K ohm resistors in series. I will use the potential difference between the last resistor and ground (2.25V) to model my sensor`s analog output, and run my tests with it. To make this more clear, an example of a test I will run is to use one ADC module to measure the voltage across the last resistor in the potential divider, and another ADC module to measure the voltage across the wires of the other end of a slip ring connected across that last resistor while it is rotating. Some more background to understand the application if you're curious is at this question. My question is can an analogue signal be modeled by a battery? What is the theoretical difference between an analogue signal and the voltage across a battery`s terminals other than the source resistance? Thanks for helping a beginner to signal processing! <Q> The idea is fine, but 220k seems rather high. <S> You should check the specs of the ADC module, there should be an indication of the maximum input impedance (typically around 50k for ADC integrated to most MCUs). <S> Use resistors with values lower than this. <S> Hint: you could also use a potentiometer instead of the lower resistors (closer to ground), so you can easily adjust the value. <S> Keep at least one upper resistor and make sure you choose values that guarantee that the ADC input voltage rating is never exceeded. <A> (there are theorems to that effect). <S> Formally, you would need a voltage source & a complex impedance, to also be able to represent the frequency behavior. <S> But it can be even more complicated than that. <S> Some of the factors that this will not be modeling: <S> Sensor noise. <S> You would need to explicitly add a noise source to represent this. <S> Sensor non-linearities. <S> This is a linear model, circuits will behave differently if current or voltage limits are exceeded. <S> Sensor instabilities. <S> Many circuits become unstable under some load conditions (for example inductive or capacitive loads). <S> Although this could probably be represented with more elaborate impedance models, the interaction with non-linearities makes it complicated. <S> But if you (1) ignore the sensor's noise contribution, (2) don't exceed loading limits, (3) don't exceed stability limits, and (4) frequency response is not a concern, then yes, you can model it with an adequate battery and resistor combination. <A> My question is can an analogue signal be modeled by a battery? <S> Yes, as a voltage source with series resistance and a noise source. <S> Batteries can be very low noise because of their low source resistance. <S> You can even buy (or used to be able to buy) source calibration batteries that are only used for voltage calibration. <S> Most analog amplifiers have a very high source resistance and pull very little current, so in most cases the source resitance of the battery can be neglected. <S> The best part about using batteries, is they are easy to shield and not connected to ground. <S> This means that you can build a source, and not worry about 60Hz or other sources of contamination from RF getting into the signal with proper shielding. <S> What is the theoretical difference between an analogue signal and the voltage across a battery`s terminals other than the source resistance? <S> A batteries voltage can be temperature dependent ( <S> some batteries more so than others), but most sources are temperature dependent. <S> Other than that, analog electronics don't care what they are attached to, if you put the same voltage and source resistance from a battery or a signal generator, the analog electronics won't be able to tell the difference if the voltage is the same and the source resistance is the same. <S> However it is the stability of voltage sources that makes the difference.
| The simple answer is: yes, in many cases a simple voltage source & resistor (Thevenin) model is enough to model the output of a circuit
|
Soldering breadboard wires onto a small chip I have some DWM1000's that I need to use for a project. One will be connected to a raspberry pi 3 B+ as the tag and 1-4 others will be connected to arduino's as the anchors. I am having trouble though with soldering wires to these chips. I want to use these breadboard wires I have and just remove the plastic covers on one end. I already took off the connectors on one chip, making it useless. Is there a better way to go about this? My background is in computer science so I'm a total noob with soldering stuff like this. Pic provided. Thank you. <Q> Use a breakout board, soldering wires to something like this is very unlikely to be worth the risk, time, and debugging overhead. <S> Particularly if you have more than one to put together. <S> If creating your own is not an option (it's something that could be put together in an afternoon, but it does imply a learning curve), you can generally find breakout boards for most widely used components. <S> Thi s is one I found with a quick Google search : <A> You should really make a carrier board for these. <S> If for a prototype you want to solder directly to them <S> use something very flexible like 30 gauge silicone wire <S> (The adafruit stuff directly or via digikey has nicely thin insulation, and won't melt when you solder right against it) and do something to anchor the wires against strain before the pads, for example affix the whole thing to some kind of carrier plate that the wire bundle attaches to first. <S> A piece of generic perfboard works well, and you can sew the wires through the holes for anchoring. <S> If your perfboard has some copper either mount the module on the non copper side or use something in between as an insulator. <S> Another crude possibility would be a junk mail credit card mockup, which you could poke wire holes in, though that may be somewhat static friendly (and hence component-unfriendly) as a material. <S> At the moment I have a BLE module stuck on the credit-card-sized carrier of a promotional mobile/IoT data SIM handout; I hadn't though about it when I did that <S> (it was just handy) <S> but in retrospect that particular plastic card is itself technically an IC package (for the SIM) so perhaps not a bad choice at all. <A> I've had to work with something like this before. <S> You will need soldering paste, heat gun, and a tooth pick or home made stencil. <S> Soldering paste application allows more control over area of coverage vs the traditional solder wire you are using. <S> Also it has enough viscosity hold things in place <S> you can see an example of application here: https://www.youtube.com/watch?v=4OYakUQmgd0 <A> However, you may not have the luxury of waiting for them to arrive. <S> What I have done in the past is to use a machine-pin IC socket as a carrier. <S> You have 24 pins on your device, so a 24-pin skinny DIP machine-pin IC socket should work well for you. <S> The PC board is simply mounted vertically on the IC socket. <S> Start by grabbing some stranded wire with many fine strands. <S> Strip <S> the insulation from the wire and separate the strands. <S> You will use those fine strands to be your conductors. <S> Solder one strand of thin bare wire to each pad. <S> Take your time and get nice solder joints. <S> The bottom pads should alternate between front and back, as should the wires on the side pads. <S> Plan where each of the wires will eventually wind up and choose the direction (front or back) accordingly. <S> Hold the circuit board vertically in place over the DIP socket and plan where each of the wires will go. <S> The wires closes to the bottom of the circuit board go to pins closest to the center of the socket, moving towards the outside of the socket as you get higher on the board. <S> Alternate the wires to each side of the socket so as to minimize any chance of wires touching each other. <S> Now lay a small bead / blob of hot-melt glue in the center section of the DIP socket and press the bottom of the PC board into the glue. <S> Move the wires coming off the bottom of the board so they line up with the tops of the pins in the IC socket. <S> After the glue has cooled and set, simply solder the wires to the socket pins. <S> Make a map of which pins on the circuit board go to which pins on the IC socket. <S> The advantage of this technique is that it is non-destructive. <S> You can snap the circuit board away from the IC socket when you receive your carrier boards. <S> The hot-melt adhesive is easily removed, as are the thin bare wires. <S> You can then mount the board onto the carrier.
| If you have no other options in terms of the board you are using to solder the components on, i recommend soldering paste. As others have mentioned, using a carrier board is the easiest and most reliable option.
|
Cooling of electronic components I have built an electronic circuit that connects to a 5V power supply. The power supply does get a bit warm when the circuit is operational. The supply is cooled using air convection method. Will buying a 12V dc fan help to cool down the supply? If not what can I use that will not break my wallet. Is there a criterion to follow when selecting an electromechanical cooling system i.e. cooling fan.? The power supply is a Meanwell RS-15. The circuit pulls about 1.8A. <Q> Too hot to handle can be as little as 45°C, which would barely register for most electronics. <S> Burnt skin starts at around 60°C which is at the low-end temperature of many CPUs in modern computers. <S> Design operational temperatures for power electronics can exceed 100°C. <S> Higher temperatures do affect lifetimes, but the effect is exponential , so not too much of an issue at the lower end of the range. <S> "A bit warm" is fine for most designs. <A> https://www.meanwell-web.com/en-gb/ac-dc-single-output-enclosed-power-supply-output-rs--15--5 Meanwell claim that this power-supply can produce up to 3A with only free air cooling. <S> The transformers used in switch-mode powersupplies have an efficiency vs temperature curve that peaks at around 80C, so it's probably supposed to be "a bit warm" <S> If you still think a fan is needed typical 12V PC cooling fans will operate at 5V (but run slower, and much quieter) <S> In the data sheet <S> https://www.meanwell-web.com/content/files/pdfs/productPdfs/MW/RS-15/RS-15-spec.pdf Meanwell say at 70 degrees ambient it can provide 60% of its rated power (graph, bottom of page 2). <S> That would be 1.8A which is all you need. <A> The temperature that your product gets to in its enclosure in free air depends on the airflow around the supply, the circuit and the case. <S> In a sealed container the temperatures could get quite high. <S> You don't describe how you product is configured, but assuming the load consuming the 5V is in the same case as the power supply: You thermal load in watts is about (5V * 2A) <S> * 1.3 <S> = 13W <S> This is actually very moderate load for this RS-15-5 power supply, but your total thermal load (assuming the supply and product are in the same enclosure) is not insignificant at 13W. <S> If your enclosure is relatively sealed you would be well advised to add a fan. <S> NOTE: <S> I would NOT suggest the use of a 12V fan on a 5V supply. <S> There is some possibility that the fan may not start when using 5V, whereas a 5V fan is guaranteed to start at 5V. Most small computer fans today are BLDC motors run a by chip. <S> These are typically NOT suitable for voltage control, and may simply not start at lower voltages. <S> Most are equipped to use either PWM, or a speed signal PWM at their rated voltage. <S> Many only guarantee performance at their rated voltage +/-10-15%. <S> 5V fans are typically small; in the 40-60mm square range and readily available. <S> Here is a pointer to an example low noise, 60mm square with ball bearings ( FAD1-06025 ) and a power rating of only 1W (200mA). <S> this would make your total thermal load about 14W. There are endless varieties of good fans available with various form factors and from various sources. <S> I'd suggest however that you get a ball bearing fan at the slowest speed possible to reduce the noise.
| "A bit warm" is not a very useful design criteria (unless you are in biological markets, where this can be an issue). I would suggest that a small 5V fan may be appropriate, powered from the 5V supplying your circuit.
|
CAN bus termination in a fully modular design I am trying to make a fully modular festo cube design that is controlled with a CAN bus, but I am trying to figure out where to terminate the bus.I am even not really sure if what I am trying to do is even possible with a CAN bus.The wiring looks something like this: [ In every box is a node that is connected to the CAN bus, but there are several open ends in every box were I could connect a termination resistor to. The box on the bottom left is the master box which has a controller connected to it which you could see as the start of the bus. My quess is that I need to connect a termination resistor to every lose end of the bus. Hope you can help :) <Q> The CAN bus is designed for a single line with nodes connected at intervals along it. <S> In this standard layout it is quite obvious that you need to terminate each end of the line to prevent transmission line effects (reflections) at those ends. <S> In your layout this is not possible as there is no single line. <S> Indeed, your layout will have some serious problems with reflections, not just at each unterminated end stub, but at junctions where multiple lines join. <S> However, all is not lost. <S> If the longest dimension of your bus is roughly 1/10 of the wavelength at the bitrate, then transmission line effects can be disregarded (basically the wave travelling time is much less than the bit time, so by the time you sample the signal value any transmission line effects will have died out). <S> However, this means that the maximum bit rate you can run is limited by the length of your network, so you'll have to check this carefully. <S> You might want to reduce the driver slew rate as well to clean up the edges of the signal, particularly if you are not running crystals in each node. <S> Edit: BTW Low-Speed and single wire <S> CAN are designed for exactly this sort of flexible bus structure, so there is nothing inherently wrong with using CAN like this. <S> The answer I provided above outlines how you can use standard CAN in this configuration as low speed <S> CAN transceivers are harder to get now. <A> It's baffling why you would consider such an arrangement over the tried and tested methods such as this: - One cable terminated once at each end minimizes reflections and allows a much higher data rates with minimal errors. <S> It also uses less cable. <A> For a fully modular design, you may consider using termination that can be switched open or closed at each node. <S> The next hurdle would be how each node decides if it needs to close/activate termination. <S> This could be done manually via a toggle switch or jumper. <S> For a more automatic solution, they could all be open by default, and decide with some sort of communication protocol that isn't the CAN bus itself.
| In this situation you can just bulk terminate the bus at the master node to ensure that the bus has sufficient load for the CAN drivers.
|
Can strong magnets cause permanent damage to RFID tags? I have an RFID-enabled ID card for work that is used for building access. While I can use a lanyard, cord-reel, etc. to carry the card and still have convenient access to it, I'd like to simply put it between my phone and its case. (I have legitimate reasons for this that are tangential to the question.) My concern is that I already have a small metal plate in there that I use to hold the phone to a magnetic mount when I'm in the car. The plate itself is of no concern to me - it's just a small piece of ferrous metal and does not impact the functioning of the RFID chip. My question is : will the regular very close proximity (1-2mm) of the strong magnets in the car mount cause any issues with the functioning of the RFID tag in the card? It would be exposed for about an hour 2x per day, plus other random trips. My guess is that it would not be impacted any more than the solid state circuitry in the phone itself, but I'd like to get confirmation. Additional notes: You now know as much about the particular RFID chip in my card as I do. You also know as much about the magnets in the mount as I do. Unfortunately, I can't provide any more specifics than this because I don't know them. UPDATE: For those who may come across this question in the future - I thought I'd come back to mention that I've been using the RFID card in the phone case since I asked this question and I've never had any issues. All the "it will have no effect" answers were, in fact, correct. At least for my particular card. <Q> The magnetic field from a magnet is static . <S> In order to induce a current and/or voltage you need a changing electro/magnetic field. <S> Typically the field that RFID uses for operation has a frequency of at least 120 kHz but typically 13.56 MHz. <S> That's much faster than you can practically generate by physically moving the RFID card around a magnet. <S> So near a permanent magnet you would be inducing a field but with a much lower frequency. <S> The RFID tags are much less sensitive for such low frequencies as the antenna will not pick up that low frequency very well. <S> So not much energy can reach the RFID chip and it should be nearly impossible to damage it that way. <A> Its all about speed of the movement of the RFID-Tag relative to the magnetic field. <S> The antenna (a coil) of the RFID chip and the magnet form a generator. <S> If you quickly turn or move magnet and antenna relative to each other <S> a voltage is induced in the antenna, which is connected to the RFID chips power supply input and antenna input. <S> Usually, a HF or UHF signal is received by this antenna and supplies any passive RFID tags with energy while transmitting the request data. <S> The RFID tag has not much control over the received signal strength (nearby or far away RFID reader) and therefore it must cope with a broader range. <S> So a higher signal strength induces a higher voltage and the RFID chip must have some over-voltage protection. <S> But, this protection most likely does not work for infinitely high voltages/energy amounts... <S> So it's all about how fast you move the RFID tag relative to the magnet, how strong the magnet is, how far away/near by the magnet if relative to the RFID-antenna and how well the over-voltage protection of the RFID-chip is. <S> Without any real data one can only guess. <S> But, I assume usual movement speeds and magnets should not induce energy that is too high for an RFID chip. <S> Also the over-voltage protection usually works by converting electrical energy into thermal energy. <S> Hence, it becomes hot... and hotter... and finally too hot. <S> If the induced energy (due to continuous motion) would not be interrupted for longer time, to let it cool down again. <S> But, the usual motion would be putting the phone / RFID-chip in place or take it away once in a while. <S> So my guess is: There should be no issues. <S> But, it's not safe to say without any real data at hand. <A> No, you're right, RFID does not work by "magnetism" in that sense. <S> The antenna picks up energy from the electromagnetic field. <S> There's nothing in a standard magnet that can emit such fields. <S> Depending on the specifics of your tag it may not work while mounted on the magnet, but I don't think that's what you're asking. <S> There is a related question about that with good answers.
| A static magnetic field caused by a normal magnet should not cause any harm to a RFID-tag. So as far as I can see there can be no issue.
|
Supercapacitor backup circuit - why seperate charge and discharge path? I am learning about super capacitors in a backup power application. I see a lot of examples using diode networks or FETs to separate the charge and discharge paths. Why is there so much effort put into separating the charge and discharge paths? Why would the example circuit I drew work? Someone mentioned the voltage drop over the limiting resistor may cause an issue? The 3V3 regulator would be powering an MCU. <Q> For your circuit as drawn, anything connected to the 5V supply would be powered by the cap. <S> This would include any regulators that may be supplying the 5V. <S> If the 5V only ever comes from a cable the situation is better, but you'd still have that 5V on some (presumably) exposed terminals. <S> To add to your pain, the resistor would just waste energy as the cap was discharging. <S> However -- for some simple battery powered thing where you just want to keep the micro going when the batteries are being replaced (or similar purpose), what you have there may actually work. <A> Power losses increase with ESR and serial Rs, so the concept is to reduce this with a current source. <S> but if you have a voltage source then a serial Rs limits the charge current to the source capabilities. <S> Energy stored = <S> 1/2CV^2 <A> one of the main reasons you would want to separate charge and discharge paths in a circuit with capacitors is safety, say your voltage supplies fails and turns into a short circuit, whatever energy your capacitor has will go that way. <S> In pulse discharge circuits there is the issue of you <S> don't want your load to draw current from your source because simply this current will be too large, even if it is for a small time it and your voltage source is properly protected <S> is not something you want to risk. <S> your circuit has the issue of the voltage drop, all the energy that from that voltage loss is also wasted. <S> Additionally, how quickly or how much of a ripple your output voltage can have, that resistor does not only limit current <S> it also slows down charging and discharging and the overall shape of the output. <S> Think of the RC constant and how it affects what I mentioned before. <A> Supercaps can have a high inrush current when power comes on and the cap is at zero volts just from draining down to zero volts. <S> The circuit needing the backup current consumes much less than the inrush current, so the series resistor has little Vdrop in terms of what the load draws, yet allows a drained capacitor to charge up full in less than 30 minutes. <S> Typical application for a few hours to a few days of backup are LCD event counters and other low power CMOS circuits that often idle at a few microamps. <S> If R is 1K ohm then the inrush/charge is at safe levels, and when power fails the resistor causes little Vdrop if the load consumes only 1 or 2 microamps per volt. <S> Many CMOS devices will retain their state down to 1.9 volts, even if 5 volts is normal operating voltage.
| If it was a battery system it would drain your batteries when you want to the capacitor to take care of it.
|
Why do some oscilloscopes have several input channels? Just was thinking to buy an oscilloscope, thus went to a shop and started to rummage all of the shops to buy an osci. which could be suitable for me. Read specifications of them. Some features was clear to me and I know them. e.g. sample rate or frequence and I know why the price of oscilloscopes. increases with increasing these parameters and I know what range of these parameters could be suitable for me but there is a question that I couldn't find any answer for it on the net. Why do some oscilloscopes have several input channels? I know with this feature you can capture several signals at the same time but the main question is that when do you need such a feature? I'm thinking that when do I need to capture several signals at the same time but couldn't find the answer. Maybe you could clear me out. I guess we cannot find any Oscope with less than two channel. Why? Please post your answer with your experience about what made you to purchase an oscope with more channels. As you know, Oscops with more channels are expensive and so couldn't find anybody ovehere that has oscope with more channels. <Q> You're right - the main reason to have multiple input channels is to be able to measure several signals simultaneously. <S> EDIT: <S> To give another example, you can use one channel to trigger on an activation signal (for example, a pin select) and another channel on an input or output pin. <S> Having both on screen at once allows you to measure the delay from one event (the trigger signal) and another (the input/output signal). <A> A use case is where I want to simultaneously observe (for example to observe time delay) <S> the input and output of a differential amplifier. <S> I could use differential probes <S> but I would need two and these can be expensive. <S> If I had a 4 channel scope I can do this: I connect scope channels A and B to the in+ and in- inputs of the amplifier. <S> Then on the scope display A-B which is the differential input signal. <S> At the output I connect channels C and D to the out+ and out- outputs of the amplifier. <S> Then on the scope display C-D which is the differential output signal. <A> I think another important aspect is that oscilloscopes have a common ground in every channel, so if you have a 1 channel oscilloscope you can only measure things that are connected to ground on one end. <S> Making you unable to measure signals from anything that is separated from the reference, with two channels you can use two probes, measure the voltage at two points and while it is a bit of a workaround you can measure the voltage of any component in a circuit. <A> Also, some scopes have digital inputs. <S> Just as having two or more analog inputs lets you see the relationship between two analog signals, scopes that have digital inputs allow you to see the relationship between digital and analog signals in your design. <S> For example, you might want to know how long it takes for a digital-to-analog converter (or analog to digital converter) to settle, once you've set up the input. <S> In addition, you can track what's happening as code executes in the digital domain, and see how it's affecting the analog domain (the output). <S> All very useful. <S> Hope this helps!
| A two-channel oscilloscope lets you compare an input signal to an output signal, and if you had a means to control the behaviour DUT (device under test) in between the two probes then being able to see the input and output simultaneously can help you make better decisions on its performance.
|
connecting panel mount switches to PCB I have a set of panel mount switches I'd like to connect - via cables - to a pcb. The pcb won't have any components on it, it will just be a wiring loom. the concept is that having the complexity on the board will improve quality. There's a mixture of DPDT, 3PDT and 4PDT switches, with a total of 60 connections required. The switches I've sourced have solderlugs, like 50212LX from Switchcraft Inc. Is there a way to achieve this other than hand soldering each connection? many thanks in advance. <Q> Unfortunately, if you have already chosen switches with solder lugs, you have little alternative but to hand-solder them to wires. <S> One alternatives is switches with PCB pins, which go directly into a board. <S> This can make alignment with the panel difficult, so the technique there is to fit the switches, then push the PCB on, then solder the PCB. <S> It does require that the panel be designed with a board in mind, so all switches on the same plane, not too far from each other, nothing big between them to foul the PCB, though board cut-outs can solve this. <S> Another alternative is switches with blade connectors. <S> You can buy ready-crimped wires with connectors that push-fit onto the switches. <S> Another alternative is switches with 0.1" pin grids for terminals. <S> These would connect with IDC connectors and flat cable, though there are quite low voltage and current limits for this type of connector. <S> As Hearth pints out, you can also get switches with screw terminals. <A> I'd suggest wires like these . <S> Cut them in half, solder the cut side to the switch, plug the other side into "crimp housings" that then plug onto male pins on your board. <S> Or use male pins to into female headers on your board. <S> I get mine from www.Pololu.com, you can easily (compared to Digikey) <S> find wire with male-male terminations, or female-female, or male-female, in different lengths, or just connectors to crimp on to make your own, and a wide selection of crimp housings. <S> 1x1, 1x2, ... <S> 1x longer, 2x2, 2x3, ... 2x longer. <A> One solution I think of at the moment, but it's a little bit expensive and suitable only for large production <S> : Create one PCB for each switch or for each group of switches which can be positioned reasonably close to each other to fit on a signle PCB, then add a ribbon cable connector to the PCB with the required number of wires. <S> Can be simple 2.54 or 1.27 pin headers. <S> This would require designing several PCBs and have the switches and the headers soldered preferably at a factory. <S> It may not be efficient if every switch is separated from the others. <S> You can also design an "universal" PCB, one board, which could fit several switches or switch combinations. <S> Kimiting the number of different PCBs. <S> Soldering everything manualy would take a lot of time. <S> Yet even manualy that would be faster and easier to solder each switch pin on a PCB than soldering each wire to each switch pin. <S> Other than that I see only this solution <S> : PCB terminals to clip the wires on, like this one. <S> https://www.aliexpress.com/item/100pcs-6-3-Inserts-Plug-Spring-Terminal-250-PCB-Solder-lug-DJ611-6-3-thickness-0/32822493330.html?spm=a2g0s.13010208.99999999.266.30453c00wu0Ad6 <A> Here are 3 switches which could be directly compatible with 2.54 mm ribbon cable or cable connectors. <S> If they are not too small for your application. <S> 1/ <S> 2/ <S> 3/
| You could mount blade connectors to the PCB as well, and push on wires with a connector each end.
|
Can I put vias on routing I2C lines traces on FR4 PCB? I am using a MSP430FR2633 micro controller and ADS122C04IPWR DAC over I2C at 100kHz. Can I route the I2C lines through vias in a 2 Layer PCB? <Q> At I2C speeds, vias will cause you absolutely no problems at all. <S> At least, no problems in terms of track resistance, capacitance or inductance. <S> However, if you have a 2 layer board, then using both layers is best done systematically, otherwise you can lead yourself into problems. <S> What many people do is to dedicate one layer of a board to ground. <S> This generally works well, UNTIL they start chopping the ground up with 'just one track' run on the other layer. <S> When this track is joined by 'just another', and another, the ground ends up looking like a lace curtain. <S> It doesn't do its job, and it's difficult to determine where and how to stitch it back together, especially if you're inexperienced. <S> Even worse, some will route all the tracks, then do a 'copper pour', in the hope that this makes a good ground plane. <S> If you're using 2 layers for signals, then it's far better to start with a plan. <S> Use a 'Manhattan' tracking arrangement, east-west on one layer, north-south on the other. <S> Start with a 'gridded ground', put parallel tracks every 10mm or so, and via them at every intersection. <S> This works almost as well as a ground plane, and at I2C speeds is absolutely as good. <S> Now you have a systematic way to run a track from anywhere, to anywhere, and can hop to the other side of the board exactly where you need to, without disturbing the existing ground continuity. <S> An alternative is to use a ground plane, but to avoid chopping it up by staying on your signal layer for all signals. <S> Cross tracks by passing tracks under components. <S> You can buy 'zero ohm' resistors for this purpose, though a 1 or even 10 ohm resistor will be as good as a wire at I2C resistance levels. <A> In general - yes. <S> 100 kHz signal is very forgiving. <S> Make sure to route both SDA and SCL in a similar fashion, close together. <S> Also keep in mind the I2C total capacitance limit of 400 pF <S> (you could run into that issue if the traces were really long). <A> We are using vias on I²C which are running up to 800 kHz without any issues. <S> The worst I have seen was a bad via, which created a series resistance in the I²C line. <S> That affected the slew rate of the board so bad that the I²C communication failed. <S> But that was on a prototype board and has never happened on a production board so far. <A> 100kHz is pretty easy to move around. <S> Our equipment has I2C EEPROMs mounted in a customer-replaceable part of the system, so that changes to calibration when that part is replaced will automatically be read by the controller. <S> The cables for this can be metres long, and so far that's working OK. <S> Admittedly we do have low-capacitance cables, but still, a PCB via is no big deal. <S> Note <S> though that whilst 100kHz is the original I2C standard frequency, it's not the end of the story. <S> I2C "Fast Mode" allows up to 400kHz, "Fast Mode Plus" allows up to 1MHz, and "High-Speed Mode" allows up to 3.4MHz. <S> Looking at the datasheet, your DAC does support "Fast Mode Plus" up to 1MHz. <S> So long as you're staying at 100kHz then you can basically do whatever you want because it's not fast enough to matter, but if you might want to go faster in future then you should investigate design rules for faster digital signals. <S> That said, the DAC itself can only manage 2K samples per second though. <S> For a 24-bit DAC plus the associated I2C overhead, data transfers will only take about 70-80% of the bandwidth on a 100kHz link. <S> If you have multiple DACs on the same I2C link then you might want to use a faster link to service them all, but if you only have a single DAC then you don't have a reason to go faster than 100kHz.
| Shouldn't be a problem.
|
Help identify the IC manufacturer: is it really possible that these ICs from the 60's just made a comeback (К155ИД1)? So we've been manufacturing Nixie Clock kits for the last few years, and for the multiplexers we used 74141 ICs, or their Soviet clones (К155ИД1). The latter were easier to find on ebay and were used more commonly. These chips have a date code like "8910" which I assume to mean year 1989, week 10. Since this is all NOS (New Old Stock), I've made a 74141 tester to verify incoming batches. I received the latest batch of chips yesterday and to my surprise about 1/3 of them failed the tests. Then I noticed that these new chips were visibly different: the plastic package is more edgy, there's some sort of manufacturer logo, and their date code is "1814"! Compared to an old one at the bottom right: I later verified that the new chips work on actual nixie clocks, but my tester rejects them for some minor reason (probably the new ones are a bit more sensitive to supply voltage, they also seem to be consuming a tad more current). Questions: Which manufacturer is this? Is it really possible that these ICs from the sixties have just made a comeback? The practical upside of this is that we'd be curious to contact the new manufacturer directly, to cut the ebay middlemen. <Q> The chips that you have are made in Belarus by a company named Integral . <S> К155ИД1 1814 <S> :- This is a recently manufactured chip. <S> (i.e. 14th week of 2018). <S> К155ИД1 8910 :- This chip is very old (i.e 10th week of 1989). <S> Link to the site of manufacturer. <A> You have К155ИД1 1814 chips made in Belarus see this . <S> Company name is Integral. <S> Unfortunately, these old chips are hard to find. <S> The Soviet Union made excellent chips; now you need to carefully verify the modern versions. <A> Consider the possibility that the chips inside these packages are NOS, from 1989 or whenever, but had been stored as bare dies, and have been bonded and packaged in 2018. <S> Which would make a lot of sense if somebody has a stash of bare dies - you could put them in ROHS-compliant packaging. <S> Coincidentally, the pin finish on these does look rather matte, typical of lead-free components... <S> Alternatively, the parts might be old stock that someone retested, refurbished, remarked and remarketed in 2018.... <S> there might be the more sinister possibility that they have been just remarked, in order to avoid ROHS-related suspicion :)
| К155ИД1 8910 is a very old chip made in 1989. Yes, it might be possible that the manufacturer is using a slightly modified architecture based on the previous one for the manufacturing of these new chips
|
Converting ADC output to real values On a custom board I have an ADC which measures -5V to +5V. And outputs in Hex via SPI like this: 0x8000 to 0x0000 = most negative to 00x0000 to 0x7FFF = 0 to most positive I just started studiyng ADCs so I am not sure what is going on. I beleive I have negative full scale of -5V and positive full scale of +5V. Questions: What this numbering system is based on? Is it two's complement? Why there is no 0xFFFF value? How can I convert these numbers to a real voltage? <Q> It looks like you have a 16-bit, 2's-complement ADC with a range of \$-5V \le V_{IN} < 5V\$ . <S> If that's the case, then the digital values can be interpreted as signed integers from -32768 to +32767. <S> The measured voltage is \$ V_{IN} <S> = \frac{DigitalValue}{32768} \times 5\$ <A> If you count most negative to 0 using integers, you count ..., -4, -3, -2, -1, 0If <S> you count most negative to 0 using hexadecimals, you count (in your case) 0x8000, 0x8001 ..., 0xFFFE, 0xFFFF, 0 <S> So, there is the 0xFFFF. <S> Now, your scale is 0X8000 to 0x7FFF, so, in decimal -32768 to 32767, the full scale is 65636 (-32768 <S> +-32767 + 1). <S> The scale represents -5V to +5V, full scale 10V.The conversion is: Vmeasured = <S> SPI_OUPUT_in_HEX / <S> 65536 <S> * 10V <S> P.S. <S> I guess using 65536 or 65535 for dividing is a matter of taste; it hardly influences the accuracy. <S> Using 65535 may be preferable/nessecary in old tiny microprocessors. <A> I'll just add to @Eliot's answer that in reality if you have floating point math available (and choose to use it) <S> you should convert DigitalValue to floating point before dividing. <S> If you only have integer math available you can multiply by five (you need more than 16 bits, obviously) and then throw away the least-significant 2 bits (or round) and place the radix point right of the 4th bit. <S> For example 0x7FFF FFFF - <S> > <S> 0x27FF <S> FFFF <S> FB -> 0x4.FFF <S> FFFF ( <S> one LSB less than 5.0000). <A> Recommended thing would be to grab datasheet for your ADC chip and see in chapter named "Data Format". <S> In this chapter it will be explained how the numeric values (data) from the ADC is represented. <S> It seems that your digital representation is something named 2's complement. <S> Here is the wiki link. <S> So, max value 0x7FFF is your +Vref (positive reference voltage) and 0x8000 <S> is -Vref (negative reference voltage). <S> 0xFFFF <S> in 2's complementary equals to -1. <S> Example: <S> Let's say your positive reference voltage is 2.5V, and negative reference voltage is -2.5V, then <S> when you see obtained value of ADC equal 0x7FFF <S> it means you have 2.5V on the ADC input. <S> To calculate other positive values you take Vref and divide it by numbers of bits value (0x7FFF equals to dec 32767), so if your ADC gives you 0x0001 <S> is equivalent of 1 <S> * (2.5[V]/32767) = 0.0000763[V] <S> BTW, this smallest step your ADC is capable to measure is called ADC resolution <S> Let's have about half range of positive ADC <S> , so now your ADC gives you a number of 0x4000. <S> This is 16384 dec, so calculation here is: true_voltage_value = 16384 <S> * (2.5[V]/32767) = 1.250099[V] <S> For negative numbers: if your ADC value is bigger or equal to 0x7FFF then your action may be: negative_measured_value = measured_value - (0xFFFF + 1) <S> So, when ADC provides you value of 0xFFFF it goes: negative_measured_value = 0xFFFF <S> - (0xFFFF +1) = <S> -1 <S> true_volt_value = <S> -1 <S> * (2.5[V]/32767) = -0.000076V <S> Huisman answers it as well. <S> Note: <S> The values calculated here are theoretical, and how precise your measurement is depend from stability and accuracy of reference voltage(s).
| From values you have provided I assume it is 16-bit ADC.
|
If the order of any component is altered in a series circuit, will the total voltage be affected? I recently posed this question to someone. I always thought that the answer was simply no because of Kirchhoff's voltage law... However, their response was... "no, the voltage is not altered due to the total resistance staying the same" ... does this have any merit? I guess I never thought about it that way, but I don't see why they'd be wrong in saying that. Would just like some clarification on this. For example, if I had two sources in a series circuit that both produced 2V as well as two 5 Ohm resistors (also in series). No matter how I positioned the sources or resistors, there would always be a total of 4V produced and 4V dropped. <Q> To a first order no, but it somewhat depends on what do you mean by component <S> and how are these connected in the series circuit. <S> Assuming all the components are two terminals and no parasitics to other parts of the circuit, <S> then yes, KVL and KCL says that nothing would change. <S> You need to add KCL as component voltages depend on their current (e.g., ohm's law). <S> If some components are non-linear (e.g, LEDs), but still two terminals and placed in the same order with respect to the circulating current, then yes, KVL and KCL says that nothing would change. <S> But, if the components are more than two terminals, even if there is no current through that extra terminal (e.g., FETs) then no, as the voltage with respect to that terminal can change the component's I(V) curve. <S> And, if there is current through that terminal or any other branch connected to the series circuit, then of course not. <A> With a 10 V source, a 2 R, 3 R, and 5 R resistor will each experience a voltage drop of 2 V, 3 V, and 5V respectively. <A> An alternative perspective comes from conservation of energy, which is more fundamental than KVL and KCL. <S> Conservation of energy implies that that power generated, must be equal to the power dissipated in a closed system. <S> Since your circuit has the elements in series, the current is the same through each device. <S> So by using passive sign convention with a common current, conservation of energy requires the total supply voltage to equal the total voltage drop summed across all the resisters, independent of ordering. <S> The fundamental idea for a closed system like your idealized circuit model, is that every watt of power that you put into a system (i.e., power generated), must come out as heat (i.e., power dissipated). <S> This is something worth remembering for anyone who designs electrical or electromechanical systems. <S> Notes: <S> Passive sign convention on Wiki: https://en.wikipedia.org/wiki/Passive_sign_convention <S> Yes, this assumes that generated power is magically there, but the concept still applies in real world systems. <S> Total supply voltage is the algebraic sum of the supply voltage with respect to their polarity (+ - symbols).
| In a circuit with constant voltage sources and only resistive components, the order they are placed in to make the circuit will not affect the voltage drop across each of the elements will not be affected.
|
Will this 555 circuit prevent a car from disconnecting power to the radio when starting? To start with I'm a little rusty with electronics, I haven't done much in 20 years and it was only a hobby back then. In my car I have a head unit with a touch screen which relies on software to run it. My problem is that it starts to load the software when you turn the key to accessories but when you go to start the car it cuts power to the head unit which hasn't finished loading the software. After the car has started (key returned to the on position) the power is restored to the head unit and it tries to load the software again. This cut in power, I believe, is causing the head unit to malfunction, it messes with the software. So I need to supply power to the unit when the key is turned to ACC or On position but held on while the engine starts.After the engine has started the key will return to the On position and power the unit as normal.When the car is turned off then the unit needs to turn off. I was thinking of powering the unit through a relay wired direct to the battery and then using the signal from the ignition, in parallel with a timer, to trigger the relay. If I use the loss of power when starting the engine to trigger the timer then the unit has already lost power and it would start loading again, so that would be of no use. I tried to use the initial connection to power (when the key is first turned to ACC) to trigger the timer. If I am correct then with my circuit the timer is put into an endless 10 sec loop while the key is turned on and when the key is turned off only the timer powers the unit till the end of the time period. If the key is returned to the On or Acc before the end of the time period then the unit will stay on but if the ignition is turned off for longer than the time period then the timer stops and the unit turns off. (This is also a diesel so needs the ignition to be turned on for a period of time before starting. I'm not sure if 10 secs is enough time but I don't think you can stretch it out further). Can anybody verify if this circuit will work? Is there a better way to do this? If this is the best choice, is there anything wrong and what should I do to fix it? (I wasn't sure if the 10k and 1k resistors connected to the base of the transistor were needed). <Q> I don't think this is going to work. <S> When your engine cranks it's typical to see a large drop in voltage supplied by the battery. <S> Even if you keep the radio from turning off deliberately you're still going to see problems with brownouts. <S> Most circuits do better if they're turned fully off than if they're dealing with this sort of brown-out <S> and I think that's what the wiring in your car is trying to protect you from. <S> Does the head unit have a manual? <A> This could just be a small rechargeable lipo battery. <S> The backup battery could be very small, as it only needs to provide power for a couple of seconds while the engine starts. <A> Here's a simple relay solution that might be worth investigation. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> The two relays hold off radio power until the starter cycle has finished. <S> How it works: <S> With ignition off both relays are de-energised. <S> Radio is off. <S> When ignition is turned on radio power is still off as RLY1 contact is open. <S> When starter is energised both relays pick up and RLY1 latches. <S> RLY2 is open and radio power is still off. <S> When starter is released RLY1 remains energised over its own contact while D1 prevents back-feed into the starter circuit. <S> Meanwhile RLY2 drops out and power is applied to the radio. <S> Turning off the ignition re-initialises the complete circuit. <S> Note that this relies on RLY1 picking before the battery voltage drops due to starter solenoid pulling in causing the battery voltage to droop and RLY1 must remain energised at the low voltage. <S> This should be possible so if it doesn't work with one relay then try another of a different type. <S> If you have an adjustable power supply you could measure the pick and release voltage for each relay before testing in the car. <A> It looks like this circuit would be pulling around 800mA through the coil and BJT which would probably blow both of them. <S> Put a current limiting resistor in there. <S> I am skeptical that the car cuts power to the head unit unless you've confirmed this somehow. <S> A new battery with a better C rating (lower output impedance) might fix your problem if it was originally designed to work under starting loads. <S> It could just be drooping more than expected. <S> Was this always a problem? <S> The simplest solution I can think of, assuming the radio is browning out and not switched off intentionally is to just place a smaller 12V lead acid in line with the head unit power with a diode preventing current from going backwards to the starter motor. <S> It will still stay charged and everything from the alternator so it only has to have enough capacity to run the radio while starting. <S> Spec it for at least 3-4 times this though. <S> Its maybe a little ghetto but should work. <S> The timer solution should would work too, although I haven't checked your circuit <A> Thank you all for your assistance. <S> Originally the unit appeared to be working correctly with the exception of erratic behaviour which I thought was software related. <S> It seems that after more time has passed some new symptoms have shown up and what the unit was doing is now happening a lot less frequently. <S> I am thinking that I was wrong as to the cause of my problem. <S> This unit has four touch switches down the side similar to to what many early smart phones had at the bottom. <S> Three of these have now stopped working. <S> The interesting thing is that the three that have stopped working are the Return to previous screen, Mute and Volume Up (the Volume Down still works fine). <S> The problems that I thought were software related were the unit going back to the home screen for no reason, the unit muting and un-muting itself, the volume would rise on it's own but never drop and the touch screen would freeze or sometimes think you were on the home screen when you were on a different screen, or it thought you were holding your finger on the screen when you only tapped it. <S> It seems too much of a coincidence that the actions of the unit just happen to be the same as the buttons that have stopped working for them not to be related. <S> I'm not sure how the screen freezing and the other issues relate to the buttons <S> but maybe the unit was overwhelmed by the signals from the faulty buttons. <S> I think the solution is to do one of three things disconnect the faulty buttons and hope the unit doesn't pickup noise due to the missing buttons (the unit has other ways of doing the same things) <S> replace the faulty buttons replace the unit (after my frustration has been satisfied :) ). <S> In any case thank you all and sorry for bothering you with something that probably wouldn't have helped.
| You'd just need some kind of circuit to detect a voltage drop on the main source, and switch temporarily to the backup battery. I think it would be more likely just being browned out when the car is starting.. Would it be possible to add some kind of isolated power source for the head unit, so it can receive uninterrupted power, even when there's a large voltage drop across the car battery.
|
What determines the frequency of network cable You have Cat6, 23 AWG, rated for 250Mhz frequency. You then have Cat6a, 23 AWG, rated for 500Mhz. Both are unshielded. How is Cat6a able to transmit data at a higher frequency then its counter part? They are both the same gauge wire, same copper, but one is rated faster. How? <Q> The “ratings” you are talking about are mostly marketing claims by the manufacturer. <S> Both cables have worse attenuation and crosstalk as frequency goes up. <S> One cable was tested to 250 MHz, one was tested to 500Mhz. <S> Perhaps their performance is similar. <S> The cable manufacturers can improve the cables, but it costs them money. <S> For example, although both cables are “gauge 23”, the lower attenuation one may actually use slightly bigger wire.(there is some slop in the wire gauge spec). <S> Perhaps the surface of the lower attenuation wire is smoother than the higher attenuation one. <S> Losses in the dielectric can be less in the better cable, due to better plastic. <S> Also, by winding the twisted pairs more carefully, crosstalk can be reduced, but then you need more expensive manufacturing. <S> Anyway, short answer is <S> cables can be and are better or worse, but there is also a lot of marketing hype in the twisted pair cabling business. <A> Inside each cable there are some pairs, each pair is twisted to make it resist noise induction. <S> Then the pairs are tested against each other measuring the arrival db and arrival time of a signal on each pair, if the db are good enough on all pairs, and there is no delay between one pair and the other one... <S> the cable is good for that signal frequency. <S> Rinse and repeat for higher frequency. <S> When the signal strenght (db) and/or the time of arrival on one pair is too late relatively to the other pairs... <S> the cable fail for that tested frequency. <S> According to the higher working frequency a category is given for sales. <S> Depending on the organization issuing the protocol, there might be other constraints (example: mandatory shielding of individual pairs, or shielding of the whole cable, or double shielding; or, again, a minimum number of turns per inch). <S> Civil authorities have some constraints, military ones have different constraints. <S> But the important things are signal and timing, because those are tested after deploy in order to get certified systems. <A> There are multiple factors that affect the behavior of transmission lines, including their top transmission frequency. <S> From here, the next topic to visit is Telegrapher's Equations . <S> These equations break down how in every transmission line there is line resistance and inductance along the wires, and conductance and capacitance between the wire pair. <S> In lossless transmission lines, the transmission speed is infinite. <S> However, lossless transmission lines only exist in theory, as all conductors will present some resistance and all insulators will present some conductance. <S> The problem is these lossless lines will degrade the signals the further they travel. <S> In the case of sending a square pulse for example, you might remember a square pulse can be simulated out of multiple sin waves through the Fourier Series , each with a multiple frequency. <S> Each of these sin waves will travel at a different speed down the transmission line, as the lossy transmission line model is dependent on frequency. <S> Thus, as the wave travels down the line, the pulse will slowly become deformed, first around the corner edges, and then until it is no longer recognizable. <S> I tried finding an image of this concept but was unsuccessful, invite anyone else to post a link <S> Thus the transmission speed is limited by the parameters of the lossy transmission line. <S> The line works up until the individual pulses are no longer distinguishable as 1's or 0's at the other end.
| To start of simple, the gauge of the wires will determine the maximum current rating, and the insulation between the wires will determine the maximum voltage rating.
|
Switch supply for external circuit I have two circuits and want to on/off the supply of other circuit supply with one. I don't want to use a relay because of the dimension. What can I use for this implementation such as a mosfet, optocoupler or ssr? Other circuit needs around 3A. simulate this circuit – Schematic created using CircuitLab <Q> If you want a high-side switch, then a P-channel MOSFET is needed; you have drawn an N-channel. <S> To turn a P-channel on, you pull its gate low. <S> To turn it off the gate is taken high, to the supply voltage. <S> MCUs generally have 3.3V or 5V output. <S> Just the way Eduardo drew it up. <A> M1 is acting as what is called a high-side driver and it is also a source follower. <S> Because it is a source follower, to adequately turn-on that MOSFET <S> you need the gate voltage to exceed 12 volts by a few volts. <S> If you look at the data sheet for the IRF530 (the one used in your circuit) you will see that it has an on-resistance of 0.16 ohms typically at a gate-source voltage of 10 volts so, this means that gate voltage has to rise to 22 volts with respect to ground to adequately turn on M1. <S> Even then you will drop 0.48 volts when the load takes 3 amps. <S> The general idea is right (if you have the ability to drive the gate as high as 22 volts) but possibly the MOSFET choice is weak. <S> If you can't drive the gate to 22 volts then your only real option is a low-side driver or a P channel high-side driver. <S> However, a P channel high side driver requires an extra transistor: - You should also be aware that a MOSFET never switches off to infinite resistance and might keep a low power circuit ticking-over when you didn't expect it to. <S> Also be aware of inductive loads and use a reverse diode across the load terminals. <A> I use the circuit below that is activated by a digital output of my microcontroller, 3.3V output. <S> When SIGNAL = 0V <S> OUT = 0V. <S> When SIGNAL = 3.3V OUT = <S> 12V. <S> With a low impedance of MOSFET. <S> simulate this circuit – <S> Schematic created using CircuitLab
| A buffer, can be as simple as NPN transistor with collector connected to Gate and a pullup resistor to 12V, can be used.
|
Switch for DC application I have two solar panels in series totaling a maximum of 80 VDC and 500 Watts. I believe that would be a maximum of 6.25 Amps. Could a 20A 120/277V AC switch handle that since 6.25 Amps is a relatively low current? If not what about a 30A 120/277V AC be able to handle it? Update: The reason I want a DPDT switch is so I can easily switch my two panels setup between being series and parallel. Each panel is 250 watts with a 40 volt output. Series would give me 80 VDC at 6.25 Amps and parallel would give me 40 VDC at 12.5 Amps. Because of shade issues on the RV, there would be times that one setup would be advantages over the other. The switch would not be switched often. I am wondering if a knife switch would serve my purpose better. <Q> NO. <S> The AC ratings of a switch don't translate well to DC. <S> The critical rating of a switch is the current it can interrupt, not the current it can carry. <S> When you open a switch, any arc that forms across the contacts erodes them. <S> With AC, the current goes through zero many times per second, helping the arc to extinguish. <S> With DC, the current is persistent. <S> DC switches need to have oversized contacts compared to their equivalently rated AC counterparts. <S> They often have contacts that open faster, or further. <S> In your particular case, 6A 80V DC into a 30A 277v switch. <S> There may be enough de-rating there. <S> The switch will certainly work a few times. <S> Whether it will still be working after 1000 operations is another matter. <A> The short answer is maybe. <S> Alternating current reverses <S> it's direction of current flow many times each second, providing a built in point of zero-crossing. <S> This means that AC switch contacts don't have to be built up as heavily as DC contacts, because any arc drawn when they open will tend to put itself out when the polarity reverses. <S> Minimizing this arc is one major factor in designing the switch to last many, many operations, so switches designed for DC will allow for that. <S> 1970s era DC motor controllers used multiple silver plated contacts with a surface area the size of a large person's thumb print - these machines still run on older American submarines with those controllers. <S> If it's a hobby project and you've put fuses in all the right places, either one of the AC switches you mentioned will work - until maybe <S> they don't anymore. <S> And remember to account for the worst case - switches occasionally fail closed, so leave yourself the ability to unload the system some other way so you can pull the fuses out without drawing an arc across them. <A> 80VDC is a fairly low voltage, so arcing may not be a major issue <S> - I don't know. <S> I was curious about the rotary-action isolation switch used on my 12-panel array (about 360 VDC, 3 kWp), and discovered that a small number of solar installation fires had been caused by the use of AC-rated switches in a solar DC installation. <S> This surprised me, because the isolation switch is very rarely operated (typically, only to replace a failed inverter). <S> So there may be a safety issue even with the switch just sitting in the closed position? <A> You can get specific DC Disconnectors which are designed for use with solar installations, and rated for DC operation. <S> My invertor has a hand operated u-link form isolator, to provide a large arc path (several cm), I believe this is also linked to as series electronic isolator (the part is designated Electronic Solar Switch). <S> Even so, this isolator is speced for a limited number of disconnects and is designed as a consumable component. <S> An 800v/35A component is rated at 50 disconnects under short-circuit operation. <S> spec
| If you are designing a system you intend to last, see if the manufacturer provides DC ratings for the switch you intend to use - lots of switches have both AC and DC ratings. At lower current ratings it can be as simple as designing the switch to have more metal available so that any given amount of damage is negligible.
|
How to decrease nrf24l01 transceiver range? I've just got my Nrf24l01 but then I found out that the working signal range for the module is 100m and above. Our real intention for the module is to let the nrf24l01(Receiver) out of range for about 5m in order to activate the alarm. The purpose of our project is for anti-theft alarm which sounds the alarm if the module is out of range. We only need a short distance in order for the alarm to work since our research project is only for personal use. We have an idea which we want to clarify if it is possible, it works by having the (receiver) to reach a fixed distance value in the coding of the arduino and when the (receiver) goes beyond that value, it will activate the alarm or beep. We want to ask if our idea is possible and if yes, can you help us how to code the arduino?. I don't have any background in making a code for arduino but I know how to program an arduino using a ready-made code from the internet. <Q> No, not a fixed distance, but you can certainly reduced the expectable range, simply by reducing the output power of the transmitter, or by decreasing the sensitivity of the receiver. <S> These systems have no way of knowing distance – they can just "hear" or "hear not" the transmitter. <S> How "loud" a transmitter is at the receiver depends on how loud it transmitted, or what happened to the signal from the transmitter to the receiver. <S> So, you need to tone down what the receiver can sense. <S> You'll find out that it's extremely difficult to achieve a reliable range by adjusting power – effects of random orientation, obstacles (e.g. humans) between transmitter and receiver and <S> so on are much much larger than a few meters of difference in distance. <S> So, you might find out that it's conceptually impossible to define an "alert" radius without accepting a high number of false alarms or being too late to alarm. <A> The way I did this was to use pre-made boards that use a circuit board j-pole antenna, lowest power setting and the highest speed. <S> Once triggered they would ping-pong each other once every 2 seconds until the primary failed to receive 5 pings, at which point it would sound an alarm. <S> This averaged out to be about 10 to 15m, which was plenty accurate for my application. <S> You could also use something like a box made from hardware cloth to reduce received power. <S> Ops, just noticed OLD question, but the answer still applies. <A> Due to the relatively short wavelength compared to wall size and reflector/absorbers like metal and wall construction material, it is impossible to avoid Ricean Fading where signals vary by 10 dB or more if you attenuate the signal, this can reduce the range but the standard deviation on range then can be as much as the short range for alarm Loss of Signal or LOS. <S> Even if you use 10m to 20m band RF signals, you can reduce this affect but you still have non-isotopic antenna loss patterns, which can span 30dB range. <S> so sorry but design concept will fail. <S> With a Return Loss bridge I made a motion sensor and short and long range telemetry design. <S> short range was with no antenna, long range of more than 500 miles line of sight to satellites, when I discovered these properties in 1977 while designing an aerospace unwrapping dipole antenna for spinning rocket telemetry. <S> the motion sensor worked only in short range with 2m bands using the modulation of reflections or echos called return loss from body reflections of the transmitted signal anywhere in the lab with a directional dipole antenna. <S> I later learned that alarm companies developed option sensors for hallways and interior space by picking up the modulation of echos in a directional coupler or splitter to an RF detection diode. <S> So your concept fails the properties of RF, physics and EE antenna theory. <S> But keep trying. <S> IT often takes 10 to 100 good ideas to get 1 great one.
| This can be either done by configuring the involved chips to use less power to transmit, or by using worse antennas, for example.
|
How is companding different from pre-emphasis and de-emphasis? What I know is that interference increases linearly with frequency and thus noise power is concentrated at higher frequency. So, to reduce noise, pre-emphasis and de-emphasis filters are used which works by boosting the input signal before modulation and at receiver, it is deemphasized to original signal and the noise signal is not boosted and eliminated through deeemphasis filter. In this way, a great deal of noise is reduced. So, why companding is introduced then? What companding achieves more than preemphasis-demphasis? Both emphasis and companding vary the input signal is some way or the other. Preemphasis-deemphasis vary the frequency and companding vary the amplitude range, to be precise dynamic range. But what advantage companding achieves when compared to emphasis? In this site , it is written that most of the noise is reduced through companding but that can be also achieved through preemphasis-deemphasis. Also, if we see companding as compressing-expansion of signal, isn't the frequncy of signal gets changed? That is principle the emphasis. So, how companding is different from preemphasis-deemphasis? <Q> Small amplitude signal variations are increased above the noise floor so that when these are decreased on the receiving end the noise floor is lowered. <S> As any non-linear operation if transmitter and receiver are not exactly matched there will be harmonics and inter-modulation distortion introduced in the signal. <S> Companding was commonly used to reduce 12-bits of voice information in phone systems to 8 bits for transmission, and the noise this addressed is quantization noise. <S> It mostly works due to the non-linear aspect of our auditory system. <S> And due to the use of complementary non-linearities, it is better applied on the digital domain. <S> Pre-emphasis and equalization are linear frequency-domain operations that do not introduce harmonics or intermodulation distortion into the signal. <S> These simply change the frequency response to compensate for the frequency response of the channel. <S> Pre-emphasis, by placing more power on the most attenuated frequencies, attempts to place the signal above the noise floor on the receiver. <S> Being a linear system, matching is not as critical. <S> Although pre-emphasis and equalization are common for data communication channels, the principle is related to what was used for Dolby B/C noise reduction technology for analog audio recordings (in this case a combination of pre-emphasis and automatic gain control (AGC) just for high frequencies was used to compress the dynamic range to remain above the noise floor. <S> An AGC is a form of companding that, although still non-linear, avoids introducing harmonic distortion). <S> You would not normally use companding for analog music recordings, as the wide bandwidth and required complementary non-linearities would guarantee the introduction of harmonic distortion. <S> However, on some analog applications in which a wide dynamic range is required, there are filtering techniques that make use of analog companding. <S> These are called log-domain filters. <S> Bark space <S> A mixed technique that underlies all of our current lossy music formats from MP3 onward. <S> Is the compansion of individual portions of the spectrum and the assignment of more or less bits (down to zero) to each portion of the signal spectrum. <S> Our auditory system is highly non-linear and masks signals that lie too close to other larger signals that occur around the same time. <S> These hearing models are used to compand the individual bands of the signal to save storage and transmission space. <A> Preemphasis and deemphasis are linear operations that boost or suppress different portions of the signal spectrum. <S> They're basically 'plain old' filters. <S> Companding is a nonlinear process, where a signal is boosted before transmission (or recording, I know of it from audio) when it has a low amplitude, and then undergoes a comparable reduction in strength on reception (or playback). <S> To my knowledge it acts on the whole signal equally across the spectrum (although one could, I suppose, compand a signal by spectral component -- ugh!). <A> A couple of things. <S> First, interference does not always increase linearly with frequency. <S> Sometimes, but not always. <S> It depends on the characteristics of the transmission medium and the signal systems involved. <S> Two systems that use pre/de-emphasis are LP records (RIAA) and FM broadcasting. <S> In FM broadcasting, the system noise floor appears to be relatively constant across frequencies. <S> However, there is much less high frequency energy in the signal being sent, so the noise appears to be a higher percentage of the received signal at the higher frequencies. <S> In an emphasis system, the gain for the signal changes as the frequency increases. <S> Again using FM, for baseband frequencies above 2122 Hz the signal amplitude increases at a rate of 3 dB per octave ahead of the transmitter, and is reduced by the same amount in the receiver. <S> So the amplitude is increased by 6 dB at 4.24 kHz, 12 dB at 8.49 kHz, etc. <S> The main difference between pre/de-emphasis and companding is that companding is applied equally across the entire spectrum of the source signal (as in telephone calls) or across separate parts of the spectrum (as in Dolby-A noise reduction). <S> Within the frequency band of interest, the amount of gain boost/reduction is based on the amplitude of the entire frequency band (usually the RMS amplitude) rather than specific frequencies making up the spectrum. <S> So if the system determines that for that instant of audio, the gain should be increased by 4.7 dB, that 4.7 dB of extra gain is applied to all frequencies, not just the highs or lows. <S> In short, a pre/de-emphasis system changes an individual frequency's energy based only on its frequency; a companding system changes an individual freqquency's energy based only on its amplitude.
| Companding is a non-linear time-domain operation that alters the signal in way that reduces the influence of perceived noise.
|
Which type of sensor for measuring height of horse step? Is that possible? I'd like to track the amplitude of a horse leg/foot height for a show. The many many difficulties: constantly moving! So ultrasound way too slow the floor is irregular, made of shavings seems that an accelerometer isn't reliable for measuring distance, as I read here. The best would be a sensor I can put on the horse's leg/foot rather than video tracking (i.e. openCV) because the stage is round, and I'd love to keep the process light (not putting cameras all around etc). <Q> More specifically, a 6 degree of freedom motion sensor unit. <S> The problem with using accelerometers to measure distance is accumulated errors. <S> Since the gait always ends at the ground, you can remove errors at the end of each step. <A> That is a complicated task, which may require a lot of post-processing. <S> For the sensor part, I would give a shot to barometers. <S> As of 2019, it's typical to find high-precision barometers having 0.01 hPa resolution, such as LPS25HB. <S> It's enough to track altitude delta of centimeters. <A> You could use a Lidar , which can very rapidly determine the distance from a given object. <S> They are frequently used in robotics and autonomous vehicles to determine their distance from walls, floor, etc. <S> Some of them are pretty small and could possibly fit on a horse's leg, and are relatively inexpensive .
| Actually, because of the cyclical motion, and the fact that each gait ends on the floor, an accelerometer might be ideal. Despite the air movement, an array of barometers could be used for maximum SNR.
|
0-2v to 0-10v amplifier tl;dr: I want to understand op-amps better! How do I choose the magnitude of resistors? 1k, 10k, 100k, etc.. I made a schematic to play with an In-amp to create a gain of 5. What other considerations should I implement in a real world circuit? - I'm trying to learn how to use op-amps to provide a gain of Av=5Eventually I would like to apply this knowledge to the use of strain gages in a Wheatstone configuration. That's a later thing :) For now, I'm diving into it by creating a instrumentation amplifier with a gain of 5. I like this approach because as I understand it, it is a more robust solution to boosting a signal for a variety of reasons. Not to mention the gain is easily adjusted, which I may be interested in using a digital potentiometer in that location for just that reason. This is what I have came up with so far in ltspice: This yields what I want: The frequency is not very relevant in my situation, the signal won't be changing very much over the course of 1 second. So my question to you guys, Why do I choose resistors in the realm of 1k? Why not 10k? or 100k? How can I add some basic noise filtering? Do you see any issues that may occur in applying the circuit above to real life? I apologize for my lack of knowledge and terminology if I butchered anything! Thank you for any time you can lend me, I do circuits by hobby and want to understand more of the analog elements. <Q> If you choose too low valued resistors, they would draw too much current and dissipate too much, which isn't needed. <S> (And when they draw really too much current, the may overdrive the OpAmps as well and/or damage it). <S> And next to that, the OpAmp draws a small input current as well. <S> With too big resistors, those currents influence the circuit too much. <S> Depending on the voltage of the OpAmps you pick the resistors. <S> I'd advise to start reading about RC filters and apply them. <S> Leaving that to others. <S> I'm not that experienced applying them. <A> Noise is just a signal you haven't understood yet... <S> It's pretty typical to filter out signals that might get in the way of making a robust measurement of your sensor outputs. <S> You'll get a better measurement if you can remove signals you wont be able to measure accurately or don't care about. <S> For example: Most systems will sample vout with an ADC at a fixed frequency. <S> Any signals that have a frequency higher than 1/2 of that frequency are very hard to understand with that ADC and conventional methods, so it's pretty typical to ensure that the amplifier acts as a low pass filter to remove most of the frequency content above 1/2 the sampling frequency (the Nyquist limit) <S> (In this case signal that would be hard to understand <S> is filtered out of the output waveform leaving only the signals that can be well measured for future assessment.) <S> Many indoor systems will measure 50/60Hz noise from the AC power systems around them. <S> If this noise is enough to saturate your sensor systems, you might want a band stop filter to remove it. <S> You might also see a high pass filter to filter this out if you don't care about the DC offset from your sensor. <S> (In this case an understood signal is being filtered out. <S> If it's small enough it will be easier to filter out these signals in software later.) <A> High value resistors easily introduce high phase shifts into the negative-feedback loop, causing peaking, ringing, and oscillation. <S> For fun, attach 1,000pF or 10,000pF to the inverting ( -Vin ) pin of just about any opamp circuit.
| If you choose too high valued resistors, your circuit may be prone to external noise.
|
Can I use a 28 VDC 10 A relay to toggle a 10 kV DC 0.03 A source? Neon signs use transformers that can output at 10 kV DC at 30 mA = 300 W. I want to toggle this power with a common 5 V relay, but the DC rating for these are about 28 V DC at 10 A = 280 W. Does the output of the transformer exceed the rating of the relay? If so, why would the same relay (with a 125 V AC 10 A rating) work on the input of the same transformer if its input is rated at 125 V AC 0.9 A? Am I confused about the power limits of the relay? <Q> No, the contacts will arc across and it will likely arc to the coil as well. <S> Very, very dangerous. <S> A commercial relay properly rated to switch 10kVAC will probably cost in excess of $1000. <S> There are some DIY approaches, but better suited for non-novices. <S> You should consider switching the primary, but being an inductive load that relay may not be acceptable. <A> It's not the power, it is the individual components of power - current and voltage. <S> The characteristics of the contacts determine the current rating of the relay. <S> Contact material, shape, and pressure affect how much current they can pass, and can break open repeatedly without degrading prematurely to the point where then cannot conduct any current at all. <S> This is why small signal relays have gold contacts (no corrosion even under light pressure) and <S> high current AC power relays have silver-based contacts (lower resistance/ <S> lower heating, and the arc that forms when the high current is broken helps "clean off" oxidation). <S> When open, the contacts are too close together for that high a voltage; it will arc continuously across the gap. <S> Change your way of thinking: <S> A relay does <S> not switch power - it switches a current at a voltage. <S> Neither rating can be exceeded. <A> One major point to consider on relay (or any switch) <S> contact rating is whether it is AC or DC. <S> With DC, the switch has to break the connection in a manner that will not arc or continue to function reliably with the resulting arc. <S> You're calculating power and comparing the resultant power (300 vs 280 watts). <S> However, in this case the voltage is part of the reason for the rating (arc distance) as well as current (contact thickness). <S> Finally, the direct or alternating nature of the current determines what the device can ultimately safely (and reliably) switch.
| Generally speaking, switch contacts can handle higher amperes if you are using AC because of zero crossings. In your case the problem is the voltage rating.
|
Beginner's second PCB circuit design: AC to DC I am designing an AC to DC converter and I just want to make sure that I am so far doing the right thing — placement of components and wiring etc. — as I am a beginner. Does this look good so far? This is my updated schematic and PCB. Do the rectifiers look correct? I tried to space out 8mm for the transformer. Do the connections look good? How does this look like? For J4 I have a 230v appliance but no power and for J5 I have the AC 230V and a relay which acts as a switch. Relay is rated for 260v. Would this work? <Q> Per IPC spec you need creepage and clearance, even if your not using this in a product (and getting it tested at an ETL). <S> Source: <S> http://www.ni.com/white-paper/2871/en/ <S> That means if you have AC mains running into your board, you need at least 1.6mm around the conductors that carry that voltage (meaning the hot and neutral lines) and any other conductor . <S> This is for pollution degree 1 which for you means no dirt (stuff that could lower the conductivity of the PCB). <S> It's generally a good idea to place a large gap between AC and DC sides of the board for good measure. <S> If this is a product it needs to be double insulated or made inaccessible to someone who would service it. <A> Here are three documents that can give you more insight. <S> IPC2221A: <S> Generic Standard on Printed Board Design. <S> UL61010-1: Electrical Equipment for Laboratory Use. <S> (if you have access to it. <S> Other "sources" might be available) UL60950-1 <S> : Safety of Information Technology Equipment <S> These documents are a lot to read and understand, so as a general rule, you should keep 2mm gap between traces/pads on the mains side, and 8mm gap between the mains and the low voltage, where you might touch with your hands the conductors. <S> You can go below these clearances, but you need to know what you are doing, and this is where the three documents listed above, come into place. <S> As yours is a hobby project, and space is not a big issue, stick to the values I have given you above, or greater, and you will know to be safe, as long as you stay below the 300VAC. <A> There is no reason for N$4 to wrap around N$5. <S> If you swap the two transformer pins, the layout becomes much cleaner. <S> As above, the net connecting the D1 and D2 cathodes is way too close to the end of the fuse. <S> Also, it is unnecessarily trapped. <S> Getting a trace from there to some other part of the circuit means wandering around the transformer pins. <S> Consider rearranging the diodes such that the two outputs of the bridge are more accessible from the right. <A> Parts placement should be done very different. <S> Distance between primary circuit parts to secondary circuit parts should be as big as possible, also between all primary to all secondary tracks. <S> So J1 and F1 at the left side of the board close to the primary transformer pins 1 and 2. <S> D1 to D4 on the right side of the board close the secondary transformer pins 3 and 4. <S> Pins 3 and 4 may be swapped when neccessary to avoid crossing tracks. <A> I suggest rotating the transformer 90° CCW and moving the bridge to the upper right of the PCB. <S> As others have said, your clearance primary-secondary is inadequate for the mains. <S> 8mm of creepage would be good for primary-secondary. <S> Swap equivalent pins (such as the transformer secondary pins) to make the layout more straightforward. <S> This is of more importance on high voltage and/or high current conductors than on traces that carry only signals. <S> You should also be thinking about mounting holes early , they're often a pain to try to fit in later.
| The first thing that is really evident, is that you have not left large enough gaps between the Mains AC and the secondary of the transformer.
|
Direct connection of UART serial port to USB type-C in my project I have a USB type-C 3.1 2nd Generation (24 pins), which should provide 5V to a charge battery/battery system. In addition, there is a uC powered at 3,3V where the maximum voltage on GPIO pins is 3,6V. The question is: can I connect directly the UART pins of the uC to the Rx/Tx pins of the USB port? If I connect the USB to a laptop or wall adapter, what will it be the voltage in Rx/Tx pins? Is there a risk that I overcome the maximum 3,6V of the UART pins of the uC? Note: I prefer to not use a bridging between a USB port and an enhanced UART serial port. <Q> From your last sentence it sounds like you actually want a working UART over USB. <S> This will not work. <S> The Rx/Tx pins in an USB-C connector has nothing at all to do with the UART in your microcontroller. <S> USB doesn't use high voltage on the signal pins even if you use a high voltage for charging. <S> This is even true for the initial USB. <A> You want an USB serial adapter with TTL levels (i.e. 3.3V levels). <S> While connecting the serial to the USB might not damage anything it will certainly not work. <S> You need to the adapter for it to make any sense. <A> Your concern is whether your device can suffer a damage if someone plugs your device into regular Type-C laptop port (with regular C-C cable). <S> Your concern is not founded. <S> The voltage level on Rx pins will occur only during "Rx detect" stage, where a common-mode pulse will be applied with 400 mV amplitude, making the voltage at 800 mV at best, and only for 10-20 ms. <S> However, the voltage will have a negative peak as well, but your UART will likely tolerate this. <S> You can use the Rx/Tx wires as you wish, however you need to keep in mind that the Rx+/Rx- (and Tx) are typically twisted in pairs, so you might incur substantial cross-talk if you don't use the wires correctly. <S> You need to use, say Rx- and Tx- as signal ground wires, and Rx+ and Tx+ as your UART signals. <S> More, to get any voltage from your laptop, you need to have 5.1k pull downs on each of CC1 and CC2 wires, otherwise the Type-C port will output no VBUS. <S> Also please be aware that USB Type-C specifications also "went an extra mile" and do define DAM - Debug Accessory Mode, precisely for the purpose of extensive debugging. <S> The attach/discovery protocol is crazy IMO, but you might want to take a look at appendix B of the specs if you are serious. <S> The transport layer is not defined, so you can use the simple propriety UART functions.
| To summarize, you are making a device that uses Type-C port to charge some inside battery when connected to a Type-C laptop, and also want to use the Type-C Rx/Tx pins as UART interface to some internal IC from some debug adapter. However, you will not physically damage the pins.
|
How can I cover PCB burn marks to identify burns later? I'm repairing a receiver from the 1990s that has a few burn marks on the PCB where a few transistors overheated. After replacing them and any other offending circuitry I'd like to be able to know if the issue has been resolved -- i.e. I'd like to know if the PCB burns again. I don't have access to a thermal camera and I don't feel like using the back of my fingers to test for overheating. Is there some way to cover up or clean the PCB so that I'll retain this indication? <Q> These for example are 14 mm diameter which should be small enough to attach near one of your transistors, or even on the transistor itself. <S> Another option could be to slip a small piece of heat-shrinkable tubing over each transistor, maybe using a dot of cyanoacrylate glue to hold it in place. <S> Standard heatshrink typically shrinks at somewhere between 70 and 120 °C, which is probably the range you would consider as overheating. <S> If you observe that the tubing has begun to shrink, you know it's seen a temperature somewhere in this range. <S> Note that anything that impedes airflow over the transistors could cause them to run even hotter, though. <S> This will increase the stress on the solder joints and PCB tracks though, in case the equipment is likely to experience mechanical shock or vibration. <A> You could use a blob of wax on the components that you suspect are getting hot to see if it deforms due to the high temperature. <S> Waxes come in many different melting points . <S> For example, beeswax melts at about 62 °C. <S> If something is getting really hot, you could use hot melt glue, which melts at about 110 °C. <S> As user71659 pointed out in the comments, calibrated melting temperature crayons are available in temperatures from 40 °C and higher. <S> See here and below. <A> Use a sharpie and draw some arrows, or circle the affected areas. <S> Then use a thermal sensor from Home Depot to check temperatures. <S> I have one like this, I think made by Ryobi, is great for all kinds of measurements around the house <S> (Found with search for "infrared thermometer"): https://www.homedepot.com/s/infrared%2520thermometer?NCNI-5 <S> Might even be able to get a phone app that shows hot spots. <S> Google "phone infrared camera app"I think most are an infrared camera that displays on your phone <S> , I don't how good they'd be for smaller area views. <S> The $35 infrared gun with laser pointer <S> I know works well. <S> Better than burning a finger! <S> (done that too ...) <A> Most error locators use alcohol to find the error. <S> If the alcohol vaporises then there is a hotspot. <S> Alcohol does not conduct electricity, and it cools down the components.
| Finally if you're concerned that these transistors run hotter than they should, you could fit them with clip-on heatsinks anyway as a precautionary measure. Non-reversible temperature-sensitive labels are available which will change colour irreversibly if they experience a defined temperature.
|
When charging rechargable battery, should we care about watts and ignore voltage and current? I know this might be a silly question, but it's from a beginner. Let's say I want to charge a rechargable lithium-ion battery, so when I charge it should I only check the watts and ignore voltage and current specific values? I mean the current is about how much electrons are flowing, and voltage is how much potential energy is stored in each electrons (this is how I understand voltage and current, correct me if my thoughts are wrong), so I thought that if I charged a battery with 10 watts, 1 amp and 10 volt, then the battery will be charged well, but if I charged the battery with same10 watts, but 2 amps and 5 volt, won't the battery be filled with weaker electrons? I mean this time the battery will be filled with less-energy electrons than the previous example, so should I look at the volts and amps when charging or the watts is enough? <Q> You may not only look at watts when charging a lithium-ion battery and ignore voltage and current limitations. <S> Charging a lithium-ion battery is very sensitive process. <S> It will explode easily with overcurrent, overvoltage, or overcharging. <A> You should probably start by understanding the final step of charging "CV" charging. <S> During this step the charger behaves as a voltage regulator, using power to push current into the battery, but as it does all that, it's controlling the voltage across the cell which is why we call it "Constant Voltage" <S> Constant Voltage charging begins at the end of Constant Current charging where current is controlled as the voltage across the cell rises from 2.7V or so to the Constant voltage limit. <S> At the time the charger changes over from CC to CV, the power into the cell is maximized. <S> Some chargers cannot achieve this maximum power and have to limit power, running a constant power mode between Constant Current and Constant Voltage. <S> In each case the charger uses some sort of switch to control the flow of electrons into the cell. <S> But whether it controls voltage or current, the other is a function of the first as determined by the Lithium Ion Cell and a number of other variables like its temperature. <S> Usually a cell datasheet will have a recomended charging profile with a charge current limit and a charge voltage limit (as well as a recommended termination current). <S> These are followed or violated at the designers discretion and will affect things like the capacity and the lifetime of the cell. <A> All batteries (in contrast with capacitors) exhibit what is called variable charge efficiency. <S> That is, when the voltage is low, forcing 1 coulomb of charge into the battery will allow you to deliver (almost) 1 coulomb of charge when discharging at that voltage. <S> This is a charge efficiency of about 1. <S> When a battery is nearing full charge, the charge efficiency will drop, often to quite low values. <S> Since power is voltage times current, the "unused" current and power near full charge will show up as excess heat, and the battery will start to warm up. <S> So power is not a useful measure of charge effectiveness. <S> In most chemistries, there exists a maximum voltage, above which all input power is wasted, and this condition usually will damage the battery, as well. <S> In "standard" lead-acid cells, the electrolyte may actually start boiling. <S> In lithium cells, trying to charge above a certain voltage (nominally 4.2 volts for a 3.7 volt cell), the cell will be irreversibly damaged, and may catch fire. <S> In addition, your statement that won't the battery be filled with weaker electrons? <S> I mean this time the battery will be filled with less-energy electrons than the previous example, shows a really major misunderstanding of how batteries work, to the point that I'm not sure how to respond. <S> For instance, if you have 3.7 volt lithium cell and you try to charge it with "1 amps and 10 volt", a charger which will supply 10 volts and a maximum of 1 amp will current-limit at 1 amp, and the battery will show a voltage much less than 10 volts (and if it shows more than 4.2, you're hosed). <S> Likewise, a "2 amps and 5 volt" will limit at 2 amps, and less then 4.2 volts - unless you're hosed. <S> Now, it's true that, as a rule of thumb, the 2 amp charge will produce a full charge in (more or less) half the time as a 1 amp unit, but that's not at all what you're asking. <S> In all cases, the useful effect of charging is to change the composition of one or more molecules. <S> In the case of lithium ions, the most common reaction is of the net form $$LiC_6 <S> +CoC_2 \Leftarrow\Rightarrow <S> C_6 +LiCoC_2 $$ and <S> the reactions use the same electron energy regardless of the input current. <A> A lithium battery is first charged with a constant current. <S> That current depends on the battery specifications (smaller battery smaller current). <S> When the voltage reach the 4.2V then the voltage is kept on 4.2V. Do not exceed this voltage! <S> When you keep this voltage the current will drop to allmost zero. <S> And then the battery is full. <A> And if it doesn't, you'll still damage it, reducing its capacity or just making it more likely to explode in the future. <S> Batteries, lithium-ion especially, must be treated with care and you should not try to design your own charging system without some serious research first into how to properly use them.
| If you do not use the correct voltage and current to charge a battery you run the risk of having your battery quite literally explode.
|
How is the fuse amp rating determined for a 12V AGM battery bank? I would like to put 3 x 130Ah AGM batteries in parallel and attach a fuse to the positive terminal of each battery. What size fuse should I use? How is the amp rating of the fuse determined? The batteries will be connected to each other and the inverter will be connected to the battery using heavy duty copper cable. Smaller gauge cable will possibly be connected to the output terminals with Anderson connectivity. An inverter of up to 1000W will be connected. EDIT: Even though an inverter of up to 1000W will be connected, the maximum power draw likely to be less than 500W. I would like to have this safety margin in case an appliance of higher wattage needs to be used so I want the system to be able to cope with a current of as high as 80A. The batteries will be charged either via solar panels or via a 20-30A 240V AC charger. <Q> The fuse amperage should be less than the maximum discharge current rating of the battery the maximum current rating of the cables and connectors <A> The fuse rating for the wiring to the inverter should be determined by the inverter's requirements. <S> The inverter manual probably specifies an appropriate fuse size and type. <S> For smaller gauge wiring, the fuse should be sized to protect the wiring. <A> I realize this is an old question. <S> I decided to answer it anyway. <S> OP is probably gone but others may benefit. <S> Fuse size cannot be discussed without discussing wire size. <S> The purpose of the fuse is to prevent wiring from getting too hot. <S> If the wire gets too hot it may cause burns to people who touch it or even cause the insulation to smoulder or burn. <S> It can even start a fire. <S> Fuses protect against that. <S> To figure out how large the wire needs to be consult an ampacity table (use a search engine). <S> From the cerrowire website ampacity table I see that for 80 Amps you would need 3 AWG wire. <S> But I have never seen 3 AWG wire. <S> So I recommend you use 2 AWG (equivalent to 6.5mm diameter) wire. <S> You can use a 100 Amp fuse. <S> There is one more thing. <S> Fuses have a maximum interrupt rating. <S> Fuses do not operate instantly. <S> The higher the fault current, the faster the fuse will blow. <S> But if the fault current is extremely high, a physically small fuse may explode when it blows. <S> Theoretically it may even fail to open the circuit due to arcing or something. <S> With lead acid batteries, the potential fault current may be several hundred Amps. <S> So you should select a fuse with an interrupt rating of at least 1000 Amps. <S> If the wire has to run a long way, you may need to use heavier wire to avoid excessive voltage drop under maximum load. <S> That is fine. <S> You can still use the 100 A fuse with larger wire. <S> If you want to learn about voltage drop, pose a separate question. <S> Note: <S> the wire to the solar panels will likely be much smaller diameter. <S> So you need to put a separate fuse in series with those smaller wires, sized based on their diameter. <A> I guess that you use 12V car batteries. <S> The voltage can drop by large currents to 11V. P= <S> U*I1000W <S> / 11V = 90A90A /3 <S> car batteries = <S> 30A. <S> You can use a 32A fuse for each battery. <S> Use real thick wires! <S> and as short as possible. <S> One voltage drop over a wire uses 83 watt, that will fry your wire for sure.
| The fault current must be in excess of the fuse rating for some period of time before the fuse will blow.
|
Cheap 12V to 5V DC-DC converter for PCB? I want to clarify: I understand there have been many questions on this site regarding this same voltage conversion; however, what I am asking is not simply how to convert 12V to 5V but how to do it in a simple way I can integrate into a PCB. My situation is that I am trying to create a 'smart' RGB light strip controller. The strip runs off 12V and the NodeMCU board controlling it runs off 5V (usually via USB). This is the guide I am following. Currently, I have the circuit wired up using jumper cables and I am using a USB car charger to convert 12V to 5V for the NodeMCU, and the whole circuit works perfectly. However, I am now looking to design my own PCB which negates the need for all this wiring, as I am looking to use this circuit more permanently. Thus, what would be ideal is to simply plug in a 12V plug to the PCB, have that feed the strip and also be converted down to 5V and offered on a pin so that I can attach to the NodeMCU's Vin pin via a jumper wire. My question is: What is the cheapest (requires the least parts) and best way of converting 12V down to 5V with through-hole components on a PCB? Other guides recommend using an 7805 converter due to its simplicity; however, I understand that this is extremely inefficient and can pose serious problems due to excess heat, which would not be appropriate considering that I am looking to have this circuit permanently on. If someone would be able to attach a schematic of a circuit which solves this problem reliably, I would be very grateful. Thank you in advance for any help. P.S. Is it be worth just simply copying the circuit inside the USB car charger, as this seems to work perfectly with barely any heat? If so, which components are required? P.S. As for the current requirements of the NodeMCU, I am unsure exactly what the maximum draw could be, as I couldn't seem to find a straight answer online. However, I assume that it must be way under the possible current of even the most basic USB ports, as it is such a small device. <Q> A buck converter is the classic way to do this. <S> It requires input and output caps, an inductor, a controller chip and optionally (depending on the controller chip - some have synchronous rectifiers built-in) a Schottky diode. <S> I am a surface-mount guy, so I don't have a particular part to recommend for through-hole. <S> I use the AP1509 pretty routinely for this job. <S> Rather than dropping the voltage with resistance (dissipating the voltage drop power as heat), they directly convert wattage at one voltage to the same wattage at a lower voltage (at 80-95% efficiency, generally). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here, the "switch" represents the buck controller (not shown is the feedback from the output to control the switch frequency and/or duty cycle). <S> The switch closes to charge the inductor, raising the output voltage. <S> Once the output voltage is high enough, the switch turns off. <S> The inductor will develop a voltage across itself as the coil collapses. <S> The Schottky diode will anchor the left side to ground, forcing it to power the output. <S> Eventually the output voltage will fall, and the switch will close, and the cycle will continue. <A> My favorite are "7805" equivalent DC to DC converters , some of them can even be used to generate negative voltages. <A> www.pololu.com carry several 12V to 5V adapters in different current ratings. <S> Digikey does also, several that will take the place of a 7805 size device, from Oki Murata <S> https://www.digikey.com/products/en/power-supplies-board-mount/dc-dc-converters/922?k=oki+murata&k=&pkeyword=oki+murata&sv=0&pv1989=0&sf=0&FV=ffe0039a&quantity=&ColumnSort=0&page=1&stock=1&pageSize=25 <S> Putting your own parts on a board: This one is good for 2A: https://www.digikey.com/product-detail/en/diodes-incorporated/AP1509-50SG-13/AP1509-50SGDICT-ND/1301653 and doesn't need much in the way of external parts <S> https://www.diodes.com/assets/Datasheets/AP1509.pdf <S> I put 9 of these, or a similar chip, on a board recently, to bring in 12V (automotive) and distribute it to 9 LED strips, vs using an 18A design and a much bigger inductor. <S> Turned out very compact. <S> I can add more details when I get home. <S> Edit:this is the chip I used TPS54239EDDAR http://www.ti.com/lit/ds/symlink/tps54239e.pdf Nice pad on the bottom for soldering the chip down for cooling, and the datasheet had good details on selecting components to use with it. <S> Also good for a high currents, up to 4+ amps. <A> So you essentially have two options. <S> A linear regulator, or a buck converter. <S> The linear regulator has the advantage of being cheaper and less complex, but will be far less efficient, and may cause you thermal headaches. <S> Dropping 12V down to 5V at 390mA means that you would be dissipating 2.7W in the regulator, which means that you would likely need a heatsink to keep it cool. <S> However, depending on how often you are in this high power transmitting state, it might be easiest to go with the simple solution. <S> A Buck converter is more complex (more components involved), but will be far more efficient than the linear solution, and will generate far less heat. <S> Unfortunately through-hole switching regulators are not nearly as common as surface mount ones, so it will likely cost slightly more than an equivalent SMT circuit, but they do exist. <S> One example would be TI's LM2575 : <S> The circuit above comes from the datasheet for the chip. <S> It can be purchased from digikey . <S> You will need to buy the schottky diode D1, the inductor L1, and both capacitors Cin and Cout seperately, but digikey will also have thousands of options for each of these. <A> If you have time, go to ebay and buy there a "step down converter" for US $0.99 you have it on your doormat.
| Buck converters are preferable to 7805 style regulators because they're much more efficient. Any cheap step down converter will do, because you only have to power the arduino with 5V. (sometimes you have to adjust the voltage counter clock wise with some chinese converters)
|
How to read a multimeter (digital) -- should I get 2A readings from a couple of AA batteries? I have been using some old multimeters in a high school science lab. We did a current an voltage reading from a AA supply (two batteries) with a variety of resistors. In some cases a 1M resistor. I noted that the only way to get current readings was to set it on the 10A circuit. OK, I thought, maybe the batteries put out that much current in toto. But I saw this happen even when the 1M resistors were hooked up -- we would get readings of 2.7 V or so which makes sense (old batteries and all) and then when we measured current we would have to put it on the 10A and get readings like 2.0 A or 1.8. Ultimately, the digits made some sense in terms of Ohm's law, it was just the powers of 10 seemed off. So I wasn't sure if I was just reading the thing incorrectly. (Like, was it actually showing milliamps?) To give one example: I put a resistor in line with a 3V supply. It's 46.4 Ohms. (The multimeter is an Extech instruments, found in a drawer). I touch the probes to the resistor and get 2.60 V, which seems reasonable enough. I connect the red probe to the 10A input and check current and I get ~1.9 A. But V = IR says that it should be (2.6V) = I (46.4) --> I = 2.6 / 46.4 ~ 0.05A We have gotten similar results with other meters, though my "good" one (a new Mestek DM91A) reads 0.04A when it's hooked up to the mA side (if I try to set it to mA range the alarm goes off, with the message "FUSE") and about 1.8 A on the 10A side. In at least one multimeter I replaced the battery. I tried similar readings with a 965,000 Ohm resistor. I got 0.03-0.04 A with the newer Mestek, and 1.8 A with the Extech, which wouldn't read it when i used the milliamp connection. I suspect there is some stupid mistake I am making in reading the things, and yes I checked that the positive was to positive and negative to negative. (I grew up with analog meters, and you could just read the scale :-) ) Anyhow I hope people about can help me figure this out. Thanks in advance. <Q> Votage readings are taken with the meter in parallel with the load. <S> Current readings are taken with the meter in series with the load. <S> A voltmeter presents a very high resistance load to the circuit. <S> The ammeter presents a very low resistance load to the circuit. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Correct way to measure voltage and current. <S> By connecting the ammeter in parallel with the load you have drawn the full short-circuit current that the battery is capable of. <S> This will have blown the mA fuse but the battery isn't powerful enough to blow the 10 A fuse. <S> If you had tried it on a car battery it would have - and if there is no 10 A fuse (as is common in the cheap meters) it would have burned out the shunt or PCB tracks and possibly destroyed the meter. <S> Never short-circuit a battery with an ammeter. <A> Which happened because of measuring the wrong way like the others suggest. <S> And that also makes sense with a 2A short circuit current for AA batteries. <A> A fast way to check if the fuse is broken is to switch the multimeter to resistor measuring mode (having the probes connected to do resistor measurements). <S> Touch with the positive probe point (likely the red probe) <S> the input of the 250mA on the meter. <S> It should be very low ohmic, 0 to 1 ohm. <S> If not, the fuse is broken. <S> NEVER replace the fuse with an AC fuse, use a DC fuse, same type, same ratings.
| Guessing those meters have a blown fuse on the mA input, open it up and see if you find a 250mA-ish fuse that looks broken.
|
Interfacing GPIO 3.3V to 74HC04 5V Will it be safe if I connect STM32F407 GPIO outputs (Vdd=3.3V) to 74HC04 hex Inverter inputs (Vdd=5V)? What is the best recommended method to add an inverter at the output of STM32F407 MCU? edit: Included schematic details about the 74HC04 circuit. <Q> A 74HC04 is not guaranteed to function correctly if driven from a 3.3V CMOS output (although it will almost certainly work as it is very unlikely the device will be at the data sheet limits). <S> The worst case Voltage required to be recognized as a high level is 70% of the supply voltage (Vih). <S> If the 5V rail is in fact 5.25V <S> then this would be 3.675V - above the 3.3V supplied by the GPIO. <S> The more normal case is that it would sense just over ~50% as a high which for a 5V rail is 2.5V and so would work. <S> The HCT family of devices is a similar to the HC devices but are designed for this type of interfacing and have a threshold that is independent of the supply rail with a worst case of 2V which is guaranteed to work with the 3.3V input signal. <S> They do have a more restrictive supply rail voltage of 4.5 - 5.5V though. <S> The 74HCT04 device is equivalent to the 74HC04 but has this better controlled threshold. <S> The HCT devices were actually designed to interface to bipolar TTL devices which had a worst case Voh of 2.4V but will work as well for your application. <A> 3.3V logic levels should not be used with 5V powered HC chips. <S> One option is to use 5V tolerant IO pin on STM32 in opendrain output mode and have a pull-up resistor to 5V so this way the logic level fed to inverter is 5V. <S> The limitation is that when MCU is in reset, the resistor keeps high and inverter output is low. <S> Another thing is, if the MCU can use 5V logic levels, why keep the inverter at all. <A> The question is whether or not the 3.3V logic high input will be recognized as a logic high or not. <S> You need to look at the specific manufacturer datasheet for your part, and see what the "Minimum High-Level Input Voltage" is. <S> An example here: <S> In the case of the above part, a little linear interpolation yields 3.5 as the Minimum High-Level Input Voltage when Vcc is 5.0V. <S> This is a worst-case scenario, and in my experience, 3.3V logic into such a part will usually work, but can't be relied upon, and you usually never want to rely upon a behavior not expressly promised in a datasheet unless you have done significant characterization testing of many parts. <S> Note that the Nexperia 74HC04 will work just fine as a 3.3V inverter if Vcc is 3.3V. <S> This part (Nexperia 74HCT04) promises a Minimum High-Level Input Voltage of 2.0V when Vcc is 5.0V. <S> This part will definitely work just fine for your application.
| It should be safe, as the hex inverter will be able to handle 0-5V on its input at least, and you will only be applying 0-3.3V.
|
Writable area on a PCB I would like to create an area on my PCB where you can write on with a pen.Is this possible, and if so how should I do it? <Q> Assuming white silk screen, just draw a solid box on the silkscreen layer. <S> You can then write on it with a permanent marker. <S> Here is an example from a board I made a while back: <S> That was before I wrote on it. <A> A common practice is to leave a square area on your PCB, filled with white silkscreen as a background. <S> If you use an adhesive label, make sure that it's applied after reflow/soldering. <A> Yes. <S> You can write on basically any part of a PCB with a permanent marker. <S> If you want, you can also use your silkscreen layer to provide e.g. a white background, or checkmark boxes etc. <S> This is <S> pretty common, eg. <S> for boards that go through manual QA. <A> The adhesive can break down over time and cause problems. <S> Same goes for some inks. <A> If you ever decide to serialize via the board house level instead of writing manually (say, for limiting the board house's marking of their manufacturing stamp, CE marking, etc.), another option we've used is to mark out a rectangle on a mechanical layer that is labelled "Note 11 on Top Overlay", then in the design notes add an annotation that reads something like "11. <S> Mark Vendor ID, Date Code, and UV94V-0 Using White Epoxy Ink". <A> Agreed on labels, the adhesive is basically made of organic compounds and can be carbonized with high voltage. <S> This HV breakdown problem is not just a problem for adhesives. <S> Often times PCBs have a keepout zone used to separate low voltage from high voltage circuits. <S> From personal experience I know that quality inspectors use either office supply stamp pads or permanent markers for their notations and usually zero in on these zones. <S> The inks generally use carbon in their pigments and I have experienced breakdowns that forms tracking paths following the inspectors marking. <S> if you must use these inks in critical area, you really should do insulation resistance testing on a sample or PCB coupon. <S> The Best practice is to provide a designated zone for notes where there are no HV circuits involved.
| If you would like to write on the PCB with something other than a permanent felt marker (e.g. a ballpoint pen), a paper adhesive label could be placed over-top of the area. If you use a sticker, make sure it doesn't go near any high voltage sections.
|
Really occasional malfunction of the Schmitt-trigger input I'm using some Schmitt-trigger inputs for debouncing the interrupt problems. Referring to above circuit, the circuit has a RC circuit and the output of the RC circuit goes to the Schmitt-trigger IC(74HC14).Also, I soldered like this circuit and the look is like as below picture. Well, the picture is not good I apologize for the view. (18k : R1, 2k : R2, 0.1u : C1) Anyway, I'm convinced of the equality of the circuit picture and the real one.However there is a very small issue. The output of the Schmitt-trigger IC is connected to the MCU interrupt pin. So, if the Schmitt-trigger output goes to low(I use falling-edge-triggered interrupt), the MCU will run some ISR. Suppose I "bring(NOT push the button)" my hand to control the device on my own. At most, the device not malfunction. However, very occasionally the device goes to interrupt mode(ISR). The schmitt-trigger goes to MCU(ATmega2560, STM32F411RE, and so forth) and the MCU has some attached interrupts that can respond for the some schmitt-trigger inputs. And if the problem occurs, MCU just execute some ISRs. I thought the problem is occurred by the ESD so I used some TVSs and collected all the GNDs as near as the SMPS input, but the issue is not entirely solved. The problem hardly happens(probably just one time of a day) so this wouldn't be important one, but what could be the cause of this situation? Could I get some intuitions for the issue and the solutions? <Q> Your awareness of stray ISR resets, tells me you have an EMI problem. <S> It can be caused by stray E-fields, modulated by your hand that acts as an antenna and variable capacitor to couple line E fields near a floating supply voltage used on MCU. <S> Although 0.1uF seems low impedance, the thin wires also raise ground impedance, so the distance to MCU can be important and RF broadcast signals can also combine with this interference. <S> Often Common MOde (CM) <S> noise converts to Dff. <S> Mode (DM) noise by unbalanced impedance of signal and reference. <S> (supply) <A> As @Sunyskyguy says, you have an EMI problem. <S> Very likely due to your perforated board construction and the distance between components. <S> Enough of a stray field can be induced into those long wires to cause some problems. <S> Without a proper PCB it could be hard to avoid, but here are a couple of things that you can do to reduce the likelihood of the problem: <S> Make sure to have a low inductance connection between the ground of the 74HC14 and the MCU. <S> Place these as close as possible, with as thick a wire as possible. <S> Make sure to have a ~100nF decoupling capacitor for the 74HC14 power supply. <S> Directly on its pins. <S> Place the debouncing capacitor as close to the 74HC14 as possible. <S> This will serve to filter any EMI coupled into the wiring. <S> Optionally, add a zener or TVS before the capacitor. <S> This will serve to dissipate any ESD energy before it hits the ICs. <A> I would avoid this whole problem by doing the debouncing in software. <S> There are several subroutines for this on https://playground.arduino.cc/Main/LibraryList#IO_buttons That, to me, is the correct way to solve the problem. <S> Requires no external components beyond the switch. <S> You can set INPUT_PULLUP for the pin to eliminate the pullup resistor. <S> "The cheapest and most reliable parts are those which are not there." <S> - Gordon Bell (designer of the PDP-11)
| To reduce stray E-fields; typically, ground planes, shielding , ferrite beads, twisted pair wires, using tighter/smaller layouts to reduce loop area are some methods used and use lower impedance sensors.
|
How can I check if two outlets are connected to the same electrical circuit? I have some internet problems, and I was thinking to buy a PowerLine , that will help me extend the WiFi range. The thing is, to get this PowerLine working, both need to be conected to the same electrical circuit. So, one of my outlets is in A room, near the router, and the other outlet is in the B room. How can I check if the outlets are connected to the same circuit? I live in an apartment, and I have a circuit breaker. Here is a photo: <Q> I have no idea <S> what a "PowerLine" is and you haven't provided a link. <S> If it is a socket strip with a built in filter then it only protects devices plugged into it and not into the other sockets in the building. <S> Update: <S> It appears that you are asking about a WiFi extender that transmits data over the mains. <S> Your fuseboard is single-phase so all circuits share the same live and neutral. <S> The device should work. <A> Plug lights into the various sockets around the apartment and turn them on. <S> Switch off the breakers one at a time, and go round checking which lights have gone out. <S> You can label the breakers to make life easier next time. <A> The device you link to will work - same circuit is loosely written by Marketing probably... <S> Same household supply is probably more accurate. <S> As I put in my comment I used a pair of these, one downstairs and one up, both on different power circuits fed from the same supply board in the house. <S> Fit them and test.
| To determine if sockets are on the same circuit switch off the circuit breakers one at a time to cut power to one socket and then check if the other one is cut too.
|
Does anybody know what this IC is? I tried different search keywords combinations, but I couldn’t find a datasheet, nor a hint of what this IC could be. Maybe someone who’s familiar with it could give me a clue. The IC is on a board that was part of the control panel of a home theater. The options were volume up and down, surround 2.1 or 5.1 and mute. The brand was National Star, as far as I can recall. Thanks in advance. <Q> Judging by the ceramic resonator and load capacitors connected to pins 15 and 16, it's almost surely a microcontroller with a custom program in OTP or flash memory. <S> And typically the program memory is read-protected for the security of the manufacturer's firmware. <S> So more than likely it's not the sort of thing you'll be able to buy from other sources than the original, such as a used board or repair part. <A> Looks like a remote control decoder. <S> LED1 though LED4 have drive transistors and LED5 <S> doesn't - I'd theorise <S> it's an IR LED to receive remote codes. <A> Every time you push each of the switch the led indicator will glow that indicate which program was used. <S> It's a signal decoder ic and not commonly failed in a circuits. <S> My advice is to resolder each components. <S> Nevertheless, check the vdc supply and test each transistor on the pcb for failure. <S> Good luck!.
| Most likely it is a led decoder for each switch on the board.
|
Why do you use a buffer when the whole point of op amp is to amplify signal? This might sound like a dumb question but I just learned about buffers and learned that they have same Vin and Vout from the amplifier providing nearly identical Voltage source value to the next op amp or sensor. But I don't understand why we use this in the first place. If you want to amplify your signal in the first place why don't you directly hook up a non inverting or inverting amplifier to directly amplify the signal right away? Why do you have to put a buffer and then pass down your signal to the next opamp to amplify signal? Wouldn't it make more sense to directly use the non buffer amp hooked to a source that you want to measure and directly amplify the signal? <Q> There are several ways to "amplify" a signal, not only its amplitude is important. <S> Buffers essentially amplify "current", or "input power". <S> They typically have a very high input impedance and low input capacitance, so they don't load much the signal source even at high frequencies. <S> At the same time they have a fairly low output impedance (typical target is 50 Ohms), so they can drive coaxial cables and "deliver" the signal over fair distances without distortions, like in "active" oscilloscope probes. <S> This the the main purpose of "buffers". <A> Why do you have to put a buffer and then pass down your signal to the next opamp to amplify signal? <S> Wouldn't it make more sense to directly use the non buffer amp hooked to a source that you want to measure and directly amplify the signal? <S> You don't have to, sometimes. <S> But there are other times when you do. <S> It depends how you want to process the signal. <S> Let's say you have a source you don't want to load at all, so you'd need a high input impedance to your amplifier. <S> If you're happy for the amplifier to have non-inverting gain, then you can build it with gain, and still have a high input impedance. <S> If you want inverting gain, let's say you want to add several signals together, then an inverting gain stage has a low input impedance, and you'd need to precede it with a buffer stage. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> A buffer is a form of servo control. <S> The input to the buffer is designed so that it consumes very little power, thus making life easier for the op-amp which provides the input signal. <S> But the output is capable of delivering lots of power to the downstream load, even if the load is difficult due to low impedance or frequency-dependent impedance. <S> It may not change the voltage, but it protects the signal source from feeling the impedance presented by the load. <S> It is similar to power steering or power brakes in a car. <S> You still control the car with the steering wheel and brake, but the power assist makes it easier to move the control. <A> Amplifiers serve many functions; making a signal larger, or smaller, orchanging from low impedance to high impedance, or from high impedance(current source) to low impedance (voltage source). <S> One useful function of a buffer (voltage gain of 1) amplifier is to prevent interfering signalsat stage 2 from propogating to stage 1 of a signal-processing chain. <S> So, a buffer between a source and a long wire connection will serve toprevent antenna-like pickup on the wire from disturbing the source. <S> Buffer outputs can be safely probed without disturbing device operation. <S> One might use a buffer before an element that has multiplestates (like, low-power sleep mode) to protect the signal from beingcorrupted by the dead element, or during power transitions, or to prevent the signal from powering-up circuitry intended to be inert. <S> Another useful function is to limit the signal; a buffer can deliver outputthat is guaranteed to be within a known range (voltage, current, slew rate limits)for input to low-speed, low-voltage, or logic elements that are intolerant of somepotential signal characteristics. <S> Finally, integrating power output onto an operational amplifier chipruns a risk of thermal feedback effects; a buffer amplifier with heatsinking can be the perfect output stage, thermally distant from thesensitive input pins of the first-stage amplifier. <A> Buffers used right after an opamp, and used within the feedback loop, are a way to minimize the THERMAL DISTORTION of the signal chain as contributed by the opamp. <S> For DC to 1,000Hz audio signals, and particularly for 100Hz signals that have onchip (in the silicon) propagation-delays THRU THE SILICON as heat, that couple from the hot output transistors to the input differential-pairs, you as signal-integrity designer need to evaluate the intermodulation between low-frequency and high-frequency tones in audio. <S> And in high-precision measurements, the settling-time will be degraded by thermal tails.
| A buffer output can drive a shield to minimize or null stray capacitivecurrents.
|
DC 3-12V Water pump connected to power source doesn't work I'm trying to make a simple connection of 3-12V Water Pump to a power supply( https://www.amazon.co.uk/dp/B07D3XS91J?ref_=pe_1365641_54848841 ). I'm using power supply which is 5V ( https://www.amazon.co.uk/dp/B01ELGR26I?ref_=pe_1365641_54848841 ). I have made simple connection, whith positive on power supply to positive on pump, and same with negative. When I turn on the power supply, it's LED turns green for fraction on second and water pump starts, but then it goes off and repeats. As Im absolute beginner, please point me in any direction. Thank You. <Q> Your power supply does not provide enough current for the motor, it's as simple as that. <S> EDIT: <S> Links to a vendor like amazon (or ebay or aliexpress) are often of little value. <S> You should try to provide links to an actual manufacturer's data sheet. <S> It's buried toward the bottom, but the listing for your motor does tell you how much current it needs...and not to bother trying to use a supply that provides less current. <A> The Amazon link for the pump recommends a power supply of 5 volts that can supply 2 amperes. <S> The Amazon link for the power supply says that the maximum output of the supply is only 700 ma. <S> Solution: get a more powerful power supply. <A> Do you have any junk wall warts, just about any would work for you. <S> Cut the plug off and connect the motor. <S> Just don't go over 12v <S> and if you get the + - wrong it will just run backwards. <S> correct your motor needs 2 amp or better, many ww will meet this.
| So that's your problem: the power supply cannot supply the current required by the pump. If you can't find the manufacturer's data sheet then don't buy the product.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.