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Does the EM radiation from AC power lines propagate through air or is it fixed around the line? Because of the AC current flow, there is a changing magnetic field around the conductor, that in turn induces a changing electric field and vice versa But does the EM radiation get "loose" from the conductor and propagate through air like with a dipole antenna for example or is it fixed around the conductor getting weaker with distance? <Q> <A> Since the summ of all currents in the distribution network is zero, then the magnetic fields caused by each conductor cancells out. <S> The complete cancellation however occurs at infinite distance. <S> But looking practically they cancel in close vicinity, some hundreds of meters from the powerline. <S> Since the poweline is not an antenna it does not irradiate the EM power. <S> Like you have mentioned a dipole, it's a setup with two wires going in opposite direction. <S> It also has to be matched, for example dipole has to be half wavelength, you can do a small calculation how distant it would need to be. <S> EDIT: <S> To sumarize your question: The field is arround the conductor, only. <S> It does not propagate through the air as antenna. <S> If you look at Maxwell equations, then this is so called near field region. <A> Suppose I'm using 10 amps average, or 100 amps peak, in an audio system. <S> The 100 amps peak comes from the need to recharge the amplifier energy storage capacitors <S> ---- perhaps 100,000 uF --- at the peaks of the 60Hz sine waves. <S> The electrician who installed the 10 amp wiring inside the wall of the house was running short of "RETURN" wiring, so he simply ran the "HOT" wire thru the wall behind the audio system, to the power-outlet on the wall, and then used a short piece of RETURN wire back to the circuit-breaker box. <S> Thus the magnetic field of the HOT wire is not cancelled by the magnetic field of the RETURN wire. <S> If your vinyl record RIAA preamplifier has a 4" by 4" loop in its STAR ground, and that STAR ground loop is located 8" from the HOT wire in the wall, how much non-hum evil-singing (because of the harmonic-content) is induced into the preamplifier's ground? <S> Use Vinduce = <S> = <S> [MU0 * MUr <S> * Area / (2 <S> * PI * Distance)] / <S> dI/dT <S> This simplified to 2e-7 * Area/Distance <S> * dI/dT. <S> Let dI/dT = 100 amps/10 microseconds. <S> What is Vinduce? <S> Vinduce = <S> 2e-7 <S> * <S> 0.1meter <S> * 0.1 meter/ 0.2 meter * 1e+7 amp/sec <S> Vinduce = <S> 2e-7 <S> * 0.05 <S> * 1e+7 = 0.1 volt. <S> All because the fluxes did not cancel. <A> Key is to calculate Poynting vector S = <S> E x H. <S> It turns out, on a closed current loop from the energy source to the load, the potentials of the +- cables and directions of currents give electric and magnetic fields that give a pulsating Poynting vector, at double the AC frequency, pulsating everywhere between zero and a maximum value and always only pointing from source to load by net value. <S> Directions of both fields are such that Poynting vector has no total average flux radiating outside from the circuit. <S> The so-called quasi-static fields do extend outside, but with no Poynting vector outwards flux.
| EM radiation will continue to propagate at the speed of light and weakening with the inverse square of distance from the lines, unless it is screened by something.
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Constant current source with MOSFET, OPAMP and Instrument Amp UPDATE Based on the information from everyone answering below, I have modified my circuit and the oscillation has completely vanished! I started performing small modifications in order of easiest to perform and the problem went away. New schematic @Andyaka and @Maria both commented on problems that could arise with using a MOSFET such as the one I was using for linear applications. I replaced the MOSFET I was using with a different MOSFET that also had a gate capacitance 10 times smaller than before. This initially helped a small amount with the oscillation but it was not the main culprit. At the mention of almost everyone I removed capacitors C3 and C6 from the schematic and tested the circuit, this actually made the oscillations worse and at much higher frequency (which is to be expected I suppose). With the oscillation still present I reduced the gain on the INA chip by a factor of 2 (I was heading towards removing the INA as per @Andyaka's suggestion). The final step was following the idea of @sstobbe to use an Error-Amplifer Compensation configuration of the op amp by inserting a capacitor between the output and the feedback input. I chose a 0.22 \$\mu\$ F capacitor as that in combination with the 1k resistor gave a crossover frequency of 700 Hz. After plugging everything back in and switching it on again the oscillation had vanished! There is still some small oscillation on the op amp output but it is almost non-existent. I will upload the updated schematic and comment further if I make further improvements.Thank you all for the help! Original Question I have been working on a constant current driver that I require for driving up to 10 A through a coil (equivalent resistance 3 \$\Omega\$ ) with current stability better than 1 kHz. Slew rate is not enormously important as any current changes I need to make will be on the order of 1A/ms. I have decided on using an Instrumentation Amplifier ( INA114BP ) to measure the voltage dropped over 3 parallel shunt resistors ( PWR4412 ) with a parallel resistance of 0.2 \$\Omega\$ . The measured voltage is then fed into an Op Amp ( LT1097 ) and compared to an input analog voltage (0-10 V) from a DAQ card. The Op Amp output controls a MOSFET ( IRFP7718pbf ) in the linear regime. I first simulated the circuit to test that it operates as I expect, and have since built the circuit. On the surface it works as expected, outputting the current I expect however it is far from stable, with oscillations at 2.7 kHz on the op amp output that translates to large current oscillations in the coil. I have done a large amount of reading today trying to track down the source of the oscillation but as I have no formal electronics background (Physics background, unfortunately avoided electronics throughout my degrees), I have reached the extent of my understanding. I previously had a capacitor and resistor in series between the output and the negative input of the op amp, but after struggling with 10 kHz oscillations I removed them. This changed the oscillation frequency to 2.7 kHz. The inductance of the coil seems to have little impact on the circuit as substituting in equivalent resistors instead of the coil has the exact same behaviour. My current theories are the following: The gate capacitance of the MOSFET is enourmous, 28000 pF, and the Op Amp has big problems driving it. I have tried a small series resistor of 20 \$\Omega\$ in series with the resistor, and have since tried adding a capacitor to ground to filter out unwanted frequencies. The low pass filters I have in the circuit are a hindrance rather than a help. The LT1907 features a compensation capacitor for driving capacitative loads. Looking at the data sheet now I can't tell what the effect of leaving it floating would be. I believe if I connect a capacitor \$\approx\$ 100 pF to the compensation pin I can help slow the slew rate and reduce oscillations. Are there any immediately obvious problems with the circuit drawing I have attached? Any feedback on my design and insight into the workings of systems like this would be greatly appreciated. <Q> However, I am not certain that the IRFP7718 is good for linear application. <S> The data sheet does not mention using in linear applications. <S> Try substituting a MOSFET that clearly states it is designed for linear application. <S> For example : IXYS has a class called "Linear L2" that is designed for use in linear applications. <S> The stated features of the IXYS are : Designed to sustain high power in linear mode operation Low static drain to source on-resistances <S> Avalanche rated Applications : <S> Current sources Circuit breakers <S> Soft start applications <S> Power amplifiers <S> Programmable loads Power regulators <S> Motor control <S> Power controllers <S> Which includes "current source" such as yours. <S> I am not endorsing IXYS. <S> I have no relation to the product nor company. <S> Other manufacturers have linear rated MOSFETs as well. <A> In this answer on a similar subject , I helped the designer improve the situation by adding a BJT between the op-amp and the MOSFET like so: - <S> This then stabilized the situation where previously the design was oscillating. <S> You might take some tips from that design and note that the problem was caused by the op-amp output being loaded from the MOSFET gate thus adding enough phase shift that the negative feedback became positive at the oscillation frequency. <S> In your design you have an extra device that could make this situation worse; namely the INA114 and I <S> would ask you to reconsider putting this in the signal chain because, it doesn't bring anything at all useful to the party. <S> For your choice of MOSFET, you do have to be careful. <S> Those devices designed for switching applications tend to be vulnerable to thermal runaway when driven in so-called linear applications. <S> This answer given by me covers the topic reasonably well and gives links to several papers that cover the thermal runaway topic in more detail. <S> In short, just because it's a MOSFET doesn't mean it can't have thermal runaway. <S> I was involved in reviewing the design for a similar job where this issue came up and, several devices were failing test but, more importantly, several hadn't failed and were installed in rather sensitive applications. <S> My name was mud for a couple of weeks because it meant a product recall that hurt the company in question a lot. <A> Personally, I would not try to search for the component values that will render that circuit stable. <S> You will have an easier time in the long run configuring the LT1907 as an error amplifier with a cross-over frequency of a few 100 Hz to start. <S> Here is a basic overview of the 3 classes of error amplifiers to choose from. <S> Source: <S> https://www.powerelectronics.com/power-management/introduction-control-algorithms-switching-regulators 1) <S> I would remove C3 and C6 from your schematic. <S> 2) I would configure the LT1907 as a Type 1 error amplifier. <S> 3) <S> Leave R8 unchanged. <S> 4) <S> Select C1 of error amplifier for a cross-over frequency of 1 kHz. <S> 5) Increase R3 to 1 <S> \$k\Omega\$ for isolation of the amplifier's output from the capactive load of the MOSFET. <A> Note that at 10A, you should use a 4 point Shunt.
| You are trying to use the MOSFET in a linear application rather than switching application (which you seem to already know). Try adding a 1nF ceramic cap in parallel to C6 (then try with and without C6), increase R3 to 10k and remove C3.
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Current sensing on negative 1.1kV DC rail I searching for a way to do high side current sensing on a -1100V DC rail. My HV is generated from the hamamatsu C9619 High voltage power supply module and it powers the hamamatsu R647P Photomultiplier tube together with the E849-90 D-type socket assembly also from hamamatsu. My current region is up to 500 μA. Any ideas or suggestions would be extremely helpful. <Q> If your need is only for occasional measurements, it would be easy enough to create a battery-powered wireless milliammeter. <S> For example a shunt resistor, op amp and a microcontroller with some kind of wireless communication (eg. WiFi or Bluetooth). <S> The Hamamatsu C9619 is capable of 2mA current <S> so you have a fair bit to play with. <S> You would have to use (continuous) high voltage <S> rated resistors for the power dropper (and, to be prudent, in series with the high-voltage N-channel transistor drain). <S> A very low power op-amp and a very low Iq shunt regulator for the power would complete the circuit. <S> Recieving it would be a transimpedance amplifier referenced to ground. <S> Pick the series drain resistor to limit the current to a few mA. <S> For example: simulate this circuit – Schematic created using CircuitLab <A> You could use a hall effect current sensor . <S> 500uA isn't much, but you can use multiple passes through the loop to get a large enough reading. <S> Another option would be to use an isolated current shunt. <S> You can either do this with a dedicated device , or simply use a microcontroller, and then pass the signal through an opto-coupler. <S> You'll need an isolated power supply for the microcontroller; or you could just run it on a battery. <A> 2 - Buy a clamp-on DC uampmeter. <S> Make sure you use it in "floating" mode. <S> 3 - Design and build your own DC uampmeter.
| Another approach would be to make a current sink circuit using a high voltage MOSFET that operates at the -1100V rail. I have 3 solutions for you: 1 - Replace the power source with one that has both voltage and current meters.
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How to power an Atmega328P and Pi from 12V? I have a bare-bones Arduino (basically an Atmega328P) and a Raspberry Pi 2. I have them communicating with each other over I2C and a 5v/3.3v logic level converter. Right now I power the Pi (micro-usb) and Atmega (5v from FTDI) separately. I have a 12V 6A DC PSU (5.5mm barrel jack) that I would like to use to power both the Pi and the Atmega (and my LED strip lighting). What's a good way to go about doing this? I have a few LM7805 chips on hand if I could use those, but I'm open to ordering new stuff if I have to. Do I convert to 5v from 12v and then use that to power both Pi via GPIO and Atmega or is there current draw issues with doing that? <Q> It looks from the comments you stated that 1.5A at 5v is not enough for you <S> so It would be better to look for some of the higher power usually isolated DC/DC converters for example : <S> 15w dc-dc converter will give you (3A at 5v) and are readily available though not cheap around 15-30 <S> $ meanwell <S> SKM15A-05 . <S> However using a linear regulator is next to impossible , since the heat loss is around (12-5)*2 = 14W!! <S> for 2A load , and (12-5)*3=21W!! <S> for 3A load . <S> and thats alot of power , you will need a heatsink the size of the respberry Pi and a fan to cool that . <S> There also the option to use 2 stage , 1 DC/DC converter 6v for example followed by a 5v linear regulator if you need to have clean power rail with less noise .. <S> have fun :) <A> You want a DC-DC converter; linear regulators will waste a lot of power and require very large heatsinks in this application. <S> You might need more power than the Recom part can provide, for example, or you might find something cheaper or in a more convenient package. <A> itA switching power supply is clearly the only option especially for the power hungry Rpi. <S> There are plenty of pre-made modules that will make it an easy task. <S> If you are unsure on the current rating, just plug your circuits to a 5V power supply and measure the power consumption, multiply that figure by 1.5, and get a module for that ballpark figure or higher. <S> Unless there is a good reason for using a 12V power supply, I suggest that you buy a 5V power supply which can deliver at least 2A. <S> This might also be the cheapest option. <A> Digikey carries these 12V to 5V switching regulators that would work well https://power.murata.com/data/power/oki-78sr.pdf <S> Pololu also carries similar switching regulators.
| A quick search on Digi-key found cheap this self-contained module (a Recom Power R-78E5.0-0.5) that will do what you need, but I'd advise browsing around a bit in the power supply section of the distributor of your choice.
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Can I convert 72V to 12V using potential divider? In our automation project, we'll be using a buck converter to convert the DC voltage from 72V to 12V. This method of conversion is quite expensive!! We are converting the voltage to power the raspberry pi 3 B+ and some sensors. So first, we need to charge our power bank through a car charger. I was thinking if we can use a potential divider circuit for this purpose or a cheaper way? What are the drawbacks of doing this? it seems like not possible or having critical drawback :) <Q> I would generally recommend an isolated converter <S> unless you know exactly what you are doing, sharing return paths between high power high voltage systems and low voltage systems can be a recipie for trouble. <S> Unfortunately 72V DC puts you in an awkward place, it's too high for most common DC-DC modules, but too low for mains power supplies. <S> So converters are going to be specialised and hence expensive. <A> A voltage divider would be a really bad choice here. <S> The Raspberry Pi, at worst case, requires about 1.8A. To provide 1.8A to the Pi through resistors, you would have to dissipate at least P = IV = (72V-12V)*(1.8A) = <S> 108W! <S> (Roughly twice the power of a good soldering iron!) <A> In essense: no. <S> You will need to use very large resistor values, elsewise the current draw from the 72V source and heat dissipation on the voltage divider will be too large. <S> By making a potential divider with huge resistor values, you make a power supply with a ridiculous output resistance, since the output resistance of a potential divider is equal to R1*R2/(R1+R2), where R1 and R2 are the resistor values in the potential divider. <S> Let's try to design a potential divider for your case. <S> The current drawn by the potential divider itself will be 72V/(R1+R2), while the heat dissipation will be (72V)^2 / (R1+R2). <S> 1W power losses is already very bad, but let's stay with that value for reduced complexity. <S> From (72V)^2 / (R1+R2) <S> = <S> 1W we see that R1+R2 = 5184 ohm. <S> Since you need 12V output, R2/(R1+R2) = 12/72. <S> From there two equations we get that R1 = <S> 4320 <S> Ohm, R2 = 864 Ohm and <S> the output resistance R1*R2/(R1+R2) = <S> 720 Ohm. <S> 720Ohm output resistance means that this potential divider PSU could handle no more than 12V/720Ohm current, which is equal to 16mA. <S> This is way too low for the Pi to even turn on. <S> Furthermore, any current ripple will kill the power from the Raspberry Pi, since the voltage ripple will be equal to 720 Ohm * current ripple. <S> You may try to increase the power dissipation. <S> Then, for example, for 0.5A current draw <S> you can reverse-solve the equations I gave before and get a power dissipation of (72V - 12V) <S> * 0.5A = <S> 30W. <S> Even a linear regulator won't suite you here, since 60V drop is just humongous <S> (I think I've never even seen one that could handle such a high voltage). <S> The power dissipation (and loss) will be the same: (72V-12V)*0.5A = 30W. <S> So, you need to use a DC-DC converter. <S> And they cost a lot because handling 72V is no easy task in power electronics.
| In general voltage dividers are a poor choice for power supplies because to acheive a stable output voltage over varying load the current wasted in the divider must be much greater than the current delievered to the load.
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Why does an LED have to be a diode? I know LED stands for Light Emitting Diode; but why does this component need to be a diode to emit light? My question assumes that the "leds" we see everywhere (for lighting, screens, etc) are actually diodes -- this assumption might be wrong. <Q> The existing answers miss the core of the question. <S> An LED needs to be a diode, specifically because the way the charge carriers recombine in the forward-biased diode junction releases the correct amount of energy to create photons in the visible range. <S> Passing a current through a chunk of semiconductor with no diode junction in it would simply produce heat. <S> It's also important for efficiency that the semiconductor be a direct band gap material, so that energy is not lost to phonons (crystal vibrations — heat) rather than photons. <S> Regular silicon diodes emit light, too, but because the band gap is too low, the photons emitted are in the infrared range, and invisible to the eye. <S> Also, silicon is an indirect band gap material, which greatly reduces its efficiency at producing photons at all. <A> why does this component need to be a diode to emit light? <S> By conservation of energy, light emission implies power input. <S> It is normal to deliver electrical power through two wires, sothe simplest electric powered light emitter has two wiringterminals, i.e. is a diode. <S> Two-terminal semiconductorsreplaced two-terminal tubes (vacuum or gas-filled)having two electrodes, which were called 'diode', and the namehas stuck. <S> Electroluminescent panels of yesteryear were also semiconductors that gave off light, but weren't produced in the kinds ofhigh-tech assembly lines that electronic diodes are made in. <S> So, those weren't called diodes. <S> White "LED" devices around you are not simple semiconductordiodes, but are structures with diodes and phosphors that giveoff useful amounts of white light, having a blue-emitting diodeand red/orange/yellow/green phosphors that convert the blue light. <S> Lenses and other features for effective light emissionare common; LEDs do not resemble other practical diodes, exceptthat they have two wires or connecting terminals. <A> LED stands for Light Emitting Diode; but why does this component need to be a diode to emit light? <S> Because LEDs are a diode which posses the same characteristics a common solid state diode. <S> My question assumes that the "leds" we see everywhere (for lighting, screens, etc) are actually diodes -- this assumption might be wrong. <S> Your assumption is correct. <S> A diode is an electronic component that has low resistance in one direction. <S> It is a dual electrode (anode and cathode) device where electron flow from cathode to anode is low conductance and the primary electron flow is the high conductance flow from anode to cathode. <S> The most common diodes are made of crystallized semiconductor materials (e.g. silicone, germanium and gallium arsenide, indium phosphide, sapphire, and quartz) which are doped with p and n type impurities which are separated by the simplest semiconductor building block, the p-n junction. <S> There are many types of diodes with various characteristics. <S> It is the properties of the p and n dopants and their affect on the voltage-current characteristics of the p-n junction that separates one type of diode from another. <S> The above applies to all diodes including LEDs. <S> When the electrons are crossing the p-n junction, many of the electrons are transformed into sub-atomic particles called photons. <S> Light Emitting Diodes are called diodes because the are indeed semiconductor diodes that also emit photons in the form of UV, visible light, and IR. <A> A diode is the simplest semiconductor device. <S> And, the simplest semiconductor device can also be made to emit light. <S> Voila! <S> Light emitting diode! <S> One could potentially call it a light emitting semiconductor device(LES), but that would just sound like a real name... <S> sooo not cool. <S> [edited] From the comments below, semiconductor devices without diodes exist. <S> However, they seem to be sensors and do not emit light. <S> The peltier module does transfer heat, but does not create it. <S> So, it is conceivable that a light emitting semiconductor device may be created in future which is not a diode. <S> However, it would likely have a new name (not a diode or LED). <S> So, thus far, in a TV screen and everywhere else, if it is called an LED, the light (visible or invisible part of the spectrum) actually originates from the LED itself. <S> It is the cheapest efficient light source to produce. <S> There can off course be "fake LED"... <S> neon bulbs or mini incandescent bulbs that look like LED. <S> I have seen those in christmas light decorations, and they are a bit cheaper than the real thing.
| In LEDs the dopants have electroluminescence properties.
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Using decapped ICs in production We were looking for a very specific type of ADC in a small package for one of our projects, and found something suitable in a TSSOP. We wanted to save more space, so looked into getting bare dies; the manufacturer confirmed the dies are 2mm square, but said we'd have to order "some millions" to make it worth providing them. We needed maybe 500/yr and the budget is not huge, so that was the end of it and we decided to do something else. But I was curious: What do people do when they want small numbers of bare dies? Does anyone decap ICs and use the dies in production? If so, can the process be made reliable, and roughly how expensive is it? If anyone has examples of products or case studies, that would be really interesting. <Q> I can't speak for all manufacturers or all product lines, but I've worked as an applications engineer at Maxim Integrated Products for 25+ years. <S> You mention that the product in question is some kind of ADC, so there will be lots of internal adjustments performed after packaging, during the final test. <S> (e.g. bias trim, reference adjustment, linearity, etc.) <S> And that post-packaging final test program uses secret "test mode" commands, which are company confidential. <S> (If you were a primary/strategic/key customer those might be available under NDA, but you'd be having that conversation with the business manager, not me.) <S> This will very likely degrade its performance, permanently. <S> Modern IC design uses MEMS technology to relieve mechanical stresses which are internal to the package, those mechanical forces on the chip would otherwise degrade performance. <S> If you're trying to get decent 20-bit (or even 12-bit) performance from an ADC chip, subjecting it to that kind of mechanical violence could ruin its linearity, making the whole exercise futile. <S> You might be able to get away with decapping a pure digital chip, but for precision analog I would strongly urge you to reconsider. <S> I just now looked at our online product selector guide (precision ADCs) and found a few 12-bit/16-bit SAR ADC which are smaller than 4mm2 (the only requirement you mentioned). <S> This includes WLP Wafer Level Packged parts, which is pretty close to bare die, but just a little bit nicer to deal with. <A> I have used de-capped IC in pico-probing for silicon debugging. <S> (Where you remove the top and passivation layer and then put probe needles on the die) <S> The decapping is done with special hot-acid pump and a special rubber 'window'. <S> The idea of decapping is to have a more or less complete package but have access to the silicon. <S> You save no space. <S> You have the whole package but just with a hole at the top. <S> The bond wires where still there, so no clean die. <S> You could try throwing a bundle of chips in boiling acid and see what comes out. <S> But my guess is the bond pads will not be usable anymore. <A> The manufacturer won't make a new package variant on its own since it has to do all characterization again. <S> It cannot guarantee the same specifications in a different package, this requires testing and validation. <S> They might be willing to do this at a smaller scale, at a higher price to offload the risk. <S> You will need to pay upfront, or sign contracts. <S> Decapping to recover dies is not the only step. <S> You also have to remove it from the leadframe, which is glued. <S> And re-do the wire bonding. <S> Removing wire-bonding is something I have not heard of before. <S> The amount of specialty equipment and skill required to develop and perform this operation will be significant. <A> I believe IS I or Quik-Pak may be able to work with you for repackaging, and they are both used to smaller volume customers. <S> Another poster pointed out a potential show-stopper, the factory tuning on the ADC. <S> The new package may require careful attention to achieve the specs of the original.
| Decapping the chip out of a TSSOP and ripping it off of the leadframe (typically a conductive epoxy bond) will definitely subject the chip to mechanical stresses beyond its design limits. Depending on the specs of the ADC, the packaging may be codesigned with the IC.
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Convert Battery to AC @ same voltage I have some LED fair lights that run on 30V @ 0.4A supplied from an AC mains supply. They have been wired to flash using some form of notification to the AC (I'm not interested in them flashing). I want to wire them from a battery I have at 30 V (off a balance board) but a direct wiring only lights half of the lights. If I switch the polarity the other half (only) light. It's going outside so I don't want the original adaptor which needs a minimum of 110 V. Is there a simple fix to make this work? It's only a simple project - so cheap and simple. <Q> Simple & cheap with stuff from DigiKey is to build a MOSFET H-bridge that's driven by a 555 timer running at 60Hz. <S> But that's a lot more complex than the other simple & cheap. <S> You need a pair of gate drives that have built-in shoot-through protection, you need to generate a +12V gate & control electronics supply, and you need four suitable FETs. <S> Then you need to wire it all up. <S> There <S> may be a simple & cheap eBay solution that'll generate 30V square-wave AC from your 30V battery, but if there is I don't know it. <A> The LED light that runs from AC is likely to be a design where LEDs are wired as pairs in antiphase. <S> This means that DC will only illuminate 50% of the LEDs and to illuminate all the LEDs you need AC i.e. 50% of the LEDs illuminate on the positive cycle of AC and the other 50% illuminate on the negative cycle of AC. <S> Here is an example: - If you want all the LEDs to light you need to supply an AC voltage. <A> You did not provide the information on the lights wiring. <S> Some of the LED lights have removable design, in case you need to replace them. <S> Try removing one bulb that does not light up. <S> If the rest of lights will go out too then simple rotating of all non-lighted bulbs in their sockets might work for you. <S> But then you'd need double voltage to light them up to full brightness.
| Simple & cheap with stuff from Wallmart is to use a 12V battery, a car inverter, and the wall-wart.
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Max voltage of capacitor? Is there a max voltage drop across a capacitor?Or does it always charge up to the same voltage as the supply?If there is a max voltage, then what would happen if the supply voltage far exceeds the max voltage of the capacitor, would the dielectric material break? <Q> The cap will try to charge to the supply voltage. <S> On its way to that voltage, if the voltage exceeds the voltage rating of the capacitor, the capacitor will eventually fail. <S> At that point it will be permanently damaged. <S> It may have even externally ruptured. <A> Is there a max voltage drop across a capacitor? <S> Or does it always charge up to the same voltage as the supply? <S> Yes, there is a breakdown voltage associated with capacitors, you must not exceed the rated breakdown voltage ever. <S> Usually it is printed on the capacitor itself, or found in the datasheet, or by identification of a color scheme if you know what company makes it. <S> If there is a max voltage, then what would happen if the supply voltage far exceeds the max voltage of the capacitor, would the dielectric material break? <S> If you exceed the breakdown voltage, the dielectric or other capacitor material breaks down and it turns into a resistor and could short. <S> I have seen some capacitors explode this way . <S> Tantalum capacitors should be derated , I usually go ~70% of the breakdown voltage. <A> A capacitor will charge up to the supply voltage.
| If you exceed the maximum allowable voltage for the capacitor, it will break (read explode) and become like a resistor/short Capacitor Resource: https://www.electronics-tutorials.ws/capacitor/cap_7.html
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Is there any difference between voltage drop and voltage across a component? Voltage drop across a component describes the reduction of energy when current moves through passive elements, is this the same as voltage across? What about for capacitors, technically current does not actually travel through the dielectric, can this term still be used for capacitors? <Q> Yes, they're the same, although I think that most people would want to reserve the phrase "voltage drop" for the voltage developed by running current through a dissipative element such as a resistor, relay, or diode junction. <S> As a counter-example, referring to a "-9V drop across the battery" would be somewhere between odd and positively strange. <A> Voltage drop across a component describes the reduction of energy when current moves through passive elements, is this the same as voltage across? <S> What about for capacitors, technically current does not actually travel through the dielectric, can this term still be used for capacitors? <S> You can measure voltage across capacitors, which is dependent on the time value of the current. <A> Along with the other fine answers, I would add that one term may be more suitable than the other, depending on how it is used. <S> For example, I was recently teaching some basics to a beginner and the conversation included statements such as: “Due to the voltage drop of R1 , the voltage at point <S> A is now...” <S> and, “What is the voltage across R1 ?”... <S> In the first case, I used “voltage drop” to help keep focus on the fact that our reference for the calculation we’re interested in (i.e. the voltage at point A ) is still ground. <S> And I used the other to imply that the reference is just beyond the element R1 .
| Voltage drop and Voltage across the component mean the same thing, assuming the same component and the voltage across that component.
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How to solder a 70 °C (158 °F) thermal / temperature fuse For my DMX project (schematics is not important), I soldered this temperature fuse which has just two leads (see picture below). The fuse doesn't work (meaning it does not conduct electricity) ... which of course is logical since my soldering station has a minimum temperature of 200 °C (392 °F) (although I soldered it with 350 °C (662 °F), forgetting about this). But how should I solder this component? <Q> This was a problem with transistors in the early days and standard advice was to use a pliers as a heat shunt close to the body of the component. <S> Figure 1. <S> Protecting a capacitor from overheating when using a big ignorant soldering gun. <S> Image source: Mother Earth News . <S> Oddly enough, the image above was the only one I could find in an image search. <S> The technique may be getting lost. <S> A long nosed pliers with a rubber band on the handles may suffice and the rubber band will hold it in place freeing up one hand. <A> I think those thermal fuses are normally connected using crimp terminals, rather than by soldering. <S> Even with the heatsinks the other answers mention, it would be much too easy to damage the device while soldering wires to it. <A> You can use a heat shunt - really just a pair of pliers on the lead, and keep the soldered joint as far away from the body as practical, but ideally use crimp joints, soldering these things is always tricky. <A> Soldering stations which can be set to much lower temperatures than 200°C are not unheard of (and wide temperature range somewhat correlates with quality). <S> If you had one of these, you could use low-temperature solder like chipquik. <S> For a one-off job and considering your space constraints, you could simply solder copper wires where the fuse needs to be, then twist these wires with fuse terminals , which is not as good as crimping, but does get the job done in practice. <S> Twisted parts can then be trimmed to save space (3-4 twists is enough) and heat-shrink could be put on if insulation is required. <A> My suggestion, given your lack of space is to attach with conductive silver epoxy . <S> This will give you a fairly robust conductive connection with no application of heat. <A> As was mentioned in other answers, crimp termination would be best. <S> Followed by a pin or screw terminal. <S> Since you seem stuck with the design and don't have space for a pair of heat sink pliers, there is one other option if you must solder it. <S> Wrap the component as much as possible with a lightly damp kimwipe or other low lint cloth. <S> Make sure it's distilled water and allow it to dry completely before energizing the circuit. <S> Not ideal, <S> but I've used it successfully in the past for tight spaces and heat sensitive parts. <S> You may even be able to chill the water provided your fuse can handle cold without damage. <A> Do as everyone else does. <S> Wrap the therm with a ziplock bag of ice. <S> Then solder away. <S> A wet wrag placed in a fridge is the preferred method. <A> Use Bismuth solder along with above mentioned heat sinking, 185 degrees melting point.
| Another idea is to bend the fuse terminal in a snake-like pattern to save space, and then solder the ends of therminals, while cooling down the therminals entering the device (e.g. by holding them with a wet cloth).
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Verilog router design and best way to handle variable size packets in verilog? I have a synthesizable Verilog/logical design question. My question is more logical than syntax. I wish to implement some sort of router that has three input/output ports of full-duplex UART RS232, that are sending packets to each other. Main points of design: The packets are with variable size and they contain the following fields: source(1 byte), destination(1 byte), payload length(2 bytes), payload. payload length is variable between 4-2000 bytes. I want to have some sort of routing table "what destination will be routed to which of the UART ports", and some sort of firewall(not sure the proper name) something like "which sources allowed to send packets to which UART ports". loopback is allowed (the packet that is received and sent through the same port) broadcasting is allowed (maybe destination address that routed to more than one UART port) the module needs to be generic so I could expand it to more than 4 ports. For me, I think the hardest part to start with - is how to handle variable size packets? If the packets were fixed size, and not too big, I would have had an input FIFO and an output FIFO for each UART port: reg [packet_size-1:0] fifo[num_of_packets] each input FIFO would have stored in each fifo[I] a whole packet that was received from the uart, and a module that reads whole packets from each input FIFO, checks for source the dest of each packet and using that routes the packet to the wanted output FIFO,and from there a UART transmitter that takes a packet and sends it byte after byte.Problem is that a packet can be really really big, so I don't want to hold a fifo that has 2000 bytes in each cell... That also brings the question of how to transfer the packets between the internal modules?As said, a packet can get really big. I realize there are a lot of questions in my post, but I will much appreciate any help. Also, if there are some reference designs or something similar that can help me, I will happily review it. Thanks for the help. <Q> Supposing that multiple input ports may receive data at once, you can't guarantee that you can transmit from one port to another in real time. <S> Because two ports could try to route to the same destination at the same time. <S> Therefore you will need a packet buffer for each input port. <S> You certainly don't want a 2000 byte x <S> N FIFO because that would waste a lot of memory. <S> The consequence of wasting memory of course is that you can't buffer as many packets. <S> In situations where there is a lot of traffic <S> this could lead to dropped packets. <S> That way you won't have wasted space when storing smaller packets. <S> To implement the circular buffer just keep a few indexes. <S> Let index P1 point to the start of the first packet in the buffer. <S> Let index P2 point just past the end of the last complete packet in the buffer. <S> Let index P3 point to the last byte received. <S> As the packets come in process their header in real time. <S> Only put them in the buffer if you have space (which can be calculated from the P1, P2, and your buffer size). <S> If you can fit the packet then put its bytes into the buffer as they come in. <S> P3 is incremented for each byte. <S> If a packet is only partially received and a timeout occurs you can reset P3 back to P2. <A> Simply put, "Worst-case analysis", and in this case, I'm guessing you have little-to-no performance requirements, and just want to receive data and transmit it. <S> 2kB is not a lot of memory, but, the preamble seemingly has all the information you need to transmit "real-time". <S> Once you know where it is going (destination), and assuming you have no need to interpret or operate on the data, all you need to do, after receiving the preamble, and setting up your transmitter , is collect a byte/word (whatever your data size is for the payload) and transmit it, over-and-over, until you have met the byte count. <S> You should be able to do this with a "ping-pong FIFO" architecture... <S> probably, with a latency close to the preamble length (i.e. a pipeline delay). <S> Minimum storage (memory and/or registers) for what I described is ~ <S> (number of bytes in preamble) + (2*number of bits for data) <A> Well, you need to use FIFOs, but they will store bytes instead of packets <S> and you'll use state machines to control transfers so that the correct number of bytes are transferred for each packet. <S> It is also possible to split the header from the payloads and store them in separate FIFOs, which may or may not be useful depending on the application. <S> But is this really a good application for an FPGA? <S> UARTs are usually quite slow, and you can get microcontrollers that have a lot of hardware UARTs (i.e. Atmel xmega).
| Instead of using a FIFO with each element being a packet, use a circular byte fuffer.
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I designed a touch PCB keypad what can I do for silking and covering I want to design a custom keypad for my board. I put my board in a Plastic box 15cm x 19cm and the box has a simple box door. I want to put a keypad and a 2x16 character LCD in this door to control the board.the box must be water resist. I want to change or modify this door so that it looks beautiful. currently I want to design a Capacitive touch keypad but what is next or how can I put it in the box. My question is what kind of keypad do you suggests and how much is the cost? I can even change the box if a better suggestion. <Q> Let's assume that the box is fully waterproof, I would still recommend using a box as the one Michel suggested. <S> The main reason for this is the transparant cover. <S> You really don't want to drill holes in the cover and try to make everything water proof <S> so what is easier then mounting everything inside. <S> With the transparant cover you can easily read the lcd <S> and you can mount capacitive buttons on the inside of the lid. <S> Example in figure below. <S> *image taken from Source image regarding the touch buttons themselves, the capacitive touch buttons are quite easy to do, they are being used extensively for this purpose. <S> I would suggest reading the "TI slaa576a" Design Document Design Document the document describes the theory of operation and the ways you can design your buttons, which materials you can use and so on. <S> As far as IC's go there are lots of options to choose from, fully assembled breakout board based on the "Atmel AT42QT1070" or the "microchip MTCH101" or a lot more options. <S> Pricing and cost will be determined by your final solution but should definitely not break the bank. <S> The IC's I listed are both available for less then 1 euro. <A> I would think more like this enclosure box . <S> You have to find one that fits your size better (this one is too small, just as example). <S> It has a rubber inlay, and more important, no holes through the box. <S> (example picture): About the LCD, just make sure you glue it very good <S> so it is waterproof. <S> Typically, 1602 and 2004 displays are used, unless you want a graphic screen (OLED maybe). <S> About the keypad, you cannot use probably ones with feedback/'moving' keys, so a 4x4 keypad might do. <S> Costs: < 1 <S> euro/$. <A> If you can do a bit of micro-controller programming, there is a very neat way to do that with capacitive touch sensing. <S> If you don't want to program or do electronics, you can use an evaluation kit which is ready to use.
| Are you sure your box is waterproof. You can sense the touch trough plastic, metals like aluminium, glass, keeping the electronic safe from water damage and making the design very simple.
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Can a resettable thermal fuse be soldered? Earlier related questions: How to solder a 70 °C (158 °F) thermofuse Using a thermal fuse in a PCB terminal block? History: I 'normally' soldered a nonresettable thermal fuse, which (of course) failed after soldering. I have ordered a resettable thermal fuse RY Tf, for 73 degrees (Celcius), after getting a comment from PlasmaHH in question 2 that such components exists. However, I'm wondering, what happens if I solder these? During soldering, the temperature will go much further than 73 degrees, but will it 'reset' or keep on resetting until the soldering process is finished and it has cooled down enough. Thus will the thermal fuse function normally after soldering? <Q> The answer to all these linked questions is to read the datasheet ! <S> This will (should) include max temperatures, hold temperatures, times etc. <S> While these may be hard to exactly achieve by hand, it will indicate the regions that are acceptable. <A> Yes, they can be soldered. <S> So long as there's no permanent damage to the insulators in the fuse it'll just reset after it cools. <S> The sensing element is a bimetallic strip that has some hysteresis by virtue of being formed slightly convex, it snaps away from the stationary contact as the inside metal expands more than the outside. <S> I've soldered a number of these without issue. <A> It's easy to use a removable fuse with clips. <S> Solder the clips with high heat for minutes on end <S> and they won't care. <S> Put the fuse in afterward. <S> You'd have to be doing something very strange with your iron to make the clips themselves fail mechanically.
| All (well made) components will give details as to how they should be soldered or attached. It may seem silly, but: depending on the size of the fuse, the easiest thing to do is to not have heat on the fuse at all.
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Difference between internal and external oscillator for a microcontroller? I am a beginner with respect to microcontrollers, so I just want to ask is there two option for choosing the frequency of the microcontroller: to be internal or external? If there is internal frequency available in the controller, then we don't need to use external crystal oscillator? <Q> It depends on what you want to use the processor for, the manufacturer, etc. <S> RC internal oscillators are "only" good for 1-2% accuracy, there are some processors that have tuning adjustment registers that allow you to shift the frequency, so with the right software/calibration, etc. <S> you can get better results. <S> For (asynchronous) serial port type communications, the RC inaccuracy can become critical at high baud speeds (generally 230kbps and above, depending on processor and baud rate clock flexibility) With RC oscillators <S> you typically get 1 or 2 frequencies unless the MCU offers PLL capabilities in which case there can be a wide range. <S> This can be used to advantage to dynamically scale the clock speed for battery powered situations where you can't go to 'sleep' due to complicated monitoring situations, but don't have to be running flat-out. <S> A crystal can get you any frequency you would want within the MCU operating limits, however you need to verify that the loading is correct so that the oscillator starts up correctly and is accurate and stable over time and temperature. <S> The RC internal oscillator is generally guaranteed to work by the manufacturer with no hardware design issues. <A> Main difference is that the internal oscillator is usually a RC type oscillator which is not very accurate. <A> Most practical differences I have encountered: Frequency stability (especially across different temperatures). <S> It depends on your application if you need a very stable clock or not. <S> I had issues once using an ATmega8 running from the internal oscillator with an UART that lost sync below zero. <S> For VGA a crystal is a must. <S> Power consumption - internal oscillators usually will be more power efficient than external crystals Available frequencies. <S> For example to get ideal 115200 <S> baud UART timing on an ATmega8 <S> you will need a clock with an "oddball" frequency of 7.3728 MHz, while the closest frequency the internal clock can only deliver 8 <S> Mhz (so some bauds will not be available). <S> Depending on the exact MCU you can also get great clock stability if you use it as an USB device. <S> The internal oscillator can lock on to the clock signal from the USB host (that is crystal-stabilized).
| External oscillators can be of type quartz crystal which are far more accurate.
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Operating a fan on UPS A short, useless intro to the question, feel free to skip it: I had a power outage in my home yesterday. Power outages aren't what they used to be - Back in the day, it rendered me completely unable to do anything, but nowadays I have my laptop, which is still connected to the internet via my phone, which is connected to a strong mobile recharger I have (10 Ah) - and I can hardly care less about the rest of the power. Except for the part about my fan. It gets really hot where I live. And I currently simply have no solution for this. Small USB fans don't do the trick. And I just can't figure out how nobody I know has a solution for such a simple, supposedly common problem. Is mankind simply not advanced enough to take on such problems? Will we be able to operate fans during blackouts? What level on the Kardashev scale must we advance to before we can solve this? [/end-rant/intro] Is it possible to operate a normal, standing fan during a power outage? How? Is connecting it to a UPS a good idea? I've read elsewhere that the waveform generated by UPS devices doesn't behave well with these sort of devices. And how long is a fan expected to last on a standard UPS? <Q> A UPS as I understand <S> it is an battery attached to an inverter. <S> I have used portable fans on an inverter with no issues. <S> Assuming they are designed for the same frequency. <S> If you have a AC motor designed for 60 Hz and a inverter operating at 50 Hz it would just alter the speed of rotation. <S> One thing I don't recommend is plugging an electric digital clock into an inverter because a minute will pass every 30 seconds or so. <A> A battery UPS will work, but a UPS has a high loss even with no load. <S> It's meant for temporary solutions. <S> Either to safely shutdown, or bridge the gap for the diesel to start running. <S> Fans for comfort are a luxury, sometimes this luxury is unavailable to those who cannot improvise. <S> Get some 12V PC box fans and fangrill that you can run from a cheap lead acid battery. <S> Either silent breeze, or tornado amounts of moving air on request. <S> Or go to the scrapyard and take the radiator fans from an old car. <S> These are also 12V. <S> Or you can go the old fashioned way, for fans and lighting . <A> It depends upon the power of the fan. <S> In Thailand a Hatari 18" fan consumes between 78 W and 114 W (depending upon model) , so at 240 V, that is a maximum of 0.3 <S> A - 0.5 <S> A (let's say 1 A for simplicity). <S> The capacities of UPS appear to be measured in kVA rather than Ah, even though it really is just a battery with an inverter. <S> Going by this kVA calculator , for (single phase) <S> 230 V and 1 A: <S> the Power is 0.16 kW <S> and; the kVA required is 0.23 kVA. <S> There are plenty of small 10-20 kVA UPS out there. <S> With respect to how long it will last: Lets say you get a 20 kVA UPS, and for simplicity's sake, the fan is 100 W. <S> So the fan consumes 100 W every hour. <S> The kW of the UPS is <S> 20 kVA x 0.7 <S> = 16 <S> kW <S> Therefore the UPS will last 160 hours. <S> That seems like an awfully long time - I may have made a mistake <S> Notes <S> Apparent Power (VA) = <S> Load (W) / <S> Power Factor <S> where Power Factor = 0.7 <S> UPS Backup [in hours] = <S> Battery Ah <S> * Volts * Power Factor/Load <A> It makes alot of noise and runs at 1/3 of the power with no difference on the speed control you select on the fan. <S> It also doesn't last 160 hours. <S> I have 3 UPS Cyper Power with 1000VACyper Power with 650 VAa Zircon with 360VA <S> They all give the same results. <S> They last less than 1 hour. <S> For some reason the electronic circuitry and the internal fan seems to drain the battery. <S> Same as when I plug my Wi-Fi router. <S> It last about 45 minutes. <S> (it uses hardly any power, about the same as a fan 1 Amp.)
| If your laptop is OK with it then your fan will tolerate it OK. A 220 volt Hatari fan running on UPS doesn't work well.
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Why is a resistor divider used on gate of P-channel MOSFET of discrete USB power path circuit? I noticed this discrete power path circuit on the design of the ESP32-A1S Wi-Fi+BT Audio Development Kit : Sorry, but unfortunately none of the designators are numbered and the resistors does not have values. I took the liberty of numbering them D1, Q1, R1 and R2. I have also explicitly indicated the body diode of Q1 (as pointed out by @Damien ). I understand the job of D1, Q1 and R2. When USB is disconnected, current is sourced from VBAT to VCC5V via the body diode of Q1. R2 pulls the gate of Q1 to 0V and Q1 switches on so that there is a much lower voltage drop than the body diode of Q1 (a.k.a. "ideal diode function"). When USB is connected Q1 is switched off and VBUS supplies current to VCC5V via D1 (with a regular diode voltage drop). The preferred current path is via D1, because VBUS >= 4.4V and VBAT <= 4.2V. Update: Here is the design that I used (I have indicated the normal voltage ranges of +5V_USB and +VBAT; Gate, Source and Drain of MOSFET explicitly shown): The designs I have seen and used does not have R1 (R1 = 0 Ohm). There must be a good reason why they included R1. Can anyone offer an explanation and what resistor values are good? Update: Thanks @Cristobol for pointing out voltage spikes (e.g. ESD) on VBUS / +5V_USB. A low value resistor for R1 (e.g. 1k) will offer some protection for Q1 with Vgs_max = +-8V (of course only if Q1 has an internal zener connected between Gate and Source to protect the Gate). Update: Thanks @Dorian ! I think your answer is excellent! A resistor divider prevents a transient dip on VCC5V during the time that USB is disconnected and VBUS (battery) must take over and supply current. <Q> I understand the job of D1, Q1 and R2. <S> When USB is disconnected, R2 pulls the gate of Q1 to 0V and Q1 switches on so that VBAT. <S> Actually it's not the case. <S> VBat always supplies the VCC5V due to the protection diode in Q1, the only difference is that when it's pulled down, it will have less loss as the mos will switch. <S> Supposedly it is supposed to work like this:As the battery is 3.7V, when you connect the USB, R2 (0ohm) will pull the gate up to 5V, and thus disconnecting the battery from the circuit if the gate threshold is below 1.3V, drawing current from the USB instead of the battery. <S> But this implementation is quite strange and if VUSB increase for some reason, or the battery voltage decrease, you might overload the battery, and have a nice fire. <S> Also the gate threshold voltage vary a lot with temperature. <A> I've seen this divider used before in cases where the gate-source voltage would be exceeded with a direct connection. <S> It seems a bit overcautious for 5V, but perhaps that FET has a protection diode from source to gate that would provide a path from VBUS straight to the battery. <S> Of course, it may just be that it was copied from a higher-voltage circuit. <S> My money's on the protection diode. <A> While VBUS drops from 5V to 0 Q1 will open to late at VBUS below 3.3V (given a gate threshold voltage of 0.7V) and VCC5V will have a drop to. <S> R1 and R2 makes a divider which raise the VBUS threshold voltage around 4V. <S> It's true as Damien noted that it's a design flaw that the Q1 reverse diode might cause trouble if VBUS is to high or the battery is severely discharged. <S> simulate this circuit – <S> Schematic created using CircuitLab See here the simulation results for R1 = 0 <S> (VBUS on X axis, battery current and load voltage on Y axis), the load voltage drops to 2.8V while VBUS is going from 0 to 5.2V and the same simulation with R1 = <S> 15Kohm <S> the load voltage is steady until D1 is conducting.
| It may also be protection from spikes or noise on VBUS.
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STM32F4 I2C read out with least overhead I have a sensor connected over I2C, where I read out data at a constant address. The uC is an STM32F410CB with 100MHz. A control loop runs at around 16kHz without reading out the sensor and drops to 5 kHz when I always read out the sensor. I2C is configured in the fast mode (400kHz). So I decided to use the DMA, where a function is available in the HAL libraries that look pretty promising: HAL_I2C_Mem_Read_DMA(i2c1,addr,addr_mem,addr_mem_size,data,data_size); So I implemented this function, which is also called every cycle the control loop runs. The frequency of the control loop increased to 7kHz now, but that's still not enough for my application. Only 3 Interrupts are generated for the read out, I don't know where this huge overhead comes from. When I only call the function every third time, the control loop runs twice at 16kHz, then once at 7kHz. I need a more stable result. Is there a way to fully autonomous read out a sensor over I2C at a constant register address, such that nearly no overhead is generated? /** * @brief Reads an amount of data in non-blocking mode with DMA from a specific memory address. * @param hi2c Pointer to a I2C_HandleTypeDef structure that contains * the configuration information for the specified I2C. * @param DevAddress Target device address * @param MemAddress Internal memory address * @param MemAddSize Size of internal memory address * @param pData Pointer to data buffer * @param Size Amount of data to be read * @retval HAL status */HAL_StatusTypeDef HAL_I2C_Mem_Read_DMA(I2C_HandleTypeDef *hi2c, uint16_t DevAddress, uint16_t MemAddress, uint16_t MemAddSize, uint8_t *pData, uint16_t Size){ uint32_t tickstart = 0x00U; __IO uint32_t count = 0U; /* Init tickstart for timeout management*/ tickstart = HAL_GetTick(); /* Check the parameters */ assert_param(IS_I2C_MEMADD_SIZE(MemAddSize)); if(hi2c->State == HAL_I2C_STATE_READY) { /* Wait until BUSY flag is reset */ count = I2C_TIMEOUT_BUSY_FLAG * (SystemCoreClock /25U /1000U); do { if(count-- == 0U) { hi2c->PreviousState = I2C_STATE_NONE; hi2c->State= HAL_I2C_STATE_READY; /* Process Unlocked */ __HAL_UNLOCK(hi2c); return HAL_TIMEOUT; } } while(__HAL_I2C_GET_FLAG(hi2c, I2C_FLAG_BUSY) != RESET); /* Process Locked */ __HAL_LOCK(hi2c); /* Check if the I2C is already enabled */ if((hi2c->Instance->CR1 & I2C_CR1_PE) != I2C_CR1_PE) { /* Enable I2C peripheral */ __HAL_I2C_ENABLE(hi2c); } /* Disable Pos */ hi2c->Instance->CR1 &= ~I2C_CR1_POS; hi2c->State = HAL_I2C_STATE_BUSY_RX; hi2c->Mode = HAL_I2C_MODE_MEM; hi2c->ErrorCode = HAL_I2C_ERROR_NONE; /* Prepare transfer parameters */ hi2c->pBuffPtr = pData; hi2c->XferCount = Size; hi2c->XferOptions = I2C_NO_OPTION_FRAME; hi2c->XferSize = hi2c->XferCount; if(hi2c->XferSize > 0U) { /* Set the I2C DMA transfer complete callback */ hi2c->hdmarx->XferCpltCallback = I2C_DMAXferCplt; /* Set the DMA error callback */ hi2c->hdmarx->XferErrorCallback = I2C_DMAError; /* Set the unused DMA callbacks to NULL */ hi2c->hdmarx->XferHalfCpltCallback = NULL; hi2c->hdmarx->XferM1CpltCallback = NULL; hi2c->hdmarx->XferM1HalfCpltCallback = NULL; hi2c->hdmarx->XferAbortCallback = NULL; /* Enable the DMA Stream */ HAL_DMA_Start_IT(hi2c->hdmarx, (uint32_t)&hi2c->Instance->DR, (uint32_t)hi2c->pBuffPtr, hi2c->XferSize); /* Send Slave Address and Memory Address */ if(I2C_RequestMemoryRead(hi2c, DevAddress, MemAddress, MemAddSize, I2C_TIMEOUT_FLAG, tickstart) != HAL_OK) { if(hi2c->ErrorCode == HAL_I2C_ERROR_AF) { /* Process Unlocked */ __HAL_UNLOCK(hi2c); return HAL_ERROR; } else { /* Process Unlocked */ __HAL_UNLOCK(hi2c); return HAL_TIMEOUT; } } if(Size == 1U) { /* Disable Acknowledge */ hi2c->Instance->CR1 &= ~I2C_CR1_ACK; } else { /* Enable Last DMA bit */ hi2c->Instance->CR2 |= I2C_CR2_LAST; } /* Clear ADDR flag */ __HAL_I2C_CLEAR_ADDRFLAG(hi2c); /* Process Unlocked */ __HAL_UNLOCK(hi2c); /* Note : The I2C interrupts must be enabled after unlocking current process to avoid the risk of I2C interrupt handle execution before current process unlock */ /* Enable ERR interrupt */ __HAL_I2C_ENABLE_IT(hi2c, I2C_IT_ERR); /* Enable DMA Request */ hi2c->Instance->CR2 |= I2C_CR2_DMAEN; } else { /* Send Slave Address and Memory Address */ if(I2C_RequestMemoryRead(hi2c, DevAddress, MemAddress, MemAddSize, I2C_TIMEOUT_FLAG, tickstart) != HAL_OK) { if(hi2c->ErrorCode == HAL_I2C_ERROR_AF) { /* Process Unlocked */ __HAL_UNLOCK(hi2c); return HAL_ERROR; } else { /* Process Unlocked */ __HAL_UNLOCK(hi2c); return HAL_TIMEOUT; } } /* Clear ADDR flag */ __HAL_I2C_CLEAR_ADDRFLAG(hi2c); /* Generate Stop */ hi2c->Instance->CR1 |= I2C_CR1_STOP; hi2c->State = HAL_I2C_STATE_READY; /* Process Unlocked */ __HAL_UNLOCK(hi2c); } return HAL_OK; } else { return HAL_BUSY; }} Edit: Here is the code for the I2C readout: Low Level: //DMA I2Cerr_t I2C_Driver_TXRX_DMA(uint16_t addr, uint16_t addr_mem, uint8_t* pData, uint16_t size) { addr = addr << 1; err_t err={.value=HAL_OK}; err.value=HAL_I2C_Mem_Read_DMA(i2c1,addr,addr_mem,1,pData,size); //Error callback called in error case return err;} Higher Layer: uint8_t buf_rx[2]={0};uint16_t pos=0;void triggerDMAMeasurementPosition() { if(getI2CErrorState()) { I2C_Error_Solver(ENC_ADDR); i2c_err=0; } else { I2C_Driver_TXRX_DMA(ENC_ADDR,RAW_ANGLE_ADDR,buf_rx,2); }}void HAL_I2C_MemRxCpltCallback(I2C_HandleTypeDef *hi2c) { pos=buf_rx[0]<<8 | buf_rx[1];} I2C Init: /* I2C1 init function */static void MX_I2C1_Init(void){ hi2c1.Instance = I2C1; hi2c1.Init.ClockSpeed = 400000; hi2c1.Init.DutyCycle = I2C_DUTYCYCLE_2; hi2c1.Init.OwnAddress1 = 0; hi2c1.Init.AddressingMode = I2C_ADDRESSINGMODE_7BIT; hi2c1.Init.DualAddressMode = I2C_DUALADDRESS_DISABLE; hi2c1.Init.OwnAddress2 = 0; hi2c1.Init.GeneralCallMode = I2C_GENERALCALL_DISABLE; hi2c1.Init.NoStretchMode = I2C_NOSTRETCH_DISABLE; if (HAL_I2C_Init(&hi2c1) != HAL_OK) { _Error_Handler(__FILE__, __LINE__); }} Edit 2: BTW, the slow down happens only when HAL_DMA_Start_IT(..) is called inside HAL_I2C_Mem_Read_DMA(..). So it is probably somehow related to the interrupts and what happens inside there. <Q> No, the STM32F4 I2C implementation does not allow a fully autonomous operation. <S> There is a state machine which has to be implemented in software for the peripheral to do its work. <S> This can either be done with interrupts or with polling. <S> From my experience handling it with polling results in a more robust operation. <S> Polling of course has the problem that your controller isn't doing useful work, but if your system is just waiting on new input values, it wouldn't matter. <S> You can implement that yourself to try and fix your problem. <S> If you go down that road, be prepared for some hardship as the I2C peripheral is not a masterpiece in terms of usability. <S> Going down that route also allows you to get the I2C clock higher than 400 kHz - it is outside of the normal specification <S> but there is a note that you can get details from ST on how to run it at 1 MHz. <S> I asked them and never got feedback, so I fiddled around until I ended up with 800 kHz running from 8 MHz. <S> I guess with 100 MHz you should get 1 MHz quite easily. <A> First of all - 16 bits + 2x address (16 bits) + memory read address 8 bits + <S> ACK / NACK * 4 <S> I think <S> + 2xSTART <S> + 1xSTOP <S> (if I did not miss anything) <S> which makes about 40 clocks. <S> 400kHz / 40 ~= 10k sequences / second. <S> Reading two bytes in using DMA is useless. <S> It will probably take more time setting it using HAL and processing the callbacks than the transfer itself. <S> It is much easier and efficient if you program it the bare register way. <A> The I2C data rate is probably not high enough to support your desired sampling frequency. <S> Looking at your code, it seems you are reading two bytes of data ( ENC_ADDR , and RAW_ANGLE_ADDR ). <S> Combine that with the address and R/W byte gives 24 bits of data. <S> Now, 400 kHz / 24 is 16.7 kHz <S> so it's tight even with no overhead. <S> I would suggest trying to sample reliably at, say, 8 kHz. <S> Alternatively, looking at the AS5601's datasheet , it supports 1 MHz I2C; if the microcontroller supports that, you could push the data rate up. <A> The HAL_I2C_Mem_Read_DMA function has blocking calls to send the slave address and memory address, and then only uses DMA for the data transfer. <S> At 400 kHz it will block 45 microseconds to send those two bytes, which is quite a problem when your loop time is 62.5 microseconds! <S> HAL I2C code: <S> https://github.com/STMicroelectronics/STM32CubeF4/blob/master/Drivers/STM32F4xx_HAL_Driver/Src/stm32f4xx_hal_i2c.c <S> Look at the function HAL_I2C_Mem_Read_DMA, which calls I2C_RequestMemoryRead, which calls several blocking functions. <A> I'm not familiar ST chip and library, but translating from what I know from other system. <S> Based on the code, particularly from the lock-unlock sequences, it seems it is being used in a RTOS, which is probably the framework of ST for their chip. <S> RTOS are great, but it needs some understanding for dealing with code that needs to run fast. <S> Sometimes, if the code is fairly simple and you need speed, it is better to avoid using the RTOS. <S> RTOS basically is a task scheduler (a mini kernel), and every time it needs to switch from a task to another, it will use some time. <S> What happens in this code, is that it request the handle on the I2C port, as it wait, it probably will execute other tasks. <S> If that port is used by other tasks, it needs to wait until that other task is done. <S> Depending on how your hardware is wired, if the I2C line is only used for this sensor, you can bypass the I2C handle request, open the I2C one time and then just keep reading it, this will probably optimize quite a bit. <S> You can also put this process on a specific task with high priority. <S> If it's used by other chip you need to access, you can also keep everything in a single task, and handle yourself when which chip is called, also avoiding having to unlock / lock the I2C bus. <S> EDIT <S> Since it's not an RTOS and added code to your question: <S> Try rewriting the I2C driver to avoid all the mutexes. <S> I'm not sure the DMA is a gain in there, since you seem to have to handle the packet byte by byte anyway, it loose sense. <S> Try avoiding interrupts, processor time switching into interrupt can be quite long and it is always faster to poll it from the main loop to avoid context switch.
| The solution would be to rewrite the I2C driver so it can use interrupts for the addresses. Conclusion - you cant reach this rate using I2C. Do not use HAL.
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Is rated current supposed to vary based on voltage? Recently I have been reading a lot of datasheets and noticed that many manufacturers specify a current rating for their products (especially connectors) without specifying at what voltage the current rating applies (except for a max voltage). This seems super counter intuitive for me as when the voltage increases you will have more power (watts) moving through the component that, in my mind, should affect the rated current. So at a lower voltage you should be able to pass more than the rated current. Some examples of this: https://www.molex.com/webdocs/datasheets/pdf/en-us/0475070010_PCB_HEADERS.pdf https://www.vishay.com/docs/31017/rcwp99.pdf (this one specifies a rated power but that is just for the voltage drop across the jumper. What is the rated voltage at that current?) https://www.hirose.com/product/en/download_file/key_name/FH34SRJ-10S-0.5SH%2850%29/category/Specification%20Sheet/doc_file_id/43118/?file_category_id=9&item_id=05801251550&is_series= These datasheets make it seem as if there is a physical property that attenuates current flow no matter the voltage when I think it is based on voltage and current. <Q> There are two aspects to the maximum allowable voltage specification a given resistor has. <S> One is the packages' power dissipation rating, which for a given value of resistance will have a maximum applicable voltage before heat dissipation values are exceeded: <S> P=V^2/R <S> The second is a safety persective because the smaller the package the closer together <S> the two terminals are. <S> Eventually the electric field gradient between them becomes so strong that you can get electromigration effects, tracking, etc. <S> that will cause a short across the terminals. <S> There is a corollary in that post manufacture cleaning of boards becomes more critical as well <S> , debris tends to become conductive over time with some soldering techniques. <S> In the case of the jumpers, Vishay RCCe3 series does have the voltage rating, could be the other datasheet is just missing the row (typos in datasheets happen all the time, most get caught some don't.) <A> Voltage rating for a connector is based on the insulation and spacing between contacts, and does not depend on the current. <A> I think that I am just confused. <S> Thinking back to USB fast charging, the current in the wire is what you are trying to avoid by increasing the voltage. <S> By increasing the voltage <S> (im pretty sure) <S> you don't change much. <S> But by increasing the current you have a larger voltage drop across the conductor with V = <S> I <S> * R. <S> So to charge at high currents fast chargers send higher voltage lower current power <S> and then the device will step the voltage down to something that the battery can accept.
| The current rating for connector contacts (and relay and switch contacts) is dependent on current, as current will cause heating of the contacts - this is independent of the applied voltage.
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One resistor in series and two in parallel for LEDs So I've read that using one resistor to limit current to two or more LED's in parallel is bad practice, so I wondered what about this kind of circuit: Is this also bad practice? Each LED has it's own current limiting resistor, but the current is being limited further by a resistor in series. Are there any differences compared to just having each LED connected in parallel directly to the voltage source with a 355Ω Resistor? Also (just to double check I am doing my calculations right as I am still a student learning this stuff), assuming each LED has a forward voltage of 2V, is each LED drawing about 20mA here (7V across resistors, Rtotal = 355Ω so 7V / 355Ω = 0.0197 Amps)? <Q> No, this is fine. <S> All LED's are not created equal, the I-V curves have a tolerance. <S> It is probably 'slightly' better to have the LED's go directly off of the rail but this configuration works. <S> The most important thing is to not have LED's directly in parallel. <S> Also (just to double check I am doing my calculations right as I am still a student learning this stuff), assuming each LED has a forward voltage of 2V, is each LED drawing about 20mA here <S> (7V across resistors <S> , Rtotal = 355Ω <S> so 7V / <S> 355Ω = <S> 0.0197 Amps)? <S> The drop across the shared resistor would be 220 <S> *40mA= 8.8V <S> Something is wrong with your calc because this would mean that you would have more than a 9V drop total. <S> I'd just solve for the node voltage for the 220Ω resistor you get <S> \$\frac{9V-x}{220Ω}=i_{220}\$ <S> Each LED would be \$\frac{x-2V}{270}=i_{LED}\$ <S> The whole equation would be \$\frac{9V-x}{220Ω}=\frac{x-2V}{270}+\frac{x-2V}{270}\$ Solving for x <S> you get 4.66. <S> Then solving \$\frac{9V-4.66V}{220Ω}=0.0184A\$ \$\frac{x-2V}{270}=0.0985A\$ or half of the amount you calculated through each LED. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Equivalent circuits. <S> Your circuit is fine. <S> The 270 Ω resistors will make up for any variation in forward voltages of the LEDs and prevent the one with the lower V f from hogging the current. <S> Your 355 Ω calculation is wrong, however. <S> The effective resistance for each LED is 710 Ω. <S> The current is given by \$ <S> I = \frac { <S> V}{R} = \frac {7}{710} = 10 \ <S> \mathrm <S> {mA} \$ . <S> That will be quite bright for most modern LEDs. <A> Your circuit with three resistors is less bad than using a single resistor for two LEDs, but a failure of one LED will still affect the current in the other LED. <S> Also, you have to remember that the 220 Ohm resistor in your circuit will be carrying the current for both LEDs. <S> If you want 20 mA in each LED (most modern LEDs will be quite bright at 10 mA or less), then you need to plan for 40 mA in the 220 Ohm resistor - with 2 volt LEDs and 270 Ohm resistor, you'd only have 1.6 volts across the single resistor, so it would have to be 40 Ohms, rather than 220.
| Because you have a resistor in series with each LED this takes care of any kind of problems that might arise from matching.
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How can I make Bluetooth (2.4 GHz) reception area narrow and controllable? There are some Bluetooth modules available on the market, like the cheap HC-05 for 3 or 4 USD delivered. This one mentioned has an internal PCB antenna which I guess has a toroid-like radiation pattern (omnidirectional): However I need this module to be receivable in the narrow area only. Like 5-10 square meters area between the module and the ground. What can I do to make this possible? Put some metal case around? Cut off the internal antenna, and connect an external directional antenna? Any other thoughts? <Q> Neither will work. <S> If you build a metal case with a narrow opening, you'll have an even wider beam; the smaller the aperture, the larger the beam. <S> It's practically impossible in this form factor to controlledly cut off the existing antenna, and still have something to attach an external antenna to. <S> But even so: at 2.4 GHz, the ground you mention will work as an OK reflector, and hence, things will still work in the area "indirectly" illuminated by ground reflection. <S> 2.4 GHz devices (such as Bluetooth devices) have to be designed to work with multiple reflections, so it being impossible to limit the area if you're illuminating a wall or a piece of ground <S> is a feature your device has to have. <S> You could try to add so much attenuation to the antenna that the signal is really really weak, but since a reflection might have relatively little loss compared to the first couple of meters of free space loss, this won't work out, either. <S> Long story short: you can't. <A> RF reception is rarely "narrow and controllable" unless you create a hard obstacle for the RF signal. <S> By a hard obstacle I mean, a metal box (Faraday cage) out of steel plates or fine metal wire mesh (like the is in the door of a microwave oven). <S> Only that can completely block the reception. <S> Putting a case around the transmitter will prevent it from working (if done correctly) or make reception bad (if there are some holes left). <S> You'd think that a directional antenna would do the job <S> , that's true if there are no reflections which there will always be unless you're in free space. <S> In the real world you cannot avoid a connection over the reflected signals unless you apply RF absorbers to all the walls in the room. <S> So in practice: this will never work as well as you want it to. <A> This is very likely, impossible. <S> Even if you would attenuate the signal, it would all depend of the sensitivity and power of the other pair. <S> While you might be able to reduce your signal and on a particular device it would be only visible for a certain distance, it would totally change with another device. <S> If you do so for security reason, it would only take someone to have a sensitive bluetooth device, or directional antenna to get the signal. <S> Also RF signal is quite unpredictable, it can be reflected off objects, absorbed or go through. <S> The only plausible way to have a RF signal locked to a certain area would be to have a Faraday cage, for example covering the walls, windows, floors with a conductive paint, or some sort of conductive shielding. <S> It wouldn't be perfect, in which the person would have to put his device in order to get the signal. <S> Something like that (don't laugh at the image <S> , you get the point :P).
| Alternatively, you could build a sort of small Faraday cage enclosure where you can control the signal, have the transmitter inside at lowest level.
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Are audio cassettes FM or AM? I know that cassettes store an analog signal, but is the signal Amplitude-Modulated (AM) or Frequency Modulated (FM)? <Q> The magnetization on the media is directly (and hopefully fairly linearly) related to the waveform amplitude. <S> There is a high frequency bias signal added to the audio signal, to get the resulting signal on the tape to a linear range of the magnetization curve, but the signal at the head is a sum of the two, not a modulation. <S> That image is from here , a good description of the process. <S> This is stressing my memory of tape decks I had 40 years ago... <S> The one instance I recall of amplitude modulation (kinda) was Dolby HX, which changed to amplitude of the bias in response to high amplitude audio signals, to keep the resulting signal from going into saturation. <S> https://en.wikipedia.org/wiki/Dolby_noise-reduction_system#Dolby_HX/HX-Pro <A> It is neither, the signal is recorded as a magnetized pattern without any form of modulation . <S> This can give an impression of how that works: <S> I got that from here . <S> See, no modulation needed, the amount of magnetic flux is proportional to the actual audio signal. <S> Some HiFi Stereo Video recorders do use FM to record the audio though. <A> Analog tape recording for audio cassette doesnt require modulation. <S> But since magnetization ability of the particles on the tape does not have linear response at low signal levels of the audio to be recorded, poor response will result during playback. <S> To prevent this, technique called biasing is applied. <S> This creates very linear response of magnetization for the entire dynamic range required for high fidelity audio. <S> When recording, magnetic tape has a nonlinear response as determined by its coercivity. <S> Without bias, this response results in poor performance especially at low signal levels. <S> A recording signal which generates a magnetic field strength less than tape's coercivity is unable to magnetise the tape and produces little playback signal. <S> Bias increases the signal quality of most audio recordings significantly by pushing the signal into more linear zones of the tape's magnetic transfer function. <S> – Wikipedia <S> So, you may have confused between this high frequency AC bias used in audio recording with modulation.
| Neither, there is no modulation involved.
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Why is my DC motor losing torque when over voltaged I have these small dc TAMIYA motors rated at 3v at around 2-3 amps. I also have a driver that I'm planning to use to drive the motor. https://www.pololu.com/product/2992 I could not find any other driver that has a very low minimum voltage at a high current. As you can see the driver has a minimum input of 6.5v, I am aware that over voltage can reduce the lifespan of the motor, but I don't mind since there are a bunch of these motors here collecting dust. At 6.5v motor does not spin (which is weird). Upon raising it to 7.5 it now spins, very very fast. So I lowered the PWM to 5% (I'm using an Arduino analogWrite with a value of 10) and slowly increased to more suitable RPM. Now as am playing with the voltage I noticed that the motor loses torque as I increased its voltage. Why is this happening ?? Also, while as I was playing with the PWM I also noticed that the torque also peaks at a certain PWM. Like at 10% the torque peaks, above and below values is just weaker. Can anybody please help explain? EDIT: Another observation i have just noticed is that the driver gives out 7v, upon connecting to the motor the motor spins weakly. but upon upon connecting the motor to direct 7v it is much stronger in torque and speed <Q> In a nutshell for a PMDC motor: <S> voltage = speed <S> current = torque <S> Note that for a given voltage there is a speed at which there is (almost) <S> no current going through the rotor, because of back emf (the rotor is inside a magnetic field generating a counter-emf on the brush inputs). <S> To get more torque at that speed you have to increase the voltage (and basically overdrive the motor, leading to early failure). <S> There are a whole bunch of other factors that can come into play such as mechanical resonances at certain speeds, etc. <S> Regarding the startup problem you had <S> /have- is it repeatable or did it just happen the first time? <S> It could have been a poor brush contact, etc. <S> Gory details: http://lancet.mit.edu/motors/motors3.html <A> It's likely demagnetization. <S> Under highload conditions, however, when motor current maybe high, the effect will cause a reduction in thetorque constant of the motor and a consequentreduction in torque output. <S> Above a certain level of armature current, the fieldmagnets will become permanently demagnetized. <S> Therefore, it is important not to exceed themaximum pulse current rating for the motor. <S> What has likely happened, as a result not of overvoltage, but overcurrent that was permitted by the controller, since it has a 13A limiter in it, is that the magnets are now substantially weaker than they started out as. <S> That means more current is needed to generate enough torque to overcome the stiction of the brushes and bearings to get it running, and also that the emf generated by rotation is proportionally that much lower - so it has to spin really fast to get the same emf to balance the supply voltage. <S> If you want to modify the controller - the shunt resistor is 2 milliohms, replacing that with a 7.5 or 8 milliohm will get you to a more reasonable current limit. <A> A DC motor produces the high pitched whine of its PWM frequency when the motor tries to start and the PWM duty-cycle is too low. <S> A DC motor's magnets can also be permanently de-magnetized if it is overloaded and gets too hot.
| The permanent magnets of aDC motor field will tend to become demagnetizedwhenever a current flows in the motor armature. This effect is known as “armature reaction” and willhave a negligible effect in normal use.
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What could be a near rail-to-rail opamp with very low voltage for this use? What could be a near rail-to-rail opamp with very low voltage for this use?LM386 seems to be out of limits for my needs. How do I choose it from datasheets? What values are important (of course limits & power) but what else? Is there something more important than power inputs & current limits? simulate this circuit – Schematic created using CircuitLab on this schematic I did not calculate component values yet. I want to make a very light and portable mic amplifier on 2 batteries.It should consume very low power and have medium/high amplification. What could be a near rail to rail amp with very low voltage for this use? I tried from this datasheet to guess that the LT1494 would fit this need. Is this correct? (If not, why?) LT1494 datasheet For audio, will 100nF be ok for C1 & C2? mic will be an electret connected device may be another amplifier or why not a speaker <Q> IMO, your best bet is to do a parametric search on a decent distributor (Digikey has a pretty good setup). <S> Note that you will want to select rail to rail as a parameter, and <S> then the power supply voltage range- read those carefully as they can be split up into single ended and dual supply types, and you can miss a good candidate such as NCS2001, NCV2001. <S> In the worst case you can divide up the gain stages (helps with oscillation problems too). <S> Most won't have a problem running with split supplies or single supplies, your choice, but you might also add an output coupling capacitor in case you connect something that has DC coupling on the input, which would drain your battery quickly. <S> The LT1494 would likely not work well due to its extremely low gain bandwidth product of 2.4KHz. <A> You can run this circuit on only a single battery, if you wish. <S> The output voltage will not be rail-to-rail, if you use only bipolar devices (even CMOS/MOS devices can only get "close" to rails and even then only if the ZLOAD is a light load. <S> 1 Kohm is not a light load.) <S> simulate this circuit – <S> Schematic created using CircuitLab <S> On 2.4 volts, assuming each gain stage is biased to VDD/2, you will get 1.2/ ( q/K*T <S> ) = 1.2 / 0.026 = <S> 50X gain (34dB) per discrete bipolar transistor To run on a single battery, adjust values of R3 and R4, to move Vcollector closer to VDD, and have Q2 operating with a higher Vemitter, <S> so you can get more output swing. <S> A better approach for a single-battery is to drive the Vout from the Q1 collector (DC-blocked, of course). <S> The load capacitance will strongly affect the bandwidth. <S> ============================================== <S> Here is a better "opamp" <S> simulate this circuit <A> LM10C is one of my favourite op-amps and will run from 1.1 Vdc to 40 Vdc. <S> The output will swing within 15 mV of the supply rails. <S> It's not particularly fast but should be entirely adequate for your needs.
| One important thing to note is that these opamps tend to have low gain-bandwidth products, so make sure the part you select has the chops for your microphone type (you didn't have that info on the schematic).
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Question about straightening the pins of a resettable polyfuse I have resettable polyfuses like the picture below. I am working on some PCB boards where I solder the wires myself on the PCB. Because it is very hard/cumbersome work to solder very small wires, I use the leads of components to make a path to a nearby component, preventing small wires or tin to be soldered. However, I am wondering about the specific shape of the leads of polyfuses. Why do they have the bent in the leads(unlike resistors, capacitors and other components)? Is it that the fuses need to 'stand off' a bit from the PCB, or can I make the leads straight to push the orange part very close to the PCB, to get some extra length of the leads if needed? <Q> For manual assembly, you don't want to distort the leads so as to crack the epoxy case seal (long term reliability problem for contamination and stress cracking of lead to body attachment points) <S> If this is a one-off project with no service costs, then those rules can be relaxed as you see fit. <S> Automated lead forming through hole components allows fast and accurate machine placement, and there are still services that provide custom forming (axial to radial for example) and re-taping. <S> Through hole component assemblies are generally run through a wave solder machine, and forming/staking <S> the leads help retain the correct position preventing components shifting, tilting, sliding up/down. <S> In the case of a PTC, the goal is get a repeatable position for thermal environment (you don't want 5% of your production lot parts leaning against a hot transformer for example) and for the aforementioned epoxy seal integrity. <A> These components as shown are indeed designed to hold the device up off the board. <S> This allows for air flow around the component and could permit a faster cooling and reset of the polyfuse after an overload has been removed. <S> It is recommended to use care to avoid cracking the coating. <A> These components are thermally-sensitive. <S> As any thermally-triggered device that rely on joule dissipation and corresponding rise of body temperature, they need certain ambient conditions and thermally conductive paths to have certain heat flow balance. <S> Obviously encapsulating these devices into some different thermally conductive environment will change their guaranteed parameters. <S> These bents are there for easy assembly to assure that the device has thermal environment for which it was specified. <S> Personally I think it doesn't matter much how to mount them, their turn-off parameters are so imprecise, so small differences in thermal resistance to PCB do not matter much. <S> With shorter leads you will have a somewhat higher turn-off threshold.
| If your hole size and spacing is correct this shouldn't be a problem; if you need to re-form the leads try to do it in such a way that the seal doesn't crack by using two needlenose pliers for example. You can reform the leads to suit a different spacing to the board and between leads.
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How could have Banksy kept an RF receiver running for 12 years? I'm talking about Banksy's "Girl with balloon" painting that shredded itself at auction. I think the general consensus is that he didn't, and that it was set up before the auction (or a couple years ago, not 12 years). But lets say he did, how could he have made it keep its power for that long? What sorts of RF receivers run on such low power? What batteries can withstand a 12 year low-but-constant-load life? <Q> A relatively large lithium primary cell would be my choice. <S> They are specified for something like 10 or 20 year life running water meters, including periodic radio communication. <S> And maybe a second cell to run the motor so it stays relatively fresh. <S> The Israeli company Tadiran makes such products. <S> As long as the temperature does not get too high the shelf-ish life of such Lithium cells is in the decades (they claim 40 year operating life). <S> You can bet that an expensive piece of art will be kept in carefully controlled temperature and humidity conditions. <S> Given that power source, power management would be important but not crazy critical. <S> You would want to keep the average draw in the << 100uA range most likely. <S> Doing the sums, their 19AH TL-4930 works out to 54uA (average) for 40 years. <S> So even if it turned on a receiver once per 10 minutes for a second it could draw tens of mA, provided the sleep mode power was inconsequential. <A> The most likely answer is that he didn't. <S> While that was the claim, it is much more likely that he swapped out the frame soon before the painting went up for auction. <S> Dave Jones from the EEVblog did a video on this: https://www.youtube.com/watch?v=SdKdQWhlNTY <S> UPDATE <S> Here's a follow-up video from the EEVblog: https://www.youtube.com/watch?v=8dbYGefDdWo <A> Why would it be a low-but-constant-load life? <S> Assuming the auction takes hours, it would be enough to awake the RF receiver once every hour for a few ms to check for an RF signal. <S> The transmitter would either transmit the RF signal:- For a complete hour, knowing the RF receiver will be awake <S> at least once- <S> Know in what time period (like every whole hour) <S> the receiver will be awake and send it within that time. <S> In both ways, the transmitter can continuously send the message, only the receiver needs to receive it only once. <S> ** Update after James Trotter's remark <S> ** New algorithm: <S> As above but with the following addition Every hour it checks for an RF signal, except it doesn't start shredding, but after this signal it starts checking every 10 seconds (for let's say the next 10 minutes). <S> This Awake Often signal can be send multliple times. <S> Directly after the auction a Start Shred command is given which is received by the painting and shredding starts. <A> The RF signal itself carries sufficient energy to power the RFID tag. <S> Now we know that a big battery had to be present for the shredder, but we can use a small wakeup circuit to keep that battery entirely disconnected for years, and only connect it when the passive radio receiver is energized. <A> Primary cells, a arming switch, and non-rf method of control, like IR, as there was a plant in the audience. <S> They could easily sit for years and still be good. <S> It was discreetly armed right before auction. <S> But as Banksy's instagram video showed, it didn't work right. <S> In rehearsals it shredded it completely. <S> This is likely due to how long the batteries sat in the frame. <S> Even top quality Alkaline primary cells would show signs of aging 10 years later, enough for large motors to only work half of their intended job.
| Passive RFID tags run on zero batteries.
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How to deal with an inconsistent / noisy power signal? I'm using a 12v power supply to operate a system of sensors that is very sensitive to fluctuations in the voltage input. Unfortunately, I can't control what goes on upstream of my components and this is causing a few issues: 1) On occasion the voltage will dip or spike to levels that are outside my usable range (I've observed fluctuation from 10.5v to 13.5v). The change doesn't last long, but it's enough to mess with my electronics. 2) Additionally, the power line seems to pick up a not-insignificant amount of low voltage interference (EMI?) that I can't control wholly via software. To that end, I would like to introduce a regulator/filter to control my input voltage - keeping it as close to 12v as possible and eliminating the outside interference. I don't have a lot of space to build up a whole new circuit or introduce custom components, so I'm trying to use as many off-the-shelf parts as possible. My thoughts: A) I've used buck converters in the past, and I'm familiar with the operation of boost converters. Could I use a buck converter (dirty 12v -> 5v) and then a boost converter (5v -> clean 12v) to deal with the large fluctuations? I'm also trying to keep the current draw in mind as the power supply has a 10amp fuse in my circuit. What kind of impact would this have on the available current? On heat dissipation in a closed environment? One of the reasons I'm leaning toward this is because I may have need of a clean 5v power line later, and this takes care of the conversion. B) Could I ground out some of the noise with a dc power-line filter? I haven't been able to identify the source of the noise, so I'm hoping to go as broad as possible - would a filter to reduce common and differential mode noise be a good start or should I investigate other options? I know that my points are rather vague - I'm still in the early stages of putting my sensor system requirements together, and I'd like to protect against as much as possible now, while I still have easy access to the raw power line. Any advice / direction is greatly appreciated! <Q> The 12V → 5V → 12V is going to take up quite a bit of space and may waste power. <S> Might be worth looking into other options first such as: a 12V regulator (78xx) a reverse biased zener diode. <S> These are simple, cheap and fairly effective. <S> If your circuit needs to be very temperature stable you may want to look at placing two zener diodes in series, one forward active and one reverse biased. <S> Generally speaking, the larger the nominal zener voltage, the larger the temperature coefficient <S> But capacitors in general are a good idea for stabilising your input voltage, especially when you turn the circuit on if EMI is an issue try checking out using twisted pairs or coaxial cables <A> I could be wrong but what I gathered from your question is that the 10A fuse is in the power supply, and not in your sensor circuits. <S> Presumably, your sensor circuit requirements would be quite low. <S> If they are below 1A, and the power supply is variable, I'd suggest 7812 Voltage Regulator IC and running it at slightly higher voltage, say 17V. I think it'll give you your required stability in input. <S> If the power supply is fixed and you cannot change that, maybe a boost converter and then 7812? <S> P.S. Remember to attach a heat sink with the IC. <A> How about powering it from 24V supply? <S> And what about power <S> - how much do you use? <S> I mean, you could have even a linear regulator down to 12V if it's just hundred(s) of milliamps, <S> waste of power will be (relatively) tolerable. <S> And with some really small caps to filter out 24V's switching supply noise in addition to some regular 1uf or maybe even more, you could probably get good results. <S> And bypass everything. <S> Anyone correct/add?
| if you have a digital circuit or any circuit with lots of switching you’re going to need a capacitor (or a few) to help counteract parasitic inductance (check out dc bus capacitors).
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Does a diode block current but not voltage? When running this test circuit on CircuitLab and my expected result was that only M1 would open but then i noticed that the M4 mosfet is also open even though there is a diode between it and the the 3V the Gate threshold on both is set to 1.5v i also noticed some strange spikes appearing on the current Does it mean a diode blocks current but not voltage ? simulate this circuit – Schematic created using CircuitLab <Q> Figure 3 from the 1N4148 datasheet. <S> Diodes have very small leakage current. <S> At 3 V this will be between 3.5 and 10 nA. <S> The <S> 2N7000 has a gate−body leakage current, <S> Forward of -10 nA max. <S> The diode also has about 4 pF capacitance. <S> When the supply voltage jumps from zero to +3 V on power-up the diode capacitance will cause the gate of M4 to jump up too. <A> The effect of voltage that you care about in most electronics is current. <S> The problem in the circuit shown is that it essentially leaves an unshielded MOSFET gate floating (no current path to or from it). <S> A floating MOSFET gate is almost always a design error, leading to undefined behaviour. <A> The idea of blocking voltage is a nonsense. <S> You need to think the problem in terms of "resistance" and voltage divider (I'm putting quotes around «resistance», because it's not reactive <S> but it's heavily non linear). <S> Since the transistor is a MOSFET, it's gate has a much higher resistance than the resistance offered by the reversed diode (because of the leaking current) and the divider made from the diode D3 and the M4's gate transmit nearly all the voltage to the gate. <S> So the solution is simple : Just put a high value resistor between M4's gate and the chassis ground. <A> Let's look at what "there is a Voltage" actually means: <S> The presence of a voltage between two points of a circuit means we measure a difference in electric potential with the value \$U\$ . <S> So, if there was a conducting path between these points with \$R<\infty\$ there would flow a current \$U=RI\$ . <S> The presence of a potential difference ( \$U\$ ) doesn't mean any electrons could get from A to B, but that one spot would accept the "surplus" from the other. <S> A diode is, in theory, \$R=\infty\$ in one way and \$R=0\$ in the other. <S> However, those approximation isn't true in reality. <S> This can lead to a small current flowing "backwards" through a diode.
| When a device blocks current, it makes voltage inconsequential, save electrostatic effects.
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why is this microcontroller brown-out level detection at 2.6V and 3.4V The default brown-out detection level for the ATmega32u4 is 2.6V (typical) in the range of 2.4V to 2.8V. The table of possible values: 2.0V2.2V2.4V2.6V default3.4V3.5V4.3V Unfortunately there is a large gap between 2.6V and 3.4V. Another possibility is "brown-out detection disabled" which is not in my table but it certainly exists. My application uses an Adafruit board at 3.3V and 8MHz. The application is non-life threatening but accuracy is always of interest so I ought to benefit from brown-out detection, if feasible. This board aside, 3.3V seems to be a popular voltage for a microcontroller Vcc level so the default 2.6V brown-out level seems to be a good choice except for the fact that the datasheet also has this graph. One interpretation is that at 2.65V (which is above the default brown-out level) a frequency of 8MHz is in some way unsafe and the safe frequency is somewhat lower by extrapolation. I don't want to run slower than 8MHz. Another interpretation is that the graph gives you information about the maximum frequency between 2.7V and 5.5V and no other conclusions should be drawn. There should be no extrapolation. My question is: What were the design considerations that caused there to be a large gap between 2.6V and 3.4V on the brown-out level table? It seems wasteful to have to choose a 3.4V brown-out level in a battery powered application. <Q> Common logic voltage levels are 3.3 V and 5 V, and many batteries have 4.2 V or 4.1 <S> V. So the brown-out levels are obviously designed to be 0.7 V below these supply voltages. <S> In general, power supplies are assumed to have tolerances of <S> ± 10 % (this is why the datasheet often says "4.5 V – 5.5 V"). <S> And when your power supply outputs 4.5 V, you want to have some additional safety factor so that your microcontroller does not reset at the slightest dip, but only when the voltage is so low that the power supply is (apparently) completely toast. <S> (The decoupling capacitors buffer the voltage, so a dip of 0.7 V indicates that the actual power supply is already lower, and that the decoupling capacitors are already partially discharged.) <S> In other words: 2.6 V is the correct brown-out level to use for a nominal 3.3 V supply. <A> It might work, but is that good enough for you? <S> It isn't for me. <S> If you need a reliable BOD reset you'll have to supply it externally or use different chip. <S> Perhaps they committed to the settings before they characterized the chip and figured out that it will not operate reliably at (say) 1.8V. <A> Actually when you use battery, your voltage drops gradually. <S> and you dont want your system to stop working while voltage drops just a little bit. <S> and as you can see in the diagram if your clock is 8mhz you can work with supply voltage as low as 2.7 and lower than that is unsafe or unstable. <S> so you'd want brown out detection to detect when your supply voltage is low enough to make the system unstable or unsafe. <S> 2.6v is a good value when your using 8Mhz clock. <S> if you decide to use a higher clock rate then obviously it's good practice to make brown out threshold higher. <S> you can also make the threshold lower if you'd use lower frequencies or want to extend battery life and you'd risk performance in non critical applications.
| You need to look at the tolerance of the brownout voltage threshold as well as the nominal value. The 2.6V setting (and really, any of the settings) is useless for a 3.3V nominal supply at any clock frequency.
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How would I drive about 440 LED's without breaking the bank? Alright, I'm trying to create I guess an LED matrix display(?) so that I can make essentially a text display using 8x5 LEDs per letter, to make 11 letters in total. I'm definitely willing to learn a lot and NEED to learn a lot as I have little experience with something like this. My goal is to be able to display individual letters to make the display show like "Hey there" using 5x8 LEDs for each letter. I'm looking to use possibly an arduino device, but as they have limited inputs, and I need about 440 inputs.. how would I make this work? I've researched a bit and it seems I could do something related to this https://www.circuitspecialists.com/blog/build-8x8-led-matrix . Would this be my best bet maybe? How would I extend this from 64 to 440 in terms of delivering it to the arduino device? Also I would like to ask what arduino device might be best for this, and what accessories I might need that aren't mentioned in that page, if the provided link is a good idea. Edit: Sorry for being somewhat vague. I'd be using 5mm Red LEDs and I only need to be able to turn on and off each individual LED. https://www.circuitspecialists.com/bag-red5mm.html <Q> For example the LE171596A allows you to control 96 LEDs, so you would need ~5 of them. <S> The chip costs around $9 at Mouser. <S> The driver allows you to turn on and off a single LED (or groups of them), so in order to display anything meaningful you need to update the driver registers all the time (think: just like dynamically driving a 7-segment LED display). <S> To achieve that you need a fast MCU with SPI and DMA . <S> It may be too much for a classic AVR (like ATmega328), but an Xmega could do it. <S> Also pretty much any Cortex-M with DMA is suitable for this task. <S> Once you figure out what to build (the components) you need to lay them out on a PCB. <S> I guess that 4-layers will be the minimum to get any decent spacing between the LEDs. <S> Fortunately you can order them easily for China. <S> You've chosen THT LEDs - <S> this brings the problem, that if you pack them tightly there will be no space for other components on the other side of the PCB (capacitors, driver ICs). <S> SMD LEDs are much more suitable for matrix displays. <S> Finally you also need a low-voltage high-current power supply. <S> These are also available off-the-shelf (eg. <S> MeanWell). <S> If the voltage is too high the drivers will dissipate too much power (and go into self-protection or blow up). <A> Some options: use a LED matrix, with your own row and column drivers. <S> 8 row + 55 column outputs needed, a common approach is to use shift registers for the columns. <S> TPIC6B595 might be a good choice because it can sink a lot of current. <S> = <S> > best choice if you want to learn a lot use LED driver chips. <S> MAX 7219 drives 8x8 LEDs, so you would need 7. <S> A module with a max7219 + 8x8 LEDs can be bought for less than $1 from China... = <S> > best value for money for a small display use serially addressable LEDs, like WS2801 or WS2812. <S> = <S> > recommended if you want to have a lot of room between te LEDs. <S> And they are RGB! <S> buy a ready-made LED panel with built-in (multiplexed...) drivers. <S> A 64x16 panel is < $10, but shipping cost doubles that price. <S> = <S> > best choice if you later want to extend your expierence to larger displays. <A> You don't say what "break the bank" means exactly, and you should realize that one of the first things an engineer will (and should) ask is, "What, exactly, are the requirements? <S> " In this case, is 40 bucks for drivers OK? <S> This trick is to get away from the multiplexed matrix approach and go with the serially-addressed approach as Wouter suggested, but using a home-brew serial network rather than LEDs with the ability built in. <S> The standard IC for this sort of thing is the 74HC594. <S> One will provide 8 bits of drive, so you'd need 55 or so. <S> Jameco, for instance, will sell them for about 65 cents apiece. <S> The idea is to hook them up into one long string, and update the entire array in a single 440 bit burst. <S> For instance, if you use an output clock rate of 1 MHz it will only take 0.44 msec and this will be invisible. <S> You'd also need 440 current limiting resistors, but since you'd be buying in bulk you should be able to get with less than 10 bucks for those. <S> Finally, you'd need a 5 to 6 volt power supply, and assuming 10 mA per LED <S> you should size this as 5 amps.
| I would start the design by looking for off-the-shelf LED driver IC (ie. google "led matrix ic").
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Is the voltage supplied by an LCD backlight inverter in a laptop dangerous in any way? It is my understanding that in a laptop, voltage from a low voltage (say 10V) battery can be converted into higher much higher voltages by way of inverter for an LCD display module. Is this voltage dangerous in any way if I were to put a load on the output, such as my wet finger? I believe these inverters have a high source of impedance and don't supply much current? Yet I was wondering if there would be a circumstance in which one could possibly feel a shock. Earlier today, as I was typing on my LED backlit Macbook Pro keyboard, I spilled water all over it while simultaneously having accidentally dug a portion of my finger under a key. Although not part of the display, I thought I would have felt something. Why didn't I? Apologies for the scattered array of questions. Thanks <Q> Macbook parts are relatively well known. <S> The backlight driver IC in a Macbook is an TI LP8550 . <S> This has a maximum output voltage of 40V. Not that dangerous. <S> Traditional CCFL backlight however, create a voltage of several thousand volts. <S> And from experience I can tell that this hurts quite seriously when you touch this. <S> Dangerous. <S> In order to get shocked there needs to be a path. <S> A wet keyboard battery powered laptop do not form a path. <S> There is a path when you're holding the monitor chassis with left, and touch the inverter with your right palm. <A> @Jeroen3 LV DC can be lethal in the right conditions. <S> Because DC does not operate on an oscillating frequency it can stop your heart surprisingly easy. <S> Look it up. <A> There are a few things going on here. <S> There can even be a reasonable amount of current, at least for a little while from charged capacitors. <S> So why then didn't you feel anything? <S> Well firstly it's possible the liquid didn't actually penetrate to any active conductors. <S> Then there is the nature of electricity, it prefers to take the lowest impedance path. <S> Water has a very high resistance, so does human skin. <S> Any voltage trying to travel that path will have to drop significantly unless it is backed up by a lot of current. <S> Water is also going to cause lots of shorts if it does get in. <S> Those shorts, while still having reasonably high resistance are still much lower impedance than going into your hand. <S> You really need to directly touch the terminals of something.
| Basically it would be pretty difficult to get a shock from spilling water on a laptop. Firstly yes there can be high voltages even though the power source was a 10V battery.
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In general, how do I know if my project needs a file system? This question gives a good high-level overview of when an operating system is appropriate on an embedded platform. The three topics mentioned are networking requirements, GUI requirements, and file system requirements. Networking and GUIs are obvious enough, but I am wondering when a file system is appropriate and/or required for an embedded system. What sort of embedded tasks lend themselves to a file system solution? Maybe something like a data collection application, where previous measurements or log files need to be stored for future reference? <Q> Operating Systems abstract things <S> so different applications on the same machine do not need to care about each other. <S> File systems follow the same logic. <S> This can be a generic mapping from a (possibly hierarchical) primary key to a single data blob of arbitrary length, as most file systems, or it can contain additional features like secondary streams, indexes (e.g. artist search in MP3 folders), a record format like in Palm OS, ... <A> The most obvious situations where I find a file system desirable: <S> Whenever you have to exchange data with an external system (eg. <S> a PC). <S> Definitely a FAT file system is the easiest solution when using an SD card or USB stick (example: datalogger, preparing config files on a PC). <S> You use NAND storage. <S> You need a file system that is NAND-aware, because all NAND chips have bad sectors and the file system has to work around them. <A> I worked on a project of remote data collection. <S> While the system were supposed to be permanently connected through RF, we had to handle the case in which either the RF, the internet down or any other reason data couldn't be sent as we couldn't afford to loose data. <S> At that time we were using an RTOS with a Fat file system on a PIC24 and SD card. <S> Unfortunately, this wasn't reliable as in case of power loss, it could happen to have the file system corrupted. <S> Also FAT is quite complex to handle with uC, especially when you start to have a lot of files. <S> Our solution at the end was to write data in raw into the SD card with the write/read sector being stored on the first 10 sector of the SD with randomisation to avoid flash wear. <S> The downside was that we had to use a special software in admin mode to read the data from a computer, although it was much more reliable than a Fat file system, and also much easier to handle on the PIC. <S> Looking back, I would probably now use an embedded Linux to handle this instead of a PIC as it saves a lots of time and headache. <S> Connectivity is also much easier to handle on a linux than a uC, downside of linux is cost, power consumption, boot time and IO latency depending on what you do. <A> The only time you need <S> a file system is when you have a requirement which mentions it, like "the system should provide the user with such and such files" or "the system configuration has to be stored in a config file". <S> In all other cases, a file system is a trade-off between software complexity and convenience. <S> File systems let you store data without dealing with low-level details, are easily scalable, and (if you pick a compatible filesystem) make data exchange much easier. <S> Sometimes you can't afford complexity because your target platform is very limited in terms of memory and/or performance. <S> Sometimes complexity gets in the way of achieving the desired reliability (as in a device which can be powered down at any time but doesn't tolerate to lose data). <S> In these cases working with fixed data records instead of files should be considered. <S> If you can afford the complexity and need (or foresee that you may need) flexibility, abstraction or compatibility, using a file system is a sound idea.
| If you need storage for multiple independent components that should not need to coordinate resource usage between each other, you need a mediating component, which would be a file system.
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How is something level trigger vs edge triggered? I understand what they both are, but how exactly is something "made" to be edge triggered? I understand some computer architecture concepts, and while I understand individual components (gates/flip flops/etc...) I don't understand what makes something rising edge/falling edge triggered or level triggered for example? Im more or less looking for a "Physical" reason that determines why one vs the other. <Q> Consider this schematic: simulate this circuit – Schematic created using CircuitLab <S> The input of the AND gate is either <S> L,H or H,L. Never L,L or H,H. <S> So, the output is always L. Right? <S> WRONG! <S> Every gate has a propagation delay. <S> A few nanoseconds. <S> What happens is the following. <S> Input is L. The AND sees H,L. <S> The output is <S> L. Input has a L→H transition. <S> The inverter has a delay. <S> The AND sees H,H. <S> The output is H. <S> The inverter finally propagated the H on its input. <S> The AND sees L,H. <S> The Output is <S> L. Input has a H→L transition. <S> The inverter has a delay. <S> The AND sees L,L. <S> The output is L. <S> The inverter finally propagated the L on its input. <S> The AND sees H,L. <S> The Output is L. <S> So, this circuit is an edge-detector for the L→H transition. <S> It creates a few nanosecond H pulse during that transition. <A> In SOC systems which I have seen they work a bit different from what most people expect: They do NOT use the interrupt signal rising/falling edge immediately. <S> Completely a-synchronous interrupts first need to be synchronized. <S> This is not needed with most internal generated interrupts as they are often generated by a circuit which is already running at the CPU clock. <S> Then the signal is delayed by one-clock cycle which give you the 'previous' state. <S> Next a logic circuit checks if the current value differs from the previous one. <S> If so you had an edge. <S> If the old state was low and the current state is high <S> you have a rising edge. <S> If the old state was high and the current state is low <S> you have a falling edge. <S> Here is some Verilog code: module <S> irq_edge ( input <S> sys_clk, input <S> sys_reset_n, input <S> async_irqA, // <S> A-synchronous interrupt <S> input <S> sync_irqB, // <S> synchronous interrupt // <S> Outputs // <S> Beware : <S> high for only one clock cycle output rising_edge_A, output rising_edge_B, output falling_edge_A, <S> output <S> falling_edge_B );reg meta_syncA,safe_syncA;reg prev_irqA,prev_irqB; always @(posedge sys_clk or negedge sys_reset_n) <S> begin <S> if (!sys_reset_n) begin <S> meta_syncA <= 1'b0; safe_syncA <= 1'b0; <S> prev_irqA <S> <= 1'b0; <S> prev_irqB <S> <= 1'b0; end else begin // Sychronise async_irqA to system clock <S> meta_syncA <= async_irqA; <S> safe_syncA <= meta_syncA; <S> // Delay for edge detection prev_irqA <S> <= safe_syncA; <S> prev_irqB <S> <= sync_irqB; <S> end <S> end <S> // <S> Compare old and current value of signals assign <S> rising_edge_A <S> = ~prev_irqA & safe_syncA; // was low now high assign <S> rising_edge_B <S> = ~prev_irqB & sync_irqB; <S> // was low now assign falling_edge_A = <S> prev_irqA & ~safe_syncA <S> ; // was high now low assign falling_edge_B = <S> prev_irqB & ~sync_irqB; <S> // was high now lowendmodule <S> And here is a waveform: <A> To create a rising edge triggered flip flop, you would first have a latch which is transparent when the clock is low. <S> The output of that latch would go to the input of a latch which is transparent when the clock is high.
| The normal way to create an edge-sensitive flip flop is to combine two level-sensitive latches which are sensitive to opposite states of the clock.
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Moving Ground in between resistors in series What happens when you move Ground to the given point? How does it affect the Voltage and the Resistors after moving it in the circuit? Missing values: .Current (I)= 30mA (V.R1)= 15V (V.R2)= 3.6V (V.R3)= 10.7V (V.R4)= 10.7V (V.R5)= 12.6V (V.R6)= 7.4V <Q> In most circuits, Gound is simply the point we choose to call Zero Volts, and use as a reference when measuring voltages elsewhere in the circuit. <S> By moving the Ground as you show, the voltages across each resistor will not change, but the voltages at the points between the resistors, relative to Ground/Zero Volts will change. <A> This circuit is just a voltage divider with a bunch of different resistors. <S> Even if you move the ground point, it's still just a voltage divider. <S> The tricky part is that usually, we use the term 'Ground' as a reference to the lowest voltage level in the circuit. <S> However, the physics is exactly the same if we label any other point in the circuit as 'Ground' instead. <S> Because the resistors are all in series, we can sum them up to find the equivalent resistance, Req (2000 ohms), of the circuit. <S> Req is the same, regardless of where we label 'Ground'. <S> From Ohm's law and Req, we know that the current through the circuit is 30mA. <S> We can find the voltage drop across each resistor by multiplying 30mA times each resistor value. <S> This gives us the voltage drop across each resistor. <S> Then, starting at GND, and moving around the circuit, we can sum up the voltage drops for each resistor, and the voltage gain from the source: (0V) <S> --R4-- <S> (-10.7V) <S> --R5-- <S> (-23.3V) <S> --R6-- <S> (-30.7V) --V1-- (29.3V) <S> --R1 <S> -- (14.3V) --R2-- (10.7V) <S> --R3-- 0V <S> The important thing is that no matter where you start, when you finish going around the circuit, the voltage level ends back where you started. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> The same circuit but with differing ground points. <S> Figure 2. <S> The CircuitLab DC solver shows the voltages with respect to the ground point. <S> Rather than calculate or measure the voltage drop across each resistor it is common to measure voltages with respect to the ground. <S> The voltage drop on each resistor can, of course, be calculated by the difference in potential (relative to ground) at each end of the resistor. <S> How does it affect the voltage and the resistors after moving it in the circuit?
| The voltage across the resistor doesn't change but the voltage with respect to ground does.
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Under what circumstances would 5V@2.1A be lethal? Besides obviously sticking an output of 5V directly to the heart. Are there areas inside the body where resistance in ohms is so low that a 5V electric potential would be fatal? I got into an argument about this with my roommate after he made a joke about sticking an lightning apple lightning cable directly up his urethra. I told him that his brilliant idea could cause a nasty shock since internal body resistance is so low. He, however, insisted that the lowest internal body resistance still sits at a few hundred ohms and 5V (5V 2.1A Charger) would not suffice to provide a dangerous current. Would it be possible for internal resistance in this area to be so low that even 5V or less would be dangerous? Say it took a path up the bladder and through the kidneys and to the heart. <Q> if applied for a long duration electrolysis can cause acid/alkali burns - see warnings about swallowed lithium button cells. <A> Except if you proceed with a surgery to apply the 5V directly on the heart, or some other vital organ, it certainly can't be lethal. <S> The resistance between any two accessible points on your body is either: Rather low, because the two points are very close together, possibly on a wet surface (e.g. the tongue... or the urethra, whatever...). <S> So a significant current goes through, and you can feel it (it could hurt), but it won't kill you because the current will not disturb or harm any vital organ because they aren't in its way. <S> Very high, because the two points are far away from each other, in which case the current will be so low you probably won't even notice. <S> The NIOSH states "Under dry conditions, the resistance offered by the human body may be as high as 100,000 ohms. <S> Wet or broken skin may drop the body's resistance to 1,000 ohms". <S> "Wet or broken skin" could be the equivalent of what you have on the tongue (urethra?), so at 5V it would give 5mA. <S> Which apparently translates to "perceptible but no muscle reaction". <S> See also: https://en.wikipedia.org/wiki/Electrical_injury <A> He may feel a localised electric shock, with 5V stimulating sensitive areas. <S> The two electrodes for the 5V supply are both within the cable though, so current will only pass through a very small portion of his urethral area. <S> It certainly will not reach his heart, unless he breaks the cable open and pierces his skin with a connector. <S> On another note, I have friends who engage in putting things up their urethra. <S> (It's known as "urethral sounds".) <S> (Which is why they do it, of course.) <S> The average USB cable is relatively large, compared to the urethral hole. <S> I assume he's planning on cutting the connector off, because that simply would not fit up there without causing damage. <S> Even with the connector cut off though, a cable that size is likely to cause some injury for someone who is not used to this. <S> Also the cut-off end of the cable will be rough and will likely also cause damage - there's a reason that people who do this use medical equipment, which is made of smooth, rounded metal. <S> The risk of infection is therefore also very high. <S> At best, your friend is in for some serious discomfort. <S> More likely, your friend is on his way to hospital, and thereby to infamy via hospital war story. <S> At worst, Mr Happy may be permanently damaged, and your friend will never pee normally or have sex again. <S> Not recommended.
| Whether adding 5V would cause significantly more pain than the already-painful act of sticking a cable up there, I couldn't say, but simply stuffing something up there is definitely not comfortable.
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Output impedance of STM32H743 MCU? How do I find the output impedance of the STM32H743 (and potentially other similar devices)? I've looked through the datasheet (DS12110 Rev 5) and the reference manual (RM0433 Rev 5) however haven't found much information. I'm mostly interested in the SPI lines, and want to match the track impedance to the output impedance using a series resistor. For now I've put in placeholders of 33R, thinking the PCB trace could be around 50R and the output impedance around 20R. <Q> The meaning of this table is that if a 8 mA load is connected to ground and GPIO is driven HIGH, the drop between Vcc and output is 0.4 V. <S> It means that the internal equivalent impedance of GPIO port, per Ohm's Law, 400mV/8mA = 50 Ohms. <S> Same calculations are valid if the load is connected to Vcc, and the pin is driven to LOW. <S> The table says the residual voltage is <S> no more than 400 mV. Again, it is equivalent to 400/8=50 Ohms or less. <S> There is one caveat however. <S> Most MCU have simplified GPIOs, where the impedance depends on load. <S> In this particular case the specifications (same Table 60) says that the voltage drop is 1300 mV if the load takes 20 mA, which makes the output to have roughly 1300/20 = 65 <S> Ohms. <S> In short, VOH and VOL data at specified load give you an estimation of output impedance of CMOS GPIO driver under 65 Ohms worst case (for STM32H743), and likely close to 40-50 Ohms typical. <S> If your track's impedance is 70-100 Ohms, you will need to add a 22-33-47 Ohm series resistor at driving side if you want a neat waveform. <A> It's not common to specify the output impedance of digital logic, because the voltage/current behavior is nonlinear. <S> The best you can do is look for graphs of typical output current vs. output voltage and estimate the slope of the line for your region of interest. <A> For ST specifically they usually share their IBIS models which are used in simulating high speed lines. <A> In the datasheet for the STM32H743xI <S> there's a paragraph "Output buffer timing characteristics". <S> They have a table with t[rise] and t[fall] into given capacitive loads at each of the four possible output pin speed settings, where The fall and rise times are defined between 90% and 10% and between 10% and 90% of the output waveform, respectively From that data you should be able to calculate (estimate) <S> the output impedance. <A> The output impedance will vary as the output FETs are turned on and then turned off; the operating region is the FET_IN_TRIODE, once the FET is solidly on. <S> In TRIODE, the FET looks like a fairly linear resistor, as long as the voltage excursions are less than 50% of the FET threshold voltage; with Vthreshold of 0.3 volts (an estimate of numerous modern CMOS processes), you can have 0.15 volts reflection and still have a fairly dependable RESISTOR action. <S> However the output impedance will VARY with temperature, and with VDD, on the PCB. <S> And well or tub or bulk noise will also affect the FET behavior, tho you may not detect this minor delta_Rout. <S> And as the semiconductor foundary implants atoms to form the Nchannel and Pchannel devices, there will be variations in the conductivity (the output resistance) because thermal annealing will not be precise. <S> Depending on the exact implanting used, the channel doping may be the DIFFERENCE of two implant cycles, and slight errors of implant-time and spatial fluctuations of implanting because of variations in the plasma-ion-cloud density will be magnified by the differencing. <S> Summary: you will have to FINETUNE that resistance for every system you ship. <S> IMHO <S> So ..... keep the edges SLOW; set the slewrate SLOW.
| Output impedance of (typically CMOS) output driver for any MCU can be calculated/estimated from VOH and VOL electrical characteristics from datasheets .
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Altium: multiple component footprints for the same component on the PCB The question is one, but revolves about: component definition in the library placement between the schematic and the PCB document so that they remain synced proper Net definitions and avoid errors captured by the compiler and proper BOM list generation. I am relatively novice altium user and I am designing a fairly simple 2-layer fanout board for a component. I have also created a library for the component. The component in focus has input, output and IO pins and its default component type is standard, which means that it will be included in the BOM list. Now what I want, is that the board has pads for this component on both sides , for test reasons. In such a way the test engineer will be able to solder the component on either, but only one at a time , side of the board, leaving the pads on the opposite side unused . Of course the two sets of pads will be electrically shorted and hence, will belong to the same Nets according to specs. This way I will be able to route the board later. What I did, is that I placed two instances of the component in the schematic: one with "standard" type and one with "standard (no BOM)", which almost brings me to the desired result. The problem is that I get an error for the output pin from the compiler that "Net contains multiple Output Pins." I understand the error, but I don't see how to resolve it, as according to the recommendation I should break my library definition, by changing the I/O specification. Another option would be to place only one component in the schematic and two components in the PcbDoc, but this brings up two new questions: How to keep the two synced, such that when I am doing a "Update Schematic/PCB document" no changes occur in the Engineering Change Order? How to place a component with body in Mechanical layer, so that this is not visible in 3D View? What is the way to do this properly in Altium? Thank you very much and cheers! <Q> Another way would be to define a connector pattern with the same fooprint, pins are passive, prefix could still be U_something_A (whatever your convention is), and no 3D body associated with it. <S> I hate ignoring netlist errors, eventually they come back to haunt you. <A> Put two parts on your schematic, connect all pind in parallel. <S> Ignore the warnings about two outputs on the same net - you know only one part will be installed, yes? <S> Place the parts on top & bottom of the board. <S> You may find you cannot place them exactly opposite each other and still be able to route them. <S> Some placement experimentation may be needed. <A> The part you don't populate for a given variant should be marked as "Not Fitted". <S> Before you can do this, however, you have to create your two variants and compile the project. <S> Once compiled, a tab will appear at the bottom of the schematic editor which you can click on to enter the sheet editor: <S> When you enter this sheet editor there is a button at the top in one of the toolbars that looks like a component symbol with a red X through it. <S> When you hover over it the label shows Toggle Part Fitted or Not Fitted : <S> If you click this you can use the tool to change the part so it is not populated for the selected variant. <S> It will either not show up on the BOM at all, or if you have "Include Not Fitted Components" checked in your BOM configuration, it will show up on the BOM with a quantity of 0. <S> This way you have two "Standard" components, each with their own footprint, and your variant determines which footprint is populated. <S> I would consider this the right way to do it.
| The most common way to do this is to add a two copies of the component in the schematic, and use variants to indicate which is populated (the part on the top or the part on the bottom).
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Is this a selenium rectifier? The item below is one of two components of an old (c 1950?) electric chisel. (The other part is a reciprocating solenoid.) The name tag on this part says "Electric Converter", with a 6 amp rating. Inside the case are 10 thin disks, separated by plastic-looking spacers. The output socket has 4 pins: frame ground; 'neutral' (connects to one AC input line); and two which (unloaded) each measure 120 Vac relative to the 'neutral'. The latter two do not show continuity to any other pins, or to each other. No DC, at least on my DVM. I'm wondering if this is a dual half-wave selenium rectifier (Wikipedia) . However, if the 10 disks are the "plates" described there, the 20v reverse voltage per plate rating seems marginal. Questions: Any idea what it is? If it is a rectifier, it's broken, yes? (AC on the output) If it is a selenium rectifier, is the 20v per plate Vrev not a problem? If it is selenium, would that material's properties (e.g., current-limiting (per wikipedia)) be advantageous in this application (large switching inductive load), compared to a diode-based rectifier with a similar power rating? EDIT: The output cord from this device feeds the solenoid section of this tool. The nameplate on that says 120 V, 60 cy [hz]. So, not a rectifier. Size: 6" long, 4" diameter Disks & spacers: <Q> Any idea what it is? <S> Your guess of a selenium rectifier seems likely. <S> Based on your analysis, it may have failed. <S> This is not surprising; selenium rectifiers were rather fragile and prone to failure already, and using one in a high-vibration environment (like an electric chisel!) <S> would have placed it under a lot of stress. <S> This device probably used a selenium rectifier because silicon rectifiers did not exist at the time it was constructed, or were not large enough to use in this application. <S> If a similar device were to be built today, it would use a silicon diode -- there is no advantage to the use of a selenium rectifier. <S> Selenium is rather toxic. <S> If you have touched the disks inside this device, wash your hands thoroughly. <S> Do not place this device in the trash; please bring it to a local hazardous waste center for disposal. <A> It is possible that the pictured component is a surge suppressor. <S> If you could trace the wiring and draw a schematic diagram, you could probably determine which it is. <A> The 120 V line input connects to the center of the disk stack, and the outputs are connected to its two ends. <S> When loaded (100 W bulbs), one output measures <S> 37 Vdc (relative to AC neutral); the other measures -37 Vdc. <S> I don't know for certain that it's selenium, but I think that's the most likely, given its design and vintage.
| A selenium surge suppressor has construction and appearance similar to a selenium rectifier. It is a rectifier, and appears to work.
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Will a DC circuit breaker trip due to high voltage? Will a DC circuit breaker trip due to high voltage (exceeding voltage rating) or will it only trip because of high current? For example, if you have a 40 Amp DC circuit breaker which also has a 32 voltage rating would you have to keep the amperage below 40 and also keep the voltage below 32? Or do you only have to consider the amperage not going above 40? The reason I ask this is because someone else on here asked if an AC circuit breaker will trip because of voltage and the answer was no. An AC circuit breaker will only trip due to high current and can't even detect voltage. I was also hoping this applies to DC circuit breakers as well. Does anyone know the answer to this? <Q> Circuit breakers do not "know" the voltage of the circuit they are used in, so cannot trip based on that voltage - they only trip on over-current. <S> However, they are designed and rated to operate correctly (trip cleanly) only up to a certain voltage - using a 32 volt breaker on a 120 volt circuit may result in the breaker arcing when it trips (likewise, using a DC-rated breaker on an AC circuit, or vice versa, may result in faulty operation.) <S> (GFCI and AFCI breakers may be more sensitive to voltage than normal breakers) <A> The problem is that if the voltage is too high when the breaker trips, it might not be able to stop the current because of arcing. <A> The answer to the question is no. <S> A beaker does not measure the voltage in the wire as it has no specified reference to measure against. <S> There is a voltage across the terminals when they are open and this would be the working voltage (perhaps higher with unexpected over voltages) and it would have a voltage referenced to the chassis or other adjacent conductors (of relevance for the insulation withstand) though these are not used as a reference. <S> The voltage ratings on breakers are due to safe switching voltages without detrimental arcing and contact erosion AND insulation withstand capability. <S> The first rating is not important unless the breaker is open or operated (manually opened or closed or opened due to fault) as it relates to the contact gap (and speed of operation, gas or vacuum fill or quenching systems) and not to the closed circuit. <S> The second rating (less often even mentioned) is the insulation resistance and may be 500V on even low voltage breakers and exceeding this with respect to external surfaces, adjacent poles or mounting hardware will result in unintended breakdown that could easily cause fires. <S> These ratings are different for DC, AC, HF and will be affected by relative humidity and atmospheric pressure unless the breaker is a fully sealed unit. <S> As mentioned in my comment: As a design goal all your components and assemblies have to have a rating greater than the actual supplied voltage. <S> So if your battery now has 48 (possibly more when fully charged) volts then you need to make sure all your breakers, controllers and motors are all able and designed to be able to cope with the higher voltage.
| You need to keep the voltage less than 32V, but not because the breaker will trip if you go higher than that.
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Altium assign one pin to multiple pads? I am using a P Mosfet in the schematic, pin 1, 2, 5, 6 are all connected together. How is this done in altium? <Q> This is long-requested feature, it has a history on Altium's bug/feature tracker that goes back to 2011. <S> It is currently listed as 'in development'. <S> If you have an Altium Live account you can view the feature request here: <S> https://bugcrunch.live.altium.com/#Bug/317 <S> However this requires customizing the footprint to suit the pin layout of the component. <S> This is not such a big deal in things like SO8 MOSFETS which have a conventional pinout, but would be a pain for ICs that use ganged pins but an otherwise standard package. <A> The easiest (and probably the most common) way of doing this is placing multiple pins on top of one another in the schematic. <S> If you ensure all of the electrical "hotspots" are lined up with one another on the grid, they will automatically be "read" as connected to each other. <S> Then, if you want to show all of the pin numbers (like what is shown in your image) you can adjust the pin number margins to separate them from one another. <S> For example Pin 1 will have a margin of 10 DXP units, Pin 2 will have a margin of 20 DXP units, Pin 3 will have a margin of 30 DXP units, and Pin 4 will have a margin of 40 DXP units. <S> They will show up in the schematic just like your image. <A> Since you haven't gotten a better answer, I'll say you should just add multiple pins to your symbol. <S> It's not as compact as overlapping the pins, but it makes your design intent, and the required connections, crystal clear. <S> It can look something like this: <A> Instead of assigning the same identifier to multiple pads in the PCB footprint you also can assign the same JumperID to pads that connected together. <S> But as you can see this is almost the same because for each type of pattern with connected pins you must have separate copy of "pad record" in library. <S> In the early days of Altium Designer there was another option. <S> Every schematic library element can have pin map for each connected model. <S> As footprint is a model you can have a mapping table. <S> Mapping was designed not only one to one, but also one to list. <S> I mean one pin from schematic library's element to list of pins from pattern. <S> ( https://designspark.zendesk.com/hc/en-us/community/posts/115000098289-Multiple-PCB-Pin-for-a-single-Schematic-Pin ). <S> This feature was slightly more powerful then dumb patterns with connected pins as this mapping table is not the part of pattern. <S> But still twice messy: this table must not be the part of logic element (i.e. element of schematic library) <S> instead it must be the part of component; pin numbers seeng on shematics never show mapped pattern's pin numbers, so nobody know did you done any remap or don't. <S> For now this type of logic-pattern connections is not supported by Altium though mapping tables for each shematic library element still presents. <S> So, of cource, AD cannot be "industry standard" at list at component's library design.
| Another workaround is to assign the same identifier to multiple pads in the PCB footprint, which will cause them all to be connected to the corresponding schematic pin.
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Equivalent Characteristic Impedance for Different Trace Widths Suppose I have a PCB trace that needs to be impedance matched, but the trace width changes as it travels from its source to its destination. Something like this: The 5mil trace would have a certain characteristic impedance, whereas the 10mil trace would have another impedance. How do these two different impedances combine? Asked differently, what is the equivalent impedance of this trace? <Q> Assuming the height above the ground plane is the same for the whole trace, the 5-mil-wide segments will have one characteristic impedance and the 10-mil-wide segment will have a different, lower, characteristic impedance. <S> The whole trace would not normally be thought of as having a single "equivalent impedance" characteristic. <S> What you could do is calculate the input impedance, or the S-parameters of the trace, as a function of frequency. <S> To do this, you could use a simulation tool like Keysight ADS, or Smith chart techniques (if you are only interested in the input reflections), or simply sit down and do a bunch of algebra. <A> For standard FR4 (pcb material) <S> the electric permeability constant is ~4.4. <S> The height will depend on your PCB stackup, and what layer the ground plane is on. <S> Source: <S> https://www.eeweb.com/tools/microstrip-impedance <S> If the transmission lines are not matched, you will get a reflection at the point of mismatch and lose power. <S> The equation for finding this would be the reflection equation: Z1 would be the characteristic impedance of the 5mil trace, and Z2 would be the impedance for the 10mil trace (after you calculate it) <A> I fear you have entirely missed the point of controlled impedance. <S> Your diagram actually looks (from an impedance point of view) like this simulate this circuit – <S> Schematic created using CircuitLab <S> where Z0 and Z2 are 5 mil lines, and Z1 is your 10 mil line. <S> The whole point of using a controlled impedance line is to ensure that the junction of Z2 and ZLOAD does not cause a reflection when the signal hits it. <S> Assuming that, with your board geometry and material properly selected, Z2 and ZLOAD are in fact matched, you now need to consider the rest of the trace. <S> The junctions of Z0 and Z1, and Z1 and Z2 will BOTH produce reflections, and this is entirely backwards from what you want to do in the first place.
| You need to know the height of the trace between the ground plane, the relative electric permeability and how high the trace is to find the equivalent impedance.
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What relay do I need Hello friendly friends. In my wood shop I have two separate 220V circuits, A and B. B is new and I want to use it to drive the dust collection (1800W). Now I want it so that when I power on one of the machines (1500W+) driven by circuit A the dust collector will power on too. Simple, I thought, just buy a relay and hook it up. So I bought an Eltako R12-100-230V relay (See amazon, I cant put a link in here). I then connected the phase wire of A trough connectors A1 and A2 and the phase wire of the dust collection to 1 and 2. Alas it doesnt work. Instead when I power on a machine, the relay will click and cut the power of A!??? I am now unsure wether I use the wrong relay or hooked it up wrongly. This is what I have: And If I understand NMF++ from the comments below right then this is what I should do: Here's what I have atm: <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) What you did. <S> (b) What you intended. <S> Relay coils have a high resistance relative to your motor load. <S> When wired in series as in Figure 1a <S> the relay finds it's 'neutral' connection through the load. <S> Its high resistance lets a small current through and this isn't enough to start the motor. <S> If you put an AC voltmeter across the motor you would only read a few volts. <S> Wiring the relay in parallel with the load as shown in Figure 1b will give the desired result. <S> Both the relay and the motor have a live and a neutral connection. <S> The ASCII schematics were a noble attempt although they might be clearer using | - and + characters. <S> N <S> P| <S> || <S> || +----+ <S> A1 <S> | 1| <S> | <S> + <S> ++ <S> || <S> | <S> | <S> | /| <S> | <S> | | /| <S> | <S> + <S> ++ |+----------+ A2 | 2| |+-( ) <S> -+ <S> Even better, there is a CircuitLab schematic button on the editor toolbar! <A> Allright. <S> I ended up buying the KEMO M103N for 20 Euros which is basically a master/slave circuit in a box with a bunch of connector wires and does what I want. <S> Thanks everybody. <A> If you have machines (plural) on circuit A, to be able to switch circuit B when any of them are on, you'd need a relay for each machine, with the coils wired across the motors on each one, and then connect the contacts all in parallel - a relay OR gate - which are then wired into circuit B as you indicate. <S> There are current-sensing relays, that you could use, which would switch on when any of the machines on A were turned on. <S> These would be wired exactly how you have the original arrangement, though most high current ones rely not on a direct connection through sense terminals, but on either current transformers or Hall current sensors so that you only need to loop one of the live wires through the core to operate it.
| For a machine on circuit A to to be able to swich circuit you'd need to connect your relay, which has a coil that operates at 230V, across the motor in that machine - after all the isolators and switches that allow it to operate.
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Is a normally high switch the same as a normally closed switch? I'm very much an amateur when it comes to electronics so apologies for the stupid question. I have a switch that is normally high (5V) and then when i hit the mechanical button it drops to 1V. I need to configure my CNC firmware for this. So are these normally open or normally closed switches? To me this would be a normally closed as with a normally open switch you wouldn't have a voltage when the mechanical button is not pressed, but the spec page for my switch says it is normally open so I am confused. Thanks in advance! <Q> Figure 1. <S> What appears to be the same switch from Biqu Equipment . <S> Their instructions say: <S> Red line connecting VCC (ramps of <S> +) <S> Connect the black wire GND (ramps of -) Green Line connection SIGNAL (ramps in s) <S> Green is actually the switch output and black probably provides the common for the LED cathode (negative). <S> I suspect that you're using red and black when you should be using red and green. <S> I recommend that you connect up the black anyway. <S> The visual indication of switch status will be useful. <S> Figure 2. <S> The schematic from Thingiverse . <S> Red, Vcc. <S> Black, GND. <S> Black, GND. <S> Green, Output. <S> How it works: <S> With the switch as shown the output is high and the LED is shorted out (so it is dark). <S> When actuated the output is pulled low and current flows through R1 and the DETECT LED to ground. <S> The LED is lit. <S> R2 makes sure that the line is pulled high during switching from NO to NC rather than leave it floating and susceptible to stray switching due to interference. <S> C1 further filters any noise. <S> The schematic on the CNC shield page is a horrible low-quality JPEG <S> but we can make out enough information. <S> Figure 3. <S> The end-stop pinout. <S> Note that CONN_3 allows selection of GND or 5V to the end-stop switches. <S> Note also that you can fit end-stop switches on both ends of each axis but that they share a common input to the Arduino. <S> (There are a limited number of pins available so some doubling up is required.) <S> The controller will usually determine which endstop was hit by the direction of travel at the moment the input was triggered. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 4. <S> (a) CONN_3 connection for switched positive. <S> (b) Switched GND. <S> (c) <S> Three-wire sensor boards. <S> I recommend that you try wiring as shown in Figure 1c. <S> Jumper 1 and 2 on CONN_3. <S> That will put out a +5 V on each of the X+, Y+ and Z+ pins. <S> Somehow connect up the limit switch black wires to any of the spare GND pins. <S> Pin 3 on the select is available. <S> Wire the greens back to the X-, Y- and Z-. <S> Note that the schematic and board top-view pinout are not very clear so you may need to do some PCB tracing. <A> An open switch means that there is no current conducting through the switch, causing the voltage at the terminals to be high, 5V in your case. <S> (this is probably non-zero because of some resistance you still have after your switch). <S> This image is similar to your situation where the switch is open and Vout is 5V. https://en.wikipedia.org/wiki/Pull-up_resistor#/media/File:Pullup_Resistor.png <A> It sounds like you have a normally-open single-pole switch, with a pullup resistor. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is a common arrangement. <S> When the switch is not actuated, the controller sees 5V through the resistor. <S> There's only a tiny amount of current flowing into the controller, so there is little drop across the resistor, and the voltage at the controller is very close to 5V. <S> When the switch is actuated, the controller input is grounded. <S> Current flows through the resistor, but it's a trivial amount that causes no harm. <S> Why does it drop only to 1V when pressed, and not 0V? <S> There's probably some resistance in the wire, or corroded switch contacts, or a loose connection somewhere.
| When the switch is closed it means that the circuit is closed, causing a current to flow through the switch, pulling the voltage to the negative terminal, which in your case is 1V
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Non-inverting output of transistor circuit (at NPN base) is 0.69V max. why? I tried to make a simple NPN transistor inverter so that I can use the strobe line of the parallel port to store data in two separate flip-flop IC's. (74HC174 and 74HC574) The IC that data will be stored in is based on whether the strobe is on the falling edge or the rising edge. I connected everything including this circuit and there is no smoke, no heat or anything. I could assume my resistor choices are terrible, but I'm not sure, especially with how many changes there are to PC port designs. For reference, the parallel port I'm using is coming from a dell laptop and I specifically disabled ACPI. When I tested the particular circuit shown, The voltages I get from the NPN base are 0.09 and 0.69V. However, the NPN collector voltages are acceptable (0.0 something, and 4.9 something). What would cause the NPN base voltage to be so low in both cases and how do I fix it? <Q> The transistor base is not an output of this circuit. <S> It is where the input signal connects to the transistor. <S> There is not expected to be any gain between the input and this node. <S> If you want a non-inverted output with gain, you can add another CE stage. <S> Or if you want inverting and non-inverting outputs with equal amplitude and equal phase delay, you will need to go to a fully differential amplifier circuit, like the emitter-coupled pair. <S> Or just buy a fully differential op-amp. <S> Or if you just want a non-inverted replica of the input signal to drive another high-input-impedance circuit, you should just connect to LPT_STROBE rather than to NORMAL. <S> If you want to connect more than 2 or 3 of these same circuits to a single CMOS output, you'll probably want to increase the value of the input resistor to something more than 1 kohm, though. <S> Note: As Bimplerekkie points out in comments, you could have much higher fan-out if you used an n-channel MOSFET in place of the BJT in your circuit. <S> Or just buy a CMOS buffer and inverter chips. <S> One-gate versions are typically less than $0.10 in reasonable volumes, so buying the one-chip solution is usually cheaper than what you pay to place all the extra resistors in a discrete inverter circuit. <A> The 0.69V is caused by the base loading (it's effectively a forward biased diode) and that is exactly to be expected. <S> There is no way to "fix it" with a BJT in that position and output drawn from the base, it's doing what is expected. <S> You can, however, substitute a different part or hack the PCB to draw the output from the input. <S> The 2N7000 suggested by @Bimpelrekkie (assuming a TO-92 package as well as the 2N3904) has a compatible pinout so you can just mount it instead. <S> 2N3904: E-B-C <S> 2N7000: S-G-D <A> The maximum voltage across the base-emitter junction of an NPN trasistor is about 0.7 volts. <S> You could use the LPT_strobe signal (before the 1K base resistor) as your "normal" signal. <A> Ok, I figured it out. <S> The #1 problem was that the resistor was connected between NPN base and VCC. <S> schematic source: http://www.dinceraydin.com/8051/index.html
| Or if you want both inverting and non-inverting outputs with equal amplitudes from a single circuit, you could add an inverting buffer circuit (perhaps based on an op-amp). All I really needed was resistance from input to base and the NPN collector resistor just like how Q2 is setup in the schematic below with the two resistors connected to it.
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Transformer Inefficiency and Heat If a transformer has efficiency of 96%. It means for a 500W capacity and load, the input would be 520W, output 500W so 20W would be emitted as heat. If load is 250W, then input would be 260W and 10 watts would be emitted as heat. But if there is no load (zero load), how do you compute the wattage that would be lost as heat during the initial stage when magnetic field and flux is being built with no load on the secondary? Is it more than the inefficient percentage lost with load? I'm computing sealed enclosure heat transfer for a design I'm having and I need to know the above. <Q> Transformers are not rated by "efficiency", they are rated by losses, although the efficiency at maximum load can be defined. <S> So you cannot assume that at 250W of load the transformer will dissipate 10W if at max load (500W) <S> it dissipates 20W, it will dissipate more than 10W. <S> This is because transformers have two types of losses , "load losses", and "no-load losses". <S> Even at zero load a transformer will run warm. <S> Load losses depend on wire resistance of coils, and are proportional to load. <S> No-load losses are independent of load, they occur due to continuous magnetizing and de-magnetizing of transformer core, with two basic mechanisms - hysteresis losses, and eddy losses. <S> These losses depend on quality of core construction and materials. <S> To reduce eddy losses, the cores are made of thin sheets, so the eddies can't spread over much of core volume. <S> Hysteresis losses depend of magnetic alloy composition. <S> Some portion of magnetic field escapes the core and induces eddy currents and warm up surrounding conductors (mounting hardware, bolts and brackets), which also adds to no-load losses. <S> Without knowing precise details of core construction and manufacturer's data, you can't "compute" these losses. <S> For example, cheap knock-off transformers for Christmas decorations can stay pretty warm even when the lights are off. <S> ADDITION: I happen to have an old 200 VA auto-transformer 115:230V, model SU-38, made by TODD SYSTEMS. <S> In idle (no load) mode, with ambient of ~25C and sitting on a pack of papers, its core gets to ~40C, see the thermal image. <S> : <A> There are two main loss mechanisms in transformers a) Core losses, which are dependent on the input voltage b) <S> Copper losses, which are dependent on the load current <S> Designing a transformer to be low loss is a compromise between these two. <S> In applications where the transformer may spend most of its time at no load, it's worth designing for low core loss, at the expense of high copper loss, by using a low core flux and a lot of turns. <A> Transformers are rated in VA rather than W, where V and A are RMS. <S> VA and W are the same only for a resistive load. <S> There is loss in the windings due to the load current, as I squared times R. <S> There is loss in the core due to eddy currents and hysteresis loss. <S> Core loss does not change appreciably with load. <S> There is also some loss in the copper due to the magnetising current. <S> I presume you are talking about a mains frequency transformer. <S> Other factors apply at higher frequencies.
| Unless the transformer has a very thorough data sheet giving you losses at several loadings, it's not possible to compute the loss at any loading, it must be measured.
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Physical effort in outputting 5V 1–2A? I have an iPhone, and sometimes recharging it from a wall is inconvenient. Maybe I am on a long bus ride. If I am willing to do physical activity/exercise to produce power, how much effort would I need to exert to act as a phone charger, supplying a steady 1–2 amps at 5 V? This hypothetical device would need to be small and light enough that I could carry it... no point in it if I have to be at home to use it. For example, there’s a simple hand exercise device that you squeeze over and over to build hand muscle. How much power could that generate? <Q> 20 watts delivered into a hand-cranked generator (enough to deliver 10 electrical watts after conversion losses) could be managed all day by a fit and motivated person. <S> This would employ all of your arm muscles. <S> However, you'd need reasonable length cranks, comfortable handles, a good secure mounting for the generator, it's not something you'd nonchalantly carry onto and use on a bus. <S> If you don't mind standing, then a mini-stepper could be rigged with a generator. <S> This would use your legs, so we're back in the sustained many 10s of watts, even 100+ region. <S> It could lie stably on the floor, and maybe fit into a small backpack, so is probably the most suitable device overall. <S> Consider that a 70kg individual walking up 200mm high steps one per second is delivering 140 watts. <A> To get an estimate I looked up such devices, apparently they usually provide about 500N of resistance over a distance of 6cm. <S> This gives us about 30J of mechanical energy per squeeze. <S> Not sure how many squeezes you can do per second, so let’s go with energy. <S> A NiMH AA battery cell stores about 12kJ of electricity, <S> so you’d need 400 squeezes to charge a single NiMH AA battery . <S> Good luck with that! <S> And this calculation doesn’t even account for losses. <S> As Neil_UK also mentions, arm and leg muscles are much stronger. <S> Even an untrained person can output 100W on a bicycle over long durations (>1h). <A> 10 watts compared to the 550 foot-pounds and 746 watts of ONE Horsepower, gives us useful numbers. <S> You need to lift 550 foot-pounds * 10/746 = 550 * 1/75 ~~ 8 pounds lifted up ONE FOOT, every second. <S> Can you lift a gallon of water, every second, up one foot, for an hour?
| A grip-squeeze device could be rigged to generate some power but, only being able to use the small grip muscles in the forearm, you'd struggle to get more than a few watts for a few minutes before exhaustion. Neil_UK’s mention of hand grip strength training devices made me curious how much power your hands could provide.
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What are these exposed copper rectangles for on the mbed NXP LPC1768? Today I noticed some exposed copper rectangles on the bottom of an mbed NXP LPC1768 dev board. They don't look like they're meant for components. I think they may just be test points, but I'm curious if there's another answer. Here's an image of the board with the copper rectangles circled in red: I'm not sure what the mbed interface chip is. Googling it suggests that it's proprietary. I can't tell where the traces are going. <Q> <A> An educated guess would be JTAG (TCK, TDI, TDO, TMS) plus VCC and GND. <S> VCC and GND could be verified with a multimeter. <S> Note that the interface chip is also a programmable microcontroller, and thus needs JTAG or SWD for production programming. <A> Sometimes, you see such elements on microwave boards on the microwave lines – then, they serve the purpose of being a capacitive or reactive component (or both); but since this board definitely doesn't look like an upper-GHz RF board: <S> These are almost certainly test points. <A> Adding to the answer of phill g: <S> What are these exposed copper rectangles for on the mbed NXP LPC1768? <S> which provides the schematic of that board at https://www.nxp.com/downloads/en/design-support/ARM_mbed_LPC1768_Schematic.pdf <S> In the schematic they're even designated as the cfg0-cfg5 pads of MBED-IF01 chip. <S> On https://os.mbed.com/questions/76861/mbed-IF01/ on a question regarding the datasheet of that part, it is stated that: IF01 is the the interface circuit of the LPC1768, which infact is an "LPC2148" MCU <S> , in short we can't open source a lot of the information for the interface <S> so this is why its hidden. <S> It seems to be the predecessor of https://os.mbed.com/handbook/mbed-HDK <S> It's a microcontroller implementing https://os.mbed.com/handbook/cmsis-dap-interface-firmware <S> The CMSIS-DAP Interface Firmware provides: USB Mass Storage Device for drag and drop programming of the target chip USB Communications Device Class for Serial Communication with the target chip USB HID CMSIS-DAP for debugging USB bootloader for updating the interface firmware itself As to what those pads really are, when you look in the datasheet of the LPC2148 <S> https://www.nxp.com/docs/en/data-sheet/LPC2141_42_44_46_48.pdf <S> we can see that those pins are in fact the tdo, tdi, trs, trst and rtck pins. <S> Those pins are used thus to flash their custom cmsis-dap interface to that chip (probably using pogo pins).
| NXP has the schematic for this board, it looks like those pads make up a programming interface for that chip, which as you say is likely custom in some way. Those look like test pads and seem large enough for soldering a wired connection.
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UART expander (5 ports to 11 ports) I have a board that I would like to design and there are 11 devices that can only be spoken to via UART . I am restricted to a Microchip uC chip product range and I have found one with 5 UART ports. I have come up with a jumper based solution whereby the user, with making the connection via jumpers to the 5 available ports, can choose 5 of the 11 devices. I have been looking for a chip that will take either UART, SPI or I2C in and give me a few additional UART ports but I have come up short. Can anyone suggest a solution or a product that they may have come across? It is not critical that all 11 devices are able to connect to the uC (that would be nice) but it is important that these devices are on the board for a modular product. However, I wish to make the product more user-friendly as the jumper based solution is a bit complicated at first glance or so I have been told. EDIT Okay sorry, I am not explaining the problem properly. With this method, you limit the user's options. Let say you link port a to device 1 and 2 and port b to device 3 and 4 and so on... But what if the user would like to use device one and two? Then the product is limiting the user heavily. With 5 port options and 11 choices for devices, the binomial coefficient yields 462 combinations for the user to choose from. That is overkill, I am not looking for that level. But with the jumper based solution port a can connect to 9 of the devices while port b can connect to 7 of the devices, port c can connect to 8, port d can connect to 7 and port e can connect to 9. The number of mux switches needed to do this would be over the limit. Is there another solution to giving the user the option to use (close to) any combination of the devices? I have thought about using two uC with 5 UART port each to connect to 10 of the devices and then using a much smaller jumper based solution for the last device, but this solution is too complicated for production and is expensive. Are there any Chips or concepts out there that will expand the UART control to 11 devices? <Q> You could use analogue switch ICs (analog multiplexers) to replace manual jumpers. <S> You need to use 8-channel switches, like DG4051E . <A> Use some UART-SPI bridge ICs, like: MAX3701 <S> FT9xx <S> HT45B0F <S> SC16IS7xx <S> You can have as many UARTs as you have spare GPIO pins to use as Chip-Select lines (or use a demux or decoder <S> and you can have orders of magnitude more) - limited only by your UART baud rate vs the max speed you can run your SPI. <A> drive a SIPO shift register (eg. 74hc595) from 2 GPIO pins <S> connect the ~OE input to the UART output <S> using the GPIO clock in a pattern that represents the device you want to talk to, then use the UART to send if a pull up is in place the port with a zero bit will echo the UART output. <S> for the receive end use an and gate/gates to combine the signal from the device UARTS. <A> I think you are using the wrong protocol and/or topology for your case. <S> I would suggest using RS-485 transceivers and build a bus topology. <S> If you have 11 possible devices, you can use 4 lines to enable one of them. <S> The caveats I see for this option are: Half-duplex communication (No simultaneous <S> Rx/Tx) <S> You can interact just with one device at a time <S> (Bus topology) Your uC will be like a master. <S> It must enable a device and start communication with it. <S> (transceivers are configured for write to or read from the bus) <S> I did not see any detail in your questions about how these limitations can affect your system or if it is worth at all. <S> I hope this could help.
| If you have 5 UART ports, you can multiplex 1 of those ports to 8 multiplexed UART ports.
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Identifying the pins of an AC lamp I have an AC indicator lamp which I have to add to a design and and here is the only information. As you see the two terminals are tied on one side and the other two pins has no continuity with any other pins. I cannot verify the correct phase and neutral pins. Is the upper two(silver like) neutral and the bottom gold like color is for line? What can be the reason there are four pins? I measure the resistance infinity but shouldn't it be 230/3mA = 76kOhm? I measure between the golden and the silver pins. <Q> The data sheet shows the leads for each termination configuration, and for all the other terminations the anode/positive is the top lead (red or longer) <S> so I would be surprised if the brass lugs are not the anode and the tinned lugs the cathode in the DC variants. <S> As yours is the AC variant, and assuming it's got two LEDS in different directions, diodes to prevent backward voltage getting too high and resistors for current limiting, it should start glowing with a few volts DC across it. <S> The forward voltage of the diodes could mess up multimeter resistance readings. <S> I'd guess that each of the tinned pins corresponds to one LED and you can either connect it for either full wave or half wave. <A> The lamp is red. <S> Source: RS. <S> Extra positive terminal for lamp test facility <S> 2 negative terminals allow "daisy chaining"0.11in. <S> (2.8 x 0.8mm) terminals. <A> I suggest you contact the (real) manufacturer (or design owner) and ask if you don't feel like figuring it out for yourself ( <S> if you are 100% comfortable poking around with mains voltage just use a series resistor and you won't harm the lamp- <S> if you are not comfortable get someone else local to help so you don't harm yourself). <S> Clearly the mains voltage will be applied on opposite sides of that barrier! <S> So that narrows it down to the commoned terminals and one of the other terminals. <S> My guess: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The manufacturer appears to be: Signal Construct GmbH Brückenäckerweg 4 <S> D-75223 Niefern-Öschelbronn Tel. <S> (49) 7233 <S> / 9531-0 Fax (49) 7233 / 9531-29 <S> E-Mail: <S> info@signal-construct.de Web: http://www.signal-construct.de <S> GPS-Daten: <S> N 048° 54´ 03,4 <S> " <S> E 008° 48´ 31,1 <S> " <S> I think the actual part number is 22H0289. <S> A pox on distributors who hide manufacturer's information.
| This is an LED lamp, probably with a high value series resistor and maybe a half-wave series diode. It could be a capacitive dropper too but the German documents show AC/DC compatibility which points to a resistor.
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Is BLE 5 backwards compatible to BLE 4.x ? Can a phone with bluetooth 4.x talk with a device with BLE5 I would like to develop a wearable with a BLE5 module. I wonder if this module would work with any BLE4 enabled mobile phone? I am aware with that it may not use the advantages of BLE5, but I don't want to develop something that will not be used for a while. I read a similar question for BLE4.1/BLE4.0 compatibility, and answer was yes. However I am not sure about major version number changes are also backward compatible. Do you have any experience with that? <Q> P.S. <S> You will probably be using GATT for exchanging data; i.e. your wearable will be the GATT server hosting a GATT table for remote phones/tablets/PCs (GATT clients) to read the data from. <S> Familiarise yourself with these terms before starting your project :-) <A> Easy answer: <S> BLE 5 is backwards compatible with the current versions. <A> just proven now myself, Mi Band 4 with Bt5BLE cannot receive pairing data from samsung A7, but A7 can read pairing from Mi Band 4. <S> So the Pair is broken. <S> "compatibility" is an utopia, or a Sales Gimmick. <S> Early Adopters will simply become the victim of "such compatibilities". <S> End of story. <S> Lets wait until all phones and BT dongles move to BT5BLE. <S> Put that Mi Band 4 in the fridge for the next 2 years, or buy a newest BLE5 phone.
| Yes, as long as you are not using the one of the new Bluetooth 5 features (2xSpeed, 4xRange, or LE Advertising Extensions), then your wearable will work with Bluetooth 4.x devices.
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For MMBT2222; How to get hfe value to calcluate minimum base current to drive transistor in Saturation? If my collector current VCC = 3.3V & Ic = 31.82uA then how to calculate minimum base current is required put transistor in saturation mode? Depending on it I will calculate base series resistor As per my understanding hfe varies drastically in saturation region. Which hfe value to put into Ic = hfe Ib formula? <Q> The Vce(sat) <S> curves on a typical datasheet for that part don't go under 1mA <S> (see figures 2 and 6). <S> Gain drops at lower current, especially on some older parts, but it's probably >> 20 at 31uA Ic. <S> The MMBT4401 , for example, lists minimum hFE as 20 at 100uA, but typically the gain does not drop off as rapidly as that would suggest (see figure 1). <S> You are probably safe using <S> Ic/Ib = 10 (as in figure 2), but <S> Ic/Ib = 5 would be even safer, since the datasheet is silent on lower collector currents. <S> The minimum is a bad number to use, it will vary from part to part. <A> I'd be more inclined to begin by looking at this diagram in the data sheet: <S> - It only takes you down to a collector current of 1 mA <S> but you can "sort of" interpolate to lower collector currents. <S> At 1 mA collector current and 0.2 volts across collector and emitter, the base current is going to be about 7.5 uA <S> but you are in a shady area where things are not well defined and I would expect that if the collector current was 0.1 mA the base current would be more like 10 uA. <S> In other words, the hFE drops off as you start to reduce the collector current. <S> It's also very hit and miss trying to define the base current for a given Ic and saturation voltage. <S> For the 1 mA example, I'd be considering a base current of more like 20 or 30 uA and expecting a saturation voltage of about 50 mV. <S> The reason is because around the knee of the graphs you have little control and temperature effects can dramatically alter saturation voltage. <S> If you are looking for a defined saturation voltage then I think you should use other methods. <A> I like to start by figuring out how much collector current it would take to bring me to a saturated value of \$V_{CE}\$ , and once there look at the data sheet to try to figure out how much base current I need to do it. <S> If you can't find nifty figures in the data sheet, like Andy points you to , An "OK" rule of thumb is to assume that \$\beta\$ <S> is no larger than 10 in a transistor driven to saturation.
| You want to use a safe number that will saturate any transistor you pull off any given reel of parts and will saturate it at the maximum Ic and over the entire operating temperature range.
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Why does a micro-controller pin use push pull when configured as an output? What is the main reason some micro-controllers form a push-pull topology when set as an output pin? My only guess is not to load the pin. But I couldn't find enough information about the reasons. What are the benefits of push pull output here? <Q> A proper output needs to be able to do both of those things. <S> [1] <S> If you're asking why the schematic shows transistors, that's a different question. <S> The transistors form a CMOS inverter. <S> They work exactly the same way as the triangle with a dot on the front. <S> They're probably drawn separately because the transistors that drive the output pin are larger and can supply more current than the transistors used for internal logic. <S> The two inverters together form the output buffer for the pin. <S> [1] You can also have an "open-drain" output that only has the pull-down transistor. <S> You would then connect a pull-up resistor to the pin. <S> The advantage of this is that you can make a "wired-OR" circuit where multiple open-drain outputs control the same line. <A> microcontrollers get used in all kind of weird places and putting in the extra transistors to enable push/pull is often worth it to enable those uses. <S> Push <S> /Pull along with a high-impedance state can combine to mimic either open drain or common collector making it flexible in that way. <S> Push/pull are not dependent on external resistors to pull the output high or low. <S> This means less external components that need planning for, faster switching and lower power consumption when not switching. <A> From the comments under ratchet freak's answer: A power supply can also be turned on or off <S> but it doesnt have push pull output. <S> That is correct but a power-supply can only source current. <S> It cannot sink current. <S> For flexibility the micro-controller can do both. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> The equivalent diagram (top) and various options available with a push-pull output. <S> Having a push-pull output gives great flexibility in driving various loads. <S> (a) <S> A low-side switch can handle a load connected to positive supply. <S> (b) <S> A high-side switch can handle a load connected to ground. <S> (c) <S> A push-pull arrangement can switch both. <S> When SW3 (high side) is closed D4 lights and when SW4 (low side) is closed D3 lights. <S> (d) More complex arrangements are available such as this H-bridge arrangement which could drive, for example, a bi-coloured, 2-pin LED. <S> To turn on D5 close SW5 and SW8. <S> To turn on D6 close SW6 and SW7.
| Push-pull means you can drive the output line high (by connecting it to VDD) or low (by connecting it to ground). The disadvantages are that the output's rise time is slower and the circuit uses extra power when the output is low.
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Can 10-100mA kill a person / be deadly? According to various sources something as little as 0.01A = 10mA can be painful, at 0.1A = 100mA you can expierence * "ventricular fibrillation of the heart". "an uncoordinated twitching of the walls of the heart's ventricles which result in death". "As well as other painful things such as extreme breathing difficulties, Severe Shock, Muscular paralysis ...".* How is that true? I've worked with an arduino and LEDs, LED Strips etc for quite a while yet never heard that 10mA can be painful let alone 50-100mA deadly. <Q> A circuit that is drawing 10mA or 50-100mA is quite a bit different than a circuit that could push that much current through you . <S> Figure out what the resistance of the human body is, then <S> calculate how much voltage you would need to force that much current through that resistance. <S> You will quickly see how your arduino is incapable of forcing even 1mA through your body. <S> Remember, just because something can source 10mA, doesn't mean that it can force that current into anything. <A> First off, its voltage not current that kills (because you have to bypass the skin). <S> The actual mechanism is current once you get under the skin. <S> A person can feel at least 1 mA <S> (rms) of AC at 60 Hz, while at least 5 mA for DC. <S> At around 10 milliamperes, AC current passing through the arm of a 68-kilogram (150 lb) human can cause powerful muscle contractions; the victim is unable to voluntarily control muscles and cannot release an electrified object.[11] <S> This is known as the "let go threshold" and is a criterion for shock hazard in electrical regulations. <S> Source: <S> https://en.wikipedia.org/wiki/Electrical_injury <S> Your not getting 10mA through your heart when you are playing around with electronics or low voltage < <S> 60V. Skin is more than 10k (as much as 100k), to get 10mA through 10k is at least 100V. <S> If you have dry skin, it is much higher. <S> Per IEC directives Low voltage is lower than 50V <S> In the European Union, the Low Voltage Directive defines low voltage starting from 50 V AC, and 75 V DC. <S> The directive only covers electrical equipment and not voltages appearing inside equipment or voltages in electrical components. <S> IEC 60364 defines it as 50 V AC and 120 V DC. <S> Source: <S> https://en.wikipedia.org/wiki/Extra-low_voltage <S> The body has resistance to current flow. <S> More than 99% of the body's resistance to electric current flow is at the skin. <S> Resistance is measured in ohms. <S> A calloused, dry hand may have more than 100,000 Ω because of a thick outer layer of dead cells in the stratum corneum. <S> The internal body resistance is about 300 Ω, being related to the wet, relatively salty tissues beneath the skin. <S> The skin resistance can be effectively bypassed if there is skin breakdown from high voltage, a cut, a deep abrasion, or immersion in water (Table (Table2).2). <S> The skin acts like an electrical device such as a capacitor in that it allows more current to flow if a voltage is changing rapidly. <S> A rapidly changing voltage will be applied to the palm and fingers of one's hand if it is holding a metal tool that suddenly touches a voltage source. <S> This type of contact will give a much greater current amplitude in the body than would otherwise occur. <S> Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2763825/ <A> To answer your question, you should be safe for most electronic work, but it is good safety practices to respect all electric circuits and take fundamental safety precautions. <S> The amount of current a person can handle depends upon many factors. <S> Current quantity, current type, body mass, broken skin, path and length of contact time, etc. <S> Many similar charts exist. <S> From Science online - Benefits and dangers of electricity <S> The only scientific study I know of is Charles Dalziel, 1961 (Inventor of the GCFI). <S> Ultimately, this is where the 100mA 60Hz AC and 500mA DC comes from. <S> As can be seen from the chart DC levels are 3-5 times safer than 60Hz AC. <S> The first 5 rows are based upon human tests (check out pictures at end of report), while the last is estimated from experiments on dogs. <S> It would be highly unethical to do these tests today. <S> A higher current will cause serious injuries but <S> the heart/lungs may start once the person is removed from the shock hazard. <S> Even currents as low as <S> 16mA 60Hz AC and 76mA DC can cause unconsciousness and death if the exposure is long enough as the individual is frozen to the circuit. <S> Your brain, hearts and lungs work, but you cannot let go. <S> All of the muscles of your body are contracted. <S> Think really, really good frozen tag.
| The minimum current a human can feel depends on the current type (AC or DC) as well as frequency for AC. Fibrillation (100mA-200mA) can be deadly because it sets the muscles of the heart pumping randomly requiring defibrillation as the only way to restart the heart.
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Changing AC amplitude and frequency I'm working on a project for school to monitor a 6 and 15 VAC signal coming in at 11KHz for both. I'm trying to find a simple way of recreating that at home for running some tests on circuits I'm developing. I have a simple function generator but it only goes up to 5VAC and has a very low current limiter so that's not an option and I'm poor so I can't afford a lab grade function generator. Ideally, I just want to breadboard a few components that I can repurpose later if I need to. Reading up on some other posts I found Changing the frequency of an AC supply and the common suggested answer was "a rectifier followed by an inverter". I found DC-AC Inverter Circuit - Toshiba Semiconductor (PDF) which explained a bit but it's still muddy for be. I understand I need to rectify wall supply 110VAC to a DC voltage then use an inverter to convert it back to an AC voltage. What I'm unclear on is how: do I convert to 6 and 15 DC ORdo I convert to a 12 and 30 DC (because the AC will do half positive, half negative)? then use an inverter to make it AC at 11KHz? would it be PFM? Any help or examples would be appreciated because most of the circuit examples I've found in my research are for converting 12VDC battery to 220VAC or 110VAC, AC to DC (guess I could just use it backward) but with no mention of frequency modification, or AC to AC but again no mention of frequency modification. <Q> 11kHz is well within the audio frequency range. <S> So if you have access to an audio power amplifier, you could just feed your existing function generator into that. <S> Otherwise, make an audio power amplifier out of common components. <A> First question: <S> Your "simple function generator" was ruled out because it has a very low current limiter. <S> So, what current do you need on the 11kHz signal? <S> Or, to put it a different way, what's the input impedance of your circuit for the 11kHz signal? <S> If your circuit has a low input impedance (i.e. draws a lot of current from the 11kHz signal), then yes, you'll need to build up an inverter. <S> I won't go in to detail with that until you update the question to say how much current you're drawing. <A> As @selvek suggested and what I found it is hard to change the frequency but not too hard to change the voltage using an op amp. <S> I had to make a lot of supplies for my setup <S> so I ended up just going with putting in my function generator that unfortunately was capped around 6VAC (around 5-5.5 but close enough) and used a Non-inverting Operational Amplifier to step it up using the + <S> /-15DC <S> supplies I had to be the +/- <S> references for it. <S> Thank you for your input.
| If your circuit has high input impedance, you can just use an op amp to bump the signal from your simple function generator up to the required voltage.
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Multimeter, outlet and a Spark: the question of simple Voltage measurement So I got a TekPower TP4000ZC MultiMeter . ( Image source ) It has COM port and two red ports: V-Ohm-etc. set at 600V Max and 500mA Max Fused 10A red fused port. Unlike other multimeters, there are no separate red ports for Volts and Amps.TP4000ZC MultiMeter has literally the inscription between COM and 1st red port, saying that using them is 600V MAX and 500mA max. I want to measure voltage in the wall socket. I presume that plugging probes into the outlet would complete the circuit, thus making the multimeter as the load of some sort. Since my first red port has 500mA MAX, I presume that wall outlet has more amperage flowing than pathetic 500mA. (Arent they set at 15A?) So I plug the red cable to 10A red port, black to COM. Black probe goes into larger opening, red probe into the smaller opening of the outlet, set the multimeter knob to Volt setting AND ..... SPARK! (in the outlet) So I disassembled the multimeter and saw that the 10A fuse connects COM and 10A Red port. I guess I shouldn't have put red cable into 10A port, but kept it in 500mA red port V-Ohm-Amp-Hz one. My questions: Why the 1st red port (600 V / 500mA MAX red port V-Ohm-Amp-Hz one) is the correct one to use when measuring voltage in the outlet? Wouldn't plugging the multimeter into the outlet complete the circuit, allowing current (under voltage potential) to flow into the multimeter (regardless of whether the multimeter is plugged in in series or in parallel)? Thus I should have "honored" the 500mA MAX warning (or 250mA on some other multimeters) on the 1st red port and theoretically plugged into 10A port? If the fuse that connects 2nd red port (10A) was set at 10A (F10AL250V), then could I presume that the multimeter just became a "load" in the circuit - e.g. just like any small appliance - then why fuse broke, why it did not act as said "load"? Thank you in advance for your time typing the answers for this.I cannot find clear answers to this despite plethora of resources that just goes around my questions tangently. <Q> Page 16 of your manual: 2.2 DC and AC Voltage measurement 1) Connect the black test lead to "COM" socket and red test <S> leads to the"VΩHz" socket. <S> 2) Set the selector switch to desired “ V [squiggle]” position, and press“SELECT <S> ” key to choose function.(DC or AC) <S> 3) Connect the probes across the source or load under test. <A> Why the 1st red port (600 V / 500mA MAX red port V-Ohm-Amp-Hz one) is the correct one to use when measuring voltage in the outlet? <S> but you need to select V AC range to measure it. <S> Wouldn't plugging MM into the outlet complete the circuit, allowing current (under voltage potential) to flow into MM (regardless whether MM is plugged in in series or in parallel). <S> Current measurement is done by measuring the voltage drop (usually about 200 to 300 mV) across the very low value shunt resistor in the meter when switched to amps range. <S> By shorting out the mains with your meter on current range you would have allowed a current of > 100 A to flow until the fuse blew. <S> This was very dangerous. <S> Thus I should have "honored" the 500mA MAX warning (or 250mA on some other MM) on the 1st red port and theoretically plugged into 10A port? <S> The 10 A port has a much lower shunt resistance value than the mA port and the fault current could have been much higher and more dangerous. <S> If the fuse that connects 2nd red port (10A) was set at 10A (F10AL250V), then could I presume that MM just became a "load" in the circuit - <S> e.g. just like any small appliance <S> - then why fuse broke, why it did not act as said "load"? <S> The load limits the current to a sensible value that the meter can handle. <S> You have been very fortunate not to have been injured. <S> You need to learn a lot more before playing with mains power. <S> Never connect an ammeter directly across a non-current limited supply. <A> The 10A circuit has its own socket so that the current doesn't have to go through the selector <S> switch - the selector does have a contact that switches the mode so that the brains in the meter knows to look at the voltage developed across the 10A shunt. <S> All the other ranges go through the one V-Ohm-Amp-Hz socket, and then through the selector to connect to the appropriate circuit in the meter, the mA and Ua circuits will connect to lower rated shunts - this is where the 500mA warning is needed, to prevent putting too much current through that circuit. <S> Often there's a second fuse that protects the mA range input. <S> The voltage ranges connect to dividers inside the meter that scale the voltage to the value that the meter itself reads, often 200mV. <S> The divider has a fairly high resistance, often in the megohms, so that the current drawn when measuring voltage is very low. <S> The current capacity of the circuit you're measuring is irrelevant. <A> Notice how the right side red port has a bunch of symbols over it, which correspond to the symbols on the dial. <S> The key is that every one of those measurements uses that port - notice both voltage and current are included. <S> Your multimeter is smart - <S> it knows that voltage measurements can only be taken with high input impedance, and current measurements can only be taken with low input impedance. <S> Thus, when you select voltage on the dial, the multimeter has a high input impedance (probably megaohms). <S> So when you connect it to a 120V outlet, <S> virtually no current flows. <S> When you set the dial to uA or mA, the multimeter switches itself to low input impedance. <S> If you were to connect it to the outlet with this setting... bzzt! <S> On the other hand, the 10A port is only used for current measurements, so it always has a low input impedance! <S> If you connect it to the outlet, you'll short circuit the outlet - as you found out.
| Presumably you wanted to measure voltage so the V and COM sockets are correct If you're curious, you can use a second multimeter to directly measure the input impedance and see how it changes as you change the multimeter settings. Current measurements are done with the meter in series with a real load.
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Advice for 25-50 Mbit SPI lines - PCB design I would like to ask some advice for my layout design. I am working on a layout design. The designed PCB will be a Beaglebone Black (BBB) extension card. It is a four layer PCB with the following stack-up: signal/GND GND Power signal/GND. On the circuit there will be four devices (SI8652) which are connected to the BBB with SPI interface. With this extension board, I want to reach approximately 25-50 MBit speed over the SPI. The SPI lines (SCLK, MISO, MOSI) are placed on one layer. No through hole or any vias on these traces. The four SPI devices are connected in a row. The SPI line's lengths are not matched. And my question is: Will I have any problem with the speed? Do I need to do something with signal integrity? Do I need termination resistors, filter capacitors... and so on for the SPI lines? UPDATE: Any advice is welcomed! Thanks. UPDATE:It is working well at around 35-40MHz SPI clock speed. <Q> When you have a signal with short rise time, you need to be concerned about SI analysis and termination to prevent from reflection. <S> A simple rule is this: all traces longer than 0.3*Risetime(ns)*7.5(cm) need SI analysis and termination. <S> 25MHz is some how high for SPI communication and you could have unwanted radiated emission due to long 17 cm traces. <S> Length matching is not the case here but adding some ground traces as guard lines could reduce the probable emission and RF immunity problems. <S> the guard traces could also reduce the return path loop then reducing the unwanted antenna gain. <S> The probable problems are depends on the distance between the SPI routing layer and the ground layer. <S> the less distance, the less problem. <A> Will I have any problem with the speed? <S> Yes, you will. <S> Capacitance on both ends of the lines will probably determine the rise time of the SPI lines. <S> In the past I have not had to worry about transmission line effects, these will not matter anyway unless the line drivers are matched. <S> If you do have line drivers (transceivers) then match the PCB trace to the transceivers. <S> (I'm pretty sure the BBB is only a regular GPIO). <S> If not then look at the capacitance of the drivers. <S> Also look at the fanout (and total capacitance of the line). <S> In my experience, it's not hard to get 25MHz, 30-40MHz is more difficult. <S> The 0.1" header pin of the BBB also contributes a few nH of inductance and creates an RLC filter which may cause some ringing. <S> Do I need to do something with signal integrity? <S> Do I need termination resistors, filter capacitors... <S> and so on for the SPI lines? <S> Probably not, any more capacitance you add to the lines will slow the rise time down. <A> First of all, I must say that I am no way a professional in high frequency designs. <S> The signal lines length should not be a problem here, as I guess; problems really kick in when the timing gets horribly wrong due to the signal propogation speed. <S> Let's assume that for the 50Mbps the frequency is 50MHz, and that the timing gets "horribly wrong" when the time shift is around 10% of the period. <S> So, dl / c = T/10, where c is the light speed, T is the period and dl is the max trace length difference; from this we get the maximum trace length difference of 60cm, which is possibly way higher than you have. <S> The real trace length thing kicks in in the GHz range for the most part. <S> However, pull-up resistors at the each end of the trace are a must according to the SPI documentation, if I recall it correctly.
| The termination resistors aren't really needed in your case also; the rule of thumb is that termination resistors should be used if the transmission line length is greater than 10% of the wavelength; since the wavelength (c * T = 6 meters) is way higher than you trace length, it should be okay.
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No current through AC optocoupler I am just testing this component MOC3021 optoisolator, its datasheet can be found here . This is my circuit in Proteus: Please forgive me for my ignorance of electronics, I have my background in computer science. I tried the whole day yesterday trying to make out what's written about it in the datasheet, the point of connections was easy to understand, rest I don't know much. However, I did learn a lot and welcome to accept more. However as you can see the LED didn't glow, now for one thing I know it's an AC optoisolator but shouldn't current flow from one of the diodes of the DIAC? Thanks a lot :) Update : Problem Solved Thanks a lot, all of you it's a great community here ( especially for noobs like me ).This is the final working circuit. P.S. All answers were correct but I could only choose one, sorry others :| <Q> You've got three problems. <S> 1 <S> : no current limit on the MOC3021 LED. <S> With 5V there the LED inside the MOC3021 will not last long, so add 220 ohms in series with the input. <S> 2: not enough voltage on the output. <S> The output is is a phototriac which has a forwards voltage drop of 1 to 1.5V - to light the blue LED <S> you want 3V or so and a resistor to limit current, 9V with a 470 ohm resistor in series would be a good starting point. <S> 3: the output is a triac: once triggered on, it will continue to conduct until something else stops the current - using an AC supply to power the output side circuit would help here. <A> You have two small problems: No resistor in series with the opto led. <S> Such LED typically require about 3 to 3.5V. <S> You have 1.5V on this side which is far from sufficient. <S> With a 100 ohm resistor, something like 4.5V would give something like 10-15mA. <A> In addition to what dim commented, Your voltage source BAT2 has the wrong polarity. <S> The LED D2 is now in reverse bias by the voltage source BAT2. <S> You could change polarity of either BAT2 of LED D2 in order to let current flow throug the LED when MOC3021 is enabled by the primary side.
| So the current is not limited, and will burn the opto if you try in real life A voltage source too low to light a blue LED.
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Are three-phase contactors rated per pole or in total? Are three-phase contactors rated per pole or in total? For example, have a look at this datasheet for a random three-phase contactor: datasheet In the datasheet it says that for AC-3 loads (1) the rated operational current [Ie] is 32 A, and (2) the motor power rating is 7.5 kW at 220..230 V. These numbers match, since 32 A * 230 V = 7360 W. On the Technical FAQ pages of this manufacturer it says: "Contactor current ratings are per pole. For example a contactor rated 40 amps AC3 could switch upto 40 amps AC3 on just one pole or on all the other poles." (see source ) What I'm confused about is that motor power rating is usually not declared per pole, but in total. So if I use 10 kW contactor for a 10 kW motor, my contactor if oversized by a factor of 3. Is this correct, or I'm missing something? <Q> I think the calculation is not that easy because: <S> the power rating for (industrial) motors is not given for the input (electrical) power but for the output (mechanical) one. <S> If you want the electrical power, you have to consider the efficiency ; you have also to deal with the power factor which is not equal to 1. <S> For example, here is a list of AC 3 phase motors from Emerson/Leroy-Somer: <S> For example, a 4kW motor at 415V (voltage difference between two phases) <S> 50Hz has a rated current of 8.2A and a power factor of 0.81. <S> We could calculate its efficiency: $$ \eta = \frac{4000} <S> {\sqrt{3} \times 415 \times 8.2 \times 0.81} = <S> 0.84$$ <S> Now, if we take the datasheet for contactors from Schneider (TeSys Green) (I think this is approximately the same products you gave the datasheet): <S> We can see that for a 4kW/415V motor, we have a 9A current. <S> This is just above the 8.2A of the aforementioned motor. <S> Also, when you read motor datasheets, you will see that standard output powers are the same given in your document: 7.5kW 15kW 18.5kW... <A> You are correct. <S> The current rating is amps per pole for the power poles. <S> If there are auxiliary contacts, they will have a lower amps per pole rating. <S> Some contractors have one or more normally-closed power contacts. <S> Those may have a different amps per pole rating. <S> There are likely other contactor configurations, but the rating specifications would be similar. <S> You are also correct about the motor power rating. <S> Since motors are usually rated in output mechanical power, there can be only one rating. <S> They are never rated in power per phase. <S> If a contactor is rated for use with either single-phase or three-phase motors, it would have a single-phase and three-phase power rating. <S> The power ratings would also be stated separately for each motor voltage rating for which it might me used. <A> You need to look at the peak current for each pole. <S> For a resistive load, this would be in time with the voltage peak; the other two poles split the return current ( <S> \$\cos 120^\circ = <S> \cos <S> 240^\circ = -0.5\$ ). <S> For an inductive load, there is a phase offset, but the peak height is the same.
| I think the powers that are given in the datasheet you write about is the one you can find for standard motors.
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Are circuits damaged by a constant 55°C environment? I know that sun like temperatures kill electronics, but what about 55°C if it's constant? I have a small commercial web camera that is officially rated 0°C to 70°C. It has no jelly capacitors. I intend to heat it to a very constant 55°C so that it's operating characteristics never fluctuate whatever the room temperate. Clearly the device doesn't explode at exactly 70°C, so there must be some form of curve around that threshold. Can this adversely affect the camera? Is there some form of general life de-rating for all components, not just jelly capacitors? Are devices damaged by high current, or high temperature? amd What exactly gets "worn out" and damaged by heat? exist but don't quite answer this one. <Q> The general working assumption is that life halves for every 10 degrees C rise, so you could guess the lifetime will be only 25% of the life at 25 degrees C ambient. <S> From your proposed operating temperature to the maximum permissible is only 15 degrees so it’s hard to actually do an accelerated test. <S> You could try pushing the rated limits in a destructive test, but you may get an unnecessarily pessimistic number (many parts will work with relaxed specs at very high temperatures). <S> It’s also possible to get an optimistic answer, particularly if it temporarily stops functioning, but that’s less likely. <S> My guess is that it would be okay for some years, but that and $2 will get you a cup of coffee. <S> A better number could come from analyzing each component and applying reliability calculations. <S> The LSI chip and any ceramic or tantalum capacitors are probably the weak links, and any stressed power supply components such as linear or switching regulators. <S> Electrolytic capacitors often have an operating life of only 2000 or 5000 hours at rated temperature (usually 105 or 85 degrees C), so they are a typical weak link, <S> so it’s <S> good <S> your camera has none. <A> This is a very complicated problem that has many caveats and qualified answers. <S> This is just a general overview copied from my comments. <S> As a rule of thumb heat will always decrease the lifetime of the device, however this may still be longer than the expected (warranted) lifetime. <S> All electronic components will fail, eventually, the question is amount of time and method of failure (Quiet fail safe, or dramatic "heat event"). <S> Your goal as a designer/integrator is to identify the effective lifetime and acceptable failure modes, then work to qualify the device or design workarounds. <S> In the worst case, if there is no suitable alternative a workaround may be active cooling. <S> Your best bet is to purchase a few samples and do some tests at 50C and perhaps some accelerated tests at 70C in order to determine if it meets the performance and lifetime characteristics you need. <S> Designing this type of test is beyond the scope of this Q&A, but there are commercial facilities that will rent you time on a thermal test chamber and expert advice, or you can try something yourself. <S> Make sure to place temperature probes on certain parts that tend to be hotter than ambient (e.g. power regulators and filter caps) <S> in order to make sure you are not exceeding their rating. <S> Additionally <S> Note that for ceramic capacitors, the class of dielectric matters. <S> Class 2 (e.g X7R) which happens to be the most common type (Best Price vs Size vs Rating) , have a large temperature dependence. <S> Class 1 (e.g. NP0 or C0G) have almost no temperature dependence. <S> These relationships may impact the operation of the device even if it does not fail or significantly affect lifetime <A> Take a Linear regulator, typically they have a maximum junction temperature of 150C. <S> Now a designer may take "credit for" the operating environment being that of 40C which produces a much larger temperature differential to help keep the junction cool. <S> Initial testing may have shown no additional cooling (heatsink etc...) was needed for the expected use cases. <S> If you then operate at a higher ambient, an ambient outside of the stated operating range, then there may be permanent damage. <S> My advice, if this is what you want/need to do, open the device up and use a thermal camera to see how hot parts get. <S> By some simple maths you would get a rough estimate as to how hot the case of some devices would get. <S> Do NOT let anything exceed 100C
| Depends... Running electronics at temperature does shorten their life but equally will damage them if taken beyond their operating temperature.
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3-phase 380 V to 3-phase 230 V I have a portable bearing heater which works with 3-phase 230 V power supply. My power supply is 3-phase 380 V. Is there any way to convert the 3-phase 380 V to 3-phase 230 V? Please note that since the equipment is portable, it is important that the solution be portable too. I added the picture of wiring diagram of the equipment.The manual indicates: The equipment is designed for 3 phase 230V power supply (Between each hot wire, 220 volts can be measured) when 2 phases are connected. it means 2 phases out of 3 phases are connected. The supply power is 3-phase 380V,which means between each hot wire, 380 volts can be measured and between the neutral and any of hot wires, 220 volts can be measured <Q> Figure 1. <S> Coloured up version for single-phase 230 V + N wiring. <S> It appears from the wiring diagram that you can just connect L3 to neutral instead with no internal modification. <S> The only concern should be that the components' insulation now has to withstand 230 V instead of \$ \frac {230}{\sqrt 3} \ \text <S> V\$ . <S> You should check, if possible, that they are rated for that. <A> There may be a simple solution to this depending on the connection, If the load is connected between the two phases and no neutral connection as you have indicated in the comments section, you can connect the bearing heater between L1 and the neutral from your 380v supply. <S> This will give you a voltage of approximately 220v and de-rate the output power by about 1KVA. <S> The only other option without knowing the internal connections would be a big transformer on a trolley. <S> Looking at your wiring diagram it appears that what I suggested above will work. <S> The only problem that I can see is if the 230v neutral wire is used by any monitoring electronics not shown on the diagram. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> simulate this circuit – Schematic created using CircuitLab <S> You have only to rewire heating elements from wye to delta connection. <S> EDIT: <S> You said: Actually the equipment use 2 phases out of 3-phase 230 simulate this circuit <S> All you have to do is to connect your load between phase to neutral, and not phase to phase anymore. <S> But 22kVA seems huge power for single phase operation. <S> You'd better dismount your device and post some pictures. <S> For exaple you could separate the electronic part which needs low voltage by means of SMPS or transformer and the power part by replacing two phase graetz diode bridge with 3 phase diode bridge. <S> Check <S> if L1, L2 and L3 are series connected as shown in schematics. <S> There is a notation: 3x6mm^2 or 3x10mm^2 at 2x220V, which means that sections are made for different voltage, combining them series or parallel. <S> You have to use phase to neutral voltage for control unit. <S> This can be easily done by rewiring L3 control supply to N. <S> The next thing is to check if the coils are really connected in series. <S> IMO, by combining L2 and L3 in series or parallel you can get 380V/220V. <S> So they shold <S> be connected parallel right now. <S> simulate this circuit Another possibility for combining multiple voltages. <S> Watch if the windings are wound in same direction or they are connected anti series, dectructing their magnetic field. <S> simulate this circuit <A> Most likely your heater expects supply of three phases at 230 V(rms) <S> phase-to-neutral, and the supply you have is three phases at 380 V(rms) <S> phase-to-phase. <S> Fortunately these are the same* thing! <S> So most likely <S> you won't need any conversion, except perhaps a plug adapter. <S> (*: Within a few percent that can be chalked up to rounding; and utilities seemingly redefining their nominal voltage by 10 V up or down every several decades without non-electricians in the populance noticing; and is dwarfed by tolerances anyway). <S> It is quite rare and nonstandard to find three-phase AC at 230 V measured phase-to-phase or 380 V measured phase-to-neutral, so it would require extraordinary evidence to believe your heater or supply is one of those.
| The best way to do this would be a 3 wire connection as shown above by replacing the existing plug with a 380v one, or if you need to keep the 230v compatibility an adaptor box that is clearly marked for use with only this unit.
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Fan speed switch: why OFF :: HIGH :: MED :: LOW? Many AC fans, including inexpensive floor models, have a rotary speed control which rotates from OFF to HIGH to MED to LOW . Since the switch does not rotate 360°, you have to rotate it backwards to shut it off. This has always slightly annoyed me, because it appears to make no logical sense to have to speed the fan up to turn it off. Why are the speed settings from fastest to slowest? This design seems counter-intuitive, but I assume there is a good reason because the design is so common. <Q> Fans of that type have induction motors with two windings with a capacitor in series with one of the windings. <S> For every individual motor design there is a certain capacitor value that allows the motor to develop maximum torque and operate at the maximum corresponding speed. <S> Smaller capacitor values are used to reduce the torque so that the load overcomes the motor's torque and slows the motor down. <S> When the motor is turned on, the motor must overcome the static friction of the bearings and get the fan moving. <S> It must then accelerate the motor inertia. <S> That means that using the highest torque setting is desirable for starting the motor. <S> Once the fan is moving, the torque can be reduced for lower speed operation. <A> It may be intentionally designed to force the fan on "high" for it to spin up before allowing it to drop back down to "low" The amount of power needed to overcome the static rolling resistance can be higher than what is output to maintain the "low" speed, thus it could end up stalling and not moving or only wiggling back and forth on a "low" setting <A> Not just fans <S> A lot of devices are designed this way. <S> It almost seems standard, even though I agree it is counter-intuitive. <S> For example, in my kitchen I have: <S> Cooktop hood fan/light - fan goes from Off to High to Low (continuous except for switch Off) Cooktop hood fan/light - light goes from Off to High to Low (discrete settings) <S> Cooktop gas burners - Off to High <S> to Low (continuous except for switch Off) <S> In the case of the fan, the other answers regarding motor startup make sense and would apply to this fan as well. <S> In the case of the light, I suspect it is to match the fan since it is part of the same device. <S> The end result is <S> all using the counter-intuitive Off->High->Low
| In the case of the burners, I think it is because ignition needs to be done when first turning on a burner and ignition is most reliable on the High setting, so starting with High makes sense.
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Fast voltage translation from 3.3v to 5.7v I need to translate a 3.3v signal to 5.7v with transition of around 60ns or less. Voltage translators such as the TXS0101 usually specify the logic high level for the input as VDD x ~0.65 (= 3.7v in my case), and their upper level is usually 5.5v. I have tried with a P and N MOSFET (see diagram below) but the P MOSFET is not saturated because 3.3v - 5.7v = -2.4v which is insufficient. Adding a resistor between 5.7v and the P MOSFET source fixes the saturation but the switching time becomes too long. Is there in fact an IC which can do what I want? I've looked around but can't find one. One other consideration is that a voltage drop at the output (to around 4.7v) is acceptable, but doing that without burning quite a lot of watts seems to be tricky. I tried using a 4.7v zener to drop the P input to 4.7v but it wastes a lot of current. Update with 2nd attempt here. It's not quite as quick as I'd like but think it can work. Thanks for the input, the baker clamp was new to me, works like a charm. 2nd attempt: Fall time: Rise time: simulate this circuit – Schematic created using CircuitLab [edit] relaxed timing requirement to 40ns [edit] relaxed timing requirement to 60ns [edit] added 2nd attempt <Q> eg. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Rise time should be in the 25ns range typically and fall time better (10%-90%). <S> If you only need 50ns typical you can increase R1. <S> Do not use a larger than necessary MOSFET for M1 or the drain source capacitance and Miller capacitance (assuming some source resistance) will kill your speed. <A> This IC is quite a bit overkill, but it would do the job. <S> I am sure there are single-channel level shifters/translators. <S> The issue is finding them. <S> You could probably also use a TTL LVDS receiver in a pinch, by biasing one of the inputs and attenuating the input signal. <A> There is one way to do this with a single transistor (NMOS), I used the BSS123 (you can get it in small sot-23 packages). <S> You could try something like this: <S> This bidirectional—you could send a signal from the high voltage side to the low one, too. <S> But say, you want to push a 0 from the DL (low side) line to the DH (high side) line, the NMOS will turn on bring the 0V level to the drain. <S> If you leave DL floating or apply 3.3V, the NMOS will be off and DH will be high due the 10k pullup <S> \$R_2\$ . <S> As far as the speed, well, I did a quick test in LTSpice and found that a 0V signal propagates through (from DL to DH) at a reasonably high speed (~5ns). <S> Of course, in practice this will depend on the input impedance of whatever you attach to the DH side. <S> The green plot shows a 0V signal going from the DL to the DH side (blue). <S> You can find readily available versions of this circuit on DigiKey for example check this out Hope this helps.
| If your capacitance loading is light (as in the 15pF load shown below) and your power budget generous, you can just use a single transistor inverter.
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Is there a wrong way to wire a potentiometer? I'm using Circuit Wizard to simulate a circuit and I'm trying to understand how to wire a potentiometer. While trying to learn how current flows through the pot, I found that it works in both directions as it's just a variable resistor. But when drawing a schematic, which is the RIGHT way that current should flow through the component? See the following two schematics: I have the following questions: Is current supposed to flow in the direction of the arrow or away from it? Is it unusual to wire a potentiometer what appears to be reverse as I've done here? Is there any problem with not using the third lead in this circuit? <Q> If you are just using it as a single variable resistor, rather than as a voltage divider, it is common practice to connect the wiper to the "unused" end of the element - <S> this ensures that there is still a resistor in the circuit if the wiper connection becomes intermittent. <A> The arrow you are referring to, which is the "wiper", is only an arrow to show that it's a wiper. <S> That is to say that it's irrelevant to direction of current. <S> Electrically it doesn't make a difference which direction the current flows. <S> But physically, there are some considerations. <S> Typically, a pot will have 3 pins, with the middle-pin as the wiper. <S> For a simple symmetrical trim-pot, it may not matter physically how the pot is oriented on the board. <S> But for some pots, you may not be able to fit the pins in a certain orientation (too close to the edge of a PCB, for example). <S> Thus pin 3 might be closer to the circuit you want then pin 1, and you simply choose to use pin 3. <S> A common determining factor is the direction of the knob's turning. <S> Say for example you want the resistance to increase as a pot's knob is turned clockwise. <S> That determines which end of the pot to connect to. <S> Because you want the resistance to increase in a certain turning direction, you must choose the other pin to use besides the wiper (1 or 3). <S> This should be detailed in the datasheet. <A> It's good to keep track of which way a pot is wired. <S> All the schematic symbols for pots that I have created have that information as part of the symbol. <S> For pots that are not 'B' (linear) taper, such as audio volume controls, it's not only good, it's necessary , because the resistance change for a given angle rotation is much different at one end compared to the other. <S> The consequence would be that the volume is very high over most of the rotation and then suddenly drops as the pot is rotated mostly CW, with reasonable volume levels being very hard to set. <S> It's intuitive (to me anyway) <S> that clockwise rotation of the shaft should increase the brightness in this case, so you would want to connect the pot so that happens. <S> (You should also add a series resistor so that the maximum current is limited and does not damage the LED and/or pot). <S> In fact, it's functional either way, and with a linear pot, not much different. <S> And, as others have suggested, it's typical to connect the "open" end of the pot to the wiper, which slightly reduces the wiper noise and has other beneficial effects. <S> As far as current flow, the simulator diagrams show conventional current, which flows from + to -, so in the direction of the arrows. <S> Electrons flow the opposite direction but you don't need to worry about that in most situations.
| A potentiometer is just a resistor - current can flow through it in any direction (including from/to both ends to/from the wiper).
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Why are there so many vias on this board? I was looking at the MMZ09312BT1 development board layout, and I was curious about all the holes they have on the board. Are these vias? What is their purpose (I heard somewhere that they are meant as a filter)? Also it doesn't say explicitly, but is it possible to tell if they have a ground plane on the bottom layer? Datasheet: http://cache.freescale.com/files/rf_if/doc/data_sheet/MMZ09312B.pdf Development board on page 8 <Q> This is generally referred to as via stitching, and it's generally used to reduce either the high-frequency electrical impedance or the thermal resistance between layers. <S> However it's easy to do, and doesn't generally hurt anything on a board as sparsely populated as this one. <S> You would need to consult the design documents to determine the stackup details if the layers aren't clearly visible. <S> Often for dev/eval boards the manufacturer will provide a full package of manufacturing documents. <A> It's a high-frequency RF part. <S> 900MHz = 30cm wavelength. <S> So even a board that's a few cm across is a significant proportion of a wavelength. <S> The vias are to make sure that the top copper is really a ground plane, and not some weird unintended resonator. <A> I assume there is a copper pour on the top as well, and the bias are stitching the top and bottom planes together. <S> Depending on the frequency of operation, it is possible that the via spacing would help to cancel out emissions. <S> But in this case this effect would not be significant. <S> What I find interesting is the different via spacing and sizes in the input and output sections of the board. <S> These must be significant, probably contributing to impedance coupling or simply filtering. <S> I’d be curious to know the relation between via spacing and wavelength in those sections. <S> Of course, these could also be attachment points to simplify test setups. <S> You might be able to get a straight answer in the manufacturer’s forum. <S> In low frequency boards, you would find prototyping sections that look very similar, but that is clearly not the purpose here. <A> This IC has 30 dB of gain; even small amounts of feedback will upset the gain flatness and the phase linearity, both of which will upset dense constellations and degrade the data-eye. <S> The IC is only 3mm across, with that footprint-octagon defining the 3mm. <S> The via spacing is about 1.5mm, so the via density has some purpose. <S> If each via is 1 nanoHenry inductance, which is + <S> j6.3 ohms at 1GHz, we can view this "PCB" as a cascade of not-very-good voltage dividers, each divider having a series element and a shunt element. <S> The series element is the low-inductance PCB surface; the shunt element is the high-inductance via.
| It can also be used to provide a low DC resistance path between layers for high current pathways. In this case the reason is certainly RF impedance, however the level of stitching shown is probably overkill even for a 900MHz RF part.
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LM324 instrumentation amplifier gain and offset problem I want to translate a Pt100 RTD to a 0-10V range, to be read by a PLC. I designed a circuit (instrumentation amplifier) using the LM324 op-amp. It is powered using a single 12V supply.The resistor R11 is to be referred to as a Pt100 RTD. I am supposed to get 0 volts at RTD resistance 100Ω, but I get either 2.2V or fluctuating mV if I adjust the pot R4. Also, I am supposed to get 10V at RTD resistance 250Ω, but I get only 6.5V. As per the gain calculations I took all flexible possibilities. However R4 (the gain pot) doesn't change the gain much. Where am I going wrong? simulate this circuit – Schematic created using CircuitLab Update from the comments: You can think of R11 as a Pt100 heated from 0 to 400 degree C, giving 100 to 250 ohm resistance. <Q> The reference leg of your bridge is at, $$ V_R = 12 \dfrac{100}{4800} <S> = 250 \text{ mV} <S> $$ <S> The RTD leg of your bridge will span from 250 mV to $$ V_{T,max} = <S> 12\dfrac{250}{4950} \simeq 606 \text{ mV} $$ <S> Say for arguments <S> sake, the RTD leg is outputting 300 <S> mV. <S> You now have a differential input voltage of 50 mV. <S> Now if the op-amps were operating in their linear region of operation, both input terminals of OA1 would be at 250 mV. <S> Similarly, both input terminals of OA2 would be at 300 mV. Ignore potenimoter R4 and treat R1 as the gain setting resistor. <S> The two op-amps are trying to drive their outputs to develop the 50 mV differential potential across R1. <S> The top of R1 at 250 mV, the bottom of R1 at 300 <S> mV. <S> Now OA1 is going to try to sink the 50 uA flowing through R1. <S> OA1 must drive its output low enough to allow 50 uA to flow through R2. <S> In this that would be -1.1 V across R1, placing OA1 output terminal at -850 mV. OA1 clips at the ground rail (assuming it had the drive strength pull all the way to ground). <S> Since the bride output is uni-polar, OA1 is always going to be trying to sink current below with its output at or below 250 mV (where the LM324 has no drive strength left). <S> Loosely speaking, getting an linear output of 0 - 10V with an LM324 on a 12 V supply is going to be a challenge. <S> I suggest you start with split 15V supplies to gain an understanding how this circuit works (though spice is an even easier place to start). <S> You may also want to add a resistor below your bridge to place the common-mode voltage mid-supply. <A> You just have too many variables floating around. <S> Tune your in-amp with some known voltages or resistors to verify it's function, Use 1% resistors (or better). <S> You also have no way to balance your bridge. <S> If you need zero output at 100 \$\Omega\$ , replace the RTD with an accurate 100 \$\Omega\$ resistor, and use a trim pot on the bridge to tweak it out (keeping in mind that the LM324 is not guaranteed to go below 20mV with a single supply -- You'd have better luck working around midscale). <S> Lastly, RTDs are not subtle devices. <S> There's no reason to use an in-amp to amplify a 250% change in a resistance. <A> The problem here is that the input buffer gain is not ideal for a single supply and that the reference I took this circuit from, did not mention this. <S> The problem is pointed out by @sstobbe (Limited Driving power of LM358). <S> This answer is for readers who come across the same problem and are looking for a circuit that just works. <S> So, here it is: simulate this circuit – <S> Schematic created using CircuitLab Advantage here is that you can control the gain using only one resistor(R6). <S> It works fine with 150mV @ 0 <S> °C and 9.98V @ 400°C Input differential Voltage(V2 - V1) is 0 to 300mV.
| Your are trying to operate your op-amps beyond the rails.
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If neutral line touches the live wire, how to protect the devices connected due to neutral fault? If the neutral line accidental touched the live wire, then due to momentary surge the home appliances connected gets damaged, what are the ways to protect the devices. Would RCCB be helpful to protect the home appliances. Initially I thought the devices are burnt due to high voltage hence installed servo stabilizer but even that got affected. <Q> If the neutral line accidental touched the live wire, then due to momentary surge the home appliances connected gets damaged, what are the ways to protect the devices. <S> If neutral touches the live wire there will be a surge in current through the short-circuit but the voltage will drop . <S> That will typically not cause a surge in voltage. <S> The regular MCB should trip. <A> They do not take up much room. <S> It protects against accidental surges in the 120-240 volts entering the home from such things as lightning strikes. <A> Circuit breakers or fuses will provide protection from short circuits; shorts quickly develop very high currents and quickly trip breakers or melt fuses. <S> An RCCB will not do anything extra to stop a line to neutral short. <S> The RCCB trips when it senses a ground fault when it detects a current imbalance between the line and neutral. <S> With a line to neutral short, there is high current but no imbalance.
| In many smart homes with sensitive electronics and appliances, I have installed a surge protector at the electrical panel.
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What exactly are low speed devices? The definition of I2C is - "A multi-master multi-slave serial bus used for communication between low-speed devices ". However, I have been searching the web for a concrete explanation for this but can't seem to find any, and other search results are troubleshooting tips for related errors. The closest I got to a definition was these restrictions listed in the explanation for how USB works : The maximum packet size for data transfers is 8 bytes They cannot use Isochronous nor Bulk Transfer pipes May only have 2 endpoints other than the zero endpoints What are some examples of low-speed vs high-speed devices and what exactly does "speed" (speed of data transfer?) mean in this case? My guess was that low-speed devices use serial transmission whereas high-speed devices use parallel transmission, but I am really unclear about this. I am a newbie with regards to electrical engineering and would appreciate if a less technical/guided explanation could be given, thank you! <Q> Low speed is somewhat subjective; in the case of I2C, the standard speed is up to 100kb/sec, full speed (previously known as high speed, confusingly) is 400kb/sec. <S> These are the two most common speed ratings. <S> The packet size and payload sizes generally are not impacted by the speed definition, although most interfaces have 8 and 16 bit payloads, but may easily have more. <S> The "low speed interface" terminology was introduced (quite informally) to distinguish these interfaces from high speed interfaces (operating at several Mb/sec such as <S> 10/100 ethernet or even Gb/sec rates such as Fibre channel , PCI Express , <S> Infiniband and the fairly recent 100GE ethernet to name but a few). <S> Many high speed interfaces are serial in nature (there are advantages at the physical layer due to signal skew issues in parallel interfaces). <S> As I noted, there is not really a hard definition, but generally, interfaces running below perhaps 1Mb/sec are usually considered low speed. <A> "Low speed" is a vague definition, on purpose. <S> You also hear about "low power" and "low frequency". <S> If you look up what they mean, "low" can be 1kHz, or 5MHz. <S> 1 Watt or 100 Watts. <S> The rule usually applied is that if it is "low" it is low enough not to worry about. <S> For instance, a high power motor means that you need to consider the cooling of the device during use. <S> Whereas low power motors means that they will easily handle the heat loss etc during normal use in normal environments. <S> To put another example, on a power grid, frequencies of 1kHz could be considered as high, as the distances the wires travel over means that you run into issues due to the frequency. <S> Meanwhile a 1kHz signal on a 5cm PCB would be low signal, as the issues with the design are unlikely to be due to the frequency. <S> Back to I2C, they are talking about low speed devices which are slow for the usual set up. <S> So they would be updating a few hundred times a second on a small PCB. <S> This compares to a high speed sensor which could be updating at thousands, or more, times a second. <S> But if you have a larger set up, over a metre or so, you will need it to update slower. <S> The reason for this is the way that I2C works. <S> The data and clocks lines are pulled high to power rail by a resistor, which means that the data rate is limited by the size of this resistor and the capacitance of the line (which is made up of parasitic in the PCB and in all the components on the line). <S> To answer you question: depends. <S> Low is relative term. <A> I2C uses (in most cases) <S> passive pull up and is only actively driven low. <S> This limits the speed the bus can operate at because it takes time for the pull-up resistors to charge the bus capacitance sufficiently to pull the bus high. <S> Usually, for I2C the standard limits are either 100kHz or 400kHz, but there is an (uncommon) <S> variant that goes faster (With an active pull up, but really SPI is the more common approach if you need speed). <S> For the sorts of things you usually hang on I2C this is fine, we are mostly talking about temperature sensors, very slow IO, configuration interfaces, power regulators, fan controllers, that kind of mostly housekeeping stuff where a hundred microseconds or so to complete a transaction is no big deal because you are not doing them that often.
| There is NO universal formal definition of "Low-speed device", and the USB one, for example, is specific to the USB specification.
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Why are SRAM based FPGA used more than NVM based FPGA? SRAM based FPGAs need to load the bitstream again after power off. Meanwhile the Non-Volatile based one don't need that. I wonder, why are more experiments and security research done on the SRAM FPGA than the the NVM based one, it seems that the volatile technology is more used regardless of its security limits (when it comes to ensuring secure boot). (PS: I have no statistics, it is a personal observation) <Q> The main driver is the fact that SRAM is highly compatible with the same physical process that is used to implement the actual logic. <S> Indeed, most FPGAs these days are based on LUTs (lookup tables), which are really just tiny bits of RAM themselves. <S> On the other hand, the process required to build EEPROM (nonvolatile memory) requires extra steps — to create floating gates with special oxide thickness, etc. <S> This process is NOT directly compatible with the logic/SRAM process. <S> This means that nonvolatile FPGAs are somewhat of a compromise in both areas. <A> In addition to Dave Tweed's answer regarding the fabrication processes involved, most flash-based FPGAs actually still use SRAM to drive their fabric. <S> The bitstream is loaded into the SRAM from flash just like in a more conventional FPGA, the only difference is that the flash is internal. <S> This architecture is evident when you look at their datasheets and appnotes. <S> In particular, some devices like the Lattice MachXO2/3 support reprogramming their flash while the device is running, which is only possible because the device actually runs from SRAM instead of directly from the flash. <S> So a "flash based" FPGA needs the flash in addition to the SRAM, which means it needs more die area. <S> Regarding security, you are right to point out that the FPGA startup process can be a weakpoint in terms of allowing the bitstream to be captured. <S> To help close this gap, many FPGAs now incorporate support for bitstream encryption, which is based on a secure key stored in dedicated memory within the FPGA. <S> A bitstream image is encrypted with this key, loaded into the configuration memory, and then when the FPGA starts up it reads the encrypted bitstream in, and decrypts it as it loads it into its (Some microcontrollers that require external memory have similar capabilities, and the principles are largely the same.) <A> More than anything, it depends on your requirements. <S> While Size, Weight and Power (SWaP) are the main drivers for ICs in-general, if you aren't compelled to develop an ASIC because of those requirements, Performance is your next consideration, which may push you back to an ASIC anyway, but, you may be able to use an FPGA if you can afford the SWaP trade-offs. <S> FLASH-based FPGAs require "no time" to configure, as they are "instant on". <S> Your design may require this. <S> FLASH technology is lower power than SRAM <S> FLASH-based FPGA doesn't need a BOOT PROM, thus one chip vs. two (or more). <S> You may have a requirement to power-up in the previous state. <S> FLASH-based offers more Rad-Tolerant solutions. <S> There are ways to deal with radiation requirements, or SEUs in-general, in SRAM-based FPGAs, but, Microsemi offers "hardened technology" FLASH-based FPGAs (Actel, now Microsemi), traditionally, have not had the density or performance one could achieve with SRAM-based FPGAs, so, if performance was the driving factor, you would choose Xilinx or Altera (now Intel), or maybe Lattice. <S> Essentially, you are driven by the requirements of your system, and your IC specifically. <S> Early-on you address these requirements and perform a trade study of the different FPGAs (spread sheet). <S> SWaP and performance, followed by recurring cost are the main considerations you want to iterate on with your team (systems, CCA, maybe even SW) that is fedback to your project Chief engineer/manager. <S> Other concerns such as reliability, manufacturability, etc. are usually provided by other team members from those respective organizations, but usually don't mean much without your initial trade, and typically won't prevent your choice. <S> There are articles on the web if you search "SRAM vs FLASH FPGAs", but you will likely learn more from a trade-study against your requirements using the data sheets than you will anything else. <A> Arguably that's just as secure as storing the configuration internally: a dedicated attacher who is able to extract the private key from a decapped chip will also be able to read out the internal flash. <S> Flash-based FPGAs are typically used when the implementation is rather simple (so a large SRAM FPGA is not needed) or when an instant startup is required.
| To address the security aspect, most modern SRAM FPGAs can be configured with an encrypted stream , usually with modern encryption standards such as 256-bit AES.
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Can a PCB circuit have through hole components on both sides of the board I'm making a temperature sensor ciruit on a 1×1 inch PCB and needed to put some components on the other side to fit everything. All the components are through ole. Is this feasible, I was able to order the boards and it said there wasn't an error, but I wanted to make sure I didn't do anything incorrectly. <Q> There is no PCB design or electrical reason this is not feasible and in most cases you would need to define a custom DRC for this to be an error. <S> Note that unless you have appropriate 3d models interference between TH parts won't be checked either. <S> With two sided TH components, generally speaking at least one side will have to be soldered by hand for commercial production. <S> So if you persue commercial production with a contract PCBA assembler or with your own production line you will likely pay much more per board to manufacture . <S> With one sided TH <S> it can be wave soldered , greatly reducing cost. <S> If you are assembling small runs by hand anyeay, or designing as a kit, this may make things more difficult, but generally not as big a problem. <A> Although possible, that seems cumbersome and you run the risk of the pins and solder colliding with other components. <S> Leading to placing components in awkward positions (Eg., your electrolytic and the connector). <S> You might be better off simply replacing at least one of the components with a surface-mount one. <S> If you stick to the larger sizes (eg., 1206 and up), these are really not that hard to solder by hand. <A> That is if everything connected to the same pad is supposed to be the same node. <S> Just make sure you pay attention to the order you solder your parts onto the PCB. <A> Placing components on both sides of a board is avoided where possible, as it makes board assembly more difficult, but it's certainly allowed. <S> What isn't allowed, however, is placing a pin inside another component. <S> The large diode on the back of the board has a pin that will stick into the resistor R2 on the front. <S> Move R2 down until it is aligned with R1 -- not only will this resolve the collision, but it should make room to put the diode on the front of the board.
| If you are able to solder both components into their place and you ensure the via has enough diameter to accommodate both components pins, I see no reason it would not work.
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How should I used two MLX90614 sensor on one Arudino Uno given that the sensor has a fixed I2C bus adress? I have an arduino uno and want to use a non MLX90614. I plan to used more than one sensor on the same Arduino but I don't know how to use two MLX90614 on the same I2C bus. <Q> Typical 2-channel device is PCA9540B . <S> The multiplexer has its own I2C address, so you will need to add a code that programs it to switch between two "channels", and then use the same routine to access your sensor(s). <S> Here is the typical connection: <S> As a bonus, the chip can perform level translation if you need one. <A> While the answers suggesting using I2C mux chips is good. <S> In the case of just two overlapping devices, I’d suggest instead an I2C address translator, such as the LTC4317 . <S> Such devices let you effectively change the address of downstream devices. <S> This was you don’t have to have the extra software to operate the mux. <S> This device works by rewriting the I2C address on-the-fly by XORing a byte into it. <S> Thus, if you want to use two devices with the same address but make one with a different LSB, you’d set up the translator to XOR 0b0000001 <S> into the bus before the second target device. <A> Probably the easiest way to do this is by using a I2C address multiplexer such as the TCA9548A 1-to-8 I2C multiplexer. <S> This particular one will allow you to address up to 8 of the same device using a single I2C address. <S> All you do to contact each individual sensor is write a byte to the multiplexer with the desired sensor number and then communicate with it just like a regular singular I2C device.
| To get two identical I2C devices on a single I2C bus you need to use an additional I2C device called "I2C-bus multiplexer".
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Converting a soldering station from 110 V to 220 V i bought this soldering station from the USA and it is 110 V. I need it to be 220 V. Is this possible without external tools? There are 3 Trimpots: V1 VR1 VR2 This is the inside of the station: <Q> First check the specs of the unit using its manual. <S> Many devices (such as phone chargers) accept a large range of voltages so that they may be used all over the world, this station may be the same. <S> Alternatively, you could find a step down converter that will convert the 220 VAC to 110/120 VAC (something like this: https://www.amazon.ca/Voltage-Converter-220-240-110/dp/B001ES8YY6 ). <S> Just want to note that simply using an adapter would not be enough because the source voltage would still be 220 VAC. <A> The controller looks as if it's an SMPS convertor. <S> The input looks to have a simple bridge rectifier and 2 supply caps. <S> The input voltage limitations could probably be gauged by looking at the voltage rating of those caps. <S> It has a transformer, and given the size the unit may be switching in the kHz range. <S> Unless the unit is marked to accept 110-240V as a range <S> it is unlikely you will find any way to alter the input voltage range based on the pots in it. <A> The power appears to be derived from a switching power supply based on a Power Integrations TOP253 . <S> It could have been designed as a doubler feeding the supply, but it does not appear to be so designed. <S> If you are outside of that, the effort to make it compatible internally would not be worth it. <S> My impression is that it probably does not have <S> valid safety agency approvals for Western countries- looks like a somewhat shoddy Asian product.
| If the power supply has been designed for 80-265VAC operation, it will probably work as-is (except for the plug). So, refer to the manual and markings on the unit for the recommended voltage range.
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Why do 0 dB attenuators exist? I am reading this paper : Upgrade of a low-temperature scanning tunneling microscope for electron-spin resonance and found that for thermalization, they use 0 dB attenuators. Page 6: Having the finite cooling budget of our cryostat in mind, and noting that the heat transfer of the RF line is dominated by the micron thick Ag coating of the center conductor, we thermalized all SR cables using 0dB attenuators I understand attenuators reduce signal intensity (power), but what does it mean to have a 0 dB attenuator? What is the general purpose of 0 dB attenuators and in the paper is it just being used for thermal coupling? <Q> A 0dB attenuator is an attenuator that is electrically and mechanically just like other attenuators in that product line. <S> You can use them in place of another attenuator when doing tests. <S> It acts just like the other attenuators (in terms of insertion loss and frequency range) and fits in exactly the same space mechanically. <S> A 0dB attenuator meets all the specifications of the others in the same product line, so swapping it in keeps all other performance characteristics of your setup the same. <S> Might not matter if you are tuning a CB radio, might be critical in high performance, cutting edge stuff. <S> The point of using one in the referred paper seems to be to reduce the amount of heat seeping into a very sensitive piece of equipment. <S> Heat causes noise in the extremely low temperature circuits the paper describes. <S> The 0dB attenuator seems to be used because it passes the RF signal (almost) as well as the cable, but doesn't pass heat as easily as the cable would. <A> JRE's answer covers why 0dB attenuators exist in general, in this specific case it sounds like they're using them to make good thermal contact between the cable's inner conductor and the thermal isolation stage as you can see in the picture on page 20, and thus to the cryogenic reservoir which is cooling the thermal isolation stage. <S> Basically, you want the cable to be at the same low temperature as the stuff down in your cryogenic can, rather than conducting all the heat from the connection up top at room temperature. <S> An added problem is that cables have anisotropic heat conduction: they conduct heat better along their axis than they do radially, because the insulator that keeps the centre conductor electrically isolated is also a good thermal insulator (or at least a lot better than the metal of the inner conductor). <S> Attenuators have better thermal conduction between the outside and the centre, and so they do a better job of thermally connecting the inner conductor of the cable to the thermal isolation stage. <S> Judging from the mounting it also provides some mechanical stability, the attenuator is held in one place while the cables attached to either end will move around a bit due to thermal contraction as you go from room temperature down to 4K or lower. <S> Cryogenics is hard, and heat transfer from the rest of the horrible 300K world is your greatest enemy. <S> Page 20 <A> The attenuator substrate is usually a better thermal conductor than the cable’s Teflon dielectric. <S> They don’t, however, help with noise coming down from 300K electronics. <S> For that you would want attenuation equal to or larger than the ratio of room temperature to the temperature of the attenuator. <S> For example, you would want a 20 dB attenuator to reduce 300K noise to 4K level.
| A 0-dB attenuator helps with physically thermalizing the center conductor of the cable.
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Interrupt debouncer for switch sensor I want to read the state of a switch sensor (magnetic contact). I'm thinking to connect the switch between an interrupt pin of my micro and ground, listening for CHANGE status. In this way I have debounce problems, so I wonder what's the best way to deal with this problem: 1 - Disable the interrupt listening when it fires and enable it after the program has read the value (0 ->ground->switch closed OR 1->floating->switch opened)I don't think a floating state could be good, but if I set the interrupt input pin as INPUT_PULLUP it's no more floating and could be good, am I right? 2 - Connect the switch to an interrupt pin and also to a digital input pin through a transistor as a switch.In this way, when the interrupt is triggered I read the value of the digital input pin. On the other hand, in this way I complicate the circuit 3 - To use an hardware debounce circuit Are these three solutions valid? Are there any more better than these? MORE CONTEXT:I'm using these switches as limit switch for a garage door.In my code I'm using a timer to raise the garage door for ten seconds OR until the limit switch interrupt is triggered EDIT:with a change in my code I solved the problem using interrupts. However you suggest to use polling since it's not a real-time application, so I wonder: isn't better the same to use interrupts so that I can use a power safe mode for my micro and wake-up it only when an hardware interrupt is triggered? If i use a polling solution the system would not be suitable for battery power supply; isn't it? <Q> The real answer is that you shouldn't use interrupts for this at all. <S> You should connect the switch to an input pin and poll the pin periodically. <S> The period of the polling should be greater than the time interval during which the switching transient (bouncing) subsides. <S> Also, you need to pull up your pin with a pull-up resistor. <S> Simply connecting the switch in-between GND and the pin is not good enough; Some microcontrollers have internal pull-up resistors which you can activate from software, so you don't need to use an external one. <A> I'm using these switches as limit switch for a garage door. <S> In my code I'm using a timer to raise the garage door for ten seconds <S> OR until the limit switch interrupt is triggered. <S> This is a low-speed application and an interrupt is not required to give adequate stop. <S> The ramp down time of the motor will be hundreds of times longer than a simple input polling and debounce circuit could handle comfortably. <A> I think of these few points <S> Do you have to care about debouncing when you simply wait until the door touch the sensor for the first time? <S> Hardware circuit could be as easy as RC filter. <S> RC Filter helps big time and it is very simple. <S> If your sensor wire is longer than one feet, I'd recommend external pull-up, to make at least 10mA. <S> It is much more robust against EMP caused by storms etc. <S> The interrupt is better for very short events, like miliseconds or less. <S> Raising your garrage door will most certainly take more time. <S> I'd recommend pin reading every, let's say, 20ms. <S> Good old ZX Spectrum does the same. <S> At least it is a good start. <S> The RC filter with external pull up and software debounce is the best option here. <S> If you read the same value after 20ms you can be sure it is correct <S> Back to your questions: I think you will spend lot of time tuning this As you say, this makes things complicated. <S> You also waste more pins. <S> Anyway, you gain experience by failing. <S> That is, after all, what counts.
| RC Filter with Schmidt gate is the HiFi option, but unnesecesary in this case. Set the limit switches back a little from the end of travel and adjust them so that the door stops at the correct open and closed positions.
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Why are FPGAs so expensive? I mean compared to ICs (ASICs) with similar complexity, speed etc. Let's compare Ethernet switches to Kintex FPGAs (note that the most expensive switch from the list is circa as expensive as the cheapest Kintex): FPGAs are well structured ICs (like RAMs). They can be scaled and developed easily. The design tools ( Vivado , Quartus , etc.) are expensive too, so I think the price of an FPGA is the price of the IC (and development) itself excluding the cost of support and the tools. (Some non-FPGA vendors give free tools whose development cost includes the IC price.) Are FPGAs produced in lower quantities than other ICs? Or is there any technological harness? <Q> FPGA chips include both logic and programmable connections between logic elements, while ASICs include only the logic. <S> You'd be amazed at how much chip area is devoted to the "connection fabric" in an FPGA — <S> it's easily 90% or more of the chip. <S> This means that FPGAs use at least 10× the chip area of an equivalent ASIC, and chip area is expensive! <S> It costs a certain amount to do all of the processing on a given silicon wafer, no matter how many individual chips are on it. <S> Therefore, to a first approximation, the chip cost is directly proportional to its area. <S> However, there are several factors that make it worse than that. <S> First, larger chips mean that there are fewer usable sites on the wafer to begin with — wafers are round, chips are square, and a lot of area is lost around the edges. <S> And defect densities tend to be constant across the wafer, which means that the probability of getting a chip without a defect (i.e., "yield") goes down with chip size. <A> Another key driver of cost is verification. <S> FPGAs need to be individually tested before sale. <S> This is partly to ensure that all of the thousands to several million routing interconnects and logic cells are functional. <S> The verification however also involves characterisation and speed grade binning - determining how fast the silicon can operate and that the speed and propagation delays of all the many interconnects and cells are suitably matched to the timing models for its grade. <S> As such the time required for verification is likely far less, and thus cheaper to perform. <A> There is one (more) important point which is usually overlooked, process technology. <S> FPGAs that have high market share are manufactured with cutting edge technology. <S> To be more specific, Kintex-7 FPGAs have TSMC 28nm process and their shipment started in 2011 [1] . <S> TSMC had started mass production of 28nm in the same year [2] . <S> [1] Xilinx ships first 28nm Kintex-7 FPGAs (By Clive Maxfield, 03.21.11) <S> [2] <S> Chang said: "Our 28-nm entered volume production last year and contributed 2 percent of 4Q11's wafer revenue." <S> I don't know the process of the ethernet switches, but most of the ASIC design companies don't follow the cutting edge technology. <S> It doesn't make sense for foundries as well. <S> The following chart shows TSMC's revenue by technology ( 1Q18 ). <S> Even in 2018, 39% of the revenue comes from technologies older than 28nm. <S> If we think about the number of chips, it is not hard to imagine that more than half of ASICs are today manufactured with technologies older than 7-year-old Kintex-7. <S> I don't claim it is a dominant factor, but significant enough to be considered. <A> I'm going to go out on a limb and say that this is by far dominated by simple supply and demand. <S> Ethernet switches are mass produced with huge economies of scale and sell at discounts over chips that are not so widely used. <S> FPGAs, I'd say, are not nearly so widely deployed as ethernet switches <S> and so they cost more because the development and infrastructure costs are spread over fewer customers. <S> This isn't about process or die size or anything like that. <S> Consider the Xilinx Virtex-7 (only because I could more readily find data for it) and let's compare to a few contemporaries : Virtex7 (2011), <S> 28nm, ~6.8 billion transistors, $2500USD (popular models) to $35,000USD <S> (higher end models) <S> NVIDIA Kepler GK110 (2012), <S> 28nm, ~7.1 billion transistors, Tesla K20 cards ~$3200USD <S> at launch (chip price some smaller fraction of that) <S> XBoxOne SOC (2013), 28nm, <S> ~5 billion transistors, $499 USD for whole XBox at launch <S> Xeon E5-2699 v3 <S> [18 core] (2014), 22nm, ~5.6 billion transistors, ~$4500USD <S> So overall the Virtex FPGA seems reasonably priced (more popular models) compared to other silicon of a similar transistor count, generation, and sales volume. <S> The XBox <S> SOC sticks out as something which was widely deployed in a consumer device and the cost is likewise much lower. <S> NVIDIA's compute <S> GK110 was much less widely deployed than similar consumer chips that ended up in gaming cards and was similarly more expensive, even given the architectural similarities and the fact that the chips were made in the same factory. <S> As for the Virtex chips, there isn't a 10x difference in the complexity of the $2500 chips vs the $35000 chips <S> - the latter are simply much less popular and, with lower sales volumes, the cost per unit is necessarily higher. <S> The market is full of this. <S> Anything you can sell a hundred million of you can always make cheaper than something you will maybe sell a hundred thousand of.
| As a conclusion, process technology is one of the factors that make FPGAs more expensive. For ASIC designs, testing is typically simpler - a yes-no does the design perform as expected.
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How to convert a wide voltage range (1 V-12 V) to 5 V? I have an unusual application where the input voltage can vary from 1-12 V DC. This must be converted to 5 V @ ~200-250 mA. Since there are no available buck-boost converters suitable for this wide input voltage range, I came up with this (simplified) circuit: simulate this circuit – Schematic created using CircuitLab The XC61CC5002MR-G is a voltage CMOS supervisory IC which: Outputs VCC when VCC > 5 V - this enables the LDO and protects theboost converter from the higher voltage. Outputs 0 V when V CC < 5 V - this disables the LDO and biases the PMOS transistor to allow current to the boost converter. The boost converter's absolute maximum voltage rating is 6 V, so it needs to be protected at the higher range of input voltage, whilst the LDO is OK up to 13 V. The circuit worked as intended from 3-12 V, however the PMOS transistor couldn't switch when the input voltage was lower than 3 V, which shouldn't have surprised me since the V GS threshold was about 2 V at 250 mA. I have also looked at high-side load switches and over-voltage protection ICs, but I cannot find any that will operate over the whole range of the input voltages after browsing for hours on Mouser and DigiKey. Lastly I have explored using a N-channel MOSFET in the circuit above with an open-drain output variant of the same supervisory IC and a charge pump to bias the NMOS transistor when voltages are low, but to my surprise I couldn't find any charge pumps that work in the 1-5 V range. I am looking for any suggestions to either make my circuit work with the lower voltages or how else I can achieve this 1-12 V to 5 V conversion without dramatically increasing the PCB footprint or cost. Unfortunately the input voltage cannot be changed, but the power supply can provide more than enough current to run the circuit. <Q> Allow me to make this circuit suggestion: simulate this circuit – <S> Schematic created using CircuitLab <S> For 1 V < Vin < 6 V <S> the boost converter converts to a 6 V internal voltage and <S> the LDO regulates that 6 V down to 5 V <S> For <S> 6 V < Vin < 12 V the voltage at the Boost converter's output will "follow" the input voltage with some voltage drop due to a (Schottky) diode, so Vin = <S> 7 V = <S> > <S> Vmid = 6.5 V and Vin = <S> 12 V = <S> > <S> Vmid = <S> 11.5 V. Remember that boost converters have this basic circuit: <S> So when Vin is higher than the configured (regulated) output voltage, the output will follow the input voltage with a voltage drop from the coil's resistance and the diode. <A> A SEPIC might be worth a shot here <S> In particular, I would use a LM2621 converter IC with a low-drop Schottky diode -- this gives the chip the best chance of starting up at or about 1.0V (the datasheet specifies nominal startup at 1.1V and maximum startup at 1.2V over temperature, but the startup voltage likely depends on the drop across the diode). <S> As to passives, I would use the coupled-inductor version of the SEPIC topology -- this gives better ripple performance while using less board space than two separate inductor cores. <A> To avoid the EXTREME switch, you need a flyback stepup with zener post-regulator, to provide the FET's gate drive. <S> Set the zener at 10 or 12 volts. <S> Many huge FETS want a low max gate voltage (from what I recall). <S> Now you can use a LARGE FET, capable of 10 or 20 amps or whatever you need, that FET gate being driven 0/10 volts for efficient switching. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> There's a lot of ways to skin a cat, and this is a particularly vicious cat. <S> Here's my thoughts: <S> Boost set to 5.5V or so. <S> I'd have to go shopping for parts to be sure, but I suspect you'll need something with an external switch to handle the 8-9A input at 1V. Follow that with a buck to 5V, to handle the case when the input's above 5V. <S> Don't mess with the linear supplies unless you need really clean power. <S> If you can't find a suitable boost chip or switch FET, use a lightweight boost that's just big enough to power the "real" boost controller (and by extension, the "big" switch gate). <S> Maybe even switch the lightweight guy out of the circuit if the input voltage gets too high.
| This situation (wide voltage range input, fixed voltage output, low-ish current requirements) strikes me as a decent opportunity to try one of the lesser-used switching topologies: the SEPIC.
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Is there a practical way to create/inject some visible interference into a nearby AC mains? I have this 230 to 6V rms step-down transformer. I use it to observe the mains voltage. But recently I needed to create some interference or spike in the mains. So I first tried the following: I used a filterless power strip. I coupled the transformer's primary side to one of the sockets of the power strip. And at the adjacent socket I plugged a heat gun. I sometimes very randomly saw some tiny fluctuation and that disappeared. My approach was to see the interference from the nearby AC socket. What type of load would create a visible interference or spike? Or is there any other practical way to achieve this. I want to create a spike or interference by a load and observe that from a nearby AC socket via a scope and a transformer. More over is there a safe way to observe the Line to earth or neutral to earth voltage on scope? I mean if I use the transformer secondary for the line can I connect the scope ground to the mains earth safely? <Q> Any powerline communication device does exactly that: inject a signal on top of the power signal, and extract that at the other end. <S> It works. <S> The fact that you don't see much with an oscilloscope just means you're using the wrong instrument. <S> set it to a medium dim <S> attach a large load, e.g. a hot plate, toaster, large lamp, <S> whatever, close to the maximum rating of your dimmer use the transformer, plugged into the outlet you want to observe, only as a source for the trigger of your oscilloscope use that trigger signal to trigger the oscilloscope, observing a capacitive divider: two (high-voltage, medium to small value!!) <S> capacitors of two different sizes in series between live and neutral of the power outlet you want to test. <S> What does a cheap dimmer do? <S> Typically, it's a phase-fired controller, which means it always turns on the load (like you, flipping a switch) at the same phase (i.e. angle) of the grid voltage. <S> That's awesome, because a powerline system has different properties, depending on which phase you look at it. <S> Now, your transformer might not even have the bandwidth to notice the high-frequency noise caused by switching on, but your capacitive divider definitely has – in fact, it should act as kind of a high-pass filter. <S> By triggering the oscilloscope with your transformer, you make the grid phases take a "fixed position" on your oscilloscope screen. <S> That's immensely helpful, since now your switching noise will also always get a fixed position (since the switching always happens at the same phase), and be much easier to spot. <S> Compare load with and without dimmer, and with no load and dimmer at all. <S> Then start playing around with the dimming ratio. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> For EMC testing to meet standards, a so-called CDN (coupling-decoupling network) or LISN (Line Impedance Stabilization Network) is used. <S> You may be able to find schematics and instructions for making such a network. <S> I have built LISNs (for measuring noise created by circuits) from such information and they worked fine for pre-compliance testing. <S> Here is the internal schematic of a commercial device . <S> As you can see it's a decoupling network on the mains input and a coupling network on the other side to allow injection of signals. <S> As far as creating nasty noise without any fancy equipment, I've found a classic 100/140W <S> Weller soldering gun to be a good informal source of noise (photo from Amazon.com). <S> Certainly they are better used for creating noise than for soldering for 99% of electronics purposes. <A> One coil lead is wired through the NC and Common, the other lead straight to mains. <S> The inductive kick-back from the coil should give you some good rough spikes. <S> You can adjust the contact clearance and the spring tension to minimize excess noise.
| You could use a 230 v relay (SPDT or greater) wired as a buzzer. Try the following: use a dimmer for resistive loads (for incandescent lamps, i.e. the cheapest dimmers you find).
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Do I need a relay for my LED bars? I installed two 4" 18 watt LED light bars on my ATV. It is currently wired to the factory switch on the ATV handlebars and there is no relay being used. I have had to replace the right LED bar twice because it has died on me. Is this because I didn't use a relay? Will the relay help? As a side note, the battery seems to be failing and not holding a charge well (old battery). When I start the ATV (can't use electric start, pull only), the lights will turn on but they will start to dim unless I rev the engine. The ATV is a 1995 Polaris Magnum 425 4x4. <Q> Change your battery first. <S> Unlike cars which use alternators, ATVs and motorcycles get power from very simple stator generators. <S> How it works is that there's a permanent magnet on the flywheel which excites a stator coil, which is followed by a bridge rectifier. <S> Unlike an alternator, regulation is done by measuring the output voltage and then shorting the stator coil to ground with SCRs. <S> How does the system convert rectified AC pulses to DC? <S> It treats the battery as a capacitor. <S> And unlike a car, the system is 1-phase, so the voltage does drop to 0 twice every revolution. <A> No, a relay won't make a difference. <S> You are either electrically or mechanically <S> (maybe they are of a s**t construction?) <S> overloading the LEDs. <S> Do they just stop working, or they start to fade and go out one by one? <S> Change the battery <S> ; Depending on their chemistry they work for different periods of time, but eventually all of them die. <S> Pay special attention to change it with one which is electrically compatible to the ATV. <A> The alternator normally puts out a higher voltage (14.6 v or more) in order to push a charge into the battery. <S> A good battery with good connections provides the load needed to stop runaway voltage from the alternator. <S> Your LED lights are probably getting too much voltage, therefore too much current. <S> The right side LED may have a shorter wire or better connections, so it goes first. <S> Often, high current LEDs used for many hours can fade to half their original brightness. <S> And of course excess current can pop the junction. <A> have the aletenator tested, it may be faulty, test the battery too, when recharged it may still be good.
| When the battery fails, the pulses and noise aren't filtered out, the regulation becomes poor, and you get overvoltages, killing the LEDs. If the ATV won't give you a satisfactory voltage you could add a wire-wound 1 ohm resistor into the circuit.
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How to isolate several voltage sources that feed a single voltage meter I need a way to isolate several voltage sources around a large vehicle that feed a single voltage meter some distance away on a panel. Each source is selected by a momentary-spst (at the source). Here is a block diagram, with 2 discrete options: All current is DC. Voltage sources run 4.6 VDC to 54.0 VDC. Desired display accuracy is 0.1 Volt. All the power supplies share a common ground. The (arbitrarily) colored areas display the core differences in the current two options. Option one (green) uses 'ultra-low loss diodes' (ideally an IC array), Option 2 (blue) uses typical signal diodes (1N4148) and a post diode voltage adjustment. For option 1, the MAX4020 came to mind, but the voltages are out of range. Is their an IC / IC array designed for this? For option 2, what is the best way to adjust the voltage up 0.7 V? We're somewhat stuck with the basic layout of using multiple momentary-spst feeding a single voltage meter, so a rotary selector switch is out. edits: updated 'mom' body text to 'momentary' for clarity. <Q> Since the switch is momentary contact, you don't have to have protection if only one switch can be pressed at any given time. <S> However, to guard against two or more switches closing, put a 5000 ohm, 1 watt resistor in series with each switch. <S> Unless the voltmeter is strange low input resistance, you won't even notice voltage drop across resistor. <A> This is an op amp circuit that provides an effectively ideal diode, by using negative feedback to adjust up its output voltage to compensate for the diode drop. <S> It's just like an op amp voltage follower, but with a diode added: simulate this circuit – <S> Schematic created using CircuitLab <S> The downside to this is that you'll need to find an op amp that can take 48V input and also output 48V. <S> This may not be cheap. <S> Another possibility is using a MOSFET-based super diode. <S> There are devices that can control a FET to act as an ideal diode, such as the LTC4359 , and while I know that fully-integrated FET-based super diodes exist (such as the SM74611 ), I'm not aware of any that can withstand the voltages you need for this particular use case. <A> What's the point of having the switches at the sources if the meter is some distance away on a panel? <S> Instead, you should have the switches at the panel, and wire them like this, so that only one source can be connected to the meter at a time: <S> simulate this circuit – <S> Schematic created using CircuitLab
| Also, even with the resistors included, you can tell if a switch hasn't released if the voltmeter shows voltage when none of the switches should be closed. It sounds like what you want is a super diode .
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How do I fix the speed of a generator? If I need to generate power at a specified frequency, then I need to make sure that the rotor of the generator rotates at a specified speed (rpm). But when I am rotating it with steam or water how do I control this speed? It seems to me that the mechanical forces that rotate the generator somehow has to balance to achieve this. How exactly is this done? <Q> Electrically Some systems do this electrically. <S> The generator either generates DC, or the variable-frequency AC is rectified to make DC, and then an inverter makes the desired AC frequency. <S> Common on more modern small wind turbines. <S> Mechanically Other systems are mechanically controlled to get the desired frequency. <S> The mechanism used would be called a governor . <S> Most simple mechanical governors are not very accurate, so this would not be good enough for a grid connected device. <S> It is also possible to make more accurate governors which work mechanically in a similar way to the paragraph below, these are commonly used on internal combustion engines. <S> With Feedback <S> Another approach, and probably the most common is to have some form of feedback. <S> For example, it could open and close a sluice gate to adjust the water flow through a turbine. <S> A more complicated system could adjust both a sluice gate and the turbine blades to keep the correct frequency while also varying the output power. <S> Grid Synchronous operation <S> In some cases, it might not be necessary at all. <S> If you have a small wind turbine, connected to the mains grid near a coal power station, you could just hook it up and forget about it. <S> The huge turbines in the power station will stabilise the grid frequency, and fix the rotation speed of the wind turbine. <S> If the wind blows harder, you'll just get more current, and a slight power factor shift. <S> Note that as more and more wind turbines get added, the folks who run the power station will get less and less happy about this, so the grid operator will eventually ban it. <A> If I need to generate power at a specified frequency, then I need to make sure that the rotor of the generator rotates at a specified speed (rpm). <S> No, that is not necessarily true. <S> A lot of wind generators use a doubly fed induction generator (DFIG) and that can regulate frequency by controlling the rotor winding: <S> - They can convert power at one frequency to another frequency i.e. they can produce 50/60 Hz even though the rotor may be running too slowly. <S> This is done by injection of an AC current into the rotor coils. <S> The control system that acheives this may also be able to alter pitch angle of the turbine as another means of increasing or decreasing mechanical rotation speed. <S> For more information please read this EE answer . <S> But when I am rotating it with steam or water how do I control this speed? <S> There can be situations when there is too much rotational speed and although the DFIG can deal with this it is best to have a combined approach such as the pitch angle control shown in the above picture. <S> However, the inevitable bottom line is that if you have a plentiful source of mechanical energy and very little load demand at that moment, then you have to un-hook the generator from the grid. <S> If you don't "have" a grid, then you have to have a back-up supply that can supply the low power demand and this usually means a diesel generator or solar power via an inverter. <A> That is what a governor is for. <S> The mechanical version is a device that will use centrifugal force or a blower to actuate the throttle/intake to slow down the engine when the speed climbs too much. <S> You can make it electronic with an RPM sensor and a electronically controlled throttle/intake. <A> In my (limited) experience of generator design, you have to look at multiple factors: <S> Speed of mechanical input (turbine, wheel etc) Power of mechanical input <S> Output voltage Output current Output power (which depends on voltage and current, but you usually want to maximise this, at a peak power point) <S> In many cases what you want to maintain is a constant voltage output, which will vary with the electrical load; higher current would make the output drop. <S> In your set up, you only care about the speed of the shaft (for whatever reason that is). <S> There are two ways to do this: control input power or control output power. <S> If you know that your mechanical power will always be more than the output power, you could use a governor or similar, which will limit the shaft power to keep the speed constant. <S> This will control the input power in a simple manner. <S> If you can't guarantee that your mechanical input is higher, you'll need to limit your output power in some way. <S> I have done it where we controlled the output current via a PID feedback controller to keep the shaft speed at fixed values. <S> But that was in a DC system, where we had a large battery to push the current into during high supply times, and draw from during lower supply times. <A> This isn't the way it is done, but it's a way it could be done: vary the load in order to modulate the speed. <S> You can imagine a (potentially very large) auxiliary load bank in addition to the normal load, with the input power source sized so that it will provide at least enough power for the normal load under all conditions. <S> This isn't completely unthinkable for hydro setups, since the head is the usual determinant. <S> Then, as the load and the input vary, the auxilliary bank is controlled so that the total demand on the generator produces mechanical loading on the input which maintains the output frequency at the desired point. <S> And no, this isn't a serious suggestion, but it could be done.
| A microcontroller monitors the frequency being generated, and adjusts the mechanical system via some form of servo to get the right frequency.
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If you have two identical but oppositely wound inductors does the inductance effectively cancel? I was just curious if the effect inductance of a branch of a circuit would cancel if you put two identical but oppositely wound inductors in series. It's not something I have come across.But looking at the change in signs I imagine this to be the case.Someone mentioned it to me regarding wound resistors in a circuits, and splitting them into 2 resistors in series, with opposite winding to reduce/attempt to cancel the inductance. I have also recently needed to place several surface mount inductors in series, and I have no way of identifying which end is which, or the winding directions. They are not all the same make or model.And I have had difficulty measuring their inductance. <Q> It depends of the extent of their mutual inductance. <S> If, as it seems to be in your case, they are separate components, they are wound on separate cores and have little mutual inductance. <S> In case of air inductors or resistors as mentioned, the distance between them and exact geometry has to be taken into account and only an electromagnetic simulation software can tell how much is canceled, but that would be much less than a complete cancellation. <A> If you put two inductors in series with a common axis, there will be little cancellation. <S> If the two were put into close proximity side-by-side <S> you'd get more, but probably not much. <S> It is possible to get non-inductive wire-wound power resistors. <S> These essentially take a single conductor, fold it in half, then wind the resulting doubled wire on a core (not a magnetic core). <S> Because the halves are physically close, the voltage rating of the resulting resistor is less than with normal construction (winding the wire on a core), but the inductance is much less. <A> In reality, it will never cancel completely! <S> The equivalent inductance is: Mutual inductance is always less than the sum of the inductance. <S> When two inductances are identical <S> (L1=L2=L) and perfectly coupled (M=L), then it will be completely cancelled <S> but I don't think it happens in reality. <S> To find the polarity of your SMD you can measure the total inductance in series connection in both polarities and if there is at least a small mutual inductance, you can see the difference in the measurement. <S> (if series connection in the same polarity, the total inductance will be L1+L2+2M)
| Roughly speaking, if they are wound on the same ferrite core, they have a large proportion of their inductance that is mutual (as in a transformer for instance), so they would mostly cancel.
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Switching a resistor ladder Can anyone suggest how i could switch this resistor ladder with the smallest possible footprint. I was looking up analog switches but not sure if thats the right way to go. The circuit tells an ecu what speed to run the engine at, and states the resistor tolerance at 5%. I am looking to control this circuit with an mcu and would like suggestions of what sort of IC could be used as a switch. it would have to have a low Ron (6ohm) to keep in spec with resistor tolerance. Thanks in advance <Q> I am looking to control this circuit with an mcu and would like suggestions of what sort of IC could be used as a switch <S> In this case your best bet would be to use digital POT, something like MCP4141 or AD7376 . <S> This one chip alone will replace all the resistors and switches and is extremely easy to control by MCU, with probably the smallest footprint possible. <S> available dirt cheap. <A> For many applications, I would say go with the analog switch. <S> That said, you have some pretty low resistor values, and you need to be absolutely sure that your on resistance of the switch doesn't do anything nasty. <S> If you can't convince yourself that it's OK, I'd consider SSR's. <S> You can get them in some pretty small packages. <S> See <S> http://www.ixysic.com/Products/SSRFormA.htm for one product line. <S> You can get quads in small packages -- <S> e.g., https://www.mouser.com/datasheet/2/389/vni4140k-32-974430.pdf <A> If your MCU has a D-A converter, you could just use that output, buffered through an op-amp to force the top of the divider string to the appropriate voltage to simulate the divider, given that you have a description of the way the signal is interpreted by the ECU, which will just be employing an A-D to interpret the control input. <S> Most likely it'll have defined transition points that set the accepted voltage range for each of the states, but whether there are guard bands between each valid state or not to dismiss unacceptable variations in the value will depend on the criticality of the operation. <S> On setups I've created for reading keys in this manner I haven't bothered beyond open/short detection, but they weren't for an engine ECU. <A> Smallest footprint is PWM and a single 120 ohm resistor. <S> The 220 nF capacitor will smooth out the lumps caused by the PWM to some extent but increasing the capacitor will probably help, from what I understand from the blurb the other end of the wire is a resistor to +5V,a another capacitor to ground, and something that reads the voltage. <S> You get a time constant of about 50 us <S> so an 8-bit PWM running at say 20Mhz clock speed should be able to get close-enough to the needed voltages to trigger the commands.
| For non-MCU control there are hundreds of miniature rotary switches
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Converting 0-5 volts (linear) to +2.5 -- 0 -- +2.5 volts ("vee") I'm trying to find a simple way to convert a 0-5 V output from a joystick hall effect sensor to +2.5 -- 0 -- +2.5 V. Here's how I want the response to look: The output voltage will be sent to a frequency converter and on to a stepper motor driver. I want to have the same voltage at both extremes so the frequency (motor speed) is the same regardless of the positive or negative direction of the joystick. I intend to use a comparator at the zero point (with an appropriate deadband) to reverse the direction of the stepper driver. I've thought of a few ways to accomplish this, but none seem to be particularly straightforward: Create a second, inverted output from the sensor and switch the input source (via comparator and mux) to the frequency converter at the 2.5 V midpoint. This would provide 2.5-0 V from the inverted output and 0-2.5 V from the original output. Shift the 0-5 V range to -2.5-2.5 V and then use an absolute value circuit to invert the negative portion of the output. Generate a bias voltage, based on joystick position, and add/subtract it from the sensor output. Ideally, I'd like to have a single input to the frequency converter instead of trying to switch it to another source, such as an inverted output. The second idea above would satisfy that preference, but it would require a negative voltage source. Also, I'm trying to use discrete components in this design, so a microprocessor isn't an option for me. Is there an easier way to accomplish this task? The problem seems fairly straightforward, but I'm having trouble coming up with a simpler solution. I appreciate the help. <Q> Here is one way to do it that uses a cheap quad op amp and a bunch of 100K resistors (and operate from a single +5V supply): simulate this circuit – Schematic created using CircuitLab OA1 provides the two reference voltages used from the supply. <S> OA2 subtracts the input voltage from 2*1.25V to give the left part of the output. <S> OA3 subtracts 2.5V from the input voltage for the right part of the output(both saturate at 0V with a single 5V supply, so no diodes are required). <S> OA4 sums the two to give the desired output voltage. <A> the circuit you want is an " absolute value circuit " but to get the vee shape you'll need 2.5V on the non-inverting input of the first op-amp. <A> However, at that point I have to ask: “is that really more efficient than a tiny micro controller?” <S> A 6-pin microcontroller with no external components can perform this function AND directly provide you with the output frequency. <S> Thus removing the need for yet another IC. <A> I'd use the absolute value circuit and a subtractor to shift the output to the 0-2.5V range. <S> Jasen's answer provides an absolute value circuit I've used before and which should work here. <S> I'd follow that with an op amp difference amplifier with a gain of 1. <S> You'll need a 2.5V reference for both the absolute value circuit and the difference amplifier. <S> You can either use a simple resistor divider from your positive power supply (which I would buffer with another op amp) or a 2.5V reference IC. <S> Here's a schematic which you can simulate (I assumed a 5V supply was available for the 2.5V reference): simulate this circuit – <S> Schematic created using <S> CircuitLab <S> I didn't pay much attention to the resistor values -- I just made sure they have the right ratios. <S> You'll know better what values work in your application. <S> Similarly, I just used a generic op amp. <S> A DC sweep of the input from 0V to 5V produces the following outputs for the absolute value circuit (in orange) and the difference amplifier (blue): <S> Since there are four op amps you might be able to use a single quad op amp IC. <S> Or, if you use a 2.5V reference IC (so you only need 3 op amps), you could use a dual op amp IC for the absolute value circuit and a single for the difference amplifier. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Opposing magnets will give a null point in the centre of stroke. <S> If your joystick is amenable to modification you may be able to generate the required response by replacing the one magnet with two opposing magnets separated by the stroke of the movement. <S> With the joystick in mid position the magnetic fields will cancel out. <S> Moving to either end of travel will increase the flux from the respective magnet and increase the output. <S> It may be simplest to adjust for an output voltage > 2.5 V and attenuate it with a resistor divider. <S> Get out the hot-melt glue gun! <A> Connect a VCO to the raw input and beat the output against a fixed oscillator, send the beat frequency to the stepper driver. <S> you could also beat agaisnt a quadrature of the fixed oscillator and get the other phase for your stepper drive that way, and not need a stepper driver :)
| I can see a possible solution with a super-diode circuit biased at 2.5V and a couple of extra op-amps to invert the signal and add it back to the input with the proper gain.
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What Reference-Potential does an Operational Amplifier use as "Ground"-Potential? Wikipedia states on operational amplifiers, that, simply speaking, they provide an output voltage at their output which is the difference of the two input voltages, multiplied by some (very very large) number.Since voltages are just electric potential differences, I'd like to know what point in the circuit is the reference point for the output of the operational amplifier? For the inputs, it essentially doesn't matter. Since the behaviour of the operational amplifier only depends on the difference of the two inputs, it doesn't matter what ground they correspond to. My simplest guess would be that the output voltage of the operational amplifier has the same common ground all the other devices in the electric circuit have. In that case: How does the operational amplifier know of this common ground, if it doesn't have a connection to the common ground? I need to know what the reference point is to even make any sense of the statement "the output voltage is A times the input difference". If I don't know what the reference point is, then the output voltage is just a number without meaning. <Q> It doesn't know nor care. <S> Opamp's internal circuitry works like this: Uout is near 0V when U1 < U2 and Uout is near the full supply voltage Us when U1 > U2. <S> Just around the case U1= <S> U2 <S> there's a transition zone <S> A. <S> Its width in practical opamps is well below one millivolt. <S> Nobody guarantees that the zone is linear or symmetrically around the zero, but normally opamps are used with a feedback circuitry which forces the opamp to output such Uout that U1-U2 is inside the transition zone. <S> People often divide Us to two parts in series. <S> The upper part is said to be <S> the positive supply and the lower part is said to be the negative supply. <S> But internally the opamp references all to one of the poles of the supply voltage Us. <S> In low cost IC designs the internal circuitry cannot accept U1 and U2 to be whatever, they must be between 0 and + <S> Us <S> and there's some margin needed. <S> The margin is needed at least in one end of the range 0... <S> +Us <S> to leave some operating room for the internal circuitry. <S> Many buyers want no margin at 0V, they want that U1 and U2 can be 0V. <S> If the internal circuitry is designed properly (=PNP input transistors), the usable range for U1 and U2 is from zero to 1... <S> 1,5V less than Us. <S> One old and well known IC design LM124 is like that. <S> Analyzing its internal schematic reveals that it actually uses the plus pole of the Us as its internal reference point, but that makes no difference to my drawings and equations. <A> I struggled with the same problem for a while. <S> The answer is not always obvious. <S> The op-amp, generally, has no idea where ground is as there is no ground input pin. <S> Often it's the negative voltage as in single-rail supply applications and <S> other times it's somewhere between V+ and V- as in split-rail supplies. <S> Almost all amplifiers configure the op-amp with negative feedback to control and linearise the gain. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Two most common op-amp configurations. <S> From basic op-amp application theory it should be clear that in both cases shown in Figure 1 that the negative feedback will cause the output voltage to go to whatever voltage is required to bring the inverting input to the same potential as the non-inverting input. <S> The result for each case is: $$ V_{\text{Oa}} = - V_{\text{in}} \frac {R_f}{R_i} + V_{\text{ref}} $$ <S> $$ V_{\text{Ob}} = V_{\text{in}}\left( 1 <S> + \frac { <S> R_f}{R_i}\right)\ + V_{\text{ref}} $$ Note <S> that \$V_{\text{ref}}\$ and \$V_{\text{in}}\$ have to be referenced to some point and that the output is referenced to the same point. <S> If \$V_{\text{ref}}\$ is zero <S> then we get our standard op-amp amplifier gain formulas. <S> $$ V_{\text{Oa}} = - V_{\text{in}} \frac {R_f}{R_i} $$ <S> $$ V_{\text{Ob}} = V_{\text{in}}\left( 1 <S> + \frac {R_f}{R_i}\right) <S> $$ <S> Lever amplifier simulate this circuit Figure 2. <S> The inverting lever. <S> The fulcrum is 1/3 along the lever so it is working with a gain of -2. <S> (a) <S> The input, fulchrum and output are all at the reference voltage. <S> Their height above reference is zero. <S> (b) <S> The input has been pressed down to -1. <S> The output has risen to +2 with respect to the reference. <S> (c) <S> Exactly the same lever <S> but with the reference moved to a different height. <S> Now input, fulcrum and output are at +5. <S> (d) <S> The same angular displacement has been applied. <S> Now the input is at +4 and the output at +7 relative to the reference. <S> Q: What is the difference in amplification of (d) versus (b)? <S> A: Nothing. <S> The world hasn't changed. <S> Our point of reference has changed so input, fulcrum and output have a fixed offset due to the new reference position. <A> It really doesn't matter. <S> It can only output as far as its power rails, and if you look at it as ideal, it would always be at one of those rails in the absence of any feedback. <S> The feedback structure is what gives it a reference. <S> For example, in a single ended negative gain configuration, the positive input supplies the reference; it's a chosen point in the range where that voltage on the input results in the same voltage on the output. <A> In practice the offset voltage of the op-amp multiplied by the gain is typically much more than the supply voltage. <S> For example the LM324 has a gain of about 100,000 and an offset of perhaps a couple of mV so hundreds of volts at the output. <S> If you need a number for calculations, you can think of it as ( \$V_+\$ + \$V_-\$ )/2 <S> if you like, which conveniently works out to zero for balanced supply voltages, but whichever number you pick within the supply rails (for example, if you pick 0 for a 5V single supply op-amp) the error should be small for typical op-amps. <S> For example, a precision amplifier with a gain of 1,000,000 and an offset voltage of <S> +/-12uV <S> typical.. <S> still +/-12V at the output based on the offset.
| The midpoint is said to be the ground and all voltages in application designs are referenced to it.
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Relationship between I2C drawn energy / power consumption and data rate Referring to just what the I2C lines draw, am I wrong thinking that the higher the clock frequency the shorter the time there will be (the same amount of) current flowing through the pullups and thus lower power consumed? side qeustion I don't think I am going to reach 100 kHz, that's way over the limit of my hardware. I am alternating between about 32 and 4 kHz. Will the same resistor value (3.3k @ 3V) be good for both? <Q> The I2C data and clock lines draw power when they get pulled low. <S> Because then power is sunk through the pull-up resistors. <S> While a line is pulled low it will draw 5V/4.7k \$~\Omega \approx\$ 1mA. Assuming 5V VCC and 4.7k pullup resistors. <S> The clock line will have a 50% duty cycle. <S> The data line is low at least 1 out of every 9 clock cycles (every ack for a successful byte) but you are rarely going to send/receive only 0xff bytes. <S> It's more likely going to be pulled low 75% of the time. <S> But indeed faster clock means shorter transmission which means less power lost through the pull-ups. <S> However faster transmission may require lower value resistors to overcome the parasitic capacitance between the lines and ground. <A> Your thinking is correct, as long as you can achieve a higher speed with the same pull up resistors. <A> Increasing the clock frequency from 100kHz to 400kHz usually requires the pull-up to be reduced with a factor of 4-5. <S> Since the power is inverse proportional to the resistance the power consumed will be almost the same. <A> As @ratchet-freak stated, In terms of time, you could have 75% of the time the bus pulled-down, hence, if you increase the clock rate, your consumption by the bus will decrease as long as you have the same value for pull-up resistors. <S> But, at higher speeds, resistor values should be reduced. <S> Having this, the consumption of the bus will be lower, but slaves and masters devices could increase their consumption depending on the clock rate. <S> Regarding your side question, if 3.3kohms suits both 4khz and 32khz, you have to check the capacitance of your bus. <S> This capacitance depends on the length of the bus, the distance between lines and the number of devices attached to it. <S> It could be difficult to calculate the real capacitance, but you can check the waveform of your data in the bus at both frequencies and see if there is any distortion of the signal at 32khz using 3.3k. <A> At high data rate power consumption will be more:- MCU need to work faster- <S> Low-value pullups will be required to get proper rise/falling edge of the pulses
| Higher clock frequency usually require lower pull-up value, thus increasing the current.
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Resetting CD4017 Counter when power source is OFF I found the circuit below online and I have two questions about the reset switch. Does it help to click the reset button when power is OFF? I mean, does it clears the memory of the CD4017 chip when power is OFF as well? How is it different between the two cases: when resetting while power is ON and when it is OFF? <Q> No, pressing RESET while the circuit is off will do nothing as the chip is not powered up. <S> Holding RESET during power up is a different matter and will result in the circuit always powering up with Q0 on. <S> If you don't hold RESET during power up you will get a random LED on every time. <S> To automate the reset you can add a capacitor - maybe 1 to 10 μF across the RESET button. <S> The result will be that when the power is switched on the capacitor will pull the reset pin high (+9 V) until the reset resistor pulls the pin down to zero by charging up the capacitor. <A> When the power is applied the state of the flip-flops in the 4017 is not guaranteed . <S> It may be anything (probably there is some small asymmetry that will make it non-random). <S> To reset it you should have a sufficiently long reset pulse applied after the power rail reaches operational numbers. <S> This is often done with a supervisory chip that contains a precision reference and a timer. <S> A poor method is to use a capacitor (for example across your switch) which will reset it if the power is applied cleanly but will fail in many other cases. <S> Amateur and cheaply built consumer devices sometimes have such cruddy reset circuits and they often cause problems for the end user. <A> Not only will it not work, it's a violation of absolute maximum conditions listed on the data sheet. <S> The sheet specifies <S> INPUT VOLTAGE RANGE, <S> ALL INPUTS................. <S> -0.5v TO VDD+0.5V <S> This means that when the chip is unpowered, you should not provide any HIGH voltages to the inputs. <S> Exceeding maximum values formally means that the chip is no longer guaranteed to function to spec.
| Pressing the reset button will not have a known effect unless the power supply is within the range for proper operation (3V or more for the CD4017).
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What do I call this wire-to-board connector? I'm repairing a toaster oven and have encountered this connector. What should I call it when I search? This is a power connector. It says "R8 JD" on the base of the male part. It is 3/16" wide. Below is the female part. It says "8 STS". Here they are before I disassembled it. <Q> AMP/Tyco calls them Fastons. <A> Spade connector/crimp/lug <S> Probably because they look vaguely like a gardening tool for digging holes in dirt. <S> The end on the wire is a crimp because of how its fastened to the bare wire, and the end on the board is a through solder-mount lug because its through the board and held on with solder not crimping. <S> Do note that the wire in your photo appears to be rated for high-temperature because its in a hot environment. <S> Please make sure your repair keeps or improves on the standards already set. <A> <A> Spade connector. <S> I'm in the UK, there may be regional differences in the most widely used name. <S> I think Lucar was a brand name, from their use in automotive products by the British firm Lucas. <S> Faston seems common in the US. <S> Check the datasheets for your chosen manufacturer, or copy the existing rating. <S> They are likely to be roughly standardized between manufacturers, but I wouldn't guarantee it. <A> These are known under many names, some of which refer to multiple types. <S> As seen above, these are known the female connectors are Push On, while the male is a Tab connector. <S> Tongue connectors is not uncommon. <S> Quick Disconnects is a common term as well. <S> Both ring and fork connectors can be used with screws.
| I've known them as "quick connects". I know them as “Lucar” or spade connectors, may well have other names though... In addition to Criggie's note about cable temperature, be aware that the colour of the plastic cover on the crimp section is significant; it indicates the range of wire size that can be crimped into it, so it is loosely linked to the current rating. Male Fork and Blade connectors can fit in female blade connectors. Also known as Blade connectors, sometimes known as Spade connectors ( But spade connectors are really Fork connectors ). They are similar to Fork connector, which look like garden forks: Ring connector, which are rings (an item that can be lost in the garden)
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Reading a Bi-Color LED on Arduino I am looking to read the status of a battery charger using an Arduino Board. The battery charger shows red when a battery is being charged and green when a battery is charged. I understand that the charger uses a Bi-Color LED. One option is to use an RGB color sensor (eg. https://learn.adafruit.com/adafruit-color-sensors/overview ) but I am looking to see if there is any way I can read the Bi-Color LED directly on the Arduino pins. Any ideas? <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Opto-isolator monitoring of battery charger LEDs. <S> Wire the opto-isolator LEDs in series with the existing LEDs. <S> The opto's LEDs are infrared and will drop about 1.4 V. Enable internal pull-up resistors on the micro-controller. <A> The red filter will allow the LED light to pass through to the photoresistor when it is red and will block the LED light when it is green. <S> The photoresistor will then show two different resistance readings for red verses green. <A> A little known fact is that all (non-opaque) diodes (or PN-junctions for that matter) are photosensitive. <S> LEDs in particular are the most sensitive to their own emission wavelength. <S> You can put together a detector by amplifying the currents in both legs of reversed-biased bi-color LEDs that are as close as possible (in distance and characteristics) to the existing ones. <A> The easy way is to run a wire from the LED to the Arduino pin. <S> At least one electrode of the LED is connected to the charge status signal. <S> By monitoring the charge status signal you will know the LED color. <S> The charger and Arduino would need to have a common ground. <S> If you wanted to go the color sensor route, you could use a photodiode like the Everlight <S> CLS15-22C/L213(RGB) . <S> See: Sharp Photodiode Application Note
| You could use a photoresistor with a red filter in front of it.
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Using an opamp with fractionnal gain with supply less than input signal Let's say we have an AC signal that can swing from -15v to +15v max peak to peak. If I use an opamp with the power rails to -5v and +5v, can I get a non clipped output signal if I use a gain of 0.333? Or will op amp failed to recognize anything above/below +/-5 V at it's input even if the required output is within range? Do I need to have my opamp's power rails and specs within the input range or the output range provided that the output have less peak-to-peak swing than the input with my gain of 0.333? The signal I'm talking about is a low, signaling level (about 1mA max) with varying frequencies, like an audio signal. <Q> Figure 1. <S> Inverting and non-inverting amplifiers. <S> All About Circuits . <S> The inverting amplifier gain is given by \$ <S> A = <S> -\frac {R_2}{R_1} \$ and <S> the inverting input is at virtual ground so it will be fine. <S> You can see that the non-inverting amplifier has <S> it's V in connected directly to the input <S> so this can't be directly fed. <S> Instead you would use a potential divider on the input to bring the voltage down to within the common-mode voltage for the op-amp. <S> The gain is given by \$ <S> A = \left(1 + \frac {R2}{R1} \right)\$ <S> and if you set R1 to infinity (open circuit) the gain will be 1. <A> If the input pins of the op-amp don’t exceed the common mode specifications or the maximum specifications (e.g., Vdd-1V, 0V or whatever these are) then there is no danger. <S> Adding protection diodes (if you don’t fully trust the op-amp’s pad protection) and using high enough resistors can make sure of this. <S> However, I don’t really understand why people use OP-Amps with fractional gains. <S> It is actually more precise and less noisy to simply use a resistive voltage divider. <A> The op-amp will have a common mode input range specification. <S> This means that there is a range of voltage relative to the supply rails in which you need to keep the inputs for the op-amp to function correctly. <S> For example, with LM324 you can take the input down to the negative rail but not within 1.5V of the positive rail. <S> If you use an inverting amplifier configuration then the inverting input of the op-amp is a virtual earth so you might have 0V on both inputs, even with a 15 V signal. <A> The input stage of a simple bipolar OP-Amp looks like this: simulate this circuit – <S> Schematic created using CircuitLab <S> If your V+ is +5V and your input signal is >5.7V, <S> the collector diode of the input transistor becomes conducting in its forward (the transistor's reverse) direction. <S> The input stage cannot operate the intended way. <S> If your V- is -5V and your input signal is <-4.3V, the input transistor will never conduct. <S> The input is at the rails. <S> If your input signal is about 5V lower than the V- rail, the emittor diode will break through and the transistor is toast. <A> I would suggest a voltage divider to get the voltage swing down to within your power supply rails, and put the resulting V-div output through a unity gain opamp buffer. <S> That gives you a low source impedance signal for whatever further conditioning you want to do.
| A non-inverting configuration where you apply the signal directly to the non-inverting input would not work.
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I need help sing a 555, op-amp integrator, and comparator to generate a voltage controlled PWM signal I have LM358p op-amps Datasheet I'm trying to base my design off this: retrieved from here .My control voltage will vary between 0 and 3.5v. I want the output pulses to vary between 2.5 and 0.5ms and at a frequency of about 50Hz. This is to control a servo using Vcontrol. My clock signal is from a 555 timer in astable mode. The square wave voltage is 0 to 8.4v. I understand that this diagram is based on the ideal op-amp, so I'll need a resistor on the non-inverting input equal to the inverting input resistor. I'll also need a high value resistor parallel to the capacitor. The output will probably not reach 0v, but that's ok as long as the triangle is at least 2.5ms wide at its widest point. I only have a single rail power supply, 0 to +12v to power the op-amp. The clock signal also never goes negative. I don't quite understand how the op-amp integrates. Is the non-inverting input the reference line, above which the integral begins to rise and below the integral falls? What should my input pulses look like and how do I calculate the R and C values for the op-amp? <Q> I don't quite understand how the op-amp integrates. <S> The left op-amp is used as an integrator, and is probably best left for you to research the theory behind how it works. <S> But for simplicity sake, you can consider it as the active equivalent of a series RC circuit. <S> For more information on how the integrator works, simply search "Op-Amp Integrator." <S> how do I calculate the R and C values for the op-amp? <S> You need to choose an RC value such that you obtain your desired triangle wave at 50 Hz. <S> As others have suggested, you will need a feedback resistor parallel with your capacitor, and that RC combination will determine your "cutoff" frequency, where any frequencies above this "cutoff" will be an integral output of the input. <S> Below is your overall transfer function of the integrator alone: $$\frac{V_o}{V_{in}}=-\frac{R_F}{R_i}\times\frac{1}{1+j\omega(R_FC)}$$ Take the magnitude of the above at 50Hz (or 314.16 rad/s), and that will be your gain at that frequency. <S> A resistor on the non-inverting input isn't entirely necessary, but is typically used to balance the bias currents and is set equal to the equivalent resistance seen by the inverting input. <S> It doesn't not affect your gain or cutoff frequency. <S> What should my input pulses look like <S> If you're talking about the clock, it should be a 50Hz with a 50% duty cycle, since the frequency of your pulses is 50Hz. <S> If you're talking about the voltage control input, then they shouldn't be pulses. <S> The voltage control determines the duty cycle of your output PWM. <S> For example, suppose your triangle wave oscillates between 0V and 8.4V. <S> For a 50% duty cycle (10ms pulse width in your case), set the voltage control to half of the max voltage of your triangle wave. <S> (4.2V). <S> If you want 2.5ms, then 2.5/20ms = 12.5% duty cycle. <S> Set your voltage control to 12.5% of your max voltage (1.05V). <A> You don`t need the opamps, you only need the 555. <S> This is from the amazing book The Art of Electronics <S> Otherwise they won ` t oscilate at all. <S> You can bias them with a resistive divider. <A> I'm trying to base my design off this: Sorry, but that circuit will never work as shown. <S> It is not intended to be a "real" circuit - it's what is called a "notional" circuit. <S> Where to start? <S> The waveforms as shown are simply impossible. <S> The integrator input is a positive signal, yet so is the output. <S> Nope. <S> Never happen without some extra biasing, since the integrator is an inverting integrator referenced to ground. <S> Next, it requires a perfect 50% duty cycle on the input clock, or or the output will drift at a rate determined by the input phase mismatch and the input voltage swing. <S> The drift will only stop (be limited by) <S> the excursion limits of the integrator op amp output stage. <S> If the input clock duty cycle is very close to 50% (and <S> a 555 output should be close), you'll get one tip or the other of the triangle blunted where it hits the stop. <S> This may or may not be an issue, depending on accuracy requirements and how stable the offset circuitry is. <S> The fact that you're limited to a single power supply will only make things worse. <S> I suppose you could do something like this: simulate this circuit – <S> Schematic created using CircuitLab <S> Here, R1/C1 are chosen to get a decent voltage swing centered around the voltage at C3. <S> This voltage is set by adjusting R2/R3, and should be in the center of the voltage swing of the input clock. <S> I've assumed that this would 6 volts. <S> OA2 serves as a low-pass averager for the integrator output, and has R4/C2 much larger than R1/C1. <S> The output is fed back to OA1 to keep the average voltage at the C3 voltage. <S> Note that loop stability is not guaranteed.
| If you are bent on using opamps then you need to BIAS your inputs when using single rail supplies.
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AND Gate Stays Open I have a very simple circuit on a breadboard with two push-button switches, an AND gate (74LS08), and an LED. I have the two switches hooked up to pins 1 and 2, while the LED goes from 3 to ground. Pin 14 is given 5 volts, while pin 7 goes to ground. I'm just trying to test to see if the AND gate works and so far it seems as though it doesn't. As soon as I plug in 5 volts to pin 14, I get current through all the output pins, 3, 6, 10, and 13, regardless of what's going on with their respective input pins, even if pin 7 isn't grounded. Obviously, the LED should only turn on when both switches are switched on, but once 5 volts is supplied to pin 14, it doesn't matter what I do to the buttons. I've tried a couple of the same AND gates from the pack, as well as some OR gates, and they all do it. <Q> It would help if you had added a schematic, but from what I can see, you are missing one vital component. <S> A pull-down resistor. <S> What this does, is it makes sure that the inputs are at 0V when there is no voltage present at the input. <S> Once the button is pressed, you will get your 5V and when both buttons are pressed, you get 5V at both inputs. <S> As it is right now, your inputs are 'floating' which means they are in a state that is unknown, which the IC could determine as a '1' state, which is why your LED is always on. <S> This will also be why the same is true for all outputs. <S> With these ICs, you should always tie unused inputs to GND via a pull-down resistor. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Look at the above schematic. <S> The top one is how I see your configuration at the moment (please correct me if I am wrong). <S> When there is no voltage applied to the input, it is left with a floating voltage and may not be at 0V. <S> The bottom one is how it should be. <S> Some pull-down resistors will ensure that when there is no voltage present at the input, they will stay at 0V. <S> Add these resistors and <S> you should see your problem go away. <A> The inputs on bipolar TTL families (74xx, 74LSxx, and others without a "C" in the middle) will source current - when left unconnected they will act as a logic High. <S> For the 74LS family, you have to draw about 0.4 mA from an input for it to be recognized as a logic Low. <S> If you want the switch between the input pin and +5V, you would need a pull-down resistor under 2000 Ohms to ensure the input was Low when the switch is open. <A> simulate this circuit – Schematic created using CircuitLab <S> This is because your circuit never put entries to 0v <S> (entries of the gate).you have to use pull-up/pull-down to manage signal on entries of the gate. <S> So when you do not use the button, the 5 volt is supplied ... <S> if you use it it connects entry to gnd the output of the gate changes. <A> TL;DR: <S> Separated from 5V by a simple switch EQUALS "pin not connected" <S> DOESN'T EQUAL " <S> no voltage applied" <S> DOESN'T EQUAL "logical 0". <S> An unconnected input of a true TTL chip (74xx, 74Sxx, 74LSxx) behaves as if it is connected to 5V DC, whereas on a CMOS chip (74HCxx, 74ACTxx, CD40xx), it behaves as if it is connected to 5V AC. <S> In any case, unless your LED has a (rare) built in current limiting resistor, YOU NEED ONE.
| Traditionally, we would put switches between the input pin and Ground to ensure that the input could be pulled low enough to be seen as a Low, and add a pull-up resistor from the pin to +5V to ensure that the input would be High when the switch was open.
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Stability of BJT circuit using Re resistor This is a post I found on Quora. I don't understand why decreasing Vbe decreases the current.Vbe=Vb-VeTo decrease Vbe you could increase Ve, Vb doesn't necessarily have to change, so neither Ib.... <Q> To decrease Vbe you can either lower Vb or increase Ve or both. <A> when someone analyses the temperature, they talk about a small phenomenon and then check that is positive feedback or negative feedback effects.(positive feedback generate instability) <S> For a simple example, if you drop a fixed voltage on NTC(a resistor with a negative temperature coefficient) then the temperature is raised( even 0.1 degrees) <S> , then a small delta(T) <S> even 0.1 degrees, generate a new R that is smaller than before: <S> R_new = <S> R_old - NTC_Coef <S> * delta-T <S> In fixed voltage, decreasing R, make current higher (V_fix/R) and then R*i^2 make the temperature higher than the last step, then makes R smaller -> increase temp -> R is smaller -> increase temp ... <S> .this called positive feedback and not good, and you should avoid instability. <S> But for your example: With temperature rise, we know the Beta and Ic will be raised, if Ic is raised even a small value, V_emitter <S> ~ <S> Re*Ic will be increased and <S> then Vcc_fixed = <S> RB <S> * IB + Vbe + <S> Ie <S> * Re ~ constant + Vbe + IC <S> * <S> Re <S> The equation shows the Vbe should be decreased and this decrease Ib( even pico amp ) then Ic decrease, inc temp - <S> > inc <S> Ic - <S> > <S> inc <S> Beta -> decrease ic -> decrease temp injunction - <S> > ... <S> This is a stable region for a circuit(all parameters microvolt, 0.1 degree, pico amp, ... ) <S> It is a negative feedback effect, the circuit tries to work on a bias point, even changing the environment temperature. <A> To bias a bipolar transistor at 1milliAmp collector (emitter) <S> current, you'll need approximately 0.600 volts Vbe. <S> For 10X less, at 100 microAmps, expect Vbe to be 0.600 - 0.058, or 0.542 volts. <S> For 100X less, at 10 microAmps, expect Vbe to be 0.600 <S> - 2*0.058,or <S> 0.600 - 0.116 = 0.484 volts. <S> For 1,000X less, at 1 microAMps, expect 0.600 <S> -3*0.058 = 0.600 - 0.174 = 0.426 volts across Vbe. <S> For 10,000X, at 100 nanoAmps, expect 0.600 - 4*0.058 <S> This relation, where each 0.058 volt reduction in Vbe causes a 10:1 reduction in collector (emitter) current, comes from the exponential equation long trusted to describe behavior of a junction. <S> Vdiode = <S> 2.718 <S> ... <S> ^[(Q <S> * Vdiode)/(K <S> * T <S> * n)] where "n" is result of the abruptness (or lack of abruptness) of the transition from N doping to P doping in the base-emitter junction (Or the "diode" junction). <S> Note this equation has a strong factor of Temperature (Kelvin).Classically, at a constant current the Vbe changes by 2 or 2.2 milliVolts per degree C (Kelvin). <S> This behavior is crucial for most voltage references.
| Decreasing Vbe reduces base current because the base-emitter junction acts as a forward biased diode and lowering the forward voltage across a diode (Vbe) also reduces the current flow through it.
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Why do designers use op-amps with fractional gains? I often find designs like the following simulate this circuit – Schematic created using CircuitLab Where the gain is less than one ( \$\ R_2/R_1 < 1\$ ) Why not simply use a resistive voltage divider? Beyond the inversion (which tends to be irrelevant in many applications), a divider with \$\ (R_1-R_2) \& R_2 \$ would produce the same output with the same input impedance. Plus, it will not have an offset problem, an input bias current problem, a transistor noise problem, or a bandwidth problem (adding a couple of capacitors can make it basically flat in frequency). Although my first instinct is to regard this as a possible instability (gain beyond the op-amp specification), I see that it is basically a stable trans-conductance amplifier with the input current given by \$\ V_{in}/R_1 \$ , so that is not a valid objection. <Q> The opamp functions as a buffer, providing a much lower output impedance than the bare divider would have. <S> This completely eliminates any loading effects created by the downstream circuitry. <S> A noninverting voltage follower configuration would provide the same benefit (and the same downsides), but if you want the inversion, this is the way to go. <S> Also, it is sometimes important for the application that the node between R1 and R2 be held at ground potential. <A> Why not use just a voltage divider? <S> It decouples the input and output circuits. <S> The output impedance is low <S> The input impedance is R1 <S> The load has no effect on the input impedance <S> The last one is particularly important, as changing the load on the previous stage can have unexpected effects, e.g. changing frequency response or nonlinearity. <S> Why use this rather than a voltage divider and unity gain follower, for applications where phase doesn’t not matter? <S> Same number of components <A> If your sensor has high output impedance, yet the ADC needs to grab a bunch of charge and the sensor cannot provide that charge fast enough to support the desired sampling rate, then I could see using this circuit.
| The opamp’s inputs are at ground which is best for performance.
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How to get a 24 V signal generated when the input circuit state changes? So I'm an IT guy with no electrical experience trying to figure this out. How do I set up a circuit that puts out a 24 V signal for a short time or puts out a continuous 24v signal that quickly turns off whenever the inbound circuit state changes? Like when the inbound turns on I need a signal and then when the inbound turns off I need a signal. I have an incredibly stupid industrial molder that will put out a 24 V signal that changes states at the end of every cycle that did not trigger a tolerance alarm. A good part is made, it turns a 24 V signal on. When the next good part is made, it turns it off. If its on and a bad part is made, it remains on, and if it's off and a bad part is made it remains off. We have a LED pace counter that increments the part count whenever it receives a 24 V signal. A continuous 24 V signal will increment it once until it is turned off and back on. I need to figure out how to get this hooked up. I know how to handle this if I had to write a computer program but I have no idea where to even start with relays, circuits, and switches etc. I've tried playing with the setting on the molder device and even contacted the manufacture and they are no help. When I use the settings they tell me to, I get a continuous 24 V out after one part is made. The only consistent cycle I can get that also uses the built in tolerance is by changing some settings and arriving at the above situation. Simply doubling the count increment and having it increment every 2 cycles is not an answer either. I currently have it wired into the circuit that gets triggered every time the safety door gets closes but when line workers get behind, they just open and close the door like 20 times and claim ignorance to what happened. Also, I'm open to any solution, IDK if there's some other product out there that can add some logic to it. Budget is around $100 but as low as possible is best. The post tags may not be appropriate, I honestly have no idea what to tag it. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> A circuit based on industrial timers. <S> Top: the desired timing diagram. <S> Centre: the circuit. <S> Bottom: the timing diagram again but showing the internal operation. <S> They are robust and have plastic enclosures and screw terminals. <S> This solution uses an ON delay and an OFF delay timer. <S> The on-delay timer output turns on some time after the input turns on. <S> It turns off immediately supply is switched off. <S> The off-delay timer output turns on immediately with the trigger input. <S> It holds on for some time after the trigger signal has been removed. <S> To maintain output it needs 24 V power. <S> The changeover contacts of T1 and T2 are wired in opposition so that the output turns on when the status of each does not match. <S> If I have misinterpreted your requirements then please add a timing diagram into your question. <S> You can do this quite well using ASCII art and the code tags as shown below. <S> _____ _______| |____| |_____ <S> _ <S> _ _ ___| |___| |__| |___| |___ Response to PLC answer: <S> You can streamline your PLC solution using the Rising Edge and Falling Edge contacts. <S> These are edge triggered and remain on for one scan of the program. <S> X1 Y0- <S> -|^|---+-------------(SET)---- <S> | X1 <S> |--|v|---+ <S> Y0 <S> +TMR-T1---+--| |-------------+ <S> 1.000s| <S> +---------+T1.Done Y0--| |-----------------(RST)---- <A> To answer your comment, yes, this is something that can be done with a PIC and some additional circuitry, but if you want to go with transistors or relays it is also possible to implement. <S> This is the basic idea (that can be implemented in either hardware or software, I chose the CD40xxx logic family as this is the highest voltage CMOS family, some gates can directly handle 15V): simulate this circuit – <S> Schematic created using CircuitLab <S> I would use Schmidt triggers as you might have to deal with relatively slow-changing signals and noisy thresholds. <S> It will also reduce power dissipation in analog elements such as the delay. <S> Note that this can be implemented also with transistors or even relays. <S> The tricky part with relays is possibly the delay element, as a long time constant is not compatible with the relatively high currents needed to operate the coils. <S> Here is a transistor implementation <S> (I might have mixed the connections on the final XOR which would invert the output pulse, but this is the general idea): simulate this circuit <A> So, I found the Ace PLC from Velocito for $60 but luckily I guess they keep some proper PLC's laying around. <S> I downloaded the do-more software and PLC sim, did a quick tutorial <S> and I think I wrote ok logic to get what I want done. <S> I just need to get it on one of the PLC's they have laying around and make sure I have the voltage right coming out. <S> Thanks so much to everyone who responded <S> , I honestly expected to get flamed for incorrect tags, posting location, and general lack of knowledge <S> and I am very happy with all of the help. <S> So thank you!
| Industrial timers are available with DIN rail mounting, etc., and 24 V is the standard operating voltages. The main "problem" is the voltage level of your signals, most circuitry works at lower voltages than that, so voltage conversion becomes necessary for both your input and your output.
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High voltage to low voltage in ps/nanosecond time (Zener/opamp?) So I have a signal that goes from 0 to ~800volts in 30ns. I would like to use this signal to trigger a camera (input impedance of 50ohm) which takes a max of 10 volts and requires the rise time to be 1ns for this 10 volts.I tried a 10 volt zener diode which works... but delays the signal a lot (not sure why) so I get 10 volts in 8ns instead of 1ns. [People said zeners are made for DC and not so much for high frequencies, so maybe thats why] I then thought of using attenuators to make the signal go from 0 to 10 volts (instead of 800V) in 30ns then use an opamp as a comparator with a 10volt DC battery connected to +Vs and Ref, Vin-, -Vs connected to the negative side of the battery. (and ofc the main signal going to Vin+) Since different opamps got different operating frequencies and slew rates I got this one: THS3491IDDATWhich got a 320 MHz operating frequency and 8 kV/us slew rate However, I was getting weird results that I wasnt able to understand and shortly after multiple legs of the op amp broke, so I wasn't able to capture the output signal Here is the input signal which is attenuated using 38dB (div 80) [time scale is 10ns/div and voltage scale is 10v/div Was I doing something wrong? or is there an easier circuit to use? Any recommendation for other opamps that are easy to handle? (this one was tiny) The source is a Rogowski coil with a resistance of 4ohm and inductance of 1.4e-7 H. (1 turn coil). Thanks in advance. <Q> Try this simulate this circuit – Schematic created using CircuitLab <A> Sounds like what you need is a "capacitance speed-up. <S> " <S> Take a look at the following. <S> Blue is the source pulse / 100 (so it fits on the scale.) <S> Red is what a resistor + Zener would output. <S> Green is what is output if a "capacitance speed-up" is used. <S> Note that the 1N755 is only a 500mW device, and the peak pulse from C1 into it approaches 650mW. <S> Not really a problem for "a few" photos, <S> but it couldn't be run like this continuously. <S> Also note that the speed-up creates a nearly -1v tail... careful. <A> You obviously have some high cost instruments. <S> I wouldn't compromise their life span with experimental add-on circuits. <S> Get an industrially built unit. <S> High Speed Pulse Generators (search for them) are available as modules and as laboratory instruments. <S> Here's one principle: https://www.overunityresearch.com/index.php?PHPSESSID=bgbfjuvodm8ekjgusgka0r4ed6&topic=1556.msg26264#msg26264 <S> The idea is to use the sudden reverse conductivity stopping of a step recovery diode to break the current in an inductor and thus genrating a pulse.
| If you want to build one, you can use Step Recovery Diodes to generate fast rising voltages.
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ground connection for isolated rs485 Is it necessary to connect isolated grounds for 2 different isolated RS485 nodes?For instance the following isolated RS485 breakout board doesn't have any GND connection in the board. https://www.mikroe.com/rs485-isolator-click If I connect them together, will it improve the performance in a noisy environment? According to my understanding signal ground is non isolated part of the supply as shown in figure. Is there anything wrong if I connect signal GND to chassis GND and connecting all isolated GNDs together(Without connecting to signal GND)? Please correct me if I am wrong. <Q> RS-485 standards mandate that a separate signal ground (Not always the same as Earth ground) wire be used to connect signal grounds at remote nodes so that the common mode of the signal wires stays close to the center voltage of 2.5 volts, or 1/2 of the 5 volts that RS-485 IC's normally use. <S> IC's as they are designed to be electrically 'tough'. <S> A better design would have offered you a 5 pin connector or a separate ground screw. <S> If just 2 nodes there should not be much drift, but 8 nodes spread out over a hundred feet would need a signal ground wire (18 gauge is good enough) from node to node. <S> As you can see from this drawing there can be only 1 Earth ground if it is used. <S> This is copied from page 6 of: Tyco Electronics RS485 and Modbus Protocol Guide Link to Tyco Electronics <A> I wouldn't be concerned too much. <S> In an industrial RS485, usually a shelded cable with two wires (A and B) is used. <S> The shield only sheilds EMI noise and it is connected to the ground. <S> For fairly long distances it is not recommended to connect two grounds together due to compensating currents that will flow trough the shield. <S> Usually there are also high value resistors connected to signal wire and to ISO_GND and to ISO_Vcc, these are called bleeding resistors and they leak some current to equalize the floating potential. <S> What I really miss are some TVS diodes to protect large potential difference before two devices get connected, but I guess the ADM2682E has inbuilt protection also, so don't worry. <A> In principle, being a differential standard, as long as the common mode potential of the interface IC is not exceeded, it should make no difference. <S> These ICs tend to be able to tolerate common mode differences of a few tens of volts. <S> In practice the different ground potentials are commonly brought in close proximity by some form of common-mode termination of the differential lines or just leakage from the ICs. <S> Keep in mind that the output driver specifies its voltages to its own ground, a split termination is all that’s needed to reference the grounds to each other. <S> That saves you the need to add a third wire to the interface. <S> If you expect large AC common-mode disturbances, the communication modules are sufficiently isolated from the equipment, and common-mode termination is not practical or adequate, <S> then yes. <S> Tying the isolated grounds together via a third “field ground” wire would reduce the chance of garbled data and damaged ICs.
| Without a common signal ground wire hopping from node to node then nodes can drift away from the common center point and data corruption will occur. The signal ground is defined as the RS-485 IC's own ground connection on the board it is soldered to. It will not hurt the RS-485 Since the two transceivers are floating, they will reach the same potential when connected together.
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Phase register in phase accumulator In the following flow diagram: Is the "phase register" a one-bit register? What does it basically do with that summation mark? What type of register is it? I basically do not know what the phase accumulator do as well but for that I need to understand what phase register is doing first. <Q> The phase-register is an n bit width register as can be seen on the in/out-puts. <S> The DDS can outputs arbitrary periodic function. <S> Your example outputs a sine wave, but you can replace the sine ROM to an arbitrary function. <S> The phase-register stores the actual phase of the periodic function. <S> What type? <S> The register can be implemented as an n-bit width D-FF . <S> The phase accumulator can be interpreted as a frequency integrator . <S> Note, that the phase is the integrate of the frequency, and the sum is the discrete equivalent of the integration. <S> t <S> =φ/2π <S> Where the φ is the phase the ω <S> is the angular frequency , the t <S> is the time and the f <S> is the frequency . <A> It is an N bit register that serves in combaination with the adder at the input to integrate the frequency tuning word (Phase being the integral of frequency after all). <S> If you look at the diagram you can clearly see that on each clock the register will be loaded with the modulus 2^N sum of its old output and the FTW. <S> In real parts N is quite often larger <S> then the address range of the ROM (32 or even 48 bits is not uncommon), and various phase dithering and Taylor series approximations are used to improve the spurious performance. <A> For example, assume the phase register starts at 0 and the frequency word is 5.After the first fclk pulse <S> the phase register become equal to 5. <S> After the second clock pulse, the phase register becomes 10. <S> This continues for every clock pulse <S> The size of the phase register and tuning word depends on the size of Sine ROM and the accuracy you need to achieve in your waveform. <S> Larger register will have more accuracy in determining the frequency that you can run at. <S> Here is a link to a datasheet for a DDS integrated circuit. <S> Page 11 and 12 go into the theory of how they work.
| The way the phase accumulator works in Direct Digital Synthesis is that every clock cycle on fclk, the frequency tuning word is added to the phase register.
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Is there a 74HCT03 equivalent I can just replace "as is" Is there a 74HCT03 equivalent I can just replace "as is"? Of course, it will be better to buy the same exact chip reference but is it always the case? Many chips are equivalents pin-to-pin, right? If so, (in limits of current & voltages of course)...is this possible in most cases or is this always a stupid idea? Thanks. <Q> That completely depends on the circuitry. <S> Both in terms of logic levels, signal delays, and transition speeds. <S> The 74HCTxxx logic family was precisely designed to be TTL-compatible high-speed CMOS. <S> That is, fast and low-power like the 74HCxxx family but capable of accepting TTL-level outputs from the 74xxx and 74LSxxx families. <S> If it is a different logic family, your mileage might vary. <A> 74HCT03 is still an active component, available for 0.5$ (USD), thru hole DIP and SMD packages available. <S> Why not use the same part to ensure IO levels will be satisfied? <S> https://www.digikey.com/products/en?keywords=74hct03 <A> Most Logic chips tend to have the same (or very similar) pinouts. <S> It is not a stupid idea at all. <S> I have had logic chips that I used <S> but when trying to get more, my supplier was out of stock <S> so I simply got a different one in the same package with the same pinout and just fitted that instead.
| Do make sure that any IC you choose to fit has the correct ratings that you need, usually found in the electrical specifications section of the datasheets. If the circuitry around it is 74HC or 74HCT, it is very likely that it can be substituted by a 74HC or faster component. You can have a look at other NAND gate chips (there are many out there) and as long as the pinout is the same and you can get them in the same package, then yes, you can simply replace it with any of them.
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What lubricant is suitable for switch contact? I have this expensive DPDT switch (20A 250V 2hp) which the only problem is dirty encumbered inside. The encumbrance is made of carbonized material and some kind of wax and dirty electrical contacts. The contact poles are in good shape after a good cleanup. I cleaned all part thoroughly. Now it all look clean and neat, ready to reassemble the switch. Since, originally, there was some lubricant inside, that looked like some kind of wax, I am on the search for a similar product. I recently dipped contacts on a relay (2HP motor) with high temperature silicon grease called Dielectric Grease 67VR. My repair lasted about 3 days (perhaps 20 ON/OFF cycles). The contacts ended up welded together and the grease was carbonated. So, obviously, the dielectric grease is Not a good option. The Question: What kind of lubricant would be suitable in my switch so it does not end up being a disaster but rather a successful repair ? <Q> The defining feature of dielectric grease is that is is a good insulator, and is used to keep moisture away from contacts where it could cause corrosion or conduction. <S> Any silicone compound will break down under arcs leaving a silicon oxide that is an even better insulator, pretty much guaranteeing damage to the contacts. <S> Switches should be greased with a high melting point, high viscosity grease on the pivots and sliding mechanisms only - where it will hopefully not migrate to the contacts, which should be dry. <A> When contacts arc they polymerize organic materials in the air. <S> If you lube the contacts you are just speeding up the process. <S> That is, the oil tends to not separate from the grease and contaminate the contacts. <S> Not sure where you would find such a thing. <S> Silicone grease is usually thickened with fumed silica, but some are thickened with lithium soap. <S> Silicone does not usually make a good lubricant unless specially formulated for that. <S> ` <A> Cool-Amp makes "Conducto-Lube", a silver-containing grease specifically designed for switches, breakers, etc. <S> https://www.cool-amp.com/conducto-lube
| When the manufacturer lubes the mechanical parts they probably use a grease with very low “creep”. The result is a sort of crud that will cause contact failure.
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Voltage across LED and source when LED is floating from one end Refer to these pictures and can you explain why is there 3.335V coming across LED and the LED isn't even turning on. Also to note is that changing the resistor value does not change voltage at D2(A2) when switch is open. (I am using the D2(A2) point as an input for a microcontroller). (This is also the same voltage coming in my real life circuit) Addition!!! Extra Points: Also is it safe to use the point D2(A2) as input to a micro-controller (PIC18F46K22) with it being brought to 0V when SW1 is closed and being brought to high logic level when the SW1 is open. <Q> You have the answer there in the properties page. <S> The drive current is 10mA. With your 10k resistor, how much current do you think is going through the LED? <S> What it will do, is change the current going through it. <S> Your properties page shows a 2.2V drop over the LED. <S> You have a source voltage of 5V. To find the current through the LED, use the formula: <S> Iled = <S> Vs-Vled/R <S> Using this, you can see the current going through your LED is: <S> (5-2.2)/10000 = 280uA which is nowhere near enough to turn it on. <S> Try changing your resistor to something like 220 or 330 ohms <S> and you will see it turn on. <S> Here is your circuit as you have it set up (using the exact same components): <S> Look at the current. <S> Not enough to light it up. <S> Here it is with the resistor <S> changed: LED on. <S> EDIT <S> It was pointed out by brhans <S> that I may have misread the question. <S> If you are wondering why the voltage you are measuring is not the 2.2V you are expecting, it is because you are not measuring across the LED. <S> You are actually measuring the voltage across the switch. <S> This means anything you change above it will not affect the voltage across it, as the switch is constant. <S> With the switch open, it will be near enough 5V, and the bottom probe will be the voltage you are measuring, which is the diode drop with zero current flowing. <S> Close the switch and you will see the voltage changes. <S> To show you what I mean: <S> This shows the expected results. <S> Close the switch and you will see the numbers change: <S> Note that the switch has an open resistance of 100Mohms. <S> As andy aka points out, this will account for the voltage level you see. <A> This looks like a simulator artefact: the software simply substracts the forward voltage of the LED from the battery voltage and displaying you that. <S> This is also what you would measure with an analog voltmeter between A2 and ground. <S> In reality, the LED has a finite parasitic resistance which would eventually bring both its pins to the same voltage, i.e. 5V. <A> When you measure D2(A2) node in the real world your multimeter will have a finite input impedance of around 10 Mohms and that will force a small current through the LED in the low microamp range. <S> Given that the voltage is about 3.33 volts, I would suggest that a 10 Mohm input impedance multimeter is taking a current of 0.333 uA. Raising the 10 kohm resistor to a much bigger value will start to reduce the D2(A2) node voltage <S> so maybe try this in order to get a better understanding of why this is happening. <S> Regarding your simulation, an open circuit switch may indeed have an internal parameter that is 10 Mohm so check the parameters of the switch. <A> There isn't 3.336V across the LED - that's the voltage at that point with respect to ground. <S> So there is actually (5-3.336) i.e. 1.664V across the LED AND the resistor, which is too low to light the LED. <S> But even with the switch on, as MCG points out in his answer, the 10K resistor won't allow sufficient current to pass through the LED to light it up.
| The forward voltage of the LED is (almost) constant (is does vary with current/temperature slightly), so changing the resistor value won't change the voltage over the LED. If you want to measure the voltage over the LED, then you need to move your probe above it.
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Opamp output sinewave is distorted I'm using an opamp to amplify a 3.5 Vpp (peak to peak) 10 kHz sinewave which seems fine apart from a tiny noise mostly visible at its' peaks which does not particularly concern me much. The blue sinewave is the input and the output is the red sinewave amplified at 6.7 Vpp. There is a distortion and that is at both no-load and loaded conditions. This opamp is meant to be used as a resolver primary coil driver which is the case of my application. The distortion is something that worries me because it could affect the accuracy of the resolver-to-digital converter. My question is what can I do to minimize the distortion as I do not see any reason for it. The datasheet refers to zero crossover distortion and secondly the opamp is "obliged", to a certain extend, to provide an output signal based on the gain factor which is not so great that could drop the BW so low to affect amplification ratio. The BW is 600 kHz and the signal being amplified is 10 kHz. Can anybody make any suggestions? Regards Images: <Q> That looks like slew rate limiting to me. <S> The op-amp you are using is specified as having a slew rate of 0.17 volts per microsecond and this is somewhat "slow" for an op-amp. <S> Even the 741 dinosaur op-amp has a slew rate that is 0.5 volts per micro second. <A> I haven't gone through the whole circuit to figure out if there are other problems but you appear to be slew rate limited. <S> Slew Rate = <S> 2π f <S> V = 2 <S> *3.14 <S> *10k*6.7 = 0.42V <S> /us Datasheet shows 0.17V <S> /us (typical) <S> Redo the test with 1kHz input signal and see if the output cleans up. <S> If it does - that was your problem. <A> I would like to thank all of you that took the time to post your thoughts regarding the issue i am facing. <S> I am attaching a few snapshots of the output waveform at various frequencies & amplitudes. <S> Despite the fact that i have been in the electronics business for 10+ years i took one thing for granted and that was the one and only thing to mess everything up. <S> This opamp was designed to do the exact same thing that i am doing but obviously there are some conditions where issues arise. <S> So, 1KHz with the amplitude very close to the pos rail <S> and i got a perfect output signal. <S> Now when it comes to 10KHz it is not a problem if i keep the output low enough as the second snapshot implies! <S> Above that amplitude, rising & falling edges of the original signal become steeper and distortion, limited by the slew rate, starts to be visible until i reach an amplitude where rising & falling edges are stuck with the slew rate. <S> I really did not see that comming! <S> I got my lesson. <S> Many thanks to all of you. <S> *The yellow dotted horizontal ruler is the positive rail @ 10V
| It is obvious that slew rate of the opamp i am using is the culprit.
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