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What kind of DC power supply has least noise I operate an MEG system (extremely sensitive magnetic sensors for measuring brain activity) and I recently found that the DC power source that feeds into our magnetic shielded room (power used for moving the MEG/chair/bed - thus required), produces 60 Hz line noise detected by the sensors. I'll detail my understanding the sources of this noise, but my question basically is: what kind of supply or what specific features should I look for to avoid this noise? Noise from currents in wires. The current supply is a typical cheap switched-mode floating DC supply, 24V 0.5A. I did a few tests to confirm that although the difference between + and - looks "clean" and constant, each wire has a large voltage oscillation w.r.t. ground. And this "voltage" seems to be the result of a current source, so even with no load, there will be an oscillation of charge density in the wires, enough to generate a magnetic field detected by our system. (In the kOhm to MOhm range between + and G, I get on the order of 0.1 mA peak current.) Noise from ground currents. Unfortunately, if I tie the negative to ground, of course the oscillation w.r.t. ground is gone, and that mostly gets rid of the detected noise, but the "offending" current is not eliminated, instead going through the ground. That also causes some noise, probably primarily because our shielded room is itself grounded. Worth mentioning that this noise is much smaller. Edit: Price is not a primary concern. I realize a well designed low-noise supply would likely be much more expensive. <Q> If you really got a cheap switching power supply, then most of the noise you're seeing is probably not due to a lack of clean regulation, but because the AC lines couple back into your supply lines after the switch mode supply; your observations under "1." match that. <S> It's really hard to avoid that – only a very dedicated device design and excellent measurement shielding can elimate that. <S> However, there's a couple of things that are often done to overcome this: <S> Make sure you don't have any power lines running into your measurement chambers. <S> That especially applies to ceiling lights etc. <S> Make your ground better. <S> If your ground level starts oscillating when you connect your supply ground to it, you simply have a high impedance <S> w.r.t. to "earth". <S> That way, you can get the "non-energy-burning, fast-to-adapt-to-load-changes" behaviour of the SMPS, and the low-noise properties of a good linear regulator (it might be worth looking a minimal bit further than the usual, ancient, 780X and LM317 regulators – there are a few more modern linear regulators on the market, and some of them have better noise immunity). <A> This parasitic current comes from capacitive coupling between primary and secondary winding of the high-frequency isolation transformer, and creates parasitic currents at AC mains frequency. <S> There could be a smaller (intentional) DC coupling, to avoid total isolation from earth ground. <S> This AC leakage will be on non-switcher PS as well, since there will be still a transformer in between. <S> It is not possible to avoid this coupling, one can only reduce it by making special isolation transformers with grounded shield between the windings at the expense of efficiency. <S> Funny, I just looked into specifications for so-called Medical Power supplies, they list the approved leakage at 0.5mA, five times bigger than you have measured. <S> To reduce the effect of leakage currents, you need to design a good grounding scheme for the entire setup, to separate return currents from power electronics from grounds on sensitive instruments, and to return the leakage current into ground before it enters your shielded room. <S> This is always a challenging problem. <S> Good luck. <A> I am going to take a slightly different approach to this. <S> Firstly, are you sure the pickup is magnetic? <S> If so, then there are essentially two things you need to do: Connect <S> one side of the DC supply to the rooms shielding right where the cables pass into the room (Both legs of the supply cable must pass thru the same hole in the room shielding), this ensures that circulating ground currents due to the EMC cap in the supply stay outside the room. <S> All the internal wiring inside the room for stuff must be tightly twisted, this will ensure substantial field cancellation. <S> Worst case use a group of four wires twisted around a common centroid (Used in microphone circuits and known as "Starquad" in that application), wiring is by paralleling opposite conductors, and it gets you maybe another 10 or 20dB of suppression. <S> If the pickup is E field rather then H, then screening the cables (and bonding the screen to the room shielding is an easy fix). <S> To answer the question asked, a linear supply built with a transformer having a grounded interwinding screen will be almost as quiet as a battery. <S> You might find that opening up your switcher and removing the cap across the isolation barrier, and then grounding one leg of the output to the room screening is useful, the supply will no longer meet EMC, but that may or may not matter to you. <S> Good luck, this stuff can be a bear to track down. <A> If you want lowest noise and are not too worried about cost ,size ,and losses then the linear supply is the best .They <S> still make them at a price for lab use .If <S> you want to make one yourself there is less to go wrong than with a switcher .If <S> you still want to do a switcher then look at a soft switching topology or some resonant scheme .I <S> did some Apples with Apples conducted EMC tests and found that my soft switch resonant circuit was no better around 150KHz but was much much better in the MHz range .The <S> conducted EMC plot fell off fast and sudden with frequency like the Cliffs of Dover while the orthodox hard switched PWM was like a rolling hill <S> .This implies that some switching loss reduction scheme will help you for radiated EMC .Remember that this is off course not as quiet as a linear ,and also note that many chinese power supplies have fraudulent EMC compliance stickers .Just <S> test one and see!
| Generally, if you want to build a somewhat efficient, cheap, low-noise power supply, you're often in for a Switching-Mode Power supply, like the cheap one you're using, followed by a relatively large capacitor that guarantees that load changes and slight fluctuations don't matter that much, followed by a low-ESR capacitor (typically, ceramic), followed by a linear regulator, followed, again, by a buffer capacitor. The "offending" current that you mention is a leakage current inside your switching power supply.
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Is it safe for a CR2032 coin cell to be in an oven? I'm designing a portable temperature logger that I would like to be able to measure temperature within my home oven. I plan on designing a mechanical enclosure to hopefully protect the design from 'frying'. Essentially I will be insulating it and I don't intend on keeping it in there any longer than it has to be. I was looking at small coin cells to power my design (it's very important that this design be very flat, only a few mm's tall). I was looking at a couple CR2032 coin cells that assumed to be appropriate for my current draw and required capacity. However, I have noticed that they are only rated to 70C. Now my oven will likely be getting up to 250C, so obviously it will be important to have good insulation because the battery will be outside of its rated temperature. But I'm curious, is this safe? Assuming my design isn't exposed to an open flame, could the battery explode? I've been hearing a lot in the news of exploding batteries. I assume the battery would eventually stop working if it got too hot, but could the battery explode? Or would it likely just stop working? Again, I'm looking at a small CR2032 coin cell. I'm primarily curious in the case my insulation fails. <Q> So, first of all, CR2032 is a Lithium Battery type, so, yes, overheat it <S> and it has a good chance of exploding/combusting. <S> I'd really just go and use some cable to get the sensor into your oven, and keep the battery on the outside. <S> Less trouble, more reliability, less isolation that could fail. <A> The highest temperature range I've seen documented by a manufacturer are the Panasonic BR-A series (carbon monofluoride type) lithium. <S> The maximum temperature is \$125\:^\circ\textrm{C}\$. You should probably consider using something like that and then also calculate the expected duration before the interior of your enclosure reaches that temperature. <S> (Corners and edges will have different behavior than broad surfaces, too.) <S> Don't assume. <S> Develop an accurate model, because you are the only person who has enough data to make such estimates. <S> In general, semiconductor-based devices will probably be in better position to operate at those temperatures and higher, but you should also do some estimations of temperature drift in their behavior, as well. <S> I don't think manufacturers specify the behavior of their batteries when operated beyond their absolute maximum specification limits. <S> So whether or not it will explode? <S> You'd probably have to contact a manufacturer and ask, or else do the experiment yourself. <S> (You asked. <S> That's my answer.) <S> But if you provide more information, create some modeling to demonstrate results, and discuss them here, then it may be possible to convince me (or others) that it may be safe to try. <S> But with what little you've provided, it's almost certainly not safe. <S> Isn't that the usual way? <S> Why aren't you doing that, instead? <S> I've even used fiber optics for this purpose, before (both in regular ovens as well as in commercial microwave oven units.) <S> I'm not sure why you aren't approaching the problem's solution from that angle. <A> No, a battery exposed to the temperatures inside the chamber would mean violations of food safety and machine safety, no matter how good you shield it. <S> Solder will also melt at temperatures well below 250°C. <S> Capacitors will also be a problem. <S> 1. <S> Infrared <S> According to this article , this would be <S> one of the use cases infrared thermometers are designed for: Checking heater or oven temperature, for calibration and control <S> However, this answer https://cooking.stackexchange.com/questions/21655/infrared-thermometer-for-oven-temperatures suggests, that you cannot just use it to "look through the glass", which makes it impractical. <S> 2. <S> Probe Commercially available high temperature loggers are battery-powered and well shielded, but they all come with a probe ( http://www.deltatrak.com/products/high-temp-data-logger ) and the battery/electronics part is always meant to be positioned well away from the heat source. <S> 3. <S> Outside the box <S> You could attach a sealed circuit without a battery to the inside of the oven window, and power it by induction, if you can get it close enough (in the way wireless chargers work). <S> But still the challenge to get the other components heat-proof remains. <S> 4. <S> Buy a new oven <S> A decent oven would be IoT integrated and should have temperature, humidity, and the blood type of the turkey readily available on a static IP. <A> Popping the cell will also expose the lithium metal inside--not a good idea. <S> Please employ one of the work-arounds people have suggested that doesn't involve putting a battery into the oven. <A> They use 1/2" thick Silicone molds for DAQ profilers with battery LiPo battery power use for reflow oven thermal profiles on bare boards to get the thermal recipe right then do a run a few with populated boards. <S> But then the entire reflow profile is completed in 5 minutes MAX. <S> I suggest you use external DAQ board and battery or invest in hotplates. <S> and as others have said , it may rupture with nasty chemicals inside the oven and <S> your Mom/gf/wife wont like it. <S> It may add an odd carcinogenic flavour to your cooking. <A> The only safe way to put sensitive electronics in hot environment is to use an active thermal control system . <S> The simplest example would be a thermally insulated container filled with a liquid with low boiling point (like alcohol), which in turn hosts your electronic component in a sealed package. <S> The insulation slows down the heat transfer, and the boiling liquid cools down the interior, keeping your device at a stable temperature. <S> ATCS are used in real life, e.g. on space telescopes (e.g. using liquid Helium to cool down the infrared sensors), but I'm certainly not suggesting to implement this in your oven. <S> Keeping everything except the temperature sensor outside the oven is by far the simplest solution. <A> In general, there is a whole engineering discipline of high temperature electronics; for an introduction, do a web search about so called downhole electronics, these are used in the oil business where exposure to a temperature range similar to a food oven is an unavoidable fact. <A> Use evaporative cooling, it'll prevent (part of) your device from going above 100 °C. <S> A recent Kickstarter, the Meater , used wireless thermometers that you could put inside an oven, grill, smoker. <S> I'm not sure why it's so great over a wired thermistor/thermocouple, but I liked their idea to bury the battery and more sensitive electronics down in the tip of the device that you'd stab into something wet. <S> As the wet thing cooked, its temperature wasn't going to exceed the boiling point (100 °C or so) until it completely dried out, in which case it would probably be inedible. <S> Designing for 100 °C ambient is infinitely easier than 250 °C.
| You might try using sensors that run leads to the outside of the oven. Given what little you've written and what I can find on such battery types, it's probably not safe. I worked for a major battery company developing lithium coin cells like the CR2032, so I will answer simply: because of the extreme thermal abuse you are thinking about, you will vaporize the liquid electolyte inside, which will rupture the cell.
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can a female A USB port be used on an OTG host? I'm designing a USB device that will act as a host all the time. Instead of using a mini-AB OTG receptacle i'd like to use a classic female A receptacle on the PCB. Is this allowed and can this potentially cause problems? Clarification: I'm asking this question because I have a device where I would like to use the OTG functionality, i.e. give the user the possibility to hook it up to the computer OR to a peripheral, but I can't use the mini/micro OTG connectors and want to use classic USB A and B connectors. Ideally I would want the OTG port to be able to communicate both as a device and a host but that is probably impossible (?) so if I can find a way to allow the user to EITHER connect the device as a host or as a device using ONE of the two ports, that would be enough. Clarification #2: I primarily want to use the OTG port as a host port, but if I can use it as a device port as well that is an extra feature on the device I'm working on which will make it more attractive. However, I do not want to use mini or micro connectors. I get that they are the new trend and all that and that they are great for smartphones and tablets but for the application I'm working on we can't and don't want to use them. I'd like to thank all who answered already whether I can use a type A host port. Any hints on how I could add a B port as well connected to the same OTG port, and let the user connect the product as a host or as a device, but not at the same time, but using a type A and B connector rather than a single micro or mini USB connector are very welcome. Confusion for the user is not an issue here since we can make that perfectly clear in the manual (e.g. you can use the host or device port but not at the same time - if you use the host port then the device port won't work or the other way around). Hope this makes it clear. <Q> The physical connector is not mandatory for it to work. <S> It's mandatory to be in compliance, and to frankly not confuse the end user. <S> That's a perfectly legit use. <S> In this case, the ID pin of your microcontroller should be permanently wired on your PCB for host mode, if you can't do it in your code. <A> You can certainly use OTG hardware in the microcontroller to do a USB host only. <S> You don't need to use the OTG extensions for anything. <S> In fact, you shouldn't even mention anything about OTG. <S> What you are doing is a host, and thus you should use a female A connector. <S> And if you intend to connect to a USB flash memory, what other connector could you use? <S> Traditional USB flash memories generally are always plugged to A connectors. <S> So go right ahead and use the female A connector. <S> You're using it exactly as it's supposed to be used. <S> Remember that you need to provide 5V power to the female A connector. <A> Yes, the standard Type-A receptacle is still a standard way to have the USB host design. <S> However, the industry is moving towards Type-C connectors. <S> The Type-C may look over-complicated at first, but you can elect to use a minimal Type-C configuration (USB2.0 only, CC pins 56k pullup) to provide the basic host functionality, yet to be fully USB compliant. <S> The OTG port has only two physical data wires, D+, and D-. <S> The bus CANNOT BE DEVICE AND HOST SIMULTANEOUSLY! <S> It either one, or another. <S> Inside your MCU there are two different logical functions/controllers, a host controller, and a device controller. <S> They SHARE a single USB PHY, by means of internal multiplexer of two different data paths. <S> However, you can put two USB connectors, Type-B receptacle, and type-A receptacle, connect D+/D- electrically in parallel, but use the VBUS from B-connector as OTG switch signal. <S> This will constitute an ugly kluge, it contradicts every USB specification, and it will cause massive user confusion. <A> Any hints on how I could add a B port as well connected to the same OTG port, and let the user connect the product as a host or as a device, but not at the same time, but using a type <S> A and B connector rather than a single micro or mini USB connector are very welcome. <S> Afaict what you want to do in this case is Use a USB mux chip. <S> use the VBUS line from the B connector as a control line. <S> When Vbus is present on the B connector you switch the MUX to the B connector and signal the processor to go into device mode (e.g. by driving it's <S> ID pin high*). <S> When Vbus is not present on the B connector you switch the MUX to the A connector and signal the processor to go into host mode (e.g. by driving it's ID pin low*). <S> * Watch the voltage level specifications for the ID input pin though, you probablly don't want to connect it directly to Vbus.
| If your device is going to be a full time host, regardless of its OTG capabilities, then go right ahead and use the full size USB A female connector. But if you do want to use OTG functionality (as per question clarification), it is not possible to accomplish this with Type-A receptacle, because the A-connector does not have the necessary ID pin to switch the function. If the two sockets are close together, you don't care about compliance with standards and you trust your users not to plug in both cables at the same time you may be able to skip the mux chip.
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What is the reasoning behind wire/cable sizing? What is the way of thinking to chose the proper wire gauge for the following system?: There are calculators in web but I don't know the reasoning behind clearly. One example here: http://www.powerstream.com/Wire_Size.htm As far as I know, the current is the only parameter which causes the heating and voltage drop. But voltage also taken into account when sizing the wires. I think they want to size the wire that it would cause %3 voltage drop. Here is my example. Is my way of calculation correct?: Lets say the intrinsic resistance of copper is rho. So for my system for the voltage drop ΔV to be 3%: ΔV = 48 * 0.03 The total resistance is: [2*L*rho/(wire area)] so ΔV = 48 * 0.03 = [2*L*rho/(wire area)] * I (wire area) = [2*L*rho/(48 * 0.03)] * I So in my case: I = 3A rho = 1.68 * 10^-8 (copper) lets say L = 10 meters wire area = [2*10*(1.68 * 10^-8)/(48 * 0.03)] * (3) wire area = 7 * 10^-9 meter^2 Is that correct? EDIT The following two calculators calculate totally different results: http://www.energymatters.com.au/climate-data/cable-sizing-calculator.php?main_page=wire_calc&calc_action=cable_size http://www.solar-wind.co.uk/cable-sizing-DC-cables.html <Q> You are doing all your calculations without understanding this goal, jumping between clearly defined terms as voltage drop, and undefined terms as conductance of unspecified copper alloy. <S> Start with a clear goal: voltage drop must be less than 3%. <S> At 48 V this gives 1.44V, which you skipped to calculate and post. <S> Now, you have 2 wires, hot wire, and return wire. <S> To get less than 1.44V at the load, you need to divide this value by two, since both wires will have the voltage drop. <S> Which gives 0.72V per wire, assuming the wires are identical. <S> This gives you the limit on how much wire resistance you can afford at 3A load: 0.72V/3A = <S> 0.24 <S> Ohms. <S> If your distance is 10m, then the wire must have 0.024 Ohm/m, a clearly identifiable parameter. <S> This concludes the first part of the task. <S> Now which wire to select for this job, is up to unspecified condition in your exam task <S> - it could be a copper-based alloy wire, it could be aluminum wire, or could be copper-coated steel wire. <S> The rated carrying capacity will also depend on wire insulation and requirements for overheating. <S> So it is a plus-minus guardbands set by manufacturer. <S> The first calculator is goofy because it does not count for allowable voltage drop, and its output is nonsense. <S> The second calculator gives 16AWG copper for 2% loss (they don't have 1.5% entry), which looks reasonable. <S> So, what is the question again? <A> There are three factors considered in wire sizing. <S> Maximum wire temperature is affected by current, wire resistivity, type of insulation and number of conductors in a cable. <S> Ambient temperature must also be considered. <S> Allowable temperature is determined by type of insulation. <S> Selection is generally made by consulting a table for the specific type of wire to be used etc. <S> Voltage drop is affected by current, wire length and resistivity. <S> The maximum voltage drop is often limited to 3%. <S> The requirements of the load and expected source voltage variation are a considered in determining the maximum allowable voltage drop. <S> Mechanical strength of the wire is also considered. <S> The selected wire is the largest wire indicated by each of the above factors. <S> Depending on location and application, electrical codes may provide all of the required information and procedures. <A> AWG choices can be many. <S> Thermal rise, %Voltage loss, %power Loss, Ohm's Law, cable cost, availability, skin effect etc. <S> Beware which criteria you NEED before making a selection. <S> Your LINK was based on %V loss and not "Ampacity" std. <S> which is defined by regional standards" and determined by a greater temp rise of an insulated conductor. <S> eg. <S> 40'C rise or 60'C etc., which depends on thermal insulation too. <S> e.g. household wiring stds. <S> Your question was due to a lack of understanding on how AWG is selected.
| The reasoning behind wire sizing is to provide less than allowable voltage drop over the distance between power source and destination.
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Can unconnected inputs make an IC get warm? I am using an ATF16V8 PLD for some simple glue logic. While testing it on a prototyping board, I noticed that it gets warm to the touch almost immediately. I checked that no outputs were short circuited, but I also knew many inputs were left unconnected. ATF16V8 is a CMOS circuit and I read that floating inputs can be an issue with this technology, unlike with TTL. Could this be the cause of the heat output and why? <Q> Yes, CMOS circuits can get hot when there are floating inputs. <S> You should always connect unused CMOS input pins to a defined voltage, usually GND or Vdd, unless the datasheet tells you otherwise (see also the end of this answer and Michael's answer ). <S> /Vdd. <S> If you leave pins unconnected, they are said to "float" and have an unspecified voltage. <S> That voltage can be from induction on the package leads, leakage currents inside or outside the package, static discharge, etc. <S> The key point is that you don't know the voltage at the gates of the input transistors to which the pin is connected (signal A in the CMOS inverter below). <S> In the worst case, this undefined voltage will be somewhere between "high" and "low", so that both transistors are conductive at the same time. <S> Thus, a high current (several 10-100 mA) flows through the transistors from Vdd to GND (Vss), thereby generating heat and possibly destroying the chip. <S> Some ICs have special circuits at their input pins to prevent this from happening. <S> This circuit is typically called bus-holder or bus-keeper , but can also be found under other names like pad-keeper <S> (e.g. i. <S> MX processors). <S> It is essentially a buffer (two inverters in series) and a large resistor connected to the input pin. <S> This ensures that the input pin is always driven to either high or low when nothing else is driving it. <S> Image sources: Wikimedia, public domain. <A> Not in this case. <S> To quote the datasheet : <S> All ATF16V8B(QL) family members have internal input <S> and I/ <S> O pull-up resistors. <S> Therefore, whenever inputs or I/Os are not being driven externally, they will float to VCC. <S> This ensures that all logic array inputs are at known states. <S> These are relatively weak active pull-ups that can easily be over driven by TTL-compatible drivers (see input and <S> I/O diagrams below). <S> The diagram shows a “>50kΩ” pull-up resistor. <S> So unless you have very long wires combined with very strong electronic emissions <S> I very much doubt it could cause unwanted toggling. <S> To quote e.g. an EFM32 microcontroller application note: All unconnected pins on the EFM32 should be configured with the GPIO->P[x].MODEL/MODEH settings to 0 (Disabled). <S> In this setting, both the input schmitt trigger and the output driver are turned off. <S> If the input is enabled (schmitt trigger enabled), floating inputs could otherwise lead to frequent toggling of the schmitt trigger and increased power consumption. <A> Question says it gets warm on touch almost immediatly under normal circumstances it should not happen. <S> Let's look into GAL16V8 datasheet because it contains some useful info: Lattice Semiconductor recommends that all unused inputs and tri-stated I/ <S> O pins be connected to another active input, Vcc, or Ground. <S> Doing this will tend to improve noise immunity and reduce Icc for the device. <S> It states that inputs and tri-stated I/ <S> O should be connected to somewhere, including power rails. <S> As PLDs are configurable device, it is possible to configure pin as input, <S> I/O or as an output. <S> In case you connect pin to ground or power rail, and pin appears to be active output because if was configured so, there will be excessive current leak and device will start heating. <S> I had such case before (found out when I was asked to troubleshoot overheating PLD), GAL device did not fry but was heating really heavily. <S> It might be your case too. <S> You should check PLD's configuration, and ensure that output pins are not connected to power rails and not connected to another output pins.
| Other devices can have increased power consumption with floating pins, but I doubt it would be enough to make it perceptibly warm. If a pin could be configured as either input or output and you are not sure which it will be, then you could place a resistor between the pin and GND
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The best way to get rid off common mode noise for remote measurement I measure DC pseudo-differentially over 1-2 m wires. In order to avoid any voltage drop in wires the built in ADC input buffers are used, thus there almost no current flow. I expect the wires to catch a lot of common mode noise from RF to 50 Hz power grid. Most materials about common mode noise that I found, consider it in the context of power line that involve common-mode chokes . Howver, I am not sure if this is suitable for my measurement. Another way is the vanilla common mode filter But again, I am not sure if it would be good enough and provide adequate level of filtering. Moreover, there might be more methods that I am not aware of. What possibilities are there to suppress common mode noise, and what is the best method for a measurement application? In addition, how much the application of a shielded twisted pair cable would help to ease the problem? <Q> You have to take care of two kinds of interference: <S> common mode interference caused by capacitive coupling differential mode interference caused by inductive coupling <S> Capacitive coupling is already handled well by shielding and by using a differential pair of signals because it affects bouth lines about the same. <S> For handling inductive coupling using a twisted pair is very important. <S> If wires are not twisted it is easy to induce quite a large voltage (acting like a AC voltage source in series of the signal source) caused by changing magnetic fields (possible sources are transformers, speakers, SWPS <S> , cellphones... <S> any AC currents).Twisting the differential pair reduces the intereferences as every other twist adds a positive/negative intereference to the signal ideally canceling out at the end (see Image below). <S> The upper part shows a non-twisted wire pair: interfernce by induction is large (proportinal to blua area). <S> The lower part shows a twisted wire pair: interference by induction is small (or even zero) as positive (+) and negative (-) interferences cancel out (partially). <A> I highly recommend trying a shielded cable. <S> Using unshielded cables picks up a lot of noise from AC power lines, especially in audio applications. <S> (I once made an unshielded guitar cable from scratch in high school, and you could definitely hear the 60Hz hum of the mains AC) <S> Also, make sure that the shielded cable is grounded properly, as this will make a big difference in regards to noise. <A> Take a look at this on how to break ground loops: Shielding, Grounding and Ground loops .
| Using a good, properly shielded twisted pair cable and avoiding ground loops can really make a difference.
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What's the point of board-internal differential pairs? Sorry if this is an ignorant question.But I just can't fathom what the advantages are of using a differentialpair for signals not leaving the PCB. I mean, the usual argument for using a differential pair is to increase common mode noise immunity, which makes a lot of sense in cables and connections off the board due to all kinds of common mode noise that can exist in such environments. I just don't see the big advantage if the signal begins and ends on the same board. For instance, noise from adjacent traces couples asymmetrically to the pair, and one can probably say the same for broadside coupled traces. This obviously introduces noise in the differential signal. So why not just use a single ended signal and then increase the minimum distance to other traces if one wants less crosstalk? <Q> A trace on a board may also be subject to radiated signals coming from elsewhere. <S> A differential pair will likely pick up less cross talk from such sources. <S> A board may be part of a system that is subject to regulations that limit the radiation it is allowed to produce. <S> A differential receiver will be less impacted by the receiver's reference level shifting due to power supply noise or ground voltage variation. <A> There are applications where differential signaling cannot be avoided. <S> For example, when you need to connect a Gigasample ADC (analog to digital converter) to a processor. <S> Gigabit links cannot be implemented without low-voltage signal swing (200-400mV), and the low-voltage signal cannot be transmitted over single-ended traces due to crosstalk, as already explained in other answers. <S> That's why the signaling schemes as LVDS ( Low Voltage Differential Signaling ) were invented. <S> Most obvious example would be DDR3/DDR4 embedded (soldered down) memory. <S> In all cases the traces do not leave PCB. <A> There are a number of reasons why differential pairs are helpful on PCBs even when the signals don't leave the boards. <S> The main benefit is that they are less susceptible to crosstalk. <S> You say that "noise form adjacent traces couples asymmetrically to the pair" which to a very slight extent is true, but if the differential pairs are run together the signal is coupled into both nearly equally. <S> There will be a slight difference but for all practical purposes the noise is coupled into both signals identically. <S> Generally speaking, differential signal routing provides twice the noise immunity as a single-ended routing. <A> Signal integrity and EMC covers a broad spectrum and dynamic range. <S> Without details of the SNR requirements , CM E-H field strength, any imbalance in CM impedance gets translated into differential noise. <S> In some cases timing skew is critical as well as path length in 100 ps increments or less for CML and PECL and LVDS signals. <S> For RF, and logic controlled impedance is critical and paired tracks are often used , even with differential video signals. <S> In the end it's all about SNR whether it is logic or analog as even high speed logic follows analog rules.
| A differential pair will likely produce less emissions. If you have embedded USB3.0 devices or any SATA or PCIexpress sockets, differential signal tracing is a must.
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Cheap and simple solution for low frequency (period about tens of seconds) pulse generator I need to generate a pulse each let's say 30 seconds. I don't need this period to be any precise (20...40 seconds fluctuation will be OK). The pulse duration should be about 1 ms. I tried to implement NE555 for this but looks like NE555 don't intended for generation this low frequencies as it would need a relatively big capacitor (I'd like to stay within small and cheap CMD technology so 10uF cap is a maximum) and many MOhms resistors. This can lead me to the topology leakage problems. So I tried to look for any counting solution but looks like that I will be able to get 100kHz divided by 64K as a maximum which is definitely not enough. Next I tried to look at the watchdog timers . However after initial filtering I see that the longest period is 1.6 seconds and the lowest price is 33 cents. Important note: I'd like to have this solution as cheap as possible . I hope to fit within 10-15 cents for BOM of this circuit. Is there anything staying out of my sight? Any thoughts? <Q> If you don't mind using a microcontroller, the pic10f200 would be a cheap and easy solution. <S> They're about 30 cents in bulk and would easily handle this task. <A> A counting solution is the way to go, if you start with a frequency of 1 kHz and use a 15 bit counter <S> you will get 30.5 mHz. <S> Using a 14 bit counter like CD4060 is also possible if you start with 500 Hz. <S> A second NE555 may be used for the puls width of 1 ms. <A>
| If you don't mind using electromechanical, rather than pure electronic solutions, the simplest way to implement this would be a gear reduction motor turning a cam, which actuates a microswitch -- then using suitable electronics (555 as a one-shot, for instance) to generate your one millisecond pulse.
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Non invasive sensor for water measurement in pvc? Background I am on a project where they want a non invasive sensor for usage. There are many sensor available but they need to cut down the pipe to place within. Specifications Pipe is PVC 1-1/4" OD, meaning 1.38" inside diameter, near 0.14 inch wall thickness. The water inside is not pure. It contains less than 20% impurities (chemical waste like salts, potash, ammonia). To cut down pipe mean to insert a water valve type sensor but it is a fixed solution. Our project needs a portable device I am not interested in actual amount in cubic meters, I just want how much it is filled. As for as temperature sensor is concerned the temperature change is a big factor because otherwise we would need to calibrate it according to ambient temperature. The pipes are placed horizontal and water is pumped by 5 horsepower motor. The Ask Can any one can tell me how can I make a sensor that can check level of water inside pipe like 75% filled. We need to detect if water approach a specified threshold . I tried magnetic coils but they are not working under varying temperature and dust. My circuit take 12 volt dc. <Q> <A> If you need a switch, then try a capacitive sensor. <S> I have no idea <S> what dimensions of your pipe is and what do you mean by check qualitatively amount level. <A> You can borrow this idea from dishwashers' salt tanks: A magnet, encapsulated in plastic to protect it from corrosion, adjusted such that it floats (styrofoam?), floats on the water inside your pipe. <S> Outside of the pipe, you attach a reed contact just high enough for your needs. <S> In dishwashers, the magnet/plastic isn't floating on the surface, but inside the salty water. <S> Once the salt solution becomes too thin, the magnet, I believe, starts to sink deeper into the tank, and activates the reed contact.
| a capacitive sensor (like a stud finder) will detect water in a plastic pipe by its dielectric constant but getting precision in a portable device could be tricky.
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Anti-static wrist strap around your wrist or around your ankle? Does it matter whether you wear your anti-static wrist strap around your wrist or around your ankle? <Q> It makes no difference where you connect the anti-static strap to your body. <S> Also, most of the resistance between any two points on your skin is getting thru the skin at each point. <S> You're just a bag of saltwater (electrically, anyway). <S> The bag has a resistance of a few kΩ to a few 100 kΩ from the outside to the inside, but once inside, the saltwater effectively shorts everything together relative to the bag's resistance. <S> So the body adds negligible total resistance to the strap, and resistance between any two points on the body is largely independent of where on the body those two points are. <S> It's really about how dry the skin is at the connection points, not where the connections are. <A> I find however that it doesn't suit my workflow round either wrist or ankle. <S> If worn on my wrist, the trailing wire interferes with stuff on my bench. <S> If worn on my ankle, it's a long way down to attach and detach, and being out of sight, it's easy to be working for a while at the bench and suddenly realise that I've not been grounded for the last few minutes. <S> Notwithstanding that I am stronger than the popper holding it together, I can still fall over if I walk away from my bench without detaching it. <S> If people come to talk to me from alternate sides, I can contrive to wrap it round the pillar of my revolving chair. <S> Fortunately my workplace is very tolerant of solutions that work, even if they are unconventional. <S> I wear a gold chain at work, with an attached wire poking out through my shirt front. <S> This is where I clip my grounding lead. <S> It doesn't get in the way, it's easy to reach for on and off, and easy to see to remind me <S> I've got it on, and not to tangle myself in it. <A> No. <S> Just be sure that it is properly connected to whatever reference you have around. <S> The table/bench you work on should also be at the same potential, and if possible also use an anti static mat and chair. <S> All those does not need to be grounded, but they do need to be connected to the same potential. <S> If ground is available ground them. <S> You probably do not even need the anti static wrist strap: grab your silicon/boards/whatever and let it sit for a while on your bench. <S> After some time... voila', everything is at bench potential. <S> If you also are at bench potential there is no harm in touching the device without wearing the strap.
| The reason is that your body is much lower resistance between any two points than the strap is. Electrically, it doesn't matter where it makes contact, with contact being the key word!
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How to make the the voltage higher after a logic gate In my circuit a have a 2 switches that connects to a logical gate and the output of gate will go to a relay but after the logic gate the voltage got lower to mili voltage and not enough voltage for the relay to work, they say that I must use capacitor but it only last for a second and I want my relay to work in indefinitely by the result of the two switches of the logic gate.Anyone have an idea or hint to solve my problem? here's a picture of my circuit design <Q> Here's a few ways of doing the same thing. <S> Circuit 1 - is a simple AND gate made from two switches in series. <S> The LED is driven directly with R1 to limit the LED current. <S> Circuit 2 - replace the LED/resistor with a relay coil. <S> Note the reverse diode across the coil which prevents the back emf spike from damaging the switches when you turn OFF. <S> The switch contacts of the relay now control the LED. <S> Circuit 3 <S> - Now we've added the 2 input AND gate. <S> Note R1 and R4 (47k) are PULL DOWN resistors to ensure the inputs stay at 0V when the switches are open. <S> This resistor value is not critical and could vary from 1k0 to 1M0. <S> The output of the gate is capable of supplying enough current to light an LED. <S> Again the 1k0 resistor is there to limit the maximum LED current. <S> Circuit 4 <S> - If we need more current output we can add a transistor, in this case a BJT <S> but we could use an N channel MOSFET (not shown) if we needed to switch a much higher current e.g. a motor or a solenoid. <S> The transistor is just used to switch the LED. <S> Circuit 5 <S> - The transistor is being used to switch the relay coil. <S> Note we keep the reverse connected diode. <S> In this case it protects the transistor. <S> R3 value has been lowered to allow for more current through the transistor taken by a relay coil. <A> 4081 logic gate, according to its datasheet, can maximally supply 10 mA current at its output. <S> If you connect relay with low active resistance, voltage will significantly sink and current will increase up to chip burnout if applied for prolonged period of time. <A> The 4081 is an AND gate <S> so, just wire both switches in series (to get AND) with the LED and a current limiting resistor to get the functionality you require. <A> Uhm , does this circuit: simulate this circuit – Schematic created using CircuitLab not achieve the same thing as what you intended to achieve ? <S> So you don't need the AND gate <S> You don't need the relay. <S> You do want to add a 470 ohm resistor so you will not destroy your LED by applying 9V to it (LEDs generally do not like that). <S> If this is not what you wanted then specify more clearly what you wanted in the first place.
| I propose you to change circuit adding transistor able supplying required current between logic gate and relay.
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USB to Microcontroller Interface I'm doing a project with the ATUC256L4U microcontroller, and it will require a USB type A connection. I've never worked with USBs before and the microcontroller datasheet doesn't offer much information on how the hardware should be set up. I'm aware the ATUC256L4U has dedicated USB pins, which I think will handle the necessary pull up resistors, but would the circuit really be as simple as the one below? I've seen other circuits with protective ICs connected, are these necessary or will the dedicated micro pins handle this? I'm also not planning on powering the device via the USB, should I still connect the 5V pin to something or just leave it unconnected? <Q> This particular processor, ATUC256L4U, has support only as a USB device, and only at Full-Speed (12Mbps) rate. <S> Therefore, as USB device, it can have either attached cable with Type-A plug, or female receptacle of B-type, either full-size B, or micro-B, since more reliable and sturdy mini-B has been obsoleted and retired by USB.org. <S> Use of Type-A receptacle is incorrect. <S> The FS mode does not require too much care about board traces as transmission lines. <S> However, depending on type of USB transceivers, a serial resistors (22-27 Ohms) with small caps (10pF) to ground might be helpful to keep the port reliable and ESD protected, maybe even without any extra suppressors. <S> The main concern for a USB-compliant device is what to do with VBUS signal (Vcc on the diagram). <S> One function of VBUS is to supply power to the device (if it does not have its own). <S> But the other important function of this pin is to inform the device that it was connected to a host. <S> If this signal is misused (not used for this purpose), the device will have no means to detect the connection, to assert the pull-up on D+ starting USB connect negotiation (if it has self power). <S> The point is that a good USB device should not source any voltage or significant current on any USB interface pins until it is connected to host and receives valid VBUS. <S> Many older devices would ignore this requirement and pull-up D+ to 3V with 1.5k resistor regardless, thus violating USB specifications. <S> For the ATUC256L4U I couldn't locate this pin right away, maybe more examination is required. <S> If no dedicated pin, a GPIO should be used to provide this functionality. <A> The answer to this mainly depends on whether you need to pass USB certification or not. <S> If not (i.e. this is a personal project), you can probably get away with connecting the DP and DM pins directly to the jack and leaving VBUS unconnected. <S> If you need to pass certification, you'll have to do at least the following: <S> Monitor the state of VBUS. <S> Only connect the DP pull-up to the line when VBUS is high. <S> The USB module should provide a way to do this. <S> If the USB module doesn't have a dedicated VBUS pin you can use a resistor voltage divider to reduce the 5V to something compatible with your 3.3V IO pins. <S> It's best to use an interrupt-capable GPIO for this. <S> Have at least 1uF but no more than 10uF of capacitance on VBUS. <S> I've passed certification with 4.7uF <S> so that's what I'll recommend. <S> This is needed to pass the inrush current test. <S> (See also this answer .) <S> If you don't do these things, you will fail certification. <S> IO/ESD protection is not necessary for functionality. <S> Whether you want it is up to you. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> In systems I've been involved with the protective ICs you refer to are usually ESD supressors. <S> I suppose whether you actually need to add protection depends on how robust you want the design to be and how suseptible the part/application is, the downside is that anything on the USB lines will potentially degrade the signal quality. <S> Check out the application note for your chip here . <S> It looks like you do need the external Rs <S> but you should check the datasheet to confirm. <S> Whether you need anything else in circuit depends very much on your application; mainly what speed of USB interface you are intending on using and how long you anticipate the signal traces to be. <S> I haven't looked at your proposed micro and am not familiar with it <S> but if you want high-speed (480Mbps) then this is another good resource. <S> At this speed your layout is likley to be fairly critical <S> I think a good rule of thumb is if your trace length is 1/10 wavelength <S> (I think thats around 6cm at 480Mhz) then you need to consider your signals as a transmission line <S> i.e. matched impedance is required on the PCB as discussed in the articles. <S> On some designs I have done where we've had several high-speed USB hubs on a backplance with very long trace lengths <S> we've had to use the common-mode chokes as described in the intel app note to get the system to pass EMC compliance testing <S> (it was for emissions in that case). <S> If this all worries you a little too much stick with full-speed (<12Mbps) and you'll probably get away with anything. <S> Good luck. <S> EDIT: <S> As pointed out your chip doesn't support high-speed <S> so my discussion about the criticality of the layout is not really that important to you. <S> A 0 to 10R, 22pF RC on the signal lines is usually sensible, often times the manufacturer will recommend something like this. <S> I'd always start with the manufacturer recommendations.
| Many Atmel's MCU do have a dedicated "VBUS_detect" input pin for USB interface.
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Triac getting current from gate I made this circuit to power up an ac appliance using a weak mecahnical switch: R1 = 1K U1 = Diac Q1 = BTB16-800 BW The appliance connects between OUT and NEUTRAL (not shown). While testing, I Disconnected LIVE from MT2 of the triac and connected it directly to left end of R1 and to my surprise, the load got powered up. Resistor was heating up as well as expected. Is this an expected behavior? How can I safeguard my circuit from such accidental connections? Will increasing the value of R1 be enough? <Q> If the triac does not turn on or the connection to MT2 is broken the resistor will heat up. <S> Power will be >10W on 120V or >40W on 240V, so more than any small resistor can handle. <S> By the way, your diac is doing nothing useful- only increasing the electrical noise and reducing the voltage to the load somewhat. <S> Also the resistor value is rather high, again decreasing the load voltage and making it more asymmetrical (DC component), which can damage some types of loads (motors, compressors, transformers etc.) <S> The resistor can be expected to burn up if the triac fails to trigger ( <S> eg. shorted MT1 to gate) or the MT2 connection is broken, so allow for that to happen safely, by using a flameproof resistor and ensuring it can't cause anything else nearby to burst into flames. <A> You should change also MT1 and MT2. <S> MT1 should be connected to Live as MT2 is connected to load. <S> Dont leave MT1 as floating. <A> I have found a LOT of examples on the internet of triac gate driver circuitry that DOES NOT WORK. <S> After a lot of trial and error, I came up with this circuit that DOES work: <S> The important thing to note is that the resistor is connected to the top of the load... <S> the load drops most of the power, so you don't need more than a 1/4 watt 300 ohm resistor. <S> I have controlled the opto-isolator with a microcontroller providing on/off and dimming functionality. <S> Im sure you could easily put your switch on the 5v side of the opto, or if you are really living on the wild side, <S> eliminate the opto all together!
| A triac 'static switch' normally would comprise just the triac and a resistor of around 100-200 ohms.
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A good way of providing 9V/5V voltage with batteries when current draw is relatively high? I have a project (doorlock) with MCU / LCD / Keypad and solenoid (lock) connected together. MCU, LCD and Keypad runs at 5V while solenoid runs at 9V. I'd like to provide the power with a battery, so I was initially thinking about getting a 9V battery to provide power to the solenoid, and use a LM7805 regulator to provide 5V to MCU, LCD and Keypad. Problem is, on idle, the 5V components draw about 100-130mA of current and the solenoid will draw about 500mA on activation (for about 5 seconds), so that's 630mA of current being drawn. I imagine the battery will get either extremely hot or limit the current. What's a good strategy of providing a consistent 9V/5V voltage with a high current while minimizing the battery size? Something like a car battery may work here but they are way too large to fit in the box. Edit: The components are normally powered by AC/DC wall adapter. I'm trying to make a battery backup system pretty similar to shown here ( http://www.electroschematics.com/6279/battery-backup-circuit/ ), so that the doorlock is kept in operation if the AC power dies out for whatever reason. <Q> For a minimum 100 mA continuous current, I think the only practical way to power the system is from AC power, with a suitable power supply to produce 9 volts at 1 amp, and a DC-DC converter to reduce the 9 volts to 5 volts for the MCU, etc. <A> If AC doesn't work for you, I remember Radio Shack used to carry some rechargeables that were capable of unusually high current. <S> Another idea would be to simply use multiple batteries in parallel. <S> Edit: Actually, a better idea might be to put a supercapacitor acrossthe battery, to handle the load when the solenoid actuates. <S> Put aresistor between battery and cap to limit battery current. <S> Figure theLM7805 needs at least 7.5V input voltage for 5 seconds, so we have: $$ <S> e^{-\frac{t}{RC}}=\frac{7.5}{9}$$ <S> where t = 5. <S> Solving for RC: <S> $$RC=\frac{t}{-\ln\frac{7.5}{9}}=27.4$$ <S> For half an amp and 9V, R = 18Ω, <S> so C = <S> 1.5 <S> F. <S> Someone check my math and reasoning? <S> I'm just getting back intoelectronics, haven't done this stuff in years. <A> If I understand the project correctly, your device will normally be powered by AC mains. <S> If the power cuts out, the battery backup will temporarily power the system so you can still get in the door. <S> A 9V battery will work fine. <S> A rechargeable LiPo or LiIon battery is better. <S> Detect whether the device is being powered by AC or battery backup. <S> If the device is being powered by battery backup, disable the display and put the MCU into a low-power sleep mode. <S> This should reduce your power consumption to the microamp range if you're using a modern MCU. <S> Switching regulators are more efficient than linear regulators (up to 40% in this case) and will massively improve battery life. <S> That way, you never need to remember to replace the battery. <S> Just remember that battery backup is an emergency fallback. <S> You can live without the display and use interrupts to conserve power. <S> Hopefully, your AC mains are reliable enough that the system will rarely need to tap the backup.
| To improve reliability, use a rechargeable lithium cell that's recharged by AC when power is on. Nearly any alkaline 9V battery or lithium cell can safely provide 500mA to power your solenoid.
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Good layout for Ethernet magnetics I'm laying out a PCB with 100Mbps Ethernet (SMSC LAN8710 Phy), a Pulse 1102FNL magnetics transformer plus the RJ45 [a MagJack isn't an option]. The transformer - probably as with most? - is 16 pin, with the RJ45-side Rx on pins 9/11 and Tx on 14/16. I appreciate that the diff pairs are important w.r.t. layout but there's also the centre taps, with their associated termination components. Is it best to keep these routed direct to the RJ45 pins, even if nestled between the Tx/Rx diff pairs, or to prioritise the diff pairs and route these taps around the sides etc?Ta Appreciate it's hard to review a screenshot, but here's a revised layout.The thick green line is the boundary between system ground and RJ45 chassis; the dotted green under the mags is a 'voided' area in the ground plane (not extended fully to the top of the mags because the right-hand diff pair is coming under and around). 'Bob Smith' components are R201/R204/R205/R206/C203. The (blue) bottom layer tracks running horizontally are low-speed - drives to some/from LEDs, and a low-current tap off of the 3v3 rail. <Q> The priority is to optimize the routing of the tx and rx pairs. <S> Run the pair 3-6 better. <S> I assume the magnetic has common chokes built-in, then the terminations to the unused pairs do not have much impact. <S> So the routing of that would have the lowest priority. <S> I remember 100Mbps Ethernet has a bandwidth of around 80MHz, so the wavelength over a trace would be like 2m. <S> The short connections do not behave like transmission lines and it is not critical to get the exact trace impedance. <A> You may consider locating each + and - differential signals' tracks as close as possible to each other. <S> I would route it differently than you shown on the picture, having central outputs through via to termination circuit located on the other side of the board. <S> I would also separated diff signals from other signals using GND polygons. <S> There're other best practices like separating ground planes under the transformer, keeping digital signals like LED activity as far as possible from PHY circuit, but it is hard to advise without seeing circuit diagram. <A> You have a 4-layer board, haven't you? <S> If not, required track width/spacing <S> makes this impractical – though I've once seen a USB2.0 Hub on a 2-layer board with 1mm track width/spacing on the signal paths. <S> Of course, impedance-controlled. <S> Keep the RX pair together, too, on the top side. <S> Connect all other components as near to the transformer as possible, not the jack.
| Use thicker traces for the node where all the terminations go to the capacitor. You also may consider using MagJack connector with integrated transformer. Put the TX pair on the bottom layer, no other components near it.
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Amplifier humming when no source is attached I've got a little problem with an amplifier board.When no audio source is plugged in, there's a loud hum or buzz (rather low-frequency, I unfortunately don't have the proper measuring equipment). As soon as I plug in a source, my smartphone for example, the loudness of the hum decreases dramatically, and when I turn on music (hence sending an audio signal) it disappears completely.I had a look on the internet, but all i could find was about the amp humming if there was something attached. This looks like the board I'm using.I would be glad if anyone could help. <Q> Sounds to me like the inputs are floating. <S> Try pulling them down to ground with a high value resistor, say 100K. See if that helps. <A> This usually happens when an audio power amp is powered by an ungrounded secondary , which may be for mobile safety reasons. <S> Even though it may have some linear or switching power supply, the current pulses easily create a voltage in the floating secondary audio circuits. <S> We know I*R = <S> Voltage and thus imagine currents coupling by mutual inductance or voltage from capacitance being induced on a floating secondary. <S> (power line I charge pulses) <S> I * high Z <S> ( floating secondary = <S> V (noticable hum) <S> this is an over simplification yet true. <S> Thus the solutions can be; turn off a laptop charger ( if using as a power amp & speaker) connect a cap from audio jack to earth ground wrap your hands around power cable and touch earth <S> ground use a Ferrite DC Line filter with Y caps to earth ground. <S> In each of these examples there is a COmmon Mode Hum field being suppressed by the low impedance ratio of the coupling impedance ( yet DC floating) to the low impedance shunting to ground. <S> The reason they do not ground mobile chargers is to prevent ground fault currents and a potential shock if you touched the plumbing while the electrical outlet had some unsafe leakage current from line to ground and ground was faulty. <S> ( open) or some similar reason... <S> In your case , the power supply or mobile charger is at fault for this EMI disturbance with poor coupling capacitance in the SMPS transformer and no active PFC to reduce AC current spikes, <S> while connecting to the power amp improves the result. <S> How they ever passed EMC , CE, FCC is beyond my interest, but that's where the problem lies. <A> I've got a little problem with an amplifier board. <S> Now imagine that this board is powered from a conventional transformer, bridge rectifier and smoothing capacitor. <S> There will inevitably be ripple superimposed on that power rail. <S> That ripple will, via amplifier biasing resistors, find its way to the input circuit (the most sensitive part). <S> Without a strong voltage source connected to the input, that ripple could be tens of millivolts and be the problem. <S> So then you connect your strong voltage source and that ripple gets kind-of crushed down and the problem significantly reduces. <S> (I can show some math if that helps) <S> when I turn on music (hence sending an audio signal) it disappears completely Finally, due to the way the ear works, as soon as the music starts, the relative loudness of the much reduced hum is diminished further i.e. you might hear a pin drop onto a table but not when someone is talking to you.
| They are picking up hum because they have no (significant) input impedance to ground, so they're just picking up any passing induced voltage.
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can we use supercapacitor in filter circuits? Supercapacitors combine the advantages of a battery and a capacitor, i.e., faster charging, slower discharging. We also know that however good a filter circuit is, there remains a ripple when converting an a.c voltage to d.c. So my question is, can we use supercapacitor instead of capacitors, to minimize, or even diminish the ripple factor, and obtain a superior DC voltage? <Q> A plain old electrolytic is better, then followed by active circuitry to flatten out the result. <S> In some cases, large power supply caps aren't used at all anymore. <S> Power supplies that do PFC (power factor correction), for example, deliberately don't have any capacitance immediately after the full wave bridge. <S> The raw rectified AC goes into a switching power supply directly, which adjusts its current draw to stay proportional to the input voltage. <S> There is still a cap after the switcher, but supercaps are even less suited for that role since the ripple frequency is much higher than the 50 or 60 Hz line frequency. <S> At 100 kHz or more, ESR and ripple current capability are important, and large amount of capacitance less important. <A> The only advantage of SuperCaps or UltraCaps or "double-electric layer capacitors" is that they can endure millions of charge discharge cycles unlike rechargeable batteries with much lower limits. <S> Batteries are superior in all other ways including ; energy storage density, initial cost, size per watt hour <S> or Ah , Joules/gram. <S> Etc Perhaps Ultracaps can offer ultra low ESR for applications like 1kW trunk audio Amps that need >100 Amp pulses with heavy power cables. <S> However better performance with a 700 CCA battery in the trunk would perform better than a 10F Ultracap, but for safety reasons with rear collisions, it is not advisable. <S> Ultracaps can also make excellent temporary storage power for portable Irons with charge bases between ironing every 10 seconds or more and last for many many years. <S> (e.g. Panasonic makes these) <S> There are many more applications but they are not useful for powering LEDs or amplifiers or any other nonlinear voltage dependent source. <S> Why? <S> because you need a DC-DC PSU that can consume the entire voltage range and the supercaps cannot handle the same charge current per unit storage (Ah) as a battery. <S> (Comparing only same physcial sizes) <S> Why? <S> because to consume all the energy \$ <S> E = \frac{1}{2 <S> } CV^2 \$ you have to drain all the charge from 100% to 0% , whereas a battery only has to change 10-20% V over <S> it's <S> State of Charge Voltage range. <S> e.g. 12.5 to 11.5 ( car) or 3.9 to 3.5 (LiPo) <S> Batteries also enable overvoltage to speed up charging ( e.g. 4.2 (LiPo) , 14.2V(car) whereas Ultracaps can blow up if you exceed their voltage rating! <S> ( with safety vents for gradual overvoltage) <S> How many Farads do you think there are in a 16850 little LiPo cell? <S> a) <S> none b) <S> 1uF c) <S> 1000 uF d) <S> 1 <S> Farad e) <S> 1000 <S> Farads f) <S> more Proof: <S> Biggest Ultracap vs Car battery <S> Largest Ultracap at D-K Nichicon 6000Farads <S> JJD0E608MSEH <S> Description (EDLC) <S> Supercapacitor 2.5V Radial, <S> Can - Screw Terminals <S> 2.2m <S> Ohm Cost $326 <S> usd <S> E=1/2CV^2= 1/2 (6000) <S> * 2.5 <S> ^ <S> 2= 18.75 kJ . <S> 12V <S> car battery 850CCA 65Ah rating $75usd Room temp ~950A CCA=5Vdrop@950A = <S> 5.3 <S> mOhm <S> E= <S> 12Vx65Ah*3600s = ~2800 kJ or 150 times more energy than the biggest ultracap at < 1/3 of the price. <A> An RC low pass filter as used in a power supply will have a ripple that is inversely proportional to the capacitance, ideally . <S> However double layer caps tend to have rather high ESR so the total impedance is higher (they add in quadrature as Resr + 1/jwC). <S> There is a practical limit as to what can be achieved in a single filer section. <S> So having a high capacitance can help, up to a point, but reducing ripple and noise is often better attacked by reducing the ripple to a reasonable amount then adding more passive (RC or LC) filter sections, adding an active regulator or a capacitance multiplier stage. <A> A supercapacitor is just a big capacitor. <S> If this would help then adding more normal capacitors would also reduce output ripple. <S> This gives us a nice easy way to test the idea. <S> Take a power supply and stick a bigger capacitor on the output. <S> Does it get better? <S> The answer is that yes, up to a certain point it helps. <S> But beyond that point other effects e.g. the inductance of the pins and the capacitors ESR start to have an impact and reduce the smoothing effect. <A> can we use supercapacitor instead of capacitors, to minimize, or even diminish the ripple factor, and obtain a superior DC voltage? <S> I think you are missing the point about filters. <S> There are NO filters that seek to totally minimize ripple. <S> Take, for instance, the ubiquitous low pass filter - you want to be able to get a voltage out of it that is somewhat representative of the incoming DC level <S> but, you don't want to wait minutes or hours for this to happen. <S> With the biggest capacitor in the world fed from a pretty big resistor your "average voltage measuring" circuit would take absolutely ages to respond to a step change. <S> Functionally this would be pants . <S> So, instead you decide how quickly you want it to respond and how little ripple you want. <S> From this you decide a cut-off frequency (maybe 1 Hz) and weigh up how much ripple will be introduced (say from a power frequency such as 50 or 60 Hz). <S> These are all easily calculated and visualized once you know <S> how. <S> From this you choose the number of 1st order stages (that's one resistor and one capacitor) you might need to cascade to obtain what you desire. <S> You might also decide to inter-twine two 1st order stages and make what is known as a peaking second order low pass stage (it has potentially better responsiveness and lower output ripple that two independant 1st order stages). <S> That's how EEs design filters; there is a steady state factor and a transient factor and both are usually equally important.
| Supercaps are not well suited to reducing ripple of something like a dumb diode rectifier power supply because they have large ESR (equivalent series resistance) compared to other types of caps.
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Heating Element (resistor) to maintain temperatures in the 80-150°F in an incubator I would like to maintain temperatures anywhere from 80-150°F inside of an incubator (an augmented cooler). Yogurt ferments at 110°F, Bacillus subtilis at 100°F, certain sprouts do well at certain temperatures, etc. I can easily whip up the scripts to control a relay based on a temperature sensor. What I need help with is picking the heating element scheme. I am interested in making nichrome coils, or cupronickel is also on my mind. I understand that what I need are VERY low temperatures, but does not that imply only that I need very low voltages? (yes i know, current is what dictates the heat produced by a resistor, but of course that depends on the supply voltage) Would DC power be applicable? Can I use the power available from my microcontroller? Am I crazy? <Q> You can just use a chassis mount power resistor or similar type as a heater. <S> You will need to size <S> the heater- typically one might figure out how much power is required to maintain the highest temperature you need, at the lowest ambient temperature expected, and then (say) double that power. <S> Here is an example resistor from this surplus dealer (but you can buy them from a distributor in pretty much any value you need): <S> The voltage does not matter much, provided you have a supply available. <S> Something like 12VDC would be safer for you to work with. <S> You might try a small incandescent light bulb of known wattage in the enclosure when it is complete to get an ideal of what power is required. <A> Since you want to maintain consistent temperatures, I'd recommend using a thermal reservoir (high heat capacity) and then heating the reservoir directly while passing low-velocity air over the top. <S> Have your sensor directly attached to the reservoir but out of the airflow. <S> Depending on the size of your incubator, your reservoir could be as simple as a few gallons of water. <S> Make sure that the top has a high thermal conductivity. <S> You could even add a few fins from an old CPU heatsink to tighten your control loop. <A> The energy required to heat depends on the energy lost in the oven. <S> I made a portable oven which I designed with a CPU fan crushed Dry Ice or a 35W solder iron suspended in air inside a plastic cooler to regulate the temperature from -40 to +70'C. <S> I suggest you work with 'C as 'F is not linear with Thermal resistance. <S> For about $30 you can instrument your range or make a portable cooler operate at any temp. <S> The trick is multiple temp sensors and force air circulation. <S> If you have a digital Home Thermostat and relocate the Thermistor inside the oven with short twisted pair, then you can regulate it to 1'F or 'C accuracy or perhaps +/-0.5 <S> depending on brand. <S> It may take up 30 minutes to rise 10'C to the set-point temperature depending on the mass that has to be heated. <S> but 1 or 2 small CPU fans are need to remove any thermal gradients inside the oven. <S> I suggest you have 2 thermistors, one for air temp and one for container temp using adhesive tape. <S> How you accomplish this can be easily solved. <S> With some experimentation on location and mounting fans near the lid you can have it up and running in hours with moderate skills or spend weeks if you wanted. <S> I did it in <1 hour and ran it for the whole day each day for a short term project.
| Use a submersible heating element (maybe scavenge from an old electric kettle) and insulate the bottom of the water container and seal the top (you don't want to add humidity). Yes you can use DC or any heat source.
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Is hot melt glue suitable as potting compound? I need some small/cheap waterproof junction boxes for some 12v garden lighting. I am considering using little (25mm x 25mm x 15mm) potting boxes and after soldering the wires, then filling the potting box with hot melt glue. However I am not sure how good hot melt glue is at protecting against water. I understand glue sticks are typically a thermoplastic - so should be waterproof. The best information I can find in a datasheet is "Water Resistance: Good": http://www.farnell.com/datasheets/1504158.pdf <Q> Potting compounds are a science. <S> If you are just protecting some spliced wires, then hot glue may work if you do a good job. <S> The worst situation is if the junction box is completely filled with water. <S> If you do a complete job around each splice, then you might succeed. <S> If there are holes, however, water will get in and be trapped. <S> If there is a PCB in there, then be careful. <S> A hard and rigid potting compound will expand and contract with temperature, and often the thermal coefficients between the compound and the PCBA are different. <S> The result is parts being snapped off the PCB. <S> The softest potting compound for a PCB is silicone. <S> It is gentle on the components and does very well at temperature extremes. <S> However, silicone doesn't adhere very well and will pull away from the surfaces it is stuck to, which will invite water ingress. <S> Typical potting compounds nowadays are urethane-based. <S> They are a balance between the adherence ("stickiness") of epoxies, but has a bit of softness to be nice to the PCB components. <A> In my experience, hot glue sticks terribly bad to rubber cables, so it's not really an option if you use those. <A> I have used direct-burial low-voltage splice kits consisting of a plastic tube filled with some kind of grease. <S> They were closed on one end like a chemistry test tube. <S> The other end had some kind of snap closure that held the wires from pulling out. <S> I saw an ad for one brand that advertised that silicone grease was the filling. <S> These should be available where outdoor lighting or automatic lawn watering components are sold. <S> The wire connection was done with a twist-on connector that fit into the tube. <S> The grease remains soft.
| The "glue" plastic itself is pretty much watertight – the question is whether water will be able to creep along the interfaces of glue and cable etc, and then corrode/shorten things.
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Can poor hobbyists utilise FPGAs? I understand that FPGAs can be used for digital glue logic on a circuit board, like NOR gates. They can also directly run AES encryption, Ethernet and Linux. I'm not very rich and not an electronics expert. Nor do I have a posh logic analyser. I could probably put together a single board 6502 computer if I copied the code from somewhere. If I wanted to rationalise my component count of AND gates, flip flops and BCD decoders, is a FPGA feasible for my position? This is the level of sophistication I'm looking for, rather than microprocessor cores. Or, is there an alternative glue technology suitable for a downbeat hobbyist? Supplemental: A better question might be "what can I replace 5 OR gates, 3 Schmitts and a Divide by 10 counter with?" Thing is, I'm also factoring in the cost of the design software. A £25 development board + £5000 /seat compiler + £1000 programmer isn't much good to the poor and deprived. <Q> You don't need to be filthy rich to use programmable logic <S> In the "glue" world -- CPLDs are what you are after if you just need a small patch of programmable logic. <S> In general, older CPLDs these days are 3.3V devices, with newer devices having 1.8V cores and 1.8 or 3.3V <S> I/ <S> O, and can be in-circuit programmed and boundary scanned through JTAG, with the "program memory" internal to the device. <S> (5V CPLDs existed, but are universally obsolete.) <S> The one downside is that they generally are only supported by a vendor-specific toolchain (such as Quartus for Altera, or ISE for Xilinx.) <S> Fortunately, free-as-in-beer, albeit limited-size, versions are available, and they're large enough for CPLDs as well as the smaller/less expensive FPGAs that are available on not-terribly-expensive devboards these days. <S> Also, for FPGAs -- most of your debugging will be done at the simulation level with the aid of a testbench, as opposed to in the silicon directly. <S> Furthermore, most FPGAs have support for internal node scan through JTAG, although the free-as-in-beer toolchains may or may not support it. <S> Finally, some FPGA technologies require the "program" to be loaded into them at startup (Xilinx, Altera, and Atmel parts are all this way; Lattice and Actel/Microsemi parts, on the other hand, are CPLD-like in that the "program" is stored on-chip persistently.) <A> Ok, after your clarification, I understand now. <S> You need the class of devices called CPLD. <S> Several manufacturers offer these devices, Altera, Lattice, Atmel, Xilinx. <S> My experience goes with Xilinx, and I would recommend CoolRunner-II. <S> The sample development board is $37 at Mouser.com, and development tools (for basic sizes) are free, including logic simulator. <S> You can find sample boards for under $5 on eBay. <S> The smallest package for CoolRunner is QFN-32 (32-pin, 5 <S> mm <S> x 5mm chip), with 0.5mm pitch, <S> and it goes for $1.40 in qty 1. <S> If you are uncomfortable with QFN, they have TQFP-44, with more DIY-style pins to solder. <S> The minimal chip has Schmitt trigger inputs, which is not common on CPLDs, 30+ configurable macrocells (flip-flops with logic around), which will let you do fairly complex state machines (sequencers) and a lot of other cool things, level translation as one, all up to 300MHz. <S> To program it for desired functionality you will need to learn a bit of Verilog (to describe your desired functions), what to latch, on which clock, which logic function to apply. <S> Then define inputs-outputs, and go. <S> The Xilinx development tool comes with Verilog/VHDL compliler and all necessary tools to partition, map, and route the CPLD. <S> You don't need to know much about internal logic architecture if your project's logic compiles and fits into the selected CPLD. <S> If not, then you need to do more investigations. <S> There are other capable CPLDs like iCE40 UltraLite Familiy, by Lattice. <S> These CPLDs a really serious, with embedded block RAMs, about 640 LUTs for smallest chip, <S> etc. <S> , but their packaging is out of realms for DYI, 0.35mm ball pitch, 1.4mm x 1.4mm total size, etc. <S> And I know nothing about their development tools or costs. <A> (Altera comes with a version of Modelsim, Xilinx has their own built in simulator in Vivado). <S> Basically you would write source code that you could ultimately target to a CPLD or FPGA, but instead of buying hardware and testing it <S> , you write another piece of source code called a test bench which stimulates (in simulation) <S> the inputs of your design. <S> You run the simulator tools and you can monitor the outputs (which appears like a logic analyzer) and make sure they behave correctly, or even monitor internal registers/signals. <S> However, I will caution you that things you are capable of doing in an HDL simulation setting is not always guaranteed to be synthesizable. <S> I would also recommend you go ahead and use the Synthesis/Build tools periodically and make sure your design is valid. <S> Most free versions of vendor tools allow you build up to a certain size. <A> Get a development kit, it will have a breadboard type arrangement for testing. <S> Try to use something in a small QFN or SO package, as you can get carrier boards off Ebay/electronics vendors and then solder them to your circuit or plug into a protoboard. <S> Get a system that has schematic capture this way you will just draw in the <S> AND / OR / NOT / <S> NOR gates & counters. <S> Easy peasy! <S> Your CPLD device will have many more pins than you need, so bring out intermediate signal nodes to spare pins for debugging, and add LED's on spare pins too. <S> Most CPLD's will usually require a clock, this might add extra complication to your circuit, although a simple 4pin metal can 10.000MHz clock is fairly cheap and simple. <S> Once you have used synchronous logic you will never look back at your prior attempts with "glitch couple logic" For a hobbyist, trying to make something work in VHDL will make your brain explode (unless you are a serious programmer).
| Though not exactly answering your question, if you want to gain some knowledge/experience with digital logic design/hardware description language (HDL) on the cheap side, you can download free versions of FPGA vender tools which come with HDL simulators What you really need, as others have mentioned, is a CPLD.
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Are there any non-ideal side-effects of putting capacitors in parallel to increase capacitance? I want a 500uF capacitor. Theoretically, I should be able to place 5 100uF capacitors in parallel to achieve 500uF of capacitance. However, are there any side effects of practically implementing this? Are there non-ideal effects that I should account for? Note: I'm looking for a 500uF surface mount ceramic capacitor. I've been able to find these, however, the tolerances are only +/- 20%. Furthermore, I've only found one manufacturer of these and I would prefer not be too dependent on a single manufacturer. <Q> Paralleling capacitors is fine electrically. <S> That actually reduces the overall ESR and increases the ripple current capability, usually more so than a single capacitor of the desired value gets you. <S> The prominent non-ideal effects are cost and space. <A> Depending on the industry you are dealing with, dormant failure modes could be a consideration. <S> 5off 100uF @ <S> +-20% means you maximum spread of terminal capacitance is: 400uF -- <S> > <S> 600uF. <S> Sure <S> what are the odds that all are at the maximum or at their minimum... <S> If one capacitor failed open-circuit (solder, mechanical etc...) <S> the total span is 320uF -- <S> > 480uF. & the nominal range lies within this, dormant failure that is not quickly detectable during any production PAT's. <A> Parallel capacitors can actually introduce resonance at high frequencies, especially if they have different values. <S> See this link for more information. <S> Especially the plot on page 3. <S> This is actually a big problem when decoupling BGAs as you cannot get the capacitors as close as you would like, and you need to use different values. <A> Yes there is a huge penalty for ignoring ESR in parallel caps at RF frequencies. <S> Murata has championed this by raising the ESR a bit in their RF ceramic caps to reduce the Series Q and flatten the overall "low Z bandwidth" in SMPS filters, which becomes critical >1MHz switching rates. <S> You must be aware of ESR*C time constant in all shunt caps, SRF and Series Q as well for optimal ripple rejection of harmonics. <S> Proof: <S> How much resistance does the capacitor itself contribute to an RC circuit? <S> For more experience on ESR vs value of C, ref my info (which I can backup) <S> What happened to electrolytic capacitors in the 21st century? <A> Your problem is not the ESR of the caps , rather it is the high ESR of the coin cell ( based on your previous questions) <S> Solution: <S> Use a better battery such as CR123A with much lower ESR <S> 3.00V <<1Ω ESR Lithium primary cells have FAR more capacitance than electrolytic capacitors at same cost or size. <S> Load regulation error % drop in coin cell voltage = <S> RL/(RL+ESR(bat) <S> Proof of ESR ( ignoring estimate tolerances from graph but for 50% SoC cell.) <S> Sample datasheet Rule of Thumb CR1025 has 30 mAh capacity at 0.1mA load and ESR of ~161 Ω CR1216 has 25 mAh capacity at 0.1mA load and ESR of ~210 Ω - thus <S> Ah capacity is inverse to ESR of battery or mAh*ESR = constant - for given family for chemistry and supplier <S> exactly the SAME is true for any capacitor where ESR*C = constant for any given family and similar size but varies between internal chemistry, quality, supplier. <S> as cap or battery wears out ESR rises sharply <S> and C drops sharply as mAh drops. <S> ESR*C < 1us for ultra-low ESR <S> ESR*C - 100us to > <S> 1 ms for general purpose alum electrolytic ESR*C <0.01us for low ESR ceramic in small values.
| Due to Resonant (//) and anti-resonant (series) behaviors in parallel caps, ultra-low ESR ceramic caps can actually amplify noise due to high series Q, even if parallel (//) Q is low. There is really no electrical downside to this.
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Direction of terminal blocks in PCB layout I am designing a PCB and I cannot find the direction of the terminal blocks. I.E on which side of the terminal block the holes are facing. I believe they are currently looking down but I cannot find any documentation on this This is eagle by the way. <Q> You need to look at the terminal block datasheet. <S> You want to pick the terminal block and use a footprint that fits the terminal block, not pick some random footprint out of the library and try to find a terminal block that matches it. <S> For example : From the drawing you can easily see that the wires enter from the 'bottom' of the recommended hole layout. <S> It would also be normal to label the terminal block from left to right, looking into the wire entry holes. <S> Make sure the hole sizes are appropriate <S> (number one cause of disappointment with through-hole boards) <S> and you can probably increase the size of the pads to make it more robust. <A> That footprint looks pretty much symetrical to me - the terminal block could be inserted in the board either way. <S> For something like this, you should get an actual part to look at (and measure) while doing the board layout. <A> The numbers on PCB headers are arbitrary. <S> If Eagle has designated numbers then choose which way you wish to have the "orientation offset tab" by your interconnect plug. <S> It probably does not matter as long as they match at both ends. <S> In which case if you are interfacing to something else look up the details for pin #. <S> Don't Guess . <S> Murphy's Law. <S> Standard layout is top view for numbering.
| You may find that the horizontal line just below the pins matches a feature on the terminal block, so it may be used as a guide as to which way the terminal block is facing.
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"Could not find Cortex-M device in the JTAG chain!" (TM4C123g) The board was working fine but all of the sudden I can't download any code on it from Keil.Whenever I try to download code on the board, I get "Could not find Cortex-M device in the JTAG chain!Please check the JTAG cable and connection devices" I tried a couple of different cables but didn't work. What can be causing the problem? <Q> This is the suggestion from the Keil website: No Cortex-M Device found in JTAG chain <S> No Cortex-M processor-based device detected (using JTAG). <S> Device is not connected, not powered, or the debug interface is not working. <S> Enable the SWJ switch in the ULINK <S> USB-JTAG/SWD Adapter section of the Target Driver Setup - Debug dialog. <S> Check <S> you have power to the device, power cycle <S> the probe, check the probe settings. <S> If the target is crashed, its possible that you'd see this. <S> Try lowering the JTAG frequency. <S> Check the probe has a correct target voltage connection. <S> The Keil forums might have some more iseas you can try. <A> Had the same issue. <S> Tried various proposed solutions, in the end what worked for me was the solution proposed by splatapus on the TI page below: https://e2e.ti.com/support/microcontrollers/tiva_arm/f/908/t/456894 For me and for the gentleman proposing that solution, it was a Windows registry issue, hence I only had to do the first step using the link provided in that answer, see: http://users.ece.utexas.edu/~valvano/Volume1/Window8KeilDebuggerFix.htm <A> I resolved the issue after I installed the Stellaris ICDI drivers : <S> Tiva™ C Series evaluation and reference design kits provide an integrated In-Circuit Debug Interface (ICDI) which allows programming and debugging of the onboard C Series microcontroller. <S> The ICDI can be used with the LM Flash Programmer as well as any of the Tiva-supported toolchains such as Texas Instruments’ Code Composer Studio. <S> Only JTAG is supported. <S> To use the ICDI, follow the instructions in the Quick Start document (literature number SPMU287) on this page to install the appropriate drivers on the host computer. <A> in case you dont have drivers yet on devices. <S> solved the problem with this error. <S> http://www.ti.com/tool/STELLARIS_ICDI_DRIVERS <A> Solved: <A> I faced the same issue and after breaking my head for a couple of days finally found the solution. <S> My PC os was Windows 10 <S> and I had the tiva c 123GH6PM board connected to it. <S> You will find the drivers Under the PORTS TAB Plugin the device now and the drivers are installed automatically,otherwise download the drivers and install them. <S> Reboot your PC , <S> clear out the temp files Plugin <S> the device now , it works .Try <S> a basic LED blink program for testing. <A> I had the same issue: my JTAG debugger was functioning well on both my boards until I made a new software release. <S> After spending hours at my new software, I realised that I had physically inverted the connector between the board and the JTAG debugger.
| In periperipheral settings under debug->settings please select connect under reset, reset after connect this will solve issues. Go to Device manager and disable all the drivers for the device.
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100 LED chaser with multiple LEDs on at the same time I'm sure this has been covered but I am not finding answers to what I am looking for. Just getting back into electronics after well over a decade...so I'm a bit rusty. :) What I need is a 100 LED sequence broken down into 20, 5 LED segments, with LEDs lit in pairs as they chase. Ideally 1/2 on, 3/4/5 off...1 off, 2/3 on, 4/5 off...1/2 off, 3/4 on, 5 off...1/2/3 off, 4/5 on. Something like that (visually represented below). Always 2 LEDs lit and 3 unlit for each 5 LED segment. I've broken out my old lab kit (was shocked at the amount of dust on it) and to my surprise I found my packet of first gen blue LEDs. I remember having to special order them and they cost me $10 each. That'll tell you how long it's been. But I have all manner of parts and was hoping that using a 555 and 4017 (I have a tonne of these) as well as some combination of transistors or inverters is what I'm looking at here. Chances are I have the parts on hand already. But I'm straining my brain trying to figure out a workable and SIMPLE circuit that can drive that many LEDs in the pattern I'm seeking. Bonus if I can regulate the speed of the chase. Any advice, input or thoughts is GREATLY appreciated. OOxxxOOxxxOOxxx xOOxxxOOxxxOOxx xxOOxxxOOxxxOOx xxxOOxxxOOxxxOO x=OFFO=ON <Q> 4017 output is 'one hot', and won't source much over 1mA, so you can't use that directly. <S> If you must employ a 4017, then the simplest way to get to a base-5 chase sequence is to use diode logic to connect the 10 outputs to the bases of 5 NPN driver transistors. <S> So output Q0 connects by 2 diodes to TR0 and TR1 bases. <S> Output Q1 connects by 2 diodes to TR1 and TR2 bases, etc etc. <S> Total bill of materials 20 diodes, 5 current limiting resistors on the bases, and five transistors. <S> If you use a driver array like ULN2803, then you don't even need the base current limiting resistors. <S> That array goes up to 50v, so you can put LEDs in series to reduce the current requirement. <S> Finally, use an oscillator to drive the clock pin. <A> Use a 556 (or two 555) and a shift register. <S> Make one go at a desired rate but most crucially set its duty cycle at 40% (that's your pattern generator) and feed its output to a shift register's data input (be aware that since this duty is less than 50% you have to actually set it to its reciprocal, 60%, and then invert it). <S> Put the output of the other timer to the clock input of the shift register and tweak the timer frequency until it's 5 times that of the first timer or until you see the pattern/speed matches what you want. <A> This isn't a full answer if this is about using all the parts you already have on hand, but you mentioned an hour ago you are flummoxed by driving two LEDs in pairs. <S> So I thought I'd start there since that's not so hard. <S> Look at the following schematic, which only uses four LEDs instead of five (for space-saving reasons only.) <S> There are three input lines, each of which turns on two adjacent LEDs. <S> This would be a 3-state case, of course. <S> But that's one possibility. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You could also consider using two different counters similiar to (but not necessarily the same as) your 4017. <S> If both were initialized to the same starting state, the following similarly short-handed case might look like: simulate this circuit Also <S> , only 3 states (for brevity again.) <S> I've no idea why you might decide to go that way over the other. <S> But you might come up with a reason, so there it is, as well. <S> The overall problem faced might be figuring out how to get 4 states out of a 10-state device, or to find a way to combine 10-state devices into a 20-state or 40-state situation where you could then divide that evenly back into 4 states without what amounts to blanking periods. <S> I'd probably sell the 4017s online in nice packages, get the money back for them, perhaps sell those all-too-expensive blue LEDs, too, and then pick up some really cheap LED modules (which can be bought with drivers included) and get a microcontroller and do it that way. <S> More versatile for the future <S> (and I think you'd imagine some fun ways to use that flexibility as time goes on.)
| A variable speed clock varies the speed of the chase.
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Why do we use negative edge trigger Flip Flop instead of positive edge triggered? The glitches due to race condition can be avoided by using a negative-edge triggered flip-flop instead of the positive-edge-triggered flip-flop used. What does this mean? Source: http://www.zeepedia.com/read.php?the_555_timer_race_conditions_asynchronous_ripple_counters_digital_logic_design&b=9&c=26 <Q> This is an issue when you have one flip-flop output driving the input of the next. <S> If both flip-flops update on a rising edge, then the second one will be sampling its input at the same time the first is updating the output. <S> If the clock has more delay (due to trace length or capacitive loading) than the signal, then the second flip-flop can miss the value. <S> Note that this problem is independent of the clock frequency, slowing down the clock doesn't fix it, only fixing the relative delays will help. <S> On integrated circuits, FPGAs, and high speed interconnects, this is handled by careful clock routing and detailed knowledge of the setup and hold times of the flip flops, as well as their propagation delay. <S> However, for low speed buses routed on PCBs, there is another solution: update outputs on the negative clock edge, and latch inputs on the positive edge. <S> That way, there is an entire half clock cycle for the signal to stabilize before it will be read, and slowing down the clock gives more time for outputs to stabilize. <S> SPI is a good example of a communication protocol that operates this way. <A> Here are the general inequalities you need to satisfy in order to meet setup and hold requirements between a launch flop that operates on positive edge of the clock and a receiving flop that operates on the negative edge of the clock. <S> Observe if you will, with great benefit that the hold inequality is now dependent on frequency (or the clock period), and that you get approximately half of the clock cycle time of guaranteed immunity to hold (assuming 50% duty cycle). <S> The drawback is that you also get half a cycle less time to meet setup. <S> But since this is typically used for slow speed interfaces, that is generally OK. <A> Having the second flip flop negative edge triggered ensures that the first FF holds its value long enough to satisfy the hold time for the second flip flop (since the clock trigger arrives half a cycle later). <A> In the circuit the text refers to, the and-gates are enabled at the same time the flip-flop is toggled. <S> The means any differences in the propagation through the flip-flop appear at the output of the gates. <S> If you use a negative edge trigger, the and-gates are disabled at the same time the flip-flop is toggled. <S> Both and gate outputs will be zero while the flip-flop is changing. <S> Then when the clock goes high the stable output of the flip-flop is allowed through the and gates.
| The second FF being negative edge triggered prevents the hold violation that could occur (race condition) by having two same-polarity-triggered flip flops back to back.
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Using a Ryobi 36v battery to power an electric kart I'm attempting to use a BPL3650 36v Ryobi battery as the power source for replica electric willy's jeeps I've constructed for my kids. The batteries charge fine in the standard chargers available off the shelf The batteries work fine in the standard Ryobi 36v tools and respond as normal to the pressing of the battery stength indicator However, when I go to use them in a circuit not in a Ryobi tool, they only appear to show 18.6v across the positive and negative terminals. I can exclude them from the circuit and just test the batteries with my multimeter and get 18.6v. Bizarrely, the actual behaviour the batteries exhibit is to start off at around 39v and over a period of about 10 seconds, drop to 18.6v (or approximately half voltage) as indicated by the multimeter. However, the batteries are actually fully charged. There are 4 terminals on the battery, the middle 2 appear to be thermistors (designated T1 and T2)and when I opened the charger to have a look, only one of them was connected to the charging circuit board. I'm 'assuming' that the battery must be getting some form of voltage across the second thermistor when in the actual hand tools that disables the battery protection circuits. Apart from dismantling a Ryobi 36v tool to see what might be triggering the circuit protection in the battery, I thought I'd post here to see if anyone has any thoughts or had already achieved this process themselves. In short, does anyone know how to get the full 36v+ from a Ryobi battery outside a Ryobi powertool? Update: Ended up dismantling a 36V strimmer to see what was going on. In the first picture below you've see the trigger switch (Jaiben DBW-2036E). The multimeter before the switch exhibits the same behaviour as before but after the swtich, a steady 41v. Would someone be able to tell me what the electrical component is that's connecting the two poles after the switch? All I can read on it is 10A so assuming amps of some form. <Q> I had the same issue right now, trying to use a Ryobu <S> BPL3640D (36V <S> x 4Ah) for a hackish project. <S> The output of the battery appears to be extremely high impedance, it floats around the actual battery voltage of 40V (pretty full) but already starts dropping when measured with a high-impedance meter. <S> My circuit wouldn't draw any current when powered with less than ~22V <S> so the voltage was just floating around that point for me without actually making the circuit work. <S> However, connecting a 100kOhm resistor in parallel to the battery made it work for me, with the battery now putting out the full voltage. <S> So maybe the output needs to be pulled below a certain threshold voltage to tell the battery something is connected before the battery's circuitry "starts up" and permits actual current to flow. <S> This is just guesswork though. <A> The charger light goes red and it will refuse to charge until the battery has cooled down to a safe temperature . <S> This is why there is only a charger connection and not a tool connection for the thermistor - it's a protection circuit not applicable to the tool. <A> It is a DIODE 1000v 10A10 , cathode connected to + for reverse voltage protection.
| The Thermistor (T1/T2) lug prevents the charger from charging the battery if it's too hot by interrupting the charge circuitry.
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Pull-down resistors on 7447 input pins I have a simple circuit with a 7447 chip driving a 7-segment LED. The 7447 input pins A=1, B=2, C=4, D=8 apparently float high as logic 1. Using momentary pushbutton switches, I short various combinations of the pins to ground, and I get the expected digit on the display-- e.g. connecting D to ground shows the digit 7, connecting B and C to ground shows 9, etc. Now I'm trying to change the circuit so the pushbuttons bring the input pins high instead of low, so I tried putting 10k pull-down resistors on the inputs. However, if I short any of the inputs to ground via 10K resistors, nothing happens-- I no longer get any digits displayed. So I'm wondering 1) why don't the pull-down resistors work, and 2) what's the simplest way to reverse the operation of the pushbuttons. I'd prefer not to have to deal with things like inverting buffers or changing the mechanics of the pushbuttons, making them normally closed instead of open, etc. TIA Carsten <Q> So what value do they need to be to do the job properly? <S> The SN7447 datasheet tells us that to be recognized as logic 0 a data input must be pulled down to 0.8V or less, and the current you have to sink could be as high as 1.6mA. Applying Ohm's Law , we get a maximum acceptable pull-down resistance of 0.8V / 0.0016A = 500 Ohms. <S> The only problem with this method is that when you switch the resistor to +5V it will draw 5V / 500 Ohms = 10mA, so when all 4 buttons are operated the circuit will consume 40mA more than it needs to. <S> If you don't mind this extra current draw then pull-down resistors are fine. <A> The 7447 (and other bipolar TTL parts) inputs source current, so appear as High when not connected. <A> 7400 series <S> TTL logic is supposed to be used so that all signals are resting at 1 and only pulled low when active. <S> These chips' pull-highs are quite strong, 10K is not enough to pull the input low. <S> Pulling the inputs low causes quite a lot of power consumption and you need to pull them quite strongly before they are considered to be low. <S> Please consult the datasheet. <S> If you want to reverse the operation, you should consider using CMOS versions of the logic chips, such as 74HC47. <S> They can be used the way you like.
| As others have pointed out, your pull-down resistors don't work because they aren't pulling down hard enough. You need to sink about 1.6 mA to ground to make the input Low.
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How to disable & enable the JTAG fuse in code on AtXmega MCU? An application sheet suggested that by disabling the JTAG interface fuse (JTAGEN) the current consumption of the MCU will drop a bit. Can this be done from within the code itself? <Q> You can disable JTAG with the JTAGD bit in MCUCR. <S> Not sure if this will have the same power saving effect or not. <A> I don't believe so. <S> The point of fuse bits is a permanent setting usually done at a factory. <S> I haven't looked into this however. <A> You can not change the JTAG fuse (or any fuse on an AVR) in software. <S> Here is code to do that... <S> // <S> Disable JTAG interface as per 4.18.6void inline disableJTAG <S> () { CCP = CCP_IOREG_gc; // <S> Enable change to IOREG <S> MCU.MCUCR = <S> MCU_JTAGD_bm; // <S> Setting this bit will disable the JTAG interface} <S> Note that I tested this code on an ATXMEGA64B3 and disabling the JTAG this way did not seem to have a measurable impact on power usage. <S> I measured 0.8uA @ <S> 3V for sleeping with the bit set and not set. <S> The resolution of my meter is 0.1uA <S> (100nA). <S> Also also tried disabling the JTAG by clearing the JTAGEN fuse bit using a an external programmer and this also did not have any measurable impact on sleep power usage.
| You can however disable the JTAG interface at runtime by setting the JTAGD bit in the MCUCR register.
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is there any problem with these via-pads? I'm designing a footprint . in this pictures we have vias and pads and i want to know is there any problem with that? i mean I've heard somewhere that something like this is a technology named via-in-pad and it is not much recommended because of some reason .i just want holes with certain radius and pads for them. <Q> Seems many of the comments to Anonymous's answer hit the nail on the head. <S> "Via-in-pad" is indeed a technology that is used but is generally not preferred for a number of reasons. <S> 1) <S> Solder can wick into the via so more solder would be required than usual. <S> 2) <S> If you use a reflow process, your stencil would need to reflect this, which would require either multiple processes (involving multiple stencils) or additional paste on other pads (not recommended if they don't have vias in them -- That could lead to solder bridges). <S> 3) <S> This would also increase the cost of the overall PCB (most board houses prefer to avoid via-in-pad technology if possible, because it's added time and resources to the build process. <S> 4) <S> Also, if the component pins are not properly soldered (solder is wicked down into the via) <S> then it would take away from the solder on the pad and would leave a poorly-soldered joint that could cause intermittent failures. <S> There are a number of reasons I would avoid this unless you're sure it's absolutely necessary. <S> I recommend not putting vias in the footprint at all, but place them during the board layout process. <S> If you MUST put them in the footprint (I have no idea why you'd want or need to) then connect them to the pads using traces that are slightly narrower than the pads themselves (to prevent the traces from making the pads larger than intended) to connect to the vias (which are not placed on top of the pads). <S> That would be my recommendation. <A> Considering this is the datasheet you pointed out and the component is SDF-009-US91-95. <S> This is what is called a Through Hole Connector or THT Connector. <S> There is no need to include a SMD pad touching the pad hole. <S> You only need to ensure the hole size is big enough for the pin and the pad hole is big enough for the solder. <S> In this case you could use a 0.8mm drill hole and 1.3mm pad size Something like this. <S> About the Via-In-Pad technology. <S> There are some misconceptions around this technology. <S> In general, via-in-pad involves some kind of via plugging. <S> This is the via is filled with a conductive or non-conductive material. <S> Otherwise there will be some problems during the assembly proccess. <S> In general, this technology is not recomended unless you are willing to pay for these extra process. <A> Via in pad is actually recommended in some designs because they form shorter path. <A> Based on the link you provide in a comment, you are making a footprint for a through-hole connector. <S> In that case, we don't consider the holes in the pads as via-in-pad - via-in-pad only applies to surface mount footprints. <S> Since this is for a through-hole part (has pins that must go through the board) <S> You should not normally include the tracks (long pads?) <S> you show when you make the footprint.
| The only benefit I can think of would be shorter signal paths which would help improve signal integrity in high-speed circuits, but its downsides are significant. these pads are simply through-hole pads, and you get the holes as part of the pad definition - no need to place additional vias. This "via in pad" design may not be recommended for reflow and other automatic soldering methods because vias are filled with air before soldering, and when soldering starts, because of temperature increase, air in via expands, and may cause defects in soldering paste application, sometimes causing smaller parts to stand up on the board.
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Using DC in transformers? I read in many websites that transformers can only be used, to step voltage up or down, with AC current (which is why AC is preferred for power transmission since they can transfer huge power in the form of high voltage over thin wires instead of thich ones then they can be stepped down again), but then I started studying car mechanics and I discovered that the ignition coil also acts as a transformer and can step voltage from the 12V battery to ~30kV, but from the DC car battery, so the question here is: Is a transformer only used with AC? And if so, how does the ignition coil step up voltage. And if it can be used with DC too, so why use AC current in the first place?? <Q> Transformers are AC only. <S> Running DC through a transformer basically gives you a heater. <S> Critically, transformers work through the fact that a change in magnetic field induces a voltage in a wire. <S> The critical portion is that the change is required . <S> In an ignition coil, the change is created by simply connecting and disconnecting the ignition coil from the battery. <S> The disconnection of power from the coil produces a collapse of the magnetic field produced by the current-flow through the coil, and results in a high-voltage pulse on the output, and subsequently a spark. <S> The connection and disconnection of the ignition coil from the battery converts (some of) <S> the DC battery voltage to AC . <A> Details of a homemade experiment in "Old Sparky": my ignition coil circuit . <S> Diagram from the link: <S> The vibrator generates the AC that feeds the primary of the transformer. <A> @Connor Wolf has the correct answer. <S> But, I would like to add one other image, since you are an automotive type. <S> An inductor is a lot like a fly wheel. <S> With a flywheel, you can't change the angular velocity (RPM) instantly. <S> The faster you try to change the RPM <S> (accelerate the flywheel), the more torque is required (or released). <S> Similarly, in an inductor, you can't change the rate electric charge is flowing through the coils (currently) instantly. <S> The faster you try to change that rate, the more voltage is required (or released). <S> So, just like you have either apply torque to, or get torque from a flywheel to change its angular velocity, so you have to apply voltage to, or see a voltage generated from an inductor when you try to change the rate electric charge flows through it. <S> Incidentally, a transformer is much like a torque converter in an automatic transmission. <S> The magnetic core of a transformer is kind of like the fluid medium in the torque converter, coupling two inductors / flywheels together. <S> And some equations... for no good reason. <S> LOL \$\begin{align}Accel_{rpm} &= \frac{torque}{Mom\ of\ inertia} <S> \\Accel_{rpm} &\leftrightarrow \frac{dI}{dt <S> } \\torque &\leftrightarrow voltage \\Mom\ of\ inertia &\leftrightarrow L \\\frac{dI}{dt} & <S> = \frac{V}{L} \\\end{align}\$ <A> But it can work with DC and vibrator/self-exciting circuit. <S> If you want to get to know more deeply this topic check how works DC/DC transformer converters. <S> (photo from TracoPower documentation)
| Like my previous speakers said transformers work with AC.
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Detect if object is lying on the surface or is it lifted into the air? Suppose we have a small box with a circutry and a set of sensors in it. Box is lying on the table, but it is light enough to be lifted into the air. What I want to achieve is to switch off the circutry in a box if anybody lifts the box, and switch it on back when box is returned to the table. Is it possible to detect if object is lying on the surface or is it lifted into the air with just accelerometer(s) and/or gyroscope(s)? If not, what set of sensors can reliably do such a job? Proximity sensors? Distance sensors? Thank you in advance. <Q> Accelerometers will tell you if it moves, gyroscopes will tell you if it's rotated. <S> Both of those can tell you if it's picked up - but will probably give you a false positive if someone picks up or bangs the table. <S> Have you considered a simple contact microswitch on the bottom of the object? <A> An optical reflection using 5mm IR emitter and detector are most sensitive in the small gap controlled by narrow beam width and angle. <S> This pulse "signal" reduces by 2x inverse squared with distance. <S> Sensitivity with daylight blocking PD is tuned by voltage gain of the Load R and logic level. <S> there are Tons of examples of IR proximity methods for < $1 <A> Ultrasonic would work regardless of the ambient light, but probably not if the surface is absorbent. <S> There probably isn't a single, uniformly-acceptable, reasonable-cost solution for all cases
| A simple light sensitive device would work in "normal" cases where the surface is opaque and there is ambient light.
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build a constant current supply I have the following problem: I wish to build a microammeter/milliammeter that will interface to an Arduino. Sparkfun sells an ACS712 breakout board with an op-amp and two pots for calibration. Suppose I'm running a 5V Arduino UNO or equivalent. I want, say, 100uA to be full range. So I need a 100uA constant-current source to use for calibration. I've read the various answers to other constant-current questions, but they are unsatisfactory. For example, an answer that explains that the output voltage will double if the load doubles doesn't help much, because the only voltage I have available is 5.0V. Or perhaps 5.2V or 4.95V. So I need a reference that will provide me with 100uA, that will allow me to tweak the gain of the breakout so that I get a reading of, say, 1000 units on the 10-bit ADC. The ACS712 has a resistance of 1.2m ohms (yes, milli, not meg), so unless there is an external current limit, the power supply would self-destruct or the chip would vaporize. It would be nice if it were independent of the power supply voltage, but I could save the calibration voltage, which I can read, in the EEPROM and adjust according to Ohm's Law, although I'd rather avoid that. Sparkfun also has a voltage reference chip, the TL431, whose datasheet shows a constant-current sink (see Figure 21 on page 7). Would this create the constant-current reference I need? Note: my goal is to have six ranges on my ammeter. To do this, I am considering the brute-force approach of buying six ACS712 breakouts, calibrating each of them at 10uA, 100uA, 1mA, 10mA, 100mA and 1A, and feeding them into A0..A5. If I choose 1000 units as my calibration point, then overrange is going to be >1000. So all I need to do is read analog inputs until I get one that is <=1000 and thus my meter becomes autoranging. But without a set of six constant-current sources I have no way to calibrate. Thoughts/suggestions? TIAjoe <Q> I need a 100uA constant-current source to use for calibration <S> To calibrate it, connect one current-sensing pin to ground on a power supply. <S> Connect the other current-sensing pin via a precision resistor to a suitable voltage on the power supply. <S> Maybe consider 10 volts as the excitation voltage. <S> 10V across a resistor of 100 kohm produces 100.0000 uA. <S> Even if you added the 1.2 milli ohms into the equation the current would be 99.999999 uA. <S> All you need next is a precision voltage source of 10V so buy a voltage reference chip with as much accuracy as you can afford. <A> The ACS712 is a 5A sensor with 1.5% claimed accuracy, so readings of less than about 100mA will have little meaning, let alone decent accuracy, op-amp <S> or no op-amp. <S> So that particular sensor is dead in the water. <S> On the more general question of calibration, you don't need a constant current source to calibrate, all you need is a known current. <S> A stable unknown voltage source with a stable series resistor and a trusted ammeter is all that is required. <S> If you're using a cheap handheld meter, a few inexpensive precision resistors (0.1%, for example) may be more accurate than the current ranges of the meter, but don't forget to account for the input resistance of the meter. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Forget about ACS712. <S> It's not close to be good at such low currents, believe me. <S> It has noise equivalent to tens of miliampers, so you can't control such low currents. <S> For current source i would say you need emitter follower circuit, as simplest option. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Something like that. <S> So the positive input is the reference voltage and the collector is your negative output to load, while the positive is the VCC. <S> Then just calibrate your reference voltages and voila. <A> I've used the first circuit here , to make a low value current source. <S> You could pick a lower voltage reference. <S> How much voltage compliance do you need?
| Current mirror is also an option, but i would start from emitter follower, especially because you can calibrate it at each working point.
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Unknown symbol on schematic (Circle with "M" underlined) I have a symbol on my schematic but I cannot figure out what is this? I looked through Internet and didn't find anything( Is this a microphone? If so give me some helpful links to learn something about it. I have to draw this element and make a pattern for it. So links for pattern would be also great.The whole scheme looks like this: <Q> Now you've added the full schematic, that's almost certainly a motor and not a microphone. <S> The four MOSFETs, VT3, VT4, VT5, and VT6 form an "H-bridge" circuit which is a very common way of driving a motor with bidirectional control. <A> <A> Looks like a motor controller circuit. <S> Uynp1 and Uynp2 are two H-Bridge controls for direction and stop and the rest of the circuit is a interesting dead time or perhaps a dynamic breaking mode control. <S> The line under the M is likely an inhouse differentiator. <S> With an AC motor it might have a ~ (tilde) character under the M instead of the - (dash) for DC motor.
| Well, this picture is a circuit drawn according to Russian standards, so that symbol meaninig is exactly "DC motor".
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Why is 10W resistor getting hot with only 6.5W running through it? I am running about 6.5W though a 10W resistor . The ohm rating is 220 ohms, which is correct for the circuit ohms which is calculated to about 225 ohms. Here is what is running through my 220 ohm 10 watt resistor: 38.4 volts 0.17 amps 225.88 ohms 38.4V * 0.17A = 6.528W Within a couple minutes it got so hot that it burned me. I'm ok though because I only touched it for a second. But I was expecting it to stay cool since the resistor is rated at almost double the power that is going through it. The electronics people told me that it shouldn't get hot with double wattage. Is this normal? Why would the resistor be getting hot? Also, is there a fire risk? p.s. The resistor is resting on brick. <Q> First let's do a quick number crunch: <S> 6.528W/10W = <S> 65% (of 10W) Referring to the datasheet: <S> There is about a 165C rise in temp. <S> Do not touch! . <S> As for "Is it a safe temperature for the resistor?", refer to the next figure: <S> I'll admit that the Derating Curve Graph kinda hurts my head. <S> But, if you follow the 10W curve over to 25C (about room temperautre), the resistor should be able to handle 100% of it's rated power. <S> Note that I'm only assuming the ambient temperature is 25C ! <S> If you have it lying on a brick, it should be okay. <S> It appears that the resistor can handle up to about 115C ambient temp @ <S> 65% load. <S> But that would be pushing it to the max. <A> Just because it is running at 50% of its rating doesn't mean it will run cool. <S> I looked at the data sheet for a similarly sized 10 watt resistor. <S> It had a curve showing temperature rise versus percentage of rated load. <S> For 50% (5 watts), the temperature rise is 125C which is greater than boiling water. <A> That means 6J of heat per second. <S> Using calorimetric equation Q=c.m.(T2-T1), where Q is total heat, c is specific heat capacity, <S> m is <S> mass and T are temperatures, one can derive P= <S> dQ <S> /dt= <S> c.dT/dt. <S> If you use your values, you can see that the resistor's temperature rises according to P/(c.m). <S> Thanks to its small weight, the rise in temperature is really fast. <S> There is also counter-process: heat dissipation. <S> The higher difference <S> T-Ta, where T is resistor temperature and Ta is ambient temperature, the higher heat dissipation. <S> There are more variables to rule the dissipation and the temperature: Heat capacity of the heatsink (air), mass flow of the air, etc. <S> Regarding your question: Yes, it is normal and it is caused by power to weight ratio. <S> If it will be resting on a brick, there is no fire hazzard. <S> If you put it between two bricks and seal it, it can melt itself and burn. <S> If left with flamable pars, there is fire hazzard. <S> Personally, I would add cooling to it, small aluminium radiator shall dissipate 10 W easilly (Foolproof roundup).
| If insufficiently cooled, it will get really hot. This is normal behavior for a power resistor of the size you are using. You are pushing 6W to the resistor.
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Is self inductance and leakage inductance the same thing? I want to intuitively understand what they are and if they are the same thing. <Q> Let's start by defining the two terms: Self-inductance is a measure or coefficient of self-induction in a circuit, usually measured in henries. <S> It is also the property of an electric circuit that permits self-induction. <S> Here is a longer definition of leakage inductance: <S> Leakage inductance derives from the electrical property of an imperfectly-coupled transformer whereby each winding behaves as a self-inductance constant in series with the winding's respective ohmic resistance constant, these four winding constants also interacting with the transformer's mutual inductance constant. <S> The winding self-inductance constant and associated leakage inductance is due to leakage flux not linking with all turns of each imperfectly-coupled winding. <S> It also might help you if you first understand what inductance is: http://www.youtube.com/watch?v=84kl_WheNtQ <S> Here is another which is also very helpful: <S> http://www.youtube.com/watch?v=NgwXkUt3XxQ <S> Essentially, inductance is when the energy is stored in an electromagnetic field. <S> Now that you grasp inductance, let's look at self-inductance. <S> Self-inductance is what happens when the current can change in a coil as a result of itself. <S> It occurs as a result of the electromagnetic field (EMF), and has a relation to the proximity and shape of the coil predicting the scope and depth of the EMF. <S> Here is a video explaining self-inductance: <S> http://www.youtube.com/watch?v=LwwSJhEfJeA <S> Here is another video on self-inductance: http://www.youtube.com/watch?v=B8CPGiK59f8 <S> Even if you don't know the math, you should be able to glean the basic concept. <S> In a coil, the electromagnetic fields overlap. <S> As a result, the overlapping EMF can feed itself (to an extent), and this is called self-inductance. <S> Leakage inductance, on the other hand, is a problem, not a property. <S> Leakage inductance happens when self-inductance occurs undesirably, and by definition, leakage inductance will result in an unwanted outcome unless it is mitigated. <S> As a result of leakage inductance, what will happen is that the magnetic field does not follow the intended path. <S> This will disrupt the flow of the EMF unless addressed. <S> Whereas self-inductance is a property of leakage inductance, leakage inductance is not a property of self-inductance. <A> This is best explained with reference to a transformer. <S> Self-inductance of the primary is the core magnetization inductance, measured when the secondary is open circuit. <S> Normally this is numerically in henries (rather than milli or micro henries) for an AC power transformer. <S> It will include the leakage inductance of the primary but as this is about 0.1% of the magnetization inductance it represents a tiny error. <S> A good definition of leakage inductance of the primary is that inductance that doesn't contribute to inducing a voltage on the secondary. <S> It's impossible to measure on its own because the secondary winding also has leakage inductance and <S> the normal measurement method is to short the secondary and measure resulting primary leakage inductance. <S> If you look at the equivalent circuit of a transformer you will see where these inductances fit into the big picture: - So, from left to right we have: - Primary leakage inductance X\$_P\$ and copper loss R\$_P\$ <S> Primary magnetization inductance X\$_M\$ X\$_S\$ and R\$_S\$ are the secondary leakage inductance and copper loss. <S> The thing that looks like a transformer labelled N\$_P\$ and N\$_S\$ should be regarded as a perfect power converter with "turns ratio". <A> @CKCK <S> Self-Inductance is the no load shunt current needed to energize the core that is proportional to voltage and affects B field and saturation limits that causes odd harmonics and is normally around 10% of rated current but reactive. <S> Leakage Inductance is the series amount that represents the amount of leakage that does not couple between the primary and secondary, which is symbolically, shown in series to attenuate source with load by this impedance divider. <S> This is a small fraction of the core inductance and is related to coupling factor losses. <S> Coupling factor losses can be 5% for small power transformers <100VA and <0.1% for MVA big power units. <S> Additional losses are resistive.
| Leakage inductance is an inductive component present in a transformer that results from the imperfect magnetic linking of one winding to another.
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When I added a resistor to a set of christmas lights where I cut off bulbs, it gets hot. Why does a shorter string of lights not need a resistor? So I have learned that in order to shorten christmas lights, you need a resistor. And unless you get a very very big resistor (which is expensive and hard to find, or use a heatsink, or string many together to up the wattage a lot), it will get very hot and you can't let it touch anything. I don't know what I am missing why I needed a resistor for shortening a christmas string unlike the shorter string. For example, someone could manufacture a string with 78 bulbs instead of 100, and I am sure they would not have a hot resistor on there. But why does a shorter string of lights not need a resistor? I'm not sure why I am needing to add a hot resistor, I wondered what they do in the manufacturing process to determine how to make the string without a resistor? Is it the type of bulb? Or is there something else involved? Is there a way to emulate the shorter string of lights without the resistor? Note: these are strings which plug into a wall outlet of 120V. The voltage therefore cannot be adjusted since the power outlet cannot be adjusted. <Q> The strings are designed to use bulbs whose voltages sum to equal the supply voltage. <S> So a string that uses 20 bulbs for a 120V power source will use bulbs designed to operate at 6 volts. <S> And a string that uses 50 bulbs for a 120V source will use bulbs designed to operate at 2.4V. <S> When you are making hundreds of thousands (or millions) of strings you can have custom bulbs made for whatever voltage you wish. <S> If you want to remove some of the bulbs and shorten the string, then you must compensate for the power the bulbs were using or run the risk of premature failure of the remaining bulbs which you are operating over-voltage. <S> A 2.4V bulb designed for a 50-bulb string is NOT "interchangeable" with a 6V bulb designed for a 20-bulb string. <S> No matter how similar they may appear to the naked eye. <A> If you want to shorten a string of 100 bulbs to 78 (actually two parallel strings of 39 for 120VAC) you can wire a 1N4007 rectifier diode in series with the strings. <S> It will reduce the RMS voltage by about 30%, which is about right. <A> Bulbs can be manufactured to any desired voltage. <S> I'd assume the strings use different voltage bulbs. <S> Did you measure the voltage across each bulb when it's in <S> it's string and lit? <S> Incandescent bulbs also don't have a rigid threshold of "working" vs "not working". <S> The bulbs in one string might be overdriven with a voltage higher <S> then they're really supposed to be run at, and the other string might be under driving the bulbs. <S> Overdriving will produce more light, but shorten the lifetime. <S> Underdriving will do the opposite, but you'll still get some light unless it's very, very underdriven. <A> If the bulbs are connected in series then their voltages will add up to whatever the power supply outputs - this means any bulbs removed must have their voltage drop made up for with resistors. <S> If you remove 2 bulbs at 0.5W each you will need to dissipate 1W of power if you want to maintain the same brightness in the other bulbs, for example. <S> Add up the combined wattage of the removed bulbs and that's how much extra power you have to dissipate. <A> The lights were designed to 120 VAC input. <S> The power output (light and heat) can be estimated by simple Ohm's law (supposing ideal circuit): <S> P= <S> VI= <S> V^2/R. <S> By leaving some lightbulbs out, you are reducing the overall resistance and thus increase the power drained from the supply. <S> With constant voltage supply, that means you are drawing higher current. <S> Every element of the circuit dissipates power P=RI^2, where R is resistance of the element and I is the current flowing through it. <S> That's why the resistor gets hot and bulbs are brighter. <S> Your note is false, though. <S> You can adjust the voltage applied to the string. <S> Add (auto)transformer between wall plug and the string. <S> This way you can reduce the overall voltage to fit the 78 bulbs instead of 100.
| In a shorter string you would either use a lower voltage power supply (if not running the bulbs directly on the mains) or you would use higher resistance bulbs to bring the current down (with more-or-less even voltage drop across all bulbs).
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Connecting two atx power supplies in series My problem is I've bought a 3d printer from china. As usual with consumer electronics the psu is a heap of shit. I've read on forums you can connect two atx supplies in series to generate 24v instead of 12. HOwever they say you must cut the earth connection. why? Am guessing the gnd is connected to the earth hence it's then no longer in series <Q> As you can guess, you cannot connect two separate supplies in series if at least one point of them is common. <S> You know, ATX PSUs' outputs are isolated from mains: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> But you don't know that whether the output return path (i.e. GND) of any ATX PSU is connected to earth (e.g. via a series capacitor) or not. <S> If return path is connected to EARTH, you cannot connect these two supplies in series. <S> But note that EARTH connection is of vital importance. <A> Don't jump to conclusions. <S> Check the continuity between the Earth line on the line cord and the "ground" ( DC Common ) on the DC output connector with an Ohmmeter to see if they are actually connected. <S> If they are indeed isolated you should see a minimum of 10KOhms, probably much more. <S> If they are isolated, no need to cut. <A> The best way is to locate the point in the power supply where the black 0v cables are connected to the metal case of the ATX supply. <S> Disconnect the 0v connection from the case thus floating the power supply but leaving the metal case with its safety earth still connected. <S> Do the same with the other P/S and you can now connect the P/S's in series in whatever configuration you like, ie 12v plus 12 for 24v or 12v plus 3v for 15v etc . <S> don't use the -12v etc as they are very low current. <A> One atx must be ground disconnect and you'll have 24 volts make sure <S> have that atx isolated <S> and that's it,put it without the case and in a place where you can use it safe, <S> ground in the 110 vac is connected to neutral in the main panel anyway <S> ,if you use the case, don't touch each other ,a soft start when you connect the load will keep both atx working if not,on will turn off,a 5 volt load in the atx will rise the output to 12volts,a 12 volts bulb will do the job , i have 2 atx connected in series <S> the main case is grounded both psu are isolated from the case ,I need positive and negative not a ground ,atx coming with the negative connected with ground ,I isolated both atx from ground everything works fine,exagerate things about to die I red in this post,atx has positive and negative,ground in atx is for the case... <S> that's all
| To avoid the risk of a failure, you should cut EARTH connection.
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Measure temperature of IC/Microcontroller I want to measure the temperature of IC/Microcontroller. There is no On-die temperature sensor to measure the heat. I need to check how much hot the controller gets. The LM35 or DHT11 Temperature will not work as they sense ambient temperature not the surface temperature. How about thermocouples or Thermistor NTC? Any other device for measuring the temp of IC? This is the device OP is referring to: <Q> If you want to measure the actual die temperature precisely you may be able to use an internal diode anyway. <S> Such parasitic diodes usually exist, for example from an input to V- on a bipolar chip. <S> You could calibrate the diode in an oven or environmental chamber on a particular chip (no power). <S> To measure the actual temperature you would remove the power and quickly (before the temperature changes much) check the diode. <S> If you use a thermocouple, double back the wires and heatsink to other parts of the leadframe to prevent excessive errors due to heat loss down the wires (while maintaining electrical isolation). <A> What IC/Microcontroller would you like to measure? <S> Like <S> Spehro Pefhany said, didoes are commonly used to measure the temperature of computer processors: http://www.ti.com/lit/an/sboa019/sboa019.pdf <S> Thermocouples are a nice idea if you want to actively maintain a certain temperature. <S> Unless if the surrounding temperature is consistant, I wouldn't use a thermocouple to measure the temperature of a chip. <S> The thermocouple would give you the temperature difference between the chip and the surroundings, this could be very high in a cold environment and very low in a hot environment, or <S> even worse, you might get a negative signal when moving from a cold to a hot environment... <A> use any thermometer, thermistor, thermocouple or temp IC <S> This becomes a thermodynamic problem Requirements: decide also if you also want to improve thermal stability and specify identify the thermal resistance of various case packages and thickness define if you want a permanent bonded temp. <S> sensor or temporary. <S> identify thermal resistance of various adhesives per mm, hotglue, epoxy, cyanoacrylate, polyurethane, silicone ( get familiar , generally thinnest is best ) <S> design your connection to be >5x thermal insulating to ambient, relative to contact with hotspot to reduce ambient effect. <S> understand the junction temperature rise above case temperature for various thermal insulators like epoxy on SOT23 and FR4 PCB vs thickness and metal cladding characterize RF interference with dielectric or conductors near critical locations on board and effect on impedance of transmission lines use your fingertip to detect changes in Performance . <S> eg. <S> RSSI, <S> Tx level locate the hotspot and choose Temp IC and attach. <S> avoid RF interference with measurement sensor, <S> (See if Tx causes false readings and determine if any fix can be done or ignore during Transmit.) <S> minimize RF coupling to Sense wires to avoid damaging sensor from excess power .. <S> e.g. use suitable ferrite sleeve over wire pair and ferrite beads on signal for both Common Mode and Differential Mode RF coupling. <S> suggestion use appropriate size wiring for sensor, such as AWG 30 magnet wire or wirewrap wire and bond with a toothpick dot of cycnoacrylate and coat with a few mm of silicone for insulation from ambient. <S> if you have thermocouples, wisely choose location considering above interference to RF and electrical shorts and bond with Kapton Tape to board. <S> use tape for strain relief on fine wires or bond with epoxy dots to sensor and dots spaced along wire over board. <S> keep in mind adding thermal ambient insulation to sensor is counter-productive to keeping board cool with free ambient air flow. <S> It all depends on the reason you are measuring T. To design a better heatsink? <S> to integrate into another product and measure hotspot rise? <S> to compute the Thermal resistance of the Rjc for junction to case inside photo of older version <S> Improved version FCC qualified. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Alternately, a small very fine wire bead or fine ribbon thermocouple to the leadframe will measure some approximate temperature, and an IR thermometer looking at the encapsulation an approximation of the package temperature.
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50 and 60 Hertz, what is the visible effect? I hope this isn't too noob of a question for this community and that the title is worded correctly... I've got a few devices from Europe I'm using in Canada that are naturally 50Hz. I have to pro level transformers (not Samsonite ones from an airport, LOL) that are doing the job for these devices perfectly and with no real visible or audible difference. So, I'm wondering for example with my hand blender and food processor... the basic science answer is that the motors are going to spin faster. But is there a way to get a quantitative sense of what the extra cycles are doing to the appliances I use the transformer with other than, the manufacturer built it to work for 50 and you're making it spin faster? Should I use them at lower speeds, in shorter bursts?Apologies if my terminology and descriptions are completely amateur.Thank you!! <Q> The higher frequency is unlikely to cause such appliances to significantly misbehave. <A> In many home appliances you will find so called serial motors. <S> These motors are designed for universal use. <S> In other words it does not matter if they are used with 50HZ or 60Hz. <S> The only thing to keep in mind is the applied voltage. <S> appliances that have no motor (power supplies) only have a transformer or a socalled inverter. <S> Both these systems are also insensitive to the difference in frequency. <S> Larger motors such as can be found in refrigerators, freezers, etc run some 20 % faster on 60 Hz then on 50 Hz <S> but there is also no problem. <S> The appliances are mostly designed in such a way that both frequencies can be handled properly. <A> Speed will increase with no load, but vary more with load and 20% more temperature rise at max torque load, but unlikely be a risk. <S> The load speed regulation error is contributed by local step up transformer higher impedance and 20% more back EMF from higher synchronous AC is not a problem unless you stall the motor. <S> No worries. <A> Your hand blender and food processor are very likely to have "universal" motors in them (as most kitchen appliances do). <S> That is why the are called "universal". <S> Few, if any, modern appliances or electronic gear is dependent on mains frequency. <S> Back in the bad old days things like vintage LP turntables and tape recorders (and old clocks) used the mains frequency for speed control. <S> But here in the 21st century virtually all mains-frequency dependent gadgets are relegated to the museum.
| Others, such as the universal motors used in small appliances don't really care so long as the voltage is okay. Some type of motors (synchronous, shaded pole, split-phase etc.) will operate 20% faster on 60Hz as compared to 50Hz, with somewhat less torque. Universal motors will operate at essentially the SAME speed whether you are powering them from 50Hz or 60Hz (or even DC).
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Using a MOSFET driver I am trying to control the speed of a 15W DC-motor with operating voltage of 12V using an IRF640 n-channel MOSFET as shown in the picture below but without the pull-down resistor RMS = 10kOhm. Instead of it, I always set the Arduino PWM digital pin to value 0, thus when the motor should turn off the MOSFET gets 0V at its gate. : : The problem I face is that even with the maximum value of PWM (255) I don't get the motor to rotate as fast as it does when it's plugged directly into 12V. I get that there is a voltage drop in the MOSFET, but the speed should not drop as much as half of its normally is. Can the absence of the pull-down resistor Rms 10kOhm case this effect? Do I need to use a MOSFET gate driver in order to achieve better performance of the DC-motor? (The maximum power the motor uses during operation is not higher than 18 W)If yes, what MOSFET gate driver should I use in this case? Arduino pins work with voltages 0V - 5V and maximum current that can pass trough them is 20 mA. EDIT: Arduino Uno's PWM signal frequency is 490 Hz on each pin except for two pins where the frequency of PWM signal is 980 Hz. Source <Q> Your problem comes from insufficient gate drive. <S> Try a logic gate MOSFET like IR L 640. <S> Its minimum gate-to-source voltage is about 2 Volts which is quite suitable to drive from a MCU. <S> Another solution would be a P-Ch MOSFET with an NPN inverter: simulate this circuit – <S> Schematic created using CircuitLab <A> You ought to consider using a different MOSFET to the IRF640 because it isn't really going to turn on that well with your limited gate drive circuit: - Your gate drive is about 4.5 volts and, as you should be able to see if you tried to extract about 0.5 amps through it the volt drop could be anywhere between 1 volts and 10 volts (or more). <S> Look at the lowest graph - the one marked "4.5 V" then regard the curve. <S> If you got rid of the 1 kohm resistor your gate drive would be more like 5V <S> but you are on the edge of something working correctly and not. <S> A 15 watt, 12 volt motor requires a normal running current of 1.25 amps but at start up this may be 5 times bigger and <S> the lack of drive voltage to the gate of the MOSFET may still cause you problems. <A> Try IRF540! <S> with pull down resistor ! <S> if you hesitate to give pull down resistor of 10k <S> then there will be some issue you have to face !. <S> This might be help you. <S> The below plot showing for IRF540 Refer the image for characteristics <A> The R1 should be some high value about 1k5. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Use a transistor as driver from gate to ground, and a resistor from gate to 12V. You have to configure the PWM to active low.
| Alternatively you can use another transistor to invert the pwm polarity.
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Convert low power Alarm Clock from 120V 60Hz to 230V 50Hz I recently bought a vintage flip-digit radio alarm clock from ebay, because I plan to refurbish it and basically use it for the rest of my life. However I only realized that the device must be connected to mains when I had already bought the thing. I was assuming it would just work on 12V like most small electronic devices, but I guess because it's from the 70's I can't live in such luxury yet. This is the clock I'm talking about. The alarm clock is rated for US mains, so 120V 60Hz, and requires only 8W of power.The ratings where I live (Europe) is 230V 50Hz, so I have a couple of questions that I would like to have answered. I searched the web but nothing could tell me the answer to my clock-specific questions. Can I just plug the thing in my 230V supply? Do you reckon it has an internal converter or is it really 120V only? I have found several cheap converters from US to EU power rated max. 100W, but they do not specify the frequency of the AC. Do you think these converters, which cost around $30, convert the frequency as well? Is it a problem if the frequencies don't match? Will my clock run 16.67% slower on the 50Hz AC? The clock is only 8W, so might the $30 adapter too imprecise (by lack of better words) when it outputs less than a tenth of its maximum power output? What do you recommend? Thank you for your time and I'm looking forward to reading your responses. <Q> If you plug it in to 230VAC it will be destroyed, without question. <S> They do not convert the frequency. <S> Yes, I believe it will run slower on 50Hz. <S> The timing motor (probably a synchronous motor) may also overheat. <S> No the 100W converter (assuming a transformer type, which is usual at that power level) is fine for converting the voltage. <S> Try to find an adapter specifically designed for clocks . <S> Even if the frequency was converted it would not likely be accurate enough for a clock by default. <S> It will not likely be particularly cheap, but your vintage clock radio is priceless, right? <A> No. <S> The plugs & receptacles do not match, the voltage will ruin the device, & the AC frequency is likely too slow for the clock to keep proper time (even if you used a common step-down voltage adapter/converter. <S> No. <S> Cheap converters will likely not provide an increase from 50 Hz to 60 Hz (which is what your clock needs to keep proper time)--although, you may be able to find a pwm or switched mode power supply that will provide 60 hz output from a 50 Hz source. <S> Definitely verify that capability before you buy an adapter/converter. <S> Yes, it will be a problem <S> if you want your clock to keep proper time (not that there is any guarantee that your old clock will actually keep proper time at 60 Hz ;) ). <S> Yes, it will most likely run slower (but, it really depends on how the system was designed). <S> If you have a cheap AC adapter/converter, then try it to see if/how much the time difference is-- <S> it won't likely be damaged by running it at 50 Hz. <S> If the time difference is too much, then look for a better power adapter/converter. <S> No. <S> The 100W spec is the max power that the adapter is rated to provide. <S> It will provide any level of power from 0-100W. <S> If your load is only 8W, then that just means it uses less power through the adapter than the adapter is capable of providing. <S> Search for a transformer based pwm or switched mode power supply that transforms 50Hz AC @ 230V to 60Hz AC @ 110V. <S> Try here . <A> It might be worthwhile taking the radio apart and examining the clock motor. <S> There is a good possibility that it is a motor combined with a gear in one case to provide 1 RPM or 60 RPM. <S> You might be able to find a replacement motor for 220 volts, 60 Hz with the same output speed. <S> You may also find that there is a transformer for the radio that can be replaced with a 220 volt, 50 Hz transformer with the same secondary voltage as the original. <A> Some of these clock use the Line frequency for time keeping and may not be a quartz clock. <S> In the past I took a 60Hz clock to India and with line frequency being 50Hz there, it ran slower. <S> Voltage adapter won't work for these.
| If it is one of these line frequency clocks, I recommend instead using a Quartz clock with Voltage adapter to convert 120-230 Volts.
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10W LED, no need for constant current? I've got few of these LEDs which are in a 3S3P configuration and rated 9-12V 900-1050mA.I tried to power one of these with this module which I earlier set at 12V and 1A with a multimeter in short circuit. The problem is that when I connect the LED in series with the multimeter, the latter tells me that is drawing less than 400mA. So, out of curiosity and willing to fry one LED, I connected it directly to a power supply (a laptop brick, 12V 3A) and the LED doesn't draw more than 600mA.More LEDs in parallel gives me these values: 2 LEDs: 1200mA 3 LEDs: 1600mA I'd like to push the rated current through the LEDs (so to take them to full brightness), also is weird that seems like they're current-limiting themselves without the need of a constant-current circuit. Does anyone have any explanation/experience with this behaviour? Thanks <Q> You have not mentioned the color of the LEDs. <S> The typical white LEDs that make up modules like yours would take more than 3V to reach maximum rated power. <S> I would say around 3.5V could be a common number. <S> For a 3.5V white LED, probably around 20% of the voltage drop is due to ESR. <S> A LED (of the same base color) requiring lower voltage and therefore has a corresponding lower ESR would be more efficient. <S> So there is an unavoidable resistor in series with every LEDs already. <S> If you use the number 12V, 0.6A, 20% ESR then the ESR = 12 <S> * 20% / 0.6A <S> = 4 ohms. <S> These are guesses and may not reflect what you have. <S> But say if these guesses are exactly right, then you would get 1A at 13.6V. <S> (The simple ESR model breaks down badly at low current). <S> Rereading what I wrote, I was assuming 4 <S> x 3V. <S> But there are only 3 LEDs in series, so the numbers do not add up. <A> rated 9-12V 900-1050mA <S> But if they are rated for 1050mA, why do they only draw 600mA or less? <S> Ask your vendor. <S> Perhaps the 1050mA is for a higher voltage (eg. <S> 14V), or perhaps you were sold sub-standard parts, or the specs are just plain wrong. <S> If you bought them from a website such as eBay or Amazon then just be happy that they work at all! <A> What you may be missing is that you are not really connecting the LEDs directly to the power supply- <S> the ammeter acts as a resistor <S> - the more current the more voltage drop. <S> A constant current circuit- or a substantial series resistance and the associated wasted power is highly recommended if you want to run the LEDs at near maximum current.
| That's a clue that your LED arrays have built in current limiting and are designed to be powered by a voltage , not current.
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Why is it required to use a register in the final stage output of a pipeline? I'm following MIT 6.004x course, where in the pipeline section, it's stated that "our pipeline convention requires that every pipeline stage has a register on its output". I absolutely understand the reason to add a register in the mid-stage, but why to use it in the final output wire. I mean, can't we just pipe the output directly, what's the purpose to add a register to remember the output result? For example, I have the following pipeline conforming to the aforementioned convention: why can't I delete the last register and output the result directly? Could anyone give some explanation? Thanks for any help! <Q> You can use your own convention if you like, but in your OP you say that MIT 6.004x says that their convention is to have a pipeline output registered. <S> But there are better reasons than 'it's their convention'. <S> Why are you adding a pipeline register into the random logic of gates F, G and H? <S> Because they are not fast enough. <S> Not fast enough for what? <S> Not fast enough to receive inputs on this clock cycle, and have the result ready to be used somewhere next cycle. <S> So by adding the output register, we are being explicit about where the timing must be right. <S> When designing a pipeline, the easy bit of the timing is 'can one register drive another, in a clock period?'. <S> The hard bit is 'can one register output ripple through the random combinatorial logic I have between stages, in a clock period?' <S> Using a pure register at the beginning or end of a pipelined process gives you the cleanest I/O timing with respect to the clock. <S> If you need to connect your process to something, that's important as it allows for easier calculation of setups and holds, and tolerance for extra delays and clock skew between processes. <S> If I go to the modules catalogue, I'm more likely to be able to successfully use the design with the cleaner interface timing. <S> If you find you have connected two processes, and that there are two registers between them when there only needs to be one, it's trivial to remove one of them. <S> So where the two processes meet, you generally have a register that you could call either the output of the previous process, or the input to the next. <S> The MIT convention is that you call that register part of the previous. <A> If you do not use a register after your last stage, you are effectively merging the last stage with any following logic. <S> Your diagram will not show the complete pipeline and it will be useless for checking timing constraints. <S> From a technical perspective, you perfectly could connect your last stage to an output driver and even route it off-chip without any register. <S> But in the design view (at least for timing), all the logic up to the next register must be considered as part of that stage. <A> What you're missing with the question is that this pipeline is only a part of a larger design. <S> Something will always consume the result of the pipeline, frequently another unit which can itself be treated as a pipeline. <S> If you look at the fragment you've drawn, you can take several of these fragments and stack them end-to-end making a longer pipeline - this also follows the same pattern. <S> If you omitted the final flop, but connected two small pipelines, you find twice the logic depth at one stage. <S> Hence the convention which makes for an easy to transform element. <S> A more real-world convention might insist on 3 flops for a 2 stage pipeline (adding a redundant input capture flop). <S> This would add some timing margin for the scenario where you have physical separation between the two pipelines - but now its not so trivial to sub-divide, so <S> this 2nd convention isn't so good to teach with. <S> Obviously, in a real-world scenario, this convention is just a starting point, you might make optimisations in some or all of the interfaces (as you have already identified might be possible once the whole picture hangs together).
| Using an output register means that all the difficult timing is under the control of the designer of that module (or the student answering the question), and the interfaces are easy timing only.
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Why is a switched-capacitor voltage replicator useful? The figure below is a switched-capacitor voltage replicator. It is from Reconfigurable Switched-Capacitor Power Converters by Dongsheng Ma and Rajdeep Bondade. You can read it here . As you can see the output voltage Vout is finally charged to Vin. So why it is useful to use this circuit while it has a voltage gain of unity?It also doesn't seem to be a voltage buffer either. <Q> You already assume that this circuit has a direct application and that it is supposed to be an amplifier (you mention voltage gain) or buffer. <S> Why would you assume that it has this function? <S> In the book this circuit is used to explain the basics of a switched capacitor circuit. <S> There is no direct mention on how to use it. <S> In general such a circuit consisting of S1, C1 and S2 behaves as a resistor . <S> Remove C1 and replace S1 and S2 by a single resistor. <S> Now how would the charging curve for charging C2 look? <S> Do you spot the similarity? <S> It behaves the same way as a simple first order RC lowpass filter. <S> This is one application of this circuit: switched capacitor filters. <S> This is used to filter / process signals which could carry information. <S> In power electronics there is no signal (carrying information) but the circuit can still be used. <S> If you load the output and control the switching of S1 and S2 in a certain way that depends on the output voltage (in a feedback loop) you can use this as a voltage regulator. <S> That is the point of the book, I believe, as it is about switched capacitor power converters. <A> As FakeMoustache explained, the switches and C1 behave as a resistor. <S> This creates the possibility of making integrators, RC filters, A2D converters and other functions needing large resistors on a chip. <S> So for example you can build an imager with the A2D built in so the chip provides. <S> Look at this one for example; Image Chip Datasheet <S> This could not be done without switched capacitor resistors. <S> It is not very useful in discrete circuits design. <A> So, I am unaware of the use of switched capacitors as a replacement for an RC circuit, though I suppose I can see the similarity. <S> However, a use <S> I am more familiar with is as an isolated buffer for transferring a voltage between potentials. <S> This voltage is then presented to the input of the op amp, but now it is in reference to ground. <S> This allows for very good common mode rejection of the differential input, and with attention to the switching building blocks can also be a decent way of allowing small signal measurement of very high voltage signals. <S> See this app note from linear for more information on this.
| In this example, both terminals of the capacitor are switched, allowing the differential input to take any voltage that the switches can handle. The usefulness of the circuit is that it it takes less real estate in an integrated circuit to make to make than an actual resistor.
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Removing 'white sticker' on PCB I want to reverse engineer the circuit from an insect swatter. However, there is this 'white sticker' at the back of the circuit. I am assuming this is for some kind of protection or just to hide the circuit. Is it safe to remove this? And what tools/methods can I use if it can be removed? Thank you. <Q> If your goal is to see the traces to reverse-engineer it, try shining a bright light behind it. <S> You may be able to see the traces as shadows/silhouette. <A> As some of the comments state you need to confirm whether this is a sticker or printed silkscreen on the PCB, the photo isn't high enough resolution to say really <S> and I haven't seen anything like it before. <S> If it is a sticker then you should be able to pick at a corner to lift it off the PCB with a decent pair of tweezers. <S> If you can't just peel it off easily then soak in IPA (isopropyl alcohol), this should help to dissolve the adhesive without damaging the board or any plastic components. <S> Note you sould take care about what parts you soak in IPA - buzzers and batteries are a no no, but assuming there aren't any of these parts fitted and you're not reusing the board this shouldn't matter anyway. <S> (nail varnish remover if you don't have access to any pure stuff) but this could also remove the solder resist (green coating on the PCB) depending on what it's made of. <S> Definatley don't do this if you are planning on re-using the board. <A> I'm not sure, but the silk screen / sticker could be used to isolate the two HV lines. <S> High Voltages? <S> That could prevent short circuits from things like water and moisture (and your hands). <S> If it is a sticker, if is a little weird that they didn't covered the 2 pins themselves <S> but I guess it's China... <S> Others have already mentioned ways to remove it <S> so I won't repeat their anwsers.
| If it isn't a silkscreen then you may be able to remove using acetone First of all are you planning on re-using it, if so I wouldn't touch it.
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How to Connect this Crocodile Clips to my Multimeter? I got crocodile clips with my multimeter, but no cable with them. The tips of the probes I have, do not fit into the plug on the crocodile clips. How do I connect them to my multimeter? What kind of cable do I need? I assume this is some kind of standard connector, I searched around but did not find anything useful. Seems I just lack the exact name of this kind of connector. Update: The multimeter is a Testo 760-3 with the default probes. They have a small ~2mm conic tip. Here a picture of the probe tip: <Q> On some probes the plastic shielding on the tip is removable. <S> It may be that you need to remove it to use the croc clip. <S> It may also be that the vendor screwed up and shippped clips that were not compatible with the test leads. <A> The probes for my meter look something like that - the plastic over the metal probe just pulls off - then the probe tip should fit into the clip. <A> The instructions for those clips state Simply put the measuring cables into the crocodile clip. <S> I notice that there is a considerable amount of insulation missing from the clips in your photo as compared to the published configuration. <S> If the leads you show are the standard ones <S> then your ones have an additional sleeve on the end <S> - it could be that someone has previously attached the leads to the clips and broken them when trying to remove them.
| My guess would be that the clip is intended to fit over the probe but it is being stopped from doing so by the plastic sheidling on the probe tip.
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How do I wire a 12V DC fan to 120V AC source? I am trying to install this miniature computer fan in a heated incubator that operates at 50 Celsuis 24/7 in order to encourage air flow throughout the box. Small fans like this run on 12 DCV power, but because it will be running constantly, a battery wont do -- I need constant power, forever. I understand that I could use a step-down transformer to convert 120 ACV to 12 DCV, but I'm not sure about how to match the specifications of the transformer with the specs of the fan. This fan says it's "rated current" is 0.06A. Given this info, I'm not really sure how to shop for a transformer that will work and is safe. Do I have to perfectly match power supply to the fan, or just make sure that it's powerful enough? So if I bought a 144 Wt transformer, does that mean there will be 12 Amps (or more like 1 Amp?), and is it OK to wire this to a fan with rated current = 0.06 Amps? Can I power two or more fans with this same transformer? I would like to wire these in parallel, so that I can unplug one or the other. Would that work, or do I have to wire them in series? <Q> In general a consumer device will only draw as much current as it needs, as long as the supply voltage is correct. <S> In this case you need to buy a DC plugpack (transformer, rectifier and regulator all in one package - much like your mobile phone charger) <S> that will provide 12V and can provide at least 0.06A. <S> Most plug packs will manage this. <S> So you could connect your fan to a 12V plugpack which is 1200Watt and could deliver 100A, but your fan would only pull 0.06A, so this would be a waste. <S> Wire them in parallel. <S> In summary, you do not need to match exactly. <S> Just match the voltage exactly and make sure it is powerfull enough to supply the sum of all the currents you need. <A> A transformer can step the voltage down, but it's still AC. <S> I recommend buying a 12VDC power supply from your local box store (or Radio Shack or Amazon). <S> You should be able to find a wall-wart-style supply for under $10. <S> This is simple and cheap, and also has the advantage of being tested and certified. <S> So, at least 0.06A per fan. <A> If you connect the fan to a voltage of 12V, it will draw automatically the correct current at steady state. <S> The back EMF will increase to say, 11.5V, and then if it has one ohm of internal winding resistance, it will draw 0.5 a. V= <S> IR+EI=(V-E)/R=(12-11.5)/1 = 0.5A <S> Fanstastic! <S> The back emf is proportional to the speed. <S> If it goes half as fast, it will have half the back emf. <S> Then it will draw more current. <S> Thus you just need to give the fan 12VDC. <S> A transformer will not give you DC. <S> It will give you AC. <S> You can't just feed a DC motor AC. <S> If you search for power supply, you will find many. <S> It's okay to buy a power supply that can source up to 12 amps at 12VDC. <S> At half an amp per fan, you could hook up 24 fans in parallel! <S> Fantastic. <S> As is, that's overdesigning. <S> Pick a power supply that can source an amp or two, and that'd be more than enough. <S> Honestly, if there's a power supply in your computer giving 12V, just wire it up to that. <S> Be careful, of course! <A> The DC motor will pull as much current as it needs as long as you supply it the correct voltage (12VDC in this case). <S> Why not use a 120VAC to 12VDC converter specifically for getting the 12V? <S> You can wire the fans in parallel like you want as long as your power source can supply enough power to turn as many as you use on.
| You can power multiple fans off the same power supply, just make sure that you can supply the sum of all the current draws when selecting the plugpack. When choosing a supply, you need: The correct voltage (12VDC), and At least as much current as your fan(s) will take. What you need is called an AC to DC inverter. Therefore, don't stick your fingers in the fan, as this will slow it down, increase the current, and exceed the current rating. Since you need DC, it will take a few additional steps to get what you need.
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Is a separate safety earth connection necessary when using an earthed power entry module? Testing the Schaffner FN1394-10-05-11 power entry module with a multimeter shows continuity between the ground pin and the body of the module. Consequently, when the module is fitted into an aluminum panel (into a non-anodized, raw metal, opening), the chassis is connected to earth. In this situation, is a separate safety earth setup necessary, as per this diagram ? <Q> the problem is that raw aluminum quickly forms an aluminum-oxide coating which is insulating. <S> This is the reason for the recommended ground connection with a toothed washer to cut through the oxide coating and contact the metal. <S> The module appears to have a plastic front through which the screws go so the mounting screws will not provide a ground path, even with a toothed washer. <S> The incidental contact of the module body to the edges of the opening will not prevent formation or break through the oxide layer so there will not be a good ground connection. <S> To ground the aluminum chassis, I highly recommend following the diagram. <A> As was already explained you must not consider assembly screws for ground connections. <S> Grounding screws should have the sole purpose of grounding the component, because they might otherwise be removed by somebody who cannot recognize the grounding purpose. <S> Depending on your market you will find standards for grounding screws, like at least an M4 screw with tooth washer on a paint-free surface. <S> And you'll need a ground symbol on the front side of the aluminum panel. <S> By the way, the front cover of your chosen mains input seems to be plastic and non-conductive. <A> Connect the grounding terminal on the back of the power entry module to the case.
| The module mounting screws must not be used as the sole grounding connection to the case.
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Bipolar signal amplifying to high voltage I have a bipolar (+-15V) square wave with a frequency of approx. 600 kHz. I want to amplify it to +-500V (not precisely), 50W. The load is around 1kOhm.Can anyone give me an advice on how to do it? <Q> A transformer with 15:500 = 1:33 ratio can do this. <S> Be aware though that while this amplifies the voltage, it also reduces the current capability. <S> Voltage times current is power, and you can't get power more power out of a passive device than you put in. <S> For example, let's say the ±15 V signal is capable of 1 A. <S> After a 1:33 transformer, it will be ±495 V with a maximum theoretical current capability of 30 mA. <S> In reality, there will be some loss in the transformer, so the actual output current capability will be less than 1/33 of the input current capability. <S> Since actively driving a signal at ±500 V is a lot more difficult than doing it at ±15 V, you should amplify the signal before the transformer. <S> The only component that touches the high voltage output is the transformer. <S> No other circuitry is required at the high voltage, and no high voltage power supplies are needed. <A> First, you need "at least" ±500V dual or 1kV single supply. <S> Anyway, building a long-tailed-pair with some high frequency power transistors (maximum collector-to-emitter voltage should be at least 1kV) <S> is a way to go. <S> Another approach can be using the same configuration with high frequency vacuum tubes. <S> If output power is not an issue, you can use a 1:35 ferrite transformer as well. <A> I would go shopping for an old surplus medium wave RF transmitters, or Ham radio in the relevant band. <S> An alternate avenue is perhaps look for information on Tesla coil sparks to music as those folks get up to all sorts of tricks.
| Amplify and then transform is probably safest, just get power to well over 60 W at low voltage and then find suitable ignition transformers or similar that can handle the power and turns ratio.
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Motor Control board -replace capacitor with higher voltage? I hope that someone may be able to help me with a problem i have- My good lady wife dusted down her trusty Kenwood mixer last night to get started on the Christmas cake (apparently it must be done this weekend or doom will ensue) On attempting to use it she found that when switched on the motor would start then stop then slowly start etc etc- generally misbehave. Here is a short video of the misbehaving motor: https://youtu.be/7Mmma936F_w I stripped the beast and checked the brushes- all ok and so took further advise from a "kenwood expert" who says my speed control is likely goosed (a common problem it seems). He recommended replacing the triac, a couple of 4.7k resistors, a 220uF capacitor and a 150nF resistor. Although i can't see any obvious signs of distress on the board such as blown / leaking cap's or charred resistors i am not capable of testing the triac with the limited equipment i have so thought it best to go ahead and replace the components suggested by "the Kenwood man" The control board is actuated by a rotary dial - the spindle passes through the board and the speed control (Potentiometer?) we have a direct mains voltage input, a motor output and a thermistor input (thermistor is attached to the motor to prevent overheat) Upon turning the dial the motor speed should increase progressively depending on dial position- The machine also has a function whereby you can turn the dial in the opposite direction for a "pulse" of full power- this full power is applied until you release the dial and it returns to 0 (stop) via some sort of spring action in the potentiometer? I only have a Maplin in my area and can't seem to find a like for like match on the capacitor which should be a 150nF 275v x2 The only available 150nF capacitor at Maplin is a Wima Polypropylene Metal Foil 1250V 0.15uF Capacitor which is obviously 4.5 x the voltage of the one i want to replace, other than that the closest match i can come up with at Maplin is a 0.1uF 275V X2 RFI Metallised Polypropylene Capacitor. Can anyone tell me if either of these two capacitors are likely to be acceptable? I thank everyone in advance for their time, i'm sure you all get fed up with people like me popping up out of nowhere asking for help but it really is appreciated! Regards, Rob <Q> The precise value of an RFI suppression capacitor like that is not terribly critical, so using a 100nF X2 cap instead of a 150nF one should get the circuit up and running for now at least. <S> You can order the 150nF/275V X2 cap from most mail-order suppliers though -- <S> I'd try Farnell, given where you're at. <A> IME, a higher-voltage capacitor would be a suitable replacement. <S> Note that typically special kinds of capacitors are used with circuits connected directly to the power mains. <S> These are "X capacitors" and "Y capacitors". <S> But using a much higher voltage capacitor with a (presumably) double-insulated kitchen appliance is probably safe. <A> I suspect the wiper on the big black VR1 pot for speed control is intermittent, which sort of matches your description. <S> slowly start etc etc- <S> generally misbehave. <S> Beside <S> it is an internal VR2 trimpot . <S> Perhaps contact cleaner or WD40... on the VR1. <S> Nothing looks like it has overheated. <S> Reminds me of a Piher brand pot <S> around 1K I once knew around that vintage. <S> Plastic caps are pretty reliable. <S> Can you describe the malfunction more clearly? <S> e.g. ( Hysteresis ? <S> dead spots? <S> late surge start at 30%? <S> then smooth then erratic 75 to 100%? <S> My Mother usually started Xmas Cake 3 mos before. <S> others got so many they were aged 1 yr +3 mos ;) <S> New Piher pots are round. http://www.piher.net/sensors/index.php?page=shop.product_details&flypage=flypage.tpl&product_id=24&category_id=1&option=com_virtuemart&Itemid=31
| Except in some cases of high-capacitance electrolytics, using a higher-voltage capacitor carries no significant risks or disadvantages. If all erratic responses come via the speed control, then a new Pot or WD40 may work.
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Driving Relay to switch 6A It's been many years since I looked at hardware. I want to build a circuit; a 16 pin GPIO that switches a 5V Coil, 6A relay to switch a 6A, 15V Peltier . So working from the Peltier, I found a relay that can switch it on, I hopefully (power supplied using PC Power supply ) - If not, then I'll need to fix this before going on. Now my question is, can the GPIO chip switch the relay? What I could find in the datasheet is that the GPIO output current on a pin is 25mA (Page 27 of 16 pin GPIO ) and the relay requires 1A? (Page 1 of 5V Coil, 6A relay ) - or is it 6A? <Q> You need to focus on the coil data itself and not the contact data for the relay. <S> 6A will be right against your limit for switching your load. <S> 15V is well under 125VDC <S> so you should be fine. <S> Just something to consider. <S> At 5VDC, the coil has a resistance of 147 Ohms which means it will require ~35mA to switch the relay. <S> So the relay does not require 1A to switch. <S> It has a power of 170 milliwatts from 5 to 24VDC. <S> By looking at the GPIO chip, we can observe that the maximum source or sink current by any output pins is 25mA. <S> This probably will not be enough to swtich the relay on and off. <S> I will leave that up to you. <A> One chip will easily drive 7 relays of that type (12V 848 ohm coil). <S> Tie COM to 12V <S> (this is important!!- <S> it tames the relay inductance <S> so it doesn't break down the drivers) and don't forget to provide a ground line that does not require relay current to pass through the ground connections of your IO expander chip. <S> Do note the electrical life at 6A- only 60,000 operations at 6A so if you run at full current and 1 cycle every 10 seconds it will last only about a week when run 24/7 (or half that if you're using the N.C. contact). <S> Life driven with the normal type of circuit will typically be a bit less again because drop-out is slowed by the diode clamp. <S> For this current and voltage, usually a low Rds(on) MOSFET or DC SSR would be preferable. <A> The GPIO spec says about 0.5V with 8mA. 1A is the max you can put it the coil and not damage it. <S> The coil is 147 Ohms so needs about 34mA when it is on. <S> You will need some kind of buffer between the GPIO and relay coil.
| Therefore, you will need some type of amplification to make your circuit working in between the microprocessor and the relay. I suggest using an ULN2003A and a 12V relay, since you have 12V available.
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How to test a CT Sensor (Alternating Current Clamp)? I recently purchased a CT clamp. I am left wondering what the best way is to test this in a laboratory environment, specifically with an AC current generator. I just want to know whether the device works properly. Could I just generate a fixed amount of AC current from the generator and short my negative and positive wires to create a current loop? From this, could I then just clamp my CT sensor around this loop? If not, I am open to more suggestions! Any help is greatly appreciated! Kind regards, Hashim Shamsi <Q> You should be able to get clamp adapters that plug into an outlet and break hot and neutral out to the two "arms" before coming back together into a socket, as depicted below in a representative North American version, at an electrical supply house or even a home-improvement/hardware store. <S> Once you have everything set up, you should be able to use the docs for your CT to convert your meter reading to a current reading, then match that up with the nameplate wattage on your load. <S> (space heater, electric griddle, what have you) <A> If you have an accurate AC ammeter you can find a reasonably steady, reasonable resistive load (perhaps a heat lamp, for example) that is somewhere between half and full scale of the probe, run one of the AC wires through the probe, through the ammeter, and compare the probe output to the AC ammeter. <S> If you can only find a load that is (say) between 1/4 and 1/2 or between 1/6 and 1/3 etc. <S> you can thread multiple turns through the clamp-on probe to multiply the effective current. <S> It is essential that the turns are from one side of the AC and are only threaded through in one direction, otherwise they will cancel. <A> I suggest you have read the manual yet or even the wiki notes. <S> https://en.wikipedia.org/wiki/Current_clamp#Clamp_meter <S> Calibration requires standards. <S> If your range is variable then try all known current sources you have ( such as your 50 Ohm generator into a 50 Ohm load) then sweep and see what errors are found. <S> Be aware of DC and degaussing.
| Take a resistive load of known value (say, a small space heater that lacks a fan), and use the CT on the hot wire going to it, hooked up to a meter.
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In which pratical cases are used parallel and thevenin terminations? I've found some theory about various type of terminations but I can't figure in which cases are used parallel(pullup with half driving voltage and pulldowns) and thevenin termination. Since I'm used to simply boards with microcontrollers, where I can encounter this two type of termination and in which cases? Are something used in "extreme" cases like very long cables, critical or high power EMI applications? This is an example of a Thevenin termination: <Q> If the termination is not there or has the wrong impedance, the signal you're trying to transport will reflect and distort. <S> This becomes an issue when the length of the transmission line is longer than the wavelength of the signal. <S> Termination is not always of the Thevenin variant. <S> It depends what the next circuit, which is receiving the signal, needs. <S> If that is a simple inverter then a DC voltage of 0 V (ground level) will not make it work. <S> Inverters need around half the supply voltage to act as an amplifier. <S> This is where the Thevenin termination comes in. <S> It terminates the transmission line but at the same time provides the proper DC biasing voltage for the next stage. <A> Since you are talking about microcontrollers, I'm only addressing logic pulses here. <S> For these kinds of pulses you can singly terminate. <S> Let's assume the transmission line is 50 Ohms. <S> If a 50 Ohm resistor to ground was placed at the receiving end the line would be properly terminated. <S> The problem is when the send end was high, the current needed to maintain this level would be Voh/50, which is hard even on buffer chips. <S> By switching to at 100 Ohm up and 100 Ohm down Thevenin terminator, the high state the current is reduced to (Voh-Vth)/50. <S> It is much easier to find drivers to do this. <S> Another way to terminate is to put the 50 Ohm resistor in series at the transmitting end. <S> Series termination is the lowest demand on the driver. <S> The way it works is on the edge of the pulse the signal is cut in half at the send end. <S> When it arrives at the receiving unterminated end, it will double providing the original pulse. <S> In this case the current seen by the driver is just the receiving gate Iil or Iih. <S> In series termination there will be a step in the pulse for the round trip delay but this distortion doesn't matter. <A> Open collectors are current sources with a pull-up are like voltage sources with a series R. <S> But more often drivers are lower than Zo <S> so series R's need to be considered if not built in. <S> You examine this by the incremental V <S> / <S> I specs like Vol/Iol and (Vcc-Von)/Ioh vs temp and Vcc. <S> When threshold is critical for noise , terminate with an active DC bias to threshold or use two R's to make the equivalent bias to improve margin to binary errors from noise so that the margin is equal for each polarity. <S> Matching the impedances from source to line to load help reduce ringing caused by mismatch. <S> To determine when to apply these rules you consider if the propagation delay is greater than the rise time. <S> Then you must apply.
| Proper termination is needed at the beginning and end of a transmission line .
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Where did Bootstrapping get its name? Bootstrapping is a clever technique where you use feedback to increase the input impedance (mostly in an emitter follower). Why is it named "Bootstrapping"? Just curious. <Q> In engineering it usually means to start with little, use that to make a little more, use that to make even more, etc, until eventually to get to full operation. <S> From that, it is sometimes used to refer to any startup process. <S> When a computer is first turned on, it goes thru a "boot" or "bootstrap" process to eventually load the operating system and be ready for use. <S> Now, "booting" a computer is synonymous with starting it up. <S> The computer actually goes thru several startup phases before being ready for normal operation, so this is a valid use of the term. <S> I remember computers from the 1970s where you had to first toggle in a small program of maybe a dozen instructions. <S> That was just enough logic to read a more complicated loader from a paper tape or something. <S> That load contained the more substantial routines to read from the disk, etc. <S> Modern computers still go thru a number of startup phases. <S> The initial one is read from a ROM, which is why you can simply turn it on and wait. <S> Booting a computer has actually gotten more complicated and takes longer than it took the much slower and smaller computers of the 1970s to boot up. <S> Personally, I wouldn't call a emitter follower a bootstrap circuit. <S> I think that's a misuse of the term. <S> A valid example of that term in electronics is referring to the circuit that provides the initial internal power of a line voltage to DC power supply. <S> Ultimately the supply will be a switching regulator that transfers significant power from the AC input to the DC output. <S> However, that switcher needs to be powered somehow. <S> In other words, you need a power supply to power the power supply. <S> The bootstrap circuit in a power supply provides the initial slug of energy for the switcher to run until it is able to make low voltage DC efficiently and at significant power levels. <S> This includes providing the switcher's own operating power. <S> The bootstrap supply is usually quite inefficient, but simple enough to run without itself needing a power supply. <S> Either that bootstrap supply uses very little power, or it is shut off once the main power supply is up and running. <A> Only if you imagine you can lighten your load by picking yourself up by the boots. <S> Unfortunately we lose ground connection when floating , so we cannot lighten our load. <S> But transistors with common gnd. <S> references can do this. <S> This is how emitter feedback ratio to the base raises the input impedance by this ratio using positive feedback with gain <1. <S> Also on the flip side. <S> negative feedback lowers both input & output impedance.. <S> You can bootstrap the Nch Gate source voltage voltage from the Pch PWM or bootstrap for input higher impedance or bootstrap yourself to elevate your mood and more.... <S> Here is a voltage amplifier from a low Z source mic via Common base to an Emitter Follower output with Cap feedback from Emitter out to base where voltage gain is slightly <1. <S> Thus raises the 1st stage collector by impedance or input impedance of the load and thus the impedance ratio of 1st stage yields a voltage gain of 100 or so with an 8 Ohm source. <S> This is a bootstrap to raise impedance by voltage positive feedback <1 and result in 1st <S> Av <S> > <S> 1000 using 22uF for positive feedback raising base load well above Rc load , <S> so Av~ 10k/8R or a bit less. <S> Bootstrap means to boost something basic like Zin or Vb " without external assistance". <S> In the case of MOSFET Vb bootstrap the feedback of PWM to a series cap and diode clamped to Vdd yields a high voltage generated than input <S> Vdd, so we say without external assistance a passive boost above supply here (without external help) is called a Booststrap. <S> Same name yet different function. <S> ... not to be confused with flyback boost or other boost topics. <A> The fictional Baron Munchausen https://en.wikipedia.org/wiki/Baron_Munchausen once fell into a swamp, and only saved his life by pulling himself out by the straps of his boots.
| The term is derived from "pulling yourself up by your boot straps" .
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How can I check that the voltage output from this voltage divider is 2.25V? I was learning about voltage dividers from here , and I decided to try a test circuit with my Radioshack learning lab. With an input voltage of 4.5V and two 1000Ω resistors, I expected the voltage output would be 4.5*(1000/(1000+1000)) = 2.25V. After looking at this , I thought that the only way to measure the voltage output from the divider was to measure the voltage drop of a resistor (otherwise I'd just get a 0V reading), so I added a 1000Ω resistor to the circuit (R3 in the drawing below). I measured the voltage across this extra resistor, but I got 1.48V for an output voltage. What I found odd was that when I used higher-resistance resistors, the voltage drop output got closer and closer to 2.25 V (the highest I did, 1MΩ, led to the 2.25V reading I wanted). Can I use resistors like this R3 to test the voltage output coming out of this voltage divider? If not, how can I check by measurement that this voltage divider gives an output of what I'm sure is 2.25V? simulate this circuit – Schematic created using CircuitLab <Q> Welcome to resistive potential dividers, if you load them, they change. <S> You performed a calculation with R1 and R2 forming the potential divider to find the output voltage. <S> However now you are adding in an extra resistor R3. <S> That means that the lower resistor in the potential divider is now actually R2||R3 (R2 in parallel with R3). <S> In the case of you schematic example, you now have a bottom resistor in the potential divider of R2||R3 = 500Ohms. <S> This is very different from the value you calculated with in the first place. <S> If you repeat the calculation again, you get: $$V_o = <S> V_i\times\frac{R_2||R_3}{R_2||R_3 + R_1} <S> = 4.5\times \frac{500}{1500} = <S> 1.5V$$ <S> close to what you measured. <S> As you make the resistor larger and larger, the affect it has becomes less and less - you can see that from the calculation of R2||R3 - the larger you make R3, the closer to R2 the combined value becomes. <S> It's worth noting at this point that if you omit R3 and simply connect the multimeter across R2, you will actually have the same issue. <S> A multimeter in voltage mode is basically a very large resistor, so if you connect it to your circuit it will still have a loading effect - in essence it becomes R3. <S> However the multimeter resistance is very large (usually >10MOhm), so it will have a very small affect on your circuit. <A> Simply remove R3. <S> The multi-meter already has a very high input resistance. <A> However, R2 is that resistor . <S> You don't need to add anything--just measure the voltage drop across R2. <A> The best way to test a voltage node accurately, is with a "high" input resistance prove. <S> This can be an oscilloscope prove or a 10Megohm voltmeter. <S> Although the voltmeter you are using is not very good, the main reason you are not seeing the voltage you expect, is that you have another resistor (R3) across the resistor (R2) that you are measuring. <S> The accuracy will improve if you remove R3. <A> You're trying to use/misuse a memorized formula, when all you need is Ohm's Law. <S> Think of it this way: a current flows from BAT1 and goes through R1. <S> Then it splits in 2. <S> Exactly 1/2 goes through R2 and the other 1/2, through R3. <S> Since R2 and R3 each see half as much current as R1, the voltage across the pair is half that across R1. <S> This means the voltage across them is also 1/2 (Ohm's Law) <S> the voltage across R1, or 1/3 of BAT1's voltage. <S> The voltage is/should be 1.5 V. <S> The equivalent resistance of parallel resistors can also be found by application of Ohm's Law. <S> After some algebra, you'll find it's equal to the product over the sum of the resistor values. <S> R2 and R3 together look like 500 ohms.
| You are correct, that you want to "measure the voltage drop of a resistor".
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74HC238 (Decoder) - Will it use VCC to power channels? I am trying to select different combinations of LED's using a 74HC238. I am testing with an ESP8266, and supplying 3.3v from a regulator LM1117-3.3v. My question is when I make the digital channels lets say put Y1 to high, will it be using the 3.3v that is being supplied from VCC, or will it go high and use current from the GPI lines? If so, that means I would need to use a transistor NPN and supply the LED's power from the Regulator. Any help is appreciated, this is the first time I am learning about Decoders. Thank you in advanced! <Q> The outputs are directly sourced by Vdd and sinked to GND. <S> That's a basic function of logic gates. <S> But be careful, some chips have open-collector/open-drain outputs which means they can only sink the output, not source it. <S> Usually, this comes with more sink current capability and/or with a > <S> >Vdd voltage compatibility at the output. <S> Look at the datasheet. <S> Check the supported currents if you want to drive LEDs. <S> Output current and sum current on Vdd/GND. <S> It will work with a 74HC238. <S> However, the usual way to do this is using an inverting decoder as the 74HC 1 38, and tie the LEDs (plus current-limiting resistor) between the outputs and Vdd. <S> Why? <S> Because most multi-LED arrangements are common-anode , which means they have a shared pin for the anode connection. <A> 74HC238 will get current internally from its power rail, GPIO (should be in output mode to be proper input for 238) will have minimal current draw. <A> From the datasheet you can see that you can use the decoder in both modes. <S> In other words the output channel can be used as source or sink. <S> Meaning you can choose yourself what you need. <S> The limitations are the max current that you can sink or supply from a channel. <S> And ofcourse the limitations from the decoder as a unit. <S> You have to calculate ofcourse the serial resistor for the led.
| You need to look at nominal current for LED (in LED datasheet), and ensure that 238's output can provide such current under normal operation (in 238's datasheet), and use limiting resistor in series with LED to limit the current to the nominal value.
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Wiring an Inverter into a Car I recently acquired a 2000 Watt Inverter with a 4000 Watt Peak from a friend who did not like they way it worked in his RV. How would I safely wire it into my car? If it helps I drive a 2005 Nissan Sentra SE-R Spec V. I have a single battery under the hood. So far my plan was to mount the inverter in the trunk and run 4 gauge cables from the battery to it. For convince I was planning on wiring a switch in the dashboard with a 200 amp relay ( https://www.amazon.com/dp/B01I9CI3VS ). Would this be a safe option to shut it off when I am not using it? I do plan on putting a 200 amp fuse close to the battery. In the front of the car I was looking at running an extension cable and cutting the ends and splitting the power to a few of these ( https://www.amazon.com/dp/B0031C11DY ). Is this the best way to do this or is there something better I can do? Also my car is around 14 feet in length so I am imagining I will be using around that much in wire. Will 4 gauge be safe enough or will I need to use 2 gauge? I am somewhat new to this and do not want to set my car on fire. <Q> You are best served with short thick input cabling. <S> You may also find that having just one battery will cause you to rethink your friends kind gift when you run your battery flat in 20 minutes. <S> Perhaps less than an hour with the engine at cruising revs. <S> If you decide to add a second battery you can mount it <S> close tot <S> he inverter solving your cabling concerns a lot easier. <S> Make sure you arrange a suitable charging system to isolate the starter battery if the engine is not running. <S> The peak current may be around 350A, that is welding currents and will need welding cables. <S> If you only plan to use it occasionally, you could add carry handles and then fit nice traction battery power connectors (the sort used in electric forklift trucks) and lug the unit to the battery when needed. <A> The 200 A relay appears to be fine to use for a shut-off. <S> Your battery will discharge very quickly if you use it to power the inverter with anything close to rated load. <S> If you significantly discharge your battery very often, it will fail before long. <S> Consider a second small inverter for things you must run with the engine not running. <S> Even the alternator will not be capable of running the inverter at rated load. <S> You need to consider what is required from the alternator to operate the normal load it has when the engine is running. <S> You need to determine what ambient temperature the inverter is rated for. <S> Look to see if the inverter power rating is for operation in normal outside air temperature or the temperature under the hood or in the trunk. <S> There are different types of wire insulation with different temperature ratings. <S> However you must adjust for the temperature of the areas where the wire is located. <S> You should be able to find tables to help with that. <A> According to wikipedia (AWG page), 4 guage is less than 100A, and 2 guage is little more than 100A. <S> If you want to protect wire with a 200A fuse, you should use 'triple aught', 000. <S> 10m of wire (5 there, 5 back) in 4 guage would drop 1.6v at 200A, which is probably too big a chunk of 12v. <S> In 000, it would be 0.4v, which is better. <S> Best would be to mount the inverter under the hood next to the battery.
| You should consider wiring it to shut off the inverter whenever the engine is not running. Wire with a higher temperature rating can safely carry a higher current for a given size. You should consider fusing the alternator at the current rating of the alternator or less.
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can somebody give an analogy of what an ampere is? I'm trying to understand the difference between amperes and watts. I have a string of LEDs that say require 5W of power and 5amps. What's a good way to memorize the difference and significance of these two values? <Q> To a first approximation, amps is the number of electrons passing a given point in a second (divided by about 6 x 10^18, <S> but it's proportional - twice as many electrons, twice as many amps). <S> I say to a first approximation, since for p-type semiconductors the current is produced by moving vacancies in the electron cloud, and no single electron can be singled out for counting. <S> Watts is how much work you do in a unit time interval in pushing those electrons, and for electronic systems <S> it's volts times amps. <S> Think of electricity in a wire as water in a pipe. <S> Amps is the flow rate, volts is the pressure drop from one end of the pipe to the other, and watts is the power needed to move the water - or the power produced by moving the water as in a hydroelectric generator. <A> Well, the definition of an ampere is quite figurative: It is the current through two parallel infinitely long and thin wires, that are 1 metre apart from each other, so that the force between them is exactly \$2 \cdot <S> 10^{-7}\$ Newton per metre (thanks to @rioraxe for checking the numbers). <S> Admittedly, this is not easy to transfer to reality. <S> In fact, it is very hard to give another example. <S> It is easier, when you think in power, rather than current. <S> At 5 V, you get 5 Watts. <S> At 230 V you get 230 W at 1 A. Concerning your case with LEDs at 5V, you need to take light efficiency into account. <S> LEDs have around 80 to 120 Lumens per Watt. <S> So if you have 25 watts in LEDs, you can think of around 2500 Lumens, which is around as much as two Lightbulbs with 100 Watts have. <A> 5 Watt is 5 Ampere at 1 Volt. <S> Or 1 Ampere at 5V. Or 2.5 Ampere at 2 Volts. <S> Or any other combination which multiplies to 5. <S> $$P = <S> U <S> \cdot <S> I$$ <S> $$ <S> (W = V \cdot <S> A)$$ <S> To make a water-analogy, electrical current (measured in Ampere) is the amount of water you get from your tap in a second. <S> But if you want to calculate how much power (measured in Watt) that stream has, you also need its speed (measured in Volts). <A> No, watts from an engine are not like amps. <S> Both are a rate; the flow-rate of energy. <S> Gut-level verbal/visual viewpoint: <S> And, in copper, one coulomb of electrons is about the size of a grain of salt . <S> When you close the circuit, all the coulombs start slowly moving along as one, like a circular drive-belt inside the wires. <S> Electric current is like the speed of the belt. <S> But not exactly the speed, since the flow rate is faster through thinner wire and slower through thicker. <S> But still, if we double the current in a circuit, the coulombs move twice as fast in every section, and the amperes are doubled. <S> Watts? <S> That's like rubbing your thumb on the moving belt. <S> GETS HOT! <S> Watts is energy flow, is the rate of performing work (and is the output wattage of heaters and light sources.) <S> The harder you push your thumb, the greater is the back-pressure on the "drive belt" made of charge. <S> Voltage is when your thumb produces a tension-difference in the belt on either side of your thumb (that's why the ancient name for voltage was "tension," as in H.T. power supplies, "high tension.") <S> To tap some wattage out of the slow-moving drive-belt, we need the belt to be moving, and also need a tension-difference. <S> Double both of them <S> and you get 4X watts: <S> joules/sec equals coulombs/sec x volts. <S> Also: the belt, the coulombs, moves slow. <S> Amperes are a slow flow. <S> Yet the energy moves almost instantly. <S> When you rub your finger on a moving belt, you instantly tap energy out of the entire belt, slowing down the whole thing as a unit. <S> Finally: the power supply speeds up the "belt," increases the current, while all the "rubbing thumbs," the loads, slow it down again. <S> The forces balance out when the belt is moving at a certain rate, where amperes equals the voltage-force of the power supply divided by the resistance of all the loads. <S> Reduce the resistance of the loads, and the speed of the charge within the wires will rise, until it hits a particular faster value (higher amps.) <A> To answer your question about a memorization device, when I was starting out learning about electrical engineering I thought of Amp sounding like "amplitude" because it is the amount of electricity flowing through the conductor in question. <S> Volt can be remembered as being like "vault" since it is like the pressure pushing the current, it can remind you that this relates to how far it can jump. <S> Pneumonic devices work for me, even if they are kinda dumb. <S> Just as a side note in case I might confuse anyone, these units are named after physicists André-Marie Ampère and Alessandro Volta respectively. <S> They have nothing to do with amplitude or vaulting.
| Watts from an engine are like watts from a power supply. "Amperes" are the flow rate of charge: one coulomb flowing per second is one ampere.
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110 AC to 12 DC with high Watt I'm looking into purchasing a winch that is powered by 12V DC.It's a 1 horse power winch, so I assume it requires just over 700 Watt to function at max load. Right? I want to use it at home, so I somehow need to convert 110V AC to 12V DC. When I look for 110V AC to 12DC coverter online nothing comes even close to 700 Watt. Am I on the right track or what I'm trying to do doesn't make sense? Why are there no converters for that? <Q> Do the math. <S> 12 V at 700 W means 60 A. <S> That's going to require a thick and expensive cable, and/or significant losses in the cable, and/or a short cable. <S> This is why winches of this power are 12 V only when they are for automobile systems where that's the only option. <S> At 110 V, it only takes 6.4 A to deliver 700 W. <S> That's much more manageable with common cables and with less loss. <S> You get a 12 V, or "12 V battery charger" power supply, and connect it to the battery. <S> The charger won't be able to keep up with the current demand of the winch when on, but it will charge the battery to catch up whenever the winch is off. <S> Let's say the charger can deliver 20 A. <S> That means the battery is being discharged by 40 A whenever the winch is running, then charged with 20 A as soon as the winch is shut off. <S> The battery should be able to run the winch for minutes at least while sourcing 40 A. <S> In the long run, you're not going to run the winch continuously, so this system will probably work well enough. <S> Another thing to consider is that even though the winch may be rated for 700 W, it's not going to draw all that except under full load. <S> When the winch isn't pulling with its full force, it isn't drawing the full 60 A from the battery either. <S> Also note that this 700 W is the continuous rating. <S> Electric motors draw a lot more current when starting or when stalled. <S> If the winch ever gets stalled, it will probably draw, or try to draw, more than 60 A. <S> You will have to decide for yourself whether a 12 V winch plus car battery plus charger is cheaper or otherwise better for you than a 120 VAC winch. <S> If you might want to use the winch with just a car, then the former is probably the better tradeoff. <A> Mobile gadgets are designed for vehicle use where the power is limited to 12V DC. <S> But for use where common utility AC power is available, it makes no sense to use a very high current 12VDC motor. <S> A power solution to convert mains power to 12V DC @ <S> 60A will cost more than simply getting a mains-powered winch. <S> And it will be more fiddly to wire and switch such very high current DC power. <A> You are on the right track. <S> I suggest you look for 12 volt DC power supply instead of converter. <S> Just don't expect it to be cheap. <S> In general, electrical power is most efficiently transmitted at high voltage and low current. <S> As a result, 1 HP motors are most commonly run at considerably higher voltages. <S> Providing wires for 60 amps is not a lot of fun. <S> Such wires are expensive, heavy and stiff. <S> A lower current is a very good idea. <S> The big exception is appliances for vehicles, such as the winch you bought. <S> Historically most vehicles have run off 12 volt batteries, so making a vehicle winch run at 12 volts also is a good idea. <S> The wiring is quite short, so the cable losses aren't too bad, and since the cable harness is fixed in place its weight and stiffness really don't matter. <S> As for finding a power supply, that is easy enough. <S> Go to any online electrical distributor such as digikey, Mouser, Allied or Newark and search on power supplies. <S> Get set for a shock though. <S> Expect something on the order of 300 to 400 dollars. <S> Alternatively, go on eBay and search for 12 volt power supplies, Getting a supply direct from China will set you back about 150 bucks. <S> Of course, you won't get brand names and you may well get junk, but that is the chance you take. <S> Caveat emptor is the name of the game on eBay. <A> If you didnt bought the whinch yet, you should at first look for prices of AC powered whinches and DC power supplies for 60 amps and more. <S> I guess the AC whinch alone is more expensive, but the power supply is so expensive that you can save a lot of money with the AC whinch.
| If you want to be able to use this winch sometimes at home and sometimes with a car, then dedicating a 12 V car battery for it at home is one way to go. No, what you are trying to do does not make much sense.
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Why does low frequency RFID have a short read range? If I want to get the wavelength of a radio wave, I must divide the speed of light by the frequency? So having a 125 kHz RFID means an estimated 2km wavelength? If its wavelength is 2km long, why does this low frequency RFID have a short read range? <Q> Because RFID doesn't work based on wave propagation. <S> It's thus not actually a radio system (despite working at "RF"=Radio Frequency). <S> Think of an RFID tag more as the secondary side of an air-core transformer, where information is transmitted by the tag changing the amount of power it draws from the primary side of the transformer, or by charging a energy storage (capacitor), and then exchanging the role of secondary and primary side of the transformer. <A> You're confusing radio communications through the air with coupled inductors. <S> To communicate through the air using a 125 kHz carrier you would need an antenna of around 1/4 lambda so about 600 meter long in order to transmit that 125 kHz effectively. <S> Obviously RFID does not work this way. <S> RFID uses <S> coupled inductors which means there are two coils (optionally with a magnetic material in its core) which couple to each other magnetically. <S> For this coupling to be effective, the distance can only be a few centimeters. <A> Simply put, the antenna does not radiate - it operates more as a coupled inductor. <S> The coupling drops with \$r^3\$ rather than \$r^2\$ (if I remember right). <S> You have calculated the wavelength, so you can see how small the antenna is as a fraction of the wavelength. <S> With larger antennas, maybe a few meters, and higher transmit powers, its possible to get ranges of hundreds of meters using near-field communications. <S> This is used for example in cave-radio - since the near-field does penetrate rock reasonably well. <A> I'm kind of surprised the correct answer has still not been given. <S> Yes, it is important to understand that RFID actually operates via changing magnetic fields and that the magnetic fields drop off with \$r^3\$. <S> However, the reason low frequency (e.g. 125kHz) works at a shorter distance than high frequency ( <S> >1Mhz) is due to Faraday's law: \$V_{emf} <S> = -N\frac{d\Phi}{dt} = <S> -NA\frac{dB}{dt}\$ <S> Where N is the number of loops in the inductor coil, \$\Phi\$ is magnetic flux, \$B\$ is the magnetic flux density (i.e. magnetic field strength), and \$A\$ is the area of the coil loop. <S> Therefore, for fixed \$N\$ and \$A\$, the generated voltage for signal recovery/circuit powering is directly proportional to the speed at which the magnetic field is changing -- i.e. the frequency. <S> So for example, a 1MHz signal would generate 8x the received voltage as a 125kHz signal at the same distance. <S> Combining this with the \$r^3\$ decay of \$B\$ means that the 1MHz signal could generate the same voltage as a 125kHz signal at twice the distance. <S> Note: this ignores the parasitic capacitance of inductors, which also comes into play at higher frequencies. <A> Low frequency RFID as others have said relies on the magnetic vector not the electric vector of the EM radiation emitted by the coil. <S> The magnetic energy falls off with the cube of the distance. <S> The energy that can be transmitted is limited by regulatory authority. <S> The tags themselves have two types passive and active. <S> Passive tags as the name suggests harvest energy from the received field, and power themselves from that field then modulate their receiver coil. <S> The signal emitted by the tag then also falls off according to the cube law and has to be recovered from the transmitting antenna in a separate circuit to that transmitting the energy to power the tag. <S> In both tag and reader there is generally one coil transmit and receieve. <S> Active tags have longer read range because they only have to detect energy above a threshold to switch on thier receive and transmit cicuitry and their range is determined by the power available from their battery.
| Because we're not talking about a wave propagating away from antenna, but about a coil coupling into a magnetic field, the decay in power is even worse than the distance² for free space loss, and after a couple of cm, practically no effect of the tag on the reader can be made.
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Why does antenna length have to be at wavelength Resonance occurs when capacitive and inductive reactances overcome each other. Thus is the familiar resonance frequency equation: As the frequency gets lower, the antenna gets longer. But if this is only for making the antenna resonant with the selected frequency, why do we not simply add more capacitance (some metal balls or some conductor with volume) to the ends of a shorter wire to make it resonant with a low frequency instead of making it 6 meters long? Like this: <Q> Only a full wave antenne length will create an optimal radiation pattern. <S> You can play with the antenne length by adding capacitors or inductors but at the same time you do no longer create an optimal radiation pattern. <S> In many situations you have to use capacitors and/ or inductors to reach the goals you have set. <S> For instance if you want to create a multi band antenna. <S> This kind of antenna is in frequent use with radio amateurs. <S> The picture shows you a multi band antenna where resonance has been reached by introducing tuning elements. <A> To make the antenna efficient at radiating energy into space (or receiving energy from space), the key trick is to make use of standing waves. <S> The shortest standing wave you can get is at half the wavelength, and relatively easy to feed in the center of the dipole. <S> Full wave is also possible, but more difficult to "feed" by a cable if memory serves (which is where you get into practical aspects, such as feedline impedance matching = optimum energy coupling). <S> Actually quarter-wave stubs are also possible and "resonant", but need to be fed at one end by a unipolar/single-ended transmission line... and the impedance match is not very good, AFAICT. <S> This is typical for short-wave radioes (CB comes to mind) or vehicle-mounted FM radio antennas etc. <S> Note that by making the antenna physically smaller, you decrease its effective reception area, meaning that it "collects less energy" = has lower gain. <S> Which results in worse SNR in the radio input. <S> For the sexy small size, you have to give up some gain. <S> I'd like to point out one difference in the resonant properties, between a "lumped component" LC resonator and a "standing wave" type resonant dipole or cavity: if observed across the whole spectrum, an LC resonator has a single peak at the resonant frequency, but a "standing wave" type resonator actually has multiple peaks and notches, spread across the whole spectrum. <S> And, if memory serves, the standing-wave notches are generally sharper than the peaks (at least my simulations in QUCS used to suggest that.) <A> The impedance of free space is approximately 377 ohms. <S> At this impedance, the electric field has to be 377 times the value of the magnetic field. <S> To effectively transmit an electromagnetic wave this is the ratio of electric field to magnetic field. <S> Resonance occurs when capacitive and inductive reactances overcome each other. <S> This applies to tuned circuits but NOT antennas - for an antenna to be a good EM emitter it has to keep the balance between the fields in the correct proportion. <S> As the frequency gets lower, the antenna gets longer. <S> But if this is only for making the antenna resonant with the selected frequency <S> No, it's to keep the magnetic and electric fields in the correct ratio <S> and you can't disturb this ratio by trying to compensate the lack of magnetic field produced (because the antenna is "short") by making the electric field greater - that just doesn't work.
| You can indeed make the antenna shorter - this is typically done by putting a coil at the base of what was originally a quarter-wave stub.
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Step down transformer for vintage tube amplifier Hi so I've just bought a 60's tube guitar amp from the U.S, I live in Australia. Through searches across the internet I've found a few potential issues with the safety of running the amp through a 240-110v step down transformer. Firstly, the amp has no power transformer, it rectifies the line voltage directly. A common safety modification is to add an isolation transformer, I figured I could avoid doing this by using an isolated step down transformer Secondly, the amp isn't grounded so if a certain capacitor inside the amp fails, the chassis of the amp becomes live. The standard fix is to add a 3 prong plug and ground the chassis, but I've read that the ground plug in isolation transformers dont actually ground. Thirdly, would the 50hz of Australian mains damage the output transformer or any other component of the amp, designed for 60hz? Thanks in advance Here is the schematic for the amp http://www.magnatoneamps.com/schematics/magnatone_401_412.pdf <Q> An isolated stepdown is nice but not needed. <S> The extra safety introduced helps to prevent electrical hazard in case the input or output transformer of the amplifier fails. <S> If all is good then the common return from the amplifier is not connected to the chassis. <S> The chassis itself can therefore be connected to earth in both situations. <S> Without isolation transformer or standard stepdown transformer. <S> Finally there is no problem in connecting the amplifier to 50 in stead of 60 Hz mains. <S> Warning. <S> If you are working on mains operated equipment be careful and respect the fact that a mains voltage can kill. <A> All you need is a step down transformer. <S> See the note next to the power plug at lower right. <S> It specifically says that it's designed to work with 50 or 60 Hz power. <S> However, my preference would be to replace the whole power supply in the lower right corner with a off the shelf modern DC supply, if you can find something that puts out 130 V or can be adjusted to that. <S> That takes care of isolation, the rectifier tube wearing out, and the inevitable hum this amp has on the output. <S> But, it's not quite that simple. <S> You then have to find a way to run the two fillaments and the indicator lamp. <S> It seems the filaments are 45 and 20 V, for a total of 65 V in series. <S> That doesn't quite add up with the 200 Ω resistor dropping 20 V due to 100 mA thru it. <S> That only comes out to 95 V instead of the expected 110 V. <S> This is probably due to filaments being rather forgiving in voltage. <S> With 110 V applied to the resistor and filaments string instead of 95 V, the filaments get a little hotter than nominal, but probably not decrease tube life significantly. <S> Another possibility is that the filaments don't really both want to run at 100 mA. <S> Putting two dissimilar filaments in series is rather a hack. <S> This was done to not need a large and expensive power transformer. <S> Perhaps the somewhat higher than stated voltage is due to one of the filaments just getting the minimum at that voltage. <S> One option with a 130 V DC supply is to just increase the 200 Ω 5 W resistor to get the indicated 100 mA thru the filaments, or the intended 65 V across them. <S> Again, check that each filament gets at least its minimum voltage. <S> A more modern option is to make a small buck converter that reduces the 130 V to 65 V to run the filaments from. <S> You can still be retro and waste a lot of power by running a LED from that with a suitable series resistor. <S> That LED replaces the neon bulb. <A> The amp is designed to run on DC, that's why it rectifies and smooths the AC input. <S> You may want to consider increasing the size of the smoothing capacitor to accommodate the 'lower' frequency. <S> It does state 50/60Hz operation but its only half wave rectified and at 50uF that could be quite 'hummy'. <S> For old equipment its common to replace the old electrolytic caps anyway. <S> You don't need an isolation transformer because the 240/110 transformer performs that function as well. <S> A good (direct) ground connection to the metal chassis is highly recommended. <A> For the galvanic insulation of the input signal there is T2, for galvanic isolation of the output of the amp there is <S> a T1.You don't actualy need a third transformer for the galvanic isolation from mains. <S> You can use a autotransformer, which is smaller and chaper but it doesn't provide a galvanic isolation. <S> The chassis has to be grounded with third wire, ground wire. <S> The 50/60hz difference would not give any problem effect. <S> As you might see, the amp works on AC or DC voltage, there is a tube that is a half wave rectifier/regulator. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Since the amplifier is mains connected you can use an ordinary step down transformer.
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For high resistance load, Battery lasts longer? After reading this answer, I got to know that battery lasts about Total charge divided by current draw. So, lets assume the voltage across 2 terminals of battery is constant.by,$$ V(Constant) = IR$$ If I apply high resistance load in series, then very low current will flow and the battery will be drained even more slowly. My question is, Is my understanding correct (or I am missing something)? (I have never made any circuit other than led in series,parallel)I was going to use LDR to turn on and off led lights. Earlier i used to think that high resistance produces heat$$H=I^2Rt$$and that would consume a lot of energy wasted in heating. so battery will be drained faster. But today i read that above linked post and my concepts changed. (I don't know whether this question was asked before. because I don't know how do I search for it. I searched battery and resistance but irrevelant results came. so I asked question) <Q> For purely resistive loads, power dissipated can be calculated with $$ P = IV $$ where I and V can be substituted using Ohm's Law, eg: $$ P = <S> I (IR) = <S> I^2R <S> $$ <S> Hence where your H equation comes from. <S> That just takes into account time as well for a total loss over time. <S> To answer your concern about high resistance equaling high heat, higher resistance means lower current. <S> Because current is the squared term in the power equation, it tends to have more mathematical "influence" on the total power consumption. <S> You'll find that a 10ohm resistor with 10V applied (1 amp, 10 watts!) will have higher power than a 100ohm resistor with 10V applied (100mA, only 1 watt). <S> So in fact, a lower resistance will result in higher current and brighter LEDs at the cost of faster battery drain and more heat dissipation via resistors. <S> The answer you linked has way more on this, but for completeness and to summarize: Battery capacities are measured in millamp hours(mAh), which is simply how many milliamps it can put out for 1 hour. <S> A 500mAh battery can output 500mA for approximately 1 hour. <S> The relation is linear too, so that same battery could put out 250mA for 2 hours or 1000mA for half an hour. <A> A battery has a nominal charge capacity \$Q_N\$ in the order of 1 to 100 Ah for usual consumer variants (AA, mobile, car battery ...). <S> The battery also has a nominal voltage \$U_N\$ related to its chemistry and architecture. <S> An oversimplified model of a battery gives a constant output voltage until its charge is depleted. <S> If you apply a constant load to this (e.g. resistors and LEDs, no switching), then the time you can run the load from this battery could be calculated by \$t = \frac{Q_n}{I} = <S> Q_n\cdot\frac{R_{total}}{U_N}\$, where \$I\$ is the constant discharge current. <S> As can be seen, increasing the Resistance \$R_{total}\$ increases the battery runtime. <S> But the Question is whether the circuit is still able to fulfill its purpose. <S> A LED which receives less current due to a higher series resistor will emit less light, possibly making it useless for room lighting or as torch light. <S> Real batteries however are far more complicated and have to be modeled at least with a series resistor. <S> There are two effects in a battery which will actually increase its useable charge \$Q_u\$ over its nominal charge capacity when you apply smaller discharge currents (~ higher load resistance) : <S> The Recovery Effect says that in periods of low/no discharge currents the reduced "useable" charge due to high current loads gets partially replenished. <S> The reasons for both are the chemical processes in the battery. <S> Therefore battery runtime is affected by a lot more factors than Ohm's Law alone. <S> These effects do not create charge magically out of nowhere, but rather show that the discharging process is more effective in the given circumstances of lower currents. <S> Parts of this answer are excerpts from an related older answer of mine . <A> While your formula to calculate the heat generated by a resistor is correct, it is not useful in this case. <S> This is because if you change the resistance, but keep the battery at the same voltage , the current will change as well as the resistance, so you can't deduce that more heat is generated if you increase the resistor that loads the battery. <S> Because you have a fixed input voltage, you should use a formula expressing the heat in terms of \$U\$, \$R\$ and \$t\$: \$H = <S> \frac{U^2}{R}t\$ <S> This formula shows clearly that if voltage and time stay the same, less heat is generated for a higher resistance. <S> You get my formula from your forma if you solve Ohm's law for \$I\$ and substitute the \$I\$ in your formula with the result. <A> Theoretical answer is YES, But it depend on where you use it. <S> If it is for HIFI system, The sound output will goes bad. <S> Higher power source resistance will lower the fast response ability.
| The Peukert Effect describes that one can get more charge out of a battery if discharged with a low (constant) current.
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Best way to electrically and thermally connect a FET to an Aluminum heatsink I am designing a drive and need to electrically and thermally connect the tab of a series of FETs to an aluminum internal heat sink that is also used as an electrical bus. It seems the best way to go would be to solder the FET tab to an intermediary copper layer and the copper layer to the aluminum. Is there some sort of special flux and/or solder paste that would be needed to perform this? Methods to deal with oxidations? Would a reflow oven of some sort be required to ensure the copper/aluminum are soldered properly? Are there processes out there to do this sort of thing? I have struggled to find much information on this. Thank you for your time. Related question that came up from comments/answers: Does bolting to aluminum for electrical contact have any oxidation concerns? <Q> Do NOT solder your FET packages to aluminium. <S> The temperature required to activate the flux is higher than the FET packages will be rated for. <S> It may be permissible to solder a copper intermediate to the alli, then reflow the FETs onto the copper. <S> However, observe the 'time at temperature' figures for the FET packages, and know whether you can heat and cool the mass of the heatsink without subjecting the FETs to 'too long at too hot'. <S> There are lower temperature solders available, which may allow you to dwell longer at soldering temperature. <S> Bolting is a perfectly sound way to connect electrically and thermally. <S> Try it experimentally. <S> It's so much simpler to do first time, and to rework, that you have to try it before dismissing it. <A> Be sure to get it all cleaned off afterward, it may not play well with electronics (I would bet on it). <S> Copper would be easier. <S> Any kind of heat sink of substantial size is going to be rather difficult to solder to, it's in their nature. <S> Bolted connections, possibly with some thermal compound, may in fact be a better solution. <S> It's what is used in most such situations from the 120W CPU in your PC to the manly hockey puck thyristors used in industrial power applications. <A> I'm not sure it is "good practice" <S> but you can solder to aluminum with the right flux. <S> (This stuff works.) <S> with all the solders I tried. <S> So you could tin the Al heat sink and then solder on the FET's. <S> I have never done this!
| You can solder aluminum with the right flux.
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Zener voltage is not stable i have a made transformer less power supply to power a arduino board and and two relays. The circuit is designed to get 6v and a current of 150mA. after building the circuit i have connected one arduino and two relays across the output. The two relays are controlled by arduino. When i activate either one relay the output voltage across zener decreases to 5v but when i activate both of the relays at a time then the voltage decrease below 4V and the arduino restartsNow i want to make the output voltage across the zener constat and stable how can i do that...? and why the voltage decreases with change in load...? <Q> This is a very dangerous power supply, as it will be "hot" relative to earth/ground. <S> For safety, you really should use a power transformer to isolate the DC output from the AC. <S> Check the voltage on C2 when the relays are on. <S> It probably drops below 7 volts, and your load doesn't leave enough current to keep the Zener conducting. <S> The current drawn will cause varying voltage drops across R2, and also across C1 and R4. <S> You may have to increase the value of C1 to get enough voltage at C2 to keep the Zener diode conducting. <A> Zeners are poor at dynamic load voltage regulators because they need to consume the difference between the max current and minimum load current plus some minimal amount to have Vf as rated. <S> This places high power dissipation requirements and may be limited by your transformerless series cap impedance. <S> To improve this you may need to define your load requirements better and have increased storage capacitance or increased series capacitance and lower series resistance and higher power dissipation in the zener regulator. <S> Having underestimated your load current, you might think a SMPS is a better solution as offline regulators get costly with simple series RC dividers above 5W as the power lost in series is proportional to the VI drop to the regulator. <S> You should have defined your current needs better initially and choose lower current higher voltage relays like 24V then regulate down to 5 from there. <S> Also note that you do not have a low voltage common or ground in your circuit and safety <S> is poor. <S> For light bulbs ok, but not for user or other grounded computer interfaced supplies. <A> To add what other folks have said about safety and Zeners being bad power supplies... <S> This supply will not produce enough current to power most relays. <S> \$\begin{align}X_{C1} <S> & <S> = \frac{1}{2\pi fC} <S> \\X_{C1} &= <S> \frac{1}{2\pi <S> * 50\rm{Hz <S> } * 474 <S> * 10^{-6}C} \\X_{C1} &= <S> \frac{1}{2\pi <S> * 50\rm{Hz} <S> * 0.474 <S> * 10^{-6}C} \\X_{C1} &= <S> 6715\Omega\end{align}\$ <S> Which means \$\begin{align}I &= \frac{220}{X_{C1}} \\I &= <S> \frac{220}{6715}\\I &= <S> 32.8mA\end{align}\$ <S> So <S> ... yeah. <A> In order to get more current you have to increase the capacitor C1. <A> This circuit can be considered as a constant current source. <S> the value of current depends on the capacitance of c1. <S> the value of current can be calculated as I = <S> Vs/Xc <S> Vs is the supply voltage (230V or 110V) <S> Xc is the capacitive reactance Xc <S> = 1/2*pi*f*c f is the frequency of supply pi = <S> 3.14 c is the capacitance now the we can use any zener diodes which can withstand these current and the maximum current that a zener can withstand can be calculated by Imax= <S> P/V p is the power rating of diode V is the rated voltage of diode for example for a 1W 5.6V diode the maximum current Imax = <S> 1/5.6 <S> = <S> 178mA <S> so if the Imax is greater than the supply current <S> then the zener will provide a constant stable voltage <S> the value of resistance can be determined by power equation P= I2R
| Decreasing of the voltage is consequence of a lack of current.
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What do you call a relay that self-opens on power loss? I'm programming the simulation of some circuitry. These are relays that will open automatically when they lose power: Do relays such as this have a particular name? "power hold relay"? "single pole relay"? "self-opening relay"? I see that its technically a diode, a coil, and a relay, but I was hoping there was a technical name for this kind of relay. <Q> There is nothing special here. <S> Normal relays are held in one state by mechanical spring action when the coil is not energized, and in the other state when the coil is energized. <S> A relay being energized and power going away is no different from you switching it off deliberately. <S> Either way the coil stops producing a magnetic field, and the mechanical spring returns the contacts to the unenergized state. <S> There are such things as bi-stable or latching relays. <S> These mechanically stay in the same state they were last driven to. <S> Of course driving them is no longer as simple as energizing or not energizing a coil. <S> There are two possibilities. <S> There can be two coils, each used to drive the relay to one of its states. <S> Or, the magnetics can be polarized so that current polarity thru a single coil determines the state the relay is driven to. <S> One way or another, there needs to be at least three different driving states. <S> Latching relays are much less common, and any such relay will be clearly labeled as such. <S> Just a "relay" has a coil that is either on or off. <S> For normal (non-latching) <S> relays, contacts are classified as normally open (NO), normally closed (NC), and common (COM). <S> NO and NC refer to the switch states when the coil is not energized. <S> A SPST relay is the simplest type, since there is only one contact and therefore two output leads. <S> This type must be specified as normally open or normally closed. <S> If you want the relay to "shut off" when power goes away, then you want a normally open type. <S> Many relays have SPDT outputs, or multiple of them, like DPDT. <S> In that case, one of the ends is normally open and the other normally closed. <S> These flip state as the coil is energized. <S> The center contact that flips between being connected to the NC and NO leads is the common. <A> It's a relay with a normally open contact. <S> Relays can have Form A (normally open) Form B (normally closed) or Form C (changeover) contacts, which act pretty much as you would expect from the names. <S> "Normally" covers the state when the power is not applied, of course. <S> Relays with moving contacts that bridge two fixed contacts (as your schematic seems to indicate) are often called "contactors". <S> They are typically designed to switch relatively high currents, and the high power contacts are invariably normally open. <S> Sometimes they have low current auxiliary contacts that may be normally open or normally closed (or changeover). <S> Relays, on the other hand, typically have a flexure or wire that carries current. <S> In cases where the flexure doubles as the spring, overloading can cause the relay to fail 'on', which is usually considered quite undesirable. <S> In some cases, relays are wired in a self-hold configuration so the power to their coil is maintained through the relay's own normally-open contact. <S> A momentary contact closure bridges the NO contact and pulls the relay in. <S> This configuration is frequently used in machine tools and similar devices where you really want the machine to shut down and stay shut down in case of a power failure and subsequent restoration of power. <S> This is not an inherent feature of the contactor or relay, however, it's just a way of using the component. <A> Related but not necessarily what you're looking for: <S> Sometimes, relays for high power switching are known as contactors. <S> Another term used in industrial machinery for motor control is "NVR Switch" (for No-Volt-Release). <S> These are used when you want a machine to return to a safe (unpowered) state in the event of a power failure. <S> They usually incorporate circuit protection (overcurrent protection) as well as start and stop buttons, possibly wired to additional remote stop buttons. <S> Internally, a NO (normally open) relay holds itself closed until either a STOP button breaks the connection to its coil, or loss of mains power allows it to open its contacts. <A> It is called normally open (NO) relay. <S> The word "normally" implies no power applied to the coil thus, its contacts are open. <S> These type are the most commonly available relays. <S> If they carry "heavy" loads, they are called contactors.
| It's called a "relay".
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How can I reduce the PWM noise of a blower motor? I've built a ventilation box which utilizes a blower motor from an old Subaru Impreza 1996, which I was gifted from a friendly mechanic. The flow volume is amazing, but it's a bit too noisy and too powerful.The motor draws about 20A peak on start, and then operates at 7A at 12V. But when using PWM, the motor basically sounds like a screaming alien, no matter if the duty cycle is 90% or 10%. I'm using Arduino to send a pwm signal and an IRF3205 MOSFET. I've tried a low frequency (using delay and delayMicroseconds, 10-1000 Hz), and high frequency (using 32bit and 8bit prescaler of Atmega328). It only changed the pitch of the sound, but it was still very audible. The least noisy was at the low frequency, but then the motor was unstable. Should I use a capacitor or an inductor? (or both). I haven't worked with inductors before. And I imagine the capacitor won't do much good... Video: https://youtu.be/kHZ0b0wFjaw simulate this circuit – Schematic created using CircuitLab Source code: digitalWrite(pin11, LOW);delayMicroseconds(50);digitalWrite(pin11, HIGH);delayMicroseconds(100); <Q> Like most power MOSFETs the <S> IRF3205 <S> has a large Gate capacitance which must be charged and discharged, requiring about 50nC of charge to turn on properly. <S> In your circuit this charge trickles through R1, distorting your nice 12V square wave into a sawtooth that gets smaller as the PWM frequency is increased. <S> The 1N4001 is a mains frequency rectifier diode with slow switching action - not suitable for high frequency PWM. <S> You should use a Schottky diode rated for at least 3A continuous. <S> Also you must not have any interrupt code running <S> (eg. <S> timer, serial) or the PWM waveform will suffer from glitches that could be audible. <S> If you can still hear it, it's not 20KHz! <S> With good Gate drive, a fast flyback diode, and true 20KHz PWM, you should get quiet motor operation. <A> I've tried the same project, but on another scale: I tried to manage computer culler speed of rotating. <S> So, the secret is that Arduino by default have low frequency on its PWM output. <S> That causes noise, that you can hear. <S> So the decision is to set the right frequency directly through AVR Registries. <S> So, have a look here: https://www.arduino.cc/en/Tutorial/SecretsOfArduinoPWM <S> under the "Using the ATmega PWM registers directly" title. <A> I hope you didn't use the 1Amp 1N4001 diode to clamp a 7A motor. <S> Surge current is closer to 8~10x or 70A max starting with PWM. <S> Although it can handle short pulse currents much higher than 1A, it is usually better to match the current rating of the motor or more. <S> The clamp diode and the PWM switch ratings must BOTH exceed the motor surge current rating to avoid over heating and efficient stable operation for speed control. <S> This is because the diode conducts the motors inductive current while the transistor is OFF. <S> You must choose a diode >10x the motor current rating as each pulse at low speeds will be up to this surge current. <S> Otherwise it will get very hot. <S> $4.19(1) <S> - But it is better to observe the response on a scope and then determine if the noise electromagnetic-acoustic or mechanical-acoustic. <S> From the video the blade noise is excessive and cabinet resonance is evident like a loudspeaker design. <S> a better design puts the fan in a furnace duct in a remote location <S> so there is no interface friction noise or eddy currents with the vent grill PFM with 30-500us delay times, is a poor way to control Fan spped, although this is usefule for a Boost SMPS regulator, not for a buck fan speed controller. <S> you should be using PWM above 20kHz. <S> i.g. 21~22Khz and check for aliasing effects with fan noise, but it should be very quiet in a plenum. <A> Indeed it turned out that the noise disappeared using high frequency PWM. <S> In conclusion, I note the following: Using Nick Gammon's work on timers , I was able to get a PWM frequency that, according to his writings, should be around 25kHz; this caused the noise to disappear completely The gate voltage of the MOSFET must be able to charge/discharge fast; the MOSFET would get very hot; using smaller resistors (100 ohm), the problem was reduced but not solved for some duty cycles (thanks @Bruce Abbott) <S> At higher frequencies, the motor would not start, presumably because the MOSFET wouldn't charge fast enough (I'm wondering if it would be possible to use two transistors (NPN+PNP), to allow for quickly and efficiently alternating between zero resistance to ground and zero resistance to 12V, thus reducing heat loss and improving motor function and allowing higher frequency PWM) <S> With the following source code, I achieved a very quiet operation, with a small air flow, at 1.3A current ( <S> which was what I was aiming for: a low constant operation with occasional high power operation) <S> Source code: <S> #include <TimerHelpers.h>const byte timer0OutputB <S> = <S> 5;void setup() { pinMode (timer0OutputB, OUTPUT); TIMSK0 = 0; // <S> no interrupts <S> Timer0::setMode (7, Timer0::PRESCALE_64, Timer0::CLEAR_B_ON_COMPARE); OCR0A <S> = 10; <S> // <S> count to 4, zero-relative OCR0B = 5; // <S> duty cycle}
| Software PWM created with DelayMicroseconds() is not very accurate as it doesn't take into account loop overhead, so the PWM frequency will be lower than you expected. To get a good Gate drive waveform at 20KHz you should reduce R1 to about 500Ω (to charge the Gate faster) and R2 and R3 to about 1k (to ensure that Q1 turns on fully). I would suggest something like this automotive power diode.
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What's the risk of leaving VPP/MCLR floating? In the microchip PM3 ICSP design guide , this diagram is shown to demonstrate a typical implementation of ICSP: Here, the MCLR/VPP pin is pulled to 5V during normal operation, but isolated when the high voltage for programming is supplied. A similar setup is shown in this document . But I've also seen quite a few circuits taht completely omit this feature and just leave MCLR/VPP floating when no programmer is connected, like this: Why should I not do it like that? What possible negative effect could this have on the operation of my circuit? <Q> It's not 'floating' in the second circuit, that particular PIC has an internal pull-up. <S> You should read the datasheet for the particular processor you are using. <S> It's only 404 pages, a trifle compared to more modern processors. <S> Edit: <S> If there was no pullup- <S> what would happen: <S> Bad things- it would float around and might reset the micro or not depending on the phase of the moon and how you breath on the PCB (leakages). <S> If you were lucky it would not work at all from the start. <S> It's unlikely to cause any physical damage in any case, but it could cause some problems that appeared to be random and might be hard to troubleshoot (especially if they got blamed on firmware). <A> If your PIC doesn't have an internal pullup and you don't provide an external one the best you can hope for is that your program will run sometimes. <S> Without any pullup at all the chip will constantly be resetting at random depending on how many cats are nearby or what underwear you have on. <A> Leaving high impedance inputs floating is bad practice. <S> A floating pin is not guaranteed to be in either state, and worse can oscillate between states sometimes at high frequencies. <S> This can cause noise, unstable operation, high power usage, and a number of hard to troubleshoot issues. <S> In this case if the pin drifts low your processor will reset unexpectedly. <S> Also with many circuits RESET must rise and fall within a certain amount of time as specified in the datasheet. <S> Too slow or too fast might fail to properly reset all portions of the device. <S> This may leave it in a non-running or unsafe state.
| Some PICs can have the pin re-purposed as an input, in which case that floating around wouldn't have a huge negative effect unless the pin is non-Schmitt trigger, in which case it could cause excessive current draw, more quickly draining a battery, for example.
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Relay Coil should get full 12V or 0V simulate this circuit – Schematic created using CircuitLab As you can see in the above circuit, I have converted AC to DC. The transformer used in the above circuit is : If I apply 230V across the Primary, then I get 12V across the secondary coil of transformer. Then I convert it to DC using 4 Diodes and a capacitor. This DC voltage is applied to the coil of Relay. This circuit works as expected. Problem: When I decrease the voltage on Primary coil of transformer. Suppose I apply 200V to the Primary coil of transformer then also the chances are that Relay coil gets energized. What I want: If there is 12V or higher applied at the Red box in the diagram, then only the coil of relay should get energized else I should get 0V across the two terminals of relay coil. I asked one of my friend about this problem. He said I should use a Zener diode in this circuit to achieve what I want. But I don't know where do I place the Zener? Can anybody give me a suggestion? <Q> In the schematic shown below, the entire secondary of the transformer has been used, two rectifiers have been eliminated, and C1 is used to smooth the full-wave rectified 16.5 volts. <S> U1 is a 12 volt linear regulator which drops the 16.5 VDC down to 12VDC for the relay and whatever else the 12 volts is being used for, R6 representing both the relay coil and the remaining parallel load, if there is one. <S> U2 is a voltage comparator with an internal 400 millivolt reference. <S> The voltage divider R3 R4 is used to set the voltage at the non-inverting input of U2 to slightly over 400mV when the output of U1 is at 12 volts. <S> That will force the output of U2 high, turning Q1 ON and energizing the relay coil. <S> If, for some reason, U1 OUT should fall below 12 volts, the voltage at U2+ will fall below 400mV, U2's ouput will go low, and the relay will become de-energized, opening its normally-open contacts. <S> U2 is supplied with about 6.5 millivolts of internal hysteresis and if more is needed it can be applied externally as shown on page 10 of the data sheet. <A> A relay is an inappropriate component for a precision voltage detector. <S> The pull-in voltage depends on the spacing of the iron components in the magnetic circuit, that can take slightly different positions every time it operates. <S> Copper, and most other pure metals, have a large tempco of resistance, a 25C change in temperature results in a 10% change of resistance. <S> As it's the coil current that pulls the relay in, the voltage sensitivity varies by 10% over 25C as well. <S> On drop-out, things get worse. <S> Possibly only a quarter. <S> I'd expect it to stay closed down to 6v, it may stay closed right down to 3v. <S> You need an electronic solution. <S> Depending how accurate and temperature insensitive it must be, you would use a voltage reference (a zener or something more stable) and a comparator, or maybe just the nominal 0.7v VBE of a transistor, it depends on your accuracy specifications. <A> Irrespective of your question, here's your basic problem... <S> OK <S> that's the warning done with so you might want to put some series resistance in with your relay coil to ensure that it receives 12 volts instead of 16 volts. <S> Taking this one step further, you could put a little too much resistance in series so that the relay is only just operating at 230 V. <S> Then when the voltage lowers a tad, the relay drops out and gives you the result you want. <S> If you want a more reliable solution you should use a 7805 voltage regulator powering a comparator - one input pin on the comparator can take (say) <S> 2.5 volts by potential dividing the 5V and the other input pin can take a potential divided voltage from the 16 V dc output to produce 2.51 volts. <S> The comparator output will switch when the 16 volts drops by maybe 100 mV to 200 mV. Use the comparator output to drive the relay.
| The AC output is 12V and, after rectification and smoothing, the DC voltage will be about 16 V DC - this might be too much for your 12V relay coil. The iron path in the relay has now closed, and much less current is needed to keep it closed. It also depends on the resistance of the copper coil, which if you've not met size of this variation before, may astonish you.
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A story behind a weird inductor This old, crusty inductor found in an old Iskra (Yugoslavia) TV intrigued me, and I was hoping that someone could tell me what's the story behind it's design. I don't know if the pictures are descriptive enough, but inductor's "head" is magnetic, and the plastic casing is "hugging" another ferrite cylinder with small square hole in the middle. It's unlike any inductor I scavenged this far, and while I do understand how inductors work, I can't figure out a useful purpose of this arrangement. <Q> It's a once very common horizontal linearity adjstment for CRT raster scan TV sets. <S> It basically uses core saturation to achieve a somehow current dependent inductance. <S> Bias magnet just changes current where saturation and hence "corrective" effect begins. <A> The part in the picture comes from the horizontal or line deflection system. <S> A line deflection system consisted of two resonant circuits where the frequency was changed by means of a tube acting as a switch. <S> These two circuits generated each half a cosine and switched over at the top. <S> So one half of the cosine was a low frequency (the visible part on the screen) and the other half cosine the high frequency part (the invisible flyback part). <S> What is needed for the visual part should be linear. <S> This is not realy the situation with a cosine. <S> Only a very small part of the cosine can be considered linear. <S> To make the cosine more linear a magnetized adjustable inductor was placed in the circuit thereby improving the image on the screen. <A> I think I found a similar inductor years ago. <S> If I remember correctly (you could check it easily), the round cylinder is a magnet, and can be rotated with a screw or a similar tool. <S> My hypothesis was that it could change the DC component of the magnetic field in the main ferrite beam, and either change the self inductance or the limit of admissible current before saturation. <S> But I cannot answer what it is for. <A> As others have said it is a biased inductor used in the non linear region of the ferrites B/H curve. <S> http://www.repairfaq.org/sam/deflfaq.htm#dshlc has a description, basically as the current in the deflection yoke rises the inductance of this part falls as it is driven further into saturation helping to reduce the effect on scan linearity of the copper losses in the yoke. <A> Resonant circuits (sometimes called 'tanks') are a combination of capacitors and inductors. <S> The resonant frequency can be tuned by either varying the capacitance or the inductance. <S> Big capacitors are expensive and prone to failure, so for high power applications, a large inductor is paired with two capacitors, usually in a "pi" configuration. <S> See https://en.wikipedia.org/wiki/LC_circuit for a broader interpretation,But the essential fact is that CRT magnetic deflection takes high voltage and the large inductor handles the higher power better than a capacitor solution. <S> Because high voltage ages capacitors, but inductors hardly ever change (unless the ferrite slug gets adjusted). <S> After tuning, slugs are routinely fastened with Loctite or hard wax.
| In old tv sets you find a vertical and horizontal deflection system realized with inductors round the neck of the tv tube.
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Switching a stereo audio signal without distortion I need to switch a stereo signal in a lossless, non-distorted manner. Simply put, I need to control ON/OFF for each channel. Should I use MOSFETs or optocouplers? I believe the signal ranges from -2.5V to 2.5V ... standard 3.5mm output. Cluebats at the ready :) <Q> None of the above. <S> Ideally use <S> CMOS analogue switches like 405x series ( <S> +/-5v <S> analogue supply and switching voltage range, logic 0/5v compatible). <A> If you're only switching line-level you can use analog switches, such as TI's TS5A2066 . <S> If you need to control each channel separately, a dual SPST type would probably be best. <S> EDIT: <S> Apologies -- I missed the <S> +/-2.5V <S> signal requirement. <S> I'll defer you to Neil_UK's post. <A> Consider that it may be more prudent to turn off the audio signal by shunting or shorting it to ground. <S> That way there is no non-linear device (like any semiconductor) in the THRU-path of the audio. <S> Consider also that hard metal contacts (like a relay) will have vanishingly low distortion compared to any semiconductor device. <S> Suitable relays are readily available in DIP packages same size as an IC. <S> Even latching relays which require no holding power. <A> Telcom-grade relays in DIP package are ideal for this application. <S> Note that the relay-contact material and plating will greatly affect the reliability of your system. <S> One of the reasons that I mention telcom-grade relays is that these are designed both for reliability and vanishingly-low resistance values with very low values of current. <S> Relays that are NOT designed for audio or instrumentation purposes will have contacts that can develop contact resistance when passing tiny currents. <S> The problem shows up as distortion or even loss of signal until the relay is exercised. <S> There is a spec in the datasheet that is often called "wetting current". <S> This is the minimum current required through the contact to keep it clean and conductive. <S> Telcom relays with gold-plated bifurcated contacts <S> are the most-reliable readily-available relays available. <S> In years gone by, I would have suggested that Mercury-Wetted relays would be a good choice. <S> Unfortunately, these are no longer readily available.
| Relays are popular for audio applications, and are useful for switching either line-level signals (e.g. from an mp3 player) or speaker-level.
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Zener diodes in glass axial package - not inherently shielded from photoelectric effect? I discovered today that a glass packaged axial leaded 5V Zener diode will become a source of about 0.450 Volts when the glass package is held in the beam of a low-power purple (405nm) laser pointer. The test setup:Scope probe (with ground clip) attached across the zener. With laser turned off, scope reads zero volts as expected. Turning laser on and aiming it at the glass package of the diode, the scope reads a fairly stable 450mv (noisy though: 30mv p-p ~100kHz). (edit: this noise could be a product of the laser-driver step-up circuit) The laser is a cheap one and purports to be 1mW rated. Interrupting the beam with opaque materials instantly stops the voltage reading from the diode. Modulating the laser with a 5kHz square wave causes the diode to exhibit a 5kHz response (in phase with the laser's modulation as far as my scope can tell). I realise that this is rather unscientific but my question is this: Is this typical of glass zeners and if so, should a designer avoid using glass zeners in sensitive analog circuits. Or is this too specific to be a real-world problem? <Q> Diodes of all sorts, including the ubiquitous 1N4148, packaged in transparent packages tend to have some sensitivity to light (both photoconductive and photovoltaic as you have observed). <S> The 1N4148 can apparently produce 10nA in direct sunlight . <S> Zeners are not terribly precise devices in the first place. <S> However, say you are using it as a noise source, say for audio or cryptography, you might want to keep it dark or use a plastic packaged device. <S> and it is exposed to light, either from openings in the enclosure or because some designer has peppered the PCB with highly luminous LEDs that are modulated or blink. <S> That includes glass MELF packages as well as axial-lead packages (photo from Digikey). <A> " Or is this too specific to be a real-world problem? " Not at all. <S> It's a problem for me as I use them for cryptographic random number generation. <S> I've recently been using BZX85C24 Zener diodes. <S> Running it at 30uA can create a noise level of 1V peak to peak (if you measure it enough times). <S> But that's in total darkness. <S> Get some sunlight on it and the noise drops dramatically to a quarter or less. <S> Even worse is getting mains powered lighting on it like incandescents. <S> You just pick up a mass of mains hum all over the signal <S> that's totally trashes the entropy output. <S> I expect that not many people use analogue noise sources for testing, as digitally generated sources are available. <S> But for cryptography, you absolutely need the analogue variety. <S> You can use light tight enclosures, but I prefer to use heat shrink tubing on the diodes themselves. <S> If you don't take precautions against the photo electric effect in these applications, the whole device can fail to provide secure random numbers. <A> Laser induced arcs are possible in small air gaps which also have negative resistance like a semiconductor during ionization.
| So if you are operating in high ambient light and low current affects your operation, simply block the light. All semiconductors ... have a photo-electric effect including LED's which can be used as ambient light detectors. I rather suspect your zener diode when used normally with several mA flowing would have negligible response to normal room light. It's worth considering such effects if you have a very sensitive circuit
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Mounting micro USB B connector on a PCB What's the best way to mount this on a PCB so that it is more reliable? Should I go with the flat tabs an no holes like they show and just solder the tabs to the dedicated pads or bend the tabs and put them into holes and solder that way? <Q> Mount it as recommended. <S> It will withstand the force needed to connect/disconnect. <S> If you have space and will solder it by hand, you could increase the size of the front pads and drill some VIAs into it. <S> This will prevent the pad from coming of and enforce the construction. <A> In my experience the mechanically weakest link is the pads trying to hold on to the PCB. <S> You could improve the strength of the pads by adding plated vias to them to pads on the other side of the PCB or to an inner layer. <S> And solder that through of course. <S> But is <S> a through-hole connector is acceptable then I would choose that. <A> Surface mount connectors are not made to withstand much more than normal insertion forces. <S> What's really holding the connector on is how well the copper is bonded to the fiber substrate. <S> Normally, an surface mount connector will withstand normal insertion forces for the life of the product. <S> If you expect this connector to see more than normal insertion forces, then a connector with mechanical mounting tabs (through hole) will be a better bet. <A> As shown, this is the least reliable mount, especially with regard to bending. <S> This one is the worst. <S> Connectors with two through holes are somewhat better. <S> The usual problem is with signal pads. <S> They break due to fatigue, and the fatigue is due to the shroud has excessive flexibility, due to very short solder footprint. <S> A connector with a bigger through-hole footprint ( solder tabs on all four corners ) is way better. <S> However, for the real product (like a test fixture), people are using aluminum machined brackets on both sides, screwed together. <A> How many cell phones have you seen where this connector breaks off the board? <A> Never ever use SMD parts that will be subject to mechanical stress. <S> Connectors like these should always be securely mounted and not rely on some pad on the pcb (which is generally very weak).
| If this will be seeing substantial strain, a connector with through hole mounting tabs is one way to go -- but encasing the board in a way so as to maximize mechanical stability is probably the more reliable approach. Obviously you should not use any ROHS solder, and should use the usual 60/40 (more malleable, mil-spec approved) solder if you want the connector to last up to its contact rating limit.
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Why AC supply is rectified first before stepping voltage down? I have opened one of my phone chargers and was trying to understand how it works.. I found IC MB6F just at AC input which came out to be a bridge rectifier. My question is why AC is first rectified instead of stepping it down with transformers ? As stepping down in DC with a buck converter wouldn't be as efficient as with transformers, why is this path chosen ? <Q> The volume for a core, needed to support the Webers, is this: $$ \left(volume = <S> l_e\cdot A_c\right <S> ) \ge \frac{\mu_0\: <S> \mu_r\ <S> : I_{peak}}{B_{max}^{\:2}} \int H\:~\textrm{d <S> } <S> B$$ <S> \$l_e\$ is the required magnetic path length and \$A_c\$ is the core cross section area. <S> In the case of a sine wave, the last factor above, for one half-cycle of applied voltage, is: $$\int H\:~\textrm{d} B = V_0\int_0^{\frac{\pi}{\omega}} \textrm{sin}\left(\omega t\right) <S> \:\textrm{d <S> } t = <S> \frac{2\: <S> V_0}{\omega}=\frac{\sqrt{2}~ V_{RMS}}{\pi\: f}$$ <S> For 60 Hz and \$V_{RMS}=120\:\textrm{V}\$, you get about \$0.9\:\textrm{V}\cdot\textrm{s}\$, which is huge. <S> Suppose I wanted \$10\:\textrm{W}\$. <S> Then \$I_{peak}\approx 120\:\textrm{mA}\$. <S> The flux density of iron cores probably shouldn't exceed about 1.5 Tesla (though we could look at various materials and come up with a range of numbers here.) <S> \$\mu_r\approx <S> 1000\$ for iron. <S> So this suggests about \$60\:\textrm{cm}^3\$ of volume for the core -- or in the area of about 1 pound of weight. <S> This is about twice the weight that is used as a rough guide for power transformers, but it is in the right ballpark. <S> (Using another figure I've seen for CRGO steel, 1.9 Tesla, the weight drops down into the normal expectation range.) <S> Note that high permeability doesn't help you in the least. <S> Look up at the equation above. <S> That term, \$\mu_r\$, is in the numerator! <S> If you increase it, all you do is increase the needed volume. <S> The reason is that you need vacuum to store energy. <S> Not matter. <S> So more permeability simply means less effective vacuum space in the solid matter lattice <S> and so you just need more matter in there. <S> High permeability here is about containing the flux lines and providing good magnetic coupling. <S> The inductance comes along for the ride, too. <S> Low frequencies just mean big volt-seconds and therefore big cores. <S> Thus, "heavy iron. <S> " It just goes with the territory. <A> When I was a kid, my father used to teach me a couple simple "rule of thumb" formulae of iron-core transformer design. <S> 45/s turns per Volt, and P = s^2, where s = <S> square of the EI iron core's central column in cm^2. <S> I.e. the available electric power grows with the fourth power of the core's "thickness" :-) <S> And, note that tiny wattages have you end up with extremely high turn counts of pretty fine wire... <S> lots of series resistance. <S> Again for small cores, this translates into high turn counts of fine wire. <S> If tiny iron-core transformers were more efficient than similar-wattage switchers, legal regulations on stand-by power consumption in consumer electronics would make the iron-core stuff omnipresent. <S> The reality is, that only switchers can be made with a good enough efficiency in the tiny form factors (with next to zero idle power consumption). <S> This is not to say that all switchers are elegant, stabilized and efficient. <S> For instance, some of the super-sexy tiny and light Nokia chargers (ex works standard accessory) contained the most minimalistic switcher I've ever seen: a self-oscillating three-legged "transistor" in the primary, cheap elyts, no filtering, no feedback! <S> Unsurprisingly, those chargers are energetically inefficient, regulation under load is very poor (compared to a similar wattage tiny iron-core trafo) <S> and I didn't really check the EMI. <S> Some cheap slightly bigger wall warts (switchers) already have a proper feedback (via an optocoupler) and some minimal filtering, but are pretty inefficient (get hot). <S> Same thing with aftermarket noname notebook adaptors. <S> I've recently been amazed at the current-generation Lenovo notebook adaptors (with the rectangular yellow plug). <S> This small "plastic brick" adaptor runs perfectly cool! <S> Amazing, compared to some notebook adaptors I had in the past. <S> I would hazard a guess at modern TrenchFET/FinFET transistors and sync rectification at the secondary side. <A> It sounds like you're assuming they could do something like this: simulate this circuit – <S> Schematic created using CircuitLab <S> But usually you want your DC supply to be regulated. <S> (Unregulated supplies tend to have a large ripple, which causes problems for many circuits.) <S> So you need a regulator at the end <S> no matter what. <S> For safety reasons, you also usually need a transformer somewhere. <S> The schematic I drew above is a perfectly valid way to do things. <S> You could put a linear regulator or a buck converter on the end and get regulated DC. <S> But (as others have said) <S> 60 Hz transformers are big and heavy. <S> And even if you don't need a transformer for safety, it's hard to reliably convert a very high input voltage down to a very low output voltage. <S> These limitations lead to the so-called off-line converters, which rectify the mains first, then switch the primary side of a small transformer at a high frequency. <S> The secondary side is still isolated, and the transformer does some of the voltage conversion, which keeps the switch's duty cycle in a reasonable range. <S> Of course, off-line converters are more difficult to design than the simpler kind I showed above.
| My naive understanding is, that in order for the transformer not to suck/waste power when unloaded / open-ended, you need the transformer to have a certain high-enough inductance for a given primary voltage (measured with an open-ended secondary).
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Do electronics distributers test each component before sending them out? I'm trying to decide whether to order components direct from manufacturers in China or from distributors in the US. If I have to test each component either way, I'm not sure it matters to me if it's a knock-off or the "certified" one, especially when prices are 100x more expensive from distributors. It's very important to me that my customer receives something that works and isn't let down either by a high price or a failure that could've been avoided. <Q> As far as I know, this applies regardless of country of manufacture. <S> Distributors don't test anything. <S> That is not their job. <S> Manufacturers may not test 100% of the features in a product, but they will test the product in some fashion. <S> At a minimum, they will test each component on the wafer and discard bad components prior to packaging. <S> (Wafer sort). <S> The extent of post-packaging test probably varies depending on complexity of the product, etc. <S> You might have an acceptance test for things like speakers or LCD displays, but not for the components soldered to the PCB. <S> If something like that is required, you would probably modify the design to exclude the troublesome part, or work with the manufacturer to improve outgoing quality. <S> It is certainly possible to be taken advantage of in China. <S> But there are quite a few perfectly good suppliers of all kinds in China. <S> Everything is built there, and there is a vast industrial supply chain in China. <S> Taiwan is maybe one step up in quality and reliability compared with mainland China (PRC). <S> But China is much bigger. <S> One thing you should DEFINITELY NOT do in China is buy major brand parts through gray market channels. <S> There is a lot of fraud and counterfeiting going on. <S> If you buy major brands, buy through authorized distributors only. <A> And if they did, you would be paying an extraordinary premium price for the privilege. <S> OTOH, ordering anything from Asia is even a bigger gamble. <S> That said, are you creating products from components (even pre-TESTED components!) <S> and then not testing YOUR OWN product? <S> Depending that all the components are good and then blindly assembling them? <S> How do you know if your product is good even if assembled from good components? <S> Certainly, you can develop a sense of which vendors (foreign or domestic) deliver components with lower failure rates and adjust your buying preferences accordingly. <A> As a distributor yourself, you will be responsible for trace-ability to source and provide certs if necessary. <S> You can pay other Disti's to test, but this is not normal practice, nor cheap. <S> Thus you can request these certs or perform your own due diligence on supplier. <S> http://www.erai.com/PartRiskMitigation <S> http://www.erai.com/ca_Counterfeit_Awareness_
| I think it is relatively unusual for electronics products factories to test incoming electronic components prior to PCB assembly. If you are talking about IC's, yes, manufacturers (not distributors) normally test 100% of the IC's as part of the production process. It is very unlikely that even a premium US distributor tests every component before shipping it. There is a huge network of 3rd party suppliers and from my meeting with many in this line of business and having personal inside sales experience from a Long Island office in NY, I can say, that Independent Distributors use a wide variety of sources, including independent Chinese suppliers.
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Cluster Raspberry Pi power supply from an ATX PSU I've asked this question within the Raspberry Pi exchange but they stated that posting the question here might be more useful. I built a Pi cluster with 6 nodes. This runs off a 90 watt astec atx-93 PSU. Also attached to the PSU is a 12v fan and a 5v ethernet switch. However it seems the 5v rail of the PSU is not supplying enough current to power everything. Only 3 Pis boot properly, the others just keep rebooting. And with the Pi's that do boot, attaching any i/o devices to them send them in a spin. There're four 5v outputs coming from the PSU. 2 Pi's to each output through the GPIO 5v pin with a fuse to protect incase of a power spike. The 4th output is used to power the fan which is rated for 12v but I have it at a lower speed at 5v. There is one 12v rail that has a 5v regulator attached. This is powering the ethernet switch. If I put a multimeter over the Pi's the voltage is 5.12/5.20 v but the current seems to be around 0.9a to 1a max. So my question is, is there anyway of getting more current to the devices without having to change to the PSU itself, or is it just that buying a more powerful PSU is the only way forward to get what is needed? <Q> According to the schematic posted on Raspberry PI stack exchange, an on-board 1.1A fuse exists between microUSB and GPIO 5v line. <S> This fuse is not supposed to see current drawn from microUSB to connected USB devices. <S> With your method of supplying 5v power to the PI, this fuse DOES PASS USB device current . <S> It is rather difficult to determine in your complex power scheme (where multiple currents can flow in opposing directions) how much current a particular fuse will see. <S> Fuses act on net current flow. <S> Seems like this fuse is running at its limit, if your current measurements are accurate. <S> its other 12v and/or 3.3v rails. <S> The PI was meant to be powered through the microUSB - it should be powered from this point, else the on-board fuse may not properly fulfill its function. <A> To me, it seems not your power supply is the problem, but the Pis. <S> You measured a voltage of 5.2V, which is more than enough. <S> If the power supply were weak (or the power cables were too thin), the voltage at the pi would drop below 5V. <S> However, the current of 1A seems way to high, for example, <S> this site states less than 500mA for loading the desktop. <S> I don't know about the other models, but the Model 1 B+ has a 1.1A polyfuse in the 5V line of the micro USB connector. <S> This means, if the Pi draws more than 1A, the fuse blows and needs to cool down to recover. <S> The Pi will reset in this case, and may be caught in a boot-loop. <S> Even if a Pi is just below the current threshold, attaching a device may push it beyond, and the Pi resets. <S> There's still a chance the power supply sends some voltage spikes, which are not visible on a multimeter. <S> But due to the high current, I don't think that's the case. <S> As a first test, I would connect the Pis to something different than an ATX supply, and disconnect any devices from the Pi. <S> Next, exchange the SD card to see if there is a software problem. <A> Looking at the label for your power supply it appears that the +5v and +3.3v supplies come from the same DC/DC convertor block. <S> Notice that the maximum power for both supplies is limited as a pair. <S> In all probability the 3.3v supply is used to provide regulation feedback for both the 3 and 5v outputs. <S> I'd suggest what is happening is that the 5v supply is actually dropping and you don't see it on the multimeter. <S> There are small differences in the Pi's and the first ones to reset immediately reduce the current drawn so a couple of the units continue to run as the voltage starts to rise again. <S> I'd suggest you need to apply a permanent load to the 3.3v output. <S> Perhaps start with 10W (1 Ohm resistor).You could also apply a 10-20W load (An incandescent bulb perhaps) to the +5v supply, you might find that that this will prove the +5v is collapsing.
| In general, the Pis need 5V and draw as much current as they need. Your power supply is adequate, with Neil_UK's caveats about properly loading
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Would these two transistors have the same Beta? If both Q2 and Q1 are 2N3904 BJT transistors, can we assume that they share the same Beta? Thanks for your help in advance! <Q> It is not possible to say without component values. <S> The Beta will vary as a function of collector current. <S> Beta has a pretty wide range around the nominal. <S> If their bias currents are different, then they would have a different nominal Betas. <S> The data sheet will have a graph that shows the nominal Beta as a function of collector current. <S> To answer your question, the beta of a transistor also varies this temperature, which you can see in the graph, and with Vce. <A> You can do this for analysis' sake, but don't count on it for your circuit to work! <S> Transistor beta is highly variable from unit-to-unit, which is why circuits that depend on transistor beta as a parameter are not used when designing with discrete parts. <S> In general, you'll see a minimum beta spec for a few given operating points, and perhaps a typical or maximum beta for one or two of those conditions, as well as a "typical" beta vs Ic curve if the datasheet provides characteristic curves. <S> For the sake of circuit analysis though, a beta is needed, and the minimum for your circuit's operating point can safely be assumed (otherwise the transistor is out of spec and should be rubbish binned!). <S> If you want to be thorough, an analysis at maximum beta could be conducted as well <S> -- I wouldn't insist on it though, as with a proper bias network, if your circuit works with a transistor that's barely in beta spec, it'll work with any other transistor of that type. <A> Unless two transistors are on the same silicon substrate, you should not expect the same beta (Hfe) from devices. <S> The forward current gain also varies markedly with temperature <S> and it's very difficult to keep two separate transistors at exactly the same temperature. <S> Read this for information on BJT matched pairs: http://www.analog.com/media/en/technical-documentation/data-sheets/MAT12.pdf <S> The schematic you show would not be sensitive to differences in beta between the two transistor devices. <S> The first device is a Common Base amplifier and the second an Emitter follower. <S> If you look at the datasheet for your devices: http://www.onsemi.com/pub_link/Collateral/2N3903-D.PDF I'd suggest your amplifier stage design should be based on a minimum Hfe about 40 and IC of 1-2mA.
| If their collector currents are the same they would have the same nominal Beta.
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Flash driver for 220V lamps according to the state of two switches I am an absolute beginner in electronics.I would like to actualize the simplest circuit capable to pilot small 220V lamps (in parallel), making them flashing according to the state of two switches and according to the following rules: if both switch A and switch B are off, the lamps are off; if switch A is on, the lamps are on; if switch B is on, the lamps flash at about 1 Hz or little more, regardless to previous rules. Lamps don't need more than 150 W in total. It would be appreciated to have some explanation on how to vary the on/off cycle of flashing, remaining in the range about 1 Hz - 5 Hz. Maybe what I need is a simple modification of this circuit, that satisfy switch B rule, but need to implement switch A functionality (to be always on, without flashing, ONLY IF A is off): Many thanks!! <Q> (eg : <S> sending a 1Hz on/off pulse). <S> Some of those boards can handle up to 10A. <S> This should be far enough for what you want (10A x 220V = 2200W). <S> Make sure you buy one with optocouplers (as on the picture above) for safety. <S> Sending the on/off pulse and implementing the logic you want can be done using an Arduino (or any other microcontroller) or a simpler circuit like a 555 timer in astable mode (and some AND/OR gates). <S> I would avoid trying to implement yourself the part that deals with 220V <S> (eg : using a triac + transistor, as found on some schematics on the web) as it can be dangerous. <A> If you're willing to plunk down $40 on a time delay relay capable of repeat cycle (flasher) operation, here's probably the simplest way to do it: <S> simulate this circuit – <S> Schematic created using CircuitLab SWA is a standard single pole switch, SWB is a changeover (2-way/3-way) switch, and U1 is a Macromatic TE-8816U time delay relay set to function B (repeat cycle, starts in OFF) and the flash rate you wish. <S> The TE-8816U was chosen as it's reasonably priced, can accept 220/240VAC input, is UL listed (vs just being UL component recognized), and has a reasonably long-lived set of contacts (70,000 operations electrical at rated load). <A> The logic is ; if B =1 , output Y=1 flashing 0.5 Hz square wave = <S> F <S> if B =0 or B!=1. <S> and A= 1 , Y=1 otherwise Y=0, let ! <S> = <S> invert <S> , *= <S> AND, += <S> OR <S> so <S> C= A*B! <S> then Y = <S> C + B*F <S> However since you are driving a 220Vac LED load with a Switched reactive regulator, it is important to get the preferred switching method from supplier such as a zero crossing Triac or 10A AC relay rated for reactive input impedance, since specs indicate poor Power Factor = 0.6?. <S> We also need to know exact Specs of A,B input signals and expected MTBF of unit to factor surge currents , switch ratings etc. <S> simulate this circuit – <S> Schematic created using CircuitLab BOM <S> Cost $5 est. <S> For 1 Schmitt NAND, ,1 opto ZCS triac, and 1uF Caps . <S> For wetting the contacts with low current logic, it is critical to choose C1 depending on contact rating and plating. <S> The RC value also debounces the switch and the cap attenuates ESD, critical for long cables.
| Perhaps the easiest solution would be to buy an Arduino relay board and then control it from outside
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A relay that can switch from one source to another - what's that called? I want to be able to use a microcontroller to select the power source, from among two, for a 100A DC application. All the relays I see are ON / OFF, rather than a selector. So, I'm looking for something with 5 connections: Signal input (say, 3.3v digital, which when activated causes it to "switch") Power input 1 (which I'll connect to a battery bank) Power input 2 (which I'll connect to an AC Adapter) Power output (which I'll connect to the load) Common ground (for the voltage of the signal input) I'm probably not searching for the correct piece of equipment, but I can't seem to determine what this might be called. <Q> If the ground/common/negative side of your circuit can remain connected, then you need to "switch" only the "hot"/active side (voltage not disclosed) side. <S> That can be done with a SPST (Single Pole, Single Throw) switch or relay. <S> However there are several things to be aware of if you want to do something like this: <S> CURRENT <S> These are becoming harder to get and more expensive as this kind of switching is done with semiconductors (transistors, thyristors, etc.) <S> here in the 21st century. <S> VOLTAGE <S> Switching a high voltage will require essentially a "mains-rated" relay which will add even more to the price. <S> DC vs AC <S> Switching DC is more difficult than switching AC. <S> Because AC naturally passes through 0V many times per second (50Hz or 60Hz) Breaking high-current DC requires a special kind of relay designed for such applications. <S> This will add even more to the price. <S> CONTROL VOLTAGE <S> @ ? <S> ?? <S> V which can be operated from 3.3V. <S> You will need probably a multi-stage relay driver circuit to allow the 3.3V logic-level to control another circuit with enough power to activate such a large relay. <S> You make it more difficult to answer your questions when you do not include the working voltage. <S> It limits us to generic answers which may not be applicable to your situation. <A> There are plenty of relays which switch a common connection (your output) between a normally open and normally closed terminal. <S> Relays generally handle lower current applications so you might be better off looking at contactors, which are just larger versions of relays with more durable contacts. <S> If you're unable to find a contactor with NC and NO contacts on the one unit, you can also find mechanically interlocked contactors which only allow one contactor to be energised at any one time. <S> This would be the safest option if you are concerned with component failure and don't want your supplies to be crossed. <S> Your proposed signal voltage to the relay/contactor coil of 3.3V could be a problem. <S> Stacking relays - having a relay with a 3.3V coil switching a 24V or 240V supply to the main relay/contactor - would get around this but brings more complexity and areas for component failure into the design. <S> Also, if you are considering this for your household switchboard, be aware that some regulators require solar panels and other off-grid power supplies to be controlled by a particular switch in your switchboard. <S> These are designed to prevent off-grid power sources from feeding the grid when the grid is down after a storm etc. <S> Linesman could go to work on the grid, thinking it has been powered down, only to find voltage there, preventing them from working safely. <S> Galco.com sells ABB mechanically-interlocked contactors with 24VDC coil and contacts rated to 105A for the low, low price of $743.42 (probably not including shipping...) <S> Item# AE75M-30-11-81 <A> 100A is fairly high for a contactor or relay. <S> You could, in theory, use two 100A contactors, but I would assume fairly bad things could happen if both sources were connected together, so you probably want a device that has a mechanical interlock to make that very unlikely (also called 'break before make'). <S> It's very unlikely you will be able to find a 100A device that will operate from 3.3V directly. <S> Most likely it will draw a fairly large amount of power (a watt or more) from a higher voltage supply such as 12V. <S> You can still use a 3.3V control signal <S> but you would need a driver. <S> There are '100A'/'120A' Chinese and automotive-style relays on offer that are SPDT such as JQX-62F-1Z/2Z <S> (Chinese part numbers are common to many makers, with varying ratings, approvals and quality). <S> I would have doubts about their ability to carry 100 or 120A continuously- <S> Chinese amperes tend to be on the small side. <S> Coil resistance (12V) is stated at 130\$\Omega\$ so you'd need about a 100mA driver and 12V supply. <S> Something with serious ratings would probably draw more. <S> Higher voltage DC usually uses magnetic blowout to extinguish an arc, but at 12-28V <S> it's not necessary.
| You will not find any relays that can switch 100A Try searching for a transfer or changeover relay. Switching such a high current will require a large and expensive relay.
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Scattering Parameters and different load I am somewhat confused as to the correct usage of s-parameters. Looking at the problem statement below, would the S-parameters given need to be "transformed" due to the load impedance being different than the reference impedance for calculation purposes? In other words, can calculation be performed using the given values to find voltages or power at port 2 directly or do the values containing a_2 or b_2 need to be "transformed" to the correct reference impedance? I hope this makes sense. I have skimmed various literature on the subject but most seem to only deal with matched loads, perhaps someone could clarify or guide me towards literature which deals with such cases. <Q> would the S-parameters given need to be "transformed" due to the load impedance being different than the reference impedance for calculation purposes? <S> No, the S-parameters don't need to be transformed. <S> Since the load \$Z_L\$ doens't match the reference impedance \$Z_0\$, it will have a nonzero reflection coefficient. <S> This will combine with the part labelled as "two-port network"'s S-parameters to produce a different net reflection looking in to the 2-port than if it were loaded with a matched impedance. <S> You could, of course, alternatively transform everything to a different reference impedance matching the load. <S> The result would be the same. <A> There are different set of matrices of parameters for different purposes on 2 ports. <S> S-parameters Scattering of loss or gain relative to a fixed impedance <S> it is always referenced to a fixed Z eg. <S> Zo = <S> 50 or 75 Ω (video) <S> any mismatch in Z causes a loss by reflection. <S> it is calibrated with f for 3 impedances Zo=50.00, <S> Short = 0.000 Ω, Open = <S> ∞ <S> Y-parameters <S> Admittance H-parameters used for Transistors <S> T-parameters used for cascading 2 port parameters easily and supported in Spice ABCD-parameters <S> V <S> / <S> I ratios aka:chain, cascade, for passive elements and transmission line <S> parameters <A> When all of your ports are specified in terms of S-Parameters, it's simple to combine them using the standard mathematics for that system, Mason's Gain Formula (wikipedia). <S> You're told the system reference impedance is 50ohms. <S> That means that all the parameters are with respect to 50ohms. <S> You are only given S-parameters for the two-port. <S> Fortunately it's trivial to get the S-parameters of the source by inspection, as you are told the output impedance is 50ohms, meaning it matches the system impedance, meaning that S11_G is 0. <S> The load impedance doesn't match the system, so S22_L will be non-zero. <S> I'm not going to do that for you, as I'm sure that grinding through the little bit of algebra to turn impedance to reflection to S-parameter is one of the exercises you will be doing as part of the course, but of course google will be there with a plot-spoiler if you want. <S> Once generator, two-port and load are all defined in terms of S-parameters, then they can be combined by Mason's. <S> The other interesting thing you can do is combine them pair-wise. <S> So for instance combining the generator+two-port gives you the output impedance that's driving the load. <S> Conversely combining the two-port and load gives you the effective load that the generator sees.
| So you do have to transform the load's impedance into an S-parameter for its reflection.
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Wiring Parallel and Series Simultaneously Is it possible to connect 3 sealed lead acid batteries in both parallel and series at the same time like in the diagram below? <Q> A 4P2T could switch the 3 batteries from (series 36v) to (parallel 12v). <S> simulate this circuit – <S> Schematic created using CircuitLab <A> NO. <S> If you look closely at your diagram, you will see that your green links form a DEAD SHORT from GND to 12V. Clearly not viable at all. <S> Don't even think about it. <A> You'd need to be careful to prevent energising both relays simultaneously, achieving a dead short as pointed out in a previous answer. <S> To do this you could wire the signal to Coil 1 through a NC (Normally Closed) <S> contact on the opposing relay. <S> When you energise Coil 2, the NC contact will open, preventing current travelling to Coil 1.
| To achieve this you would to wire all the connections back to a couple of relays and just switch between them as needed.
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Structure of a usb 2.0 connector What is the role of the outer metal connector? (the one with two square holes)Does it have a potential? Is it insulated from the 4 pins? Are the 4 pins insulated from each other? <Q> The metal shroud around USB connectors is called "shield". <S> The shield serves two purposes, (1) To protect from over-the-air ESD events, and (2) to shield internal high-frequency noise from being emitted out and meet emission regulations for EMI levels. <S> These two processes have different electrical characteristics, so the treatment of shield connection must have somewhat more sophisticated handling than just grounding. <S> It also depends whether this is a portable device, or a stationary equipment. <S> (1) ESD event is a high-voltage one-time discharge pulse about 50ns long. <S> The purpose of shieled is to intercept the plasma filament, and route the discharge current (10 <S> A-50 A) <S> aside from the signal ground at pins of ICs. <S> Therefore the best way is to ground the shield solidly to system ground plane and chassis. <S> (2) EMI: USB uses high-speed signaling, which employs internal switching frequencies of 480MHz and higher harmonics of it. <S> Even if the external signaling is differential over a shielded bi-axial cables (which is supposed to cancel direct emissions), inner workings of digital electronics and unbalanced return currents create so-called "ground bouncing". <S> In essence, the digital signal ground in the device is noisy, bouncing. <S> If the shield is directly connected to this ground, the conductive braid along the entire USB cable will emit as a good antenna. <S> Therefore, the shield should be disconnected from digital ground. <S> As one can see, these requirements are contradictory. <S> The industry solution is to use a de-coupling circuit between the shield shroud and signal ground. <S> Different manufacturers recommend slightly different solutions for this filter. <S> Googling for something like [EMI shielding and ESD protection of computer interfaces] will give a lot of recommendations, LC, RC, etc. <S> My preference is to use a 0.1uF ceramic cap 0603 size, with a resistor of about 330 Ohms to provide galvanic path. <S> The filter works on specifics of this particular size of ceramic caps. <S> These caps have an impedance of capacitive type at frequencies up to 10-20MHz (so a 50ns pulse gets well-coupled with ground plane and dissipates in power supplies), but at frequencies above 50MHz it becomes an inductor, so if provides a good de-coupling of cable shield from the noisy digital ground. <A> The cable will have either a woven wire sheath that is connected to the metal connector parts OR the cable will have a conductive foil wrap that is intermingled with a single wire bare wire that connects to the connector shield part. <S> When the cable is plugged into the mating equipment the cable connector metal shield contacts with the like shield of the connector on the equipment. <S> The shields of the equipment connectors are connected to the chassis or internal ground system of the product. <S> One of the purposes of this shielding connection system it to keep the two attached systems at equal potential. <S> Most often the connection of the equipment connector shield is to the same internal circuit node as the USB GND wire. <S> Careful design is often used to arrange things so that the shield shunts external electrical disturbances, such as electrostatic discharges (ESD), directly to the chassis in the shortest path possible to a possible earth wire connection. <S> This shunting attempts to keep as much of the high voltage ESD out of the internal circuitry as possible where it could cause damage to the electronics. <S> Another purpose of the shielding is to help prevent internal RF signals in the products from being emitted out on the cable. <S> It is often necessary to arrange the product USB connector shields a part of an EMI containment system to prevent RF leaks from going over legal limits. <A> The outer metal connector is the USB connector shield. <S> It will shield the wires from EMI. <S> Note that the shield should not serve as connecting wire, i.e. should just be grounded by one end.
| The outer metal shield of the USB connector is part of the cable screening or shielding.
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How hot should things be getting when I measure them with an ammeter? I recently got a little LED voltage and current meter and I've been playing around with it. When I hold down both of the current measuring leads to the battery, the negative terminal heat up significantly - to hot to touch! Since I've been testing lithium based batteries my guess is that the heat is due to the high current, or due to me messing something up, maybe even making a short circuit. Does anyone have any insight into why this could be? <Q> Don't do this! <A> A meter in current mode looks like a short circuit. <S> So connecting it across a battery with no other components will be bad for battery and meter. <S> Try connecting it in series with a load such as a lightbulb / LED. <A> The only way to test capacity of a battery without internal sensor(s) is to charge and drain it and observe how long you can extract a given current out of it. <S> More sophisticated battery chargers do this for you. <S> As others pointed out, shorting battery is double minus ungood. <S> Even when used in a reasonable way a multimeter will get hot in current measurement mode. <S> So you shouldn't leave it alone for a long time if you're measuring significant current. " <S> Significant" depends on the meter but whatever the meter is rated for (10A etc) will most likely overheat it over time.
| If you're using an ammeter across a battery then you are essentially shorting it and it will get very hot, and possibly cause damage.
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Can I increase current using AC resonance? In a resonance, specially if we resonant at the natural frequency, the generated waves' intensity will add up thus resulting in a wave of ever increasing intensity. This phenomena is responsible for taking bridges down, when the wind reaches the resonant frequency of the bridge material, and the wind force starts adding up to the already resonating waves and the bridge material fractures and collapses. I wonder if we could do the same for electric current. Given an AC voltage source (not a very high voltage), with constant upper and lower boundaries for the source AC voltage, could I feed it to a "resonant circuit" such as the current measured by the amperemeter over time would be like: And use this as an artificial way to increase the current in a load? (knowing that this would probably be the recipe to burn down everything for excess of current) <Q> If the passive resonant circuit has a high 'Q' then energy can build up in the circuit over many cycles. <S> It will similarly die down naturally over many cycles. <S> The intensity is not "ever increasing", however. <S> As the intensity increases, so do the losses and at some point the losses equals the input power, so you have an equilibrium. <S> Imagine a tuning fork. <S> If you keep exciting it, the vibrations will build up (as will the losses) and at some point the metal shape might change if the excitation is powerful enough, but it's unlikely. <S> Most real circuits have a rather lower Q than the mechanical Q of a tuning fork. <S> In a real LC resonant circuit, the losses are usually due to inductor resistance, to core losses (if a core is used) and to electromagnetic radiation, especially at higher frequencies. <S> Capacitor dielectric losses contribute too. <S> Superconducting circuits can have huge Q's (in the thousands), resonant circuits made with parts from your favorite distributor, much more disappointing. <A> Yes, resonance works with voltage or current too. <S> Look up something called a tank circuit . <S> This is a inductor and capacitor in parallel. <S> With ideal components, the circuit stores energy as a sinusoidal voltage and current together. <S> It constantly sloshes the energy between the cap and the inductor. <S> This happens at the resonant frequency, which is the frequency where the impedance magnitude of the two parts are equal. <S> Again, for ideal components, this energy has no place to go once put into the tank. <S> If you keep adding more energy, then the total energy in the tank increases, and the amplitudes of the voltage and current have to too. <S> The total energy is proportional to the square of the voltage or current. <S> The magnitudes of the impedances of a capacitor and inductor are: Z cap = <S> 1 <S> / ωC <S> Z ind = <S> ωL <S> By setting these equal, we find that ω = 1 / sqrt(LC) <S> f = 1 / 2Π sqrt(LC) <S> When L is in Henries and C in Farads, then f is in Hz. <S> So far this has been with ideal components, which are unfortunately difficult to obtain. <S> Real components have real losses. <S> These are mostly due to the resistance of the wire in the inductor, and losses in the ceramic of the capacitor. <S> One way to quantify the lossiness of a tank circuit is with the Q factor , which stands for quality factor . <S> Higher values mean less loss. <S> The ideal tank has a infinite Q factor. <S> The reciprocal of the Q factor is related to the fraction of the energy lost each cycle. <S> Real tank circuits can have Q factors in the several 100 range. <S> You can build something yourself rather easily with a Q factor in the 10s range. <A> You can make the current in a coil very high compared to the input current if the coil is part of a low-loss tuned circuit: - <S> Here we have a 1 uH coil resonated at about 300 kHz. <S> The 1V RMS voltage source feeds through a 10 nF capacitor (C1) and <S> the peak voltage on the inductor is 36.5 dBV (66.8 V RMS) at resonance. <S> The voltage across the 1 uH coil implies a current of 35.4 amps. <S> This current must also flow proportionally through C1 and C2 but, because C2 is 27.2 times bigger than C1, the current supplied by the voltage source is only about 1.25 amps. <S> At the end of the day, the AC voltage source is stil supplying power to warm up the resistor that is inevitably in series with the coil. <S> Also, if you tried to extract power from the coil's magnetic field, that power would also need to be supplied by the AC voltage source and you would find that the resonance peak would start to drop because of the Q of the tuned circuit falling. <A> I found something: http://www.richieburnett.co.uk/resonant.html <S> I'm not a scientist, but to say it is impossible to achieve free voltage gain by resonance <S> it's limiting. <S> We should accept this as an idea, and try to prove it right, rather than wrong <S> (it's easy to say anything is impossible and wrong). <S> This would be huge discovery, so saying that "it cannot be done" should be replaced by, "we don't know how it could be done", and scientists should keep experimenting.
| The author says that this is possible but, it would require some engineering to prevent voltage burst in milliseconds, and to achieve stable voltage gain.
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Difference between 110V and 220V power outlet? What exactly is the difference between a 110v and a 220v labeled power outlet? As far as I can see, the outlet is just a piece of plastic with some metal connecting to the power cables. Both types of outlets seem to be identically built. Does it pose a problem to mount a 110v power outlet on a 220v power cable? To clarify, the plugged in devices are using 220v as the power system provides, the question is really just about the power outlet itself. <Q> Get the right outlet, so the device-frying dunderhead isn't you! <S> The difference between a 110V ( <S> NEMA 1 or NEMA 5) and a 220V (NEMA 6) outlet is in the arrangement of the contacts. <S> Most 110V outlets can survive 220V across them (and are sometimes expected to in normal operation i.e. when the two outlets are on opposite legs of a multi-wire or "Edison" branch circuit), and certainly a 220V outlet will survive having a paltry 110V across it, but they are configured differently to keep dunderheads from plugging the wrong thing into the wrong outlet and having it fry due to the wrong voltage -- this was obviously a much bigger deal in the days before universal input power supplies, of course, but there is still plenty of stuff around that'd emit magic smoke if fed grossly wrong mains voltages. <A> Yes, the outlet connector is just plastic and metal. <S> And it is quite likely that most any outlet made for 110V could handle 220V. <S> And yes, you could connect 220V to an outlet designed for 110V (and vice-versa). <S> HOWEVER, the problem comes from the people who are USING the outlet. <S> If an outlet has two flat slots for a 110V standard <S> but it is wired to 220V <S> , then when a user comes along and plugs in a 110V gadget, it will blow up (or burn up) from double over-voltage. <S> But for practical purposes in the Real World, it is a REALLY REALLY TERRIBLE IDEA. <S> And it is likely illegal as well. <S> Conversely, it is also not a good idea to connect a 220V outlet to 110V. Because many devices don't tolerate 50% UNDER voltage very well, either. <S> It is not clear WHY you are even asking this question, but I would recommend that you not pursue this line of thinking as it can only lead to disaster. <A> It's definitely a bad idea to wire a 110V outlet with 220V. <S> In particular, 220V is usually accomplished in the US using two hot wires, a neutral, and a ground. <S> But 110V outlets only have a connection for one hot; you really don't want to be putting that extra hot on the neutral or ground connections of a 110V outlet because very bad things will happen. <S> But even if you somehow have a single wire with 220V to ground, if you were to wire that up, the device could melt, short circuit, burn, and just generally be a major safety hazard. <S> (This depends a bit on what country you're in. <S> I'm guessing North America, since that's the main market with this combination of voltages.) <S> The correct outlets look very different depending on the voltage. <S> In particular, the most common 220V outlets (NEMA 6-15) have horizontal pins where normal 15A/110V outlets (NEMA 5-15) have vertical pins. <S> Maybe you're mistaking a 20A outlet for a 220V outlet? <S> It would be safe to put a 20A outlet on a 15A circuit, but you'd get circuit breaker trips if your device tries to use 20A. <S> But, again, it's not safe to put a 15A outlet on a 20A circuit because the outlet (and presumably anything you plug in) isn't designed to handle the current. <S> In any case, you may get more useful discussion on the diy stackexchange . <A> As voltage increases better insulation is needed . <S> In an outlet you have two kinds of insulation. <S> One is a number of air gaps between various parts. <S> Another is a number of parts made of non-conductive materials - plastic or ceramics. <S> Both must be designed for higher voltages. <S> Distance between the outer surface of the outlet and the closest contact parts and stripped wires and screws or other parts holding those wires must be larger. <S> Distance between any parts which are not electrically connected must also be larger. <S> This also implies that the smallest possible outlet must also be larger. <S> Outlet parts made of non-conductive materials must be made thicker to account for higher voltage. <S> It's unlikely that you actually face a situation where these differences cause problems in any reasonable scenario but possibility still exists. <S> If anything goes wrong because of those differences it'll be your fault just because you used a part with the wrong voltage rating.
| So mechanically, electrically, it is probably no hazard to connect a 110V outlet to 220V. Obviously the difference is that the two are designed for different voltages .
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Can connector pins oxidize while mated and lose contact? Can oxides build up enough to force contacts to disconnect? <Q> It depends on the connector. <S> So-called gas-tight connectors should remain free from oxidation indefinitely. <S> Turned-pin IC connectors are a typical example. <A> I've seen it happen, on occasion. <S> Faster, when the humidity is high. <S> If the alloy or metals are dissimilar, in the sense that their anodic indices are different, there will always be a galvanic potential present. <S> In the presence of an electrolyte (tiny amounts of water from the air, for example), this potential can drive ions as well as provide energy for chemical changes. <S> Each metal is a special case, though. <S> So there is no specific "bright line" that will tell you exactly what happens in every circumstance. <S> Aluminum, for example, oxidizes into what amounts to "transparent aluminum" or sapphire. <S> Abrade any piece of aluminum in the presence of oxygen in air <S> and it will almost immediately form a thin layer on its surface. <S> Given enough time to act, it grows thicker and thicker and it can become strong enough to resist scratching to a degree as well as to insulate. <S> Iron, on the other hand, develops a very soft iron-oxide with entirely different characteristics. <S> (Not that pure iron is ever used in circuit connectors!) <S> So... <S> It's possible. <S> I've seen it happen with instruments kept untouched in a remotely located electronic shack where it was no more than re-seating of a board or two (to remove some of the accumulation) to get things back running. <A> Yes. <S> There are several kinds of conditions where mated connectors, even mechanically un-disturbed can develop conditions (not just oxidation) that cause them to become intermittent or even fail completely. <S> Even satellite-grade connectors sometimes fail.
| There's no question that pins can oxidize and that the build-up, perhaps combined with minor movements, can lead to an eventual increase in resistance, leading to possible additional heating at the connector, leading to more oxidation, etc.
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Why does this 7-Segment Display not function properly? I've provided details in image. Although, I'll once again provide in text: IC : 74LS47N (Binary to 7-Seg Display) 7 Segment : Common Anode M.S.B : D L.S.B : A This is how absurd OUTPUT is displayed by 7-Seg Disp: INPUT (D,C,B,A to IC) ___ OUTPUT (7-SEG DISP) DCBA 0000 _______________ 2 0001 _______________ 3 0010 _______________ 2 0011 _______________ 3 0100 _______________ b 0101 _______________ 7 0110 _______________ b 0111 _______________ 7 1000 _______________ c 1001 _______________ ɔ <--- (-_- !) lol I've edited the pics, so that you may see the wirings clearly.As you can see the problem in the image itself,(i.e. the output of 7-Seg does not matches with the input to the IC). Note: There is no problem with my connections. They are exactly as shown below: I did the connections 5 to 6 times all over again, but all it shows is absurd and illogical results <Q> The 74LS47 shows the following:$$\begin{array}{ccccl}D & C & B & A \\0 & 0 & 0 & 0 & \rightarrow 0 \\0 & 0 & 0 & 1 & \rightarrow 1 \\0 & 0 & 1 & 0 & \rightarrow 2 \\0 & 0 & 1 & 1 & \rightarrow 3 \\0 & 1 & 0 & 0 & \rightarrow 4 \\0 & 1 & 0 & 1 & \rightarrow 5 \\0 & 1 & 1 & 0 & \rightarrow 6 \\0 & 1 & 1 & 1 & \rightarrow 7 \\1 & 0 & 0 & 0 & \rightarrow 8 \\1 & 0 & 0 & 1 & \rightarrow 9 \\1 & 0 & 1 & 0 & \rightarrow <S> c \\1 & 0 & 1 & 1 & \rightarrow ɔ \\1 & 1 & 0 & 0 & \rightarrow u \\1 & 1 & 0 & 1 & \rightarrow <S> 3~horizontal~lines \\1 & 1 & 1 & 0 & \rightarrow broken~6 \\1 & 1 & 1 & 1 & <S> \rightarrow \end{array}$$ <S> You show: $$\begin{array}{ccccl}D & C & B & A \\0 & 0 & 0 & 0 & \rightarrow 2 \\0 & 0 & 0 & 1 & <S> \rightarrow 3 \\0 & 0 & 1 & 0 & \rightarrow 2 \\0 & 0 & 1 & 1 & \rightarrow 3 \\0 & 1 & 0 & 0 & \rightarrow <S> b \\0 & 1 & 0 & 1 & \rightarrow <S> 7 \\0 & 1 & 1 & 0 & \rightarrow b \\0 & 1 & 1 & 1 & \rightarrow 7 \\1 & 0 & 0 & 0 & \rightarrow <S> c \\1 & 0 & 0 & 1 & \rightarrow <S> ɔ\end{array}$$ <S> This makes me think your B term is hard-coded to a 1 and that you are misinterpreting a 6 as a b. <S> Check your B term going into the 74LS47. <S> If it is working, then I think your 74LS47 has a damaged B input. <S> EDIT: <S> Good to hear that you got it working! <A> From your table, it's pretty obvious that bit 1 is stuck on. <S> Note that 0000 displays like 0010, 0101 like 0111, etc. <A> It looks as if your breadboard is not wired properly. <S> The Vcc and GND rails are in the middle of the board connected horizontally only once. <S> You have to remember, that the two rows of the board are not connected by default. <S> You have to place a wire between the two rows of each supply potential. <S> Especially your Vcc and GND connections in the lower left of the board seem not to be connected to ground.
| This is likely a wiring error, probably a floating input.
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Ultrasonic Sensors and Pets I've been eyeing using an Ultrasonic distance sensor for an upcoming project; either the Parallax Ping or the OSEPP Ultrasonic sensor (I think they are identical other than branding). Ex: https://www.parallax.com/product/28015 My concern is, this will be used at home where I have pets. I don't wish to injure nor annoy my pets. Will the ultrasonic sensor cause problems for the pets? Should I instead use something different such as an infrared sensor? (I'm mainly concerned about cats and dogs, although would be curious about other types of pets such as birds, etc...) <Q> That part appears to use a frequency of ~40 KHz, fairly normal from what I've read for ultrasonic detectors or that sort. <S> According to this chart on Wikipedia , that falls within the hearing range of quite a few animals, including cats and dogs. <S> Does that mean using that part will be painful? <S> Bothersome? <S> Just slightly annoying? <S> I dunno. :) <A> You're concerned about it being too loud for things that can hear it. <S> Well, at 5V/35mA, the sensor that you mentioned consumes an average of 175mW while operating. <S> Not all of that gets converted to sound, of course, but we can get a rough idea of what we're working with by assuming that it does. <S> According to this site , 1W of purely acoustic power is 112dB at one meter away from a point source. <S> That is, 1W spread over the surface of a 1m radius sphere would measure 112dB at that surface. <S> Considering that we only have 175mW instead of 1W <S> , that's \$112dB+10dB*log(0.175) <S> = 104.5dB\$, still on the surface of a 1m sphere. <S> Now we have two conflicting modifiers: <S> There are other functions besides the transducer, which itself is less than 100% efficient, so we actually have less than 175mW of acoustic energy to convert to dB. <S> It's actually a spherical section (cone with a round base), not a complete sphere, so the same power is spread over a smaller area and is therefore more intense at the same distance. <S> (in RF, this is called "antenna gain") <S> The relative strength of those modifiers is up to you to figure out, with a few guidelines: "One-note" or resonant design goes a long way towards ideal efficiency. <S> A large transducer compared to the wavelength being produced increases both efficiency and directionality. <S> This is why subwoofers need ridiculous amounts of power and can be heard everywhere while tweeters are pretty much the opposite. <S> What you've got is practically a "supertweeter". <S> For a point source (expanding sphere or spherical section), you lose 6dB for each doubling of distance, simply because the area that it's spread over increases at that rate. <S> For completeness, a line source (expanding cylinder or cylindrical section like a highway or line-array speaker set) loses 3dB because the area increases more slowly, and a plane source (large surface like a wall, ceiling, or floor) doesn't lose any because the area stays constant. <S> But at some distance relative to size, all finite-sized sources in open space eventually reduce to point sources and <S> so they all ultimately revert to the 6dB rule. <A> I knew a bat researcher. <S> When our facility installed occupancy sensing light switches, the bats went absolutely bonkers, and they had to remove all the sensors in that hallway. <S> That said, this is ultrasound, and not many animals hear into the ultrasound. <A> https://www.ncbi.nlm.nih.gov/pubmed/4066516 <S> That says cats can hear up to 85 kHz <A> I'd say, do not use it for anything where usage is continuous. <S> Even a quiet barely-heard noise can be disruptive to sleep patterns. <S> (I should know, I have been left with mild tinnitus by a recent ear infection. <S> I'm told I'll either recover or get used to it, and it could of course have been very much worse) <S> If usage is for brief periods, then if the pets don't react to it by showing fright or leaving the area, it's "just a noise" (if that). <A> I've been using the HC-SR04 sensor to turn on a water supply so my cat can drink from the mains water (he'll only drink running water...). <S> It uses 40khz pulses and doesn't appear to bother him. <S> He'll happily go over to it, trigger it and drink without any signs of distress, his ears don't turn any differently when near it.
| If they aren't clearly distressed, just let them go elsewhere while you use the gadget.
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Using a barrel jack with higher than rated current but lower voltage? I have a 5V/10A power supply (with a 2.1mm DC plug) and want to use it to power four or five 5V/2A devices. But I see that no barrel jacks are rated for 10A (they're all around 5A max - see this Digikey search ). Is this really an issue? I would have thought parts likes jacks, that are basically little more than a metal conductor, would be rated for e.g. a max wattage. But I see on places like Digikey that instead they're rated for a max voltage and a max current (12V and 5A seem typical values). It seems odd to have a power supply with a 2.1mm DC plug if you can't actually get a jack that can handle the supply's current. So is it OK to use a 12V/5A rated jack with a 10A power supply as long as the voltage is half the max rated? If it's not OK then what are my options and if it is OK is there anything else I need to watch out for? I was planning to just twist the wires of my various power cables together and screw them down in a 2.1mm jack-to-screw terminal block. Or solder down a barrel jack on a proto board and then solder each power cable on separately to the same board. These kinds of setups have always worked out for toying around with typical 5V/2A setups but will there be issues once I start dealing with higher currents? Parts I'm using: DC Plug Cable Assembly 2.5mm http://dn.odroid.com/homebackup/201401132011213194.jpg <Q> No, it doesn't work the way you think. <S> At these levels, the voltage and current ratings are independent. <S> The current rating comes from the contact resistance and the power the actual contact (way smaller than the whole connector) can safely dissipate. <S> The voltage rating is what the insulation is good for. <S> Multiplying the voltage and current ratings tells you the maximum power that can be pushed thru the connector. <S> It has nothing to do with what the connector will dissipate. <S> That's soley a function of the current and the contact resistance. <S> In fact, it's exactly the square of the current times the contact resistance. <S> If you want to use the full output of your 5 V 10 A power supply, you have to use a connector rated to 10 A. <S> You might ask the manufacturer of the power supply to tell you a specific mating connector it is intended for. <S> Surely they know. <S> The wire you show looks suspiciously thin for 10 A, at least with reasonable safety and loss. <S> Basically, your supply is like a 400 horsepower V8 for a Honda Civic. <S> You can't just plug in either one without dealing with other issues. <A> Current rating is mostly based on the power dissipated by the contact resistance. <S> P = <S> R <S> * i 2 <S> So, the heat generated on a connector is about the current, not voltage. <A> The voltage rating depends on insulation strength, while the current rating depends on contact resistance and construction. <S> You may be able to use a 5 Amp connector for somewhat higher currents, but it may overheat. <S> I don't recommend it. <A> In short, no. <S> Use it to power at most two 5V/2A devices, or split the power supply before feeding it to your devices. <S> Also, careful when using breadboards: the current shouldn't exceed 0.5A there, see the post: <S> How much current can Solderless Breadboards handle? <A> To add another perspective, the 2 main tests ran at safety certification labs such as UL are the temperature test and the dielectric test. <S> The temperature test requirements rely strictly on the current rating; the connectors are loaded with the maximum rated current until they are thermally stable (usually about 4 hours). <S> Temperatures along the connector are measured and checked against their maximum thermal rating. <S> The test voltage is irrelevant and is often any convenient voltage for the test lab to get to the test current; a 50 V connector and 5 V connector may both be ran at 5 V. <S> The dielectric test is performed at a voltage strictly calculated using the rated voltage of the device.
| The voltage and current ratings on connectors are independent. It also has to be rated for 5 V, but it would actually be difficult to make a connector that isn't rated to at least 50-100 V, so the voltage spec is not a issue here. So hopefully this shows that the ratings are independent of each other from a safety perspective. This not only includes things like not melting the insulation, but also oxidation and other heat-related degradation over time. Here, the current rating is irrelevant to the test parameters; a 1 A device will be ran at the same voltage as a 100 A device. Yes, this is really an issue. Or they know there isn't one, but that consumers aren't going to think of that, and think a 10 A supply is "better" than a 5 A supply.
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Female Lug connector in Eagle I'm working on a pcb design on eagle, which will be used to test some sensors. The design is made for three different sensors. You connecet the sensor to the pcb, do the test and disconnect. Here's one of them: All three sensors have different connectors, two of them use female headers, therefore i was able to find male headers the size i need. The other one uses lug fork terminals. I need to connect the lug terminals to the pcb to be able to test the sensor. My question is; How could i do this? could i use a header? if so what type? Thanks!! <Q> You can use a PCB-mount screw connector such as this one from Keystone (photo courtesy of Digikey): <S> They are available in M2.5/M3/M4/M5/6-32/8-32/10-32. <S> Refer to the relevant datasheet for the recommended hole sizes. <S> You may also find this type useful if it is available in your desired size: <S> This kind is mounted right to the PCB <S> so it requires more area, but it is made with an anti-rotation tab that engages the open end of your spade lug. <S> Also by Keystone. <S> Here, from this ebay listing, is a typical application (HVAC PCB) of the second type <S> (some other connections are male spade connectors soldered into the board). <A> You could use a terminal block of the appropriate pitch to secure your lugs. <S> The latter option won't last for as many insertions as the terminal block, I would think. <A> Use Images on any search engine for the correct keywords screw Terminal Block. <S> They come in all sizes and shapes. <S> Then you must add to the library with schema and pads <S> What size are the lugs? <S> http://www.ebay.com/itm/300V-30A-9-5mm-Pitch-4-Pin-Pluggable-Type-PCB-Screw-Terminal-Block-20Pcs-/321445165890?hash=item4ad79ffb42:g:lNcAAOSwRgJXjTYm <S> I would use compact terminal blocks for automated stripped fly-lead wire, to save cost, PCB space. <S> improve appearance and risk of shorts with no final assembly tools required. <S> (<=$0.10/contact) <S> http://www.digikey.com/product-detail/en/phoenix-contact/1990038/277-1797-ND/950925 <S> Although quick disconnect has not been defined as a requirement yet. <A> For a Sensor Test Jig , I would rigidly secure Pomona Alligator clips to an external fixture and interface to board with std ribbon header with interlaced ground for reducing crosstalk and for ease of maintenance and connection to spade lugs with the best orientation for the operator.
| Alternatively just add a hole for a screw with enough exposed copper around it to make contact. You will have to leave sufficient space between the connectors since there is nothing preventing the lug from rotating as there would be with a barrier terminal strip that others have suggested. Be sure to pick the correct size for the lugs you want to use them with.
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Anyone know the premise of this PCB assembly note? The assembly note states: Bake bare PCBs in a clean and well ventilated oven prior to assembly at 125*C for 24 hours. Why is this necessary? <Q> It's clearly to get rid of moisture, probably to keep the steam from pushing BGAs or CSP packages off the board (maybe from moisture trapped in tented vias, microvias or other places), but I didn't look into all the possible reasons deeply. <S> You might want to be a bit careful with this- 24 hours exceeds the baking guidelines in <S> IPC-1601 , so solderability may be adversely affected unless it's ENIG. <A> Usually baking for long periods of time like that is to eliminate moisture in the product. <S> If there is moisture inside the PCB, the quick ramp up to ~250 C during the reflow process can cause this moisture to vaporize quickly and, if this moisture is trapped anywhere, explode. <S> This is bad. <S> Its best to follow the instructions, especially since it is such an easy step to follow, to prevent damage to the PCB, and to ensure a longer lasting board. <S> Related, but not the same. <A> Contamination due to copper oxide and moisture make solderability poor. <S> This is why PCB's are sealed in plastic in storage for production, otherwise take heed with notes.
| A dry PCB also makes better connections that are more resistant to corrosion.
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Why does an LC-filter need two different capacitors and how do we calculate their values? I am currently tasked with reverse-engineering a buck converter based on the LM2596 adjustable switch. On the PCB that I have, there is a following filter circuit (redrawn by me): I did some research, and it seems that the two capacitors are needed for better filtering . However, I cannot find any information on how those capacitor values are related. Thus, the question - how can we calculate those capacitor values if only the switching frequency (150 kHz) is known? Are they really needed for filtering or did I make a mistake? Thanks in advance! EDIT: Here is the full schematic: <Q> Large-value electrolytic capacitors are very good for bulk smoothing to produce DC. <S> However adding a small-value poly or ceramic capacitor in parallel is often used to shunt away ("filter") high-frequency ripple as from a switch-mode power supply (SMPS). <S> It is rather common that those values are empirically derived rather than theoretically calculated. <A> In order to perform a remotely representative simulation you will need to chase down the part numbers of the capacitors used and find the ESR for each. <S> You will find that the ESR of the smaller ceramic capacitor is much less than that of the electrolytic capacitor, however the capacitance value will be much lower. <S> So you really have two hidden resistances and two capacitors. <S> You will find that the two caps cooperate in reducing the output ripple. <A> The LM2596 recommends this: - An output capacitor is required to filter the output and provide regulator loop stability. <S> When selecting an output capacitor, the important capacitor parameters are the 100-kHz ESR, the RMS ripple current rating, voltage rating, and capacitance value. <S> For the output capacitor, the ESR value is the most important parameter. <S> So, in your example it looks like they have used a lower quality electrolytic shunted by a ceramic to ensure that at high frequencies the LM2596 operates correctly i.e. the output is stable.
| Low impedance or low-ESR electrolytic or solid tantalum capacitors designed for switching regulator applications must be used.
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Output frequency unstable in 555 oscillator I'm trying to make a simple astable NE555. The output frequency increasing alone. For example it start from for example 4.2 kHz and immediately change to 4.3 kHz, 4.6 kHz, 5kHz, then play around 4.5 kHz up and down. I tried different NE555 IC like TLC555 and a different 9V battery, same problem. I am using breadboard. C1 is a ceramic capacitor. I measured the frequency using a PIC microcontroller based frequency counter. <Q> And this is why Class II ceramics make lousy timing caps <S> What you have discovered is one of the major application limits of what are known as Class II ceramic capacitors -- they can't be used for timing. <S> This is because the dielectrics used vary in dielectric constant, and thus the capacitor varies in capacitance, as the voltage across the capacitor changes <S> -- this effect is known as voltage coefficient, or just voltco for short, and is the main reason for voltage derating of these capacitors in their application domain as well. <S> How do you identify a Class II ceramic? <S> And what do you use instead? <S> Class II ceramics are easy to identify from their temperature coefficient rating -- <S> all X?? <S> (such as X7R and X5R) <S> ceramics are Class II, while C0G (NP0) ceramics are Class I, i.e. temperature and voltage stable. <S> However, C0G capacitors are not available for values greater than 10nF as a practical matter, so for larger values, film caps (polyester/Mylar, polypropylene) are used instead when capacitance stability with voltage is a concern. <A> 100uF/10V is good. <S> The 555 derives its switching thresholds with a resistive divider from the power supply and any noise (internally or externally caused) during the cycle will change the timing. <S> Use a 'clean' power supply with low impedance, such as from a dedicated regulator. <S> It has a low voltage coefficient with static voltage changes, but not if it's bouncing around during the cycle. <S> There is also the voltage coefficient issue which @ThreePhase points out, which will mostly lead to non-ideal waveforms but will change the timing if the supply voltage changes. <S> Secondly, the X7R type caps have a fairly large temperature coefficient <S> so small changes in temperature can lead to timing changes (graph from here ). <S> As you can see, a 1°C change can lead to almost 0.1% change in frequency. <A> This might be way too late for you. <S> There is a capacitor missing in your circuit - <S> there should be, at least , 100nF over from #1 to #4/#8. <S> The 555 has a quirk in which it consumes lots of mA <S> when switching its output. <S> The internal resistance of the 9V battery can't cope with that. <S> Better is putting a 10uF in parallel with 100nF, near the 555.
| First, you must have a bypass capacitor across the 555 power pins, especially for the non-CMOS type.
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multiple planes (fills) in eagle I have some connections in my single-sided PCB which could benefit from a plane especially all connections to power and all connections to ground. I can easily create a polygon for only one plane (for example to intelligently flood the remaining bare spots (areas with no copper) of the PCB with ground), however I want to also be able to fill applicable bare spots with power (vcc). I tried creating a polygon named gnd then on top I tried creating a polygon named vcc and after running rats nest, it seems that the gnd and vcc are shorting each other. If I play around with the rank values, then either gnd or vcc will fill the bare spots, but not both. Is there a way I can have both a gnd and vcc plane without resorting to non-overlapping of plane creating and without making each plane use only a small portion of the board? <Q> I have some connections ... which could benefit from a plane ... especially all connections to power Quite unlikely. <S> Ground planes are certainly useful, but most instances of power planes are due religious convictions rather than physics. <S> What makes you think you need a power plane? <S> Most likely you don't need a power plane at all. <S> Instead make sure <S> that each use of power is well bypassed to the ground plane. <S> That gives you low impedance at high frequencies. <S> Fat enough traces give you low resistance at DC. <S> However, to answer your question, the RANK setting in Eagle is exactly for the purpose you seem to be describing. <S> With the right relative rank settings, the power polygon will dominate in its small region within the larger ground plane. <S> From the Eagle help page for the POLYGON command <S> : Defines how polygons are subtracted from each other. <S> Polygons with a lower 'rank' appear "first" and thus get subtracted from polygons with a higher 'rank'.Valid ranks are 1..6. <S> Polygons with the same rank are checked against each other by the Design Rule Check. <S> The rank parameter only has a meaning for polygons in signal layers (1..16) drawn in a board and will be ignored for any other polygons. <S> The default is 1. <S> This really is quite clear. <A> You can't just draw them both as a box and have them fill as required because Eagle won't know which to put where. <S> If you give them both the same rank it will simply fill both in the same area and hence you get shorting. <S> Once you give one a different rank, the one with higher rank (a.k.a. lower number) will take priority wherever they overlap. <S> In other words if you draw two boxes over the top of each other, the lower ranked plane will simply cut out around it. <S> What you need to do is draw the higher ranked plane to roughly the required shape so that it doesn't overlap and hence cut away the lower ranked plane(s) in areas where you want the lower ranked plane(s) to fill. <A> Just remember to NAMe the planes and vias the same.
| In some cases of high current, it may be a suitable way to deliver that current without much of a voltage drop, but then you still have to consider the high current connection points to the plane. You can use vias to add 'disappearing' planes.
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Audible noise coming from Inductor when using LT3760 driver circuit I have an audible noise from inductor when I use LT3760. When PWM is 20kHz, the noise is at 10kHz. Whenever I switch PWM, the noise is at half of PWM. Where does it come from ? <Q> Here is the standard circuit provided by LT: <S> - I note that your circuit does not have 5x 2.2 <S> uF capacitors on the output of your boost regulator (after the schottky diode, D43). <S> I've drawn a red circle around them. <S> This could easily be a significant factor in causing your problem. <S> Try it out. <S> If they are present and you haven't shown it on the diagram then you are naughty. <A> the fact that a 20kHz PWM registers with a 10kHz tone might be due to your measurement device being somewhat sensitive to 20kHz, but interpreting it as 10 kHz (for example, due to aliasing, or intermodulation) by the fact that a PWM has a mean spectral shape with a lot of side lobes that spectrally might contain 10 kHz mathematically – if you have a 20 kHz PWM set at 25% duty cycle, I think I can see a spectral product at 10kHz (too lazy to do the Fourier transform of the signal in my head). <S> the fact that some things just happen every other cycle mechanical nonlinearities <S> In any case, I don't really know what to tell you. <S> As said in the comments, coils tend to squee a bit. <A> Firmly pressing the inductor (with a properly isolated finger) might damp the noise, and allow to confirm that indeed the inductor is the source. <S> Now, if the PWM is at 20kHz and there is a 10kHz component to the noise, it means that the PWM conducts significantly longer at odd cycles than it does at even cycles; that's might be visible at TP56. <S> BUT the PWM is not supposed to be operating at 20kHz in the first place. <S> RT of 60.4k is supposed to give about 700kHz, and 20kHz is off-chart! <S> Most likely, there is some serious instability in the control loop. <S> As rightly pointed in that other answer , the capacitors at the output of the rectifier are missing, and that likely contributes.
| Usually, the component generating the noise indeed is the inductor; that happens by the phenomenon known as magnetostriction . It's there to smooth the current through the inductor when the LEDs are being switched with PWM. I fear it also means that the circuit radiates much more EMI than it should.
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What dictates voltage levels on a PC's serial port pins? I'm currently seeing an inconsistency between the serial ports of 2 PCs I'm working with, and can't seem to hone in on the cause. I'm trying to control an external scanner, which works perfectly fine on one machine, but not at all on the other. All of the pins on both machines have the same voltage level ( 0V on pins 1, 2, 4, 5, 8, 9, and 5V on pin 6), except for 2 -- pins 3 and 7. On the "good" machine, pin 3 is normally at -5V , and pin 7 is normally at 5V , whereas on the "bad" machine, pin 3 is normally at around -11V , and more surprisingly, pin 7 is at around -11V too. I've actually taken the PCI Express to RS232 card out of the "good" machine and tried it in the "bad" machine, but still no luck -- it seems as though the problem is independent of the expansion card (I've tried many). Does anyone have any insight into what is causing the different voltage levels? I'm not as worried about the magnitude, as one PC may be offering more power in this regard, but the RTS pin has different states on both machines. <Q> RS232 has a wide range of operation, anything from 3 to 15V should work, although running at a lower voltage is risky and some poorly made receivers may prefer some voltages over others. <S> The different polarity is a problem. <S> Pin 7 is Request to Send, and it is a control line that means the computer requests the device to send data. <S> It is often ignored, but it is reasonable for the line to idle low when the port is not trying to communicate. <S> Pin 3 is the transmit line. <S> If that had a different polarity it would be a problem because the polarity changes to indicate the start of a byte, but I see that this is not the issue. <S> It could also be a baud rate problem. <A> I would take a terminal programm and test each PC and serial card alone and both together. <S> To test the serial card alone you need an external loop back connection between RxD on pin 2 and Txd on pin 3. <S> If the connection is present, you should see every character you type in the terminal programm looping back to screen. <S> If no external connection is present no character should appear on the screen when typing. <S> The internal loop back option of the terminal programm should be switched of. <S> Configure the terminal programm to disregard the CTS input on pin 8 and DSR on pin 6 and use no flow control. <S> If each PC works alone you could connect both PCs together. <S> Connect TxD of one side to RxD on the other side with two wires for both directions. <S> GND on pin 5 should be connected with the third wire. <S> Start the terminal programm on both PCs. <S> What you type on one PC should go to the screen of the other one and vice versa. <S> Of course both USART must be configured the same way for baud rate, number of data and stop bits and parity bit. <S> Good luck. <A> It would appear you have an OS serial port configuration error. <S> You don't say what sort of OS is involved, but if it's a Windows (7,8,10) <S> variant <S> then you can change the settings in the control panel/devices <S> /Ports(Com&LPT).The other potential problem is that some PC's have BIOS settings that enable/disable Serial and LPT ports.... <S> it could be a setting there.
| It could be that the driver has options for "flow control" which need to be adjusted. I put a lot of blame on the driver here because it sounds like the device works on one PC but not another. It can always be a baud rate problem.
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Why does an analog chip have two ground connections but not two vcc connections? I'll be using an TLC7528 dual DAC IC in my circuit, and it has two pins for ground named AGND and DGND (which to me means analog and digital ground). It only has one pin VDD for power. Why would it only have one pin for power and two for ground? Why don't they make them so I get the AVDD and DVDD and GND pins instead? <Q> The different GNDs are not for powering the different parts of the chip, but for providing different 0V reference points for the chips operation. <S> This way the digital operation of the chip can have wildly varying current flow out through DGND without affecting a rock solid (if you did your PCB layout properly) AGND 0V reference. <A> I will convert my comment to an answer: <S> The actual power paths are in the datasheet: <S> The DAC ladders actually have their own power domain (they share analogue ground): <S> That means that the power paths for the entire device is this: The analogue power path is in green. <A> It has nothing to do with trying to match the number of supply pins with ground pins. <S> A separate ground for the analogue allows a single point connection to digital ground and therefore digital currents cannot enter the localized analogue ground. <A> If you look at the equivalent internal circuit, you will see that the actual D/ <S> A resistor ladder is switched using Agnd so the the output signal will be free from any digital noise on the Dgnd node. <S> It is easier to filter voltage on the power pin than current on the ground pin. <S> If you want a DAC with SEPARATE power for digital and analog, then you need to select a different chip. <S> There are hundreds to chose from.
| If you only had one ground connection, all the currents taken by the digital IO pins would flood into the chip and impregnate the analogue circuit grounds causing noise on those circuits and the chip would not achieve the performance levels it claims.
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Does AC OR DC flow? I'm reading about transducers and have a related question. Is the audio signal flowing through the voice coil of a loud speaker AC or DC? <Q> If there is DC bias, the bias will pass to the transducer coil, which makes heat in the speaker consuming more power, so analog audio should be pure AC coupled. <S> That's the why a series capacitor used in the audio path. <A> WE try to never put DC through a speaker coil. <S> The design intent is for only AC to flow through a speaker and never DC. <S> See: <S> Why are DC signals bad for loud speakers? <A> Pure AC is a time varying electrical signal which periodically reverses it's polarity, crossing zero twice per cycle. <S> Fluctuating DC is more appropriately termed AC with a DC bias; if you remove the DC bias (by passing the signal through a capacitor, for example), you are left with a pure AC signal. <S> Thus, an analog audio signal can be either pure AC or AC with a DC bias.
| Pure DC is a constant, unvarying signal.
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Two resistors in series I know the summation equation for two or more resistors in parallel or series, and I know two parallel resistors will give more power. But sometimes I saw some circuits that used two resistors in series , and I am wondering why that method was used and why they didn't use one resistor with a higher value (equal to total series resistors)? Such as the following circuit diagram, two 33 kΩ resistors used in series. So why doesn't it use one 68K resistor? Give it better results? I mean, noise filtering or something else? Note: This circuit is an AC dimmer for a microcontroller. <Q> It's the voltage rating on the resistors that is important here. <S> They are powered from rectified 230 V AC and they need to have the correct voltage rating to suit their application. <S> Two resistors in series having an individual rating of 200 V gives a total voltage rating of 400 volts (near enough if you ignore tolerances on values). <S> Take a look at the good old MRS16 and MRS25 range from Vishay: - With 230 V AC present, the peak could be as high as 325 volts without even considering line transients. <S> Clearly two resistors should be used. <S> And, for SMT resistors this might be useful to consider: - <A> Reasons someone might put two resistors in series in a volume design: <S> A bit higher power was needed than what the commonly stocked parts can handle. <S> Let's say a company standardizes on using 0805 resistors unless there is a good reason not to. <S> They therefore end up with many 0805 values in stock, with only a few values of other packages. <S> Now you need a 200 mW resistor. <S> You could specify a 1206, but overall it is better for the company to use two 0805 resistors that they are already buying and stocking anyway. <S> I have done exactly this a number of times. <S> To spread out the power dissipation. <S> Two resistors spaced a little apart will cause a lower max temperature than a single resistor dissipating the same power. <S> To get higher voltage capability. <S> This is most likely the reason in the particular example you asked about. <S> To get lower series capacitance. <S> This can be a useful trick in high frequency applications. <S> To be able to tweak a value. <S> In this case one of the resistors accounts for most of the value, like 90%, and the other the remaining 10%. <S> For low volume hand-tweaked products, the smaller resistor can be changed out for calibration. <S> To be fair though, this sort of calibration adjustment is usually done with a parallel resistor, not a series one. <A> There are conditions where you must use resistors that don't have enough voltage rating (typically because they are small like SMD, etc.) <S> So you use two of them in series to get the voltage rating to operate safely. <A> The reason may be the power or voltage capability of the resistor or even cost. <S> The schematic you show has two 33k resistors being fed from 300V peak (rectified 230V mains). <S> They dissipate a bit less than 1W worst case (the lamp off). <S> You could use a single 66K 1W resistor (it would get quite hot) <S> but two 33K 1W resistors would be cooler (larger dissipation surface area and PCB area for each resistor). <S> You could also potentially lower cost by using 33k 0.5W resistors, which might be cheaper than a 66K 1W resistor <S> You also see this done where extremely high voltages are used (ie you see this in high voltage multimeter probes) where the individual breakdown voltage of a resistor might become a problem. <A> 66K resistors aren't easy to get hold of. <S> 33K ones are. <S> You can get 68K very easily <S> (it's one of the "basic" resistor values - E6), or 62K (which is part of the E24 range). <S> The closest in a standard range is 66.5K, which is in the E96 range. <S> Generally more expensive and harder to find, since they are used less often. <S> So to get 66K it's easiest to use two readily available 33K resistors. <S> You can read more about the standard resistor ranges here . <A> Another reason for using two resistors in series is safety. <S> A resistor can fail short or open. <S> If one resistor fails it can cause a catastrophic failure. <S> With two resistors one failure does not have to bring the whole design down. <A> You have two resistors - only half of power gets dissipated on each. <S> But apart from that - there might be some other strange/weird reasons that are linked to the easability of reworking on the PCB , engineer's preferences or in extreme cases - engineering laziness . <S> easability of reworking on the PCB : if you know that resistance required is at least 33k - you put 33k and another one in series; you may want to tweak the other resistor at some stage to fine-tune the current you need. <S> It might be pretth handy once you get first field returns of your product. <S> It is pretty handy to have more resistors stuffed and sometimes 0-ohm resistors in case you want to disconnect some circuitry. <S> engineering preferences : in some cases it might be linked to BOM limitations or savings; If you use a lot of 33k on all different boards produced - why would get one 68K; This happens especially when you need precise 66k 1%. <S> You can put two 33k 1% and save some costs. <S> finally engineering laziness <S> - you have 33K symbol handy in your schematic editor and don't bother about creating another 68K component. <S> Please note that this answer has to do with low-frequency signals and resistors <S> placed close to each other . <S> It is a different story when you have a transmission line and resistors at both ends acting as termination. <A> By using two resistors you gain three things in this circuit. <S> At 230 VAC you get twice the breakdown voltage for the resistors and you can handle twice as much power loss in the resistors. <S> If you want to do 120 VAC you simply jumper one of the resistors.
| A fixed ratio change of the smaller resistor results in a smaller ratio change in the overall resistor, so this method allows tweaking resistor values with higher resolution than the standard parts are available in. Power dissipation would be the most common reason (it seems like in this case).
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Why no resistor is connected in open collector TTL logic? What will happen if we connect it internally in IC itself?Why do we need to manually connect it outside?Is there any advantage of doing so? <Q> Is there any advantage of doing so? <S> The wired-AND gate: - <S> You can also make a wired-OR gate. <S> In many situations you can switch a load that has a higher voltage level than the standard 5V for TTL: - Here's an example of a TTL logic level amplifier: - <A> It's historical, resistive pullup can be very slow and power hungry where buses must be driven. <S> TTL logic replaced RTL (where it was resistive pullups to high in the chip). <S> In the first computers using TTL 7400 logic, most buses (in the 70-120 Ohm range) had external terminations (resistor to VCC and resistor to ground) and the bus drivers were OC. <S> Eventually tri-stateable logic made bus design much easier. <S> So there were multiple reasons open collector outputs were popular in the original 7400 series logic. <S> It allowed wide Wired 'OR' functions to be easily created (eg. functions using buses). <S> It allowed high voltage drive (15-30V OC outputs) and higher sink currents (40-60mA) with minimal change to the chip design and retaining the +5V VCC. <S> It was useful for driving direct I <S> /O functions such as multiplexed displays. <A> Having ten identical outputs, ten identical inputs, and one pull-up resistor, will generally be more convenient than either having one output that has a pull-up and nine that don't, or having one input that has a pull-up and nine that don't.
| The TTL chips themselves were not able to drive stable high's with large fanout and there were no tri-state chips available. When driving any number of open-collector outputs are combined and used to feed any number of inputs, there should generally only be one pull-up resistor.
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