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What is this canister at the top of a utility pole? I'm currently in Manuel Antonio, Costa Rica. I'm living in a relatively off-the-beaten-track house and at the top of some of the utility poles leading to my house are these canisters (the light grey thing with "IC" on it): What is that? Is it a super localized transformer for the light fixture? Is it something else? On the 5m walk from the main road to my house, there are many instances of these. NB: this is in Costa Rica, not the US. So, a second question is... what is the equivalent of this in the US? Or is it common in more rural parts of the US? (I live in a city, normally). <Q> It's a distribution transformer , aka 'pole pig'. <S> It lowers the voltage from the higher voltage in the supply lines to the power used in your house. <S> The distribution lines that run down your street are probably 5-15kV. From the length of the insulators, probably towards the higher end of that spectrum. <S> They contain a transformer and oil used for cooling (in earlier days that oil would contain hazardous PCBs - PolyChlorinated Biphenyls, not printed circuit boards, though I do have a warning sticker on my computer that says it contains PCBs). <S> Since Costa Rica uses 120VAC/60Hz <S> the secondary voltage probably is center-tapped 240VAC 60Hz as used in Canada and the US. <S> At a higher level in the distribution food chain, the electricity is all 3-phase <S> but it's common to only distribute the 3-phase along major thoroughfares and then bring a single phase down a smaller street. <S> These are quite common in North America - in rural and in suburban environments. <S> Only in major cities where the utilities are kept underground are they not seen. <S> Edit <S> : The IC is probably part of the ICE logo <S> (Instituto Costarricense de Electricidad), the state-run electricity (and telecom) company. <A> The large grey cannister is a transformer lowering the high voltage from the long distance power line into the voltage you use at home. <S> Most probably you have 110 /220 V ac. <S> if you are in the USA. <S> The other element with glass is your street lightfixture. <S> If you follow the 3 thick wires comming from the transformer they will go to your house connection. <S> In western Europe however you hardly find them because there most house connections are under the ground. <S> The transformers are there placed in little stations. <S> Now we are to it accros <S> the insulator as mentioned in the other answer <S> you find mostly high voltage fuses. <S> They also enable the technicians to disconnect the transformer from the supply if work needs to be done. <A> There is a transformer (voltage converter), some insulators, and a lamp. <A> Does Costa Rica have 50Hz power? <S> Many European and island countries do. <S> In any event, that is a Single phase primary, with one (the top) ungrounded high voltage conductor (sharing the grounded/neutral with secondary. <S> The secondary is single phase, three wire, 240VAC/120VAC. <S> Power companies do this in residential areas because it is more practical/cost wise than three phase systems. <S> BTW Three phase Wye systems are 208VAC/120VAC, with phase-to-phase voltages being 208VAC, and phase-to-ground being 120VAC. <S> This is great for buildings that require more 120VAC circuits because all three phases can be used. <S> A Delta 240VAC/120VAC with 208VAC <S> Hi Leg is great for commercial industrial because you get the full 240VAC phase-to-phase for motors, etc. <S> But that 208VAC to ground Hi leg is problematic! <S> I have seen (never did it, of course) what happens when someone connects a 120VAC appliance or luminaire to the 208VAC Hi Leg and a neutral.
| They are there to disconnect the transformer from the High voltage supply in case something goes wrong. Such socalled pole transformers are widely spread in many countries in the world.
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Can we use battery in reverse to step down a dc voltage Lets suppose, i have a 12 v battery and i want to give 9v to my variable load.can i use a 3 v battery in reverse to get that output ?.If yes, what are pros and cons of this method?Why can't we see such method implemented anywhere? <Q> No A "perfect" battery might work for this, but any real life battery is going to have big problems. <S> It will work OK to begin with but the current is flowing through the battery in the wrong direction, i.e. backwards compared to when the battery is being used normally. <S> This is the same as if the battery is being charged, so the battery will quickly become overcharged. <S> The consequences vary depending on the amount of current and the battery chemistry but may include: Battery voltages much more than 3V Producing dangerous/explosive gas Fire, thermal runaway or explosion Other damage to the battery <S> If efficiency is more important, look for a switching regulator. <S> Regulators are very cheap - my usual supplier has a bag of 10 suitable 100mA regulators for sixty pence (less than a dollar US). <A> As others have already said, this is a bad idea since it would be charging the 3 V battery with no mechanism to ensure it isn't overcharged. <S> Note that trying to charge a non-rechargeable battery is essentially "overcharging" it. <S> I want to point out yet another problem with this method, which is additional series resistance. <S> Even if you carefully monitor the 3 V battery and ensure it is not overcharged, the series resistance of this 3 V battery adds to that of the 12 V battery for the purpose of the combined 9 V battery. <S> For example, if the 12 V battery has 1 Ω series reistance and the 3 V battery 5 Ω, then the effective 9 V battery will appear to have 6 Ω series resistance. <S> As others have said, use a regulator. <S> That's what they are for. <S> The common and cheap 7809 will work in this case with up to about 300 mA output in free air, and to it's full 1 <S> A capability with proper heat sinking. <S> At this relatively low voltage range, there are a number of switcher chips out there with integrated switches. <S> You add the inductor and a few other parts, and you get 9 V from 12 V at around 90% efficiency relatively easily. <S> These are usually quite pricy compared to a switcher chip and the few parts around it, but are really simple to use. <S> This can be appropriate for one-off projects, or the less-skilled in electrical engineering. <S> One side advantage of these is that the output is usually isolated from the input. <S> That means that the 9 V supply is floating relative to the 12 V battery. <S> Sometimes that can be useful. <S> When you don't want that, just tie the grounds together. <A> If you connect your battery this way it will work. <S> And you start to charge the 3 V battery at the same time. <S> Therefore dont connect a battery to lower your supply voltage.
| Most probably that is nota good idea since overcharging a battery connected in this manner, damages the battery and can be dangerous in case of a Li Ion battery. To drop 12V to 9V for a variable load, use a regulator. To get the most stable voltage, use a linear regulator. You can also buy off the shelf DC-DC converter modules.
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Soldering SMD by soldering iron I need to solder header connector dedicatedfor SMT on my control PCB. My problem is thatI have only standard soldering iron and I don'thave a rework station. Please, can somebody tellme whether it is possible to solder SMD bysoldering iron? If it is possible what is thebest practice how to do that? Thank you in advance. The part I want to solder is: <Q> My method is used widely. <S> I use a 25W iron with a 'standard' tip. <S> Position the item to be soldered accurately, and 'tack' diagonally opposite pins, <S> 1st one diagonal, then the other. <S> Then starting from one end pin, run solder over ALL of the pins on one side. <S> Apply de-solder braid in line with the pins and re-heat with the soldering iron, moving the braid away from the item. <S> Repeat using fresh braid, until all excess solder removed. <S> Pause briefly to let the item cool a little, then do other side(s). <S> Buzz out adjacent connections to check for shorts. <S> Practise on some scrap/non-essential items before attempting soldering expensive ones! <A> This shouldn't be a problem. <S> The key is to solder down ONE and ONLY ONE pin, and then reheat the solder until all the pins and pads line up correctly. <S> Be careful about trying to adjust position when the solder is solid, as you have enough of a moment arm to torque the pad right off of the PCB with very little force. <S> Once all the pads and pins are aligned, solder the rest. <S> Use flux, and don't dally with the iron on the pads. <A> Beyond that just about any other SMT part can be soldered by hand. <S> You need a fine tip on the soldering iron, one of those bucket sized ones that is great for through hole work isn't going to cut it. <S> Some solder wick, extra flux and cleaning solvent of some sort (e.g. isopropanol) are also a big help. <S> Beyond that it's mainly practice, don't expect to be able to do a neat job first time.
| If there is a central ground pad or inner rows of pins then you can't use a soldering iron.
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Should the AC-Live or AC-Neutral wire be 'broken' by a relay? I have a 230VAC relay that will switch my bathroom light as well as the extractor fan. The question I have is whether the Live/Phase wire or the Nutral wire go through the relay? <Q> Surely this is specified in whatever electrical code applies to your jurisdiction. <S> Put the switch in the hot line, not the neutral. <S> Even without the electrical code, a few seconds thought should have revealed the answer too. <S> Purely from the device's point of view, it doesn't matter since the device and switch are in series. <S> The switch in the neutral line will work correctly. <S> However, consider the voltage the device will be at in both cases. <S> With the hot line open, the whole device is at the neutral potential. <S> This isn't supposed to matter, but stuff happens. <S> With a lot of moisture around, for example, external surfaces could be connected to line voltage thru the moisture leakage path when the neutral line is open but the hot line not. <S> There are ganged switches (DPST) just for this purpose. <S> This doesn't apply to something low power like a bathroom light and fan. <S> I only mention it for completeness and to better illustrate the point. <A> The norm DIN VDE 0100-460 states that a SPST switch must not be placed in the neutral path. <S> The norm DIN VDE 0100 <S> -470 adds that a DPST switch that switches both neutral and phase must have a lagging pole for neutral during switch OFF and a leading pole during switch ON. <S> You should check whether there is a similar norm for your country. <A> It is more safe to use a DPST relais. <S> So switch both line and neutral. <S> There is however more. <S> Where are you going to place the relais <S> and how are you going to install the wiring. <S> Unless it is allowed by the protection level. <S> So must all metal parts in the bathroom be connected to earth and the breaker is most probably a so called earth leakage breaker or RCD. <S> Since you are living in the Netherlands. <S> The NEN 1010 electrical code applies. <S> The bathroom is divided into zones each with their own requirements. <S> 230V outlets are not allowed and the bathroom electricity system must be protected with an RCD. <S> So it all depends where you want to place your relay. <A> If you're only switching one leg, it should definitely be the HOT wire. <S> Suppose you have a lamp <S> and there's a problem with one of the socket contacts. <S> You turn off the lamp, take out the bulb, and reach in to pull the tab contact up a bit. <S> POW! <S> You get a zap. <S> The device still has power even when the switch is off if you're only switching the neutral line. <S> You should ALWAYS break the hot line, and as mentioned it would be even better if you switch both of them.
| With the neutral line open, the whole device is at line potential. In cases where there are two hot lines going to a device, like 240 V devices in the US, you should be switching both of them. But remember the electrical code in your area could be different. The installation should not become wett. Maybe it's bent down and won't contact the bulb anymore (I have had this problem). Remember the bathroom is a wet area where specific rules for the installation apply.
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Are there different meanings behind digital to analog in digital audio and PWM? I'm getting into producing sound waves with various controllers and the two ways DACs are referenced is confusing me. The first kind of DAC conversion would be from PWM. If a an audio signal is built up of extremely fast changing voltages, PWM isn't doing that. It's just pulsating extremely quickly at different rates so that a speaker will effectively output a wave built up of those simulated voltages. Presumably, to get better output quality, you would convert that to an analog voltage for each small step in the wave. The other kind of DAC I can only differentiate by calling it a "signal" DAC. If I have a wave made of all of these steps in voltage, a DAC might be used to smooth the transitions to a true analog wave without a sampling rate. It seems to me like a DAC can be used to convert PWM to analog voltages while still being a digital audio output, then a DAC can also be used to convert that digital signal to an analog wave. If I want to buy a very fast DAC to just convert a PWM signal to true analog voltages with a sampling rate, what would you call that device when not referring to a DAC that changes a digital audio signal to a completely smooth wave. Being a beginner, I just don't know what to google. There's all sorts of answers about DACs and solutions involving op amps, and the term DAC is everywhere. I'd just like some clarification on what the term refers to in audio and if the definition is somehow changing based on the context. <Q> A DAC is mostly refered to as a device that converts a digital signal (a number) into a value discrete voltage signal followed by a zero order hold. <S> A DAC doesn't use PWM. <S> Class D amplifiers use PWM to drive the speakers and do already provide very smooth currents. <S> If you want to smooth these currents further, you can add a two stage lowpass T-Filter right after the amplifiers output. <A> Converting digital audio signals to analog signals can be done in a number of ways. <S> The methods that achieve the highest signal quality usually involve signals "pulsating extremely quickly". <S> The waveforms are often generated by either using pulse-density-modulation (PDM) or pulse-width-modulation (PWM). <S> Both signal types contain an almost perfect replica of the desired audio signal along with components at higher frequencies. <S> Using a scope or a spectrum analyzer these components can be seen, on the scope they don't look smooth. <S> In order to get a smooth audio signal the higher frequency components have to be removed. <S> In many cases the loudspeaker alone would be sufficient because it can't reproduce "non-smoooth" signals very well due its inertia. <S> However, for a few reasons this method can't be used and filtering has to be applied which results in a clean audio signal. <S> Delta-sigma D/A-conversion is mainly done in the digital domain with a final step in the analog domain. <S> PWM signals are often often considered to be digital signals but this is just a common misconception. <S> They are generated by a pulse-width modulator that takes an analog input signal and by comparing it to some carrier waveform a signal with usually two or three voltage levels is generated. <S> It contains a perfect replica of the input signal and components at higher frequencies. <S> PWM can be used for D/A conversion as well, but for high-quality audio signals this is a non-trivial undertaking. <S> So most often PWM is not used for D/A-conversion but for the amplification of the signal alone (class-D amplifier). <S> The D/A conversion is often left to an delta-sigma modulator. <S> So PDM and PWM are used together. <A> When all is said and done, a PWM output properly integrated and of sufficient rate is indistinguishable from a more ordinary DAC style output. <S> They BOTH produce (in a different way) <S> a step-wise approximation of the original analog audio waveform. <S> You are creating a false dichotomy here. <S> Some of the most expensive audiophile power amplifiers have used PWM and indeed most of the most modern consumer gadgets use PWM outputs to drive speakers (and headphones/earbuds, etc.) <S> One of the reasons that PWM is becoming so much more popular is because it is very easy to generate analog (audio) outputs of any arbitrary precision with only digital circuitry which is dirt-cheap to produce. <S> It used to take special digital/analog chip technology to produce traditional digital-to-analog converter (DAC) circuits. <S> But welcome to the 21st century. <S> We have learned a lot in the last decade or two.
| PDM signals are generated by so-called delta-sigma modulaters and the result is a waveform with two or more voltage levels that contains a high quality replica of the analog signal.
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Is there an "additive manufacturing" method to make an ASIC? Reading questions like this one " How much does it cost to have a custom ASIC made? ", I was wondering if there's some sort of equivalent to additive manufacturing that would lower the cost to getting prototype chips made. For example, can you "write" a chip with an electron beam or something? Carve it out with an AFM ? If not then why not, and who's working on this? What if we restrict ourselves to analog chips / large geometries, are there any methods that then become feasible? <Q> The software tools needed are usually much more expensive since the market is quite small. <S> From this point of view it does not make much sense to work on that problem. <S> It is possible to produce wafers and put them "on hold" at a certain stage during production. <S> Doing this before the final layers of metal interconnect are made it is possible to continue production at a later point in time and produce a few variants of a chip. <S> As a final resort focused ion beams can be used to remove and add tracks <S> but this a very slow and expensive process. <S> A more interesting approach could be organic electronics <S> where basically an ink-jet printer is used to generate electronic devices. <A> An method often used is the multi project wafer : <S> Using the wafer masks for different designs and sharing a wafer. <S> Only a few dice are guaranteed to be manufactured. <S> This is maybe the least expensive method to prototype designs. <S> But there is no additive feature, you have to be sure what you do when you tape out your design, even using a MPW. <S> Hold Wafers in stop allows you to change minor things in your design, for example wiring. <S> See here the XFAB MPW time schedule . <A> There's no such thing and there probably won't be anytime soon. <S> Reasons include: N-wells and P-wells are are creating using dangerous (toxic/explosive) chemicals at high temperatures (often on the order of 1000C). <S> Safety would obviously be a huge concern. <S> IC feature sizes are ridiculously small, even on older processes. <S> Additive manufacturing normally doesn't have very high precision. <S> It's possible to edit ICs using a focused ion beam (FIB). <S> You can both add and remove material. <S> However, this process is expensive, failure-prone, and very-slow. <S> Changing a few wires is a big deal. <S> Connecting hundreds or thousands of transistors? <S> Forget it. <S> Unless you can do everything in one step in a single machine, you'd still need a clean room. <S> Additive manufacturing doesn't solve the problems of design, testing, or quality control, all of which are nontrivial. <S> Most importantly, there's already a cheap and effective alternative -- FPGAs. <S> Aside from being cool, it's not clear what problem additive manufacturing of ICs would solve. <S> Who needs a tiny number of low-quality ASICs instead of a commercial ASIC, an FPGA, or an off-the-shelf chip available in quantity? <S> If you just want to hook up a few dozen logic gates, you can do that on a circuit board. <A> Yes, you can share space on a single die with others. <S> A couple of fabs offer this. <S> For example, MOSIS calls their program MPW, or Multi Project Wafers. <S> The wiki link below has links to ~5 companies that offer it towards the bottom. <S> MOSIS MPW and the wiki . <S> Our VLSI class did this for a final class project in 0.5um MOSIS technology. <S> Obviously a very, very old tech; <S> but you can do this on newer techs as well. <S> Plus for our needs 0.5um was good enough. <S> The price is proportional to the size you need (both area and I/O ports) and the technology node.
| Having a protoype ASIC manufactured is not very expensive when older technologies from 130nm up are used.
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How to power 12V 1A device with two 6V lantern batteries (1209)? I have 2 EverReady 6 volt batteries model number 1209.I also have a 12 volt 1 amp screen that I need to power. How can I power the screen with these batteries? <Q> Unless your screen requires you to limit the current itself, it draw the amperage it needs, it's the volts you have to match yourself. <A> This data sheet suggests that, if you need 10.5 volts or more, you simply will not be able to operate at all. <S> If your screen is able to operate at 9 volts, it might last about 2 hours. <S> So you need to find out the characteristics of your screen. <A> You can use a boost regulator to get 12V from the 6.4-12V battery voltage (two batteries in series). <S> Many such regulators will drop very little voltage when the input voltage is similar to the desired output voltage. <S> To get a reasonable fraction of the energy out of the batteries you need to keep the panel voltage near 12V even when the battery voltage has dropped to 800mV/cell (8 cells contained within two batteries). <S> I suggest you buy a module with a chip and other components already assembled, and capable of at least several Chinese amperes. <S> Chances are the panel actually draws less than the rated current, but this is still an expensive way to run an LCD panel. <S> Incidentally, contrary to your title, this will actually increase the current drawn from the batteries compared to the panel. <S> Because, say, 0.8A at 12V is 9.6W and <S> the batteries even nearing exhaustion at 6.5V must still supply 9.6W plus maybe 1-2W for the converter <S> so the batteries will have to deliver 1.7A. Because of conservation of energy and thermodynamics and such like.
| The more complicated answer has to deal with the question of the minimum voltage your screen will work at. The obvious answer is that your hook them in series to give you 12 volts.
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How do two UARTS know which baud rate to use? I am reading about the standard protocol for UART and I think that if the receiving UART does not have any idea on what baud rate the data was transmitted, there would be lots of problems. If the assumed baud rate is lower than the baud rate in which the data is transmitted, there will be bits that would not be 'seen' by the receiving UART. On the other hand if the baud rate used by the receiver is higher than than the baud rate in which the data is transmitted, there will be bits that will be counted twice and would result the data being 'read' incorrectly. My knowledge around UART is that when the line is idle, it is kept to a '1', the Start bit is a '0' and the Stop bit is a '1'. Also, the Stop bit being '1' does not have any difference with the '1' when the line is idle or is there a way to differentiate? Do two communicating UART's first agree on which baud rate they will use? If yes, how do they do it? <Q> Ordinary UARTs have to be pre-configured <S> with the desired baud rate (as well as word length, stop bits, parity, etc) traditionally by a human. <S> Early versions needed a known character to be transmitted, but more sophisticated versions might be able to find the rate from more arbitrary data. <S> A receiving UART typically has a local clock that runs at a faster rate - typically 8 or 16 times the baud rate. <S> This is used to sample the incoming signal and detect the bits within a word in a way that can tolerate a few percent of error. <S> Even two crystal oscillators wouldn't match rates perfectly, but the error tolerance can permit use of some less precise sources, sometimes including trimmed on-chip oscillators, etc. <S> It can also help accommodate the fact that dividing down popular oscillator frequencies may only produce an inaccurate approximation to certain baud rates - in the old days, UART master clocks sometimes needed particular frequencies to access popular baud rates, for example 11.0592 MHz on the 8051 family. <A> Two UARTS "agree" on baud rate by means of documentation and by operator/user setting the baud rate by hands, including handshake protocol, stop bit size, etc. <A> Yea, everything is set up manually, which is often a bit of a pain, especially when systems are poorly documented (I'm looking at you, every embedded system ever). <S> I know USB, SATA, and most other modern data protocols start off after some reset or initialization event at the lowest speed with some standardized default configuration and negotiate with everyone else (or the just the master, depending on the protocol) up to higher speeds. <S> Some also use pull-up or pull-down resistors on their data/power lines to indicate supported speeds. <S> See this website on USB negotiation if you're interested in delving a bit further in other protocols.
| For several decades now though there have been implementations of "auto baud" detection found in some settings, which typically works by timing key features of the waveform to deduce the baud rate.
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Running a 12V 1.6A fan on a battery I am building a portable A/C (really it's just a box with a fan and ice packs) and I want to use a 200CFM case fan. It's rated 12V 1.6A. I was hoping to direct connect this to a solar panel and keep it running (with a slower speed during cloudy conditions), but from what I've researched, I'd need a 25W-35W panel and the one I found was relatively huge in comparison to my cooler box. So now I'm thinking I'll just use a battery and charge the battery at night. So if I have a 30000mAh battery with 2 5V 2.1A port and 1 5V 1A, is there some way I can hack it to 12V 1.6A? Or should I just Amazon a 12V 1.6A drill battery? Any suggestions are welcomed, even negative ones. <Q> There's a few caveats here. <S> Are you sure the battery is 30Ah? <S> If it's cheap and less than about a kilo (based on lithium's specific energy; Wikipedia gives a pretty wide range of values), chances are that it's not really 30Ah, and that's not even including a case, charge management and output boards, etc. <S> Do you care if it runs at full power? <S> The 30Ah pack has a higher capacity in Wh, but is apparently limited to ~10W on its output, per channel. <S> Your fan, would, at full load, draw approximately 20W. It may be possible to parallel the power bank's outputs, but this is dependent on how the pack is constructed. <S> Even so, the pack may have a protocol or sorts for charging at the full 2.1A, usually dependent on data line voltages. <S> It will take 5V and "boost" it to 12V, but drawing proportionally more current at the 5V side, plus a bit for losses. <S> The good news is that these are dirt cheap (>$1). <S> eBay and Aliexpress have a wide selection, and for a low power load like a fan, these will suffice. <S> More data on the 30Ah battery would help us help you make a better choice. <A> I'd use a surplus automotive battery, but for portable weight a LiPo box <S> 72W 132WH 12V/11000mAh $65 <S> For a fan , dual VW Passat turbo fans with Buck Reg speed control using 15A 50kHZ PWM choke, 10Amp ripple cap. <S> to run it quiet at slow speed with a mini vortex maniform to get rid of eddy current (fan blade interface) noise <S> Surge current on a 1.6A fan is at least 10A <S> and if you stepped up from 2.5V to 12V that becomes at least 4x battery surge current... or as much as the boost reg can handle without overheating. <A> It might be a single 3.7V "30Ah" cell which translates to about 20Ah at 5V and much less once stepped up to 12V - ideally, 8Ah for 5 hours operation at 1.6A. <S> But even if its internal boost convecter can supply more than 5V 2A (which will translate to 12V 0.8A), that'll reduce its capacity further. <S> Another latent problem is that fast charge and discharge will shorten the life of cells not designed for it - you may reduce the life from several hundred cycles to tens of cycles. <S> Every few days someone has the bright idea of using these under-priced, over-spec'ed products that are cost-engineered to keep a cellphone charged, imagining they can power a golf cart, small motorbike (or today, an AC unit). <S> It's going to be disappointing. <S> Use a battery pack designed to run a motor in the first place. <S> It might cost a little bit more, but there's a reason for that.
| Given that you still want to use this battery, the item you're looking for is a boost converter.
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Automotive chassis is positively charged compared to earth, is this a problem? I am working on a vehicle which has electronics equipment inside it. A neutral electrical charge of the ground connection is desired. When I measured the negative terminal, which is connected to chassis and batteries negative, compared to the ground/earth wire in the wall socket, the voltmeter read ~ +6V DC. I also tried to measure the current from chassis to wall socket earth. The (cheap) voltmeter showed ~40 mA for a very brief moment, then zero milliamps, and the voltage difference was neutralized. Immediately after removing the current meter, the voltage difference would return to ~ +6V. The system is: vehicle, batteries, solar charger, inverter. (and appliances but those were disconnected) Sometimes I feel a spark when touching the vehicle while grounded. The system is 12V. I'm not quite sure where the 6V comes from. a) Why isn't chassis ground at zero potential? Isn't it being earthed via the rubber tires? b) Is there a way to achieve a better earthing, e.g. a loose wire that hangs and touches the ground? c) Is there something wrong with my configuration, or is this completely normal? <Q> Aswer to a: <S> The car is not earthed trough the tires. <S> Remember that sometimes someone get a static shock when leaving a car. <S> Answer to b: <S> Some people purchase and install a special flexible wire (carbon strap) <S> that lays on the ground removing static charge. <S> You can most probably find it in a car shop. <S> Answer to c: <S> Since the current drops to zero <S> everything seems to be ok. <S> So dont worry. <A> In addition to the earlier answers, don't assume that the ground terminal of a wall socket is at 0V relative to true ground. <S> If you want a true ground reference, hammer an earth spike into the dirt (such as a flower bed or lawn), well away from any electrical installations. <S> (If you don't have a proper earth spike handy, try a large screwdriver, garden fork or anything else that's pointy and metal) <A> Rubber tires are excellent high voltage insulators and can even store a large static charge. <S> You will need to make differential measurements. <S> Keep in <S> mind chargers use line filter caps to ground (Y2) with up to 0.5mA of ripple current which also have DC leakage to ground and transformers even SMPS have coupling capacitance and DC leakage (but very low). <S> These Y2 caps and your current meter are producing the observed effect.
| Dragging a chain or carbonized strap is often used to prevent static buildup on cars.
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Design of Window Comparator Circuit w/ Mutually Exclusive LED Indicators The functionality I seek is as follows: All LEDs off: Vsignal < 0.118VDC LED1 fires: 0.118 < Vsignal < 0.1216 LED2 fires: 0.1362 < Vsignal < 0.146 LED3 fires: 0.151 < Vsignal < 0.1607 LED4 fires: 0.1853 < Vsignal < 0.1999 Between these firing windows, all LEDs should be off. The two ICs are an LM339 and a CD40147BE. The noise from the signal voltage is <0.0002VDC. The accuracy should be to 0.001VDC. I need to recalculate the resistor values for ref voltages, but I don't know how to connect multiple comparators together; all examples I've seen have been for a single window, not multiple. <Q> Fortunately this is easy with the LM339 as it has open-collector outputs which can be joined together to form a wired AND function. <S> Figure 1. <S> The internals of an LM339 comparitor. <S> Note the open-collector output. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> By wiring the outputs together as shown the LED will turn on when the voltage is inside the window set by VR1 and VR2. <S> How it works: If both comparitor outputs are high their output transistors are off and the collector is floating. <S> R1 will pass current to LED D1. <S> If the input voltage goes above the upper window voltage set by VR1 <S> then CMP1 will turn off and short out LED D1. <S> If the input voltage goes below the lower window voltage set by VR2 <S> then CMP2 will turn off and short out LED D1. <S> You need four of these circuits. <S> I recommend 10-turn pots rather than a precision resistor divider chain. <S> Add one more pot for calibration and it should be a simple job to set everything up. <S> Don't skip the decoupling capacitors on each chip. <A> As Transistor has answered, your best bet is to make 4 independent window conparators. <S> If you are determined to do things the hard way, the following simulate this circuit – <S> Schematic created using CircuitLab actually makes use of the 74147 as a priority encoder. <S> Unfortunately, what you want is a dot-mode bar graph, and the 147 doesn't do that. <S> You can, however, add a 2:4 demultiplexer (I've used half the inputs of a 3:8, just because it's convenient) to convert the encoded address to drive the LEDs. <S> Additionally, LM339s have as much as 3 mV offsets, and although you expect the 4 on a chip to track with temperature, the unit-to-unit offset is not correlated on a chip. <S> This means that your LEDs 2 and 3, which have windows less than 10 mV wide, will potentially have 30% errors on each end of the windows. <S> To get around this, I've added a front end which converts (0.118 to 0.1999) to (0.1 to 3.1) volts. <S> The two op amps should be rail-to-rail output units capable of running at 5 volts. <S> The choice of 3.1 volts as an upper limit comes from the LM339 common mode input limits of 0 to 3.5 volts for a 5 volt supply. <S> Also note that the accuracy of the windows depends critically on the accuracy of the 5 volt supply. <S> Any variations will directly impact the thresholds of the windows. <S> Finally, my suggested circuit needs, and does not have, hysteresis. <S> However, I don't know your threshold accuracy requirement nor did you specify the amount of noise which may be found on the input. <S> Without these, there is no way to specify what's needed. <A> Use the LM339 quad comparator IC to form a bargraph display. <S> Alternately use an LM3194 bargraph IC.
| You need to make four window comparitors. This can be achieved with the LM339 by using diodes at each output to short out the previous LED, however unless you fully know what you are doing, you could end up "blowing" the chip altogether The LM3194 will allow bargraph (all lit) or dot mode (only 1 led lit at any time).
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Is it safe to drive multiple 'strings' of LEDs in parallel? I am thinking of driving COB LEDs (rated at 10 W normal power) at 3 W. I want to drive 32 LEDs from one 100 W constant current driver. The driver has a maximum voltage of 144 V. The LEDs are 35 V. I can drive a string of 4 LEDs maximum in series. Is it safe to drive 8 of these strings in parallel? <Q> The LEDs are designed for a constant current supply which produces whatever voltage is necessary so reducing the current would be effective in dimming the LEDS. <S> Rohat calculated that the supply output would be a constant 700mA and for 3W per light you need 86mA each. <S> Putting 8 lights in parallel would give 700/8=87.5mA per light which is about right, but that assumes they will share the current equally. <S> Unfortunately, whichever takes the highest current will heat up most and that will probably drop the voltage futher giving thermal runaway. <S> What you would need to do is add something to equalise the currents, a series resistor as shown would do <S> but I suspect the value needs to be higher than shown in the circuit, <S> at least the maximum forward voltage variation divided by the current variation you can tolerate. <S> As the LEDs can take up to 350mA, that is fairly easy but you probably want to keep the brightness of the bulbs fairly similar so for example 10% brightness variation implies no more than 8.6mA difference between the bulbs. <S> Higher values of resistor will maintain more even brightness but will increase heat dissipated in the resistors. <S> The advantage is that including 8 resistors is a lot cheaper than separate drivers. <S> P.S. <S> You probably realise that the supply voltage need only be 35V plus the resistor drop so <S> a 144V-capable supply is much more than you need, it is really intended for driving multiple bulbs in series so that all get exactly the same current and hence brightness. <A> If you have a 100W CC (constant current) driver with maximum output voltage of 144V, then the output/rated current should be about 0.7A. <S> If your LED has a nominal power of 10W and load voltage of 35V, then the nominal drive current should be about 0.3A. <S> If you want the LED to have 3W output power, then the drive current should be about 3W / <S> 35V = <S> 86mA. <S> As you might guess, you cannot connect all the LEDs in series. <S> Let's think about this: simulate this circuit – <S> Schematic created using CircuitLab <S> It's clear that LEDs will break down because of 350mA constant current per strip. <A> The datasheet given indicates that the current consumption of the LED system is given at a nominal value. <S> In other words it can be more or it can be less. <S> Voltage driven LED systems have an internal arangement to keep the current at the desired level. <S> If you put such systems in series the voltage distribution might not be uniform. <S> This leads to the risk that some units have an overvoltage and become defective. <S> What you can do is give each system each own driver and supply all the drivers together with a PCM modulated signal ( maybe dmx? or 0-10V) to come to a desired result of 3W per led (30% pcm duty cycle). <S> It is possible to place more then one module in parallel to a driver. <S> You could look at the LT 8048 DMX driver. <S> This driver can supply a max of 4A per channel.
| So it is not a good idea to put this type of led in a series parallel combination as asked. The problem is that LEDs have a fairly sharp turn-on so whichever bulb has the lowest forward voltage will take almost all the current and die, then the next lowest would blow and so on.
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Can an infrared thermometer (IR gun) be used to measure the ambient room temperature? I know the principle of an IR thermometer, but have one question regarding that are as below. As an IR thermometer senses the radiated IR waves from external bodies coming in their field of view and gives surface temperature reading, can we point in air or random far space and measure the temperature of the surroundings/room? Will the reading, whatever comes, be near room temperature, or will it just be a garbage value? One example of an IR thermal sensor: https://www.sparkfun.com/products/9570 <Q> In the relevant part of the infrared spectrum for this application\$^1\$, air has a high transmittance: <S> (taken from here licensed under Creative Commons Attribution-Noncommercial-Share <S> Alike 3.0 United States License ) <S> So wherever you point an IR temperature sensor, the output will be dominated by the surface you are pointing at and not the air in between. <S> If you try and point an IR sensor in the sky (no clouds) you will get a a very low reading (mine showed −40 °C but that is the lower range end), and not the air temperature. <S> So you need some surface at room temperature to measure the room temperature. <S> Do not use shiny metals, as those reflect IR radiation. <S> Normal glass can be used, so you could hang a decorative looking thing in the room and measure the surface of that – provided you have a small enough measuring cone (so it only hits your wanted surface and not something behind it). <S> \$^1\$: For −40 °C the peak in IR is around 15 µm, for 50 ° <S> C the peak is around 8 µm. <S> The sensor you linked uses a thermopile (a stack of thermocouples). <S> These sensors require knowledge of the ambient temperature to calculate the targets temperature. <S> So integrated into the IR sensor is an ambient temperature sensor, and the measured value can actually be read out, so you get the ambient temperature of your sensing element. <S> Depending on the location where you place it, that might be close enough to room temperature already. <S> But in that case the IR part looses its use and you could just use a normal temperature probe. <A> You could hang a sheet of (non-gloss) printer paper from a string in the middle of the room. <S> The paper will respond quickly to changes in room temperature, due to its low thermal mass and because it will allow air to circulate around it. <S> Plus it provides a target surface which will be less likely to reflect IR from elsewhere in the room. <S> Compare to a "normal" thermometer nearby and apply any compensation to make your readings match, then just use the IR gun from that point onwards. <A> As I interpret your statements, you are making 3 questions. <S> yes, you can use an IR gun to measure ambient room temperature. <S> Whether the measurement is any good, that is a different question. <S> Again, whether the readings you get are "usable/reasonable," is a different question. <S> the reading might or might not be "near" room temperature. <S> It depends on how representative , the temperature of the surface hit by the beam (if any), is of the "room" temperature in question. <A> Infrared thermometers cannot directly measure room temperature, because they are generally used to measure solid heat sources like the temperature of the screw of the molding machine, the temperature of the heating coil, the human body, etc. <S> The infrared thermometers sense the temperature through the reflected laser beam. <S> The link above includes some detailed working principles of infrared thermometers. <S> Hope it helpful to you. <S> The wavelength above 5um cannot be measured through quartz glass. <S> The glass has very special reflection and transmission characteristics and does not allow accurate infrared temperature readings. <S> But the temperature can be measured through the infrared window. <S> Infrared thermometers are best not used for temperature measurements on shiny or polished metal surfaces (stainless steel, aluminum, etc.). <S> To locate the hot spot, to find the hot spot, the instrument aims at the target, and then scans up and down on the target until the hot spot is determined. <S> Pay attention to environmental conditions: steam, dust, smoke, etc. <S> It blocks the optical system of the instrument and affects accurate temperature measurement. <S> Ambient temperature. <S> If the thermometer is suddenly exposed to a temperature difference of 20 ° C or higher, allow the instrument to adjust to the new ambient temperature within 20 minutes.
| In addition, when using infrared thermometers, you 'd better pay attention to these issues: Only the surface temperature can be measured, and the infrared thermometer cannot measure the internal temperature. yes, you can point the IR gun anywhere and to/at anything.
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Driving a 15V, 150W AC halogen I have a 15V, 150W Quartz Tungsten Halogen and would like to make it work. So far I have found that it is driven using AC, so I believe I need an AC/AC step down transformer (if 220VAC is used)? If so, do I buy a specific one or do I make one myself? Though, I am not entirely sure if it is the correct approach. Any relevant information is appreciated. <Q> IF it is a 150W Tungsten filament operating at 15Vrms or dc, then the current at 4000'K is about 150W/15V=10A and the resistance is 15V/10A= 1.5 Ohms. <S> We know that that PTC characteristics for tungsten are 1:10 <S> Ohms cold: <S> hot thus the cold resistance should be 1.5Ohms/10= 150 milliohms which results in a surge current of 15V/0.15 <S> Ohms = <S> 100 <S> A or 10x the hot current. <S> In order to minimize transformer losses the secondary winding DCR should be about the same order of cold filament resistance or less , which help to reduce the surge current in half or less at the expense of 5% additional conduction losses. <S> side note <S> But if you need a regulated current limiter, these bulbs make great series current regulators with 10~20W power LEDs or <S> 4x3W LEDS operating at 15 Vdc and work like PTC active Polyfuses when used with less than half of its voltage rating. <S> So you can replace the 150W wasted halogen power with 20W 12V LEDs using 15Vdc ( if you had such a supply) <A> If you use only a 220/15 transformer, any fluctuations on the power line may cause the lamp to blink or even to break it down. <S> Instead, you can buy a 15VDC/10A (Or with higher current rating) SMPS. <S> This can be more efficient and reliable. <A> You can buy a mains to 15VDC 200W supply fairly economically. <S> I would not suggest using a 150W supply. <S> Since it will be regulated DC it will probably be a wash or better. <S> You will need a ceramic high temperature socket for those bulbs- <S> they run really hot. <S> We use similar bulbs (with dimmer) in fiber optic microscope illuminators for inspection. <S> Also note that the bulb life is fairly short ( 50 hours rated) at full voltage. <S> Life increases rapidly with lowered voltage (but light output drops rapidly too). <S> Most switching supplies in that class have a pot that allows the voltage to be trimmed slightly (you can just see the top of the orange single-turn pot in the above photo, to the right of the terminal block0.
| The bulb will run okay from DC, though life will be slightly shorter than on AC due to notching effects.
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Where to go with an idea for a chip? I tinker with AVR, 6502 and z80 designs, just muck around really, nothing overly serious. But as a result I had a thought of an idea for a new chip. Specifically a SoC built around a z80 core. But who would I talk to too even see if what I'm thinking is feasible? And if it is who would I talk to too, to get a price estimate to have the thing designed into a chip? Would the cost of doing something like this be prohibitively expensive? <Q> It would be prohibitively expensive to get an ASIC made. <S> In the order of a million dollars. <S> What you would generally do is develop the chip on an FPGA first to see if it would work without such costs. <S> But to get it finally developed it's exceptionally expensive <A> Almost anything along those lines is possible, and many are even economical if the volume is large enough. <S> If I were in your situation, I would then talk to a trusted friend who works in the semiconductor design and/or fabrication business to see whether it's feasible and get an idea of costing. <S> But I'm fortunate to know a few such people. <S> If you don't, I'm sure you can find a suitable consultant and pay for a few hours of her time for an initial assessment. <S> Be sure to have her sign an NDA protecting your potential IP, she might have a suitable template you can use <S> so you don't necessarily need to engage a lawyer at this point. <S> You would ultimately end up either partnering with or becoming a fabless semiconductor manufacturer . <A> Some things: You will need to obtain a license for the z80 core. <S> This would be in the form of soft or hard intellectual property (SIP or HIP). <S> SIP is the RTL code. <S> HIP would be it implemented in silicon for a particular technology node. <S> There are more ambiguous in-between IP's as well <S> (maybe the critical areas are hardened but the glue logic is provided only as RTL for you to synthesize). <S> Based on which of these avenues you choose to pursue, you would need: - SIP : You will need a synthesis engine to translate the RTL into gates, such as DC Synthesis. <S> You will need access to full chip timing analyzers, such as Synopsys PrimeTime or ICC. <S> - HIP: You will still need the timing engine but may be able to get away without the synthesis engine depending on how you build the circuits around the core. <S> Either way you will need a foundry willing to construct the chip, and a tool to run design rules checks on the layout, and to run schematic vs. layout (Virtuoso comes to mind). <S> In the SIP approach you can implement the core in whatever technology node you want; for example .5um <S> from MOSIS is fairly affordable and used commonly in academic settings. <S> In HIP you would need the same technology that the HIP is delivered in. <S> Since this will most likely be more advanced technologies, the cost will go up significantly. <S> Masks in newer technologies run upwards of millions of dollars per mask, and you need around 20-30 masks for a 5-10 metal layer IC. <S> Overall the cost is going to be quite prohibitive unless you can line up buyers and you will most likely need a small team of engineers to properly design, validate, and manufacture this. <A> First point to go to is Zilog, there're other organizations who have license for the core. <S> What I got from interacting with them is that they may not be interested in the deal unless it will bring significant dividends for them. <S> You should take it into account because before you even start to discuss numbers and deal in overall you should be able to convince them that your implementation is worth talking and thinking about. <S> Next, as I see from the comments to the question, you are going to make DSP device. <S> Some DSP capabilities (e.g. ADC/DAC) may not be properly implemented in FPGA, thus FPGA has relatively limited application here, and you will need another entities (on the board or in silicon) providing these functionalities. <S> However even with multiple chip design it will be cheaper and will provide speed required for video DSP. <S> For that you can use any modern FPGA, and select ADC/DAC carefully, and, what's most important, make proper board design which accounts for high digital speed. <S> If you need 32K buffer for your video processing - FPGAs are having such memories which can work @200 Mhz, should be enough for more or less sophisticated digital signal processing (given, again, you select RAM organization properly). <S> Finally, you can make your own implementation of Z80, without reference to the original design, only utilizing command set and operational structure, and even not accounting for native Z80 timing making you implementation running at much higher speed than 20 <S> Mhz which is a maximum for Z80. <S> It is not a hard work, but after you make such implementation you should be able to write microcode in Z80 assembler, which may be harder given capabilities of the debugging utilities which you design.
| So a good first step would be to make sure you have an understanding of the size of the potential market for your idea and what price it might bear.
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How can I obtain 12 V DC, 3.3 V DC and 5 V DC from a single Li-ion cell? I am working on a project in which I require three separate DC voltages from a single Li-ion battery (cell) of 2600 mAh capacity. I want my Project to be portable, and I only have the single battery to work with. How can I get voltages of 3.3 V, 5 V, and 12 V from my battery? I measured the battery voltage and it is 4.2V but I understand that the battery voltage will range from about 3V when fully discharged to 4.2V when fully charged, so the power supply needs to accommodate this battery voltage range. I am using an Arduino Pro Mini 5 V 16 MHz for my project. <Q> For 3.3 volts a standard linear or buck regulator is the simplest approach. <S> For 12 volts use a boost regulator and for 5 volts tee-off a linear or buck regulator from the 12 volt output. <S> Depending on current taken on each power rail this may change. <A> Before even thinking about implementation I want to ask you question: How much current will be consumed through 3.3 V, 5 V and 12 V? <S> If we suppose that efficiency of power level converters is 100%, what is the maximal time your appliance will work properly using the battery? <S> You also should take into account that chips require A specific level of voltage to operate, e.g. datasheets should state range, for example 5 V ± 10%, meaning if your main power level will drop below 4.5 V, the whole appliance may malfunction, because its main brains will fail to operate. <A> However, you should reconsider if you really need all 3 voltages. <S> I did it once on one of my design and regreted it, as debugging the circuit was very exasperating. <S> Also, if one of the power buses is very low power (say, you only need a few miliamps on the 3.3V) <S> Also, if you could provide more details we could come up with several proposals ;)
| you can use a cheap voltage regulator to obtain said voltage from one of the others if power efficiency is not key. I would use SEPIC or buck-boost converters adjusted individually for the voltages you need.
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what connector/cable for 22pins connection In one of my project, I want to connect 2 PCB. I need to be able to have 22 wires minimum between the 2. on those lines, 19 are meant for +3.3V signals, 1 for a +5V power line, 1 for a +3.3V power line and 1 ground I'm looking for a small connector with easily buyable cable or connector but the big number of pin make it hard to find what I need. I thought about using multiple micro USB3 or micro USB2 sockets to connect all but it would need at least 3 which meen 3 separate cables. I also throught about using IDE connector as IDE wire are widly available but they are a bit wide to my taste. TL;DR : I'm looking for a connection to connect 2 PCB board With at least 22 pins For a small distance (10-20 cm) Somewhat uplugable (I may plug/unplug those a few times a year) <Q> For board-to-board connections, I'd probably use ribbon cable and IDC connectors, mating with standard post headers on the boards. <A> I actually am working on a project at work with very similar requirements. <S> I have selected some ready-made ribbon cables from Samtec ( FFSD series ) that mate with the ESHF series board receptacle. <S> You can then cut the ribbon cable to the length you need and attach the connectors. <S> Digikey also offers to make a cable for you, if that is your preference. <S> I'm not sure if Mouser has the same service or not. <A> I think this is a very good question for this board. <S> I suggest a DB25 connector, the old serial connector on the IBM PC. <S> They can be used for all sorts of configurations and you get male and female in PCB (through hole and SMD, straight and right angle) and wire termination (solder and crimp)
| You can also find other standard IDC connectors on distributor websites like Digikey and Mouser, and separate ribbon cable.
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What transistor to use? I found a schematic, but it requires 2N2222A transistor, is there any alternative? I have only KC637, can I use it? I'm new to electronics, and I'm not from US/UK, so sorry my English. <Q> Yes, it will probably work although the KC637 is lower frequency than the 2N2222. <S> This circuit probably operates at ~1MHz which is well below the 50MHz ft. <S> The other parameters look better or okay enough. <S> The pins are in a different order compared to the 2N2222 <S> so you will have to account for that. <S> The 2N2222 is <S> E-B-C and the KC637 is C-B-E. So, just flip the transistor over. <S> Also both are different from the BC637. <S> Pay attention to the direction of winding on the coils. <A> The KC637 seems to be reasonably similar to the BC637 , might be easier to find documentations for that. <S> It has a lower hFE but everything else should be ok. <S> Be aware of the pinouts! <S> The KC637 apparently looks like this: <A> it requires 2N2222A transistor, is there any alternative? <S> Yes, the 2n3904 is directly equivalent in almost all cases. <S> Most suppliers will have large numbers of this and the 2222 in stock at all times due to their popularity. <S> https://en.wikipedia.org/wiki/2N3904 <S> (page mentions the 2222 also)
| In general, if someone specified 2N2222 in a circuit, chances are high that just about anything will do.
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Do COB LEDs usually need electrically insulating from the heatsink? I have some Vero series LEDs from Bridgelux. I want to put multiple LEDs on 1 heatsink. I was planning to use thermal epoxy, so there is the possibility of the aluminium base of the LED touching the metal heatsink. I have looked through the data sheet and all application notes . I can't see any reference to electrical isolation between the LED power and the heatsink. Is this because it is standard for these type of LEDs to have electrical insulation between the power connection and it's aluminium base? I have tested with a multimeter and it is open circuit, but I will be driving the LEDs at a much higher voltage than the multimeter is using to do the test. <Q> Do COB LEDs usually need electrically insulating from the heatsink? <S> The LED chips inside the module are already isolated from the heatsink on the back. <S> They must be because these modules contain many LEDs in series and that is why they need around 20 - 30 V to work. <S> Each LED needs around 3 V so there must be many in series. <S> This can only work if all LEDs are isolated from the heatsink. <S> I would not make the voltage difference too large though but up to 50 V should generally be OK. <A> Working on a project now to connect several COB modules in series mounted on a single piece of aluminum channel. <S> Finding on about 5 out of 20 that I tested there's a small current running between a connection lead (either anode or cathode) and the base. <S> At 11 volts I'm getting 10-14 milliamps (enough to cause it to partially light up.) <S> Due to the forward voltage threshold of the diodes, simply checking for resistance with my DMM reads open circuit - you have to check for current while running them at operating voltage. <S> To add some context, I bought several dozen of these 10 watt, 12v COB modules from a couple different ebay suppliers. <S> Based on the price - $5-$10 for 10 pieces - I assume they're production seconds that failed to meet the manufacturer's QC requirements <S> so I'm not completely surprised. <S> Long story short, you may want to double check just to be sure. <A> Although I suspect they don't want you to zap <S> the chip centre with 1kV <S> , I am sure the thermal seat can withstand at least 1kV isolation. <S> Yes, electrically insulated heatsink surface is standard. <S> It claims to be somewhat ESD protected <S> but it is not clear to what levels, so avoid large negative voltage and negative discharges to +. <S> Here's another thermal reference http://www.bridgelux.com/sites/default/files/resource_media/AN31-Handling-and-Assembly-of-Vero-LED-modules.pdf
| So you can safely assume that the heatsink is already isolated from the LEDs .
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Why is the current in this circuit 100mA? I am just learning electronics and trying to reason some figures I measure in simple circuits. I need a bit of a hand with this one. I have a 3.2V power source, and 2, 1 ohm resistors wired in series. I want to determine the current in the circuit. I know that I=V/R So, I know the voltage and the resistance, so assume that after I use my multimeter over the start and end points of the circuit I should see I=3.2/2I=1.6amps However, my multimeter reports 102mA! So how does this make sense? What am I overlooking. It sounds like I may be plugging in the wrong figures somewhere..... Help please! <Q> I use my multimeter over the start and end points of the circuit <S> This is not how to measure current with a multimeter. <S> Current flows through a circuit, so to measure the current you need to break the loop and insert your multimeter in series with the current you want to measure. <S> This explains how to do it right. <A> I assume you have checked and double checked the value of your resistors and voltage source, and that you are are correctly handling your multimeter. <S> Without more details it is difficult to guess. <S> If it was sightly less that 1.6A <S> I would have said it is because of the output impedance of your source and wires. <S> But I am quite sure it is because you have hit the maximum power rating of the 3.2V source. <S> You see, if you wanted to put 1.6A at 3.2V on your load, that would be 5.12W. <S> But if your power source can not deliver that much power (a small power source of a battery), it will saturate and reduce its output voltage, and consequently, output current. <S> But you should test it yourself. <S> Connect you load to the source and measure the voltage as well as the current. <S> You will probably see that once the load is connected, the voltage should drop. <S> Nevertheless, if you could provide more details we will do our best to pinpoint what is going wrong on your circuit :) <A> Did you make sure the resistors are 1 ohm? <S> Did you measure them with your multimeter? <S> Moreover, check if your power source is actually capable of sourcing 1.6A of current. <S> Measure the voltage across the source with the resistors still connected. <S> Chances are that your load is too heavy for the source and the voltage it provides drops because of its internal resistance.
| Also check the maximum power rating of your voltage source.
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Using US lamps in Europe I've recently moved to the Netherlands from the US and have researched the use of our lamps here. In the US, we only used LED lights. From what I have researched, I can use my lamps here with a power converter or an adapter. I purchased several adapters and used some US LED bulbs and one worked for a few minutes and stopped working. The other two seem to be working. I have more that I have not set up yet. My question is shall I purchase EU LED bulbs to use with the US lamps and EU adapters? <Q> The simplest and most reliable conversion is to purchase and install new 230 V bulbs. <S> LED type should be fine. <S> Figure 1. <S> Twin cable with single layer of insulation (left) and <S> European double-insulated cable (right). <S> One thing to watch is the quality of cable and fittings in your lamps. <S> I have seen North American lamps with single-insulate cable. <S> This might be OK at 110 V but for 230 V use You will be running at higher than designed voltage and insulation should be rated accordingly. <S> You may need to get an electrician to inspect the lamps and fittings. <S> Since these are portable you can bring them to the electrician and avoid a call-out charge. <S> At worst he may recommend rewiring the lamps with suitable cable. <A> There are several different kinds of circuits used for LED lamps. <S> There are also different kinds of "adapters". <S> Many combinations of lamps and adapters are NOT compatible. <S> Your question is over-simplified and cannot be properly answered as a generic generalization. <A> What ever is directly connected to mains, should be able to run on 230V. <S> Now, there are typically two types of lamps: Those which provide mains at the sockets. <S> Here, the bulbs themselves need to accept 230V. <S> Though I didn't care about this, I think I've never seen bulbs which accept 110V and 230V. Those which need some kind of transformer/convertor. <S> They provide either 12V at the sockets (for halogen bulbs and LED replacements) or a constant current for built-in LEDs. <S> In this case, check the lamp. <S> It is possible that it also accepts 230V. <S> There should always be a label <S> / imprint about the accepted voltage range. <S> Do never connect made for 110V only to 230V. It could start a fire.
| You must do more careful and detailed analysis of how each different lamp is constructed and what kind of converter or adapter is suitable.
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How do cheap phone chargers charge li-ion batteries? I read about li-ion battery charging and it requires monitoring for the voltage and current, How does those extremely cheap phone chargers do that?Or Is it just an adapter and the charging circuitry are embeded into the phone? <Q> The device (phone, etc.) has the special charging circuit built-in. <S> And the battery pack typically has a special circuit inside to prevent excessive discharge (or charging). <S> Li-ion battery cells are inherently unstable/dangerous and require special circuits somewhere (often built into the battery pack) to make them safe to use. <A> The charger doesn't charge the battery. <A> The key to how it is possible to have cheap phone chargers made, is a combination of economies of scale, and cutting corners. <S> Cheap chargers (and some expensive chargers) will generally work but the compromises made to reduce costs are generally at the cost of safety, particularity in poor isolation between mains power & the regulated output (ever had a bit of a tingle while touching charging devices?). <S> They will generally also offer a shorter lifespan due to running components much closer to their maximums. <S> However these cheap phone chargers, simplify provide a constant voltage supply (at 5V to the device being charged) <S> it is then the devices responsibility to regulate the current and control the cut off point. <S> In general phones are a more expensive and better engineered device <S> but faults do still occur. <A> Chips that handle common, simple logic cost pretty much the cost of the materials. <S> There are few things more common than charging logic, so any logic circuits you use cost mere cents. <S> Of course phones charge through USB, which in this case is nothing little more than a glorified 5 Volt DC wall socket. <S> That means any charging logic has to be on the device side. <S> But even standalone battery chargers are dirt cheap to manufacture. <S> (I'm not endorsing that specific product, all I know is it's a standalone battery charger <S> and it's dirt cheap). <S> So everything is true: <S> Charging logic is in the phone. <S> Protection logic is in the battery. <S> Battery chargers are dirt cheap to manufacture.
| The actual charging circuit is in the phone; the so-called "charger" just provides a constant-voltage power supply to the charging circuit (and the rest of the device).
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Which type of MOSFET I am trying to set the RESET pin, which is set to high via internal pull up, of my MCU (3.3v) to low based on an external pin (3.7v). When the external pin is low, the RESET pin should be Gnd and if the external pin is high, the RESET pin should be disconnected (will get back to high because of the internal pull up). I am new to electronics by i think i found a good solution with a p-type MOSFET. Source = RESET Drain = GndGate = External pin (0-3.7v) Right now i am struggling with the wide variety of MOSFET's out there (Farnell has 1000). Could anybody please help me to calculate the right values or give me other advice for solving that problem? <Q> Rather than use discretes, I'd recommend to just use a digital inverter with 5 V tolerant inputs, but powered from the micro's 3.3 V supply. <S> If this is the only circuit driving the RESET pin, you can use for example 74LVC1G04 . <S> The cost of either of these devices is low enough that the placement of the part on the PCB is likely to cost more than the part itself (so that using a slightly pricier part to replace 2 or 3 cheaper parts pays for itself). <S> Edit: <S> I just noticed you wrote <S> When the external pin is low, the RESET pin should be Gnd and if the external pin is high, the RESET pin should be disconnected (will get back to high because of the internal pull up). <S> For non-inverting logic with open-drain output, try 74LVC1G07 . <A> Alternatively for non-inverting logic that swings from 3.7 to say 0.5v and the CMOS datasheet indicates VIH and VIL for 3.3 logic is 2.0 and 0.9V worse case (respectively) <S> you can use a Schottky diode directly between the two interfaces to perform the reset.(with internal pullup) Check and confirm both worst case output swings, and supply tolerances as well as RESET input VIH and VIL. <A> You did not provide a schematic for the scheme you have in mind, but in any case it won't work. <S> If you are using discrete parts you will need at least two transistors. <S> Here is an example using BJTs, they could be replaced with complementary MOSFETs (or you could use a dual transistor with base resistors in a single package): simulate this circuit – <S> Schematic created using CircuitLab <S> When the input voltage is less than about 2.5V enough base current flows to turn turn Q1 on, which provides base current to Q2 turning it on and pulling the /RESET input low. <S> R3 prevents leakage in Q1 from being amplified by Q2.
| If you need to do wired-OR logic to allow other circuits (like a mechanical pushbutton) to pull RESET low, you can use 74LVC1G06 .
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Calculating voltages of output and feedback voltage divider in OpAmp circuit I've been trying to calculate Vout as shown in the diagram, but I am struggling to understand the way that the circuit is wired. I need to calculate it for switch S in the lower and upper position. I realize that there is no current flowing through the virtual ground. I am confused as to why Vout, Vin, and R4 are not connected, and what "upper" and "lower" position means for the switch. <Q> Two cases: simulate this circuit – Schematic created using CircuitLab <S> When the switch is in either the lower position or the upper position, \$V_{in}\$ sets up a current into the virtual node that is \$\cfrac{V_{in}}{R_1}\$. (Call that current positive if \$V_{in}\$ is positive.) <S> When the switch is in the lower position, the opamp must pull all that current through both \$R_2\$ and \$R_3\$. But to do that, the voltage at the output of the opamp has to be of an opposing sign. <S> So the opamp output, and output voltage, will be: $$-\cfrac{V_{opamp}}{R_2+R_3}=\cfrac{V_{in}}{R_1},\:\:\:\therefore V_{out}=V_{opamp}=-V_{in}\cfrac{R_2+R_3}{R_1}$$ <S> However, when the switch is in the upper position, the opamp must still pull all that current through \$R_2\$. But not necessarily through \$R_3\$, as the load may also be sinking or sourcing current. <S> So now, the output voltage will instead be: <S> $$-\cfrac{V_{out}}{R_2}=\cfrac{V_{in}}{R_1},\:\:\:\therefore <S> V_{out}=-V_{in}\cfrac{R_2}{R_1}$$ <S> The output at the opamp will have to be: <S> $$-\cfrac{V_{opamp}-V_{out}}{R_3}=\cfrac{V_{in}}{R_1}-I_{load},\:\:\:\therefore V_{opamp}=-\left[V_{in}\cfrac{R_2+R_3}{R_1} <S> + I_{load}\cdot <S> R_3\right]$$ <S> You will need to make sure the opamp has sufficient voltage compliance, in either case. <S> With the values shown in your diagram, you will have \$V_{out}=-20\:\textrm{V}\$ with the switch in the lower position and \$V_{out}=-8\:\textrm{V}\$ with the switch in the upper position. <S> You can also see that in the upper position, and with a positive input voltage and with a positive load current, the opamp will have to have an output voltage with a larger negative magnitude than if there were no load current at all. <A> I am confused as to why Vout, Vin, and R4 are not connnected <S> \$V_{out}\$ is the voltage between the right-most terminal of the switch \$S\$ and ground. <S> \$V_{in}\$, on the other hand, is a stimulus <S> and so you can place a voltage source across the terminals where the label \$V_{in}\$ is and then label the source \$V_{in}\$. <S> Both terminals of \$R_4\$ are connected <S> so I'm not sure if that's a typo <S> or if you don't see that they are. <S> Finally, a better home for questions like this is the EE stack exchange site which includes a schematic editor which would allow an answer to easily include a schematic. <A> This is a circuit diagram of a circuit element, in this case an inverting operational amplifier that has been configured to amplify a specific signal (Vin) and output it at a certain level (Vout). <S> Firstly, R4 is connected to ground. <S> Vin and Vout are not connected, because the circuit diagram is only concerned with the operational amplifier configuration. <S> When solving for the properties of the circuit you simply assume that you are applying a signal at the Vin terminal and receiving an output at the Vout terminal. <S> Upper and lower position refers to the path the output signal would take. <S> If the switch is in the upper position the connection will be made to the path that includes R3, if the switch is in the lower position the signal will pass by the R3. <A> The op-amp inverting input is at a virtual ground (as you seem to know). <S> With a 20mV supply at Vin, then you know that 20mV/1\$\Omega\$ = 20mA is flowing through R1. <S> The value of R4 is irrelevant for an ideal op-amp (and non-ideal for an non-ideal op-amp- <S> it should be closer to 1 ohm). <S> Since an ideal op-amp has no input current, that same current must be flowing through R2 + R2. <S> We are going to assume that nothing is attached to Vout except an ideal voltmeter that has infinite input impedance. <S> The switch is a two position switch that can be set the upper position as SW1 below or the lower position as SW2 below. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Now, the switch does not change the way current flows, it merely selects a different connection to Vout. <S> Since a known current of 20mA flows through both R2 and R3 in series <S> and we know what voltage is at the left side of R2 (0V <S> ) we can easily calculate the voltages at the Vout in either position. <S> The only thing a bit tricky is to notice what the sign is (the output voltage will be negative for positive input voltage). <S> In the upper position, the voltage output is that across R2 only. <S> In the lower position the voltage output is that across R2+R3. <S> So the output voltage Vout is -20V in the lower switch position (and in fact that's always the voltage at the op-amp output) and Vout is -8V <S> with the switch in the upper position. <S> Here is another way to represent the circuit: simulate this circuit <S> If you run the Circuitlab simulation you can move the switch to either position and read the output voltage. <S> It will be almost exactly the predicted values because the particular op-amp I chose is almost ideal under DC conditions.
| If it helps you conceptually, place a voltmeter across the two terminals where the label \$V_{out}\$ is and consider the voltage measured to be the voltage \$V_{out}\$.
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How does a laptop power adapter automatically "adapt" to world voltages and frequencies I would be interested in understanding the inner workings of a laptop power supply. In particular how does it automatically "adapt" to world voltages and frequencies. I suppose there must active components to achieve this. In particular I would be interested in schematics explaining the principle behind this. Thanks! Edit: By power supply I mean the components that are inside the "brick" with AC in-DC out <Q> First, most (maybe all) laptop power adapters are offline flyback converters . <S> Here's a simplified flyback converter: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> When SW1 closes, D1 gets reverse biased so no secondary current flows. <S> This results in an increase of primary current linearly and then energy storage in the primary of XFMR1 according to \$\frac{1}{2}LI^2\$ (\$I\$ is the peak value of that linear-ramp-shaped current waveform). <S> When SW1 opens, all the polarities on the transformer reverse, D1 conducts and the energy stored in the primary winding of XFMR1 is transferred to the secondary (i.e. load). <S> NOTE: <S> Actually, XFMR1 acts as a coupled inductor, not a transformer. <S> The voltage across the load is sensed and stabilized by the feedback & control unit by controlling the on-time duration of SW1. <S> Let \$t_c\$ be the on-time duration of SW1 and \$f\$ be the switching frequency, so the duty cycle can be defined as \$D = t_c f\$ <S> (Note that the switching frequency can be between, for example, 20kHz and 300kHz) . <S> The output voltage can be calculated as \$V_o = <S> D <S> * V_{DC <S> } \$. <S> So, if \$V_{DC}\$ gets too low then the FB&C unit increases \$D\$ (By the way, \$D \$ cannot be higher than 50%, theoretically. <S> In practice, most designers limit it to around 45%) . <S> Likewise, if \$V_{DC}\$ gets too high then the FB&C unit decreases \$D\$. <S> The transformer is designed according to minimum and maximum input voltages so that the circuit can work between those input voltages (Input voltage adaptation) . <S> Since mains AC is rectified and filtered to obtain a DC voltage (because the flyback converter needs DC on its primary) , the mains frequency does not matter a lot (Mains frequency adaptation) . <A> You might want to look into SMPS power supplies or offline switchers. <S> It works like this: 1) ac voltage (85-265 Vrms) <S> is rectified using a bridge rectifier. <S> 2) <S> Ripples are smoothed out using a capacitor essentially giving you a high voltage DC. <S> The DC voltage (120V to 375V) depends upon the input voltage. <S> This is the point where you lose frequency information and hence the power supply can adapt to any frequency (50 Hz or 60 Hz). <S> 3) <S> If you have a lower DC voltage to start with, the circuit will simply adapt to it by increasing duty cycle. <S> Here power supply is adapting to multiple voltages. <A> One simple answer is that it keeps accepting input voltage - up to a maximum of 250V - and "builds up" to the desired output voltage, charging a capacitor for when the AC cycle doesn't have enough power to power things. <S> As soon as the output voltage is reached, it stops doing the conversion. <S> If the input voltage never gets to the maximum, it just uses more of the input voltage until it gets to the desired output. <S> Thus for a 240V it may use (say) 10% of the AC cycle, while for 120V it may use (say) 20% of the AC cycle. <S> And this is why it is insensitive to the AC's duty cycle - it frankly doesn't care, since it's going to produce DC output. <S> There were some older devices that used the AC cycle rate to drive something else: maybe a clock, or a refresh rate, or something else. <S> It was there, so why not? <S> The answer was simply that it made it harder for worldwide acceptance. <S> Now most devices (re)generate their own cycles as necessary from a pure DC source. <A> The powersupply of a laptop has a primary (mains) part and a secundary (laptop) part. <S> The incomming voltage on the primary part is rectified and the dc is then fed to some kind of (flyback) inverter. <S> The output side of the inverter is connected to a rectifier and filter to produce the output power for the laptop. <S> The inverter is governed in such a way that the incomming voltage is adjusted for and the desired output is obtained. <S> For more information on the subject of switched mode power supplies you can look on the internet <A> Start from the output - say it is 12 V DC. <S> This is produced by a process of voltage regulation. <S> That voltage regulator might take an input dc voltage range of 14 V to 30 volts and still produce a regulated output voltage of 12 volts. <S> So, imagine instead that the voltage regulator could work with an input DC voltage range of 120 V to 375 V DC. <S> It's the same principle but uses a few more components. <S> That DC voltage range arises when you rectify (and smooth) <S> an incoming AC range 85 V AC to 265 V AC.
| Whatever DC voltage you have is switched at high frequency using a specialized circuitry to generate a fixed voltage.
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How to stop plastic granules sticking to the iron wall of their silo? In our factory, we have some large (8 meters tall and 3 meters in diameters) iron silos for storing the plastic compounding material . Every time the silos are getting unloaded from the granules, the plastic powders stick to the inner metal layer due to the static electricity, so that the operator needs to knock on the wall severely in order to make them fall. I want to know: 1- How can I instantly make this static load discharged and get rid of this problem (Right now, the resistance value between the earth system and Null is almost 0.03 Ohm) I would appreciate if you mention the standards and the detailed answers. 2- How can I measure the quality of the grounding system and what would be the standard range of this value? <Q> These are not uncommon in the plastics industry. <A> You can't easily do what you ask since the problem is charge gets built up on the plastic particles. <S> Since they are all enclosed in one uniform conductor, it won't matter what the voltage of this conductor (the tank wall) is. <S> There will still be charges on the plastic particles relative to the tank wall. <S> Changing the voltage of the tank wall will only effect the electric fields between the tank and the ground and other things outside the tank. <S> It won't do anything to the electric field inside the tank. <S> If you really wanted to counter the attractive force between the plastic particles and the metal wall, you'd have to introduce a different conductor into the tank and drive it with a voltage relative to the tank wall. <S> One polarity will drive the particles more towards the wall, and the other will drive them away from the wall. <S> Experimentation is the easiest way to find the right polarity. <S> However, even if you put a narrow cylinder in the middle of the tank and drove it with the appropriate voltage so that the particles fly off the wall, you'll have the same problem as now they'll be stuck to the center cylinder. <S> A different strategy would be to reduce the charges built up on the plastic particles in the first place. <S> Plastic can shed or grab electrons as it is rubbed against other material. <S> This is exactly what is happening as the tank is drained. <S> Since the plastic is a good insulator, those charges stay on the particles for a long time. <S> The first possibility that comes to mind is to increase the humidity in the tank during draining operations. <S> The plastic still sheds or grabs electrons, but the humidity makes surfaces a little more conductive so that these charges bleed off faster. <S> Whether adding humidity is feasible, and whether it decreases the charge bleed time sufficiently is something you have do decide. <S> There are also various chemicals that essentially coat surfaces with something at least a little conductive. <S> Again, whether you can tolerate those is something you have to decide. <A> If you simply google for "ionized air static control", you'll find many commercial products. <A> Plastics have the tendency of giving up or collecting electrons (depends on the kind of plastic) and become positively or negatively charged. <S> If this charge is disturbing a proces then you need to treat the plastic by adding surface conductivity thereby preventing this kind of behaviour. <S> Such an additive could be moisture or anti-static spray. <S> Such an anti-static spray is often made up from a soap based material dilluted in a solvent (mild alcohol). <S> A fire retardant is then added to combat flammability. <S> The plastic has now become conductive and as long as the coating is not disturbed it will be difficult to generate static electricity in this material. <S> If additives in whatever form are not possible then it might be a solution to make the plastics adhere to a place where the unloading of the silo takes place. <S> In that case you are not changeing the plastics and even make use of the fact that they are charged. <S> I would have to see more for further assistance. <A> What about if a metal rod was added to the centre of the tank? <S> If that rod was charged with an AC potential, that went both below and above the outer wall's ground potential, then you may get what you are looking for. <S> Not very much - I'm not trying to electrocute anyone - but if the attractive force changed between the tank and the rod then the plastic would in turn be attracted to the tank wall and the rod. <S> Given the dimensions involved, it woudn't take long for gravity to win overall, and have the plastic fall to the output feed. <A> Ionized air seems to be the best solution to your problem. <S> Moisture is verboten for plastic moulding granules and chemical additives, ammonium acetate in alcohol, as used in many anti static sprays would not be tolerated also, fire risk, etc. <S> Delivery of ionized air into silo during discharge may have some issues, pellet dispersal, ozone production etc. <S> Further thought. <S> Recirculating portion of pellet discharge through deionizer and back into hopper would reduce problems with ecessive ozone and reduce amount of ionization required during discharge process.
| If your plastic pellets are getting charged, you can flood the inside of the silo with ionized air using a commercial generator . Ionized air is frequently used for static control in industrial processes.
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Stiffness control of DC motor using PWM Is it possible to control the output torque of a brushed DC motor using PWM? From my understanding the speed is directly proportional to voltage applied and torque is proportional to current drawn. So if the load of the motor increases it will just simply draw more current (if the load is within its torque limit). Since I do not have any current controller in my motor driver board there is no way I can control the output torque of the motor. How can I solve this problem? I read that a hardware feedback loop using an operational amplifier can be used to keep the current proportional to the reference voltage. Is this correct? And is it possible to make such a circuit for a 90W, 24V DC motor? <Q> "Is it possible to control the output torque of a brushed DC motor using PWM?" <S> Yes - torque is proportional to current. <S> If the PWM source is a micro-controller then you need to measure the current (consider ACS712 or similar which will output a voltage: V=Voffset + k*Imotor), capture with an ADC and then write a PI control loop to adjust the PWM duty cycle. <S> Here is a example control loop that might be called in an interrupt (of course removing the while(1) then) <S> while(1) { Ierr = <S> Idesired - Imeasured; err_sum <S> += <S> Ierr; PWMcommand = <S> Kp <S> * Ierr + Ki <S> * <S> err_sum;} You'll need need to scale the variables, work out units, and signs, etc - but that is the basic structure. <S> For tuning, set Ki to near zero, lock the rotor and apply a step in Idesired and look at the response. <S> The step current needs to be similar to that which you'll drive the motor. <S> I'll assume its 1amp. <S> Increase Kp until the step response has overshoot then back off a bit. <S> With Kp alone, there will always be an error in Imeasured - esp once the motor starts during. <S> To correct this you need the integral term. <S> With the rotor locked, continue your ~1amp current step and increase Ki. <S> Again you want Ki large - but not so large as to cause oscillation. <S> If Ki is too small the PWM response will be slow when the motor speed changes - and thus the torque will change. <S> The PWM frequency also effects the control loop coefficients (Esp the Ki) <S> So use a constant PWM frequency. <S> The frequency is typical at least 20Khz to not be audible, but ultimately it more about the motor inductance, current ripple and the rotor inertia. <S> There are countless university classes dedicated to the topic of control theory. <S> These are needed to make a surgical robot - but what I said will get your first torque controller working! <A> I get the impression that this question is partially about speed control of a DC motor rather than just trying to keep the torque constant to a demand setting. <S> From my understanding the speed is directly proportional to voltage <S> applied <S> This is only true if torque isn't changing. <S> With increasing torque at a constant voltage supply, speed will drop. <S> This effect is caused by the finite DC motor armature resistance. <S> If this resistance were zero then speed is proportional to voltage. <S> So if the load of the motor increases it will just simply draw more current <S> And, unfortunately slow down a bit. <S> I read that a hardware feedback loop using an operational amplifier can be used to keep the current proportional to the reference voltage. <S> Is this correct? <S> It's not great <S> but it does significantly improve speed regulation compared to not having it. <S> Try looking up a document by Texas (Burr Brown) entitled DC MOTOR SPEED CONTROLLER: <S> Control a DC Motor without Tachometer Feedback . <S> This explains how it is achieved. <A> You can control the torque of a brushed DC motor by controlling the current through the rotor respectively by controlling the voltage over the rotor. <S> You won't be able to control the motor current if you don't have a current sensor, but you will be able to control the speed of the motor if you have a speed sensor. <S> Your target is to command a speed of zero rpm and the controller will adjust the rotor voltage in order to achieve this target; it will control the "stiffness" of the motor. <S> Read about motor control <S> and you will find the answers you're seeking for.
| You can use a series resistor (monitoring motor current) and op-amp to create a near-constant speed characteristic across a decent range of load torques.
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LED intensity changes with time I want to operate a UV LED at continuous 20mA for one week. However, when I measure the LED intensity with a photodiode (see image), I find that the light intensity changes continuously in time. I would like to ask you how I can reduce the light intensity drift that I am observing. The LED data sheet is: http://static.vcclite.com/pdf/VAOL-5EUV8T4-LED-5mm-UV.pdf . I am operating it in series with a 461.6 ohm resistor and a 12 V AC-DC power supply. The voltage drop around the LED is 3.2 V, the current is 19 mA. I am measuring the light intensity with a photodiode connected to an Arduino. I attach a picture of the circuit and the graph of light intensity (photodiode voltage output) vs time (hours). <Q> If you look at the data sheet for the LED it says that at 20 mA the forward voltage may be between 2.8 V and 3.6 V. <S> Importantly it states that this is at an ambient temperature of 25 degC. <S> So, how well regulated is the ambient temperature in your experiment? <S> I ask this because you seem to be relying on generating a constant current by using a resistor and a 12 V supply and, if the volt-drop on the UV LED changes (maybe due to temperature or aging) then the current into the device will change and the light output will also change. <S> Gradual self-heating of the LED cannot be ruled out and <S> neither can drift on the 12 V DC power supply. <S> Now, onto the photodiode - its "gain" will be somewhat sensitive to temperature <S> but less so than for infra red types however, dark current will approximately double for every 10 deg C rise so this might start to affect the accuracy of your experiment if you are using the device in photoconductive mode. <S> If you are using the device in photovoltaic mode there may be temperature effects that make inaccuracies worse compared to photoconductive mode. <S> Also, what about local ambient lighting conditions changing? <S> In a darkened room I think you should be OK <S> but if there is any "spillage" of the sun through a window, then significant errors could be caused. <A> Thank you all very much. <S> I have provided a constant current supply with LM317 to the LED and a constant voltage of 1.5V to the UV photodiode with another LM317. <S> This voltage is counteracted by the photodiode current <S> and I measure the voltage across the photodiode using Arduino as described here <S> http://provideyourown.com/2012/secret-arduino-voltmeter-measure-battery-voltage/ <S> I have also used two pipette tips to align the LED and the photodiode and maintain them rigidly in place. <S> The light intensity measured by the photodiode is now constant and the LED is operated at 20mA. <S> Thank you all very much for your help. <A> Your photo and graph are useful. <S> In your graph, at about 7 hours, a large step appears in your photodiode output. <S> What was the cause? <S> It may give you a clue that will help debug your experiment. <S> Your breadboard setup may look mechanically stable, but very small perturbations in diode or LED position can make a large change in photodiode current. <S> You can monitor LED current by measuring the voltage drop with a multimeter across that 470 ohm resistor in series with the LED. <S> Ensure that it remains constant during your experiment. <S> Your photodiode will likely be a silicon photodiode whose wavelength sensitivity is maximum at much longer wavelengths (like 0.9 microns). <S> Your LED on the other hand, puts out most of its light power at a much shorter wavelength (0.38 microns). <S> Your photodiode response at this short wavelength will be far, far lower than at its peak-response. <S> This means that any stray light will have a significant influence on the photodiode. <S> Even a weak incandescent source (whose peak wavelength corresponds to the infrared peak of your photodiode) can contribute current to your photodiode detector. <S> Your experiment suggests that light output falls with time. <S> Is this experiment repeatable? <S> Does photocurrent continually decrease with time? <S> To do this experiment properly, the photodiode - LED combination should be very firmly fixed into a rigid tube so that they cannot move about. <S> Then the experiment should proceed in a constant-temperature environment, in a light-tight box. <A> Since this is a test circuit make a small current supply with lm 317. <S> Use a resistor of 1,25/20 = 625 ohm. <S> You can use 680 ohm for the test. <S> If this turns out to be stable you can change the circuit to a permanent one. <S> Diagram can be found on the internet or in the app sheet
| You should consider a precision constant current supply - there are circuits you can build around simple op-amps that will achieve this.
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Is it safe to use 1/8 watt resistors in this circuit? If I have the following circuit is it safe to use 1/8 watt resistors? My details are as follows : 5V at 800mA == it is an old phone charger! I assume the voltage is constant but the current will depend on the load ( I don't know how to prove this, just am assuming this to be the typical behaviour of phone chargers? ) The charger itself is a Nokia AC-15X - https://www.amazon.co.uk/Nokia-AC-15X-Compact-Travel-Charger-BLACK/dp/B004I647J2 The resistors are rated at 1/8 watt. The led draws I think 50mA. simulate this circuit – Schematic created using CircuitLab I know that P=IV , so if this is the case, P=0.050A x 5V == 250mW - suggesting that this is over the 0.125 watt rating of my resistors. If this is the case, can you recommend a decent set mixed value 5 watt resistors (just so I am safe) -- seems to be loads of little selections on 1/8 or 1/4 watt but I cannot find a nice range on higher power ranges! <Q> P=0.050A x 5V = <S> = <S> 250mA <S> dropped into consideration. <S> You need to reduce the supply voltage by the forward voltage of the LED to determine the voltage dropped across the resistor. <S> But if we assume a forward voltage of 1.2V (typical of an IR LED, one of the few that would use 50mA) then we get a power dissipation of 190mW. <S> So either move up to a 1/4W resistor or use two 1/8W resistors in parallel. <A> Don't guess. <S> confirm or show parts and ask " <S> The led draws <S> I think 50mA" <S> 5mm 20mA rated for Red/Yellow @2.2V <S> = <S> 44mW <S> thus R = <S> (5-2.2V)/20mA= <S> 140 Ohms and 3.8V*20mA = <S> 74mW dissipated in R <S> so <S> 1/8th W is ok <S> 5 <S> mm 20mA rated for Blue/White @3.2V <S> = <S> 64mW <S> thus R= (5-3.2V)/20mA= <S> 90 <S> Ohms and 2.8V*20mA = <S> 56mW dissipated in R <S> so ok. <S> If really 50mA rated or a 350mW rated LED then R will need bigger W ratings. <A> Power is measured in Watts, so P=0.050A x 5V = <S> = <S> 250mA is incorrect. <S> It would be milliWatts, or P=0.050A x 5V == 0.250W . <S> As 0.25W is 2x 0.125W that resistor would get very hot, and likely fail. <S> The actual power dissipation in this situation is less than you think. <S> That LED will 'use' (drop voltage) according to its datasheet 2V, so there is only (5V-2V) = <S> 3V across the rest of the circuit, in this case the 100 ohm resistor. <S> So the current is V/=I, (5- <S> 2)V/100ohms = 0.03A, hence Power = 0.03A <S> * 3V = 0.09W. <S> That is a bit close for 0.125W resistor, but 0.25W would be fine. <S> Or two <S> 200ohm 0.125 resistors in parallel, or, if 100ohm 0.125W is all you have got, 4 resistors, two parallel pairs of two in series. <S> Edit: A resistor doesn't draw current. <S> There is 5V across the resistor and LED, and that 5V is all driving the current. <S> Looking at the LTL-307EE datasheet , practically, the LED will not conduct until it has 1.6V across it <S> (Fig 2 in the datasheet); with a voltage below this, it looks like a large resistor, 1000's of ohm or much more. <S> At 2.0V it will conduct 20mA. <S> As the voltage increases it will 'conduct like crazy', needing little more voltage (data sheet suggests about 2.6V) to drive so much current that it will be destroyed. <S> So for voltages above about 2V, the circuit needs something to limit the current, and so stop the LED from destroying itself. <S> That is what the resistor is doing. <S> However, all of the power, P=VI, <S> 3v*0.03a = 0.9W, is being converted to heat by the resistor. <S> So the power rating of the resistor is statement of how much power it can convert to heat. <S> But it's temperature will rise the closer to its power rating it must dissipate. <S> So generally a resistor with 2x the required power rating is used. <A> I looked up the specs of the led listed in your schematic at digikey . <S> All other answers explain quite in detail how to calculate the power dissipation of all the circuit, so I will be very brief here. <S> LED <S> forward voltage: <S> 2V, max 2.6V <S> Continuous <S> forward current: <S> 30mA <S> Peak <S> forward current (if you are doing some fancy PWM or the like) 120mA Taking your 5V supply, lets proceed to calculate your resistor. <S> We want 30mA, at 5V, with a forward voltage of 2V. <S> So the voltage at the resistor will be 3V, meaning that it has to be 100Ω <S> to obtain the desired 30mA <S> (Just as you have placed it). <S> If your particular LED would have a higher forward voltage, the resistor will limit the current below the recommended (by the manufacturer) <S> 30mA. <S> And now to your question. <S> Using Joules equation (P = I²·R): P = <S> (30mA)²·100Ω = 90mW. <S> Which is less than 125mW.
| A good rule of thumb is to use resistors rated for 2x the power. No, the power only takes the voltage Therefore, your circuit will be just fine (as long as the ambient temperature in which your device operates is below 70ºC as recommended in this thread ).
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How can I know what parts a manufacturer uses in their schematic? I have an old printer, and I have the schematics from HP. The image is below. They have names on the parts, but if I look up something like "PM706," I can't get anything. How can I find out what a manufacturers parts are from the schematic ? Is it possible? I don't want to take things apart, if I don't have to. <Q> Many years ago HP, as a big volume manufacturer, used IC components (and transistors, and other parts) from huge variety of manufacturers, but all of them had proprietary HP-specific markings. <S> Without a special cross-reference list ( like this one ) it is not possible to establish correspondence with common parts, although they almost all have exact original substitutions. <A> The image provided is not a full schematic. <S> Circuit schematics have two (at least) designations for each part: a part code and a part number. <S> If both are not shown directly on the schematic, a separate document (the bill of materials, or BOM) will provide the cross-reference. <S> The part number is what you want <S> if you are looking for replacements - in other words, you want the designation for the part used by its manufacturer (e.g. Texas Instruments, Murata, Intersil,...) <S> The letter prefix follows some semi-standard codes, and the number is arbitrary (but unique for a given system). <S> Q100 = transistor #100 R34 = <S> resistor #34 J324 = connector #324 <S> In other words, part codes do not exist outside of the given assembly/circuit board and you won't find them in a part number search. <A> The image you show is NOT a circuit diagram. <S> It is a block diagram that shows on a higher-level what the blocks are and how they connect to each other. <S> There is a lower-level, more-detailed drawing that shows the actual circuit details down to component level. <S> "PM706" is NOT a part number. <S> It is a Reference Designator. <S> Like R7 or C3. <S> Very likely: "PM" means "pickup motor" (or similar). <S> And "7" is the likely the major section of the printer (like the input paper handling part), and "06" designates that this is motor #6 in that section of the printer. <S> There are other documents that actually identify the motor called "PM706". <S> However, it is certainly a custom part made for HP and <S> the part number on the motor is a "house number" from HP. <S> You don't reveal WHY you want to "look up" the motor. <S> But the chances of researching anything useful ranges between slim and none. <S> You have no practical choice but to actually measure and experiment with the motor yourself.
| The diagram you are showing only has the part codes, which are designations generated by the design engineer for his/her own bookkeeping purposes.
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What is the (most likely) technical reason behind temperature specifications? Im talking about electronic devices, which may be handheld or similar, so nothing exposed to "artificial" temperature (welding, soldering, cooling with liquid gas). Devices designed to be operated by ("unprotected") humans, indoors or outdoors, such is the temperature range. My question is: What are the especially temperature sensitive parts? I guess Im interested in the lower end (0°C and below). Are the batteries sensitive? What is also likey to fail when it gets too cold? - Which in turn also mean: which part must be kept warm if one has to attempt to operate such device in too cold a enviroment? Example: I have recently bought a laser distance meassure, a UNI-T UT391A+. The specified operating temperature range: 0°C to 40°C. Many similar devices (even the UT391A, without +) specify -10°C or lower as their lowest. Why is that? The most obvious difference between UT391A+ and others is that the former has integrated bubble levels. Could that be what bumps the temperature tolerance by 10°? <Q> The most common reason for electronic limits on freezing is the moisture absorption in molecule sizes large enough when frozen and expand <S> cause microscopic wirebonds to shear. <S> (popcorn effect) <S> Although there is no problem with storage to -40°C, the rapid self-heating with microscopic amounts of frozen water vapour can cause the most damage. <S> Over the decades, plastic has improved in black encapsulated epoxy to improve the grade of moisture seals. <S> But for clear plastic epoxy encapsulation or lens, the seal technology is not there yet. <S> I do not know which laser is used, but here is an example of such a specification. <S> You might be able to keep it running and go outside in sub-zero weather without the thermal shock of heating frozen H2O molecules in a well, sealed case, but operating when stored below 0 <S> °C can be a problem, but not guaranteed to fail. <A> Usually the semiconductors are the least temperature sensitive devices at low temperatures, a far as functionality goes. <S> They may be rated for 0-70°C (commercial temperature range) <S> but in fact many will operate fairly well down to liquid nitrogen temperatures (77K). <S> Capacitors (especially electrolytic and many ceramic types) will drop in value, sometimes precipitously or the internal resistance will increase greatly. <S> LCD displays can have problems with speed and temperature compensation. <S> Aside from the ceramic caps, some of which are just very temperature sensitive <S> , most of these devices have liquid contained within them. <S> There is not a lot of reason to specify devices over a very wide range (and test them with margin to ensure that) if they are not going to be used in an automotive or military environment, and the testing and spec'ing allows the semi makers to segment the market and charge quite a bit more money for wide range devices. <S> Many commercial semis will typically operate over a much wider range with relaxed specs- <S> for example if you clock the chip at 250MHz rather than 300MHz. <A> You might test a batch of prototypes at -10°C, and claim 0°C as the minimum supported temperature. <S> Testing at -40° <S> C costs more, in terms of time and test equipment. <S> Its not unknown for the same product to be labeled with different specs and sold at different prices (assuming that there are no yield factors to consider). <S> Equally, fairly trivial features can limit the temperature range if they require specific components. <A> There are three main temperature ranges for electronic devices: commercial : 0°C to 70° <S> C industrial : <S> -40 <S> °C to 85°C military : <S> -55 <S> °C to 125 <S> ° <S> C when the temperature range increases, so does the price because the design complexity increases. <S> Pretty much <S> any physical parameter depends on temperature: metal changes in size, things burn, and transistors do funny things. <S> A circuit that works perfectly at room temperature might start misbehaving if brought too far from it, either increasing or reducing temperature. <S> A common resistor would change in size due to metal expansion/contraption, changing its resistance value. <S> Such an effect can even be used to measure temperature. <S> Guessing what starts to fail in your device is hard if not impossible without having a look inside, <S> but there is a big catch <S> : it is a measurement instrument. <S> I guess laser measurement sends a light pulse and measures how long does it take for it to bounce back, then converts the time into distance since light speed is known. <S> How can you measure time? <S> For example you can use a counter. <S> You reset the counter, then when the pulse is sent you start it. <S> When the pulse gets back you read the value, and knowing clock speed you can find out how much time has passed. <S> And what about the clock? <S> A crystal with some fancy analog thingies generates it. <S> Crystals are pretty stable, but they do change with temperature. <A> There are three main drivers for a specified temperature limit: 1. <S> Physical considerations <S> Material science is non-trivial. <S> Component manufacturers must deal with expansion coefficients. <S> A typical IC will be packages with silicon + aluminum bondwires + some form of plastic housing. <S> All these materials expand at different rates and while an attempt is made to match their expansion as close as possible, at some temperature it will literally rip itself apart. <S> 2. <S> Material considerations Coupled with expansion is the ability for a compound to still be viable at temperature. <S> Silicon semiconductors stop behaving as expected <S> around 175°C. <S> Polyester (capacitor dielectric) starts to break down over 100°C. <S> Epoxies go through a glassification stage beyond a certain temperature. <S> 3. <S> Stated component characteristics <S> Likewise they are "bound" to what is stated on their datasheets. <S> Due to production tolerances and spreads not all parts are constructed equally. <S> If a particular batch meets the datasheets for the Military temperature range <S> (-55°C to 125°C) <S> they are marked for such usage. <S> If they fail Military they are checked against Automotive (-40°C to +125°C), then Industrial (-40°C to 100°C) and finally Commercial (0°C to 85°C). <S> If a particular batch cannot meet the datasheet spread they then scrap the parts. <A> At the top of the list of suspects for the temperature range limits would be the battery. <S> Many/most battery chemistries do not do well in very cold temperatures (or very hot, either for that matter).
| Batteries often perform very poorly at only moderately low temperatures- internal resistance increases and capacity shrinks. One simple explanation of a conservative temperature spec is just the cost of testing, compared with the value-add to the product. Due to component spread and tolerances, over a wider temperature range, the range of behaviour that you should expect will increase. Component manufacturers want as high a yield as possible from their manufacturing process.
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Capacitor emplacement during routing I'm actually designing a circuit with decoupling capacitor. I'm done with the schematic and now I'm working on the routing. Here come the question : In schematic you can put the capacitor anywhere between the supply line and the ground. While in routing the capacitor have be as close as possible to the chip. Is there any more requirement? For exemple : simulate this circuit – Schematic created using CircuitLab Please, look at it as if it was a layout. Is there any difference if the decoupling capacitor is before or after the IC? Normally these two circuits will have the same behavior but I got a doubt.' Don't hesitate to comment if I wasn't clear enough. Thanks :) <Q> On the PCB layout, decoupling capacitors are normally placed in close proximity to the power pins of ICs. <S> This allows local current transients to be evened out without disrupting current paths elsewhere on the board. <S> Other capacitors intended for general "bulk smoothing" can and should be distributed across the board where feasible. <S> These techniques are generally good practice in most applications. <S> Often decoupling caps for digital ICs are all grouped together and designated as "decoupling for IC1", "decoupling for IC2" and so on. <S> Doing it this way keeps a complex design with many ICs from becoming untidy due to the large number of decouplers that accompany the ICs. <S> So, on the schematic it really doesn't matter where the caps go as long as they are attached to the correct nets. <A> You asked a broad question, and many factors come into play. <S> First, you need to understand why capacitor takes place, and what, from physics point of view, happens to all components involved. <S> In your example decoupling capacitor "protects" U1 chip from noise generated by the power supply and other components connected to power rails. <S> And, at the same time , it protects other components from noise generated by the operation of U1. <S> Your task, as designer of the circuit and layout, to ensure as many components as possible are protected from interference. <S> In more complex and sophisticated circuits you may place capacitors between pins, or between integrated circuits - this all depends on noise frequency which could be generated by operation of the IC (or any other component) and possible interference between these components. <S> And last, but not least. <S> Be careful putting decoupling capacitors with large difference in their values close to each other. <S> I did not experience it yet :) <S> but I heard that in unfavorable situations capacitors and board may form oscillation circuit which may affect power quality badly. <A> In the schematic there will be no difference, other than clarity of how you intend to connect your capacitors in your layout. <S> Anyhow, if you have special requirements regarding decoupling or bypassing, you should specify so with a brief written note in your schematics. <S> Actually, the best way to place your capacitor in your schematics is the one that helps most with readability regarding what your circuit does. <S> Meaning that in your example circuit 1 is much clearer to me, as I clearly see that the capacitor is intended four decoupling. <S> For more information about how to place your capacitor in your layout (once you start the next step), consider also reading this post about decoupling capacitors .
| A well drafted schematic will include annotations clearly indicating where there are special requirements for part placement in relation to other parts. In simple, low frequency circuits like yours, having only two components, it may not matter where you place cap to, but I would anyway recommend to place it "before" the chip, or exactly between its pins if possible.
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What is the current capability of the common "C" wire for house thermostats I'm not entirely sure if this belongs here, so if it belongs somewhere else let me know. Home thermostats (in the United States at least, I'm not sure about other locations) for newer homes have a few different wires (depending on capabilities for the home) that are used for turning on the heater, turning on the AC, and turning on the fan. There is typically two more wires, one that is the hot (connect this to the AC wire will turn on the AC, connect this to the heater to turn on the heater, etc), and one that is the "common" wire and is used for powering the thermostat itself. These all run at 24VAC that comes from a transformer stepping down the mains voltage somewhere in the system. Are there any building codes or product standards that would tell me what the minimum current carrying capability of this circuit is? How much current can the transformer handle? How much current is the wire supposed to handle? <Q> The only wire currently sold by US big-box stores for thermostat use appears to be AWG18 x <S> whatever number of conductors. <S> Most of the transformers are 20-40VA (some are 10VA though) and are so marked, along with Class 2/Class 3. <S> Many are impedance-protected as well as fused <S> so they may just limit the current and get hot if you short the output (and if a primary fuse does not blow). <S> They probably have terrible regulation as a result. <S> Anyway, a 10VA transformer would be be able to supply about 400mA RMS at 24VAC or <S> if you full-wave rectify it and filter it about 250mA at 30VDC or so. <S> Scale up for the VA of the transformer. <S> If you do test the short-circuit current of the transformer you might want to find out where the primary is fused and make sure you have an equivalent spare on hand, because the primary fuse (often something like a glass or ceramic fuse on a furnace control printed circuit board behind a service panel on the furnace) may well go. <S> It won't blow a mains circuit breaker, more likely a somewhat hidden fuse. <S> Note that it's normal to have the supply for the central air control in the furnace and if you want to remove power from the 24V and the 240VAC main power you have to turn both off. <S> Also, keep in mind that the 'C' wire is not always available at the thermostat location, especially in older systems. <S> For that reason smart thermostats often have a battery inside and may cycle the heat or cool to keep the battery charged, <S> even if unneeded-- this is so they can have the largest possible market. <S> Homeowners may be fine with rewiring an existing thermostat, but fishing a new wire through various finished interior walls is a bridge too far. <A> The transformer installed inside an HVAC unit may be sized to operate only relays in the unit assuming a "dumb" thermostat that needs no power. <S> The thermostat wire is sized for mechanical durability, not for current carrying capacity. <S> Codes may require the circuit to be Class 2, 30 volts and 100 VA maximum, usually 24 volts and no more than 4 amps. <S> It could be much less than 4 amps and most of the capacity may be required for the relays in the HVAC unit. <A> Normally voltage distribution losses use cables for Ac designed with not exceed 5% of the cable voltage at max current. <S> But 24Vac used for furnace relays and doorbells <S> are quite low current far less than the ampacity of the wire gauge specified for a 10'C rise or higher. <S> Installation standards are regional so there is no universal answer. <S> Do you have a specific region or wire gauge and extended application in mind?
| If you are asking how much current you can depend on getting from an existing transformer to use to power a smart thermostat, I believe the answer may be not much.
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Should wires be tinned to under the insulation? When tinning a wire with solder, is it better to allow the solder to go under the insulation (so long as it doesn't bulge or burn the insulation)? I've heard arguments for both, but never really got a final answer. <Q> NASA, in the NASA Training Program Student Workbook for Hand Soldering , page 9 et seq. <S> , say: ... adding solder to the wire until the tinning has reached no closer than 0.5 mm (0.020 in.) to insulation. <S> I think NASA is a reasonably authoritative source for that. <S> At a guess , I would say that one of the reasons is to avoid embrittlement of, or other undesirable effects on, the insulation. <S> Also, allow the tinning to flow all the way to the insulation could push flux inside - flux should be cleaned off after the soldering operation and that can't be done if it is stuck inside. <S> My suspicion of wire embrittlement through heating looks like it could be unfounded, as thermal strippers can be used to melt insulation to remove it without damaging the wire. <S> That can be achieved by clamping the insulation to ~something~. If it is in an environment where there is high vibration and/or thermal cycling, it may or may not be desirable to leave a curve/loop in the wire, but that would require an analysis of the operating environment. <S> I refer you again to the aforementioned NASA document where they repeatedly refer to "Proper insulation clearance" after making a soldered connection with tinned wire. <S> If you want insulation over the tinned portion of the wire, you can use heat-shrink tubing. <S> It is available with an adhesive inner coating which is activated when the heat shrink tubing is heat-shrunk, and with the adhesive can provide a watertight seal. <S> However , Asmyldof points out that: <S> The point where the insulation stops over the soldered wire, be it PP, PVC or heatshrink, will become the "focal point" of bending stress. <S> A soldered strand of wire breaks easily. <S> If you leave some length of stranded wire, that length will be the most bendable point in the wire, thus taking the risk somewhat away from the solder. <S> Adding heatshrink to the soldered wire only moves the problem down. <S> [I changed "room" to "length" in the quote.] <S> So heatshrink tubing is not a substitute for proper support of the wire. <A> I must take this chance to advertise my absolute favourite soldering video: Basic Soldering Lessons 1-9 , made in 1980 by PACE. <S> Some of it may be slightly dated, but the basics are still very relevant, and the intro music is awesome. <S> These are training videos for soon-to-be professional repair and manufacturing engineers. <S> Lesson 2 deals with tinning wires, and explains that you should not let the tin creep under the insulation: When done this way, no solder will be pushed up under the insulation. <S> Remember: There should always be a gap left between the end of the insulation, and the beginning of the tinning. <A> For screw/spring terminals: Use of a decent quality ferrule, properly crimped is by very far the best way to connect. <S> All other cases make sure there's no notable stress on the soldered wire. <S> Like in the NASA advice of 0.5mm clearance, though I'd suggest at least double the copper conductor's radius (2*r is of course d, <S> but I always think and calculate in r). <S> Or by clamping-down (properly) the insulated part of the wire with something rigidly connected to the other side of the solder joint. <S> Like happens in most quality circular connectors with solder-on pins. <S> If soldered wire is allowed/likely to bend, it will break in the near future. <S> EDIT: <S> To clarify: If the solder goes into the insulation and the insulation is a standard mildly damage-resistant type like PVC, PP, etc, the weakest point will be the soldered wire. <S> Meaning bending stress will be put more onto the soldered wire than the stranded wire inside the insulation. <S> Which is the wrong way around.
| It is desirable to have no stress or strain on the tinned portion of the wire (because tinned wire is more brittle).
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Can enough motors generate the current to turn a separate one? I'm trying to keep a battery on an aircraft charged, but an alternator won't do the job. Can a lot of DC motors keep a battery in a state of sufficient charge to run two others? And, even more importantly, can a DC motor generate the same level of current and voltage output as is required for an input on that motor? For example, can a motor with a 12V 2A input create 12V 2A of output? <Q> What you are looking for has been known for a long time - it's called a perpetual motion machine. <S> If it were possible, you could do the following: 1) Take a 12 volt, 2 amp power source, and connect it to your desired motor and generator. <S> Just as a start, let's say that the generator will put out 12 volts at 4 amps. <S> 2) Connect the generator to 2 motors. <S> 3) Connect the 2 motors to 2 generators each, or 4 generators. <S> 4) Continue the process until you have megawatts of power being generated (16 generations) or gigawatts (another 20 or so after that). <S> 5) <S> At some point connect one of the generators back to the original motor and turn off the power supply. <S> Do you see the problem? <S> If you use a motor/generator with a gain of 1.5 (12 volts and 3 amps) it takes more layers to get to a specific power, but the principle remains the same. <S> So your electric airplane has the energy stored in the battery at takeoff. <S> And no more. <A> You can get power by rotating the shaft of an ordinary brushed DC motor. <S> I did that experiment many times as a kid, and was able to power a small light bulb. <S> When I did it I had to turn the motor very fast to get any significant power. <S> I usually used a set of gears to increase the rotation speed 50x what I was cranking at with my hand. <S> The amount of power you get out will always be less than what you put in. <S> How much less depends on the specific motor and circuits attached to it. <S> But don't expect the efficiency to be very high. <S> You may be able connect two identical motors and have one turn the other. <S> But you would most likely have to turn the first one very fast to only get a little motion out of the second one, mainly because of the poor efficiency. <A> Others mentioned the paradox of 'perpetual motion'/'free energy' you brought up (you can't generate more power from the battery than you are using). <S> I wanted to mention another theme you mentioned, which is turning the turbines with wind. <S> This will be difficult since the wind must be doing the work (as opposed to the original motor which is now dragging these "generators" through the wind) or else there will not be any positive net power generated. <S> I am not sure if this is even possible at all <S> (but I am certain it won't generate enough power to cover the cost of flying). <S> I can't say for sure about the details though since I am an electrical engineer and don't know much about airplanes. <S> Just assuming you could somehow harvest the wind energy (maybe by going up into a jetwind and just idling while it spun your "generators"), it is now going to take a lot more battery power to drive the airplane (since it weighs more for each motor -- the heaviest part most likely). <S> So now you need even more battery power for the extra weight. <S> What about solar panels as the wings or a more efficient alternator, making all of the plane as light as possible? <S> Still, even the best alternator will not get back the same electricity as you put in.
| Any motor/generator which puts out more power than is put in is impossible under the current understanding of how the universe works.
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How to read TI's mechanical datasheets? I'm having difficulty interpreting the dimensions on TI's mechanical datasheet. The part below has 16 pins spaced .5mm apart. However, I cannot determine what the length and width of the pins are. Width is written as "16X 0,30/0,18" and length as "16X 0,50/0,30". I don't know what the dimensions in the numerators and denominators represent. I thought they might be uncertainties, but the values seem too large. I haven't been able to find a guide to reading the mechanical data on TI's website. <Q> They are upper and lower tolerances hence "50/30" has an upper limit of 50 and a lower limit of 30. <A> First, see note A: all measurements are in millimeters. <S> A comma is commonly used in datasheets in place of a period to represent a decimal point, because it's considered easier to read. <S> With this convention, "0,50" is 0.50mm. <S> The fraction represents maximum and minimum measurements: maximum / minimum. <A> These are all metric pitch of 0.5mm from [1.50] divided among 3. <S> so 16X 0,30/0,18" refers to max/min with a computed average pad width of 0.24mm or slightly LESS than half the pad pitch, presumably from etch-back <S> In order to prevent tolerance stack-up , this is the correct method to specify it.
| The "16x" just means that there are 16 instances of the dimension being described.
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Why doesn't AC motor slow down when connected in parallel to a capacitor? Why doesn't a motor skow down when a. capcitor is connected to it in parallel? Here is my understanding. During the first half cycle, the capacitor charges up while the load is getting power. During the zero point, no power from the source goes to the load and the capacitor is holding a charge, but not charging. During the second half of the cycle, the load gets power in reverse and the capacitor starts to discharge in the opposite direction. So there should be two overlapping waves one wgoing go the motor and the other leaving it. Essentially, the motor would slow down or stall because it is getting negative voltage, but it doesn't. Very confused and please no MATH in your answers. It will only frustrate me further. <Q> During the zero point, no power from the source goes to the load and the capacitor is holding a charge, but not charging. <S> This is your misunderstanding. <S> Since the capacitor is directly across the mains with very low resistance anywhere in the circuit the capacitor voltage will follow the mains exactly in sync. <S> There will be no lag (as there is in an RC circuit, for example) because there is no resistance. <S> The capacitor has no affect on the operation of the motor. <A> The reason this is true is that the AC line is intended to be an AC voltage source (and in fact what you have drawn is literally <S> one) - up until the point where you break something, you can to a first order approximation consider that the voltage on the line terminals is the same no matter what you connect. <S> Of course while this is absolutely true for what you have drawn, there is a degree to which this is not strictly true for actual machinery connected to an actual AC line of non-zero source impedance. <S> For one, the capacitance will be an admittance at AC, which is to say that it will load the line during part of the cycle and so (assuming finite wiring resistance) both reduce the voltage and shift the phase a little. <S> But the motor will only see a different phase - which it doesn't care about at all and a slightly lower voltage, which if an induction motor not critically loaded it doesn't particularly care about either, since the speed absent substantially loaded slip is determined by the line frequency. <S> The practical reason why someone would actually build the circuit drawn has nothing to do with speed control , but rather with a desire to balance the inductive reactance of the motor with local capacitance, so that the load to the AC grid is more nearly resistive. <S> Although neither inductive or capacitive reactance consume real power, the cyclic AC power in and out of them has to flow through the real resistance of the distribution wiring where it does dissipate real power as heat, so industrial customers are billed a surchage for reactive loads, and may find it desirable to locally compensate out the reactance. <S> So if you want a model where something actually happens, you need to insert a resistor between the motor / capacitor assembly and the AC source, to model line losses. <A> You need to think of the motor having two parts. <S> One part is an inductor or electromagnet that provides a magnetic field but doesn't burn up any power. <S> The other part takes power from the source and transmits it as mechanical power to whatever the motor is driving. <S> The electromagnetic accumulates energy from the capacitor during one half cycle and returns it during the next half cycle. <S> The electromagnet and the capacitor can be thought of as being completely separate from the source and the operation of the motor. <S> You can think of power going directly from the source to the rotor of the motor where it is converted to mechanical power and delivered to the load. <S> There are losses in the electromagnet, the rotor and in the capacitor. <S> You can think of power coming from the source and being converted to heat in those components. <S> Induction motor operation is usually analyzed without considering back emf. <S> A very simplified equivalent circuit only includes the electromagnet and a variable resistor. <S> The variable resistor represents power converted to mechanical power and some of the losses.
| To a first approximation, the motor is irrelevant to the capacitor, and the capacitor is irrelevant to the motor.
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Using IR thermometer for measuring component temperature I'm trying to measure the temperature of various steel components being heated using power frequency induction heater. The IR meter being used can be set for emmissivity value of 0.1 to 1.0. For the same type of material EN353, I'm not able to get the satisfactory reading of temperature of different components keeping the emmissivity and measuring distance same. What other factors need to be considered- is it the mass / profile / surface finish of various components, etc..I'm comparing the reading of IR meter with a hand held calibrated K-type temp. indicator <Q> What other factors need to be considered- is it the mass / profile / surface finish of various components, etc.. <S> Only surface finish and material as that's what's emitting the IR. <S> Some reading: <S> Most high-quality electrical tape has an emissivity of 0.95. <S> One must be careful, especially with mid wavelength cameras (3-5 μm), that the tape is opaque. <S> Some vinyl tapes are thin enough to have some infrared transmittance, and are therefore unacceptable for use as high-emissivity coatings. <S> Scotch <S> ™ Brand 88 black vinyl electrical tape has an emissivity of 0.96 in both the short wavelength (3-5 μm) and long wavelength (8-12 μm) regions, and is recommended. <S> Source: FLIR . <S> Measure the temperature of the taped or painted surface. <S> That is the true temperature. <S> Source: Raytek . <S> For relatively low temperatures (up to 500°F), a piece of masking tape , with an emissivity of 0.95, can be measured. <S> Then adjust the emissivity value to force the indicator to display the correct temperature of the material. <S> Source: Omega . <S> Each of the above articles is probably worth a read. <A> Emissivity not only depends on the material itself, also from the surface. <S> For iron and steel, I find electrolytic iron, high gloss <S> polished: <S> 0.05... <S> 0.06forged <S> iron, high gloss polished: <S> 0.28rusted iron 0.69 With lower values, the material starts to act as mirror, and the IR meter starts to see the thermal radiation from other stuff nearby. <S> If you put your IR meter into a tank made of polished iron, you'll be fine with a value of 0.28 or so, but if you aim it to smaller parts, you get wrong results. <S> It would be best to cover the parts with something of high emissivity. <S> An easy solution is soot, directly from a yellow flame, which has an emissivity of >0.95. <A> You haven't mentioned the temperature that you are wanting to measure. <S> There are Infra-Red (IR) temperature sensors that use "2 color", or "Dual wavelength" process to reduce effects of emissivity differences. <S> See this article. <S> Also, when steel (and other metals) are heated to higher temperatures, scaling (oxidization) of the surface occurs, which can cause errors in temperature readings using a single wavelength IR measuring device. <S> In other words the emissivity changes as scale builds up on the surface of the heated object. <S> A 2 color (or dual wavelength) <S> IR measurement system may provide you with correct temperature measurement.
| If you are using a thermometer with a fixed, preset emissivity of 0.95, and need to measure a shiny object you can compensate by covering the surface to be measured with spray oil, flat black paint or masking tape .
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Looking to create a switch box for multiple audio aux inputs I'm looking to create a circuit that will switch a 3.5mm audio input between 3 or more sources using multiple switches. I've made a quick schematic to try to explain my idea, but I'm fairly sure this won't work as the audio will use the ground of which ever input is switched on. Thanks for the help & sorry for the beginner question. <Q> You should tie all the ground together, then switch the audio lines (right, left), not the other way around. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> The switch circuit. <S> Figure 2. <S> A suitable switch from CK. <S> The 12-position rotary switches are available in configurations of 1 x 12-way, <S> 2 x 6-way, 3 x 4-way, 4 x 3-way or 6 x 2-way. <S> Any except the first or last will do the job you require. <S> Figure 3. <S> Switch pattern options. <S> The switches feature an adjustable stop to restrict numberof positions. <A> In addition to Mr Lathrop's answer:If OP doesn't want some mechanical switching, he/she can go for an analog multiplexer.
| A multiplexer simply a device that allows the user to select an input from different inputs and route it to one (or more) output(s).
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Gain of an Operational Amplifier This is a question I found in one of the textbooks I have. The textbook didnt explain thoroughly about the concept or theory of the op-amp, and that left me confused. So please correct me if I'm wrong. I was just wondering if the voltage of the noninverting pin of the op-amp is zero, as shown in the picture. Does that mean the gain is infinity? (Assuming that the op-amp is ideal.) Or how can we find its gain? Any help is greatly appreciated. <Q> There are Three gains involved in your diagram (I'm assuming you mean the DC gain). <S> The gain of the circuit with respect to the V1 input, the gain of the circuit with respect to the V2 input, and the gain of the op-amp itself. <S> From that diagram, if we assume the gain of the op-amp to be infinite, then we assume the + and - opamp inputs will be the same voltage, and we can calcuate the circuit gains as R/R1 and R/R2. <S> We cannot calculate the opamp gain from that circuit, there is not enough information, which is why we assume it to be infinite. <S> If we have more information, like the part number, say TL071, then we can look up in the datasheet, where it says gain is typically 200V/mV, or a minimum of 25V/mV, depending on loading and type. <S> 1v output would typically need a few uV across the input to support it. <S> Given that there could be several mV across the inputs due to input offset voltage, the contribution from finite gain is irrelevant. <S> Having a finite gain also means the calculated gains will be lower. <S> If the ideal (infinite opamp gain) circuit gain is 10.00000, then the gain with a real opamp would be 9.999.... <S> whatever, you figure the exact number of 9s! <S> With 1% resistors, the gain could be anywhere from 9.8 to 10.2, so the error from finite opamp gain is irrelevant. <S> That's why for most purposes, we assume infinite opamp gain. <A> An ideal op amp is assumed to have infinite gain. <S> Basically, it will drive the output voltage to whatever is necessary to make its inputs equal. <S> The way to solve op amp circuits like this one is to assume the op amp inputs are at the same voltage and the current flowing in to them <S> is zero, then figure out what the op amp output voltage has to be to satisfy those conditions. <A> You have I 1 , and you have I 2 . <S> None of either is going through the op amp input, so they must be going through R. <S> Both op amp inputs are supposed to be held at the same voltage when the op amp is operating correctly. <S> Since you have the voltage on one side of the resistor, and you have the current flowing through it, you can calculate the voltage on the other side of the resistor. <S> Now just put it in terms of the input voltages and you get the gain of the circuit.
| Obviously having a finite gain means that the inputs will not be at the same voltage, if there's an output voltage.
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Standard way for novice to prevent small round plug from rolling away while soldering wires to it This very trivial task is to solder wires to the terminals of a small round 3.5 mm audio plug. I can do this myself, but I need to show to others and this is one of the first times they're trying any kind of soldering. The problem is the connector rolling away during the process of soldering. Students use one hand to hold the wire and another hand to hold the soldering iron. The plug itself lies on the table and rolls away after being touched. Is there a standard, widely accepted way to do this kind of soldering easily? Do I need to provide some kind of vise for them, or there is some simpler approach? The problem does not require soldering with one hand, using both hands is ok and preferred. <Q> Figure 1. <S> Image source: iFixit . <S> This works well for me. <S> Increased grip is possible by increasing the number of "wraps". <A> There are many ways to fix things during soldering. <S> One popular method is to use so called helping hands <A> it won't suffer from solder drops or brief touches with the iron, and has more friction than many other sprung solutions so is less likely to ping at the wrong moment. <S> If you're equipping a classroom, they cost next to nothing. <S> I used to keep a couple with my portable iron in a travelling toolkit; the soldering parts of that were mainly for cables/connectors. <A> I have a block of wood with mating connectors for everything I frequently use mounted on it. <S> This has the virtue of not only holding the connector, but when dealing with the cheap trash end of the market, it holds the pins in place so softening the plastic is less of an issue (Certain knock off copies of XLRs looking at you!). <A> There are tiny plastic vices which are too wimpy for most mechanical work, but are ideal for holding small connectors while soldering. <S> Typically ~ US$5 or less. <S> And there is the more traditional "Panavise" which is great for larger connectors (like XLR, etc.) <A> in addition to the "helping hands" and "mini vices" already mentioned, I use hemostats and needle nose vice grips often. <A> I make a 'third hand' with a croc-clip on the end of a bit of stiff copper wire. <A> While not overly attractive in this particular case (because small parts get hot when soldering and that does not make it particularly pleasant to hold them steady while the solder sets), it's worth learning to make better use of your hands: two fingers are sufficient for gripping a part, and moving something held in thumb and index finger against something held in ring finger and (depending on the arrangement) pinky or middle finger is quite feasible. <S> This takes practice which probably explains that electronics are these days often assembled in countries where children are taught eating with chopsticks. <A> I use ACCO Binder Clips... <S> Cheap, easy to store, don't die with age (like rubber bands do...) <S> These have plenty of other uses in the shop, too. <A> Perhaps overkill, but I use a small (2-3") precision-ground toolmaker's vise which is heavy enough not to move around with wires or whatever pulling at it. <S> I prefer that to vises which are mounted on the bench and cannot easily be moved under a microscope or to a rework area. <S> It has 'V' grooves in the jaws that allow you to hold round objects vertically or horizontally without crushing them unduly.
| The wire can be held by screwing it to a block of wood, holding in a bench vice, standing the soldering iron transformer on it, and then bent into a suitable position. You can always rest a book on the handle end to add more weight. There are similar products from other vendors. A wooden clothes peg (clothes pin in American English) works very well:
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Help identifying a varistor Hi I am trying to identify what I think is a varistor. The problem I am having is the markings, other than the voltage don't seem to corrolate with anything. So I'm not sure if it is a varistor or if it just has obscure markings.The markings are :05K3851146I know the k385 is the voltage from what I can find, but have no idea what the current is or what part number it matches up with. It's currently part of the 12v input on an electric fence energiser. <Q> They happen to make varistors which makes it even more plausible. <S> Why is your sample stuck in a tortilla? <S> I can see the herbs. <A> It's an EPCOS <S> (now TDK) <S> brand MOV. <S> The 05 is not the Joule rating, it indicates the series name which is S05 . <S> Here's the datasheet . <S> Measure its diameter and lead distance, you can find it in the datasheet. <S> I think its part number is this: B72205S0381K101 <A> Metal OXide Variators (MOV) or now generically referred to as SIOV's are commonly rated by their RMS withstanding voltage or 385Vac in this case, then above this by a Joule rating 05 (which seems low). <S> then followed by a date code YYWW for year and week. <S> These must follow a transient filter and current limiting, otherwise they may fail frequently or sooner than expected with limited number of Joules protection cycles. <S> This one is <5 yrs old in photo, consider the largest Joule rating of radial thruhole 385V MOV for replacement. <S> (low cost) <S> http://www.digikey.com/product-detail/en/littelfuse-inc/V25S385P/F7181-ND/2297519 . <S> .. <S> good choice for lightning protection. <S> 22kA <S> 625 Joules 385Vac working voltage
| That looks like the TDK-EPCOS logo (found on a Google image search for "varistor logo").
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Is the main purpose of cascoding is to increase gain in FETs? In my recent studies, I have come across the concept of cascoded amplifier using Common Source and Common Gate Configuration. What exactly is the reason of cascoding? <Q> It is mainly done to increase the output resistance and or reduce the Miller effect. <S> An example of increasing output resistance is the cascode current mirror. <S> An example of Miller effect mitigation is a normal cascode amplifier. <S> Of course this also increases the output resistance, but a driving reason is to mitigate Miller effect. <S> The gain does not increase (although the bandwidth due to mitigated Miller effect does go up) because the first transistor has no voltage gain (voltage gain of 1) but provides a current gain. <S> The next transistor provides the voltage gain. <S> If you do the math it ends up (ignoring parasitics and making approximations) matching the gain of the regular common-emitter or common-source amplifier: \$R_L/R_{ee} \$ <A> In a common base circuit, the Miller capacitance no longer has any effect on the high frequency gain being rolled off by negative feedback. <S> So the 2nd stage of the cascoded pair is the best you can get. <S> The clever thing is that the emitter of this output transistor is held near enough at a constant voltage and, this means that the first transistor also has miller effects dramatically reduced. <S> I know of no other reason other than reduction of Miller effects on both transistors for using a cascode amplifier <S> be it bipolar or FET based. <A> used in current sources and as non-linear loads where the output resistance of a single BJT isn't enough. <S> able to nearly eliminate Early Effect in one of the BJTs by holding its collector at a fixed voltage. <S> As smaller devices tend to have smaller Early voltages, so cascoding these smaller/faster devices provides an option for recovering some otherwise lost performance. <A> I have created a cascode vs. single transistor amplifier comparison circuit in this free on-line simulator: <S> https://www.systemvision.com/design/compare-cascode-vs-single-mosfet-amplifier <S> In the simulation results shown, you can see the increased bandwidth of the cascode amplifier circuit in both the time domain (with 5 MHz input to both circuits) and the AC or frequency domain. <S> The transistor models are all identical and have reverse transfer capacitance Crss set to 10pF. <S> In the case of the single transistor amplifier, the Miller effect increases the effective input capacitance by a factor of (1.0 + gfs*Rload), or 16 times its nominal value. <S> This 160pF capacitance, along with the small residual Ciss, combine with the 500 Ohm input source resistance to create an RC low-pass filter that rolls off the amplifier gain with a pole at just under 2 MHz. <S> The cascode circuit avoids this Miller effect because the lower transistor m2 has almost no voltage gain, so the effective input capacitance is just Ciss, 20pF in this example. <S> You can see the correspondingly greater amplifier bandwidth. <S> Note that the circuit probes can be moved to look at other time or frequency domain signals, and component parameter values can be observed by double-clicking on any part. <S> The circuit can also be copied and saved, so the user can make any desired changes and re-simulate to see the effects of those changes.
| Cascodes are: used as the gain elements in amp stages when the Miller effect is an issue.
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Is this SSR a good choice for switching 16V AC/~8mA with an Arduino? I'm quite new to actual physical "engineering" and I am just getting started. My current project is to build a light switch, which triggers my room light via wifi. I have two cables in my wall (like this ) which - when connected - trigger a relais in my flats fuse box which switches the light. The two wires run on ~16V, and when connected have a short spike of current of about 4-8mA. This spike lasts for about a second and drops to 0 afterwards. So my plan is to use this SSR with an ESP-01 3.3V (which has a maximum current of 12mA per pin, so that should be enough), wire up the two cables and the esp-01 with the SSR and let the ssr switch the light. So a few questions: Is there something obvious I missed? I'm still a beginner and trying to make sense of all these information, so there might be something I missed. If I read the datasheet right, I need 10mA@6V to trigger the SSR, is that right? How can I map this to 3.3V? And what does RL under conditions @ minimum trigger conditions mean? Is this a resistance I put before the Pin? Concerning the Pin-layout of the module: I guess the pins 1,3 & 4 are the ground pins for my ESP-01, a load on pin 2 triggers the optocoppler. I'm not quite sure about the pins on the AC side. What are the Pins T1, T2 and Gate for? Thanks in advance for your help! <Q> It might work, but it's not designed to work at such a low voltage. <S> The zero crossing inhibit is as much as 35V which means it could never trigger at all from a 16VAC source (22V peak). <S> It is typically about 12V <S> so it could just work badly. <S> TLP2222A <S> The LED needs ~7.5mA at ~1.15V <S> so a series resistor of about 240 ohms is probably about right- not sure what the voltage drop is of that chip so you might want to check the voltage across the resistor to make sure the current is high enough. <A> In your case the load current is actually less than the SSR trigger current. <S> In this case, one would review the LED requirements for AC or DC and the arduino power and any other supply voltages available. <S> Since the load is quite small and noise may be a problem with a long line, one can solve this in many other ways. <S> But it also could be powered other ways . <S> I would suggest this driver reed relay for your 16Vac power relay coil.6mA @3V , contact 500mA http://www.digikey.com/product-detail/en/standex-meder-electronics/SIL03-1A72-71D/374-1322-ND/3131688 <A> There's a schematic button on the editor toolbar. <S> Schematics are better than words. <S> Meanwhile ... <S> Datasheet Electro-optical Characteristics says "Forward voltage @ <S> 20 mA= 1.2 V". <S> This is typical for an IR LED. <S> You just need a series resistor to limit the current to 20 mA. <S> The standard circuit on page 10 shows how to connect it up. <S> General notes: 0.6 <S> A is the max rating. <S> Incandescent light bulbs have a current surge at switch-on of 5 to 10 times their steady state current as the filament is cold. <S> This may blow your triac. <S> You need to design properly for mains isolation. <S> Figure 1. <S> Good design practice. <S> Note complete isolation between the sides of the opto-isolator and slot machined out of the PCB to ensure no creepage due to moisture, etc. <S> Source: Vishay .
| The problem with SSR's is they have a minimum load current to turn off reliably. I suggest using a MOSFET output SSR such as the Toshiba Such a solution exists for small signals that fits this range is a DC optoisolator switch and small signal bridge rectifier if you have a 12Vac source.
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In this Amplitude Modulation circuit, what is the purpose of the diode load (L2)? In this circuit i know that the diode mixes the signal. I know that after the diode it is a half-wave. I know that the capacitor reproduces the lost signal. I have a feeling L2 is maybe plays a part in this signal regeneration? <Q> A picture is worth a thousand words. <S> The carrier and audio modulating signals are simply added but do not form a modulated signal. <S> The diode rectifies this signal forming a crudely modulated signal which contains a DC component, low frequency component and high frequency component. <S> The inductor acts as a low impedance for the low frequency component and high impedance for the high frequency. <S> The capacitor blocks the DC component but passes the high frequency AC signal The tuned LC circuit filters out all but the the desired AM signal. <A> It could be a resistor also. <S> If not present the input signal would charge the capacitor to the maximum value and than the whole process of mixing would stop. <A> You want to develop the modulated RF signal as a voltage across an impedance, but you don't really want the baseband audio signal there. <S> So the best choice of load there is a high impedance at RF frequencies, but low at DC and audio frequencies - hence, an inductor. <A> The diode is essentially a thresholding device here. <S> The instantaneous voltage of the carrier plus modulation signal has to be above some voltage to make it thru the diode. <S> Think of the modulation signal as slowly varying DC. <S> When the level is low, most of the carrier will be cut off. <S> Think of the limiting case where the DC is substantially negative. <S> The diode will always be reverse-biased and nothing makes it thru to the inductor. <S> The DC level sets where in the carrier waveform the cutoff will be. <S> The average of the modulation signal is adjusted so that half the carrier is clipped off. <S> The modulation signal can then be up to the same amplitude as the carrier. <S> At the top end, all the carrier is passed, and at the bottom end none of it. <S> The result is a cut off sine wave. <S> That's not the same as a amplitude-reduced sine wave. <S> The cut off sine has a lot of harmonics, with the desired amplitude-modulated sine being in the mix. <S> The remaining inductors and capacitors are set to resonate at the carrier frequency. <S> This filters out all the components of the messy cut-sine signal except those close to the carrier frequency. <S> The result is then a amplitude-modulated sine, as intended. <S> Note that this kind of primitive AM modulator is not all that linear. <S> It's good enough to be useful for voice, for example, but quality AM transmitters use more sophisticated mixers. <S> Added <S> I just realized the question is actually about the inductor, L2, not the diode. <S> As it says, the diode needs some impedance to work against. <S> A resistor could have worked too. <S> However, a fairly high Q filter must still be applied to the clipped-sine signal. <S> L2, together with the other inductors and capacitors form this filter. <S> L2 therefore serves two purposes, to provide a load to the diode, and as part of the filter that attenuates all the unwanted crap in the clipped-sine signal.
| The diode load completes the path for DC return.
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Night light, schematic and functioning I recently bought an el-cheapo night light for $1 just to see how they manage to get the costs so low. I expected to meet an el-cheapo voltage regulator at best or even a bridge rectifier but alas! None exist here. I just can't figure out how or why the circuit here works with mains (240V) voltage. It does get warm during operation but I wasn't going to use it anyways so it is just a learning prop for me. I have no idea what the SOT part labelled "J6" is and if it's a transistor, what kind. Please help me figure out how it works and what that "J6" could be. edit: R2 is the LDR, the other resistors are SMD resistors and the capacitor is an electrolytic cap. The board looks like this: and I've drawn the schematic as is: simulate this circuit – Schematic created using CircuitLab <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Redraw of the OP's reverse engineering. <S> At 240 V the current through the resistors will be \$ <S> I = \frac {240}{8k2 + 8k2+8k2} = 10~mA \$ <S> but with the rectifier it will average half that. <S> It's not clear from your schematic <S> but I suspect that R2 is the light sensor - an LDR. <S> When light is sensed the resistance will drop and Q2 will turn on. <S> This will "shunt" the DC on C1 to ground and turn the LEDs off. <S> This will give comfort to the user giving the impression that the unit is not wasting power when, in fact, it is running constant power whether it's on or off. <S> It would make no difference to power consumption if R1, 2 and Q2 were omitted! <S> Power dissipated in each of the resistors will be <S> \$P = I^2R = <S> (5m)^2 \cdot <S> 8k2 = 205~mW \$ which may be a bit on the high side for those SMD resistors. <A> The reason for using a wasteful shunt to turn the LEDs off instead of cutting power is probably this: in both "on" and "off" states, the business end operates at low voltages, only R3,R4,R5,D4 need to be rated for high voltages. <S> This is slightly cunning : if you attempted to cut the current off during daylight, to save power, the transistor would have to be rated to the peak mains voltage (350V or more) adding some expense as well as (possibly) more safety concerns. <S> Searching for "J6 SOT23 transistor" yields the S9014 : a perfectly ordinary NPN transistor, rated at Vce <= <S> 45V <S> and Ic=100mA. <S> If any of the LEDs fail open circuit, the transistor will probably fail over-voltage next time it gets dark, unless the capacitor fails first. <S> I expect it has been tested and shown not to start a fire in that failure mode - actual functionality and repair aren't an issue given the price. <A> The LEDs and D4 create a simple half wave rectifier. <S> The resistors R3, R4 and R5 provide the necessary current limiting. <S> C1 provides very simple decoupling. <S> When the LDR has light on it, it's resistance is very low and the base of transistor Q1 gets enough current to turn on, likely to saturation. <S> This effectively shorts out the LEDs, so they turn off. <S> When the ambient light goes out, the LDR is high resistance, and the base of Q1 receives almost no current, making it more like an open, so current flows through the LEDs. <S> It's interesting that when the LEDs are off, the resistors and D4 are still just wasting power. <S> Cheap cheap cheap! <S> I assume the designers used three different resistors in series instead of just one for power dissipation reasons, but it could also be a cost thing. <A> There will be greater peak currents to charge the Cap than than the average LED current. <S> The peak LED current is defined by resistance total,series R in which we can neglect ESR and voltage drop of LEDs <S> The cap only reduces the flicker 15% from 100%, which we can determine from the LED ESR. <S> Neglecting the LDR/NPN disable circuit we have; <S> 240Vrms half wave 50Hz input. <S> Load appears from photo to be 75mW rated white LEDs which have an ESR=1/Pd = 13.3 <S> +/- <S> ? <S> times 3 LEDs in series, = <S> 40 <S> Ohms <S> Thus peak current is 1.414*240V/(3 <S> *8k2)=14mA and conversion from half wave peak of RMS to DC equivalent is root2*rms/2 <S> thus avg LED current becomes Vrms/Rtotal or 10mA with the Vf changing only 10% over the brightness range of 10:1 and 100uF * 40 Ohms= 4ms or 25% of the line pulse current interval and using half power intensity instead of 10:1 <S> we expect the LED flicker current to be closer to 15% duty cycle <S> ON <S> and the cap peak charge current 10x <S> the average 10mA discharge. <S> bigger cap would reduce flicker, but then raise cost due to RMS ripple current ratings for small cheap caps. <S> we also expect the resistors to,flash over with > 1500 V peak voltages and burn up if there is any lightning nearby
| If your circuit is correct then we can see three voltage dropping resistors, R3, 4 and 5, and a half-wave rectifier, D4.
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Coil current changes while working If I wind a coil around a hollow plastic tube and apply voltage it will create a magnet that is capable of pulling an iron based plunger into the tube. I imagine the plunger will stop in the magnetic field's center. What current fluctuations in the electromagnet can I expect, and where in axial relation to the the coil? In other words will the magnet's current change as the plunger position changes? Will it increase or decrease as the plunger moves towards the magnetic center? How big a change can I expect? Thank You <Q> Interesting question. <S> Also, because of the mass of the plunger - unless you've critically damped its energy against the friction between the plunger and the ID of the tube - it'll oscillate for a while before it settles down at - more or less - the magnetic field's sweet spot. <S> If you exercise the coil with AC it gets more complicated. <S> Do you want to go there? <A> I've never even considered the question before, but here are my initial thoughts about it. <S> Lots of things are changing as you insert an iron rod. <S> One thing that's certainly changing is the magnetic path length, as iron acts as a kind of short-circuit and reduces the effective magnetic path length. <S> As that happens, the inductance of your solenoid rises (changes.) <S> Your solenoid starts out as an air-core, with lower inductance and a magnetic field that spreads out into space quite a distance. <S> I'll assume that the resistance in the wire (and any series resistance in your voltage source) has become dominant in limiting the current at the start. <S> So the value of \$I\$ is set at some value to begin and there is a certain energy stored in the magnetic field at this time. <S> Once the plunger starts to be inserted axially, the ampere-turns (each turn contributes some force and the force of each combines) in the field will exert a force on it. <S> Since you aren't asking about the force, I won't mess with that part. <S> The usual equation for an inductor looks like: $$V= <S> L\frac{\textrm{d}I}{\textrm{d}t}$$ <S> But in this case we happen to know that the inductance also changes with time. <S> So, it's more like: $$V=L_t\frac{\textrm{d}I_t}{\textrm{d}t} + <S> I_t\frac{\textrm{d}L_t}{\textrm{d}t}$$ which, with \$V\$ constant, becomes something like: $$\frac{\textrm{d}L_t}{\textrm{d}t} I_t +L_t\frac{\textrm{d}I_t}{\textrm{d}t}=V$$ <S> And, if you can estimate \$\frac{\textrm{d}L}{\textrm{d}t}\$ <S> as a constant rate, becomes a simple 1st order diff eq. <S> But yeah, the current will vary while motion is taking place. <S> But ultimately, the final condition for the current remains limited by the same coil resistance. <A> It is important to recognize that the coil inductance has energy E=1/2Li^2 <S> but the force is only proportional to the Change in Energy. <S> Unless the speed is great, the current will only slightly reduce and increase with motion oscillations about center and reduce as the copper temperature rises from excessive current. <S> Winding Resistance is the main trade-off affecting current with turns ratio , wire gauge , fill factor and diameter affecting force and thus velocity and thus reduction in current. <S> But when it has stopped with no change in energy applied V*I, there is no net force when unrestricted. <S> It is also important to recognize that adding turns in fact reduces force due to the rise in DCR of the current relative to the effect of increasing B field when operating in a fixed voltage, which may seem counter-intuitive. <S> Of course the current reduces with more turns. <S> But reducing turns is a very inefficient way of increasing force when you get down to only a few turns.
| If you're exciting the coil with DC, and the plunger is free and not at the center of the coil, then as the plunger is pulled into the tube, the inductance of the coil will increase which will limit the instantaneous current that can be supplied to the coil by the driving constant voltage source, which will limit the speed at which the plunger moves toward the center of the magnetic field.
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Running 100 24v 1 amp solenoids with low amperage power supply I am trying to run around 100 24v solenoids with a low amperage power supply. I originally thought when I began trying to run the solenoids I would go with a 100 amp power supply as each solenoid is about an amp. High 24v Power supplies are not only hard to come by but also usually very expensive and big. Each solenoid is on frequently, never are they on at the same time, however I would like it to be possible to turn all on at the same time, in case me or anyone who is trying control my solenoids has full control over what they are doing and are not limited. I would say most of the time 70 of the solenoids are on. My question is, how low amperage power supply could I use? Thank you <Q> Typically you can get away with using full current to energise your solenoid, but can be reduced to a much lower holding current. <S> (Application dependent) Depending on how simultaneously you need all your solenoids to come on you could conceivably have a slight delay between turning each solenoid on so peak current draw is kept lower. <S> Of course this can only be implemented if you are controlling your solenoids through a device that would allow pulse width modulation on the solenoid. <S> Luckily solenoids have a pretty high inductance and can generally be driven at a low frequency. <A> If there is any possibility that all solenoids will be energized at the same time, then you must have a power supply capable of supplying the full power. <S> It may be desirable to split the load among several power supplies. <A> Since you will be using a lot of 24VDC solenoids, I think it's a large scale industrial machine. <S> So you can use a 2.4kW transformer with diode bridge. <S> Best option is to have three phase transformer with 6 diode rectifier, you don't need any capacitors for supplying solenoids, neither a stabilized power source. <S> For the rest of your electronics use a 24V SMPS. <S> Solenoids can also produce large spikes, so it is smart to use a separate PSU only for them.
| As an extension to peters answer, the total power required can be reduced if you get clever with powering your solenoids. You could, for example, use four 25 amp supplies, each providing power to a quarter of the solenoids.
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Driving 3w leds in a car Im thinking of upgrading some drls on my car. LED SPEC: 4V max 750mA max.Daylight white. http://www.ebay.co.uk/itm/141679345010?_trksid=p2060353.m1438.l2649&var=440967222198&ssPageName=STRK%3AMEBIDX%3AIT There is going to be 4 leds per headlight. Im wondering which is the best way to drive these, since the car will give out anything between 12-14.4v. Ive been thinking about just putting some resistors, but these leds output goes down substantially when voltage is dropped. Ive also thought about building a C.C circuit to the leds, is it possible to use LM317t for these leds? Would these leds have the same brightness when the car is on / off? Maybe the best thing is to buy a finished driver from ebay? I'm sorry for the noob question. update: Thank you for answering There are leds in it right now, however, they arent that bright, they consume around 3w per headlight, they look to be driven with resistors. Im planning on having the electronics inside the headlight, so, only limited space avalible. I do have 5 dc to dc step up laying around, maybe i should use those. i dont have the leds at home yet. The leds are placed in the bottom, one pointing each direction. <Q> You are best to use some sort of Switchmode Current source .If <S> you use parallel linear regs <S> you will get an efficiency no better than the filament lamps that you replaced. <S> One option is to boost up to 16V with all the leds in series .This <S> keeps current down making the DCDC convertor cheap and easy to build .The <S> LED currents are equal despite spreads in the individual terminal voltages .You can protect your series string from LED open circuit failure by using an amplified Zener arrangement .If <S> you buy a driver from ebay <S> you will learn nothing . <A> Using resistors to drive LED for lighting applications is something I wouldn't advise for. <S> This video can provide you more perspective. <S> LM317t in itself doesn't control current, depending upon load(LED in your case) the current will vary, but on a quick search, I found some schematics in datasheet itself (on pg 11, fig 9) which probably you can test out for your application <S> and I believe should work for 3W LED's. <S> And there are many such small current controlling circuits. <S> (for eg CC using 2 transistor ) <S> But for stable and long run purpose the above may not work and drivers specially designed for driving LED's <S> would be a better option. <S> That will help increasing life of your LED's and with better brightness over time. <A> http://www.mouser.com/ds/2/308/NSI50350AD-D-118972.pdf <S> If you can find a 700mA driver it's better. <S> Do two series of 3, or 3 series of 3.As for street compliance, use it only as "extra" light, when no other car comes in front of you. <S> ;)
| Put 3 LEDs in serie (3*4V=12V) and then add a 700mA constant current regulator (CCR) or two 350 mA CCRs like this one in parallel (but it's not as efficient):
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Using a relay for retro clicky sound - how do I make it louder? I'm building a battery operated retro device that I want to sound really clicky and old fashioned. I want to include the noisiest relay (sound noise not electric noise) possible in my project. Can anyone help me identify what characteristics are correlated with more noise? Actuation time? Release time? Weight? Amperage rating? or alterternatively recommend a relatively small relay that is known to be especially noisy? These relays are not intended for switching any loads. The usual google searches for relay and decibels or noise doesn't come up with anything useful. I have considered using a turn signal but those devices are quite large. <Q> Try using an inexpensive power relay such as the T90 series , and attach it to a resonator board which will act like the cone of a speaker. <S> They draw about 1W each. <S> Since you're after an aesthetic effect you'll have to fiddle with it, I think. <S> Now if you really want buzzy clanky sounds, you can consider an AC powered "definite purpose contactor" as used in HVAC equipment which will be very noisy and retro sounding. <S> They are not even that consistent in their sound (because of the AC power), which I suspect is useful in your application because they don't sound like a fake synthetic sound. <S> You'll need something like an SSR or a small relay to control the contactor. <S> They're huge and don't mount on PCBs. <S> You can get a similar effect (at less sound level) by feeding the DC relays with some kind of randomized power (such as a ramped sine or triangle wave started with random phase) to make them pull in less consistently. <S> Again, you'll probably have to fiddle with it. <S> You can consider the relay as a special kind of speaker with a really weird response (once they pull in and the magnetic circuit is closed the current has to be backed off a lot for anything much to happen). <S> You could use the contacts to feed back the armature state to a micro. <S> Use something related to your mains frequency such as 120Hz or 50Hz to get authentic mmm.. <S> buzz-click sounds. <S> Lots of fun possibilities. <A> You might want to consider an auto starter solenoid. <S> That is something you could get as salvage. <S> If you want a big sound from something very small, you will need to reproduce the sound. <A> Consider using a "pinball knocker", essentially a solenoid that strikes the inside wall of a pinball cabinet (or a strike plate of its assembly) to make that rapid clacking/knocking sound an various points in the game. <S> Tuning the control of that solenoid and the object it is striking can get you anything from a rapid series of clacks to a single dull thud that you can feel. <A> A lot of relays may advertise how quiet they are. <S> As an example see this datasheet: random relay datasheet <S> You can see that there are a few models Low Noise Models: G5RL-1A(-E)-LN <S> High-Inrush Models: <S> G5RL-1(A)-E-HR, G5RL-1A-E-TV8 etc Avoid the low noise models, obviously. <S> Also, you could hit the relay a bit harder when activating it <S> : drive it with a higher input slope (stronger/faster driver). <A> It will probably be smaller, cheaper, and easier to implement to mimic the sound of a relay than to use a genuine relay. <S> Relay actions only take a few 100 ms at most, even for really big klunky ones. <S> Even the sound simply recorded and played back thru a small PC mount speaker wouldn't take much memory. <S> Let's be pessimistic and say you need 12 bit samples at 20 kHz sample rate (most likely 8 bit samples at 10 kHz is sufficient). <S> That comes out to 30 kBytes/s, or 3 kBytes for 100 ms duration. <S> You probably have a spare 3 kB of program memory in your microcontroller already. <S> Even if not, upgrading to the next larger micro is going to be a lot cheaper, take less space, and require less power, than adding a relay. <A> Solenoids usually produce a loud click so use internally a solenoid that activates when any of the relays activate. <A> clunky solenoids with a battery just slightly above the Vmin for the coil will be the slowest chatter clunk, and then the max rated voltage for the coil will have the loudest but solid clunk. <S> for small relays <S> here's 4.5V one that can be mounted glued to a panel to amplify the sound <S> ,that will,operate from a LiPo cell down to 3.1V <S> http://www.digikey.com/product-detail/en/panasonic-electric-works/APF3034H/255-4085-5-ND/2202482 <S> If using a 9V battery then use matching coil voltage give or take 50% To make it buzz and stick and buzz when you hit it if it sticks... <S> Use the above and wire the coil in series with 3.7V Lipo battery and NC <S> ( ormally closed ) contacts <S> and then it will buzz and vibrate albeit with some electrical noise... <S> (don't touch contacts). <S> As battery wears out in 10 hours. <S> it will,buzz slower and slower until it sticks , you shake it <S> and it buzzes some more. <S> until battery is drained. <S> But for sound bytes... just put the mic in contact with the relay and record it. <S> or record an A/C solenoid <A> At least as important as picking the right relay is the mechanical design of the part the relay is attached to. <S> In order to emitt sound efficiently you need acoustical impedance matching so the tiny (but strong) mechanical osciallation of the relays moves as much air as possible. <S> This can be accomplishged, e.g. by placing the relays onto a large board that is free to oscillate or the chassis of the device. <A> We regulary use an inexpensive 200mW relay to obtain the authentic click sound. <S> In a plausible but futile experiment, we tried to shortly overpower the relay, connecting 5V coil to 24V supply. <S> Unfortunately, the sound level was not noticeably increased. <S> However, different relays behave very differently, so the choice of a right type is a key decision.
| As a rule of thumb I would say that the higher current the relay can handle, the louder the click of the relay. So you can check the datasheets for louder relays, or those that do not specifically advertise so. The solenoid doesn't have to do anything useful of course.
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Derate heatsink when used passively I want to reuse my old computer processor heatsinks for LED heatsinking. I think I can calculate their K/W value approximately by dividing the temperature difference vs TDP at their maximum load. I ignore the CPU packaging thermal resistance and take it as an additional safety margin.This way I get about 60/100 = 0.6K/W for a beefy heatsink. The problem is, that I want to get rid of the active component and derate the heatsink accordingly. Are there any approximations I can use for derating? I know I'll have to add a healthy safety margin, but I still need a general figure as to decide what wattage can they sink. <Q> It is a challenging problem to de-rate a heat sink. <S> The problem is that if a heat sink is designed for forced air cooling (using a fan), the spacing between fins is optimized for forced air, and usually is fairly dense, 1.5-2mm apart. <S> For cooling with free convection however, the fin spacing must be at least 1/4" (6mm) or more, for the ambient air to flow between fins with any reasonable velocity, and for thermal boundary layers not to overlap along the fins. <S> Needless to say, you should maintain vertical orientation of cooling surface and vertical orientation of fins to have any positive effect. <S> In other words, forced-air heat sinks are usually unsuitable for free convection cooling. <S> The best safe estimate would be to consider the entire heat sink as a brick. <A> CPU heatsink ought to be in the 0.2 to <0.1 'K/W range with silver thermal paste if using same surface area used, otherwise derate according to useage. <S> Then derate further if ALumClad substrate is not coplanar to within a few microns ... or just measure case rise. <S> I have used 100W LED on a CPU heatsink with a small fan for < 85'C case temp <A> After following IgnacioVazquez-Abrahams recommendation, I found a way to approximate the thermal resistance of the heatsink. <S> By searching for similar heatsinks, and checking their free-convection rates I get from 2x to 10x derating. <S> By looking at the geometry of the said heatsinks it is clear that this mainly depends on fin spacing(as already predicted by Ali Chen). <S> By comparing my heatsink geometry with the data I have, I think that about 3x/4x derating should be appropriate in my case. <S> Something like this(without the fan) gives me a value of about 1.5-2.5 K/W: <S> I am going to test this later and report my actual findings.
| Some manufacturers of heatsinks specify both the forced-air and free-convection rates of their heatsinks.
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How does this latch relay work? What is the function of the two diodes and the resistor? Why does not the SET coil have resistor? <Q> The diagram shown is for an AC-coil (or AC-DC) latching relay. <S> The series diodes are there to rectify the AC. <S> They're typically rated at something like 1kV PIV. <S> The resistor is for "ampere-turn compensation". <S> Generally the reset coil needs much less current than the set coil. <S> For higher voltage relays (>50VAC) it may be present inside the relay. <S> For lower coil voltage relays it may not be present (just a coil). <S> It's more of a problem with high voltage relays because the wire would have to be very fine to get the optimal resistance <S> so it's easier to just throw away some of the power in a resistor. <A> Partial answer: <S> It's likely that if you were to push reverse current through the "set" coil, it would actually reset the relay, and vice versa. <S> The diodes prevent this from happening. <S> Pure guess: The resistor in the "reset" lead is there to make sure that if you pulse both coils simultaneously (with the same voltage), the "set" function "wins". <A> I added an RC snubber to what appears to be a half wave rectified latching relay coil. <S> R is a resistor for ampere-turn correction. <S> This resistor is included in models for 50 VAC or 48 VDC or higher. <S> For DC models, check the coil polarity for both the set and reset coils and wire all connections correctly. <S> If the connections are not correct, unintended operation may occur.
| The diodes are required to make sure that the "set" and "reset" coils perform only their labeled functions.
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Can I use the center tapped transformer as a non-center tapped transformer? I am building a single DC power supply from scratch. I use the transformer to step down the 120 volts wall line. I want to use a non center tapped transformer. I found a center tapped transformer (one that has 3 connectors on the secondary side). I just want to ignore the middle one and treat it as a non center tapped transformer. Is this possible to do? just simply ignore the middle connector? What are the "product-wise" consequences of being me ignoring this middle connector? I mean, in my schematic, it will appear as a non tapped transformer, but in reality, the product may behave in an unexpectable and unacceptable way? I am worried about this. <Q> (If your transformer has wire leads, make sure to insulate the loose end so it doesn't short to something.) <A> If there are only THREE terminals on the secondary (output) side of the transformer, then YES, you can safely ignore the center-tap. <S> HOWEVER, note that many transformers have "split" secondary windings where you can connect the windings in series for "double" the voltage and "half" the current. <S> Or you can connect them in parallel for double the current and half the voltage. <S> So, if your transformer had more than 3 output terminals, you would likely have to connect the windings in series. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> The schematic symbols for the regular transformer and centre-tapped version hint that they are the same except for the tap. <S> One could tap off in multiple places on the secondary coil without affecting the end to end voltage. <S> Is this possible to do - just simply ignore the middle connector? <S> Certainly.
| Simply ignore the center tap -- it does nothing if you leave it open circuit.
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Could the software cause permanent damage to microSD? We are using microSDs as a main memory in our devices (raw mode, SPI interface, STM32 MCU), and in these years we had many problems with them. We have upgraded our drivers many times, and we added many hardware protections to boards too. until now we have solved many problems but there are some remaining problems too. We know that our microSDs have a poor quality and sometimes they find permanent damage (bad sectors, write protected , ...). My question is that theoretically is it possible to do permanent damage to microSD by software (low level driver)? <Q> Low level software always gets in the way whether you want it to or not. <S> The SDCard itself contains firmware that makes the final decision whether or not to write data to a given location. <S> Your OS level drivers are unlikely to have much influence on the health of the disk. <S> The total amount of data written will ultimately govern the rate of wear to the card. <S> Do not buy cheap consumer grade SDCards, you should consider buying the more expensive industrial grade cards and go with ones rated for a wider operating temperature range. <S> My company did a destructive test of the high end industrial cards and it took a full 3 weeks of constant overwriting to see the first write error <S> , that was after about 18 terabytes had been written. <S> The cheaper cards gave up completely after only a couple of days - that is to say that it became impossible to write anything at all to the card. <S> The OS itself refused to understand the geometry meaning that it was not possible to recover the card back to normal usage - the card was worn out to the point of uselessness. <S> We concluded that SDcards are a pretty crummy storage medium but for low-rate usage, although they will probably be fine as long as you are willing to over-spec them to provide maybe four times more storage than your project really needs. <A> There are some ways to brick your sd memory cards with software only: <S> Use a proprietary 'interface' commands that can enable encryption or lock something other. <S> Request switch to 1.8V SDIO, while still having 3.3V on the MCU (not sure what could happen here and how wide damages could be) <S> Using wrong cluster allocation units (formatting with windows tool instead of dedicated one), or writing to the card in single 512 byte sectors instead of larger blocks (forcing to erase single 32k "flash sectors" 64 times), will reduce card life significantly. <S> http://elm-chan.org/fsw/ff/en/appnote.html#fs2 <S> In case of using consumer grade SD cards. <S> There is a lottery in behaviour after wearing out the flash. <S> Some cards may go into read-only mode, some will allow to overwrite badblocks and some will just brick itself. <S> Even more, there is a chance that SD card will brick itself "just because" without any attempt in wearing it out. <A> You could continuously write to the SD card to its death since they have a limited number of write-cycles. <S> Although, if you're not deliberately trying to kill the card, or constantly writing huge amounts of data, it's unlikely you'll ever hit that limit with normal use.
| Yes, it is possible to render an SD card completely inoperable simply by using it "normally".
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Question about power drawn from a current limited supply when it is short circuited Consider the schematic below: When the switch is open, the 10 ohm resistor will draw 0.5 amps of current and dissipate 2.5 watts of power. No issue there. When the switch closes: We know the power supply output current is limited to 1 amp. So when the switch closes, the resistor is bypassed and the entire 1 amp of current from the power supply will flow through the short to ground. What is the power being drawn from the power supply? The short circuit current is 1 amp. Power is \$P = {I_{short}}^2 \cdot R_{short}\$. \$I_{short}\$ will be limited to 1 amp, and \$R_{short}\$ (the resistance of the short) is essentially 0 ohms. So is the power drawn from the supply \$P = 1^2 \cdot 0 = 0\$? <Q> I think this is a sort of brain teaser. <S> The trick is in definitions, what does it mean "drawn power", versus "power delivered to a user's load"? <S> The paradox resolution is yes, the power delivered to the user load is ZERO (assuming ideal short with zero resistance). <S> However, the power drawn from this power source is 5W, which will be all dissipated inside, on internal current-limiting circuitry. <S> In reality this power supply will be pretty hot. <S> You can try this with any AA battery to see the effect. <S> CORRECTION: <S> As Chris Stratton pedantically commented, the dissipated power can be different from 5W <S> if the power supply, say, has switching topology, such that it can dynamically change internal effective EMF and effective internal output impedance. <A> In engineering, we must apply real world solutions. <S> As has been said before, the actual load will not be zero. <S> The contact resistance of your switch and any and all connections to ground will have a resistance around a few milliohms. <S> Power supplies react to shorts/overcurrents differently depending on type and quality. <S> If a power supply output is shorted (in my experience), it will either clamp back the voltage, so that the power delivered does not exceed the maximum rating or it will enter a hiccup mode in which the power supply will disable itself and periodically attempt to re-enable itself. <S> In the second condition, if the shorted load is removed, the power supply will re-enter regulation. <S> In your ideal case, the load/switch will consume zero power, and all power will be burned in the output/source impedance of your supply. <S> You should always draw source impedances into your diagrams to ensure accuracy. <S> If this is a physics problem. <S> The ideal conductor/switch could not have a voltage drop >0, so your voltage supply would attempt to suppy an infinite amount of current while never exceeding a ∆V >0 <A> A practical voltage source will have a series resistance to it. <S> So when the 2 terminals of the voltage source is shorted the power delivered by the voltage source would be V^2/R. <S> Where R is the series resistance. <S> So there will not be any power dissipated from the short-circuited wire. <S> But the power delivered by the voltage source will by dissipated obviously as heat from its own series resistance.
| So, technically speaking when u short the terminals of a practical voltage source there is no load at all. Your power supply will also have a source resistance/impedance that must be taken into account.
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Can't identify these elements in this schematic could anyone help me identify these elements I have highlighted in red in this schematic? I'm trying to learn how to read schematics and am making a lot of progress but since this document is from the 70s early 80s it uses some non standard symbols. <Q> The N.C refers to "Not Connected". <S> You may wonder why the emitter of Q1 is connected to a voltage divider (R10/R11) when the base and collector are not connected to anything. <S> It should be connected to a point with a relatively negative potential. <S> The 8 dots and wide black shorting bars (that short two pair of pins in one of three positions) is a 2 pole 3 position slide switch.. <S> The '0.00V' indicates the voltage at that node of the circuit- <S> it is a virtual ground since U2A drives the long-tailed pair to force that node to 0.00V. <S> -12V is a supply rail, as is +12V. <S> The arrow is the wiper of a 500\$\Omega\$ trimpot, used to adjust the voltage <S> very slightly- <S> it's nominally about 0.63V, so very similar to the voltage on the base of Q5. <S> However it is fixed, and the voltage on the base of Q5 will drop by 2mV/degree C of the CA3046 die temperature. <S> The purpose of this is to ovenize the CA3046 (maintain an elevated constant die temperature). <S> To this end, Q4 is used to heat the die, and Q5 is the temperature sensor with Q6 and U2B forming a feedback controller. <S> If you don't have details of the adjustment procedure I would suggest not messing with Mr. trimpot. <A> N.C. = not connected top center is 2-wire 3-position <S> sliding switch http://www.digikey.com/product-search/en/switches/slide-switches/1115393 , there's actaully a label saying "octave switch" 0.00 means that there's a potential of 0 V on the wire (for some reason) <S> R14 is a potentiometer <S> https://en.wikipedia.org/wiki/Potentiometer <S> -12 <S> V means that this wire is connected to power supply -12 Volt power rail <A> In this circuit schematic, NC : <S> Here, NC stands for Not Connected i.e. floating terminal/connection <S> SW1 is sliding switch 0.00V : <S> Potential on the wire on that point <S> R14 : <S> Arrow at R14 indicates that the resistor is variable. <S> i.e. it is a potentiometer with maximum resistance of 500 (unit is not visible due to red marking on image) <S> -12 : It is negative power supply of magnitude 12V. <S> There are power supplies available with negative and positive voltages with three terminals like +12V, GND, -12V. <S> Voltage across -12V <S> and GND is -12V <S> which is given to the terminal shown in your circuit diagram.
| Q1 is part of a CA3046 transistor array and the emitter of Q1 is also the substrate for the chip.
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Why is it not advisable to fuse the neutral I have been doing some research on this question. The answer I get most of the time is that this is for safety reasons. That there can be no current in the neutral that didn't come from the live. If anyone has a clear answer, please tell me. <Q> Cost : <S> One fuse in the live is adequate to cut current to the circuit. <S> Isolation : If the neutral fuse blows first the circuit would stay live. <S> It's generally best to disconnect the circuit from mains. <S> Polarity : Many countries don't use polarised plug and sockets on single-phase plugs. <S> This means that the fuse in the appliance may indeed be in the neutral. <S> So, to answer your question, " Why is it not advisable to fuse the neutral? " <S> - we do it all the time in many countries. <S> Figure 1. <S> Unpolarised American (120 V) and European (230 V) mains plugs. <S> Update: <S> Note that with the unpolarised plugs the fuse can only be guaranteed to protect against over-current in the device itself. <S> e.g., a motor short-circuit will cause the fuse to blow. <S> With the fuse in the neutral wire a short to earth would not cause high current to flow through the fuse. <A> Having a device appear to be electrically dead while its components are electrically live can be dangerous; if the neutral were fused, an overcurrent fault could easily create that dangerous condition unless the fusing assembly ensured that an overcurrent condition would disconnect both hot and neutral simultaneously. <S> While it's possible to construct fuse assemblies in such fashion, such assemblies are generally much more expensive than those which only disconnect the wire through which excessive current is flowing. <A> Fuses are not recommended in neutral. <S> To ensure complete isolation with the removal of fuse. <S> To make the total current always flows through the fuse. <A> things get hot, fires happen ...
| If you put a fuse in the neutral and there's a short to ground the current will bypass the neutral and not blow the fuse ...
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A way to make temporary connections to battery holders? I often need to make connections to these types of battery holders. For example, to power it from an external power supply, and/or to measure current draw in various modes of operation. I can use my hands to hold probes to the battery-holder terminals, but then I can't push buttons on the device or adjust the multimeter. Alligator clips can't grab onto the flat surface of the positive terminal. Is there a way to make temporary hands-free connections to these types of battery holders without soldering? <Q> Don't laugh, there are battery cell size adapters . <S> Solder wires to their contacts on the inside. <A> Magnets. <S> Solder the power connections to the magnets. <S> It will stick to most battery contacts. <S> I made some with a wood dowel (well, squared) and some copper. <S> I added a diode on one and usb connector so I could power a battery led string from usb. <S> (4.3V instead of the new batteries 4.8V). <S> If you cut or etch the copper to shape, you could use a pair of alligator clips on it. <S> It goes between the contact and the battery. <A> You can buy battery compartments for projects and they look just like the one in your picture <S> so, buy one and remove the end <S> faces containing the spring and flat connector. <S> Take these end faces and glue to a block of compressible foam rubber that can be squished up while inserting into the target battery holder. <S> Oh, and solder wires from the terminals to some kind of socket of your choice to make the external connections to the power source. <S> Spring contact on the target mates with flat contact on the squishable insert .
| Cut to size and shaped to fit in the recess for the battery contact. You could make a small shim from double sided copper clad board.
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What happens to resistor when it "breaks" due to over-current or over voltage? What happens to resistor when it "breaks" due to over-voltage or over-current? Does it short or Open? or it merely ends up outside of specification? <Q> Anything can happen, but most seem to go open circuit. <S> You can buy 'fusible resistors' that are guarranteed to fail open circuit. <A> I'll be a lone voice in the wilderness saying that under overloads that don't involve kilovolts metal film resistors and wirewound resistors tend to behave like a fuse (but not a guaranteed fuse) and some metal-oxide-film and some wirewound (eg. <S> rectangular ceramic type) <S> resistors may be guaranteed to act as a fuse. <S> However types involve carbon such as composition and carbon film resistors <S> can arc at relatively low voltages (such as mains voltage) and the entire resistor becomes a glowing ceramic rod that ends up being lower resistance than it started. <S> It's called carbon tracking. <S> I'm not suggesting anyone try this, hear me, but if you do, note that low value 'carbon film' resistors are often actually metal film. <S> I don't remember where the transition was for 1/4-W resistors, but below 100 ohms, I think. <S> It's not something they feel they need to disclose because even a low end metal or alloy like pure nickel will outperform the tempco of carbon (it is just unsuitable for high resistances, as carbon is unsuitable for very low resistances). <A> The most common failure mode I have seen in resistors is that they open up. <S> When too much current flows through the resistor (which can be caused by over-voltage as well) <S> it heats up the material, causing it to melt. <S> When it melts it acts like a fuse, breaking the circuit. <S> If only slightly too much current flows through it <S> I have never seen a failed resistor that was shorted.
| it could change its resistance, either increasing or decreasing, but ultimately it will most likely fail open.
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Is XOR equal to XNOR when odd number of inputs are considered? I was attempting a question which said that you can convert a Full Adder into a Full Subtractor with using just one inverter. Later I researched on this XOR and found that that for even no. of inputs it is the invert of XNOR and for odd number of inputs it is equal to XNOR. Is it true? Also Can we say that the XOR gate gives true or 1 value only when there are odd numbers of 1 in the input. <Q> The behavior of a a XOR gate with multiple inputs needs to be defined, it is not just a simple extension of the two-input XOR gate. <S> As always with definitions it is possible to have different, even contradicting, ones. <S> In my opinion this definition seems to be useful: <S> This definition is quite common in computer science, where XOR is usually thought of as addition modulo 2. <S> In this context, it arises in polynomial algebra modulo 2, arithmetic circuits with a full adder, and in parity generating or checking. <S> While this means that the multiargument "XOR" can no longer be thought of as "the exclusive OR" operation, this form is rarely used in mathematical logic and so does not cause very much confusion. <S> The XNOR would just be the inverse of the XOR. <A> I didn't quite understand your first question. <S> You said both are odd, yet one is inverted and one is same? <S> For the second question, yes. <S> XOR is true when there's an odd number of 1s in the input. <A> The output of an 2-input XOR is clearly defined. <S> An XNOR is simply an XOR with an inverter after it. <S> But now for multiple inputs... <S> As far as I can find, for the rectangular (IEC) symbol, the definition is that the output is 1 when exactly one input is 1 , otherwise it is 0. <S> The 'rectangular' symbol shows this by stating "=1". <S> For the classic (US) symbol the behaviour for more than 2 inputs is not defined. <S> Sometimes this "=1" is even used in the classic symbol, in which case it has the same behaviour as the rectangular symbol. <S> But the only 3-input XOR chip that I found ( 74LVC1G386 ) implements the funtion ( A xor ( B xor C ) <S> ): the output is 1 for any odd number of inputs that are 1.
| For multiple arguments, XOR is defined to be true if an odd number of its arguments are true, and false otherwise.
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What options are there for good Low DC Voltage, High Amperage Alligator Clip connections? I have a 12V DC, 20A compressor for my offroad vehicle to inflate its tyres after driving on the beach. I also have high current jumper leads with large alligator clips at each end for starting cars with a flat battery. The compressor has flying leads with bare copper ends. What options are there for a solid connection (without the risk of a short circuit) if I want to power my compressor from the car battery using the jumper leads? Here is a diagram of the setup: One option is some high-current jumper terminals, but these are quite costly. I could possibly make my own out of a block of wood, and two bolts. Does anyone have any other ideas or advice? <Q> Honestly, 20A isn't high-ampere in car electrics. <S> The starter has 250A. <S> My advice would be to neither use starter crocodile clips nor the terminal blocks. <S> They are both suitable for 300A minimum <S> , that's 15 times overkill for your application. <S> Pick a connector from a car parts store which is rated 25A and you should be all right. <A> Check your cigarette lighter fuse. <A> 20A compressor probably has motor winding resistance yielding 100~200A peak currents or < 0.1 Ohm DCR. <S> To make efficient connections for low V drop choose connector with 1% DCR or 1 MilliOhm contacts. <S> Higher is OK but <S> >10 MilliOhm starts to affect compressor. <S> Jumper cable clips work sometimes but surface oxidation and high spring force are needed with sharp edges to jump start car with similar 200A~800A currents especially in cold weather requires experience and good jumper cables. <S> This is what I would consider using. <S> (Welder waterproof connectors) <S> CDN$ 5.65 <S> 110 x 22mm/4.33 x 0.87"(L <S> x Dia) <S> "BQLZR IP68 Grade 2Pin Waterproof Plug Socket Electric Cable Wire Connector M19 Welding Type " <A> Look at marine (or fishing?) <S> supply stores for trolling motor battery connectors. <S> They should be rated for higher currents than your compressor requires.
| If that's not an option, any connector of the kind you'd find on starter cables will do; you can buy those with a screw terminal to attach your compressors' cables. It might be 20A-rated, in which case, that is an appropriate connection.
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Flip between motors each time circuit energized I want what I hope is a simple circuit but I've googled and searched this forum, not found it. When a SPST digital programmable timer completes a circuit, I want one motor to run. On the next circuit completion from the timer the other motor. Next time, the first motor, and so on. I can do this with two timers, each set on different times. But I would like to know if there is a simple way to do it with one timer. Do NOT want a circuit that must be energized all day long. Over time that increases greatly the battery and solar requirements. Something like a latching relay? A bi stable flip flop maybe? The application: A chicken coop door. Two water reservoirs. To open, fill the bottle on the door pulley. To close, pump from the door pulley bottle to the bottom reservoir. I can use an up timer and down timer but I think this can be done with some kind of latching relay or something. The goal is to be very low-cost and simple. If the alternative to two timers costs more than the second $5 timer, I'll go with two timers. <Q> You need circuit, that has two parts: 1. <S> H bridge to be able run motor in both directions. <S> I will not go to details about that, Mr. Google knows all about it, just ask him. <S> 2. <S> "Button with memory <S> " You have timer, which closes circuit, just like a button. <S> I'm a simple person, I like simple things, I will speak like it is a button. <S> You need different initial conditions when this button is pressed. <S> In fact you have different conditions alredy <S> , you just need to take advantage of it. <S> One condition is with doors closed, the other with doors open. <S> I would suggest using relays to implement it. <S> To detect initial condition, there are many possibilities, one of them is magnet and reed relay. <S> Place magnet on the top of doors and place reed relay in position <S> , that matches with magnet when doors are open, or closed, whichever fits you better. <S> You need two additional relays to remember initial condition after doors will start moving and that's it. <S> You have signal direction signal, which you will use as input for H bridge. <A> Some are simple, some need very specific signals to toggle. <S> Read the datasheets. <S> Aside: have you considered <S> not using electronics? <S> Suppose the weight on the pulley was slowly leaking back into the lower container? <S> Then you would power the pump to open the door, and it closes after some time by itself. <S> This way you would only need one pump, a bit of aquarium tubing, and something to set the flow for the leak. <S> Granted, you would need to pump more often. <A> Going to try an ESP8266, which is similar to an Arduino but with Wifi. <S> As mentioned in the comments, I was initially intimidated by what I thought was high cost and complexity and difficulty to learn and fragility and high current usage; looks like none of these are a concern. <S> The ESP-07S uses 0.9mA in light sleep mode, costs $3, and is fairly straightforward to program. <S> With the low cost and current, I can use two solar panel/battery/controller/pump setups for redundancy. <S> Thanks everyone.
| A bi-stable relay would be fine, but you will need to be careful choosing one.
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Resistor that changes resistance with clock input I am making a 555 a stable connected to a 4510 and 4511 and an led row. I need the led row to get faster and therefore the 555 astable to get faster, the more times a button is pushed. I am thinking a resistor that changes resistance with a clock input and a reset function but I am new to electronics so anything is appreciated! A video of the circuit so far is here: https: https://goo.gl/5luzl1 , and data sheets for the 4510 (I can only post two links so no 4511 pinout) is here: https://goo.gl/1UMyLg . Any ideas? Thanks <Q> There are 'digipots' which are designed for a simple up-button/down-button type of interface (rather than SPI or I2C). <S> Unfortunately the ones which pop up in a quick search are packaged in small SMD footprints, but you could probably 'dead-bug' them if you're careful. <S> Digikey Search Example: <A> You can use an analog multiplexer such as a 4051 to select one of 8 resistors using a 3-bit input that you can connect to a counter or whatever strikes your fancy. <S> If the resistances are relatively low then you might have to use a more expensive analog multiplexer, but the concept is the same. <A> I am thinking a resistor that changes resistance with a clock input and a reset function <S> Your thinking is good, but how to convert that idea into a realizable circuit? <S> Break it down into elementary functions. <S> You will need:- <S> If 10 steps are enough then a CD4510 would do the job, otherwise choose another IC which counts to the number of steps you want (eg. <S> CD4024). <S> A method of switching in different resistor values from the outputs of your counter. <S> The CD4066 has 4 switches which can be controlled from the four Q outputs of a CD4510. <S> As the count increases you want the 555 capacitor charging time to decrease, so the individual resistances should be switched into the timing circuit in parallel . <S> Since the counter outputs are in binary, the resistors should be binary weighted , with eg. <S> 100k switched in from Q1, 50k from Q2 etc., halving the resistance value for each 2^n counter output. <S> Clock and Reset pulses generated by push-buttons. <S> The reset circuit is obvious, just connect a push-button to VDD or GND (depending on whether the reset pulse should be high or low) with a pull up/down resistor to the opposite supply rail. <S> To make the counter reset on power up, wire a small capacitor (eg. <S> 100nF) across the push-button. <S> The clocking circuit is not so easy. <S> mechanical switches bounce . <S> If you use a simple push-button with pull up/down resistor, the counter may advance several counts each time you push the button. <S> So you need some kind of debouncing circuit that guarantees only one clock pulse per button press. <S> An SR latch would be ideal, except that it requires a switch with contacts for both up and down positions. <S> Alternatively you could use an RC circuit to filter out the glitches, and a gate with Schmitt trigger input (eg. <S> CD4093) to square up the output.
| A counter with clock and reset inputs. One way to do that is with analog switches.
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Why are there no BGA chips with triangular tessellation of circular pads (a "hexagonal grid")? Ball grid arrays are advantageous integrated circuit packages when a high interconnect density and/or low parasitic inductance is paramount. However, they all use a rectangular grid. A triangular tiling would allow π⁄√12 or 90.69% of the footprint to be reserved for the solder balls and the surrounding clearance, while the ubiquitous square tiling only allows π/4 or 78.54% of the footprint to be used. Triangular tiling would theoretically allow either reducing the chip footprint by 13.4% or increasing the ball size and/or clearance while maintaining the same footprint. The choice seems obvious, yet I have never seen such a package. What are the reasons for this? Would signal routing become too difficult, would manufacturability of the board somehow suffer, would this make adhesive underfill impractical or is the concept patented by someone? <Q> Unless you use via-in-pad, which costs more, you need room to put routing vias in between the pads, like this <A> What happens if you have to route a trace from the center of the BGA to another part of the PCB? <S> On a square grid you may simply route a straight line, but on the hexagonal grid you need a lot of bends. <S> Working with a very fine routing grid within the hexagonal array of balls is no fun and will need a lot more time. <S> Routing with 0 °, 45 ° and 90 ° only will not be possible, you will need the angles 30 ° and 60 ° too. <S> PCB auto routers may not work very well if designed for square pin grids only. <S> It is possible that a multilayer board will need 2 or 4 additional planes if such a dense hexagonal packing is used. <S> If there is no space for vias between the pads of the BGA grid even more layers might be necessary (only vias within pads are possible). <S> Exact placement of the pads will take a lot of time. <A> Mainly because we need space to route from <S> those pads: <S> In the first picture you show, some 6 layers or more would probably be needed for a decently sized BGA (~400-ish balls). <S> Packing stuff even tighter means that you absolutely need via-in-pad and probably need more layers. <S> This costs more money because it's harder to manufacture. <S> Some smart guy at Texas Instruments came up with a technology they call Via Channel, to simplify this routing process (often called fan-out) and also reduce the size requirement you speak of. <S> An interesting presentation can be found here (This is also where I got that picture). <A> Some packages appear to use the hexagonal packing for the exact reason you describe. <S> I'm not sure why they don't do it everywhere, but at least near the edges they are here.
| Designing the library pcb symbol for such a hexagonal array will be difficult and time consuming and error prone if there is only a square grid for placement of pads.
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Possible battery solutions for 1000mAh capacity and >10 year life? I'm looking into the possibility of running a wireless sensor node for up to 10 years on a single battery. I think the average current draw is likely to be around 70uA, with the peak current draw possibly being 20mA(the average takes this into account). System voltage can be between 2.5-3V. This seems like it might be a requirement in more and more projects as the area of IoT and remote sensors continues to grow. I was hoping to get some suggestions for a few different possible solutions that balance cost and capability. I thought about a CR123A as a possible battery, or even just a couple of AAA batteries. A li-ion rechargeable cell was also a consideration, but I think maybe the self-discharge might be a factor after that length of time and recharging/energy harvesting might be difficult or costly. <Q> First, let's see how much energy or capacity you need. <S> 70 µA <S> for 10 years comes out to 6.1 Ah, and you say this is at 3 V. <S> That's a lot more than a CRxxxx battery can do. <S> Another problem is that you need a battery that is good for 10 years regardless of discharge. <S> Many batteries aren't specified for such a long shelf life, let alone service life. <S> For this kind of application, you need to have conversations with field engineers from battery companies. <S> Take a look at Tadiran. <S> They are aiming for this kind of application. <S> Another option is a lithium rechargeable with a small solar panel. <S> At 70 µA average, a 18650 cell is way more than enough. <S> It could easily ride out a few weeks with no sun at all. <S> Just a few hours of sun a week would be enough with a modest panel. <A> Long lasting (very low self-discharge) batteries usually doesn't have a very good discharge characteristics. <S> For example Tadiran SL-360, 14500 form factor offers 2.1AH at 6mA discharge but can not be discharged above 20mA. In the datasheet, you can see it can discharge 10 years at 20uA. <S> One way to get around it is to use a energy buffering solution: slowly discharge the battery and charge a large capacitor, and use the capacitor to burst out the peak current you need, when you need it, like this energy buffering reference design targeting 15 years battery life for remote sensing application by TI . <S> I think if you choose your battery based on your peak current requirement, your design will need a very specialized battery that offers both low self discharge and good discharge current, which probably makes it expensive and hard to source. <S> Energy buffering relaxes the peak current requirement on the battery significantly. <A> As someone posted in the comments, Lithium Thionyl Chloride primary cells are a solution to this problem. <S> Such cells exist in various form factors, from 1/2AA through D, and various battery pack configurations. <S> Certain manufacturers (Saft, Tadiran batteries) guarantee 10 years of shelf life. <S> These cells provide 3.6V, take into account that it is extremely hard to gauge the remaining level of charge with these cells, and there are other phenomena related to the batteries that you will have to consider in your design. <A> The self discharge on these is roughly 10% in 5 years - all rechargeable batteries are much higher than this. <S> This is the solution used by Electronics companies for long life wireless sensor nodes. <S> This link has some info on battery discharge, while This one describes non rechargeable battery types. <S> Finally, this Farnell link <S> has a selection of non-rechargable lithium batteries (of various types!) <S> that should suit your application <A> Nearly 30 years ago I was responsible for the design of sealed non-repairable electronic equipment which was sold with a 5-year unconditional warranty. <S> I expect battery design has improved over time, so I would suggest this as the likely solution. <A> Buy sulfuric acid and plumbum (Pb), make a lead battery large enough. <S> Come into the room and recharge the battery when needed with another battery - non-stop solution (as you want). <S> Size will be big, but you don't have any limitations of that. <S> Lead batteries are used up to this days when you want very low self-discharge. <S> It stores reserve power for cell networks, computers etc. <S> Because it is reliable, simple and does not harass environment (all components are convertible back to new battery).
| I believe your best possible solution is non-rechargable (or "primary") Lithium batteries. We used Lithium Thionyl Chloride batteries (Tadiran and Saft, as I remember), in sizes from AA to D, and had no problems with battery life; we had test units which achieved over 8 years operational use (at which time I left the company and so lost contact).
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is soldering mobile processors something that i could do at home? I want to connect a mobile processor such as a Samsung Exynos or Qualcomm Snapdragon, which both have their connectors on the bottom to a PCB but im not sure if this is something i can do on my own or needs a pick and place machine, to be honest im not entirely sure what the method of soldering used is called. Can you please inform me if a socket is available for these also? Sorry if this came across as a dumb question, thanks for the help :) <Q> It is not possible to solder components with contacts on the bottom (called Ball Grid Array, or BGA) using a soldering iron. <S> You would need either a reflow oven, or a BGA rework station. <S> So the answer to your question is yes, you can do this at home if you have the right tools, but if you only have a soldering iron then you can't. <A> You can do it dead bug style: <S> Is it practical or is it going to work, especially in an 1000+ balls device? <S> I don't think so... <S> (Image from the comments area from Hackaday website ) <A> I doubt it, the bottom of a mobile processor has a lot of pins that are too small for you to handle and would need a reflow oven.
| Some people do have reflow ovens at home, or have converted toaster ovens to do the job, but even then it's a bit tricky and easy to mess up.
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Maximum Transistor current and voltage Assuming we have a transistor that can bear 400v 0.3A, is that means it can bear 30A if we used it for 4v only or it always withstand 0.3A? (since P1 = P2 = 0.3 * 400 = 30 * 4) <Q> Many power transistors have a SOA (Safe Operating Area) graph in their datasheet, here's an example: <S> The safe area is the bottom left side <S> so: Vce = 5 V at Ic = <S> 20 <S> Amps is just OK but <S> Vce = 50 V at Ic <S> = 2 Amps is not Also note that it states Tc = 25 degrees <S> Celcius meaning you have to cool this device such that its case will not exceed 25 degrees Celcius. <S> Also not that for a limited time, the device can handle a little higher value. <S> It is not clear how much time it needs to recover after such an event. <A> Power dissipated, Pd, in the transistor = Vce*Ic <S> and the thermal resistance Rja [0.2'C/mW] makes the junction hotter, which affects reliability Power increases from 0 at the origin of this graph up to the right. <S> In order to exceed this Safe Operating Area (SOA) only low duty cycle pulse widths are permitted, which reduces average power. <S> These are individual max limits similar to yours. <S> Vceo=400v <S> max <S> Ic = <S> 0.3A <S> max <S> Pd = 0.625W <S> max <S> Pd is much smaller than Vceo*Ic product due size of the part. <S> Ic max is often used as a switch with low drop voltage but rising with current like a resistor but faster. <S> Normally one uses short pulses or only uses less than 50% of Ic max. <A> Transistors have at least three limits. <S> a) <S> Voltage limit, for instance your 400v. <S> This is governed by thickness of doped regions and doping densities. <S> Exceeding the rated voltage will break down the junction, usually short circuiting it. <S> b) <S> Current limit, for instance <S> 300mA. <S> This is governed by the area of the junctions, and the thickness of bond wires. <S> Too much current will overheat the junction, or fuse the bond wires. <S> c) Power limit. <S> I strongly doubt that a transistor rated for only 300mA would be able to handle as much as 120W, or that a 120W transistor would be rated for only 300mA. <S> The power limit is governed by the thermal conductivity of the junctions to the mounting base. <S> Exceeding the power will overheat the die. <S> It's normal for a transistor to have a power rating much less than the product of its rated voltage and maximum current. <A> You don't link a datasheet <S> so I'll make this short: <S> No . <S> For a MOSFET there are three limits you need to check: <S> Maximum current Maximum voltage <S> Maximum power <S> You have already identified that Voltage times Current equals Power, but keeping within the maximum power doesn't allow you to exceed the maximum current or voltage . <S> All parameters must be below the limit. <S> It's slightly more complicated for a BJT, because its SOA is not as simple as for a MOSFET. <A> No. <S> 400 V and 300 mA are independent maximum specs. <S> The C-B junction will still break down beyond 400 V whether you run 3 mA thru it or the full 300 mA. Conversely, just because you are only applying 4 V <S> doesn't mean you get to violate the other spec of 300 mA maximum. <A> Always use 70% from the max voltage or current And select the transistor RDS on depends of your voltage . <S> More higher voltage you need select the high RDS on Low voltage you work ( select RDS on lower ) <S> 100w <S> max transistors with 20 amp max and 100v max <S> It mean the transistor will be good to work on between 50v 2amp and 10v 10amp
| Exceeding any one of voltage, power or current individually will destroy the transistor.
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Voltage drop across capacitor in a DC circuit? (Simple question) I'm trying to find the voltage drop across a capacitor in a DC circuit. We can assume that the capacitor is fully charged and in a steady state. Therefore, the capacitor is now an open circuit. So this is really an open circuit voltage problem. My question concerns the following section of the circuit: Node a -> resistor -> node b -> capacitor -> ground Is the voltage at node b zero because no current flows through the capacitor \$(V=0R)\$? Or does the voltage at node b equal voltage at node a? <Q> The voltage across a resistor is proportional to the current thru it. <S> When there is no current thru a resistor, the voltage across it is therefore zero. <A> The voltage at node b is equal to the voltage at node a. <S> No current is flowing through the resistor and therefore the voltage drop is zero. <S> The capacitor is an open circuit, therefore any voltage drop is possible. <S> The equation V=0*R cannot be used since this equation is only valid for a resistor. <A> You are interpreting the results of your calculation incorrectly. <S> You wrote: V=0*R <S> (Therefore Voltage is zero.) <S> And assumed this was the voltage at the node between the resistor and the capacitor. <S> In fact, this is the voltage drop across the resistor. <S> You are actually proving the voltage is the same with respect to ground on either side of the resistor. <S> Which is what you would expect if using an fully charged ideal capacitor. <A> Being pedantic (but accurate) <S> the capacitor (even an ideal one) never fully charges to the supply voltage <S> so there would always be a very small current flowing and so a very small (and decreasing with time) <S> voltage drop between node a and node b. <A> By following Ohm's Law, you can see that there is no voltage drop across the 50ohms resistor, therefore the voltage is the same at node a and node b. <S> But considering the fact that no electronic is EVER ideal, the capacitor will never charge fully. <S> So there will be a small amount of current flowing through the resistor and there will be a voltage drop a node b.
| Since the voltage across the resistor is zero, the voltage of the two nodes at either end are at the same potential.
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Thermometer type in electrical water heater I'm working on my home monitoring system and already succeed with electricity consumption measurement and got closer to water system. Measuring water consumption is almost solved, but as we are using boiler almost 365 days a year I would like to measure water temperature in a tank as well to have more info on it. As the tank itself is tightly isolated I've got an idea to intrude into boiler and use it's temperature sensor. Water heater has electronic ( buttons ) control (model ARISTON VELIS 50 ) Under the hood it looks like this ( 2 tanks ) And temperature sensor for this model is the following As it seen on the picture the left thermostat has 3 wires which I suppose are: GND VCC DATA whereas the right tank has kinda 4 wires going out the same place. I'm wondering: if its possible to know how to read the temperature data from 3 wires sensor or how to know the "protocol" its using for data transmission? is it possible to intercept this data not interfering with the boiler board? Maybe there are some traditional/known approaches for such kind of tasks or this is regular temperature sensor, but I don't have a clue what to start with bec. organizing a workspace under the boiler not knowing what to do is quite uncomfortable. Unfortunately I don't have an oscilloscope to read raw signal, but I have both Arduino and RaspberryPi3 and multimiter in my arsenal. <Q> I believe that @TurboJ is correct that there is no galvanic isolation. <S> That would make the signals very difficult to deal with safely- the sensor appears to be tied to one side of the mains. <S> As it's clearly (well, after the good research done by user222030 and Majenko) some kind of dual NTC resistance probe it will be very nonlinear and referenced to some voltage such as 3.3V or 5V on the board. <S> You would have to reverse engineer the circuit and then you'd still have an nonlinear nasty voltage connected to one side of the mains, ready to blow the living <S> **** out of your home automation system, damage the boiler controls, or electrocute the hapless user. <S> While you could attempt to introduce another sensor into one of the protection tubes it would have to be insulated sufficiently for the mains to be safe.. which may be non-trivial <S> and I don't think it would be responsible to try to make suggestions remotely on how to do that. <A> From your photo it appears that each sensor is inserted into a closed-ended tube that projects into the boiler, also known as a 'thermopocket' or 'thermowell'. <S> Because the boiler electronics are not isolated from mains, you then need to use isolated signal conditioning to measure the temperature from the sensor and transmit it to your home monitoring system. <S> A type J or K thermocouple would be a reasonable choice of sensor: you could get better accuracy with a platinum RTD <S> but you don't need it for this application <S> and it'd cost more. <S> You could measure this with an analogue transmitter , which typically converts the output to a 4 - 20mA current signal. <S> Typically these are loop-powered: you connect the device like this simulate this circuit – <S> Schematic created using CircuitLab <S> where the 'meter' could be a resistor across which you measure the voltage. <S> \$V_{supply}\$ needs to be enough for the minimum voltage requirement of the transmitter plus the maximum voltage burden of the meter. <S> Or you could use a data acquisition device from which you would read the measurements digitally - the type I've linked uses an RS-485 bus connection (and 'clones' of this type are widely available from different manufacturers) but you may be able to find other types. <S> As another answer notes, there are some important safety considerations here: <S> you must make sure that your temperature sensor doesn't electrically contact or interfere with the existing sensor you must treat anything that could come into electrical contact with the boiler electronics, even by accident, as mains voltage wiring <S> you must satisfy yourself that any signal conditioning unit is suitable for your purpose - the ones I've linked to are examples of what is available and <S> not necessarily a recommendation <S> you must make sure you understand what is isolated from what by the signal conditioning unit, for example is the power supply connection on the input side of the isolation or the output side <S> If you are unsure about any of those, don't do it. <A> No. <S> Those aren't digital sensors. <S> Those are two NTC resistors, with a common lead. <S> I have somewhere resistance/temperature table. <S> I can try to find it if you still need it.
| Depending on the size and shape of the thermowell, you may be able to insert your own sensor into it alongside the existing one.
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Why don't switches appear in circuit diagrams of DC motors? DC motors are typically represented in circuit diagrams as follows, This sort of diagram seems to show up in controls and system dynamics courses. Most common DC motors have commutators, switches that change the direction of current in the armature coils. Without a commutator, a DC coil would just act like an electromagnet and align with the stator field. With a commutator, the induced magnetic field changes direction, twice per revolution. The faster the rotor turns, the faster the current in the inductor should change direction. This switch never shows up in the textbook diagrams I see. What gives? Is the inductor in this diagram the actual coils of wire in the armature, or does it represent an effective inductance that is averaged out over time? Also, is back voltage constant at a constant angular velocity, or does it, too, switch on and off? Is all of this a pious lie disseminated by controls professors? <Q> There is no need to show all the details for a good understanding of the schematic. <S> If the subject changes to understanding the actual construction of a DC motor or generator and the types of winding involved more details are given. <S> The inductor in combination with the rotor in your diagram represents a socalled serial DC motor. <S> Here only the principle of the motor is shown and not the actual coils in the armature or stator (L). <S> Depending on the design there could be many coils and brushes. <S> In an electronic equivalent the brushes are replaced with semiconductors. <S> When an armature turns in a magnetic field <S> (L in your figure) <S> a back voltage is generated. <S> That voltage is in average constant at constant angular velocity of the armature. <S> For each individual winding this is not the situation. <S> In practice the commutator "switches on and off" and changes the action to a different winding. <S> In fact the voltage induced in the winding has a sinus form. <S> Since there is a fixed field in the drawing this type of motor is a socalled shunt motor. <S> The field has thereby a constant voltage applied although it is possible to make the exitation of the field adjustable to regulate the velocity of the rotor. <S> Less exitation results in a higher speed. <A> You seem to have an improper concept of what a schematic diagram is and what it is used for. <S> The INTERNAL DETAILS of components (like motors or integrated circuits) is NOT shown because it is not important to understanding how everything interconnects and works together. <S> Certainly, if you want to know the INTERNAL DETAILS of anything from a resistor to an integrated circuit with literally billions of transistors, you can find the documentation. <S> But none of that is important to a schematic diagram. <S> The textbook models are NOT "conceptually flawed", because they are trying to demonstrate the CONCEPT, not show details of how a brushed DC motor is designed and constructed. <S> To be sure, there is information about how brushed DC motors work, how they are made and what are the principles of operation and design and construction. <A> Most common DC motors have commutators <S> Which is shown in the diagram! <S> The symbol for a brushed DC motor is a crude representation of its commutator and brushes viewed end-on. <S> Actually in that circuit the symbol represents a DC generator . <S> Since a DC motor is a generator when the shaft is rotated mechanically, the same symbol can be used for both. <S> And when the motor is running it is also acting as a generator, producing a 'back-emf' voltage which opposes the supply voltage. <S> Is the inductor in this diagram the actual coils of wire in the armature <S> Yes. <S> L is the inductance of the armature windings, and R is the resistance of the windings and brushes. <S> These are shown in series with the generator because voltage induced into the windings is effectively in series with them. <S> The faster the rotor turns, the faster the current in the inductor should change direction. <S> This switch never shows up in the textbook diagrams I see. <S> What gives? <S> In this model it is not considered. <S> The winding inductance is so low that it is presumed to have negligible effect at normal running speed, and the rectifying action of the commutator turns the AC into DC so the changing current direction inside the armature is irrelevant.
| The main reason that your "switches" are never shown in a principle schematic is of practical nature.
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Algorithm for testing EEPRM What are the steps or the algorithms to verify that an EEPROM is working ? Right now I just write 10 bytes and I read 10 bytes from 10 addresses, and that's naïve approach. Is there an algorithm or an approach to verify that the eeprom is correct ? or also the problem for flash memory testing ? <Q> Function of EEPROM and flash is to hold specific data or code. <S> You can use checksum to know if it has correct or corrupt data. <S> Writing to flash too many times will wear it out. <S> Thus you need to ensure that: chips are physically present in the board (soldered into their places); you may want to have chip programmed one time in full and compare to pattern you program to ensure that chip is not DOA; you may need to check for proper data using checksum on each power up, and act accordingly (e.g. display message, use other data source etc); you will need to check for properly programmed data only after programming. <A> Your flash may or may not require a complete buffer to be written. <S> The walking ones test is well known and is one of the fastest tests to check proper operation of a memory device. <S> Update in response to comment. <S> I implemented this specific test to determine the presence or absence of a flash device in the early 1990s, so yes, it works with flash. <S> [Update for full buffer write] <S> Some older (and many serial access) devices require that a complete write buffer be filled; <S> this simply (internally) increments a counter <S> so you can programme the data word using the above algorithm and simply fill the rest of the buffer with 0xFF for every other byte - that way only one word actually gets truly programmed (flash devices are 0xFF on a byte basis when not programmed). <A> Flash outwears by individual cells leaking too much of the charge they have been loaded with at programming. <S> This could be due to heat or irradiation or manufacturing flaws. <S> It's hard to detect those pretty common errors, as when you read the data you notice nothing until it's too late and <S> when you write it, you refresh the charge so you cannot see the leaking either. <S> In the distant past, best practise was to read out EEPROMS every few years or so and re-write the contents. <S> That wouldn't let you see the leaking but at least the leaks wouldn't lead to data loss.
| For an initial test, you could use the walking ones approach.
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Schematic Confusion: How to connect the Ground Terminal? GND means ground. How am I supposed to connect this GND ? Does it means that I have to simply hook up a wire and leave it floating? What is it's significance in the circuit?I am designing a PCB and I want to know how to connect this GND in a PCB??? <Q> GND in an electric circuit is nothing more then a reference. <S> This is the position from were all potentials are given unless specifically indicated otherwise. <S> In this case GND has nothing to do with Earth or mains GND.If <S> you have a powersupply then in your case <S> the negative line of the DC goes to GND. <S> and the positive line goes to VCC. <A> You connect it to the net called "GND". <S> If there's nothing else in the schematic that connects to the GND net <S> then nothing connects to it. <A> This is really an instruction that you measure any voltages in the schematic with respect to this node. <S> If you are building a board to this schematic, this is the node that connects to the plane that floods the PCB, and connects to the case if it's metallic.
| Most simulators will require one node to be labelled 'gnd', so they have a reference to measure and display the nodal voltages against.
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How to safely dispose of a damaged lithium ion battery? I replaced a lithium ion battery in a device and the old battery is swollen/bulging but no otherwise damaged. My municipality (NYC) requires many retailers to accept batteries or devices containing them but I am concerned that none of them can safely handle a damaged battery. I asked one retailer if they would accept my damaged battery and they told me that they would simply throw it away (which is expressly prohibited by city law) because they have no way to safely store or handle damaged batteries. I have read from various sources that immersing the battery in salt water for an extended period of time (ranging from 1 day to 2 weeks!) will discharge the battery and that lithium (ion?) batteries that have been completely discharged can be safely thrown away with household trash. But then other sources claim that the salt water can corrode the contacts of the battery's cells sufficiently to prevent complete discharge, thus rendering the battery both not completely discharged (and thus capable of thermal conflagration) and not capable of being discharged by any means. So how could I safely dispose of a damaged lithium ion battery myself? The NYC Special Waste Drop-Off Sites seem like my best bet in terms of someone else safely disposing of the battery but the info about those sites on the city's websites are worryingly vague or silent about whether they accept damaged batteries. I don't want to assume that any of the organizations or institutions that are legally required to accept batteries are capable of safely doing so for my damaged battery. Given that one retailer has already told me that they would (illegally) dispose of the damaged battery, I'd like to know how to safely dispose of the battery, or at least render it (mostly) inert, myself. Here's a very related question: batteries - Storing a possibly damaged Li-Ion battery A bunch of example links: Safe Disposal | 2BFly No one wants my damaged lithium-ion battery | sqwabb How would one dispose of a ruptured or exploded lithium ion battery? : electronics <Q> Be clear there are two levels of "damaged": bulging, and breached. <S> Li-Ion batteries will have single thermal event: once the airtight case is breached they'll rapidly discharge: if there's enough energy in the battery they'll heat up perhaps to the point of catching fire. <S> Once that's happened there's no more energy <S> , it's just toxic waste. <S> In my experience the batteries just get hot and smolder. <S> Thus you have one more option: (1) discharge the battery as much as you can, (2) move into a completely safe space, puncture the battery, <S> let it heat up (or not), and then dispose of it like any other hazardous waste at a Household Hazardous Waste facility. <S> At that point it's not a battery, it's chemicals. <S> https://www.epa.gov/hw/household-hazardous-waste-hhw <S> Practically what I do is tape over the terminals of the old battery, and gently deposit in a battery collection facility. <S> They're the experts: it's up to them to determine where each battery goes. <S> Every collection center already has to deal with the potential for fires, and the need for safe transport. <S> Frankly the advice to use a special $100 bag to dispose of such batteries is bad advice. <S> Almost nobody will ever do it: the battery will either remain on the shelf or get chucked in the general trash, an outcome worse than the problem you set out to solve. <A> Yes discharging bare tabbed cells to less than 1v at around 1 amp ie automotive bulb is safe but the cell may get hot. <S> This procedure is best done outside and after checking with meter 12 hours later if negligible voltage found when open circuit 1 hour later then proceed with careful slash / salt water method. <S> Cell will bubble for a bit but then is moderately safe and unlikely for water saturated layers to then combust. <S> Leave 24 hours minimum to ensure maximum safety and avoid contact with contaminated salt water or battery components. <A> Lithium Ion batteries cannot be disposed of in your trash either! <S> Although the back of a trash truck is usually wetter and less likely to feed the sparks from these batteries and start a fire-they are still considered a toxic material known as household hazardous waste. <S> They pose numerous health and environmental hazards. <A> Nilesh Dattani linked to this site in a comment : Call2Recycle <S> specifically this page: <S> Damaged, Defective and Recalled Batteries <S> | Call2Recycle | United States <S> From the above-linked page, on handling damaged batteries: Call2Recycle offers a recycling service that meets the U.S. DOT packaging, handling and transportation requirements under a special permit granted by the federal government. <S> The service provides the appropriate solution based on the battery type and applicable shipping requirements. <S> We also provide services for defective and recalled batteries when, for example, a manufacturer or the Consumer Product Safety Commission have identified a performance or safety issue and the battery needs to be safely transported. <S> You can read more about our battery recycling services below. <S> ... <S> Consumers : <S> Place the battery or device in a non-flammable material such as sand or kitty litter as soon as possible. <S> Check the Consumer Product Safety Commission or manufacturer web site or visit the retailer you purchased it from to see if the battery or device has been recalled. <S> If it has, follow the instructions. <S> As an alternative, place the battery or device (one per bag) in a clear plastic bag and take it to your municipal household hazardous waste (HHW) recycling center. <S> You can also contact a local Call2Recycle drop-off site to see if it accepts damaged batteries. <S> Do not place them in the trash for any reason. <S> On the Call2Recycle store page are listed small and large "Damaged, Defective or Recalled Lithium Battery Kit[s]": <S> Call2Recycle Store | Call2Recycle | United States <S> Unfortunately, the kits are somewhat expensive, $75 and $100 respectively. <S> Fortunately, the site seems like it offers a very reasonable, and hopefully very safe, way to dispose of a damaged lithium ion battery.
| When you discard them, they must be disposed of at a household hazardous waste collection point (check with your local landfill) or battery recycling drop off location, NOT placed in the trash. If you think you have a damaged, defective or recalled battery, visit the store or for more information or help, contact Customer Service , who will discuss the appropriate solution.
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Using PIR 3.3 V 0.33 mA to drive 12V 4A valve coil I've been looking at ways to drive a 4 A, 12 V solenoid valve from a tiny, 3.3 V, 0.33 mA (max) current output from a PIR module. Whatever I try (MOSFETs, transistors etc), there never seems to be enough current or voltage from the PIR unit to drive enough current through the load. The PIR has 3 connections: 0 V, 12 V and output. Can anyone suggest a suitable MOSFET I could use, or an alternative circuit? <Q> "Normal" mosfets wont work in this application. <S> The devices you have likely tried have a typical gate-source voltage of 8 or so volts to get them to fully turn on. <S> Logic level fets are far less common and and you will likely have to order one from an online supplier. <S> IPP60R385CPXK from Mouser would work as its Vgs is only 3v. <A> example <S> 3V min PIR Analog sensor : <S> Panasonic EKMC1601111 with suggested one shot solenoid or relay driver from PIR sensor. <S> must not interfere with 3.3V supply and ground return, otherwise hysteresis effects or false switching may be seen. <A> There are opto isolators that will drive with very small currents. <S> I had an application that was 3.3Vdc @ <S> < 1mA. <S> In that case IXYS's CPC1301 was suitable. <S> It might be possible to use it or a similar opto in your application.
| What you are looking for is a so called "Logic Level" fet. Choose "Logic Level" MOSFET Driver with RdsOn or BJT as shown from One Shot and power source ESR to be < ~2% of coil DCR for good efficiency, load regulation and low heat loss.
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How to calculate the time to empty a battery I am a computer developer, excuse me if the question is basic. If a consumer takes 1600 mA, how long will it take for it to empty a 12 Ah battery? Is it as simple as 12/1.6? <Q> To a first approximation, yes that is how many hours. <S> To get a more accurate number you have to look at the cutoff voltage (limited by the minimum voltage the device being powered can operate at, or perhaps the minimum voltage that you can discharge a rechargeable battery to before damage to the battery occurs). <S> The capacity changes with the rate of discharge (usually getting worse at higher discharge rates, but at very low discharge rates the self-discharge becomes as factor). <S> Temperature also plays a role- at low temperatures capacity can be significantly less. <S> Note also that it many loads do not draw even a somewhat constant current for the entire discharge time. <S> An example of such a load would be a fixed load on a linear regulator. <S> Since the output voltage of the regulator is fixed the load current is independent of the input voltage. <S> Even so, many times the load will vary significantly as transmitters or motors start and stop or the device goes into sleep mode. <S> A resistive load draws less current as the battery voltage drops. <S> An incadescent filament has less variation in current as the voltage drops. <S> A modern switching supply will draw more current as the voltage drops to maintain constant output power, and even a bit more than that since efficiency is typically lowest at minimum input voltage. <S> So it's actually fairly complex to get a real number for the battery life. <S> One good way to confirm the numbers is to measure the battery capacity under standard datasheet conditions using a programmable digital load, and then measure the device operating time on the same battery in simulated usage. <A> The simple answer 12/1.6 is not entirely true. <S> The 12ah rating is likely the nominal capacity based on nominal 20 hour discharge rate, and includes recommended cut-off voltage. <S> The 20h discharge rate of a 12Ah battery is 0.6A, so 1.6A discharge rate is almost 3X larger. <S> Here is a battery specification example , from today's questions. <S> From the example, 3x discharge rate over the nominal rate will result in about 80% of nominal capacity. <S> So the estimate is 12/1.6 * 0.8 = 6 hours. <S> However, batteries differ in purpose and design, and may have different definitions of nominal discharge rate, so the estimation will vary depending on specifications. <A> 12aH battery means that it can provide 12amps of current for one hour. <S> That means if you have a device which is consuming 12amps, so your battery will get discharged in an hour. <S> So as you said, a consumer takes 1600mA (1.6A), it will last long for ~7hrs <A> Due to the numerous variables in amp-hour <S> [Ah] discharge time, the battery industry has standardized on a 20h discharge rate at 25'C for a constant current discharge rate. <S> As discussed in other answers and comments if the expected discharge time is less than 20h, then the Ah capacity reduces and visa versa. <S> As @Sphero mentioned, there many nonlinear load types and thus not always constant current. <S> As @Ali refers to actual battery SLA specs this is the most accurate way and one sees that other variables such as Temperature and discharge profiles effectively reduce the efficiency and capacity. <S> @gbarry states accurately that this is most likely the best case. <S> Increasing battery temperatures, although increases capacity somewhat in most battery chemistries, it also rapidly decreases life. <S> This stackup of uncertainties for capacity reduction are dominated by battery brand quality, chemistry and life cycle expectancy in total lifetime <S> Ah. <S> This life cycle can vary from 500 to over 10,000 cycles and 10 years such as batteries used in Mitsubishi e-cars. <S> For LiPo's for example you can almost double the life cumulative Ah capacity by only using it between 50 & 75% which is inconvenient for most people yet partially adopted in the charger profile for Lenovo laptop chargers.
| So for a Rule of Thumb you can use the Ah calculation to get an approximations for time from Amps then reduce by x% for discharge times less 20h and y% for lower temperatures and z% for age and condition of battery. If you have a device which needs 1amps, then your battery will last long for ~12hrs.
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Hardware sources of entropy on an FPGA I'm building an FPGA feed handler and one of the problems is I want to do Monte-Carlo simulations which require a high-quality entropy source, i.e. a pseudo-random LSR implementation won't do. I've read that one can try using metastability as a source of randomness, does anyone do this, is this a viable option? Also, since FPGAs have external clock inputs, can those be leveraged by some hardware "true RNG" solution? Do such solutions exist on the market? <Q> Metastability is really not a viable option in a modern FPGA technology because the metastable timing window is tiny - many orders of magnitude smaller than the setup/hold timing uncertainty window, which is dominated by things like clock skew, routing delays and variations with voltage and temperature. <S> Unfortunately there's a lot of confusion about metastability and this timing window is sometimes called the metastable timing window, as other (sometimes unexpected) sources of uncertainty in an output due to clock timings are loosely lumped together and incorrectly called metastability. <S> While this wider window generates uncertainty in the output, it's highly correlated with the above causes, not entropic (outside the tiny true metastable window). <S> If you need the mathematical and practical details, search the Usenet comp.arch.fpga newsgroup for "metastability" and "Peter Alfke". <S> TL/DR : look elsewhere for entropy : an avalanche diode, (i.e. a zener well above 6V, say 12V) amplified, sliced to logic levels, fed to an input pin would be one good choice. <A> One option could be to use a higher quality random number generator such as mersenne twister and then seed that from something external. <S> Other options may not be able to provide enough bandwidth for running a simulation. <S> Here is one such implementation: https://github.com/alexforencich/verilog-mersenne/blob/master/rtl/axis_mt19937.v <S> Another advantage to using this type of generator is that it's relatively easy to seed with a known value for debugging <S> - i.e. it makes it straightforward to seed the hardware exactly the same as the functional simulation to ensure you're getting the same result. <A> One way to do it is to have as many XOR gates as bits you want in the output, where each gate is the XOR of itself and its two neighbours. <S> An odd number of the XOR gates should be inverted at the output to prevent it falling into a stable pattern. <S> This generates a non-uniformly distributed set of random numbers (some bits will be 1 significantly more/less than 50% of the time). <S> By feeding this into a pseudo-random generator like a LHCA you can generate high quality random numbers. <S> There is much more information here: <S> http://forums.xilinx.com/xlnx/attachments/xlnx/EDK/27322/1/HighSpeedTrueRandomNumberGeneratorsinXilinxFPGAs.pdf <S> With their implementation you can get arbitrarily wide random numbers, at a high clock rate, that pass the diehard tests for randomness for the price of 2 LUTs and 2 registers per output bit. <S> Code for 8 bit random numbers, output in rnd: <S> (* ALLOW_COMBINATORIAL_LOOPS = <S> "TRUE", KEEP = <S> "TRUE" *) wire [7:0] ring;(* ASYNC_REG <S> = "TRUE" *) reg [7:0] ring_reg;assign ring = <S> {ring, ring[7]} <S> ^ ring <S> ^ ({ring[0], ring}>>1) <S> ^ {3{2'b01}};always <S> @(posedge clk) begin <S> ring_reg <= <S> ring; rnd <= {rnd, 1'b0} ^ <S> ring_reg <S> ^ (rnd & 8'h06) <S> ^ ({1'b0, rnd}>>1);end <S> the 8'h06 is the diagonal elements of the LHCA from the paper. <S> You might need to change the attributes to stop your synthesis tool screwing it up by 'optimising' the logic. <S> This works for me on Vivado 2016.3, and generates what appear to be uniformly distributed random numbers (I haven't run proper statistical tests).
| You can get some randomness from an unstable combinatorial circuit like a ring oscillator.
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Need some help picking a relay Recently my electric clothing dryer stopped working. Blew the thermal fuse. replaced. Later the heat still wasn't working and I found a lack of continuity on a couple of terminals on the timer, the heating element circuit could not be closed. There is just too much corrosion even after cleaning to get a good connection inside the timer. Thing is this is a very basic mechanical timer/switch that costs 75$. I feel like for the same price I could run this with an ardunio instead. More fun and way cooler. So now on to my problem. I cant seem to find a relay suitable for the load generated by the dryer. I need a relay that can handle 240 volts and from what I can see would need to handle about 4k watts? Amazon, adafruit, nothing as far as I can find. I am wondering if someone with more electrical experience can suggest an appropriate relay to use. PS Notes:Looking a bit harder I found this under furnace parts on amazon. https://www.amazon.com/dp/B019132A1I?psc=1 Thanks all for your input. I suspected the reasoning behind a very simple (now I know what to call it) centrifugal switch. The heater cannot physically be a completed circuit without engaging the blower. I realize this now looking at the order of the copper bars. I don't know that is really intentional as a "safety" feature, but fair enough, you're right I don't know for sure. I've decided to NOT do this project simply considering the risk/reward as someone pointed out. I strongly believe the risk truly isn't that great, and ill explain my reasoning below, but in any case, not a great first Arduino project. Maybe in the future if I have a spare dryer and more time to work out the safety aspects. I'm pretty sure we could, with time, make an even safer dryer that the manufacturer. So here is why I think the danger of this is pretty exaggerated. I still strongly believe that in the event the blower does not engage the thermostat and or thermal fuse would blow very quickly, stopping heat production. Lets say for whatever reason the element does get stuck on. Not ideal right? but is it truly going to start a fire? I am certain lint in the exhaust is infinitely more dangerous, and ONLY when the blower is running. The heating element itself is well sealed away from anything flammable, it is completely encased in a few layers of metal, the floor is cement and the basement wall is cement. Consider that only fresh air is pulled into the element. It is NOT full of flammable lint, the lint is no where near the element. Only when the blower is on can the heat reach anything in the exhaust section. Having said that there are a few factors here. 1. my wife wants the dryer fixed sooner than later :O and 2. I can't deny that SOME risk at least potentially exists. 3. I'm a coder with limited electrical experience. My first project should likely not involve 240v or 4k watts :P Never the less that's to you guys I did learn a few things anyway. <Q> The timer has three contacts- two of which are high current and act as a permissive chain such that the blower motor is powered preferentially. <S> some safety margin- maybe 35 or 40A. These contactors are widely available, but they do tend to create a lot of electrical noise when they switch (which can screw up your micro, but is outside of the scope of this answer). <S> Personally, I would not do this, given the risk-return ratio. <S> I would certainly not use a solid-state relay, given their propensity to fail 'on'- this thing only has to ever operate a few thousand times in its life and any contactor can do that reliably. <S> The one you linked to is not really an SSR- <S> it's something else and is unsuitable. <S> I believe this is an old (and now considered unsafe) connection method that was grandfathered for a period of time. <A> I would certainly not replace the timer switch by an arduino or anything self-built, because in dryers, the drive motor also operates the blower fan. <S> Which means, if by any mistake you make your drive motor isn't on while the heater is on, you burn your house down. <S> When this happens, you want to explain your insurance company the blower manufacturer is guilty. <A> As others have responded you could potentially put yourself and family in a dangerous and life threating situation. <S> (no electrical experiment is worth this) <S> Please purchase the correct parts and install them properly. <S> (use them!) <S> I've used the one below with success. <S> www.appliancepartspros.com
| If the NEUTRAL becomes disconnected the USER can be electrocuted, generally considered to be a bad thing. You could replace the timer with an electronic device controlling a contactor with at least three poles, rated for the maximum current the heater and motor draws together, plus I also note that your schematic shows the neutral connected to the chassis. I know this idea might sound good to you and sparks your technical interest on the surface but believe me this is not the place to implement an Arduino work around. There are several great online appliance parts suppliers just waiting to solve your problem.
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Best way to hand-solder this TQFP (SMD) component So I'm working on a small project right now that requires a ATmega32U4RC . I'm trying to test/prototype with it, but it's a surface-mount component, so I can't just place it in my breadboard. So I found this , which is perfect for what I want. The problem is that I haven't soldered in quite some time. I've also never soldered a component with such tiny leads. Needless to say, I ended up with this: Yea, I know it's bad :(The solder-wick didn't really seem to want to do it's job (anyone know why this is?), so I couldn't really clean it up. Are there any tips for doing this (tools that might help me, etc.)? Is there an easier way to do this than hand-soldering? <Q> As you hit each one, it'll leave just enough solder on the pin to attach it. <S> A video will do a lot to help explain here: https://www.youtube.com/watch?v=wUyetZ5RtPs <S> There's a few things that help make this work: <S> Apply flux - lots of it - before you start soldering. <S> Yes, I know the solder you're using probably has a flux core, but that's not enough here. <S> (If you use paste flux, the paste can also help keep the part in place.) <S> Using the right kind of tip helps a lot. <S> A small tip is not what you want here! <S> I use a Hakko "hoof" tip (T18-C2); there's probably equivalents for most other good irons. <A> Get better (fresher?) <S> solder wick. <S> Digikey carries it, you can have it next-day. <S> It will clean that up very nicely, in seconds. <S> You may have to add more solder then suck it away. <S> Looks like you are already using 63/37 <S> but it behaves a bit more nicely than lead-free solders. <S> Before soldering, make sure the board is clean (your ENIG board should be fine). <S> For a prototype I like to use a flux pen (like a Magic Marker/Sharpie <S> but it contains liquid flux) on the contacts- <S> some use liquid flux from a bottle. <A> The QFP-44 with 0.8mm lead spacing is far from the worst case. <S> There is no way to escape micro-miniturization of electronics, 0603/0402/0201 components, and 0.4mm - 0.5mm QFN packages. <S> Even BGA is in line next for DIY folks. <S> First, get a good iron. <S> It is always worth to have a good tool even if you have this for hobby, to avoid frustration from bad solder joints and non-working projects. <S> JBC is probably the best, next is/was Weller, but they looks like having business issues. <S> I use mid-end Weller WX1011 with WXMP-MS iron (40W), with 0.1mm cone tip (RT1-NW) and several other interchangeable cartridges (knife shape). <S> Tip temperature - 380C. <S> Next, get a spool of regular 63/37 solder wire with rosin core, I use 0.010", smallest available. <S> And 0.032" for bigger jobs. <S> Get a rosin-based mildly-activated flux, Kester flux-pen 186. <S> Get a cheap stereo microscope, 10X-30X from eBay, they can be in $120/$170 range, to check for bad joints and solder bridges and parts alignment. <S> Get a cheap ($30) <S> Chinese hot air pen from eBay <S> , it helps a lot in finishing solder joints on SMT components. <S> Soldering rule #1 - Use More Flux. <S> Soldering rule #2 - Use More Flux. <S> Don't use "no-clean" stuff or ROHS solders. <S> "Drag soldering" does not work well for pin spacing under 1mm. <S> Solder wick is your friend too.
| Basically, instead of trying to solder each pin individually, you just get a bead of solder on the tip and roll it across the pins. Learn to use "drag soldering" technique.
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Digital gain control with a digital pot? I want to make an audio amplifier with digital volume control. I figure that by adjusting the feed back path that I won't change the input impedance and won't affect the frequency response. (other than the gain bandwidth product) Does the following setup look reasonable? Would a digital pot be good for this application? Are there any drawbacks of digital pots that i'm not aware of that would make them undesirable for this application? simulate this circuit – Schematic created using CircuitLab <Q> I can see two things that are not optimal in that design. <S> For the discussion I will use a random digital potentiometer I found from this article: The DS1881 . <S> Most are similar in construction. <S> Wiper resistance <S> A digital potentiometer has a pretty high wiper resistance , especially compared to a mechanical potentiometer. <S> Not only is it high, possibly in the order of hundreds of ohms, but it is also not very stable or well-defined. <S> The DS1881 lists it as 160 Ω <S> typical , but up to 250 Ω maximum . <S> It can vary with the signal voltage and such. <S> It is not a problem if it is in series with a high impedance input, for example the input of an op-amp. <S> Then there is negligible current through it, and thus, a negligible voltage drop. <S> In your case, the wiper is in parallel with a section of the potentiometer: simulate this circuit – <S> Schematic created using CircuitLab <S> Since some amount of current will flow here, it will affect the signal. <S> How bad? <S> Not much, it is somewhat swamped by the 10 kΩ series resistance and the rest of the potentiometer section, but that leads to the next problem: Absolute resistance variation <S> A digital potentiometer has a well-defined and trimmed ratio , but its absolute resistance can vary a lot: <S> Note the ±20% resistance tolerance, and the 750 ppm/°C temperature coefficient. <S> Since you want to use it as a rheostat, you are hit by this. <S> If it matters, that is up to you. <S> It will really only make the gain fluctuate a bit by temperature and so on. <S> A digital potentiometer should really be used as a voltage divider, so you can take advantage of the close matching between the steps. <S> The article I linked to, from Maxim, has a few audio circuits you may want to take a look at, with pros and cons. <A> Even once those are resolved, the gain vs. code characteristic of an amp with a digipot controlling its gain is quite annoying (i.e. nonlinear in not-good ways). <S> A better solution would be to use a dedicated Voltage Controlled Amplifier IC such as the THAT2180 driven from a cheap-ish voltage output DAC. <S> This gives you linear-in-dB control with much better noise performance (no zipper noise!) and THD good enough for the professionals. <A> I'd just go and use the pot on the in- or output as voltage divider, but this should work, too. <S> Using a digital poti to change the amplitude behaviour of analog systems is exactly what they're designed for. <S> By the way, there's also PGA (programmable gain amplifiers); maybe these are of interest to you. <A> Digital potentiometers can have up to 50 - 100pF of parasitic capacitance to ground, which may alter frequency characteristics of the amplifier when used in the feedback net. <S> And the shape will change with trimpot value. <S> If used as input divider with a fixed-gain amp, you will have more unnecessary noise. <S> That's why they have dedicated gain control ICs in the audio application space. <S> But for a low-end hobby audio it is ok to use them, <S> provided that the amplifier remains stable.
| Digital potentiometers are indeed a suboptimal solution for changing amplifier gains -- the parasitics of a RDAC/digipot are much worse than that of a mechanical potentiometer, and issues like zipper noise can ruin your day as well.
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How to improve atmega328p board for solenoids control I'm designing a board based on atmega328p microcontroller in order to control some solenoids valves (2 proportional solenoids and 9 ON/OFF solenoids valves). This is the valve datasheet .These are the solenoid details: V = 12Vcc, R = 3.7 Ohm, I = 1.80A This is my first PCB schematics and my first board .I power the board with 12V, 40Ah.I use Eagle and I never did something like this before so the board has many errors. In fact, I tested my first prototype and these are the problems: When I try to control proportional solenoids (VSX, VDX outputs), the board crashes randomly after few seconds and get stuck. I need to power off and on the board to make it works again. The datasheet says that the valve should draw 1.8A when powered at 12VDC, but I measured 3A when PWM is 100%. This is very strange! In order to solve the problem, using the community suggestions , I added these modifications: Added a big 2200uF 35V capacitor at the main power connector Added a filter cap between VCC (7) and GND ( 8 ) and AREF (21) and GND (22) and AVCC (20) and GND (22) on the atmega328p I cut the trace on pin 22 and I directly connected it to the L7805CV GND Added a diode across each valves pins. By doing this, the board can work well for 1 or 2 minutes, but then it crashes again. Now, I'm going to design again my board, so I would like to ask you how I can fix my errors and improve my design in order to make my board works correctly. IMPORTANT : After few experiments, I notices that if I put in series, between the mosfet output and the valve, a 20R 10W resistor, the board works OK.The problem is that the resistor becomes hot in no time.The board works fine also if I put a voltmeter in series between the mosfet output and the solenoid to measure the current: in this case, the board works fine for several time. This is very strange for me!! I was thinking to: use optoisolator (like ILD213T) to control the MOSFETs use an isolate 5V regulator (like NME0505SC or AM1S-0505SZ) in order to isolate the atmega328p from the board adding filter caps on main power and on atmega328p What do you think about this? Can you give me some suggestions, please? How can I limit the current to 1.8A when the PWM is at 100%? I know that the board design is not correct, how can I improve it? Should I use the second layer for GND traces and the upper layer for power traces only? Please, help me! EDIT : I was thinking to use LM25011 with this schematics . I did this schematics by using the online TI calculator. Can you tell me if this design could work for my board? I set the Rsense to limit current up to 1.8A at 12VDC. <Q> I faced a similar problem when using a relay circuit to actuate pistons. <S> The back EMF generated due to the actuation of the piston would short the control circuit thus resetting the micro-controller. <S> I would recommend adding a Optical Isolation feature to the circuit. <S> Opto-Isolators such as ILD1, ILD2, ILD5, ILQ1, ILQ2, ILQ5 are cheap,small & come in DIP Packages . <S> Signal information, including a DC level, can be transmitted by the drive while maintaining a high degree of electrical isolation between input and output. <S> The ILD1, ILD2, ILD5, ILQ1, ILQ2, ILQ5 are especiallydesigned for driving medium-speed logic and can be usedto eliminate troublesome ground loop and noise problems. <S> These Isolators will protect your microcontroller & control circuit from the back emf generated by the actuation. <S> These couplers can be used to replace relays andtransformers in many digital interface applications such asCTR modulation. <A> Before re-doing the PCB design again, spend more time on analyzing all the problems to ; <S> Define the problem better, then describe solution with a spec ( very important) Such as root cause of heat dissipation issues. <S> Performance issues of PWM flow control vs pressure there is a difference in source impedance using SPST vs SPDT PWM <S> SPST is an Open Drain or Collector BJT switch with a clamp diode vs <S> SPDT action of a complementary driver where source Z is always low except during transient or crossover with deadtime during which there is a transient for V=L*di/dt Having designed hundreds of boards in my career <S> , I know it is possible to do what you are attempting designing a PCB before understanding all the issues, and you WILL learn from the process that there are better ways. <S> These ways require much testing and scope analysis using all the principles you have been taught. <S> This is my simple advice above test parts of the circuit until they meet all your criteria; <S> e.g. heat rise, performance, EMI, cost and time then try to measure this performance in terms of a specification that you can later Test and Verify ( or instruct someone else to do the same) <S> When you have good specs ( that meet customer expectations or yourself) and pass them, with your own devised Design Validation Tests (DVT) or compare with datasheet you end up with a perfect example design good luck and remember these principles. <S> compare your Linear Solenoid waveform under load pressure and see if there is room for improvement. <S> to do this properly , you need a scope, and a variable lab supply and good shielded pair cable and a selection ferrite beads for single and wire pairs. <S> Remember this: the solution is much easier after you understand all the problems by looking for them. <A> I don't think you need any special current regulators. <S> At least not to start with. <S> I am concerned that you have written -12V and Gnd together connected to all the FET sources which are switching the solenoid coils. <S> Get a multimeter and measure the open circuit DC resistance of the solenoid coils. <S> Then add a series resistor to the coil to limit the current to 1.8A with 12V. <S> Pay careful attention to peak power conditions as well as average power. <S> This will tell you what sort of power supply you need. <S> I suspect your power supply is not meeting your peak power requirements. <S> Edit: OP is using a car battery which is ample power in this case. <S> Design it with a little bit of headroom with respect to PWM control so that you never need to go to 100% (say 95% max duty) <S> Hope this helps.
| The 12V supply for the solenoids can be separate from the PSU for the controller.
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How can I tell if my building plumbing provides an adequate ground? I work in an old building with two-wire wiring, i.e. no separate ground. Is there some minimum-equipment-purchase way to tell if a wire soldered to a building plumbing pipe is an adequate ground? I have a standard hand-held multi-meter. <Q> http://homeguides.sfgate.com/plumbing-system-electrical-grounding-94939.html <S> This author above suggests plumbing can supplement and earth ground rod, but not be a primary connection. <S> Ask relevant city authorities for reference to building codes for commercial ground methods or consult an electrician. <S> Certainly deep water lines are better than no ground for equipment that needs to protect you from stray leakage from dust and humidity on metal shell electrical tools. <S> I might measure the AC voltage from the plumbing to Neutral and see how much voltage drop there is. <S> ( must be < 5% line) ) ... <S> this is grounded outside at the transformer but current can cause a voltage drop. <A> Adequate is a bit of a vague descriptor <S> but I'll assume that you are trying to make a GFIC work. <S> You don't even need a multi-meter - a voltage/continuity tester will do. <S> Meter from Neutral to ground (the box). <S> If you have continuity there, meter from hot to ground (still the box) and see if you get 120v. <S> If you have good voltage and continuity to ground on the neutral the GFIC will hold. <S> You should/can also check your main water line to ensure that there is a jumper wire in place around any water meter. <S> You may have good ground without the jumper <S> but it isn't a sure thing. <S> Hope that helps. <S> Chris <A> That one doesn't depend on the age of the building because your water supplier may have decided to replace the plumbing outside the building by plastic tubes. <S> Tools you need: documentation. <S> When in doubt, you have to install proper earthing outside the building yourself.
| You first have to check whether the plumbing has sufficient earth connection at all. Building codes in your area determine the ultimate answer.
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Should I use a current limiter on a battery powered system I'm designing a small battery powered system, actually I'm battery powering a module which already has all the switching regulators it needs. This provides me with a wide input range of 5-10V. I was thinking of connecting the batteries directly to the module instead of using a regulator because I don't want to pay an efficiency penalty, especially in low power mode. I was worried about inrush current though when the batteries are first turned on. I know the internal resistance of the batteries should limit it some but should I add some kind of current limiting or soft-start circuit to protect the batteries from damage or overheating? That'd be easy to do with a regulator, not sure what I'd use without one. Maybe a high side switch with some time based control. Should I even be concerned about this? Target batteries are LiOn that won't be recharged in the device itself. <Q> You haven't said anything about the current draw of your system, how much inrush you expect, nor the type of battery, so the question can't be answered directly. <S> However, it is unlikely that inrush current from occasionally turning something on is going to matter to the battery or the power supply input circuitry. <S> If you are just powering a few 10s of mA of ordinary electronics, then you are overthinking this. <A> I do not think that there will be any problem when connection your batteries. <S> Even an inrush current of 20 or 30 A does not hurt given the time required to charge your input circuit. <S> Overheating of the battery will therefore also not take place because therefore the energy loss in the battery is far to low. <A> Just beware that high Capacitance or low ESR loads may cause extra inrush currents. <S> Add a curent shunt R to monitor current if in doubt on ground side with 50mV=5A or 10 mOhm shunt and monitor I with a DMM. <S> ( or higher R to suit desired range) <S> If you buy something with No Specs and No idea how it works, How can we tell you how it will perform? <S> e.g. Fusing the input may be too slow if your output has a very low ESR load. <S> For example, this "looks" like a good variable bench supply using Series LiPo cells that are pre-balanced, but we can only guess. <S> THe good news is it's incredibly cheap. <S> $6 <S> so you can reverse engineer it, measure hot spots, and look up datasheet for obvious tips on current limits. <S> e.g. <S> (For step-up To prevent damage to the switch, its current must be externally limited to 6.0A.) <S> Input voltage :3-34V Output voltage: continuously adjustable (4-35V) <S> Output Current: 2.5A (MAX) <S> ( IMHO... not at all voltages) <S> Module size: 49 (mm) x26 (mm) <S> Input mode: <S> IN + input positive level, IN-input negative <S> I would try to design a precision external current limiter using a 50mV shunt drop R and Small RdsON MOSFET switch with comparator. <S> or do not try to run non-linear loads such as high current 2A motors or 3000 Farad supercaps or heavy pulsed loads without active current limit protection. <S> Usually the part will get hot and give some warning of this before damage unless very high surge into a low ESR load.. <S> You can evaluate the IC performance doing load drop regulation and source voltage regulation tests and efficiency vs Vin-out drop. <S> Hint buy 1 spare.
| Be careful how it is used with nonlinear loads. I think it is better to use in step-down mode with Vbat greater than output.
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interfacing buttons with python/c++ As a preface: I have 0 experience in interfacing with peripherals other than keyboard and mouse, but am willing to learn, just don't know where to start. I would like to write a software in python (if C/C++ would be needed or is the easier way to go I can interface that) running on a PC, that can be manipulated by two external buttons (it will be an experimentel setup for small children, so we need big easily identifiable buttons, that are easy to push, so the keyboard/mouse is out of the question). The easiest would be to buy two USB buttons, that can be easily accessed, but I've been told these do not exist, and that I would have to build an interface myself (microcontrollers and stuff). If this is not true that would be great, if it is, I would like to invest my time into something that is not horribly hard and long to learn to use, and both the knowledge and equipment easily reusable if other issues like this come up. What would you recommend? Thanks in advance:) <Q> People who told you they don't exist sorta lied. <S> What you will need: The cheapest mouse you can find. <S> Really, the cheapest. <S> Your Big Easy <S> To Use Whatever You Need buttons. <S> Grab an SPDT type. <S> Some fancy tripolar cable to connect them, grab the size and color you like A soldering iron, and somebody able to use it <S> First of all, crack open your mouse. <S> You will see something like this: By Job at English Wikipedia - Transferred from <S> en.wikipedia to Commons by Ansumang using CommonsHelper., GFDL , Link See those tiny little guys on the right, with red little thingies on top? <S> These are microswitches, something like this: By Benjamin D. Esham / Wikimedia Commons, CC BY-SA 3.0 us , <S> Link without the metal thing on top. <S> The red thingies correspond to the black piece of plastic on the top left of the switch case. <S> You now need to identify the terminals of the switch. <S> Odds are that they are labeled as C or COM, NO and NC, as COMmon, Normally Opened, Normally Closed. <S> Now you need to start soldering. <S> Your Buttons will have three similarly labeled terminals: you need to remove the switches and solder your cables where the switches used to be, then solder the cables to the terminals on your buttons. <S> Really, it's easier said than done. <S> Finally, enjoy making your rig nice and all so that a young human fancies using it and is not endangered by anything. <S> And... Bingo. <S> You're done. <S> Now, this is not really useful to you, in the sense that you do not learn much. <S> But this solution has many pros: it is dirty cheap, easy for you to interface to, no worries about safety for the youngsters, great flexibility, you name it. <A> Have you looked into Arduino? <S> The buttons can be used to light up LEDs, or move a servo motor, etc. <S> https://www.arduino.cc/ <A> This can be done multiple different ways. <S> You could simply hook an Arduino board with a few wires and resistors to the push buttons so you can read back which button was pushed and when it was pushed, and so on. <S> This interface would use C/C++. <S> In terms of programming, this is not hard at all <S> but I do not really know what you are trying to do with the pushbuttons so I cannot help you there. <S> In short, the keyboard of your computer is reading which pushbutton has been pressed by simply assigning each pushbuttons an ASCII character or an address that would tell you which character has been pressed. <S> The mouse performs the same way. <S> The program will deal with what has been pressed.
| You can look into many tutorials, write the code in C++ and have the kids press the buttons.
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Delay is calculated wrong in embedded C I am new to embedded C and been struggling with it for a while now.The desired output for this project is:When SW1 is not pressed, the blue LED must be on.When SW1 is pressed the blue LED should turn on and off every 100ms.I wrote this code but it doesn't seem to be working properly.I tried it on the simulator and the LED toggles but the delay is more than 100ms, it's like a second.And on the real board, I get random results, sometimes it doesn't turn off and sometimes it changes color to purple.Why isn't this code behaving as it should? Why am I getting random results? Code: pastebin.com/ShE9rDCG // BranchingFunctionsDelays.c Lab 6// Runs on LM4F120/TM4C123// Use simple programming structures in C to// toggle an LED while a button is pressed and// turn the LED on when the button is released. // This lab will use the hardware already built into the LaunchPad.// Daniel Valvano, Jonathan Valvano// January 15, 2016// built-in connection: PF0 connected to negative logic momentary switch, SW2// built-in connection: PF1 connected to red LED// built-in connection: PF2 connected to blue LED// built-in connection: PF3 connected to green LED// built-in connection: PF4 connected to negative logic momentary switch, SW1#include "TExaS.h"#define GPIO_PORTF_DATA_R (*((volatile unsigned long *)0x400253FC))#define GPIO_PORTF_DIR_R (*((volatile unsigned long *)0x40025400))#define GPIO_PORTF_AFSEL_R (*((volatile unsigned long *)0x40025420))#define GPIO_PORTF_PUR_R (*((volatile unsigned long *)0x40025510))#define GPIO_PORTF_DEN_R (*((volatile unsigned long *)0x4002551C))#define GPIO_PORTF_AMSEL_R (*((volatile unsigned long *)0x40025528))#define GPIO_PORTF_PCTL_R (*((volatile unsigned long *)0x4002552C))#define SYSCTL_RCGC2_R (*((volatile unsigned long *)0x400FE108))#define SYSCTL_RCGC2_GPIOF 0x00000020 // port F Clock Gating Control// basic functions defined at end of startup.svoid DisableInterrupts(void); // Disable interruptsvoid EnableInterrupts(void); // Enable interruptsvoid portF_init(void);void delay100ms(unsigned long time);int main(void){ unsigned long volatile delay; // activate grader and set system clock to 80 MHz TExaS_Init(SW_PIN_PF4, LED_PIN_PF2); portF_init(); EnableInterrupts(); // set PF2 GPIO_PORTF_DATA_R |= 0x04; while(1) { delay100ms(1); // if switch PF4 is pressed and LED is ON (00000101) if( GPIO_PORTF_DATA_R == 0x05) { // turn LED OFF (clear bit) GPIO_PORTF_DATA_R &= ~0x04; } // if switch PF4 is pressed and LED is OFF (00000001) else if (GPIO_PORTF_DATA_R == 0x01) { // set PF2 - turn LED ON GPIO_PORTF_DATA_R |= 0x04; } else { // set PF2 GPIO_PORTF_DATA_R |= 0x04; } }}void portF_init(void){ volatile unsigned long delay; SYSCTL_RCGC2_R |= 0x00000020; // 1) F clock delay = SYSCTL_RCGC2_R; // delay GPIO_PORTF_AMSEL_R = 0x00; // 3) disable analog function GPIO_PORTF_PCTL_R = 0x00000000; // 4) GPIO clear bit PCTL GPIO_PORTF_DIR_R = 0x04; // 5) PF4 input, PF2 output GPIO_PORTF_AFSEL_R = 0x00; // 6) no alternate function GPIO_PORTF_PUR_R = 0x01; // disble pull-up resistor GPIO_PORTF_DEN_R = 0x14; // 7) enable digital pins PF4 & PF2 }void delay100ms(unsigned long time){ unsigned long i; while(time > 0) { i = 1333333; // this number means 100ms while(i > 0) { i = i - 1; } time = time - 1; // decrements every 100 ms }} <Q> The delay function is broken. <S> First of all, you should avoid writing delays with busy-loops. <S> Such delays are unstable, unreliable and non-portable. <S> They create a tight coupling to the system clock frequency. <S> Also in case of sleep modes, they make the CPU consume current needlessly. <S> Always use on-chip hardware timers instead. <S> Now if you insist on writing such a delay loop, you must declare the actual loop iterator volatile . <S> Otherwise the compiler might remove the whole loop, since it contains no side effects. <S> It isn't enough to make the variable in the caller volatile , since the function is using a local copy of the passed variable. <S> The variable inside the function must be volatile . <S> Another issue here is the lack of switch de-bouncing. <S> You can only get away without it if you have a hardware RC filter on the switch. <S> Also, does the switch have an external pull resistor or does it assume that you enable one internally? <S> You need to share the schematics. <A> Double check which switch you're intending to use and which bit that corresponds to. <S> The comments near the top suggest that SW2 is associated with bit 0 and SW1 is associated with bit 4. <S> In portF_init() you set <S> GPIO_PORTF_DEN_R = 0x14 <S> ; , which suggests that you intend to use SW1. <S> But then in the while loop in main <S> () <S> you test if( GPIO_PORTF_DATA_R == 0x05) , which reads SW2. <S> Regarding the delay issue, I wouldn't expect the simulator to exhibit the same timing as the real hardware system. <S> It does not surprise me that the simulator runs slower than the hardware. <A> I believe you need to enable the internal pull-up resistor for PF4 (SW1). <S> But I suspect the following line does not do so. <S> GPIO_PORTF_PUR_R = <S> 0x01; // <S> disble pull-up resistor <S> Should it be this instead? <S> GPIO_PORTF_PUR_R = 0x10; // enable PF4 pull-up resistor Also, the following conditional relies on status of every bit of Port F when you are really only interested in two of the eight bits. <S> (BTW, it's not clear why you expect PF0 to be set when you didn't configure/enable PF0.) <S> // <S> if switch PF4 is pressed and LED is ON (00000101) <S> if( GPIO_PORTF_DATA_R == 0x05) <S> You should rewrite this <S> so it's not dependent on the value of the bits you don't care about. <S> Use the bitwise-AND operator to isolate the bits of interest. <S> And I think it would be nice to use helper functions that make it obvious what your intention is. <S> Maybe like this: <S> #define BLUE_LED_BIT <S> (0x04)#define SW1_BIT ( <S> 0x10)bool IsSwitch1Pressed() <S> { // <S> Bit is clear when switch is pressed. <S> return ((GPIO_PORTF_DATA_R & SW1_BIT) = <S> = <S> 0);}bool IsBlueLedOn <S> (){ // <S> Bit is set when LED is on. <S> return ((GPIO_PORTF_DATA_R & BLUE_LED_BIT) ! <S> = <S> 0);}int <S> main(void){ ... <S> while(1) { delay100ms(1); if( IsSwitch1Pressed() && IsBlueLedOn() ) <S> { // <S> Turn LED off (clear bit) <S> GPIO_PORTF_DATA_R &= <S> ~BLUE_LED_BIT ; } <S> else { // Turn LED on GPIO_PORTF_DATA_R |= BLUE_LED_BIT ; } }} <A> You are changing incorrect port register bits. <S> Port F register should be 0x10 (0001.0000) and not 0x05 (0000.0101) or 0x01 (0000.0001), if you want to change PF4. <S> To change PF2 port register value will be 0x02 (0000.0010), register value in your code is 0x04 (0000.0100). <S> Adding both together will make register value to be 0x12 (0001.0010). <S> This is for the case when SW1 (PF4) is pressed and PF2 (Blue LED <S> ) HIGH. <S> Similarly, register will be 0x10 (0001.0000) for the case when SW1 (PF4) is pressed and PF2 <S> (Blue LED) LOW. <S> Check datasheet page 660 and page 662 for more information about how these register bits are changed for the above problem. <S> For delay issue : The delay is around 76 ms and is incorrect. <S> You need to use a precise timer for the same. <S> Probably, ask your professor about why the function is written like that. <S> Depending upon the IDE you are using this delay calculation might vary. <S> Same will be observable when you program the board with the same code. <S> It is also suggested that you separate reading input and writing output, if you are a beginner. <S> Implement both separately and then try to club together.
| Your code might be behaving inappropriately as incorrect bits are being manipulated.
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What is the purpose of the ground in this diagram? I'm trying to read a throttle position sensor with my Arduino Mega. This is the schematic for a normal TPS. It's a basic potentiometer. http://autolabscopediagnostics.com/images/tpscircuit.gif However, this particular TPS only has 2 pins coming out. There's no ground, there's just the 5 volt reference and the signal return. I have one pin hooked up to the 5Volts on my Arduino and the other hooked up to an analog pin, which I am reading with analogRead. I'm getting strange readings at different positions. At position zero, it jumps around at around 300. As soon as I turn the sensor at all, it automatically jumps to 1023 and stays there. What is the point of the ground in that diagram? What am I missing here, and how does that tie in to the odd TPS readings? <Q> A proper 3-terminal TPS is a potentiometer. <S> It produces some proportional voltage between ground and the reference voltage at the top (5V or whatever) <S> A linear pot will output a voltage proportional to the distance between ground and the top. <S> But if you have only two terminals, then you only have a variable resistance (or you could call it a rheostat). <S> You must know the TPS resistance in order to select an appropriate value for a pull-down resistor. <A> It could be a 3 pin TPS or more likely a mistaken logic diagram for a 2 pin TPS unless case grounded. <S> My educated guess is a 2 pin TPS will have a fixed load R to ground so that a bridge voltage can be measured relative to some stable Vref. <S> But there is NO STD. <S> I believe loss of contact is zero throttle and internal short is full throttle !! <S> Take a look at some defective TPS waveforms here and notice the 0 throttle voltage above ground. <S> http://autolabscopediagnostics.com/tps.html <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Some metal component of the case may have a screw hole with an ground marking nearby. <S> Simple resistance measurements will tell you if this is true : if there is a fixed resistance from the "+V" terminal to this ground point, and a variable resistance between the wiper and gnd, then you just need a wire from the Gnd terminal to your Arduino's 0V.
| You could kludge a way to read your rheostat by first measuring the total resistance of the rheostat, and then selecting a "pull-down" resistor to "work against" the variable resistance in your rheostat TPS. As @duskwuff comments, it is possible that it is a 3-terminal part, with the three terminals being the 2 pins and the case.
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How deep underwater can Bluetooth reliably transmit? I know that Bluetooth operating at the standard 2.4GHz has a hard time penetrating water (I guess the water "absorbs" the signal). However, at what depth does this really start? In my case I am thinking about a device that operates within 1 meter of the surface and would transmit to someone nearby on land. Is a reliable pairing possible to attain and maintain? If so, is it doable with standard transmitters? That is, would there need to be any special signal amplifiers or the like? Furthermore, if this is possible, is it only possible with Bluetooth classic or would BLE fare similarly? <Q> The question has been studied extensively. <S> According to this lecture , at 2.4GHz (microwave frequency as well) the 1/e (63% loss) occurs in 1.4cm. <S> You probably can do better with 9600 <S> baud <S> UART. <S> :-( <A> Water absorbs high frequencies like 2.4GHz very well. <S> That is how your microwave oven works so fast. <S> Your chances of using ANY protocol at 2.4GHz (not just BlueTooth) are essentially zero. <S> That is also why submarines use sonar and not radar underwater. <S> Extremely low frequencies (10KHz and lower) are used to communicate to submarines when submerged. <A> If you absolutely have to use BT/BLE (or other 2.4 GHz tech) <S> and your depth is limited (say 1-2 m max) <S> then I'd suggest using a floating antenna. <S> This is the only way a submerged device may be able to maintain a stable BT connection with another system sitting out of water. <S> I'd not bet on any communication with a submerged antenna except for a military budget project.
| Even if you're not using 2.4 GHz, absorption would be too highfor any practical communication if transmission has to go through water.
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What does "bank" mean in these context? I am reading some specs about Intel CPU. The word bank appears in the following contexts. I am not a native English speaker. Please help me understand its meaning. Thanks. From the Intel Manual Vol. 3B Chapter 15.1 : The Pentium 4, Intel Xeon, Intel Atom, and P6 family processors implement a machine-check architecture that provides a mechanism for detecting and reporting hardware (machine) errors, such as: system bus errors, ECC errors, parity errors, cache errors, and TLB errors. It consists of a set of model-specific registers (MSRs) that are used to set up machine checking and additional banks of MSRs used for recording errors that are detected. And... 15.3.1.1 IA32_MCG_CAP MSR : MCG_CMCI_P (Corrected MC error counting/signaling extension present) flag, bit 10 — Indicates (when set) that extended state and associated MSRs necessary to support the reporting of an interrupt on a corrected MC error event and/or count threshold of corrected MC errors, is present. When this bit is set, it does not imply this feature is supported across all banks . Software should check the availability of the necessary logic on a bank by bank basis when using this signaling capability (i.e. bit 30 settable in individual IA32_MCi_CTL2 register). I understand the bank in first quotation just means a bunch of MSR (Model Specific Registers). But what does the bank in the second quotation mean? <Q> bank, n. " <S> a set or series of similar things, especially electrical or electronic devices, grouped together in rows." <S> Examples: bank of switches, bank of lights. <S> Basically, 'bank' means 'group' or 'set'. <S> When it comes to things like memory and registers, there is usually some addressing related connotation, perhaps having to switch between different register banks, or having different register banks for different modes of operation, or having different banks of RAM within a chip, etc. <A> It gives you a big clue in that last sentence. <S> If you search for that particular bit 30, and then back up a bit, you will find some useful information here: 15.3.2 <S> Error-Reporting Register Banks <S> Each error-reporting register bank can contain the IA32_MCi_CTL, IA32_MCi_STATUS, IA32_MCi_ADDR, and IA32_MCi_MISC MSRs. <S> The number of reporting banks is indicated by bits [7:0] of IA32_MCG_CAP MSR (address 0179H). <S> The first error-reporting register (IA32_M C0_CTL) always starts at address 400H. <S> See Appendix B, “Model-Specific Registers (MSRs),” for addresses of the error- reporting registers in the Pentium 4 and Intel Xeon processors; and for addresses of <S> the error-reporting registers P6 family processors. <S> Which amounts to about what you already guessed, earlier. <S> There are groups of error-reporting MSRs and some systems may have more or fewer of them depending on the functional units present. <S> There is one of these sets for each distinct hardware unit (sometimes just one set of registers may represent a group of hardware units.) <A> A bank in CPU architecture is a range of registers in some addressable space <S> (memory mapped, <S> I/O, or configuration), a set of registers, together with one register that serves as "bank selector" . <S> Changing the selector allows to address a different set of registers at the same address space. <S> This mechanism allows to "stack up" several sets/pages of registers in the same address space, saving a lot of linear addressing space.
| That pretty much tells you that 'bank' in this context means "error-reporting register bank," more specifically.
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Drive a MOSFET via BJT? My intention is to drive a MOSFET (say, a IRF840 or some logic-level) from a low-current (<10mA) SoC output pin. Using appropriate resistors from gate to pin and ground, this works... but the gate voltage change could be faster. So I considered putting a NPN (to VCC) & PNP (to GND) BJT between the pin and FET simulate this circuit – Schematic created using CircuitLab and at that point I thought I'd ask if that setup has a name and/or any known problems to avoid (feedback, etc). <Q> This is called "push pull output configuration" and provides unity voltage gain but significantly low output impedance and high output current. <S> This makes push/pull output ideal for driving high capacitive loads such as power MOSFET gates at relatively high frequencies when the driver (IC or MCU) cannot source enough current --widely used in switching converters <S> (Personally, I'm using this config in my 2-sw Forward and 2-sw Flyback converter designs) . <S> No need to place two separate resistors, btw: simulate this circuit – Schematic created using CircuitLab Note that VCC and Vi should be at the same level for proper on/off switching. <S> If, for example, VCC = 12V and Vi = 5V, this circuit cannot perform its job. <S> PS: IRF840 is not a logic-level gate MOSFET, IR L 840 is. <A> If you want to keep your high voltage FET, then the following is a common way of the doing the job, and can be driven from 3v/5v logic. <S> Q1 pulls the signal low through D1. <S> Even with the drop through D1, it gets to more or less as low an output voltage as an emitter follower. <S> Q2 amplifies the current through R1, so drives the FET much harder than a simple pull up resistor would. <S> Trim R1 to be as big as possible (lowest quiescent current) consistent with driving the FET hard enough. <S> Using a darlington for Q2 may help here. <S> If the quiescent current in R1 is an issue, then you'd need to use a different configuration. <S> You can buy a CMOS driver that drives a FET gate with several amps, for less than 1£$Euro. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Therefore I don't think the complicated NPN and PNP design is required.
| MOSFETS are voltage triggered and therefore they are perfect as they are with your microcontroler or SOC, as long as you use a pull-down resistor to pull the input to ground, you should be fine.
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How much power am I consuming from using an AC-DC converter? I am duped on how power supplies work as I am not an electrical engineer but rather a hobbyist or a DIYer. I am looking for cooling solutions and discovered the inverter technology. If I understood it correctly, the compressor is run by DC, running at variable speed for whatever the system needs. If I want to assemble a phase-change system, using a DC compressor, and I have an AC power source (residential) and convert it using an AC-DC converter, I do not know how much electricity I am consuming. This is what I am thinking, and these are just examples for simplification: suppose that I use a switched-mode power supply, rated at 10 A, and connect it to 220 V AC - the power consumed is 2,200 W. If I connect it to a load, DC, for example it outputs 12 V, and the load uses 5 A, so that is equal to 60W... My question is, how much electricity am I consuming, and consequently have to pay for? Is it 2,200W or 60W? <Q> Something between 60W + efficiency of power supply, likely between 10~20% for a decent switching supply. <S> So 66~72 Watts. <A> I expect that the power supply is rated to deliver up to 10 Amps at 12 volts = <S> 120 Watts. <S> If you only draw 5 amps, the supply is only delivering 60 Watts, and will only draw 60 watts, plus a little for inefficiency, maybe a total of 70 watts, from the AC power source. <A> My question is, how much electricity I am consuming, and consequently have to pay? <S> Is it 2,200W or 60W? <S> Full load power will be somewhat less than 2.2 kW, probably about 80% of that meaning that about 440 watts are wasted in the power supply. <S> Worst case scenario is that you will be paying for 440 watts even if the power supply feeds no output power at all. <S> It's always best to target a design of power supply that can cope with maximum demand of the load and not go overboard <S> or you end up paying for wasted heat. <S> For instance, a 100 watt power supply might waste 20 watts supply with no load connected and can quite easily supply a load demand of 60 watts. <S> Of course, if the current specified is for the output load <S> then this changes things somewhat. <S> For instance that would make the power supply a 120 watt device and would probably waste about 30 watts in heat. <S> Alternatively, if you are looking at (say) the CE marking label <S> the 10 A specified may be when the power supply is being run from an 85 volt AC supply i.e. the power supply might have a universal input range such as 85 Vac to 265 Vac. <S> You need to be clear on this. <A> if you use a switching AC to DC converter with an efficiency of 90% when the load is dissipating 60 watts, then the converter will require an extra 6 watts from the mains in order to do its job. <S> If the power the load requires is less than 60 watts, then the efficiency will drop since the converter needs to eat VA just to keep itself alive.
| Best scenario is probably about 100 watts.
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Does LED matrix scanning reduce overall brightness? Using scanning (or multiplexing) of an LED array, such as below: Such that only one row is powered ON at any given time, would this cause a reduction in LED brightness? My rationale is: In this 4x4 matrix, row 1 will only be turned on 25% of the time. This seems awfully similar to PWM control, which is also a common method of controlling LED brightness. In this example, would all the LEDs be operating at 25% of their maximum brightness? <Q> Yes, if the LED current remains the same, then scanning will reduce the duty cycle and the amount of light. <S> You can offset this by increasing the LED current. <S> But, you need to watch out for the LED current ratings (continuous and pulsed), of course. <S> edit: As an aside, keep in mind that human perception of the brightness is a nonlinear function of light output. <S> (from here ) <A> Yep, the maximum duty cycle goes with the inverse of the number of groups that get cycled through. <S> This can be compensated for to some extent by increasing the drive current, but then you may have to take some precautions to ensure that the LEDs do not get damaged if they are accidentally run at 100% duty cycle at the increased current. <S> I picked up an LED matrix assembly a while back that has integrated drive electronics (decoders and shift registers) and it actually includes a timer component that limits the length of the enable pulse so that it's difficult to leave one row illuminated indefinitely. <A>
| Yes, LED's brightness depend on the current supplied; the bigger the array, the more current you will need otherwise brightness with decrease.
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Is there a functional difference between square, round and octagonal via pads? The reason why these different shapes exist is probably historical, but I'm interested whether there's any reason today to not only use a single kind of via. The ones with round pads are the most effective in terms of area, so those are the ones I usually use. Is there any scenario in which a via with an octagonal or square pad could be a better choice, even if it's just for routing purposes? Or are there maybe some PCB manufacturers only able to make certain shapes of pads? <Q> The shape of the pad has no functional significance for a plated-through via. <S> Indeed, on high-density boards, the pad around the via is sometimes as close to zero as they can reliably manage within the tolerance of the manufacturing process. <A> square vias make it easy to ensure that the copper will join through the gaps based on your design rule set. <A> Decades ago when vector photo plotters with an aperture wheel were used you could only use those pads sizes and shapes that were present on the wheel. <S> With only about 32 different apertures you could not chose between round, square and octagons. <S> But nowadays raster photo plotters are used and you may programm the sizes and shapes you wish. <S> You may also use pentagons and hexagons if you like as long as your PCB CAD system supports them. <S> Even heptagons and nonagons are possible. <S> Triangles are possible too but their area is too small compared to a round pad. <S> A round pad leaves more space for routing as a square pad. <S> The octagon is a compromise between square and round. <S> Routing traces not only with 0° and 90° but also with 45 ° fits better to octagon pads than to square or heaxgon pads. <S> You might use those different pad shapes for marking of some pins, for instance pin 1 of an IC square, all others round. <S> If you want to use those different shapes for marking use only those easily recognized without a magnifying glass.
| The one thing I have found advantageous with using octagonal or square via ring shapes is when copper pours need to fit through via groups. For polarized capacitors you may use different pad shapes for the + and - pin, same for diodes.
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Purpose of Having More ADC channels than ADC Pins on a Microcontroller I have been doing some work with the TMSC5515 DSP chip and noticed in the process of configuring ADCs for potentiometer control that the chip supports 6 ADC channels (VIN[5:0]), but only has 4 pins that can perform the conversion (GPAIN[3:0]). I've noticed this on other embedded processors as well. What is the point of doing this? If the chip is only capable of reading 4 values at a time, the extra ADC channels don't seem like they would serve any purpose. Discription of ADC on TMSC5515: http://www.ti.com/lit/ug/sprufp1c/sprufp1c.pdf <Q> In the special case of the TMSC5515 you have 3 channels for 3 pins and 3 channels only for the pin GPAIN0. <S> Pin GPAIN0 allows higher voltages and provides a built-in voltage divider. <S> You may select the undivided voltage or the divided voltage by selecting one of the three channels. <S> (See page 10 of the data sheet.) <A> but there may also be internal on-die analog values worth measuring, like temperature. <S> The other channels not broken out externally could be connected to an on-die thermal sensor, for example. <S> For some applications, the same die is packaged say into a QFN and a TSOP. <S> Based on available pins, some ADC pads may get wire bonded to pins while others are left floating. <A> An ADC input channel costs very little. <S> A chip needs logic to generate an active-high and active-low select signal for each channel, and it needs a pair of small MOSFETs for each channel to connect it to the analog input circuitry. <S> Where there would be no point including circuitry to select among e.g. 128 inputs <S> if an analog subsystem would be unlikely to ever be used in a chip that size, it's easier to design in extra input channels (which are cheap) and not use them, than to add input channels to a design that didn't include enough. <S> Note, btw, that analog output channels are often much more expensive. <S> Input channels generally don't need to carry significant current and can thus use small pass transistors. <S> Output channels need to carry enough current to drive external circuitry, and thus require larger transistors to amplify the signal. <A> The other answers are valid but frequency and resolution is not mentioned. <S> If the ADC channels can't be configured to a PIN the other answers are valid. <S> When more ADC can be connected to one PIN they can be multiplexed. <S> There are a few interesting things this allows us to do: Increased sample rate or Decrease noise (average) or Increased resolution
| Often the same ADC IP core is re-used among chip families to save cost (or package options for the same die)
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How to attach a paper screen to a DIY PCB? For poor people who can only afford paper and not silk, a home brew silk screen can be inkjet printed onto paper and then glued onto a PCB. Component legs will be poked through the paper. So I have a PCB like:- and want to add my paper screen like:- The large multi pin package towards the middle is digital, clocked at 16MHz. There are also some analogue voltages about in the mV and kHz ranges. Photo mount type stuff is my instinctive choice of glue, but might there be issues with capacitance /resistance? What type of adhesive might be suitable to glue the paper to the PCB? Is there an electronics glue ? <Q> Paper can have a pretty low resistivity value if the relative humidity is higher. <S> I read somewhere that it changes downward by a factor of 1000 going from about 20% to 60%, for example. <S> Still may not be any problem. <S> But since you are worrying about the glue, I thought I'd throw that in just to give you something more to worry about. <S> For oddball electronics situations like this, one place I consider is <S> Cotronics Corp. I use their products already and they are good stuff to have around, at times. <S> In this particular case, you might consider their Electrically Resistant Epoxies or their Electrically Resistant Ceramic Adhesives . <S> If I were buying, I'd probably would try out the first link to start, though I'd be curious about how well those from the second link might also work. <S> You should consider contacting them directly to ask. <S> Chances are, they can recommend something more specific. <A> Rather than glue the paper to the perf board, consider using a toner transfer method to transfer just polymer toner from a laser printer or photocopy of an inkjet print to the board. <S> There are some approaches that use inexpensive coated paper. <S> I have tried this and it works, however the toner will come off with solvent rather easily. <S> There are guaranteed electrically insulating adhesives, mostly silicone or epoxy based but probably not easily or inexpensively available in consumer packaging. <S> based spray adhesive such as 3M 77. <A> The electric properties of glues are much better than those of paper. <S> The pcb base material is made of glas fibers and epoxy, you may use an epoxy glue. <S> But I would prefer a heat resistant plastic film to print on. <S> You have to drill holes in the film for the parts pins.
| You would want an "electrically resistant" epoxy or adhesive. You could try small amounts of two-component epoxy or you could test a solvent (not water) I fear making holes in the film is too difficult and time consuming, therefore this idea is not practicable.
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Why are vacuum tubes more resistant to electromagnetic pulses than solid-state devices? I heard that equipment that uses vacuum tubes are generally less susceptible to electromagnetic pulses than those that employ solid-state devices. I don't know if its true, because I didn't find any detailed research about this topic. If its true, is this because of the physical size difference between these devices, or is there another reason? I searched about this subject, and I found an article from the Science Magazine. I searched for the relevant parts, and it says: Most important, the U.S. military itself had not experienced problems, since most of the field equipment and ships exposed to EMP dated from the 1940's and 1950's, their electronic systems relying on vacuum tubes. In the 1970's, it was discovered that vacuum tubes have about 10 million times more hardness against EMP than integrated solid-state circuitry (2). As you can see at the end, it references another article: M. A. King et al ., An overview of the effects of nuclear weapons on communications capabilities," Signal (January 1980). After 2 hours of searching for this article, I couldn't find any traces of a Signal magazine or anything like that, released in the 80's. I found other citations of the same article, and it has the additional author not present in citation part of the Science article, P. B. Fleming, but still, no info on these guys apart from people with the same name but entirely different professions and other irrelevant research papers. I have some doubts about the science behind the claim that vacuum tubes are 10 million times more resistant against EMP, it kinda sounds like an advertisement from that time. Source: Broad, William J. - Nuclear Pulse (I): Awakening to the Chaos Factor Science 29 May 1981: Vol. 212, Issue 4498, pp. 1009-1012 DOI: 10.1126/science.212.4498.1009 <Q> Most semiconductor devices are designed to operate at low voltage, and voltages significantly higher than their design operating voltage can destroy the components very easily. <S> Vacuum tubes are just pieces of metal inside a glass tube. <S> Not much there that can be damaged. <S> And they are already operating at high voltages in most cases. <S> Electromagnetic pulses just generate large voltages within insufficiently shielded equipment. <S> The same induced voltages are far more likely to cause trouble in solid state electronics simply due to the nature of the components. <A> You just need to consider two things: (i) <S> The level of induced voltage (ii) Mode of breakdown. <S> Semiconductors used in modern integrated circuits have a relatively low breakdown voltage . <S> This breakdown also tends to leave permanent damage in the form of a 'punch through' failure (insulation layer or PN junction). <S> Also consider that the modern intergrated circuit contains millions of semiconductor devices but it only requires one of these to breakdown rendering the whole chip useless. <S> Valves, on the other hand, operate at much higher voltages . <S> A flash over between electrodes is unlikely to produce any permanent damage as the insulation is a vacuum. <S> The EMP may temporarily effect the operation of a circuit. <S> Internal structures of the valve are likely to be screened from the EMP by the outermost metal plates acting as a Faraday cage. <A> Following up on the search for the 1980 article by King et al in "Signal" magazine:Here is the complete citation to the King article: Capt . <S> Michael A . <S> King , US Army and Paul B . <S> Fleming, "An Overview of the Effects of Nuclear Weapons on Communications Capabilities," Signal . <S> Vol 34, No 4, January 1980, p. 60 . <S> Many university libraries own "Signal" magazine. <S> You can locate these libraries by searching in Worldcat.org. <S> Your local public library can request a free copy of this article from any university library, using what librarians call "interlibrary loan". <A> I'm no expert, but I am fairly sure that it's because of the size; solid state circuitry uses very small integrated wires that are easily overloaded to the breaking point by a burst of E.M.P.. Vacuum tubes, however, are definitely much bigger (the transistor was a breakthrough because of it's size!), so it stands to reason it would take a much more powerful E.M.P. to overload it's circuitry.
| You're correct that this does have a lot to do with the size - the components inside of a modern integrated circuit are so close together that even relatively small applied voltages can damage thin oxide layers and cause semiconductor junctions to break down and conduct when they aren't supposed to.
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Problem with reading values from an analog accelorometer I have a Teensy 3.2 MCU ( datasheet ) hooked up to an ADXL326 analog accelorometer breakout board ( datasheet ). Visual examination of the board reveals a few SMD capacitors, which are 0.1uF. I have successfully used a scale value to convert a value to a G value using advice from the manufacturer's forum. For example, when the accelorometer has no force on an axis, it will read 0G. When I apply gravity force, it will change to a ridiculous value like 7G. When the force is removed, it will not go back to 0G -- it will go to a random value from -4G to 4G. Research has revealed that the accelorometer has an impedence of 32kOhms. See the below image for details on the MCU impedence: Is there anyway to reliably go from a analog input to a G value using this setup? This is going to be used in a vehicle for data recording during impacts. I have spent $18 on this analog accelorometer, and would not like to purchase a digital accelorometer unless absolutely necessary. Disclaimer: The truth is, I have no clue what any of this impedance stuff means. I just determined this was the most likely issue from here. They stated that the likely cause is difference in impedance combined with interference from the MCU clock(s). I am not an electronics engineer, I am a software engineer with some basic electronics experience. <Q> The output impedance of 32 kohm can be a significant problem when attached to the ADC input on the MCU. <S> Table 24 entitled "16-bit ADC operating conditions <S> " states that the input impedance is typically 2 kohm and this will screw up any measurement from a source that has an output impedance of 32 kohm. <S> You might get away with a parallel capacitor - <S> the sensor data sheet specifies there must be a minimum of 4.7 nF on the sensor output <S> so have you done this? <S> The table in the question suggests that you might be using a PGA in front of the ADC - please confirm that you have enabled this because it makes things easier <S> but you will still need a 4.7 nF capacitor and normal decoupling across the power rails. <S> Reading further down the PGA table suggests that the analogue source impedance should be typically 100 ohm and this is way lower than what the ADI device is. <S> I think a buffer might be a good idea. <A> Yes. <S> Impedance will be a problem as well as DC offset and sampling range. <S> From the documents you linked, the Teensy wants the impedances on its inputs to be below 5K. <S> The ADXL326 has 32K resistors on its outputs, so you won't meet the Teensy's requirements just connecting the two together. <S> You will need to buffer the outputs from the ADXL326 before sending them to the Teensy. <S> Using a capactor to ground from each output from the ADXL326 might help. <S> If they are large enough, and you don't need to detect fast changes in acceleration. <S> The 32K internal resistance and the external capacitor makes a low pass filter as well as lowering the impedance. <S> The cutoff of the filter can be calculated using equations from the datasheet for the ADXL326. <S> You may find that isn't enough for proper operation. <S> In that case, you will need to look into buffering. <S> That would be something like this: <S> simulate this circuit – <S> Schematic created using <S> CircuitLab <S> I didn't give a type for the amp because you'll have to determine which you can use/need to use. <S> You'll need to find a rail to rail op-amp that can run on your voltage rails (3.6V) <S> There are a lot of details involved that can make the choice difficult. <S> I'd suggest looking around and picking one, then asking here if ithe one you've selected is a good choice. <S> Be sure to think about such details as precision and sampling rate, and how often you really need to read the acceleration (bandwidth, in other words.) <S> Once you get reliable measurements from the ADXL326, you should be able to use the gain settings, ADC values, and the reference voltage to calculate the acceleration. <A> From the question, it sounds like it's defective... <S> the output should give always the instant (after low-pass filter) value of the acceleration experienced by the chip, independently from the values previously sensed! <S> You should: <S> know or measure the Vcc provided to the accelerometer and ensure you have little noise on it; read the voltage V_pin from the output pin of the IC; acceleration=(V_pin - Vcc_accelerometer/2)*max_g <S> max_g = 16 in your case. <S> Do not expect a perfect 0g when acceleration is removed, since the Teensy has some noise in the ADC pins.
| Alternatively you might need an op-amp buffer.
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How can I use this HP power supply as hobby power supplly I have an HP notebook that is dead. So I thought abount using its power supply to play around with my arduino and my DC projects. So I removed the jack of the power supply This is my power supply (18v 4A) The problem is that when I removed the jack came out 3 wires: black, white and blue. I already tried making BLACK and WHITE as + and - and leaving blue disconnected, but my circuit turns on for 3 seconds, after that it turns off for 1 second, and start again and again. What is going on? Cant I use a notebook power supply as my hobby power supply for arduino, LEDs...? <Q> It may be one of those "smart" power supplies that is expecting communication with the laptop it is intended to power. <S> It powers up the laptop, but if nothing is heard within a few seconds, it shuts down. <S> The third wire is the communication channel. <S> If it's a popular model, someone may have reverse engineered the protocol or somehow obtained the specs and published them on the web. <A> I believe it is the remote sense voltage to the converter output. <S> It is usually fed back to a 2.5V ratiometric feedback control to keep the output voltage constant under any load with light lossy cables and 4 Amps of current , that otherwise have voltage drop. <S> If the supply detects a short circuit or an open circuit on sense feedback, it shuts down then the hiccup circuit retries to see if the fault condition is fixed. <S> ( such as a glitch during slow insertion of plug) <S> p.s. <S> I have used this feature in universal chargers to devise an adjustable output voltage with a trimpot and a few other parts like in an LM317 regulator ADJ. <S> method. <S> BTW FYI and FWIW <S> A connector that is on a cable is called a PLUG <S> A connector that is fixed and cannot move was called a JACK until the industry (during the bra-burning women's Liberation days) of the late 60's/early 70's decided to officially change the name of JACKS to RECEPTACLES. <S> ( since they could be either FEMALE or MALE) <S> true story.... <S> this means you butchered your FEMALE PLUG instead of getting a matching MALE <S> coaxial power RECEPTACLE to interface properly .... <S> like Dave T suggested. <S> conclusion connect extra sense wire to + Dc out <S> and it should work . <S> make sense? <A> It puts out 19.64 volts. <S> I got the same reading before I connected the blue wire to white. <S> But then I accidentally shorted the white and black wires and the power supply shut off and its LED went out. <S> So I tried touching the blue wire to white <S> (why not?). <S> The LED came back and on the power reading was the same 19.64 volts. <S> So I soldered the blue to the white, covered everything with shrink tubing and it is running my little LM1875-based speaker just fine. <S> I use a simple virtual ground circuit to create a positive and negative 9.x volts from the Dell supply for the LM1875. <S> So from my actual experience, here's what works: you can either leave the blue wire unconnected to anything or connect it to white.
| I have a similar Dell laptop power supply (CN-ON6M8J) and connected the blue wire to the white wire with no problems. Look around and you might find directions on how to hack the supply. If this is really what is going on, then no, you won't be easily able to use it for hobby projects.
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How much resistance does the capacitor itself contribute to an RC circuit? Are the effects of the internal resistance of a capacitor in a non-repetitive RC circuit with an RC time constant of less than a nanosecond significant? I have only been able to find equivalent series resistance ratings for capacitors, which are usually measured at 100 kHz. <Q> You want a rapidly rising magnetic field at ns timescales, and you worry about cap esR?? <S> ESL is more likely to be the gotcha, but even then.... <S> You may want to start by thinking about the inductance of your magnet, and then how much voltage you will need to make the current change quickly in that inductor (Inductors really do not do step changes in current). <S> By the time you are playing in ns speeds you really want to be thinking EM <S> not just E or H as often your circuit geometry means you have to think fields and waves, not lumped element. <A> I'll assume you're asking about an RC low-pass filter like this: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I've drawn in the capacitor's ESR as R2. <S> The ESR contributes a zero at some high frequency. <S> The result is the circuit becomes a resistive divider (you can also see this just from the simple "capacitors <S> become shorts at high frequency" analysis), and the attenuation stops increasing above the zero frequency. <S> Whether this is "significant" depends entirely on what your requirements for the circuit are. <S> However it's likely that your capacitor (or the wires connecting to it) also has a parasitic series inductance (ESL). <S> This will contribute yet another zero, leading to the filter having a high-pass characteristic at even higher frequencies. <S> Again, whether this is significant depends entirely what your requirements are for the circuit. <A> This is an "ultra low ESR, low ESL 1uF capacitor with ESR= 5 mΩ , <S> ESL = <S> 120pH. <S> This extremely low ESR is only in the Murata LLC series, which they compare to X7S series with 100 mΩ in a 4V rating shown in the table below. <S> ( 20x higher)) <S> Your question regarding 1 ns is better than Murata's "best cap" shown here with an ESR C=5ns . <S> Smaller ESR's can be obtained with smaller value parts but not lower ESR C time constants. <S> That requires a special dielectric and conductor design not commonly found in discrete parts unless in special microwave design or high voltage dielectric. <S> Note 120 picohenries is really small and dominate this discrete capacitor impedance above 10Mhz and ESL and capacitance determines the series resonant frequency. <S> Smaller C values in shunt raise the resonant frequency and extend the low impedance bandwidth. <S> With more work and actually raising the ESR and lowering the Q, it is possible to eliminate the high Q parallel or anti-resonance as shown below. <S> What is your application? <S> Microwave?
| It depends on how high a frequency you need your stop band to reach and how great an attenuation you need to be able to achieve.
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Are plain diodes uncommon? When I search online for buying diodes, I find much more zener-diodes than normal diodes. Is there a reason for this? Can you use every zener-diode as a normal diode or is it uncommon to use diodes in circuits and some other circuit is used to replace the stuff you usually would do with a diode? <Q> "Plain" diodes (silicon diodes, germanium diodes, etc) are very common. <S> As Gerben mentions in his rather good analogy, with Zener diodes you have a wider range of options. <S> The actual number of discrete options for Zener is greater than that of silicon due to the wide range of voltages. <S> However, the quantity of silicon diodes on the market and in use in the world is far greater - you just don't have such a wide choice, since there's no need for it. <S> Schottky diodes, too, give a greater selection due to them having more parameters of interest (switching speed, leakage current, forward voltage, etc) that you don't really care about with a boring silicon diode. <S> To put it another way and use another analogy - imagine walking into a builder's merchant's. <S> There you have piles of similar items: <S> Bricks Breeze blocks <S> Concrete Slabs <S> Bricks <S> there is only maybe one or two different types. <S> However there are huge piles of those couple of options. <S> People use a lot of them, but there's not much to distinguish one brick from another. <S> Breeze blocks there may be a number of different sizes, some hollow, some solid, etc. <S> There's less of each type, but there's more piles. <S> Finally there's the concrete slabs. <S> You have different colours. <S> Different patterns. <S> Different sizes. <S> In short, there's many many many different options. <S> So there's many many different piles. <S> However there's only maybe one pallet of each, compared to twenty pallets of one kind of brick. <S> There is actually far more silicon diodes around in the world. <S> You'll be sat within a few feet of tens or hundreds of silicon diodes right now, but maybe just a small handful of Zener diodes. <A> You can get zener diodes in different voltage ratings. <S> So you naturally you get more different zener diodes than normal diodes. <S> Just like in the supermarket. <S> There are more types of flavored yogurt than plain yogurt. <S> As there is banana-flavored yoghurt, strawberry-flavored yogurt, and so on. <A> If you just want general-purpose diodes, buy a bunch of 1N4148 diodes for low current and 1N4004 diodes for higher currents, and maybe a handful of 1N5819 Schottky diodes for when forward voltage drop matters. <S> The 1N4004 are 1A 400V PIV diodes, so you can use them at lower voltages and currents. <S> The 1N4148 are at least 150mA/60V, and the 1N5819 are 1A/40V diodes so good for lower voltage circuits. <S> They're all <S> very cheap generic through-hole parts and any junk box should have a good supply of them.
| So although it may look like there are more Zener diodes around, all there actually is is greater choice of parameters for Zener diodes.
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Controlling speed of motor using VFD Motor is 0.75Kw, 3Ph, 400V and the rated RPM is 1380. How much speed I can reduce using the VFD Can I operate at 300rpm while it is designed for 1380rpm with worm gear assebly. <Q> For 1380 RPM rated speed at 50 Hz, the motor must have a synchronous speed of 1500 RPM. <S> Synchronous speed = 120 x f / poles. <S> The number of poles is an even number, usually 2, 4 or 6 for a small motor. <S> The possible values of synchronous RPM would be 3000, 1500 and 1000. <S> Rated speed is synchronous speed minus slip. <S> Since slip is usually 1.5 to 4 percent of synchronous speed, the usual speed would be about 1440 to 1480 RPM. <S> Your motor is close to that at 1380 RPM, but 120 RPM slip makes it an eight percent slip motor. <S> That would mean that it is a high slip motor rather than a standard motor. <S> High slip motors are usually selected for loads that have a very high starting torque. <S> That may be of some concern when using a VFD. <S> You should certainly select a sensorless vector VFD rather than a V/Hz control VFD. <S> It should be a VFD designed constant torque operation rather than for fan and pump duty. <S> It might be a good idea to select a VFD rated for a 1 kW motor. <S> For 300 RPM the operating frequency would be between 50 X 300/1500 = 10 Hz and 50 X (300 <S> + 120)/1500 <S> = 14 Hz. <S> That should be no problem for any VFD. <S> Motor cooling at 22% of rated speed may be of some concern. <S> That might be enough to allow it to operate at the low speed without overheating. <S> If the low-speed operation is not continuous, that would also help. <S> Using a high slip motor does not provide higher starting torque when using a VFD. <S> Another approach is to use a 1000 RPM motor and operating it between 15 Hz and 75 Hz. <A> Yes you can. <S> The output from the VFD is input frequency independant and can vary between 0 rpm ( external cooling required) and some 10 to 20% over the design speed of the motor. <S> Modern VFD's allow you to make a fully automated setup. <S> However if the requirement become very specific manual setup might be needed. <A> It is not reccomended to spin the motor bellow 50% rated speed, because it will overheat. <S> You can mount an additional blower fan to continuosly run the motor at low speed. <S> The VFD with V/f control won't work bellow 15Hz (50hz nominal), you may need vector contol VFD to run down to 5Hz - but this is also some comercial data, it is reccomended that you don't go lower thazn 15Hz in either way.
| For VFD operation with an unusually high starting torque, it is generally better to select a standard motor of the next larger size. Since it is a high-slip motor, it may have a larger than normal housing an enhanced self-cooling because of the increased slip losses.
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Voltage/Temperature in Steel wire Im trying to heat up a steel wire of d=0.05mm with a specific current. The length of the wire is about 5cm. By measuring the voltage over the wire, how can I find the temperature? <Q> I think the only problem is to find the resistance-temperature dependency of the wire. <S> I wouldn't rely on published coefficients or tables unless you are very sure they refer to <S> exactly the same steel alloy of your wire <S> (I can imagine that you don't know it exactly). <S> I'd measure the resistance-temperature relationship by myself as follows (see picture): <S> Build a device using a hot air gun to create an air stream (red) of known temperature. <S> You can measure the temperature of the air stream by a conventional electrical sensor. <S> The temperature of the air can be controlled by the power feed to the hot air gun. <S> You probably have to take some care that the air is mixed enough after heating so temperature is homogeneous. <S> Use the air to heat a sample of the same type of wire (brown; maybe of larger length to get a more accurate result) and measure the resistance of that wire in the interesting temperature range. <S> Of course the resistance of the leads connected to the sample wire must be negligible (e.g. by using thick copper wire) and the connection must be heat resistant at the interesting temperatures and resistant against oxidizing; at least for short time. <S> I.e. you can't use soldering; I suggest crimping or welding. <S> Once you have determined the resistance at several different temperatures in the interesting range you can fit coefficients of an appropriate model function <S> (the most simple model would be a linear regression with only two parameters: slope and intercept) and use it backwards (inverse function) to determine the temperature when resistance is given (measured). <A> The wire will have a certain resistance and that resistance will change with temperature to a degree, and with time as the wire corrodes and oxidized. <S> If you know the relationship between temperature and resistance you can get an idea of the average temperature by measuring the current with a constant voltage, or by measuring the voltage at a constant current. <S> You can look up temperature coefficients of various steel alloys and you can measure the wire you have at a known temperature (preferably close to the operating point) or you could calibrate the wire in an oven over the expected operating range, depending on what kind of accuracy you expect. <S> Unless the wire is operating at a low temperature and/or <S> a vacuum it won't be very stable because of oxidation. <S> This works much better with a platinum wire or a tungsten wire. <A> How does a blacksmith know the colour of a piece he is working on. <S> Most blacksmiths will cite experience <S> but it's based on Plank's law and black body radiation. <S> Here's a simple chart you can use <S> but you need to match any chart with the emissivity of the material being "measured". <S> This is for steel <S> but you should be able to find one for iron <S> (it's probably very similar): <S> - And here's some more information on black-body radiation: - <S> So, in principle you can get a fair degree of accuracy without resorting to trying to calculate the temperature based on the voltage across the wire (given the many problems this method has). <S> In other words, use your eyes like a smithy does! <S> Or, use a pyrometer: - http://media.fluke.com/images/572-2-c-04a-600x402.jpg
| If you measure voltage at points along the wire you could get an idea of the temperature distribution.
|
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