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The fields and waves associated with a transformer Induction in a transformer happens via a fluctuating magnetic field in coil A inducing a voltage and hence current in coil B. An electric field is associated with a fluctuating magnetic field. Is the electric field at right angles to the magnetic flux lines? Is this magnetic field actually an electromagnetic wave? What part, if any, do electromagnetic waves/fields have in the operation of a transformer? Do electromagnetic waves travel down the wire? <Q> An electric field is associated with a fluctuating magnetic field. <S> Is the electric field at right angles to the magnetic flux lines? <S> The simple answer is yes as per this diagram from here : - For a transformer, the changing flux is contained within the core and the windings that surround the core limbs "collect" the field that is at right angles to the magnetic flux lines. <S> Two turns of a winding collect twice the voltage etc.. <S> Is this magnetic field actually an electromagnetic wave? <S> What part, if any, do <S> electromagnetic waves/fields have in the operation of a transformer? <S> No. <S> Electric and magnetic fields are not forming an EM wave because they are temporally shifted; it is the rate of change of magnetic flux that delivers voltage; BUT in an EM wave, electric and magnetic fields rise and fall with time syncronicity: - This doesn't happen with a transformer because the induced voltage is \$N\dfrac{d\Phi}{dt}\$. <S> Do electromagnetic waves travel down the wire? <S> If you dig deep enough you will find that all currents and voltages can be thought of as EM waves and, as such, all wires can be thought of as transmission lines. <S> Do you need this to understand "regular" transformers? <S> No! <S> Do you need this to understand RF transformers? <S> Sometimes when in the upper VHF region and beyond and certainly as you approach the microwave range it is important to understand what effects can happen. <A> Read Nikola Tesla's patent on the AC transformer from 1890. <S> ok too much "legalese" from the Attorneys. <S> Basically the Vac, and f with primary inductance creates the mutual flux in the core that couples the secondary turns and produces the voltage ratio. <S> This generates a small magnetic current with no load but is necessary to couple the primary and secondary in the shared core. <S> E(Electric) <S> fields from voltage to H(magnetic)fields from current are both at right angles to each other and the direction of the wire. <S> The Alternating voltage induces alternating current sinewaves are not the same as travelling waves on a long transmission line. <S> The load also draws a current wave from the secondary voltage. <S> If is a linear load, it is also a sinewave current. <S> Don't confuse the two distinct uses here of the term "wave". <S> Since the wavelength for 50Hz is far greater than the length of wire used there are no travelling-wave effects from the transformer itself. <S> Also read Maxwell's Treatise on Magnetism and Electricity at archive.org <S> (search and <S> if you fail re-search it again that's what your education is all about. <S> Learning how to learn. <S> For the 11 to 19 yr old Is current exciting? <A> It's hard to answer. <S> Each point in the space (or near the transformer) has its own electric field ( E ) direction and magnetic field ( B ) direction, so it's hard to say in general whether E and B is perpendicular. <S> No, the EM field itself is not an EM wave. <S> An operating transformer do emit EM wave. <S> However, due to most transformer works at extremely low frequency (50Hz/60Hz), the emitted power is extremely small. <S> The low frequency EM wave does not travel in metal conductors. <A> Although this video is about antennas, I think the same principles could be applied to a transformer regarding the magnetic field, electric field and electromagnetic waves. <S> Electromagnetic Wave Propagation
| It is a magnetic field (and an associated electric field) and no part of understanding the operation of a transformer is relied upon by using electromagnetic wave theory.
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How does an ATX power supply turn off? It's well-known that for powering up a standard PC power supply, its PS_ON# line must be momentarily pulled low (i.e. connected to GND). But how does the motherboard actually turn it off ? <Q> The motherboard doesn't turn the power supply on and off. <S> It controls it between a low power standby, where only the 5v_STANDBY line supplies power, and full output. <S> What controls these two states is the (perhaps mis-named <S> ) PS_ON# line. <A> There are two converters inside an ATX power supply: The main converter which provides +12V, +5V, +3.3V, -12V and -5V. Low power standby converter which provides 5V with relatively lower power ( <S> e.g. 5V/1A). <S> This output is called +5V_SB. <S> The latter one is entirely independent from the main converter and is always on. <S> This output supplies some of the logic blocks, BIOS and some USB ports. <S> Besides, some motherboards have a feature for powering up from mouse, LAN or keyboard. <S> +5V_SB is needed for this as well. <S> The main converter can be activated by pulling low the input called PS_ON (green wire) or can be deactivated by either pulling that input high or leaving open. <S> That is what actually the motherboard does. <S> This task is shouldered by a logic block and this block is supplied from +5V_SB. <S> So, the ATX power supply is never turned off. <S> Only the main converter is disabled by the motherboard. <A> After reviewing a few (of dozens different PSU) i've here in the bin <S> i see most had a bridge (gray/purple cables) from the SG6105D to an optocoupler through a 100R which enables ground pin from 431A through the optocopuler's led. <S> Unshorting the PW_ON ground connection at the optocupler's leg would cut the signal feedback-loop driving the MOSFETs, halving the large transformer's high current pulsed signal instantly to zero.
| Logic on the motherboard, powered by the 5v_STANDBY line, pulls PS_ON# low permanently to enable the power supply main output, and releases it high to disable the main output.
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What is the logic of running a NiMH battery in charge-discharge cycles everyday? I have an 1500 mAh NiMH battery for powering a 2250 kV motor and four 9g servos for my radio controlled plane. I had bought the battery 8 months back and I had used it to fly my plane which drained my battery after 20 minutes and 8 months hence when I drain it, it runs for like 9 minutes. So I haven't used it during that span of 8 months nor have I charged it. After a lot of googling and research as to why that happened, the reason I got was that I should be running my battery on charge - discharge cycles everyday but didn't get any answer as to why we have to run it. <Q> That's bad advice. <S> As mentioned in the comments, for most cells, including NiMh cells, the number of cycles they go through determines a large part of their service life. <S> For many NiMh cells this is often stated to be 1000 cycles. <S> The 1000 cycles is usually only valid when the cells are charged and discharged at a nominal (low) current like C/10. <S> So for a 1500 mAh cell that's 150 mA. <S> Then charging and discharging lasts a full day. <S> If you use a higher current to make this go faster, you might not reach the 1000 cycles as the cell will wear out sooner. <S> For long-term storage of NiMh cells I think the best solution is to either trickle charge (with a very small current) <S> the cells to keep them full or just let the cells discharge and charge them only before use. <A> Every day seems very excessive, unless you have a daily use of them of course. <S> I would advise to charge and discharge at least once a month. <S> That's what I try to do, I don't notice cells going much less than 1.2V in that amount of time. <A> Eneloop, Energizer and Duracell sell Ni-MH batteries that hold their charge for one year <S> so you do not need to trickle charge them and do not need to frequently charge them.
| NiMH batteries tend to have a higher self discharge then other technologies (they loose their charge even when not in use), hence why its advised to cycle them regularly. I should be running my battery on charge - discharge cycles everyday If you would cycle every day then the cells might be worn out after only three years.
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Why the earth attracts charge and where does the charge go when it goes into ground? Self-explanatory question but to add what I know, by this I will also know if I am right or wrong. I read somewhere that Earth is positively charged, but is not the state (whether positive/negative) of a body relative to the near body/object. If Earth is positively charged, why is it so? Is there any negative charge around it which helps Earth retain it's positive charge? Also, what happens to charges that flow to the ground ? <Q> As a planet overall, the Earth is electrically neutral. <S> However, the surface of the ground can have a localized charge in certain areas, such as under storm clouds. <S> ( Source ) <S> The action of wind and droplets of water within the cloud causes the charge at the top of the cloud to be positive with respect to the bottom of the cloud. <S> The strong negative charge at the bottom of the cloud also induces a corresponding positive charge in the ground directly below it. <S> It is this lower charge distribution that is responsible for cloud-to-ground lightning. <A> If Earth is positively charged, why is it so, is there any negative charge around it which helps Earth retain its positive charge ? <S> It's actually a little more complicated than that. <S> The earth around you locally is part of a giant circuit! <S> Thunderstorms generate a negative current which then flows back to the ground wherever there aren't thunderstorms. <S> So it really depends on the weather. <S> It also depends on a variety of other factors like solar storms and ionization of the upper atmosphere, but this gives you an idea of what goes on. <S> I coudln't find the graph but the local electric field also changes when the sun shines from ionization. <S> The electric field above also contributes charge, thunderstorms and other effects all contribute. <S> All in all, the ground's net charge can be considered zero and a reference for all other charges. <S> Source: <S> https://slideplayer.com/slide/6192933/ <S> Also, what happens to charges that flow to the ground ? <S> There isn't really a good way of determining the earths total charge as you would have to account for all of the factors. <S> Just call it 0V for now. <A> Richard Feynmann explained lightning as the charges arriving from cosmic rays, collected in high-altitude clouds, and then building up potentials high enough to cause lightning bolts down to earth.
| The ground also has conductivity (or functions like a resistor), so it distributes the charges (as in a lightning bolt) or the really low currents that come from the air to the ground to keep it's potential the same.
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Why does WiFi have a shorter range than LTE? It seems confusing that my phone receives -87 dbm LTE signal and shows a full 4 bars with the speeds of UPLOAD: 20Mbps DOWNLOAD:13.6Mbps But my WiFi is showing 1 bar at -89 dbm and I get disconnected as soon as I move slightly away, and speeds are very low.Why does this happen? It happens with all my phones. <Q> As with any radio receiver, if it can handle a higher data rate, then it is usually burdened with having a higher RF bandwidth and this inevitably means more received background noise i.e. a wider BW lets in more noise and hence, you need a higher received signal level to operate with a decent SNR (signal to noise ratio). <S> Therefore WiFi is at a significant disadvantage because it has a wider RF bandwidth than LTE (normally) and needs a higher signal level to operate at a decent bit-error-rate (BER). <S> This is embodied in the following empirical but commonly-found relationship. <S> Power (dBm) needed by a receiver <S> is -154 dBm + \$10log_{10}\$(data rate) <S> For example, if the WiFi data rate is ten times your LTE data rate, then you need 10 dB more signal to operate at the same SNR. <S> Basically if you double the RF bandwidth you "collect" 3 dB more noise. <S> This means that WiFi is usually the first to suffer as signal levels drop (compared to LTE data rates). <S> Why WiFi has shorter range than LTE? <S> This is related to the Friis transmission equation <S> but, more simply, you can think about the same effect with light bulbs; consider a 1000 watt lamp and the distance <S> you could see this at night time - you would probably see it fairly clearly from 10 km away <S> and, if you walked a further 100 metres, it wouldn't look significantly dimmer. <S> Compared with a small 1 watt lamp, you might see it glowing at 100 metres <S> but, if you walked away a further 100 metres, it would be noticeably dimmer. <S> There are a bunch of other factors too such as operating frequency - WiFi <S> can operate at a higher carrier frequency and <S> the Friis transmission equation informs you that as frequency rises, the path-loss increases: - Path loss (dB) <S> = <S> 32.45 + \$20log_{10}\$(F in MHz) + \$20log_{10}\$(D in kilometres). <S> In other words at ten times the frequency, the path loss increases by 20 dB. <A> in addition to Andy's answers, WIFI is usually limited in power, 30dBm in north America, lower levels in most of the world. <S> LTE usually can transmit up to 4W (36dBm) <S> And the towers transmit at a much higher power. <A> For comparison, these are the transceiver part of a LTE <S> network(from two different vendors)(antennas are mounted on towers/poles and connected via cables to the eNodeB), equivalent to the antenna part plus the transceiver circuit of a wifi AP. <S> https://www.motorolasolutions.com/en_xl/products/lte-broadband-systems/broadband-systems-equipment/enhanced-node-b/rbs6101.html#tabproductinfo <S> https://www.scribd.com/document/204866576/RBS-6000-Spec-Sheet <S> A LTE eNodeB can handle hundred of users(depending on configuration) in a controlled manner. <S> How many can a wifi AP handle??? <S> You cannot really compare Wifi to LTE. <S> It is two completely different kind of system meant to be used in quite different scenarios.
| Also, LTE has much better network management capabilities (automatically finding the best channel and data rate), the towers have much better clock sources than WIFI (this affects receiver sensitivity) and the towers have higher placed antenna (10-30m) than the common wifi router.
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How to measure leakage current? I have an AC powered device, and a customer asks for exact leakage current. Current that can trigger a leakage current circuit breaker, so it's obviously something flowing through Y capacitors to the ground wire. The question is, what is usually measured and what actually matters? Peak to peak? RMS? Pulse width? <Q> The required insulation test uses high voltage DC where one clamp is put on both the AC inputs and the other probe is put where people can touch (earth wire, metal chassis, low voltage DC components, ...). <S> Because it is <S> DC class-Y capacitors will not be leaking current. <S> You can frankenstein an extension cord for that purpose. <S> This is how a GFCI works, when that current exceeds the rated value it will trip. <S> There are more advanced devices that will used high voltage AC instead which will include the effect of the class-Y capacitor. <S> If that leakage current if because of high frequency noise on the line you can add a common mode choke on the AC input as filter. <A> How to measure leakage current? <S> I have an AC powered device, and a customer asks for exact leakage current. <S> Of your device? <S> I believe you are referring to current to ground that should be on the neutral. <S> If you want to find how much current is leaking to ground in a device, electrically isolate it and measure current to ground through a known path. <S> Current that can trigger a leakage current circuit breaker <S> , so it's obviously something flowing through Y capacitors to the ground wire. <S> Hmm in retrospect there may have been something wrong with that first sentence and maybe it was supposed to be part of this one. <S> Anyway I think you're talking about an AC GFCI circuit breaker. <S> Most GFCI devices work by passing both live and neutral wire through an induction loop. <S> If all of the current flowing out on the line wire flows back on the neutral, the magnetic fields will cancel out and no current flows in the induction loop. <S> If any of the current flows to ground instead, it won't return through the loop on the neutral and a portion of the magnetic field won't cancel out, causing a current to flow in the sense loop and breaking the circuit. <S> This article goes into detail about what's legally defined as a class A, C, D, E device and why trip level requirements are what they are. <S> The question is, what is usually measured and what actually matters? <S> Peak to peak? <S> RMS? <S> Pulse width? <S> The circuit <S> I'm aware of mentioned above is basically measuring the absolute value of the total lost current, meaning both that it is measuring quantity and that it measures current regardless of direction of flow. <S> The legal requirements for NEC specifications are clearly based on both current and exposure time, but because faster is always better in this case it matters more that it measures up to the minimums set by the government and performance may be significantly better in reality. <S> At any rate, Class A GFCIs must trip at 5mA and class B GFCIs at <S> 20 mA, class B GFCIs are for old underwater fixtures in swimming pools where leakage current exceeds 5 mA and can cause nuisance tripping. <A> If someone is asking for an AC current , and they don't specify how it is measured, then it will be RMS. <S> RCDs (GFCIs) are set to trip at a specified current imbalance between phase and neutral. <A> For an AC powered device, finding leakage value in mA can be performed using a flash tester of some kind. <S> You can use a 500V or 1400V (depending on TVSD) <S> AC HIPOT test to highlight if there are any issues with flash over, leakage and clearance between earth and live/neutral. <S> In the event of having a high leakage I usually look for damaged/nicked wiring, poor clearance and take a look at any filters that may present at mains input. <S> It's good practice to perform an IR test after a flash test, in case any harm has been done.
| The simplest test to test for AC leakage is to put it on a non-GFCI protected circuit (to avoid trips) and put a clamp meter around live and neutral (but not ground) and set to measure AC.
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Is it possible to interrupt the copy process of a struct by an interrupt in embedded C? Inside the driver I have got a function to copy the data from the internal struct into a struct from the application. Can this process be interrupted by a microcontroller interrupt trigger? uint16_t getRawData(struct Data *Data_external){ if(Data_external == NULL) { return ERR_PARA; } else { *Data_external = Data_internal; // the copy process. Could this be interrupted? } return ERR_NONE;} <Q> Yes. <S> Pretty much everything in an MCU can be interrupted by an interrupt request. <S> When the interrupt handler completes the previous code will just continue so it is usually not a problem. <S> In a special case the interrupt handlers can be interrupted themselves by interrupts of higher priorities (nested interrupts). <S> If a series of instructions must not be interrupted then you need to implement a critical section (basically globally disable interrupts, do the job, enable again). <S> Remember that depending on architecture of the target CPU a single line of C can be compiled to many assembly instructions. <S> A simple i++ on an AVR is compiled to multiple instructions if i is for example uint32_t . <A> Any operation that is not atomic can be interfered with by an interrupt. <S> This kind of programming is often very different than most other programming and can be confusing to people who haven't studied processor design or computer architecture. <S> You may think to yourself " <S> This will never actually happen, how long does this code take to copy and <S> how likely is an interrupt?" <S> But with most production embedded applications it will happen because the product is on for years without updates. <S> The other issue with struct copies like this is when they do happen they are extraordinarily difficult to debug because they only happen when the interrupt occurs at just the right time (which can be as little as one cycle). <A> The whole point of interrupts is that they can (and do) happen all the time, and are designed to have zero impact on any code that happens to be running when they occur. <S> All the registers are saved, and depending on the CPU architecture, a completely different register set may be swapped in, the interrupt does its thing, and then the original registers are restored and the code continues to run as normal. <S> Problems can occur when the interrupt service routine itself tries to access memory that is being accessed by the running, interrupted code. <S> Even more subtle errors can occur when a time-critical I/ <S> O process is interrupted. <S> These problems are commonplace with older, simpler, less secure architectures where there may be little separation between "user" and "supervisor/kernel" mode code. <S> This kind of problem can be hard to identify, and often difficult to reproduce, but once identified they're often fairly trivial to fix using defensive programming, mutexes/semaphores or simply by disabling interrupts in critical sections of code. <S> The general class of problems has been studied extensively, and modern multi-core CPUs and even multi-tasking operating systems would not be possible if multiple solutions were not already tried and tested. <A> I'm just going to go ahead and assume you asked this for a very good reason. <S> *Data_external = Data_internal; <S> Can be split (barring some edge cases that are unlikely to be in play here). <S> I don't know your CPU <S> but I haven't seen a CPU yet <S> that can't do the moral equivalent of: cli(); / <S> * mask <S> all interrupts <S> */*Data_external = Data_internal;sti(); / <S> * restore interrupt mask <S> */ <S> Now it can't be split on any single core CPU because nothing can interrupt while interrupts are off. <S> Whether or not this is a good idea depends on a lot of things that I'm simply not qualified to evaluate. <S> If you are multi-core (and I finally remembered there is a multi-core embedded CPU on the market) don't do this. <S> It is worthless. <S> You would need to develop proper locking. <A> However, before you start making critical sections all over the place, you should check a few things: <S> You say this function is "inside a driver". <S> Are interrupts already disabled when this function is called? <S> Or is it called inside an interrupt handler which prevents other interrupts from triggering? <S> If yes, the operation cannot in fact be interrupted. <S> Is Data_internal <S> ever accessed inside an interrupt handler? <S> If not, there is no harm even if the operation can be interrupted. <A> [Not enough rep to comment] Another gotchya with that kind of struct copy is that it is a shallow copy . <S> You may need a deep copy instead. <S> A shallow copy could possibly but not likely be atomic, depending on machine architecture. <S> A deep copy is almost certainly not atomic on any architecture. <A> A quality implementation suitable for embedded-systems use will document how volatile -qualified reads or writes of various types will be performed in sufficient detail to indicate whether and how they may be "split" by interrupts, and also whether or not volatile reads and writes are sequenced with regard to non-qualified reads and writes. <S> While some implementations may behave as though all reads and writes are volatile -qualified, <S> implementations are generally expected to be free to process sequences of non-qualified reads and writes in whatever fashion would be most efficient when there are no intervening volatile accesses. <S> On a typical 32-bit microcontroller, reads and writes of volatile -qualified integer types that are 32 bits and smaller; an assignment will consist of an atomic read followed by an atomic write. <S> If you want to ensure that a 32-bit structure is copied atomically, place it in a union with a uint32_t , and read or write that member to read or write the structure as a whole. <S> Quality implementations that are configured to be suitable for embedded systems use will allow unions to be employed in such fashion without regard for whether the Standard would require implementations that aren't intended for such use to do likewise. <S> Note that gcc and clang will not reliably behave as quality implementations suitable for embedded systems use unless various optimizations are disabled.
| The code as you presented it indeed can be interrupted.
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Are Civilian GPS signals cryptographically signed? As far as I understand, receiving enough GPS signals at the same time enables to deduce the position and the time. So I guess it is possible to use an offline receiver as a very precise clock? If so, is it possible to flood this offline receiver with fake signals to make it believe it's 12:00:01 when it's actually 12:00:00? Or more specifically, is it possible to design a receiver who can't be attacked this way? If GPS signals (or Galileo's, ...) are cryptographically signed, it's easy to reject non-signed signals by saving the public key of the satellites beforehand. But are the GPS signals cryptographically signed ? Edit: My question is not about the civilian signal being encrypted or not (meaning unreadable for people not having a secret key), but signed or not (meaning the authenticity of the signal being verifiable thanks to a public information: the public key of this satellite). <Q> Therefore, the system cannot be made secure by cryptographic signatures. <S> The conceptually simplest way to spoof is to erect a number of highly directional antennae and point each of those at a GPS satellite, such that it receives exclusively signals from that satellite. <S> Then feed those signals through a bank of delay lines, mix them back together and use another directional antenna to send the result toward an enemy aircraft. <S> You can then sit in front of the delay lines and force arbitrary position errors upon the unsuspecting enemy. <S> If you introduce a delay for the satellite that is south of your position, the enemy's receiver will consider itself further north than it actually is, about 30cm per nanosecond of delay. <S> Cryptography doesn't help you to detect or prevent those attacks, as the signals are only delayed but never changed. <S> The only defense a receiver can mount is radio direction finding. <S> If all satellites' signals come from the same direction, it's probably a spoofer. <S> All modern military receiver employ this method, more sophisticated ones also crosscheck the directions of arrival against the known position of the satellites. <A> So I guess it is possible to use an offline receiver as a very precise clock? <S> Yes, you can use the GPS PPS (pulse per second) signal available on most every receiver to sync up your clock an maintain better time accuracy. <S> If so, is it possible to flood this offline receiver with fake signals to make it believe it's 12:00:01 when it's actually 12:00:00? <S> Or more specifically, is it possible to design a receiver who can't be attacked this way? <S> No, if someone spoofs a GPS satellite signal (which is really sophisticated kind of attack ) then you are at the mercy of that signal. <S> You could possibly work to build a system that could detect spoofing. <S> If GPS signals (or Galileo's, ...) are cryptographically signed, it's easy to reject non-signed signals by saving the public key of the satellites beforehand. <S> But are the GPS signals cryptographically signed? <S> With GPS, there are two codes, and depending on which receiver you have access to (unless your military your probably not going to get access to p-codes). <S> These determine the accuracy, but so does the receiver, so I would worry about what receiver your using and not how the signals are getting to it because that is what determines the accuracy. <S> There is no 'cryptographic signing' of the signal and it can be spoofed. <S> There are two codes that are used: C/ <S> A code: <S> The C/A code on the GPS signal is the one used for general or Civilian Access. <S> This code is transmitted at 10.23 million chips per second, Mcps. <S> P code: <S> The P code is the precision code that can only be accessed by the US military. <S> The P code transmits at a rate of 10.23 Mcps. <S> Source: <S> https://www.radio-electronics.com/info/satellite/gps/signals.php <S> The accuracy of the PPS signal is also determined by the receiver , and how it generates this signal from the GPS data. <A> Yes, the last generation of GPS and Galileo systems use cryptographic authentication of geolocation transmissions, called navigation message authentication or NMA . <S> But there are limitations to these systems. <S> Here's a more detailed look into NMA . <S> In the GPS system, Chimera, digital signatures are transmitted at most once every 3 minutes. <S> The Galileo system, OSNMA, uses symmetric MAC -- with the MAC key also being broadcast -- making it possible to forge alternate contents for a message after it has been received from the real satellite. <S> Crucially, it seems there is nothing to prevent an attacker from intercepting real navigation messages and re-transmitting them with a carefully chosen delay to give a particular receiver a different understanding of its location. <S> If an attacker was able to cut off a receiver from real navigation messages entirely and make it listen to just retransmitted ones, it's still possible to credibly mess with someone's location (e.g. a man-in-the-middle attack). <S> How effective attacks are will also depend on how GNSS receivers are implemented and how they will deal with conflicting data. <S> For better accuracy, consumer receivers listen to signals from all systems, including GLONASS and BeiDou. <S> If they receive authentic messages from just a few Galileo satellites and lots of spoofed (unverifiable) messages from other systems -- are they going to reject the unverifiable messages? <S> I suspect not.
| GPS can be spoofed without decrypting or creating signals.
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Infrared LED and Photodiodes I am currently working on a project that uses an IR LED and a photodiode in a proximity sensor configuration (shown below). My question is, I have read some sites that state that your human body reflects IR light transmitted by the IR LED hence when you touch or move your hand close to the proximity sensor the circuit will activate. But other sources say that the human body does not reflect IR light but rather refracts it. So which is correct does the IR light emitted by the IR LED gets reflected or refracted towards the photodiode when your finger or hand moves closer to the sensor? The second part I noticed is that even if the IR LED is removed the sensor still activates because our bodies do also emit radiation which the photodiode can detect. Provided the threshold voltage is low enough. So then what's the point of having an IR LED in the circuit if the circuit already activates when a finger gets closer to the circuit, or am I understanding this wrong? N.BThe circuit I built is going to be used to detect when a finger or hand gets closer to the sensor. <Q> To what's already been said here, I want to add a bit more. <S> Human body temperature is about 310 K. <S> The blackbody spectrum at 310 K peaks at about 10 um: ( source ) <S> The photodiodes used in this kind of circuit is typically a silicon device, with a bandgap energy corresponding to about 1.1 um. <S> The responsivity falls off strongly at longer wavelengths: ( source ) <S> So this detector will not have any practically useful (or confounding) response from the radiation from the human body. <S> In fact, if it did, it would also be detecting radiation from other nearby objects of similar temperature, such as the PCB it's mounted on, the metal or plastic housing around the actual photodiode, etc. <S> The reflectance of human skin depends somewhat on the color of the skin in question, but at the near-IR wavelengths that will most affect your photodiode, we're all about the same. <S> We are not great reflectors, but probably good enough to have a measurable effect on your sensor: ( source ) <A> The detector must have a black lens which passes IR and blocks stray visible light. <S> aka <S> "Daylight Blocking Filter" Skin reflects some IR, even if you are black-skinned. <S> It also conducts IR <S> so it is semi-transparent, which is good for finger sense heart rate and oximeters. <S> But since the dielectric constant at this wavelength is greater than 1 it also refracts. <S> ( plastic is higher than glass so they can use thinner lenses for eye-frames) <S> But skin is also an irregular surface compare to the wavelength, so it also scatters light. <S> We also generate IR as heat, but this is a longer wavelength than most IR LEDs and PD's can detect. <S> FIR or Far InfraRed is (>5um) <S> but I'm not sure of the peak wavelength. <S> The body also absorbs infrared heat both short IR or SIR from tungsten heaters and far IR or FIR from our partners bodies. <S> It gives " black body effect" a whole new meaning to some, but some researchers, including myself, know of the healthy benefits of FIR and some say married couples live longer if they sleep closer. <S> All Doctors of Acupuncture know this because they use FIR lamps. <S> Another play on words is "Skin effect" because we know the longer wavelengths penetrate deeper, an important quality of FIR for therapeutic reasons. <S> Often it is synthesised with many mineral phosphors including Germanium with a short IR back heater. <S> I would call it a "diffuse surface" which gives lower resolution monochrome images or scattering of incident light at and below the surface. <A> Without giving you the actual numbers here I can tell you from my own experiences with such IR circuits that the IR light is reflected from your body and that your body does not emit enough IR itself to fail this type of circuit. <S> But sunlight IR is a big problem that you need to consider. <A> The second part I noticed is that even if the IR LED is removed the sensor still activates because our bodies do also emit radiation which the photodiode can detect. <S> Provided the threshold voltage is low enough. <S> So then what's the point of having an IR LED in the circuit if the circuit already activates when a finger gets closer to the circuit, or am I understanding this wrong? <S> Provided the threshold is low enough. <S> When that happens, the circuit will also trigger from things like stray sunlight. <S> So the threshold is set higher than the highest ambient illumination you think is likely, and this makes it impossible to trigger off your body's emitted IR. <S> Then you need an LED to provide even more illumination so the circuit can trigger at all. <S> Note that putting an IR filter in front of the detector helps, since it eliminates ambient radiation in the visible, and improves the amount of unwanted power falling on the detector. <A> I don't, off the top of my head <S> know the ratio -- but even if most is absorbed, some is still reflected. <S> Thus, if the ONLY optical path between the LED and the detector involves a reflection off a person, the detector will become MORE activated -- assuming the person is not blocking a bigger source of IR light, such as the sun. <S> Many detector circuits modulate the LED, and then you look for a change in the detector at the modulated frequency. <S> This way, you know you're only looking at changes involving the light source of interest.
| Like any imperfect mirror, some IR light will reflect, and some will be absorbed.
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Using Raspberry Pi to turn TV on and off directly (not using infrared) I am creating a device that requires a TV to be powered on and off when commanded by the code in my Raspberry Pi 3 B+. The Raspberry Pi also needs some sensing ability to know if the TV is on or off, so it can execute the turn on function when necessary, or turn it off when necessary. I am going to eliminate the IR sensor on the TV so it can't be messed with by remotes. I am going to be controlling two TV's in this fashion, independent of one another. My TV has the generic button circuit board that is typical on TV's and monitors. All in all, I am not sure how I will do this from a hardware perspective. <Q> You can use CEC (via HDMI) - I use libcec to turn my TV on and off with a RPi running a media centre. <S> It also allows you to monitor the power state of the TV with the RPi. <A> This would prevent other remotes from working by physically covering the sensor, and would avoid having to modify the TV internally. <S> Existing RPi tv control code can work. <S> The opto sensor is how you tell of the tv is on. <A> You did not specify which TV you want to control, so I don't know if it possible for you. <S> I did same for my TV which has RS232 communication for command interface. <S> So I opened TV and disconnected PCB with IR module (and hardware buttons) and connected to RS232 via level shifter, it worked perfectly.
| Simple method is using an IR Blaster cable, a IR transmitter tapped over the Tv's IR sensor, and a opto sensor over the Tv's power led.
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Can I connect DC12v on VAC output (2A 100-240V) relay? Solid State Relay G3MB This relay's output says "2 A at 100 to 240 VAC". If I wanted to connect something like DC12v 2A to this relay output, It is not going work? It might work, but it will damage the relay circuit? I know this is a very stupid question, I googled but couldn't find a page explaining the differences between VAC and VDC output relay. Hope somebody can clarify the differences.Thanks in advance. <Q> First be sure this relay is built with MOSFETs or SCRs. <S> If built with triacs they will NOT work with DC. <S> Worse case is you have to buy a compatable or conventional relay, else: <S> As long as you do NOT exceed the voltage or current rating of the output, it is a "Don't care" scenario. <S> With an AC rating you could change polarity of the DC voltage and it would still be "Don't care". <S> Be aware that you maybe running the relay at its current limit, so if it gets hot put it on a large heatsink. <S> Since it expects a much higher voltage it may have some internal resistance that will cause a voltage drop under full load. <S> If less than a 1 volt drop I would just live with it. <S> If the drop is several volts than you need a source greater than 12 volts to make up for the loss. <A> NO you cannot. <S> Nada, no way <S> Jose. <S> 決して、今日ではなく、決して <S> This is a Triac controlled SSR and these latch ON with DC. <S> Read the fine print in the 1st page summary. <S> "The G3MB-202PEG-4-DC20MA crosses directly to the Motorola M0C2A-60 series power triac." <A> Voltage and current inside a relay works in a slightly different way. <S> First of all, current will create heat inside the relay. <S> In your case: 2A. Therefore, no matter what the voltage is, you will need to be bellow 2A on your load. <S> For the voltage, you need to picture how a relay is made inside your head. <S> A metal piece is moving back and forth creating or cutting a contact between two port. <S> That metal piece is controlled by a electric coil that clamp the contact on. <S> When the contact is not clamped, there is a distance in between the contact and the metal piece. <S> While air is a nice dielectric (insulator) there is a fixed voltage that it can withstand before it begins to be a conductor. <S> Think of lightning, electric ark, welding. <S> Therefore, if the voltage is above the rated voltage, you will most likely create an ark inside your relay and this is not a desirable thing at all. <S> In the case of a solid state relay (your case) the switching is done with a semiconductor. <S> Therefore, the dielectric will be different and there won't be any moving parts inside it. <S> But, exceeding the voltage may break down the device or cause it to ark in surprising ways.
| Depending on the conductor size, you will likely have a maximum current that can pass through the relay without burning it up.
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Using latching Hall sensor to toggle a LED - brightness issue I'm using a U18 latching Hall sensor ( datasheet ) to toggle a LED. I'm driving it off a 6V L1016 battery. My issue is that the LED is very dim. My understanding is that the U18 limits the output current to 20~ish mA. In theory my LED wants 20mA forward current . So I am not sure why it is not very bright. Testing the LED directly with the same battery both 150 & 300 ohms resistance it's much brighter. simulate this circuit – Schematic created using CircuitLab Original and tidied-up schematic. How can I make this brighter? Would the best solution would be an op-amp? Is there something else I've overlooked (in terms of setting it up incorrectly)? <Q> You need to use a small mosfet or transistor so that you can have the right current on the LED. <S> If the output is an open collector output, you can do it with a PNP (supposedly you want the led on when it activates.) <A> Output current at 5V is 5.5 mA. <S> It's also an open collector output, so it pulls to ground when triggered. <S> Your schematic as drawn would not work. <S> At minimum the led is backwards. <A> That device has an open-collector output - the LED and its current-limiting resistor must be connected between the positive supply and the device output terminal - like so: simulate this circuit – <S> Schematic created using CircuitLab
| The hall sensor is not designed to have enough current to drive an LED.
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Trying to identify threaded round 2-way power connector on a radio I'm trying to find a power connector for a Maxon DM200 Mobitex radio terminal. As far as I'm aware, the part of he manufacturer who made this radio closed down some years ago, and there's no support available. I've uploaded some photos of the connector to Imgur: https://imgur.com/a/1rztAaS There's a logo on the back of the connector which looks like a letter "W" in a circle. Here are front and rear views of it. Here is the top view of the connector. Some facts and figures: Connector mating section thread -- I measured this at 12.7mm, it could be a loose-tolerance 13mm. There are only 3 complete threads. The back mounting nut isn't shaped (eg hex), it's completely round with two notches. I guess you'd use a pin spanner to install and remove it. The panel hole seems to be around 12mm diameter. There's a foam or rubber gasket (hard to tell which, it's perished) between the socket's bezel and the front panel. It clearly pushes in from the front. Pins are 1.5mm diameter and 2mm diameter. This is the only polarity keying I can see on the connector; there's no bump or notch polarising aid like the GX16 connector and most screw-in DIN plugs have. Pin spacing is about 4mm. Does anyone know what this connector might be? I've ruled out GX12 and GX16, but either might work as a replacement. <Q> By the looks of it and the thread I'd say it's a 2-pin aviation connector (such as the one depicted here: https://www.ebay.com/itm/Hot-Sale-1-Set-GX16-2-Pin-Screw-Type-Electrical-Aviation-Plug-Socket-Connector/201465426325 ). <S> You could give one of these aviation connectors a try and if it doesn't work out just unscrew the whole connector from behind and solder in something more standardized instead (perhaps even the said aviation connector). <A> It is a 2 position and is threaded. <S> MFG part number SL102M <S> https://www.digikey.com/product-detail/en/switchcraft-inc/SL102M/SL102M-ND/413399 <A> This is the only information I have found its a custom cable and connector made only for that radio and is £50 hope it helps maybe contact the person in the thread and see if he knows where to get one from, https://www.eevblog.com/forum/projects/need-help-iding-a-12-7mm-2-pin-round-connector/
| However this one has one of the pins wider than the other (and the center notch is missing too), so this could be one of those lousy "proprietary" designs derived from the (more widely used) aviation connector design. I know it is not exact, however switchcraft has a similar part.
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Is "humming" normal in an AC relay? I have a 240V relay that uses a 120VAC coil. When I switch power to the coil, the relay makes a faint humming sound. It isn't very loud, and sounds like a transformer almost. A normal speaking voice or small fan in the room is enough to drown it out, to give you an idea of the volume. To be clear, this isn't a situation where the coil isn't getting enough power and contacts open and close rapidly (described as "buzz" in other questions). I have verified that the voltage is correct, and I have observed that the armature is still (not vibrating) when the coil is getting power. So I am wondering if I have a bad relay, or if this is normal for AC relays, which I haven't used before. I am used to 12VDC relays for the record. <Q> Figure 1. <S> Source: Machine Design . <S> Since the coil is powered by alternating current the magnetic field collapses to zero at each mains crossing and the armature tends to start to release. <S> Its inertia is high enough that the contacts remain actuated long enough to maintain contact through each zero-cross of the mains. <S> The buzz is normal. <S> It is caused by the vibration of the armature on the yoke on each half cycle. <S> Just a note on relay terminology <S> : "I have a 240V relay that uses a 120VAC coil" is a little confusing. <S> "I have a relay with 240 V contacts and a 120 V AC coil" would be clearer. <S> Update: Spehro and Tony's answers both address the use of 'shading' poles on the armature to help maintain force during zero-cross. <S> This in turn will reduce the vibration. <S> Figure 2. <S> The yoke of an AC 'contactor' (high-powered relay generally used for AC motor circuits, etc.) <S> showing two shaded poles. <S> Image source . <A> AC-coil relays use a shading ring that helps smooth out the force from the varying field. <S> Photo from this website. <S> The shading ring acts as a shorted turn that causes an out of phase current so the total field does not drop to zero. <S> It still varies though. <S> It's normal for AC-coil relays to hum acoustically at a multiple of mains frequency. <S> They can excite resonances in mechanical bits causing the noise to become objectionable. <S> Large industrial contactors can make quite a racket. <S> DC-coil relays do not generally hum noticeably even when carrying AC <S> (though if the contact currents are very large there can be some effect from the magnetic fields. <A> This is the shading pole effect or moving iron type of an AC Relay which draw 2x power of DC relay or more. <S> (Normal) <S> But contacts must snap for long life from short arc duration. <S> If " I have verified that the voltage is correct," then disregard/// get the right Relay. <S> Everything sounds normal.
| The relay coil, when energised, pulls in the armature to actuate the contacts.
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Practices for using a scope, scope probe and termination resistor for a proper measurement setup In a setup, assuming a function generator has 50 Ohm output impedance and a coaxial cable has 50 Ohm characteristic impedance. Therefore to prevent reflection of a signal a 50 Ohm termination is used. But for many people I know they dont use 50 Ohm termination for basic use(non RF) and they also use 1X probe with scope's 1X setting. And if we use a 50 Ohm termination then our signal's amplitude halves. How is this problem compensated? By setting the scope to 2X? How about the probe? 1X or 10X? What is the proper probe/scope and termination resistor setup depending on the signal frequency? Can we roughly make a category? I want to learn the proper measuring technique with scope and probe depending on the waveform and frequency. Since these are based on experience I cant find a compact info about it. <Q> Ideally, an oscilloscope would give you an option between 1M ohm input resistance (in parallel with perhaps 15pf capacitance) and a 50-ohm input termination - often cautioned not to apply too much power, lest it over-heat. <S> This 50-ohm internal termination is quite vulnerable to burning up into an open-circuit, so it is wise to do a measurement check to see that it is still there, and <S> still 50-ohms. <S> Many 'scopes give you no option - the default 1M ohm input applies. <S> In this case find a BNC "T", and add a 50-ohm termination right at the 'scope input. <S> Use the 1X scale, and use 50-ohm coax to connect to your signal source, not the 1X probe. <S> Not as good as the internal termination described above, (a short unterminated section often remains), but its about the best you can do. <S> Don't forget that its there - when you go to measure that +24V DC supply, you may smell smoke. <S> Most 1X/10X probes have cable impedance higher than 50 ohms between probe tip and BNC connector, so don't use a probe in a 50-ohm measuring system to make a careful amplitude measurement. <S> A good function generator takes care to drive its output through a 50-ohm resistance, so its open-circuit output should be twice as high as its 50-ohm terminated output. <S> It is common for function generator outputs to be 20V p-p open-circuit, and 10V p-p when terminated with 50 ohms. <S> Such a generator might say in its manual, "Will deliver 10v p-p to a 50 ohm load". <S> A source calibrated to deliver 0 dBm power will deliver that power into a proper load (usually 50 ohm) - of course it delivers no power to an open-circuit, and very little more to a 1M 'scope input. <S> Your oscilloscope (set to 1X) might make <S> an RMS amplitude measurement - power is simply \$ \frac {{ <S> V_{rms}}^2}{50} \$ <S> when you've got that BNC-T with 50 ohm termination attached. <A> I simple terms: Your default <S> everyday setup with a low-mid range oscilloscope should always be to use 1M input impedance probes and scope set to 10X if available. <S> This setup is non-invasive, and gives you good bandwidth. <S> If you are a beginner, you can basically stop reading here. <S> 1X mode is meaningful basically only when you need to measure voltages that are too low for 10X mode. <S> Which is never. <S> Almost. <S> 50ohm input mode is only used in certain more advanced cases, where you want to push the bandwidth or have the scope simulate a missing load to avoid reflections. <S> If it is connected usign a probe, then you shall use 1M mode. <S> If you connect the signal generator to a load, and hook up the scope halfway on the coax using a T-connctor, then 1M is the appropriate mode.50ohm is not "better" than 1M. <S> It is "different", and shall only be used when appropriate. <S> On a side note, if 10X mode does not offer sufficient bandwidth for a measurement you need to make, and you can't afford high frequency active probes, then there is a neat trick you can do with the 50ohm mode. <S> By using 50ohm mode, a 50ohm coaxial cable, and a series resistor at your measurement point, you can get the bandwidth of an active probe. <S> With the disadvantage of lower impedance. <S> But for digital signals this is often acceptable. <S> The method is explained here: https://www.youtube.com/watch?v=9GqfZMcrAFY <A> And if we use a 50 Ohm termination then our signal's amplitude halves. <S> How is this problem compensated? <S> There is no "problem" at all. <S> Signal is whatever it is, and scope probe has nothing to do with the reality of it. <S> If you don't use 50-Ohm termination, the signal "doubles" as compared to what the signal generator dial assumes (although there could be scaling options depending on assumed termination setup. <S> But is done for dial convenience only). <S> If you load the other cable end with 50-Ohm terminator, the voltage will be half of that. <S> It is what it is, and you should always take measurements with high-impedance probe ( <S> > <S> > than 50 Ohms). <S> There are coax probes with 2K input impedance, or even with 500-Ohms, which are still qualified as "non-invasive" for 50-Ohms signaling environment, technically. <S> There are however situations when you want to use INTERNAL 50-Ohm IMPEDANCE input mode that can be found in modern high-speed digital scopes, for direct connection to coax cables. <S> In this case everything gets matched automatically, yet the signal is still what is is, and no "2x correction" is necessary.
| If you connect a signal generator straight to the scope using a coax with no other load on the coax, then 50ohm mode is appropriate.
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Is a modern radio wave broadcasting station a very fast spark gap transmitter? In a spark gap transmitter, is that a "real" radio wave that is produced? I know you can hear it on an AM radio, if you are close to the gap, but is it a radio wave the same as from a radio station? Except from a radio station or modern transmitter the "gap transmitter" would be very, very fast? <Q> In a spark gap transmitter, is that a "real" radio wave that is produced? <S> It's the antenna that converts the spark induced carrier wave (controlled by an LC tuned circuit) voltage to a radio wave. <S> It does so by matching the impedance of the cyclic voltage and current oscillations (low impedance or low V to I ratio) to 377 ohms where 377 ohms is the impedance of free space or a vacuum. <S> All radio antennas do this and, without a proper antenna, you will create a "radio" signal that doesn't travel effectively through the medium of space or air because it won't have E and H fields in the correct proportion for an optimized transmission. <S> is it a radio wave the same as from a radio station? <S> It will be a radio wave <S> but it will be an on-off carrier where the on and the off are determined by a morse key that is used to "code" language into dots and dashes: - <S> After the spark gap circuit and with the invention of valves/tubes, proper modulation (as opposed to on/off keying) could be implemented and radio receivers became much more sensitive. <S> That marks the big difference in tech that emerged and basically the same principles exist today. <A> Modern transmitters generate the signal electronically. <S> The signal from a modern transmitter occupies a very narrow range of frequencies, so many transmitters can operate without interference. <A> Radio waves are electromagnetic waves and are the same now as in the spark-gap transmission era. <S> The spark gap was a means of generating fast rise time <S> current pulses that would cause EM radiation. <S> Before electronic oscillators were available the highest frequencies of alternating current (AC) <S> that were available were supplied by alternators that were rotated rapidly with many poles. <S> These were large and cumbersome and had lots of practical limits to their maximum frequency. <S> The tuned circuit spark gap and the rotating spark gap were a period solution to a problem that had no other solution. <S> Tuned circuits were added later to control the transmit and receive frequency selectivity to achieve better range and bandwidth utilisation. <S> If you were to make a rotating spark gap (that makes an arc very rapidly) and couple it across a high-Q tank circuit (using superconducting inductors say) you could generate presentable on-off (CW) <S> EM carrier waves that look just like those generated with modern electronic oscillators and amplifiers. <S> They would however be very limited in modulation and tuning capabilities and very large and cumbersome to fabricate. <S> Might still be an interesting experiment.
| The transmissions from a spark gap transmitters were "real" radio waves, but very noisy, and a single transmitter would cover a wide range of frequencies.
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Two TVs use the same antenna, but one affects the other I’m having a problem with my TVs. I will call them TV1 and TV2. TV1 is Asanzo brand, and TV2 is LG 4K, which is much more expensive. After replacing the antenna wall socket of TV2 (The TV is near to me in the video), the silly thing happens. Every time I turn off the TV1, TV2 has very weak or no signal. To watch in TV2, I have to turn on TV1 (even when the two TV have different channels) . But when I turn off TV2, TV1 still works normally. (update: disconnect TV1 from power supply and antenna wire, TV2 still won't work.) Video: https://www.youtube.com/watch?v=COGERGH66oU The two TVs have integrated DVB decoder, use the same antenna with USB power for each one. The antenna has amplifier circuit. I’m not sure if it’s caused by the new wall socket, because I had replaced it very carefully. Hope to hear your explanation about this and of course the solution. I don’t want to have both TVs on all the time just to watch one. Thank you! <Q> It seems to me the problem is your antenna is only ever powered through the USB adapter at TV1, which is plugged into TV1 . <S> If you turn off TV1, the USB port has no power and neither your antenna gets any power. <S> The power adapter at TV2 is redundant. <S> It can't power the antenna because there's no power pass through the power adapter plugged into TV1. <S> Power <S> the antenna through a mains-USB adapter instead. <A> Something that you mentioned caught my eye. <S> When both TV's are on same channel <S> one TV doesn't work. <S> Both TV's on different channels both TV's work. <S> That in itself rules out any antenna issue. <S> It could be the IF frequency. <S> Google superheterodyne radio design . <S> When a TV or radio receives the stations frequency it is mixed with the local oscillator to generate the intermediate frequency. <S> Same channel = same IF , different IF channel = different. <S> If the TV's are in relative close proximity this could be the issue. <S> If the antenna amplifiers are phantom powered you should have a DC block on the input of each TV. <A> It's quite possible is that you need to correctly split the antenna signal into two TV feeds and do so without disrupting the antenna impedance and/or creating standing waves. <S> This is what the problem sounds like. <S> For instance doing this: - IS WRONG!!! <S> Dong this on the other hand <S> : - Is much more likely to work because it splits the power to both TVs correctly and provides an impedance to the antenna that remains at 75 ohm. <S> If you want the theory read this website . <S> If you think you can treat an antenna feedline like you can (say) an AC power line and plug in appliances all to one circuit then think again. <S> We're talking about transmission line theory and impedance matching when we are dealing with high frequency signals on a line that is many times the wavelength of the signal.
| It could be a simple RF interference issue.
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Is a 2 Ampere power supply dangerous? I'm sorry if this question seems extremly dumb or/and pathetic. I'm quite new to electronics.For one of my electronics projects, I want to buy a power supply like this one , but it says Output: 12V 2A and I know that 2 amps is dangerous for the human body.So what does it mean? Will I hurt myself if I touch the two cables or is it fine? I don't really understand this.I'm sure you all know the answer and if so, please explain why.Thank you! EDIT: My question has been marked as a duplicate but it's different. The other question is about what voltages / currents are dangerous but I do now that. However, I'm asking if the 2A power supply always outputs 2 amps or if it is the maximum current it CAN output. <Q> Since you have clarified your question, the answer is as follows: <S> 2A is the maximum current it can deliver. <S> It will always deliver 12V - as long as you don't put a load on it such that it would have to deliver more than 2A. <S> Then, the voltage will drop. <S> Either because it over heats and burns out or because the engineer who designed made it so that it safely limits the current so as not to burn out and be a fire hazard. <S> The current that really flows depends on the resistance of the connected load and the voltage. <S> Ohms law says this E=IR, where E is voltage, I is current, and R is resistance. <S> It can be rewritten like this: I= E/R. <S> This is what we want, since it is the current that can kill you. <S> Your body can be thought of as a resistor. <S> It has a resistance of several thousand ohms. <S> It can vary from 1000 ohms to 100000 ohms. <S> Lets stick with the middle field: <S> 10000 ohms. <S> I= E/R I= 12/10000 <S> I= 0.0012A <S> I=1.2mA <S> So, your power supply can deliver 2A, but it can only force 1.2 thousandths of an ampere through your body. <S> Always buy your power supplies from trustworthy sources. <S> Switching power supplies must be properly made in order to isolate the ouput from the high voltage input. <S> A cheaply made powersupply might skimp on the isolation, and allow the output to be at a high voltage compared to the ground. <S> There will be the rated voltage between the two output terminals, but the full line voltage between one of the outputs and ground. <S> THAT can kill you, but has nothing to do with the rated output current or voltage. <S> That kind of thing is poor design and manufacturing. <A> Power supplies are labelled with nominal voltage but maximum current. <S> The current, up to the limit of the supply, is determined by the load. <S> If you have no load (open circuit) you will get no current. <S> Just like the many mains sockets in your house that don't have anything plugged into them, no current is flowing out of those sockets. <S> If you connect a 1000 ohm resistor across a 12V supply you will get 12/1000 <S> = 12mA flowing, which is less than the 2A that your supply can supply <S> so the voltage will remain at 12V. <S> If you reduce the value of the resistor, more current will flow, but when you get close to 2A, the voltage will fall towards 0. <S> You won't get 2A through your body from a 12V supply because of the resistance of your body. <A> Everything you need to know can be found here <S> This is one of many Youtube videos made by a real character who goes by the handle of ElectroBOOM. <S> He clearly knows a good deal about safety, and is willing to bend the rules to the breaking point in order to both educate and entertain. <S> Another useful demonstration he provides is here . <S> And yes, touching 120 VAC hurts. <S> A bunch. <S> At about 1 minute into the video he applies 25 volts to his tongue with a 30 amp supply. <S> Just so you won't need to. <A> Product Safety is supposed to be by design and adherence to IEC Compliance and OEM company credibility to adhering to good safety designs. <S> If the company appears to have longevity and good standards, you can trust it is safe, but verify by looking for standards compliance specs. <S> First impressions look ok for this LED driver. <S> While Asian e-skateboard designs are not. <A> According to many lectures, anything beyond 100mA can produce an hypothetical heartattack. <S> beyond 300mA you can easily get molest superficial skinburns. <S> and using switching powersupplies theoretically can really burn your fingers and salty skin beyond 300mA. <S> Thus. <S> Yes.
| 2 ampere can be relatively dangerous (eventually not lethal) to a person depending if it's switching or not according to us. So, your powersupply is safe for you to use and to touch with your hands. 2A can damage skin and muscle and produce severe involuntary muscular movement and cause dramatic laboratory damage.
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Calculate how much heat an IC produces I'm building a device containing a large number of ICs. Now, I'm 99% certain heat won't be an issue. (Considering the volts and amps involved, I doubt you could accurately detect the amount of heat involved.) But, just for argument's sake, how would you go about calculating this stuff? My first thought was to look at the datasheet. But it doesn't seem to say anywhere "this chip will produce X units of heat in normal operation". (Presumably because there's too many different variables that affect it, so they can't easily come up with a definitive number.) The only relevant thing I can see is a section on "thermal resistence". If I'm understanding this currently, this is a measure of how quickly any heat generated would be able to escape the casing. (Presumably depending on how hot inside vs how hot outside; it seems to be expressed in units of °C/W.) Clearly thermal resistence is part of the equation. But without knowing how much heat per second the IC produces in the first place, I'm not sure where to start with this. <Q> Here is rather simplified answer: with few exceptions (like light emitting or RF radiating) the electronic circuits convert all incoming power into heat. <S> Of course, if you want to calculate it in advance or predict temperature in various conditions you need all those things described in @SpehroPefhany's answer. <A> Power is measured in watts (joules per second). <S> You calculate the dissipation by understanding the IC and how much current it draws from what voltage(s) under the conditions you will be using it. <S> Some parts draw current just sitting there <S> that is significant, others use almost no power until you clock the part, then the power may be almost proportional to the clock frequency. <S> Once you know the power, you have to understand how it is transferred out to the environment. <S> One component of that is how it gets from the die to the outside of the IC package. <S> Eventually it has to find its way to the air, water, earth, be radiated into space, etc, or the temperature will continue to rise until the part fails. <S> Whole careers are built on thermal management, and of course books are written. <S> Intuitively you can get an idea by looking at devices around you. <S> Something the size of your fist that draws a few watts may be okay if the power is not too concentrated inside, for example. <S> It also may fail in the blink of an eye if the power is concentrated in a small chip that is poorly coupled to the outside (too much thermal resistance). <S> Thermal resistance implies a linear relationship between temperature rise and power, which may or may not be true. <S> Radiation and convention are far from linear, but conduction is and mass transport (eg. water cooling) may be. <A> My first thought was to look at the datasheet. <S> But it doesn't seem to say anywhere "this chip will produce X units of heat in normal operation". <S> Remember that $$P = <S> I <S> \times V$$ <S> That is, power (and, thus, thermal dissipation) is equal to current multiplied by voltage. <S> The datasheet may not explicitly list thermal dissipation, but it should list a maximum current draw; multiply that by your working voltage <S> and you have your number.
| So, if you measure the power consumed by your assembled device you can get pretty good approximation of the heat to be dissipated.
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How could using an ungrounded appliance with a grounded extension cord be a fire hazard? My local fire department, @Barrie_Fire, recently tweeted (then subsequently deleted) this: Don’t even THINK about using a 2-prong plug in a 3-hole slot! Use only the required number of slots in an outlet or power strip. Below was a picture of a burned-out grounded extension cord. I'm hesitant to argue with anyone in the business of keeping our food, shelter, clothing and loved ones from combining with oxygen, but this seemed quite strange; I can't think of any possible way this could be a fire hazard. The NEMA 5-15 wall receptacles in Canada are grounded by default, for reference. <Q> The Fire Dept is wrong - it is perfectly normal to plug a device with a 2-pin plug into a 3-hole socket. <S> If a high-current load, like an electric heater, was plugged into that burned outlet, and the contacts made poor contact, that would cause the overheating and resulting fire, whether the heater had a 2 or 3 pin plug. <A> The statement by "Barrie" is nonsense, many devices don't have a ground connection and thus only have a 2-pin mains plug. <S> Such devices are "double insulated" and have the 2-squares logo and possibly some text like: <S> The damage of that power strip was very likely caused by a short circuit and/or overload. <S> There is no ground/earth connection required for that to happen. <S> Likewise it is unlikely that a ground/earth connection would have prevented that damage from happening. <A> Connection a 2-prong device into a 3-prong socket is OK. <S> Properly designed 2-prong devices are isolated and don't need protective earth. <S> What is dangerous is plugging a 3-prong device into a 2-prong socket, or using a 2-prong extension cable with a 3-prong device. <S> That will cut the protective earth wire on an appliance which needs it, exposing the user to electric shock upon failure inside the appliance.
| Breaking the ground pin off a 3-pin plug, then plugging that into a 2-hole or 3-hole socket may produce an electrical hazard - possibility of a shock.
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Is it normal/common to use this way of connecting point to point wire? I want to make a circuit and am uncertain if the following is feasible/common and if there are better solution. From the pin headers (the light blue circles), e.g. GndIn (upper yellow circle), I want to use 24AWG stranded wire to three other places on my protoboard (and one other), see the lower yellow circle. Should I: Solder it like in the picture (one pin header with three wires soldered directly to the pin header); I can imagine it's a bit hard to solder three wires at the same time', maybe I should connect them first and solder them as one wire Make a vertical solder line until three pins below the GndIn pin header and solder each wire to a separate protoboard hole? However, this takes more space. Use another solution? And I have actually the same questions for IC pins (not clearly shown in my example picture). The protoboard I use is this type: <Q> With so many unused holes, you can be more creative in routing jumper wires to holes and expand the number common pads. <S> Use the adjacent row if necessary. <S> Just route then as neat as you can, like a PCB layout without clusters of overlapped wires to avoid signal crosstalk and so solder joints can be clearly inspected. <S> It doesn't take any longer and if the wires are tight and routed tight or bent in right angles, it will look better. <S> Snake wiring looks a bit suspect to prospective clients or employers. <S> Flush rectangular routing without overlap looks well planned. <S> ( as long as you know it is permanent) <S> Don’t use excess solder. <S> Get the right solder temp to allow you to solder in 2~3 seconds by preheating and then add solder wire in a smooth sequence then release. <S> Magnet wire is popular for thinner appearance <S> and I just burn thru the varnish without inhaling or use a fume extractor. <A> My recommendation would be to only have one or two wires per pad and to daisy-chain your connections. <S> It will make soldering easier and fault-finding less confusing. <S> On the other hand, if you can get three wires through the hole then there's nothing to stop you soldering all three. <S> Figure 1. <S> The underside of a project made on Veroboard. <S> Note the neat solder joints and the holes drilled through to break the tracks. <S> Image source: Wikipedia Veroboard . <S> Figure 2. <S> A Vero tool is rotated by hand to remove copper around any hole thereby isolating the strips either side of that hole. <S> This tool doesn't go right through the board as in Figure 1. <S> Figure 3. <S> Rather neat circuits can be built on Veroboard if the components can be mounted perpendicularly to the direction of the copper tracks. <S> Here there are no flying wires and <S> all cross-connections are done with components or wire links. <S> Image: Wikimedia Commons . <S> Other options such as solderless breadboard patterned versions exist also. <S> Figure 4. <S> The breadboard-patterned version. <S> Source: Wikimedia Commons . <S> There are many stripboard designers available which allow you to place components, links and break tracks. <S> These can help to generate a plan for a very neat layout. <A> If you have a component inserted in a hole, it may be easier to solder the wire to that hole/component lead, rather than putting the wire in an adjacent hole, then trying to connect the two holes. <S> (You'll find that solder will bridge large gaps where you don't want it to, but won't bridge a microscopic gap where you do want it to).
| Trained assemblers will use instant adhesive dots sparingly to prevent long loose wires. Unlike stripboard or the plastic breadboards, that board appears to have no connections between holes, so you have to make any between-hole connections yourself. The standard strip-board or Veroboard is much simpler to use and very versatile. I've only used that type of board a couple of times and found it very difficult to make a good looking job.
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How to measure voltage of a floating GPIO (Raspberry Pi) I would like to measure that voltage of a floating GPIO on a Raspberry Pi 3. When I use a multi meter or an oscilloscope it looks like the pin is pulled to ground and shows 0. Another way of asking the question is, 'How can I determine if a GPIO pin is floating using an oscilloscope or multi-meter?" How can I use an oscilloscope to measure the voltage of a floating GPIO pin on a Raspberry Pi? <Q> One simple way is to measure the voltage to ground and then measure it to Vdd (which is 3.3V on the Raspberry Pi SOC). <S> Since most voltmeters have something like 10M input resistance you've effectively got a resistor across the meter input. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If the meter reads about zero <S> both ways then it is relatively high impedance (if there was 10nA leakage it would read 0.1 Volt). <S> Leakage in the high impedance state is typically very loosely specified but it is normally extremely low at room temperature, so this method does work. <S> If it reads non-zero <S> but both voltages add up to Vdd <S> then it is probably fighting another output or is being switched at a high frequency <S> (an oscilloscope will tell you the difference). <S> This is the 4th combination of measurements, the first two are obviously high and low. <A> How to measure voltage of a floating GPIO (Raspberry Pi) <S> The most practical answer is "You don't". <S> A floating CMOS input is very high impedance. <S> That is usually 10s of MΩ at least, often 100s of MΩ in reasonable humidity with clean boards. <S> Your question then really comes down to "How do I measure a voltage with a few 100 MΩ impedance?". <S> Again, the realistic answer for someone that has to ask here is "You don't". <S> Typical voltmeters and oscilloscopes with 10x probes have 10 MΩ input impedance. <S> It should be obvious that connecting 10 MΩ to a 100 MΩ source is going to drastically change the source voltage. <S> There is specialized equipment and specialized techniques that can measure voltages at such high impedances, but these are not easily accessible to amateurs. <S> Then there is the question of what the purpose is? <S> After all, it's floating , so can pick up any random voltage for unpredictable reasons. <S> What information would you get out of knowing what the voltage is at one particular time? <S> Your whole question therefore makes little sense. <A> With " floating " I think you meant the Z mode (high impedance) of a tri-state pin. <S> This is difficult to measure, since if you touch it, it stops being high impedance <S> Ideally you need a SMU (source measure unit) , a fancy power supply, to measure the leakage current of the GPIO. <S> With those numbers you can calculate the expected "floating voltage". <S> ( image source ) <S> Since a SMU is expensive equipment, few people have access to this high end equipment. <S> Broadcom has measured these numbers in their characterization, and these are published in the datasheet of the BCM2837B. <S> However, it's is generally recommended to avoid the high impedance mode on any gpio, except analog. <S> Either use external pull-up/down resistors, or enable the internal weak pull resistor when you're able to. " <S> Floating" pins can cause sporadic issues in the field when the pins are used by hardware or code. <S> Pins that are not in use are sometimes left "floating" to save energy, but on an RPI3 this is insignificant. <A> Assumptions for the purposes of this answer. <S> Nothing is connected to the pin (open circuit). <S> You have set the pin to float in the code. <S> You want to check if the code has actually set the pin floating. <S> How can I determine if a GPIO pin is floating using an oscilloscope or multi-meter? <S> If the voltage follows the pull, then the pin is floating. <S> If the voltage stays low when you try to pull it high (or stays high when you try to pull it low) then the pin isn't floating. <S> I'd start with a 10kΩ for the pull resistance. <S> Be aware that the microprocessor may have its own internal pull resistors that are controlled by software. <A> You can try to measure the voltage by any means you have. <S> But consider that there is no such thing as truly floating. <S> Modern GPIO pads on MCUs have many dozens of semiconductor structures attached to it, in buffer, output (disabled) buffer, ESD structures, etc., and every structure (diode, transistor/switch) would have certain leakage (conductance), which can be in both directions, to Vcc rail, or ground. <S> So the sum of leakages is hardly predictable. <S> Manufacturers only guarantee that the leakage won't exceed, say, <S> +-1 uA, which comes to about 3 MOhms and above. <S> So you would need a voltmeter/probe with 100 MOhms input impedance, but the measurement won't make much sense due to the above considerations. <S> When you try to use a 20-K DMM or even 1 MOhm scope probe, the instrument input impedance will drag the pin down and you will read zeros. <S> Many MCU vendors have "weak pulls" to create definite voltages on unconfigured pins during power-up sequence, to avoid uncertainties. <S> For some details on GPIO leakages see this answer . <S> The simplest way to determine is a GPIO pin is "floating" <S> is to put a 10-MOhm standard scope probe on it, and touch the pin with finger. <S> If you see 50-60 Hz rail-to-rail waveform, the pin is floating.
| To check if the GPIO pin is floating, attempt to pull the pin up to the logic voltage supply (or down to ground) using a resistor.
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Is an unenclosed cable modem circuit board a fire hazard? I've come up with a clever hack to save space and reduce # power adapter eye sores by bolting my cable modem (Netgear CM500) under the Wifi router (WRT 3200ACM). The heat sink over the Broadcom baseband processor is very hot to touch. I measured ~70C for the processor and ~55C for the RF (de)modulator enclosure (had to set mode to glossy or else readings for shiny surfaces are wildly underestimated). Modem consumes 8W max. As clever as it seems, I was worried if there's any fire hazard or other problems if I were to let the heatsink lie on top of a a wood surface or carpet. I had a read some of this guide, https://incompliancemag.com/article/electrically-caused-fire-in-multilayer-circuit-boards/ and the main take away was that for fire, it had to generate a temperature beyond the ignition temperature. But I didn't see any way to calculate the temperature. That would be a function of the power density and thermal resistance of the heating element. Also, it said to distinguish between normal conditions and fault conditions.Normal condition of ~70C already feels worrisome. But most developer boards for embedded platforms are unenclosed. I guess they would've thought about fire hazards also? <Q> 70 c is not enough to spontaneously ignite paper let alone wood. <S> The carpet depends on the material. <S> But there is no reason you have the heatsink touching anything. <S> Use stand offs. <S> That said, of you do catch your house on fire <S> , I'm sure your insurance would rule that cable modem as voiding your insurance. <A> I doubt that it is a fire hazard <S> but it will likely put both the cable modem and the router in a higher temperature environment than they were designed to handle. <S> This will potentially increase error rates, cause random crashes, or even cause the devices to fail entirely. <S> The modem is normally vertically oriented so that convection currents help cool the heatsink. <S> The trapped heat will cook both the modem and to a lesser extent the router. <S> If you really want to do this, sit the whole thing on top of a mini laptop cooling fan. <A> It is easier to mechanically damage the circuit board. <S> This could cause a short circuit, and thus fire. <S> You have removed the fire retardant plastic casing, meaning you can place other flammable materials near it. <S> Yes, you have <S> increased the fire hazard.
| It is not an direct fire hazard, it's still a low power device using an AD/DC adapter. With the modem sitting under the router, the heat can't escape as easily.
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Understanding the Maximum Speed that can be Transmitted over a Cable I am trying to source a FFC/FPC cable for USB3.0 (+5gbps). This brought me to the question of signal transmission. I'm somewhat of a beginner in this topic. I know that you should match impedance on your PCB with your connector/cable to minimize reflections. I was wondering how to tell how fast of a signal you can transmit over a wire. Specifically what kind of cable parameters affect transmission speed?Any help is appreciated. <Q> The maximum frequency is mostly related to the frequency-dependent loss characteristics of the cable. <S> Eventually you get to a frequency where you simply don't get enough signal at the other end to use. <S> Resistive losses in the conductors (including skin effect) <S> Dielectric losses in the insulating materials Radiation losses if the cable is not fully shielded <S> All of these tend to increase with frequency. <S> This is why we generally switch to other technologies at very high frequencies: <S> waveguides for microwave radio equipment and optical fibers for high-speed data. <A> You can't just "source" FFC/FPC cables for USB 3.x. <S> These cables (and corresponding connectors) are not qualified for USB 3.x channels. <S> The cables for USB 3.0 have to meet many more requirements than just a wire parameters and not just having certain differential impedance. <S> To use non-standard (not within USB defined configuration) cables you will need to run all USB cable qualification tests on your own, ensure limits on insertion loss, NEXT/FEXT crosstalk, differential impedance across mated connectors, etc. <S> etc., if you want your product to work with any reasonable degree of reliability. <S> To run your own qualification you will need at least a 8-16GHz oscilloscope and a 20-GHz TDR (Time-domain reflectometer) instrument, plus make a dedicated break-out fixture to access signals in correct way. <S> The list of electrical requirements for USB 3.0 transmission lines is given in the following USB-IF document . <S> Although the document is mostly for qualification of standard cables and mating connectors, appendix to the document shows the general electrical requirements to meet. <A> The frequency that can be used inside a wire is highly dependant on the skin effect. <S> At low frequency, the signal will be equally distributed through most of the wire, with a higher frequency, the signal will be predominantly distributed around the perimeter of the wire (the ''skin''). <S> The wires that allow the best characteristics will always be really small and with multiple conductors to reduce the skin effect. <S> Despite this, the higher you go, the more loss you will get. <S> Then the protocol steps in and will increase the voltage, use twisted differential pairs and push the boundaries to the maximum until you need to switch to a different transfer technology altogether. <A> I was wondering how to tell how fast of a signal you can transmit over a wire. <S> Specifically what kind of cable parameters affect transmission speed? <S> Any help is appreciated. <S> The Comcast XB6 Cable Modem will do over 1.5 Gbps using your standard cablevision coax. <S> The speed is limited to your last-mile speed, otherwise it would be higher. <S> PCIe 5.0 does ~4GB/ <S> s (or x16 @ <S> ~128GB <S> /s). <S> A x1 connection, the smallest PCIe connection, has one lane made up of four wires. <S> It carries one bit per cycle in each direction. <S> So 2 pieces of wire can do ~2GB <S> /s in practice , in theory <S> you could squeeze some more out of it. <S> For plain wire coax cable is the fastest because it's shielded. <S> Along with shielding length is the next most important factor, with shortest distances (inches) being the best.
| Simply put, the bigger the wire, the lower the frequency it can carry without having signal loss caused by an increase of its impedance.
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240v Inline Fan running On 120v? I have a 240 inline fan that can be speed controlled, would it be safe to run the 2 hots to 2 seperate 120v outlets that are both on seperate breakers? or could I run the two 240v hot wires to the one hot wire of 120 with just less power? the fan will be used 3 hours per week <Q> The fan may or may not work on 120V <S> , it's impossible to say without more information. <S> It might be possible to rewire the fan to run on 120V--many appliance motors are dual-voltage, depending on how the windings are connected--but again, it's impossible to say without more information. <S> Your country may be different on the code front, but it's still not safe. <S> The unconnected male plug will be connected to 120V from the other circuit through the fan motor. <S> The motor is almost certainly low enough in impedance that you now have exposed contacts sitting at 120V with the ability to apply a lethal amount of current to anyone who comes into contact with them. <S> So don't do it. <S> Have an electrician run a proper 240V circuit or get a 120V fan. <A> AC consumer fan motors that are designed to be speed controlled can usually be speed controlled by voltage reduction. <S> However, you are asking about a 50 Hz, 240 V fan that should be supplied with 288 volts for full magnetic excitation with 60 Hz. <S> That makes 120 volts at 60 Hz which would be 42% of full voltage. <S> A fan and motor combination running at 42% of rated voltage is not likely to provide much air flow. <S> In addition, the motor may stall or run erratically. <S> If it runs smoothly and provides enough air flow, it is probably pk to run it on 120 V, 60 Hz. <S> Another question here mentioned a product with safety features that provides a means of using two outlets to get 240 volts. <S> As Transistor mentioned, being on two circuit breakers is not sufficient, the two circuits must be derived from different phases. <A> One option that you can try is to use a series capacitor with a value picked such that the capacitor resonates with the motor inductance at the power-line frequency. <S> This has worked very well for me when re-purposing large box-type fans removed from surplus equipment. <S> Fans are rated for 220 Vac / 60 Hz and I see about 180 - 190 Vac RMS across the fan winding. <S> Be sure to use a non-polarized capacitor(s) with a voltage rating of at least 200V. <S> If I recall correctly, capacitor value for my particular fans was about 12uF. <S> What you do is to grab an assortment of small capacitors and simply try various capacitor values while monitoring both motor voltage and air flow. <S> Add capacitors in parallel to build up towards the desired value. <S> The voltage will peak with a particular amount of capacitance. <S> As you increase the capacitance beyond that value, the voltage will start to drop again. <S> The fans that I performed this trick on have been working reliably since the early '80s.
| Running a 240V device from two separate 120V branch circuits is not safe or compliant with electrical codes in the US. Best not to risk it. Consider what happens when you have one (male) plug connected to one circuit, and the other (male) plug not plugged into anything.
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USB type-c power supply (up to 20V 5A) from DC source I have a custom made 5s power bank (30 x 18650 cells) and I want to make a power supply to charge my laptop using this power bank. It's not a big deal to make a 5V charger because I just have to hard wire a 5V step down regulator and it will work, but what about higher voltages? Of course, I can use a standard AC power supply and connect it using a car power inverter, but efficiency in this case will be quite low. I tried to find such power supply/controller on aliexpress or ebay, but it looks like only 5V versions are available. What is the best/easiest way to do it? Thank you! <Q> Form the set of keywords "USB", "type-C", "power supply" "up to 20V 5A" and the literal question, <S> I want to make a power supply to charge my laptop using this power bank. <S> It's not a big deal to make a 5V charger ... <S> [snip], but <S> what about higher voltages? <S> it is obvious the that the question is about implementation of Power Delivery protocol over Type-C CC lines and design of power supply (controllable DC-DC converter) with variable output in accord with Power Delivery handshake protocol. <S> Answer: there are several semiconductor companies who offer a portfolio of ICs and design guidances on how to build a PD over Type-C. One is Texas Instruments, see their presentation on USB Developer Days 2017. <S> The other is ST Micro, <S> see their guide . <S> This is from just the first page of Google search. <S> There are several other companies like Maxim, Fairchild, NXP, ON Semi, Renesas/NEC, Rohm, and maybe many more Chinese sources with similar products. <A> I am aware of two such devices. <S> ChargerLAB POWER-Z DC to USB PD Converter Board JGL-DCPD review at https://kknews.cc/digital/e8mezky.html <S> Magicfox <S> PD180. <S> It takes a 12-28V DC via 5.5 * 2.5/2.1mm barrel (or USB C PD 100W) and provides two USB C ports and an USB A port. <S> A review <S> http://bbs.chongdiantou.com/thread-57387-1-1.html <S> here and the same is https://kknews.cc/zh-tw/digital/rorpmmv.html here as well. <S> Google Translate helps in translating the Chinese reviews. <S> The first one seems to be available from an Asian reseller in English, shipping worldwide (Google is your friend). <S> The second is on jd.com <S> but there are various Chinese shopping proxies to help. <A> If you are still looking for a solution, I found a DC to PowerDelivery converter which has reviews on Amazon. <S> I have not used this myself I a searching for a Solar Panel DC output to PowerDelivery converter. <S> I may eventually try this myself, but none of the reviews talked about solar panels, just AC power supplies with this device. <S> https://www.amazon.com/BiXPower-Type-Power-Converter-Delivery/dp/B07BC5TWJB/ref=sr_1_3?keywords=USB+Type+C+Power+Converter+bixpower&qid=1581808832&sr=8-3 <S> I thought power delivery devices would be more prevalent now, but only a few solar panels with PD output are available which are very expensive.
| Cypress Semiconductors offers a portfolio of reference designs as well.
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Methods for confirming on both sides that a signal was received in one direction Say there are two devices (A & B) that talk to each other (full duplex). For the sake of this example, let's say the medium is RF (wireless). If device A sends a message to device B, what methods/strategies are there for both devices to confirm that B received the message, and for both devices to be sure that the other device knows it? This is perhaps best explained through a real-world example: Say you have a full duplex RF key fob/remote for your car that controls the locks. The remote has an LCD screen. If you command the locks to close, the remote will send a signal to the car. If the car receives the signal properly, it can lock the doors immediately and then send an acknowledgment (ACK) back to the remote to say "the doors are locked" and the remote can show this on its screen. However, the remote may or may not receive the car's ACK. If the remote does not receive the car's ACK, the remote and user do not know if the doors are locked. At this point, the user can either attempt the process again, or give up and walk to the car, either of which are not desirable because it was unneeded. It may also cause further issues, such as the user expecting the car to be unlocked, and thus not bringing their keys to get an item out of the trunk. Thus there is desire for both remote and car to be "on the same page". Now let's consider the same sequence, except that the remote received the ACK. The remote now knows that the car received the message. It can display to the user that the doors are locked... But the car doesn't know that the remote knows. From the car's point of view, it's possible that the ACK was not received, thus the doors should not be locked yet or else we run into the issues mentioned in the last paragraph. Thus the car can wait for the the remote to send an ACK of its own to the car. If that is received, the car knows that the remote knows that the car knows to lock the doors... But does the remote know that? No! Seeing a pattern here? Both devices continue to pass ACKs back and forth. Neither can quit and say "I know 100% that the other device knows I'm going to perform my action" . Is there any way for both remote and car to be sure that the doors are locked and the remote shows it? If that's not possible, what strategies other than simple brute force (car repeatedly sending ACKs) are there for being pretty sure that the doors are locked and the remote shows it as such? <Q> Mmm! <S> The usual way mimics the way we exchange information ourselves. <S> "Here is a message to which I will need a response. <S> The message is "lock the doors when you get my OK." <S> One good way for the car to respond would be to send back the same message but with an ACK1 code attached. <S> It is waiting now for the remote to repeat the command perhaps with a slightly different ack2 code. <S> However I think you have in mind something much more critical than car door locking & unlocking in mind. <S> How about "landing gear locked down" on an aircraft? <S> These systems are called open & closed loop. <S> An open loop just carries out the action. <S> There are many ways in which a message can be checked for its authenticity. <S> The addition of check characters at the end is once such method. <S> Cyclic Check Character (CCC) or Block Check Character (BCC) are ways of counting say, the number of "ones" in a digital message. <S> A bit like counting the numbers of "A's" in a text would give you a better check on integrity. <S> At the end of the day if of vital importance the message can be sent twice or more times and compared to make sure it is faithfully received. {faithfully received - faithfully received}! <S> It will not be actioned until a previously agreed number of similar messages are exchanged. <S> I hope this opens up the possibilities of how it is possible to proceed with pretty near bombproof security and authenticity and answers your query sufficiently. <S> This in no way considers the question of interception or rogue generation of commands. <S> That is more within the sphere of cyphers & codes - a rather different ballgame. <A> As mentioned in the comments, this is known as the Two Generals' Problem , and is has been well studied. <S> It has been proven that there is no perfect algorithm which will always lead to both parties agreeing and knowing the other has too. <S> But if you are willing to send a lot of messages, you can get arbitrarily close. <S> The "best" strategy will be determined by the specifics of the problem. <A> In computer networking, the requesting side usually resends the request until it gets an ACK. <S> If the ACK is lost, then a duplicate request will be received. <S> Example exchange: <S> Remote: "Lock the door." <S> Car: (locks door) <S> Car: "ACK" (message lost in transmission) <S> Remote: "Lock the door." <S> Car: (notes door is already locked and does nothing) <S> Car: "ACK" (message lost in transmission) <S> Remote: "Lock the door." <S> Car: (notes door is already locked and does nothing) <S> Car: <S> "ACK" (message received successfully) Note that to use this method, the protocol must be designed so that duplicate requests do not matter. <S> So the remote can't tell the car "change door status". <S> Instead, the remote has to first ask whether the door is currently locked (with retries) and then set it to the opposite status (with retries). <S> After a certain number of failed retries the remote has to give up (maybe the car is not in range). <S> In that case it doesn't know whether the car is out of range, or whether the car failed to transmit the response. <S> This can be avoided by sending a no-op message first (which causes the car to send an ACK, but not do anything) to see whether the car is in range. <S> As noted by other answerers, there is always a nonzero possibility that the remote will fail to receive the ACKs from the car when the operation was successful. <S> This is unavoidable. <S> At best, the probability can be reduced so that it's unlikely to occur in normal scenarios.
| For something like remote door closing flashing the car lights is a good way in practice of indicating that the message has been actioned. It will depend on the cost of sending the messages (which might be different in both directions), the cost of failure, the likelihood of messages going missing, and whether or not missing messages are correlated.
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How to replace this connector My friend accidently stepped on a wire for his light and the tension broke the adapter's connector - I told him I'd try to fix it. I've already cut the connector off and have readied my replacement. I plan to use solderless connectors but as the original connector colors are unfamiliar I want to check with this site about how to check for which wire is power and which is ground. Can I use my multimeter to check which wire is power and which is ground? If so what setting do I need to use? The adapter: The new connector: <Q> The power supply has the output label. <S> Center positive for the barrel connector, 15V DC. <S> Also use your multimeter's continuity test or ohmmeter feature to confirm which wire on the replacement connector goes to the center pin and outside of the replacement connector. <A> Yes red = <S> + , Blk=- or 0V. Measure the positive voltage and match to Red. <S> Connect and use strong (packing?) <S> tape to mechanically secure the pair of connectors. <S> A common test is a 5lb pull test for reliability. <A> If you do not want to solder, at least use something more reliable than that flimsy plug with tiny internal contacts. <S> Something like this , for example. <S> Extremely easy to connect and has polarity markings for center-positive connection (same as your AC adapter). <S> To find output polarity put multimeter in DC voltage mode, measure adapter output. <S> If the reading is positive (should be around 15V) <S> then the wire connected to multimeter's red input is "+". <S> If the reading is negative, then the wire connected to red wire is "-".
| Yes, you would/could use your multimeter to measure the two wires, and determine which one is positive using the voltmeter feature (In DC mode if it has both AC and DC).
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Tool for Measuring Low Current I currently have a power supply that will be able to supply around -60 kV DC and 10 mA under the load that I will be hooking up to it. However, the internal resistance of the load will fluctuate and may not be pulling the full 10 mA at all times, thus, a means of monitoring the current flow will be needed. The tricky situation is that I will need to connect the negative power line from the power source directly to a multi meter that will measure the current. Please look to the graph below to reference the design: simulate this circuit – Schematic created using CircuitLab The negative side of the power source will provide the -60 kV DC, which must be manageable for the multi meter. Thus, I was curious what I could do to inexpensively allow a multi meter or ammeter to handle such high voltages and such a high power while still being able give an accurate measurement of the current. <Q> The only sure fire way of measuring low current on a high-voltage line is with differential probes rated 80 KV (4 each 500 <S> M 20KV resistors, Caddock 0.025%) each across a precision (1%) <S> 1K 100W-300W resistor which is in series with your load (the high wattage prevents arcing across the resistor). <S> The source common mode voltage of 60KV is divided by 10,000 to get a safe 6 volt common mode voltage. <S> 10mA would give you 10 volts across the current sense resistor and 1 mV at the TL052 buffer inputs and 10 mV at the AD524 inputs. <S> The AD524 gain is set to 1,000 and can be read with any DVM at the output pin as 1 volt/1 mA. <S> The AD524CD runs on +/- <S> 15 volts. <S> Because you are dividing the input by 10,000 the input impedance of the AD524 is an issue. <S> I added a TL052 dual op-amp wired as a buffer with a gain of 10.00, to the AD524 inputs. <S> The AD524 also has input and output offset trim and fine-gain trim. <S> Recommend using battery power to avoid common mode errors but safe for PC running LabVIEW if accumulating results. <A> I am not sure where the problem is. <S> If you use, say, 1-Ohm shunt resistor (of any size) as on the diagram below, and connect a DMM in mV-measuring mode, you will have 1 mV/1 mA. <S> The DMM is safely grounded, and no need for any kV-grade probes. <S> The voltage applied to your fluctuating load will be just 10 mV less than the 60,000,000 mV, or just like 0% error. <S> I don't see why you can't connect your current monitor this way. <S> You can add two regular diodes to protect DMM an shunt if you want in case or problems with parasitic capacitance on the LOAD during power-on/connect sequence. <A> that has Bluetooth or other RF communication of the voltage across the shunt at high potential. <S> Again, the shunt resistor can have a TVS or gas discharge tube across it to protect the meter input in case of a short or surge. <S> If you want to put the meter at high potential you will likely want to surround it with an electric shield (as much as possible while still letting the RF out) to keep corona discharge from causing problems. <S> Putting it deep in a box with one side open should work. <S> Here is an example of the kind of currents that can flow from a pointy electrode fairly well spaced from a grounded plane: Graph from Diguang, Z., & Dexuan, X. (1990). <S> Analysis of the current for a negative point-to-plane corona discharge in air. <S> Journal of Electrostatics
| Either use a shunt at ground potential (you can put a TVS or gas discharge tube across it to take care of surges) or use a battery powered meter at -60V
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Is it possible to use DC-restoration as a half-wave rectifier? A simple DC-retoration circuit: seems to do the about the same job as a half-wave rectifier, is it standard to use these two circuit instead of each other ? I tried to increase the capacitance of the capacitor to make this circuit serve as a full-wave rectifier but so far my attempts led me to the exact same results as above, Is there anyway (change diode type ...) to make a full-wave rectifier of this circuit ? <Q> You capacitor value is too low to get good results with most diodes- <S> the diode capacitance can be much larger than 10pF, especially with larger diodes. <S> This circuit is similar to the voltage doubler used in many older microwave ovens. <S> A mains-frequency transformer steps up the input voltage to a few thousand volts. <S> A diode and capacitor gives you double the voltage for the negative half-cycles. <S> Full description and animated gif from this website. <S> Don't pay too much attention to the arrow or the direction of the arrow, it's the opposite of conventional current flow through the magnetron. <S> Note that you do not get smooth DC out of this, you get pulses that are always positive (your circuit) or negative (the above circuit). <S> That's fine for a microwave oven magnetron. <S> It may not be fine for some other load. <S> So, yes, it's possible and the above application has been proven practical in hundreds of millions of microwave ovens. <A> could say that the CUK convertor uses this for its output stage .I <S> have used the villard in pushpull as a fullwave rectifier .So <S> the answer is YES . <A> Yes Old vacuum tube NTSC TVs used a tube diode to clamp (DC restore) on the sync tip.
| The Villard rectifier is a good example of using a DC restorer as a half wave rectifier .This is sometimes used for AGC generation in vintage AM car radios .You
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Can an OpAmp be built without an IC or transistor or a Vacuum tube? Is it possible to build an OpAmp without using an integrated circuit or a transistor or a vacuum tube? I wanted to see how it would look on a larger scale before miniaturization on a chip. I was hoping for a high view of what that would entail, not only that it is logically possible. @HarrySvensson Thanks. That is what I wanted. I didn't mean to cause a ruckus. I only wanted to see what was going on outside of all the YouTube videos and all the rest. I figured if I could see it with big chunky non-IC parts, I could understand it. Apparently I miscalculated there. What I should have asked for was an equivalent circuit. Now I know the keywords to search and have found many. <Q> Opamps have power gain, so one way or another you need a external power source and active components. <S> You have ruled out the usual active components used for amplification in electronics, which are transistors and vacuum tubes. <S> You therefore need to get clever and find other ways to amplify. <S> That means you need to be able to control a large amount of power by variations of a small amount of power. <S> Some possibilities: A motor driving a rheostat. <S> LEDs controlling LDRs (light dependent resistors). <S> Pressure-controlled water flow valves. <S> Pressure-controlled pneumatic valves. <S> A gasoline engine with the control being the throttle. <S> A mechanically controlled adjustable transformer, like a variac. <S> A electrical generator where the power input is the mechanical rotation of the shaft, and the control is done by changing the field winding current. <S> Once you have something that can amplify, you still have to use it, or several of them in the right configuration, to make a opamp. <S> This is just like <S> a transistor isn't a opamp, but multiple transistors arranged the right way (with some passive parts) <S> can be a opamp. <A> I think I understand what OP really wants. <S> If OP wants to understands how Op-Amps works, then maybe OP should play around with a simulator . <S> In this particular online simulator, OP can actually see the current and the voltages and how everything behaves. <S> Things which are difficult/tedious/impossible to do without an oscilloscope or other measurement devices. <S> After you've clicked the link, then you can browse many other Op-Amp designs under circuits > Op-Amps. <S> The one in the link was the circuits <S> > <S> Op-Amps <S> > Amplifiers > Inverting amplifier. <A> In principle you could design and build an op-amp with magnetic amplifiers or one that operated mechanically. <S> Amplifiers are amplifiers, so at a high level they'd be designed on identical principles. <S> The devil, of course, would be in the details. <S> Mechanical analog computers have been around for centuries, and mag-amps pre-date vacuum tubes and silicon ICs. <S> The literature on them is mature, and designing an op-amp with these technologies could be done. <A> The US Navy used magnetic amplifiers to servo the aiming points of 16" guns even as the battleship underwent mild roll/pitch/yaw. <S> I recall the power levels were 100,000 watts or about 1,500 horsepower. <A> You could always build circuits using fluidics . <S> You would use hydraulic fluids instead of electrons, and it was a hot research area before the integrated circuit was invented. <A> I'm not sure why you would want to do this, but there is a reason op amps are used so widely with chips. <S> But the problem is <S> even two components like that of the same model can have pretty varied functionality, and it changes with temperature. <S> One advantage of an op amp on an IC is that (1) the components are more likely to be very well matched, since they are on the same die, and (2) they are at exactly the same temperature. <S> Look up "long-tailed pair", a classic op amp input design, which takes advantage of these. <S> If you are wondering why you don't see op amps that aren't on a chip, or <S> at least a couple of (matched) transistors, the answer is that op amps became far more practical to build with the coming of analog ICs. <S> If you are asking is there some way to build an op amp (equivalent) without silicon involved, people have given some ideas above.
| You could build one with vacuum tubes or transistors.
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How to identify, if possible, a battery I have these two batteries, 18350 form factor, without any label.As far as I know they are rechargeable because I've salvaged them and put in my salvaged "li-ion to be cecked" box. Now I've completely forgot where I took them and they doesn't seem to be li-ion. They were at 2.7V more or less, then not too low to be used if li-ion,but then I've tried to charge them with my li-ion charger at 0.1A and the charger blocked the charge because no current were delivered. I thought they were already fully charged (quite strange, since they were in this box since january) and in fact after 24h they show 2.85V. There's a way to understand which kind of batteries are them?PS: I can charge/discharge under constant load/voltage and log the current/tension. (in a safe container, of course) <Q> This could be due to a problem in your charger's constant current mode operation. <S> May be you can try increasing the 0.1A current limit to a higher value like 500~800mA. Also monitor the voltage and current externally. <A> CR123 is 2/3 the size of an A size battery. <S> (Type A batteries are obsolete). <S> And a CR123 is a primary battery that cannot be recharged. <S> The RCR 18350 battery is not really a replacement for CR123. <S> The RCR123 has a 1 mm larger diameter than the CR123. <S> Typically a diode is added in series with the Li-ion 18350 battery to drop the 3.6V by about 0.7V. Other Li-ion battery chemistries (e.g. LiCoO2 (ICR) or LiMn (IMR) chemistry) can be used without a diode as well. <S> A 3.0V-3.2V <S> LiFePO4 <S> (IFR) <S> Li-ion can be used becasue <S> it is less vulnerable to over charging but is a horrible replacement. <S> See Types of Li-ion <S> Some RCR123 batteries do not have the protection circuit, and as a result, will not be compatible with many chargers. <S> Li-ion chargers should use a protection circuit because LI-ion batteries have a tendency to start on fire and explode when misused. <A> For starters, you should post a photo of your battery in question, to see what kind of tabs or other connector does it have. <S> It might give some clues. <S> Second, you should try to refresh your memory and provide some idea of where these two cells came from, how old the equipment was (it might narrow the identification and exclude new commercial developments). <S> However, when attempted to "charge", it should consume full charging current and its voltage should creep up a bit. <S> The charger should not disconnect. <S> It also should be able to drive moderate loads (say, 100 Ohms) for measurable time. <S> If you report that your standard Li-Ion charger gives up (I would guess very shortly, in a fraction of a second), it is likely a dead Li-Ion cell which has all internals vastly deteriorated, has nearly zero capacity, and developed a very high ESR, such that the charger voltage immediately goes above 4.3 terminal voltage, and charger disconnects. <S> To check if this variant is true, you should apply a moderate load (100 Ohm), and see for how long it will keep the 2.7 V level, if ever. <S> Again, if your cell doesn't charge and has no capacity, throw it away. <S> If it holds some charge and has 2.7 V, throw it away as well. <S> So, you have only one outcome :-)
| Generally if this is a CR123A non-rechargable cell and it has 2.7 V, it is nearly discharged. A rechargeable CR123 should be referred to as an RCR123.
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What is the Gain, Input and Output Impedence in this circuit? So I'm trying to know what the input, output impedence and gain is in this circuit To begin, I'm kind of confused since I don't know if those two resistors at the right are in series or in parallel. <Q> With an ideal op-amp the input is infinite impedance- <S> like it isn't there. <S> So Vin is effectively just connected to two resistors of value 'R'. <A> One rule you can go by to know what the input impedance is to ask yourself what the current is through the input? <S> How much is it? <S> Impedance is directly related to the current through the input (because V = I*R) <S> So looking at the diagram again, where is the current going? <S> How much? <S> You could even calculate the input impedance by only knowing the voltage which is Vin and the current through R (we'll call it I <S> ) \$V_{in}/I_{R}= <S> Impedance_{in} \$ . <A> Opamp input impedance is assumed to be infinite - so the 2 resistors at the input (in series) as all the input 'sees' = <S> > input resistance is 2R <S> The 8R can be considered to be part or the opamp due to being inside the feedback loop (so consider it a short) - so gain due to opamp is 12 (1 + 11R/R)Gain due to the 6R and 3R <S> will be 3 / (3+6) = <S> 1/3- <S> So overall gain is 4 <S> Now consider the - input to have infinite impedance, and the opamp output impedence (including the 8R) is effectively zero, so the output resistance will be 3R and 6R in parallel = <S> > 2R <A> Input Impedance: <S> Is 2R, since it's just the series combination of the two single R valued resistors connected to the positive input of the op-amp (you assume the input impedance of of the op amp inputs themselves is infinite for an ideal op-amp). <S> Take 11R and R in series for 12R, and this is in parallel with 8R for a combined impedance of 4.8R. <S> This is in series with 6R for a total of 10.8R. <S> This is then in parallel with 3R for a total output impedance of roughly 2.348R. Gain: <S> You need to find the overall Vout/Vin, not just the gain of the op-amp. <S> to make the math easier, assume 2V at Vin, which then puts 1V at the positive input of the op-amp. <S> As an ideal op-amp, it will strive to make it's negative input 1V as well, so the voltage at the negative input is now set to 1V. For this to be 1V, the node at the top of the 11R resistor would have to be at 12V. <S> This 12V would divide divide across the 3R and 6R to make Vout 4V. Since the input that was assumed was 2V, and the output is 4V, Vout/Vin is then 2. <S> So the overall gain of the circuit is 2.
| Output Impedance: Looking back from the output, and again assuming the ideal op-amp characteristics of infinite input terminal impedance and zero output terminal impedance.
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Can I use a Step-up Power Converter with Arduino instead of a Relay? I want to make two projects that use a 12v input, a solenoid valve and a electrical lock. Both need a 12v current to work. I've seen some exemples for both projects that use a relay and an Arduino to activate mechanisms that need more than the 5v Arduino´s output. I was wondering, however, if I could use a Step-up Power Converter connected directly to Arduino´s Digital Pin. Is it possible? And is it safe? <Q> A relay (solid-state relay or FET) is the best way to switch high voltage from low voltage. <S> Arduino has low current output not meant to do what you're describing. <S> It's not possible. <S> Use a relay or a FET. <S> It is even safer to use an optocoupler for isolation. <A> if I could use a Step-up Power Converter connected directly to Arduino´s Digital Pin. <S> Is it possible? <S> That will not work, the Arduino's outputs cannot deliver enough current. <S> An Arduino's output can deliver around 10 mA and that's enough to light up an LED but little more than that. <S> In theory a 5 V to 12 V Step up converter can be designed and made but at 10 mA input current the output current needs to stay below 5 mA and that's also very a small current. <S> It is by far not enough to power some mechanism with a motor. <S> And is it safe? <S> Define "safe". <S> Safe in my definition is "will it harm <S> /kill people" the answer is no. <S> But you're looking at the wrong solution. <S> You need to see the output of the Arduino not as the source of power but as a signal to control the power . <S> The signal is the output from the Arduino, VCC and GND simply connect to the 5 V supply that the Arduino uses. <S> When connected like that the Arduino can control the switch in the relay, you can connect that in series with a 12 V power supply to power almost any module that needs 12 V with a high current. <S> The relay shows how much voltage and current it can safely switch. <A> You really should provide some sort of sketch of what you're trying to accomplish. <S> If you're trying to drive a high-current device with a logic pin, the terminology for what you need is a "driver". <S> There are a variety of approaches, the simplest probably being a simple bipolar transistor.
| You can use a module with a relay like this: The GND, VCC and signal connect to the Arduino. If your definition is, will it damage anything (like an Arduino), the answer is no because an Arduino's output can only deliver such a small current that nothing can get damaged, the current is too small.
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How to minimise trace fracturing in PCB design - multiple thin traces or one thick? Designing a board with strict and inconvienient requirements. Currently the boards are breaking frequently - the aim is simply to reduce this frequency as much as possible.. The board is a 100mm square but only supported with standoffs in each corner. On the board are 20 throughhole 4 pin sensors. Effectively they are analogous to DIP switches. The PCB experiences a lot of vibration as well as intermittent force (perhaps up to 100N point load) randomly, suddenly applied on a random sensor. Board is FR4 1.6mm. The environment is also humid and moderately corrosive. I would love to redesign/support the board etc etc to be supported better, not loaded heavily, with a backing plate etc - but this is not an option. Unsurprisingly, the board traces breaks frequently under this flexing. This generally causes unreliable performance. Electronically the board is simple slow digital signals so noise etc is not a concern. I'm unsure as to how best to prevent/mitigate the traces breaking. I could either route multiple narrow traces for the same signal, thus my thinking being that stress is more likely to break one and the crack not propagate through to the others, or one wide trace. I'm also thinking curved traces might be worth a thought and I'm interested in how best to connect the trace to throughhole connection to minimise risk of fracture. Also interested in if using extensive ground planes may help and if the most critical traces should be routed on top or bottom of PCB - so that they are either predominantly in compression (top) or tension (bottom). I have read around flexible PCB design but not got much conclusive information and am unsure if a rigid PCB that flexes is a significantly different case. Thanks <Q> Tricks we have used with some success:Make your feedthru holes big enough to insert a small piece of bus wire and solder top and bottom. <S> Also solder top and bottom on your sensors if possible. <S> We have seen lots of breakage between traces and the plated through holes and this is pretty effective, especially if you use 2-sided boards. <S> Multi-layer boards are prone to cracking between trace and plating on the plated through hole, so stick with two-sided if you can. <S> If you must use multi-layer, be sure and watch the heat on your joints. <S> Overheating can stress the joint because the board material expands at a different rate than the copper. <S> The idea is to prevent cracking at the outset. <S> Trim component leads before you solder; Trimming them afterwards can stress and start a crack which will propagate. <S> No sharp corners on the traces. <S> Cracking often begins on the corner where a trace runs into a pad outer radius. <S> Fan out the trace to where it matches the pad tangentially. <S> Of course, you are probably already mounting your heaviest parts nearest to the mounting holes where they have the most support. <S> Good luck! <A> Answers to questions: Compression vs. Tension <S> This is only applicable when there is continuous flex on a board. <S> In your case, you describe a strong, intermittent force. <S> When that force begins, your board will flex one way. <S> When the force ends, your board will not immediately come back to its original shape, it will release the stored energy and flex back in the opposite direction before the oscillations damp down. <S> So any trace you have on the board will both compress and stretch. <S> Meanders <S> When stretching, the copper will be deformed but is largely fixed to the rigid FR4. <S> This limits the benefit achieved by meandering traces over the board. <S> Suggestions: <S> Component placement <S> The amount of flex will be directly proportional to the torque on your board. <S> This is minimized by moving the components as close to the supporting struts as possible. <S> Try to keep things away from the middle of the board. <S> I would also remove as much copper from the middle of the board as possible. <S> Wiring <S> We used to call them printed wiring boards. <S> :) <S> In your case, it might make sense to go back to jumper wires. <S> Rather than having copper traces affixed to the board and likely to break, you could solder jumper wires across the board for signal lines. <S> Power should still be provided using power planes but your signal lines could be soldered between through-holes located close to the edge of the board and close to the component lines. <S> These wires should be tacked down using a glue in the middle of the line. <S> Usually you would tack them every 1" but with your scenario, one tack in the center will reduce the stress on the solder points. <S> Pick the thinnest wire you can get away with for your signal. <A> There are several things you can do. <S> Apply as many of these as you can: <S> Thicker PCB. <S> Your boardhouse should have several options. <S> Heavier copper. <S> Instead of 1oz. go for 2oz. <S> or 3oz. <S> Wider traces <S> Larger annular rings around through holes Add space between components and the board. <S> Place something behind components that take force. <S> Non-conductive foam, for example. <A> I would love to redesign/support the board etc etc to be supported better, not loaded heavily, with a backing plate etc - but this is not an option. <S> Ideally you would add additional supports in the middle of the card. <S> However... you have stated a redesign is not an option. <S> This means options like Thicker PCB ( I only use 3mm thick cards in the environment I work in, high Vib) <S> Thicker tracks <S> Heavier weight copper ie all the things to stiffen the card, are not possible If these are not possible the only real option is the equivalent of Anti-Vibration mounting on whatever you screw the board downto. <S> Can you place rubber washers down to help mechanically isolate the card?
| For long parts like through hole resistors, mount them parallel to the expected flexure so that the leads are not getting stretched and compressed. If you have leaded parts that mount flush, leave some space so that they are not sitting directly on the board. Equally you would want the high mass components near the edge of the card.
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Can an EPROM be "refreshed" without UV erasing? I have lots of AM27C512 65k x 8b EPROMS (with the UV window) which are quite old (1980's.) Using a Xeltek SuperPro 3000U, most can still be read and verified. A few have failed outright and were replaced. Question: Can these be successfully "refreshed" by reprogramming the same bit pattern back to them, without UV-clearing them first? They seem to take a reprogramming fine, but was curious whether this is moot without an erasure first. Edit : Wanted to mention that so far, I've tried "refreshing" about 10 of these 256kB EPROMS. All of them read fine, were programmed (with the same data) fine, and verify fine. So "refreshing" hasn't seemed to hurt them at all. I'm still a little sketchy about whether this will make any lasting difference in their longevity however - answers are quite mixed. <Q> When the EPROM is erased, all the bits are typically read as ‘1’, so each byte is 0xFF. <S> That is the way they will tend to age, with time and high temperature or radiation. <S> If you don’t have access to a suitable high intensity short-wavelength UV lamp to erase the chip <S> you may able to refresh the EPROM, there is nothing in the EPROM itself preventing it. <S> A given programmer may or may not permit this- <S> for example the OTP micro controller programmer we use for NXP parts performs a non-optional blank check before programming, so <S> it’s not possible to (say) add serialization to a blank area of the memory of a device that has already been programmed. <S> Of course this only applies if you are programming the exact same pattern into the EPROM. <S> Even if the data is equivalent but not exactly the same, the attempt will almost surely fail. <S> Germicidal lamps can be found in water treatment units but take care of exposure to the light which can cause skin burns including corneas. <S> EPROM erasers, usually equipped with proper interlocks and timers, are cheaply available on eBay etc. <S> You could also try leaving it out in direct sunlight (no glass or plastic over top of the quartz window) for a few days. <S> I'm told that will will work, never tried it. <S> It would be better to monitor the EPROM for erasure and then expose it for several times longer than required to read all 0xFF to ensure full erasure , otherwise those '0' bits could come back under different conditions of voltage or temperature. <A> If you are repairing equipment it's best to buy a new, functional EPROM and program it with the right data. <S> You can't damage the EPROM by reprogramming already-programmed locations, so there are no worries there. <S> The main issue is that bits that have flipped from '0' to '1' can be fixed by reprogramming, but bits that are stuck in the '0' state can't be restored without UV erasure. <S> So reprogramming a programmed EPROM can only fix one of the two failure modes. <S> Anecdotally in this situation I have found erasing and/or reprogramming old faulty EPROMs doesn't always restore the EPROM to a fully functioning state and some locations remain incorrect no matter what. <S> That's why getting a new EPROM as a replacement is always best, and it extends the lifetime of whatever you're servicing quite a bit. <A> Refreshing it is basically recharging the previously charged cells. <S> This requires no discharge operation which requires UV-C light. <S> Second law of thermodynamics implies that no cell will accumulate charge without any force affecting it, so all the cells that discharged before will remain in this state. <S> So, it has to be fine to refresh it this way. <A> The short answer is no. <S> The longer answer is that, sometimes, if you're very lucky (as in unreasonably, insanely lucky) it can happen. <S> An erased EPROM of this sort has all bits set to a logic one. <S> As PlasmaHH has commented, programming simply flips bits - in this case, it leaves all ones as ones, and flips the others to zero. <S> What it cannot do is flip zeros to ones - <S> that is what erasing is for. <S> So, if you just happen to have a new data set which which has no ones where the existing PROM has zeros, you're golden. <S> The programmer will change the necessary ones to zeros, and will successfully attempt to produce zeros on bits that already are. <S> As you might guess, having this happen on a 512K PROM is, how shall I put it, wildly unlikely.
| As a quick hack for some equipment that isn't going to a customer I think reprogramming an EPROM that is suffering from "bit-rot" with the correct pattern is an OK short-term fix.
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Voltage increase on regulator load I have a simple adjustable power supply (1.8V - 16V / <1A) that I put together with an LM350, works great for the various small bench-top things I use it for. Typically on a load being applied, I see the typical small voltage drop as current spikes, all expected. Last night I hooked it up to a hacked together transistor-as-audio-amp circuit I threw together on a breadboard to test the DAC on a new MCU. Eveything went as expected, sound played fine, but I noticed that my voltage went from 8.4V to 9.75V, around a 16% increase! I'm not completely new to building low-ish power circuits, but I'm trying to get better at the theory, and I'm not seeing why this increase occurs. It is constant, not a spike, and a jump of 1.35V seems like a lot. Of course using a BC547 as an audio is non-optimal, and the transistor only does an ok job handling the current, but I only turn the circuit on for a few seconds at a time. I was thinking that there is some transistor-related phenomenon that I am unaware of that would account for the voltage, but it does occur in circuit simulators, which leads me to believe that it's a real parts vs ideal parts phenomenon. Can anyone point me to a source for understanding why this happens? simulate this circuit – Schematic created using CircuitLab <Q> Probably you don't have a large enough capacitor on the output of your regulator to keep the voltage relatively steady. <S> Try something like 1000uF. <S> The capacitor gives the relatively sluggish regulator time to catch up. <S> Your peak current could be <S> relatively large- 8.4V/4 ohms is over 2A so the transistor will be limiting. <S> Note also that the regulator cannot sink current- <S> it can only source <S> it- <S> so without some kind of passive load (called preload in the regulator datasheet) the regulator will tend to err on the high side. <S> A capacitor will smooth it out so that's less of an issue. <A> Also check for high frequency noise from the audio transistor. <S> You could even consider adding some smaller capacitors (like 0.1u or 0.01u) in parallel with your large electrolytic cap to help with noise suppression. <A> You have inadvertently created a bit of a boost converter! <S> Look at a generic boost converter circuit and recognize that your speaker is the inductor. <S> The LM350 will not sink current by itself.
| Even with a large bypass capacitor, if your audio circuit is generating a lot of noise, that can also affect the feedback loop of the regulator.
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How many capacitors are needed for a step down circuit? I'm tyring to build my first step down component using TPS562219 to get 5V output from a 12V input (instead of 1.05V output shown in the example). But as I'm new to this electronic stuff, I have some problems of understanding. Input capacitors In this example circuit there are three capacitors. Why should I use three of them and how do I know the values of these. In the text of the datasheet (page 15, section 9.2.1.2.3) they are talking of two capacitors: "A ceramic capacitor over 10 µF is recommended for the decoupling capacitor. An additional 0.1 µF capacitor (C3) from pin 3 to ground is optional to provide additional high frequency filtering." Output capacitors For the output capacitors table 2 on page 15 gives a range of 20-68µF. How do I choose the correct value? And why should I use three of them? In another circuit which is using the same TPS562219 they are using just two input capacitors and one output capacitor - with different values: So I would be very thankful for some explanation, so I can understand that and build some own things. How many capacitors do I need and which values are recommended? <Q> Input capacitors <S> Ideally, the input to the regulator has 0 impedance. <S> That's not possible, of course, especially if there are long wires between the regulator input and wherever the input voltage is coming from. <S> The capacitors guarantee a low impedance at the frequencies where this really matters. <S> Since 0 impedance is ideal, infinite capacitance would be great. <S> The datasheet gives you guidance of how much capacitance is good enough for what the chip needs. <S> The extra 100 nF is to provide a lower impedance at even higher frequencies than the larger caps are effective at. <S> The extra 100 nF by itself is inconsequential when added to 20 µF. <S> Output capacitors <S> The output of this switcher is pulses of current. <S> That requires capacitance to smooth out to a reasonably steady voltage. <S> In this case, the capacitance does two things <S> It smoothes the pulses to make a flat-enough output voltage. <S> It is part of the overall system the controller in the regulator is working with. <S> Both too little and too much capacitance can cause control instability. <S> Stick <S> to what the datasheet says. <S> Values towards the higher end of the range will result in less output voltage ripple, but also allow less additional capacitance to be distributed among the consumers of the output voltage. <S> Multiple capacitors <S> Often multiple identical capacitors are used in parallel on both the input and output of such switching regulators. <S> These capacitances need to be low ESR (equivalent series resistance) to the point that they pretty much need to be ceramic with today's technology. <S> Ceramic caps can only be so big before mechanical problems make them unfeasible. <S> Multiple caps in parallel allow for higher capacitance at the same voltage. <S> If a new technology came along that could fit the combined capacitance into a single part without mechanical problems or higher ESR, then it would be fine to use such caps. <S> The technology tradeoff keeps changing, within the lifetime of many datasheets. <S> Look around. <S> Perhaps you can get the total capacitance with the right specifications in a single package today. <A> Output capacitors: - <S> You'll probably find that using 3 x 22 uF (66 uf in total) has a lower parallel ESR and higher self-resonant frequency (SRF) than just using (say) a bulkier 68 uF. Input capacitors: - This can be problematic if the source impedance of the power feed is sloppy / too high or has resonant impedances involved. <S> Add to this a cable feeding the regulator <S> and you need to use belt and braces to cover the likely eventualities. <S> 100 nF will have a reasonably high SRF and ensures the incoming supply impedance is "made" fairly low above a MHz or so and anything added in parallel is there to bulk out the capacitance should there be the possibility of drop-out or poor transient behaviour under load. <A> They are using multiple capacitors to save money. <S> MLCCs above a certain size get are only available as stacks of smaller parts and these stacks are crazy expensive <S> the two 10s make up a 20uF capacitor on the output <S> they use 3 22s in parallel as a single capacitor. <S> as the switch is the other end of the inductor there's no need for a RF filtering capacitor here.
| The purpose is to get more capacitance than a single part of the right type can provide.
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How to test a CPU watchdog on board? The watchdog of an ATMEL ATXMega128 should have been enabled with fuses. It triggers a reset, if the timer was not reset within the configured time span.I want to be sure, that it is enabled and working properly . What is a good method to verify that the watchdog is enabled and working on the finished PCB? Is it common practice to prepare a spare input signal in order to trigger the watchdog artificially? Should the watchdog trigger, if I pull the external oscillator to GND, or will this stop the watchdog too? <Q> You can test through a sequence. <S> Use an existing non critical output like an led as a test signal. <S> Program the board with a test sequence that will toggle the led and loop, And does not pet the watchdog. <S> Test for the loop. <S> Then program it to do a second loop, where it will not reach if the specified watchdog timer does not reboot. <S> Test to make sure it passes the watchdog petting. <S> These two steps will make sure the hardware part of the watchdog are good. <S> The rest is ensuring your final code does not affect the watchdog by disabling it or petting it when something that shouldn't lock up gets locked up. <S> As to your question about the external clock being stopped, I'm not familiar with any microcontroller that would have two independent clock sources that would keep the watchdog ticking when the main clock source is disabled. <S> In a externally grounded clock situation, a watchdog reset wouldn't fix that anyway, unless your code can switch to an internal clock for the main code after that reset. <S> The watchdog is meant to clear software related hangs. <S> But I'm sure those exist for mission critical applications. <A> When your watchdog expires, it will call a specific function. <S> The normal action of that function will be to take whatever remedial action is required under those circumstances. <S> However, before it does that, it should check a variable called something like WDT_test, and only do the remedial thing if it's false. <S> If WDT_test is true, it should instead log a report (perform some action) that the WDT is working. <S> To test the WDT, set WDT_test true, stop kicking the WDT, and wait for the report within a certain time. <S> That's the non-intrusive way to do it. <S> To do a more complete end to end test, just stop kicking the WDT and wait to see whether it eventually performs the remedial action. <A> AtMega MCUs have a watchdog with dedicated 32kHz oscillator, so yes: putting the main oscillator on pause will trigger it. <S> Dedicating a pin triggering a fault makes less sense. <S> I'd never include that in the final software, and if you're going to do a test software anyway, you don't need the pin.
| Writing a test software which intentionally fails to service the watchdog is another possibility, and it will enable you to validate not only the fact that the watchdog in enabled, but easily measure the timeout as well.
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In what kind of circuit does current flow? In what kind of circuits do current flow? For a circuit to be "closed" does the end point have to return to the start point like in the first image? If you have a straight line conductor between two components of opposite charge like in the second image, will there be current flowing between them? If it's true, will current then flow from one pole of a battery to the opposite pole of another battery(like the third picture) and why/why not? <Q> 1) YES - <S> The current will flow in the loop and <S> it's value will be decided by the voltage of your power supply and the resistance of the path. <S> 2)YES <S> - The electrons will flow from negatively charged body to the positively charge body until both bodies are at same potential. <S> Once that happens, no further electron flow will take place and hence no current will be flowing. <S> This case is sometimes referred to as a transient current. <S> 3) <S> NO - Current won't flow because both batteries are uncharged objects. <S> Also, net potential difference the bodies of those batteries is zero. <A> Diagram 1 <S> Yes current will flow. <S> The magnitude of the current will be limited by the internal resistance of the battery and can be calculated by \$ <S> I = <S> \frac <S> {V}{R} \$ (Ohn's law). <S> Diagram 2 <S> Note that if there were more of one charge than the other that the excess would be split between the two spheres. <S> If the static voltage is high enough the equalising current will flow through the air when the gap is reduced enough to cause the air to breakdown. <S> Diagram 3 <S> There is a potential difference between the ends of each cell but there is no overall charge on the batteries. <S> That is, there is no static electricity build-up on the batteries <S> so there is no equalisation current when connected as shown. <A> Those pictures appear to show physically what's there. <S> However, they miss other circuit elements that are important. <S> The second picture appears to show two separated objects, one carrying positive charge, one negative. <S> For instance, one is the cat you've just stroked, the other is you. <S> If they are conductive, and if you connect them with a wire, a current will flow briefly between them. <S> Many times I've drawn a small spark (much to the cat's discomfort) off its nose, after stroking its fur, which has charged it with respect to me by the tribolectric effect. <S> What your picture doesn't show is the capacitance between the two bodies, which for cats and people is in the order of a few pFs to a few 10s of <S> pFs. <S> That completes the circuit. <S> Given that typical charging voltages can be thousands, even tens of thousands of volts. <S> That's a significant amount of charge, noticeable to fingers and cats. <S> Your third picture shows two batteries. <S> What it doesn't show is the capacitance between all the electrodes which completes the circuit. <S> When you make the connection between the batteries, this capacitance will charge up, and a tiny current will flow for a moment. <S> However, as the voltages involved are just a few volts, the amount of charge that moves is so insignificant that you won't notice any effect.
| Yes, if you can charge up the two spheres with static electricity of opposite polarity then when connected a current will flow until the charges are balanced.
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What kind of tool is used to record a voltage over time to a PC file like a waveform? I don't have any training but am doing some hobby electronics. I would like to record a change in voltage over time, sampled at around 100+ kHz. Ideally, I'd like to record around 20-30 min of two different channels, so I guess that works out to be at least 240 million samples. I'd like to view this recording on my Windows computer similar to a stereo .wav PCM waveform. (In fact, if I could save it as a .wav file, that'd be ideal.) I considered simply using a sound card to make this recording, but this seems to not be possible with a DC signal . Is this something that a digital oscilloscope can do? Or is there a tool designed for this specific task of recording to a file, similar to the line-in on a sound card? Converting to digital and streaming over USB to record on a PC seems to be what I'm ideally looking to do. Edit: Giving some context as requested... This is for a latency test. I am wanting to record a variable latency of a video game system plus an HDMI adapter. Using a photodiode I can create a voltage stream that will show when the result appears on screen and compare it to when a button was pressed. (Hence why two channels are needed.) Thanks! <Q> This is called data logging. <S> Your application is a bit out of the ordinary only because of the large data files generated. <S> However, the data sheet for the software warns that files over 1 M "may suffer performance issues". <S> Other software packages are available, and I'm sure one of them will do the job. <A> Using a photodiode I can create a voltage stream that will show when the result appears on screen and compare it to when a button was pressed. <S> (Hence why two channels are needed.) <S> You can use an unmodified soundcard if you make your signal AC. <S> I presume you will have the device output a voltage pulse or step on a port, and at the same time change the brightness of an area on screen. <S> Then you detect the change in brightness on screen with a photodiode, and you are interested in measuring the latency between the two events, correct? <S> You're only interested in detecting the edges. <S> You don't care about the flat parts of the waveform which would be preserved by a DC acquisition, so in fact your signal's information content is not diminished by acquiring it in AC. <S> So all you have to do is adapt the detection software to detect positive and negative going peaks... <S> that's all... <S> Another option is to use a microcontroller with a USB or serial link, and have the micro do the latency measurement. <S> This simple to do with two comparators (one for each signal) with proper analog thresholds... by the way you'll probably want to AC couple the input in order to only detect changes in brightness and voltage. <S> Feed the comparators' outputs to the micro's Timer Capture inputs, configure the peripheral, and you're set. <S> Write an interrupt handler to grab the timer values and output a latency measurement to the PC over USB/Serial. <S> Note that you can use a digital scope and set it to measure the delay between two edges. <S> If your scope is too dumb to do this, use two comparators and combine the outputs into one signal, maybe with a XOR gate, and set the scope to measure the length of the resulting pulse, which is your latency. <S> Then you will either have to read the value on screen, or convince the scope to forward it to a PC over Ethernet or USB. <A> This is what bench meters are great for. <S> A number of them can sample well into the 100's of kilosample/second. <S> The problem is that these can be expensive (for a hobby). <S> Digital oscilloscopes cannot do this. <S> They take snapshots, based on the trigger. <S> Usually, inside, the ADC is constantly writing to a buffer, that has a certain length. <S> When a trigger even occurs, this buffer is then stored and processed. <S> Once that happens, the ADC starts writing to the buffer again and the process starts from the front. <S> As a result, a scope will not be able to measure everything. <S> Some get very close though. <S> However - some oscilloscopes might be able to capture the entire thing you ask for in their buffer, at low sampling speeds. <S> However, they are not going to be cheap. <S> I do believe there are modules from National Instruments and others that can do this, but these will be very expensive, and require a PXI chassis, which again is very expensive. <S> UPDATE: <S> As first pointed out by Marcus Müller, and confirmed by Matt, apparently some Picoscopes do support continuous sampling, which would make them a suited for the application of a 100 kS/s continuous sampling digitized (I am aware that the scope of the question has changed since posting this answer, but I am keeping it here for people in the future who might indeed require continuous sampling) <A> This sounds a lot like the requirement for the baseband part of software defined radios! <S> Disclaimer: I'm slightly affiliated with them, but Ettus does sell the USRP series of devices, and that has the job of sampling analog signals at multiple (to hundreds) of Megasamples/s and transport the resulting digital signal to a PC using USB, Gigabit or 10Gig Ethernet, or PCIe. <S> But even a ~USD 10 rtl-sdr TV dongle can be put into a direct sampling mode, which might, if you can live with only about 2 MS/s and 8 bit samples, be used in the manner you describe. <A> Do it with the sound card, the DC level will not be recorded, but the edges will, (when the level goes up or down) and those are where all the information is!
| Measurement Computing will sell you a 12-bit, 500 kHz digitizer for under 100 bucks, and the free software wil theoretically do what you need.
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Shimming loose AA batteries with aluminum foil... Will this cause corrosion? If you folded up a piece of aluminum foil and put it between the nickel plated terminal of a AA battery and the nickel plated contact in the battery holder.... Would galvanic corrosion cause either the battery or contact materials to become damaged? Does it matter whether you put the aluminum foil on the positive or negative end of the battery? Are there any other risks such as safety risks, exploded batteries, or as I've mentioned, any expected corrosive effect that might ruin your device contacts after a period of time? <Q> The battery will be fine. <S> Would galvanic corrosion cause either the battery or contact materials to become damaged? <S> While galvanic corrosion is possible between aluminum and nickel (or nickel and any other metal), in most environments, it is not a problem. <S> Yes, it is possible, but it would depend on the life of the battery (e.g. replaced monthly, yearly, or decennially), the environmental moisture content (e.g. air humidity) and the galvanic component potential. <S> If the humidity were 100% then the following may apply (submersed in tap water). <S> Aluminum has a galvanic component potential of about -375 Nickel has a galvanic component potential of about -125 <S> The difference between the two is only 250. <S> The difference between the two in sea water is about 450. <S> So it depends upon the environment. <S> For example if it were used at the beach (with high salt and humidity) would be slightly (insignificant) worse than in an air conditioned office. <S> Source: <S> THE EFFECT OF ELECTROLESS NICKEL COATINGS <S> Does it matter whether you put the aluminum foil on the positive or negative end of the battery? <S> Good question. <S> I do not believe that current flow has any effect on galvanic corrosion. <S> I have never see this said in any papers I have read on this topic. <S> Intuitively it would sound reasonable the current would exacerbate the corrosion. <S> But I have not seen any papers written proving this to be true. <S> I first learned about galvanic corrosion back in 1972 about joining copper and iron water pipe when I was doing plumbing. <S> Joining copper an iron requires a fitting called a dielectric union. <S> There are three things necessary for galvanic corrosion. <S> Current flow is not one of them. <S> Electrochemically dissimilar metals must be present e.g. iron and copper <S> These metals must be in electrical contact, like pipes screwed together <S> The metals must be exposed to an electrolyte, like water in the pipe. <A> Without an electrolyte , there would be no galvanic corrosion. <S> Aluminum is less noble and will become the cathode on the galvanic corrosion process. <S> Even humid air near the sea can do this. <S> The drier and enclosed the metals are, the less likely it happens. <S> Google "lasagna cell" for examples. <A> @Misunderstood did a nice job explaining if corrosion can occur. <S> This answer explains the consequences of where the corrosion occurs. <S> The outside of a typical AA battery is a steel can, positive and negative terminals, and seals between them to keep the liquid electrolyte inside. <S> (The may also be a printed label, which is not important here.) <S> Normally when we think of battery corrosion, it occurs inside of the cell. <S> It attacks the seals, which causes the electrolyte to leak out. <S> In turn, the electrolyte can corrode the outside of that cell, nearby cells, the battery holder contacts, and the circuits of the device itself. <S> Yuck. <S> With an aluminum foil shim, the corrosion will occur on the foil itself, and any terminals or contacts the foil touches. <S> It will not attack the seals, so it is no more likely to cause the cell to leak. <S> It will increase the resistance of the battery (i.e. like an additional Thevenin series resistance), quite similar to using that device with dirty battery terminals. <S> The amount of corrosion will be quite small, although it will increase with use of the device, so you may not even notice any change in the performance of the device. <S> You can fix it by cleaning the contacts and replacing the battery and foil.
| Say you're in a very salt water environment, then the aluminum foil will corrode before the nickel does. There is zero chance of explosion. Humid air is the most likely electrolyte.
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I need help understanding constant voltage model for diode So I'm confused about the constant voltage model (which assumes diode has .7 volt drop). In the pic below, I'm confused about the third case. What if the voltage source has voltage less than .7 V but the diode is in forward bias? Does that mean it's open circuit and hence the voltage drop across the diode will equal that of the voltage source? (In the last two cases---in the circuits to the right--I just made up numbers to illustrate the concept) <Q> If you are modelling the diode as a constant voltage, then yes, if the source voltage is less than 0.7v, that will be the voltage across the diode. <S> It looks like you understand it fine. <S> With reverse bias, and with forward bias < 0.7v, it's open circuit. <S> Otherwise, it looks like a 0.7v battery. <S> A model as simple as this is adequate for some purposes, and not for others. <S> Remember, all models are wrong, but some models are useful George Box <S> If a constant 0.7v is too wrong for your purposes, let's say you want to estimate the diode voltage drop at 1nA, then you would use a better model. <S> A popular one is the Shockley Diode Equation . <S> Watch out for the 'ideality factor', a fiddle factor which allows you to align the model with measurement over a wide range of parameters. <S> At high currents, you'd need to add some series resistance. <S> At high frequencies you'd want some parallel capacitance (voltage dependent), and perhaps some series inductance as well. <S> And you still haven't taken account of the charge storage, which is as important for stopping 1N4004s from being used in SMPS as it is for making GHz step recovery diodes work. <S> You stop elaborating the model when it's good enough for your purposes. <A> I think you might be confused about a basic concept with how these simple diode models are used. <S> You must use a different model for the diode depending on whether it is forward biased or reverse biased. <S> In forward bias the diode looks roughly like an ideal voltage source of 0.7V and in reverse bias the diode looks roughly like an ideal current source of 0A (an open circuit). <S> If you are not sure which to use, then just pick one. <S> If you assume that the diode is reverse biased, analyze the circuit, and determine that the voltage across the diode is greater then 0.7V <S> then your assumption was incorrect and the diode is actually forward biased. <A> Consider this, where the plot is a log-vertical-diode-current, and linear-horizontal-diode-voltage <A> Simple answer is that diode can't act as a voltage source. <S> If external voltage (Vext) is greater than 0.7V <S> then drop across diode is 0.7V and if Vext < 0.7V then the drop across the diode can't be greater than Vext. <S> So, if you see the I-V chart of this approximation you can see that before cut-in voltage(0.7V) current(Id) is zero. <S> Similar Analogy: <S> Assume a box on ground (with <S> "MAX" frictional force say 10N) <S> Case 1 <S> : If we push it with 20N then frictional force acting is 10N to oppose motion. <S> Case 2: If we push it with 3N force then frictional force acting is 3N only else if its greater than 3N then box will be in motion. <S> So frictional force is for opposing the motion, it can't make box move (and similarly a diode can't be a source).
| If you assume that the diode is forward biased, analyze the circuit, and determine that the diode current is negative then your assumption was incorrect and the diode is in fact reverse biased.
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USB device with 2 ports - how to protect against user plugging in both I've got a Full speed USB device which I want to provide 2 ports for - one each on opposite ends of the enclosure. This is so the cable can be connected wherever is most convenient. My MCU (atmega32u4) has only one interface, so the physical ports will shared, but only one should be plugged in at one time. Of course users can't be trusted to not plug in both sides at once. How to protect against this? Ideas I've come up with: Just wire them up "as-is" and hope the host can deal with joined D+/D- pins NAND gate with separate 5V pins as input, output to a MOSFET that disconnects Vcc when both are plugged in. Does the second option sound reasonable, or is a more complex solution required? <Q> As pointed out, you can't directly connect two hosts to a device - so if you just wired them together and somebody plugged it in on both ends, you'd have a problem. <S> However you also have a problem if only one end is plugged in. <S> USB, especially high-speed (480Mbps) mode is controlled impedance. <S> If you wire both connectors data lines together you end up with what is known as a stub in high frequency design. <S> The cable going to the unused connector will degrade the performance of the active connector. <S> You can buy dedicated USB2.0 multiplexers designed specifically for this sort of application - something like the TS3USB30 . <S> That would allow you to connect the data lines from both ports to the mux inputs, and connect the output to your device internally. <S> The mux will disconnect the unused connector which will disconnect the transmission line stubs. <S> For power I would probably use a power multiplexer such as a diode OR-ing circuit. <S> The VBUS line from one of the ports (before the power multiplexer) can then be used as the input to the data multiplexer. <S> Ground would be common (connected) between the two USB ports and your device. <A> Multiplexer as suggested by Tom Carpenter is a good solution. <S> But for full speed USB (12 Mbps), the stubs in the signals are not particularly important. <S> If the distance between the stub ends remain below 1/10th of wavelength, i.e. below ~2 meters, the reflections will not distort the waveform much. <S> Also, the voltage levels on D+ and D- pins will remain within the acceptable range, so there is not much risk of electronic damage to either host. <S> The main problem that remains is if you connect the +5V pins from both hosts together, there could be large currents involved. <S> You could use a diode from each USB connector's +5V pin to only let current come in, never out. <A> The lazy electronical design way would just be to use a physical multi connector switch to choose between which one is used. <S> Note that this would have the added benefit of being able to leave them both plugged in and use the switch to select which input is to be used. <A> There is no USB compliant way of doing this. <S> At best, you can use a USB multiplexer or switch IC, with GPIO to determine which connector is connected. <S> You'll have to decide which USB connection has precedence, because both cannot be connected to your USB interface at the same time. <S> Or look at alternatives. <S> Have one computer communicate to the other. <S> Or use Bluetooth or use Wi-Fi or a different connection type. <A> I have a Wacom tablet with this feature. <S> They handle the problem you're describing by using a mechanical interlock. <S> There's a plastic slider that blocks one port, or the other, depending on it's position. <A> Then the enclosure can be rotated so that the USB port is facing the connector, while all other elements are still conveniently accessible. <S> Another option is to locate both ports on one side of the enclosure (at the opposite ends) and add a sliding stub inside the enclose which the user can move to open one port or the other. <S> 12 MHz USB has a good chance of working with an extra disconnected port attached to it.
| Perhaps it would be easier to leave the USB connector at one side of the enclosure and duplicate and / or relocate all other user-facing elements symmetrically. To do this properly, you want a multiplexer IC. USB is not designed to be shared between two hosts and attempting to do so will lead to potentially disastrous situations.
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18650 in series while charging using TP4056 and connecting two buck regulators LM2596 in parallel, is it safe? I'm building a robot which contains a battery system. The 3x18650 batteries are connected in series and are charged using TP4056 boards. The voltage is then fed into the circuit shown below. The voltage dividers and power resistors assist in obtaining the voltage and current of the system. My questions are: Is connecting the batteries in series while charging a bad idea? The buck regulators have a maximum current output of 3A, that is why im thinking of connecting two of these boards in parallel to share the load, is it safe to do this? Any thoughts of improving the system. What is the simplest way to get the Arduino to know if the batteries are being charged or not? <Q> This will short out the battery. <S> This is because the ground is not floating on USB hubs. <S> Gnd will short out the lower V+ connected to it. <S> You would need 3 isolated USB wall chargers capable of 5W or 1A or more to charge in series. <S> You may add 2 LEDs with the anode (+) to 5V to indicate charging with the cathode (-) to 330 ~ 1K series R to CHRG low when active and STDBY low when done. <S> An actual current of 10mA is adequate which is V/R= <S> I <S> The charge current depends on the onboard PROGramming resistor into a current amplifier. <S> \$I_{BAT}=\dfrac{1V*1200}{R_{PROG}}\$ thus <S> 1.2k = 1A <S> The battery chargers are linear regulators and only need 4.2V in but 5V is ok but gets warmer with 1000mA and may need heatsink from 3V discharged cell. <S> (5-3)*1A=2W If your USB Hub cannot drive 1000mA then it is slow to charge up. <S> Your LM2596 will not generate split +/-5V supplies. <S> Yes, it is true the intermodulation or having each Buck supply interference with simultaneous charge currents is not a good idea. <S> Although it is a good idea to decouple the motor power from logic Vdd and Vss but not with this PS design. <S> Go to ti.com and register your email and choose a triple output design from their "Webbench" online in the browser with Java approval or Analog Devices. <S> Often a 3 secondary winding <S> forward converter <S> works best on a custom coil form. <S> By tight mutual coupling of coils, the ratios of voltage are maintained while regulating off one for feedback. <S> This is how PC PSU's generate all their voltages. <A> [ We are missing a schematic for the little charger boards. <S> In lieu of that, I'm going to assume that this charger board doesn't have galvanic isolation (or something equivalent) between the USB and charger output. ] <S> The proposed charging arrangement isn't going to work, because the USB ports aren't floating 1 with respect to each-other. <S> The grounds of the USB ports are connected together at the hub. <S> The battery ground of each charger board is connected to the USB ground. <S> That's okay when you have only one cell. <S> When you have two cells each with its own charger board like that, you will be connecting the node between the cells to ground. <S> So, you will be applying a short across one of the batteries. <S> At best, a circuit breaker on the hub will open. <S> By the way, do your cells have built-in protection? <S> I would recommend to abstain from using unprotected Lithium-ion and Lithium-polymer cells. <S> 1 Isolated USB hubs do exist, but they are an exotic. <A> It won't work. <S> TP4056 modules connect ground between input and output, so they cannot be "stacked".
| The batteries can NOT be connected in series when charging. At worst, you will toast the charger chip, or the USB hub.
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Why put a step-down transformer after a vacuum tube in an a valve amplifier? I'm researching valve amplifiers. I found this schematic for one: So the input is amplified by the first valve, and then the amplified signal is amplified again by the second valve, right? My question is, why is the voltage being stepped down before going to the speaker? It seems pointless to me, increasing the voltage with the valves and then decreasing it again. All the schematics I can find online do this. Why? (Is the 300V rail at the top related to the transformer? If not, what's it for?) <Q> It's a question of impedance. <S> The anode (plate) voltage of the tube varies over a wide range, while the current varies over a much smaller range. <S> If you define output impedance as $$Z_{out} = <S> \frac{\Delta <S> V}{\Delta <S> I}$$ <S> This usually works out to a fairly high number for a typical vacuum tube, on the order of thousands of ohms. <S> On the other hand, most speakers have a low impedance — on the order of 4 to 16 Ω — which means they want a relatively higher current change coupled with a relatively smaller voltage change. <S> Note that in both cases, you're talking about the same amount of power (voltage × current), which is what the amplifer is really achieving — an increase in signal power from input to output. <S> The transformer provides this impedance change. <S> It trades off a high voltage swing for a high current swing. <S> Without it, you'd get only a tiny fraction of the available signal power actually delivered to the speaker, limited by the relatively low current in the tube. <S> From a comment: <S> Any idea what the 300V rail is for? <S> Is it simply a power supply for the valves? <S> Why is it so high-voltage? <S> The 300V power supply is required for much the same reason: The output of the impedance of the tube is inherently high. <S> The 6V6 tube is rated for 50 mA plate current (average), which means that the signal current swing must be less than about ±40 mA (peak). <S> Similarly, the tube is rated for a plate voltage of 250 V (nominally, but it is frequently overdriven in this respect), so the signal voltage needs to be less than about ±120 V (peak). <S> The signal power available at the output is therefore the RMS current multiplied by the RMS voltage, or: $$\frac{40 mA}{\sqrt{2} <S> } \cdot \frac{120 V}{\sqrt{2}} <S> = \frac{4.8 W}{2} = <S> 2.4 <S> W$$ <S> Note that this works out to an output impedance of: $$Z_{out} = <S> \frac{120 V}{40 mA} = 3000 <S> \Omega$$ <S> To drive an 8Ω speaker, you'd use a 3000Ω:8Ω transformer (19.4:1 turns ratio), which would give you 4.38 V RMS and 548 mA RMS at the speaker. <A> In addition to what Dave Tweed said (+1), the transformer in this case also eliminates the DC bias current from going to the speaker, and decouples the common mode input and output voltages. <S> The plate current of V1 sits at a center value when idle. <S> The input signal causes the plate current to go both up and down from the center value according to the peaks and troughs of the input signal. <S> Even if there was a speaker that was impedance-matched to the plate of the 6V6, the DC bias current thru <S> it would not be desirable. <S> The transformer also blocks DC, while passing the relevant AC parts of the signal. <S> Note that impedance matching is still the primary reason. <S> Since a transformer is required for that anyway, the designer of the circuit made use of the fact that it also blocks DC, and that the common mode input and output voltages are decoupled. <S> This latter fact allows one side of the speaker to be grounded, even though the transformer primary is tied to 300 V. <A> Short Answer: <S> Reduce output impedance to prevent significant voltage loading For good bass response the speaker is a linear motor/generator with back EMF on kick drum pulses. <S> Thus the output impedance must be much lower than the speaker. <S> THis is also called the Dampening Factor= Zspeaker/Zout and is only 20 on cheap low power amps , 100 on good amps and 1000 on great power amps. <S> So what is it on a Vacuum Tube Amp? <S> THat depends on the Tube Zout divided by turns ratio of transformer squared. <S> So the impedance reduction of turns ratio n² reduces the high output impedance to somewhat lower than the speaker impedance. <S> Without specs, its hard to guess but never as good as soldid state but infact the harmonic distortion from back EMF, not just the tube's soft limiting but of the poor damping factor may be "pleasant" to some guitar players but "muddy" to audio experts playing broad spectrum. <S> Since turns ratio also reduces voltage by <S> n, the tube voltage swing must be n times bigger than what the speaker sees <S> e.g. <S> thus maybe 9 times bigger swing and Vdc and /81 reduction of the high output impedance.. <S> .perhaps <S> more turn ratio... <S> 20;1 Voltage ratio is 400:1 impedance ratio possibly giving a dampening factor of <10 <S> ie poor <S> D.F. <S> so they often used 16 Ohm speakers. <S> BTW Many Tube amp designs are much better than this one. <A> I need to correct your misleading terminology. <S> It is an impedance matching power transformer , not a step-down transformer! <S> In order for you to understand the answer, you need to know: 1) <S> The purpose of an amplifier, is to amplify power (not current or voltage). <S> 2) <S> Vacuum tube devices could only provide "small" currents, but could handle high voltages. <S> 3) <S> Vacuum tubes had impedances of K ohms , while speaker impedances were in the order of ohms . <S> Since P = VI, to provide the max power amplification with small current devices, one has to use the max voltage that the device can handle (this is the answer to your "why high voltages" question). <S> Since max power transfer between two devices occurs when their impedances match, the impedance matching power transformer was the ideal solution to this problem (and the other problems mentioned in the other answers). <S> The voltage rails of any circuit, are required because of the "conservation of energy law." <S> Although signal power is being amplified, it comes at the cost of the power supplied by the voltage rails.
| If you use a lower plate voltage, the available power is reduced proportionally.
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Add constant DC voltage to a variable DC input I'm designing a PCB and I need a voltage rail, roughly 3 volts above the main input (of 3 - 30V). This rail is to power some CMOS devices, so it won't need to supply much current. Currently I plan on using 2 AA batteries in series with the input, but those batteries are huge compared to the PCB so I want to get rid of them. I have searched for multiple solutions, but none match my requirements: Charge pumps probably won't work, because of the big voltage change in the input. Boost converters deliver a steady voltage, but I need an addition to the input. I have looked at isolated DC/DC converters, but these usually run on 5V or higher, or they are too expensive. Do you have any ideas? EDIT: I don't necessarily need a higher voltage. I need a stable 5v from the input. I'm not willing to use a buck boost converter, since that will take up too much space. <Q> All you need is a roughly 4 V square wave somewhere. <S> It can be between ground and +4 V, for example. <S> The square wave goes directly into a capacitor, so there can be a arbitrary DC offset across it. <S> As long as this offset changes little during individual pump cycles, it can be considered as "constant" for this purpose. <S> Here is the basic circuit: <S> Vout will be the amplitude of the square wave, minus the two Schottky diode drops. <S> For example, it would be about 4.3 V open circuit with a 5 V square wave as input. <S> C1 allows any arbitrary offset voltage between the square wave and Vout. <S> Of course C1 needs to be rated for the voltage. <S> C2 only needs to handle Vout. <S> Either side of Vout can be used as the reference. <S> If Vout- is tied to your 30 V rail, than Vout+ will be a little higher. <S> If Vout+ is tied to your 30 V rail, then Vout- will be a little lower. <S> The current capability of this charge pump is proportional to the pump frequency. <S> A convenient place to get a square wave from can be a clock output of a microcontroller that is there already anyway. <A> 3-30V is a big input range. <S> Assuming you're not up for custom magnetics, you could do it with two or three devices- <S> a buck-boost regulator (eg. LM5118 ) to 5V then an isolated 5V:5V DC-DC, or an LDO + boost regulator + DC-DC. <S> (Or use a charge pump in place of the DC-DC once you have 5V, but the 5V:5V isolated DC-DC converters are cheap). <S> The first would be much more efficient, but substantially more costly. <A> A 74HC gate will work from 3V (or even lower) to 5V so you could simply use a 5V LDO to limit the input 3-30V to 3-5V and supply your 74HC gate with this. <S> Since static supply current is very low, you should be able to find a LDO with tiny dropout voltage. <S> Unless the gate supplies something that draws current. <S> If all you need is a NOR gate though, here is one that will work with your 3-30V input voltage. <S> ( source ) <S> This will not be as fast as a 74HC gate but it could be what you need. <S> It is also small and cheap. <S> Logic can also be implemented with open-collector output comparators like LM339. <S> The outputs can be combined. <S> Note that the output voltage is the same as the supply voltage, <S> so if you want to limit it, simply add a zener.
| You can also make a square wave from a Schmitt trigger inverter with a R-C low pass filter feeding its output back to its input. This sounds like a good application for a charge pump.
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Why doesn't my computer USB port break when I plug in a 2A device? For typical AC to DC power adapters, I understand that the device draws the current it needs from the adapter. If it pulls more than the adapter can supply, this can cause overheating and lead to breaking the adapter. Following this... I am always charging my phone or tablet on 1A USB chargers or in normal PC USB ports, yet I have never experienced a broken USB port resulting from this. My question is: Do USB ports or devices typically have a mechanism to regulate current to avoid overloading ports? I'm mostly thinking in the context of devices which charge via USB. <Q> To be compatible with the original standard, USB devices should not draw more than 100mA (which is plenty to power the logic interface), until they have negotiated with the host, to find out what it can supply. <S> After successful negotiation, they can draw up to 500mA. <S> This is to protect the operation of a 4 port hub, should it be plugged into a PC with all its downstream devices already attached. <S> Not all USB devices are compliant to the standard, but just draw full current anyway, USB toys commonly do this. <S> Most PCs provide 500mA anyway, so it all generally works. <S> Dumb power supplies generally hold their data lines in particular states, to signal to the device being charged that they are a power supply, with a certain capability. <S> Later standard revisions allow USB-C and PCs higher currents, and higher voltages (eek!) <S> to be negotiated. <A> Historically: really old mainboards connect USB power pins to the 5V power rail, with no protection power on by keypress was added, which added a jumper or a BIOS setting that decided whether USB ports would be powered from the standby power or from the regular 5V rail. <S> Since standby power was introduced in ATX, this does not exist on AT mainboards. <S> USB port power control was added to root hubs, allowing the host to turn off power to ports programmatically (with the hub controller switching an external "power" FET). <S> These supply a lot of power to the ports, and have no meaningful protection besides using the FET as a current limiter, which is generally a bad idea for prolonged time as these are not usually cooled. <S> the control FET was later integrated into the root hub as manufacturing processes improved to a point where you could run a few hundred mA through what is otherwise a logic IC. <S> This setup has lower current carrying capability, and shorting USB ports will usually destroy the southbridge IC and/or cause a reboot. <S> current monitoring and emergency shutdown were added to the controllers as well as processes permitted. <S> There are also some older mainboards that implement current monitoring as discrete components (increasing board cost, but giving robustness), but on consumer boards, expect the cheapest possible approach. <S> Some modern boards also use the same kind of integrated voltage/current controller that usually provides CPU and chipset power to control other circuits, as these ICs are sufficiently cheap that duplicating the logic around them saves enough engineering effort to make them a good contender to "dumb" FETs. <S> On such boards you'd probably be able to draw exactly 2.000A, but current monitoring and reporting may be limited as communication between the USB root hub and the power controller is just "enable" and "error" signals. <A> My question is: Do USB ports or devices typically have a mechanism to regulate current to avoid overloading ports? <S> The answer is yes, they do. <S> USB includes a fairly elaborate protocol that allows devices and hosts to negotiate the amount of power that the device can use. <A> Why doesn't my computer USB port break when I plug in a 2A device? <S> and Do USB ports or devices typically have a mechanism to regulate current to avoid overloading ports? <S> Yes, USB ports do have a mechanism, but not to "regulate current", but to protect USB ports from overcurrent . <S> Overcurrent leads to port disconnect, not to "regulate it". <S> the device will take from a port. <S> If the port is a regular USB data port, connected devices will limit their consumption to 500 mA (and to 900 mA if the device can detect USB 3.0 port). <S> That's why your computer ports don't break, because the "2-A device" can detect this, and becomes a "500/900-mA device".
| And (not "or") USB devices (with internal batteries) typically have a mechanism called "port signature detection", which makes a USB device not to "regulate current", but "limit current"
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Battery shorting out in ESD mat? So we are having a discussion on if an Anti-Static ESD mat would conduct current across the leads of a battery leading to a short. The overall consensus is that the battery would short out. <Q> No, the resistance of a typical anti-static mat is far to high to cause significant current to flow from a typical battery. <S> Did you try measuring the resistance? <A> Indeed a current would flow but it will be a very small current, you would be nowhere near the current that would flow if the battery was shorted with a copper wire. <S> An ESD mat is conductive but not that much. <S> For ESD a 1 Mega ohm resistance path is already enough for the charge to flow away in a safe and controlled way. <S> The not-that-low resistance of an ESD mat is high on purpose. <S> Suppose you're working on some mains live equipment. <S> That's also why an (already high-resistance) ESD mat should be connected to ground via a 1 Mohm resistor. <S> That's for safety. <A> Literally, yes, the battery would conduct across the mat, in essence a short. <S> But in practice, it's not a dead short, where high current draws and high heat can occur, which is what most people would consider a "short". <S> The short on a ESD mat is on the same level as leakage current or self discharge rates for a low voltage battery.
| If you touch a live connection you will suffer a much more severe shock if part of your body is touching a low-resistance grounded plate (like a grounded metal plate) than if you would be touching a high-resistance ESD mat.
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Why is the time constant 63.2% and not 50% or 70%? I am studying about RC and RL circuits. Why is the time constant equal to 63.2% of the output voltage? Why is it defined as 63% and not any other value? Does a circuit start working at 63% of output voltage? Why not at 50%? <Q> Other answers haven't yet hit upon what makes e special: defining the time constant as the time required for something to drop by a factor of e means that at any moment of time, the rate of change will be such that-- if that rate were continued --the time required to decay to nothing would be one time constant. <S> For example, if one has a 1uF cap and a 1M resistor, the time constant will be one second. <S> If the capacitor is charged to 10 volts, the voltage will fall at a rate of 10 volts/second. <S> If it's charged to 5 volts, the voltage will fall at a rate of 5 volts/second. <S> The fact that the rate of change decreases as the voltage does means that the voltage won't actually decay to nothing in one second, but the rate of decrease at any moment in time will be the current voltage divided by the time constant. <S> If the time constant were defined as any other unit (e.g. half-life), then the rate of decay would no longer correspond so nicely with the time constant. <A> It's built into the mathematics of exponential decay associated with first-order systems. <S> If the response starts at unity at t=0, then after one "unit of time", the response is \$e^{-1} = <S> 0.36788\$. <S> When you're looking at a risetime, you subtract this from unity, giving 0.63212 or 63.2%. <S> The "unit of time" is referred to as the "time constant" of the system, and is usually denoted τ (tau). <S> The full expression for the system response over time (t) is $$V(t) = <S> V_0 e^{-\frac{t}{\tau}}$$ <S> So the time constant is a useful quantity to know. <S> If want to measure the time constant directly, you measure the time it takes to get to 63.2% of its final value. <S> In electronics, it works out that the time constant (in seconds) is equal to R×C in an R-C circuit or L/R in an R-L circuit, when you use ohms, farads and henries as units for the component values. <S> This means that if you know the time constant, you can derive one of the component values if you know the other. <A> The decay of an RC parallel circuit with capacitor charged to Vo v(t) = \$Vo(1-e^{-t/\tau})\$ , where \$\tau\$ is the time constant R\$\cdot\$C. <S> So v(\$\tau\$)/Vo is approximately 0.63212055882855767840447622983854 <S> In other words, the time constant is defined by the RC product (or L/R ratio), and the seemingly arbitrary voltage is a result of that definition and the way exponential decay or charging occurs. <S> Exponential decay is common to various physical processes such as radioactive decay, some kinds of cooling etc. <S> and can be described by a first-order Ordinary Differential Equation (ODE). <S> Suppose you want to know the time when the voltage is 0.5 of the initial voltage (or final voltage if charging from 0). <S> It is (from the above) <S> t = <S> -\$\ln(0.5)\tau\$ or about 0.693RC <S> Either way you do it, some irrational numbers pop up and dealing with RC=\$\tau\$ is the "natural" way. <A> Just as a complement to the other excellent answers by Dave Tweed, supercat and Spehro Phefany, I'll add my 2 cents. <S> First a bit of nitpicking, as I wrote in a comment, the time constant is not defined as 63%. <S> Formally it is defined as the inverse of the coefficient of the exponent of an exponential function. <S> That is, if Q is the relevant quantity (voltage, current, power, whatever), and Q decays with time as: \[Q(t) = <S> Q_0 \ <S> ; e^{- k t} \qquad <S> (k>0)\] <S> Then the time constant of the decaying process is defined as \$ \tau = 1 / k \$ . <S> As others have pointed out, this means that for \$ t = \tau\$ <S> the quantity has decreased by about 63% (i.e. the quantity is about 37% of the starting value): \[\frac{Q(\tau)}{Q_0} = e^{-1 <S> } \approx 0.367 <S> = 36.7 \%\] <S> What other answers have only marginally touched is why that choice has been made. <S> The answer is simplicity : the time constant gives an easy way to compare the speed of evolution of similar processes. <S> In electronics often the time constant can be interpreted as "reaction speed" of a circuit. <S> If you know the time constants of two circuits it's easy to compare their "relative speed" by comparing those constants. <S> Moreover, the time constant is a quantity easily understandable in an intuitive way. <S> For example, if I say that a circuit settles with a time constant <S> \$ \tau = 1 <S> \mu s\$ , then I can easily understand that after a time \$3\tau=3\mu <S> s\$ <S> (or maybe \$5\tau=5\mu s\$ , depending on the accuracy of what you are doing) I can consider the transient ended ( \$3\tau\$ and \$5\tau\$ are the most common choices as rules of thumb for the conventional transient duration).
| In other words the time constant is an easy and understandable way to convey the time scale on which a phenomenon occurs.
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How can I modify this circuit for some hysteresis and better design? I need to make a circuit even when the push button pushed and released very fast, the final output will stay ON at 12V for roughly a second. %10 accuracy for the timing is fine. An LED will indicate the ON and OFF time. I tried to combine couple of circuits to obtain the goal. Below circuit uses C1 and R4 for adjusting the pulse ON time. LM2903 is used as a comparator. The output of the comparator is buffered by a transistor buffer to obtain enough current to the load and the LED. The need for buffer is to obtain better results for different loads: (left-click to enlarge) And here is the voltage and current waveforms from the simulation for the nodes marked as n , p , op , out , LED current I(D1) and switch current I(S1) : My problem in this circuit I cannot add hysteresis. Because if I add positive feedback it will be connected to the RC timing circuit. I might overcome this by changing R4 but I cannot even create hysteresis by adding a resistor between the nodes out and p . How can I add some hysteresis to this comparator without affecting the function of this circuit? Is there anything fundamentally wrong or missing with this design? <Q> I haven't fully analyzed your circuit. <S> You can always add a buffer between stages to avoid loading effects (such as positive feedback) between stages if that's what you are worried about. <A> @Jonk showed many of hundreds of great ways to make a pushbutton power one-shot. <S> Schmitt Trigger CMOS is very useful and comes in many supply voltage ranges. <S> The old CD4000 series could drive a few mA max. <S> while the 74HC series is limited to 5V. <S> THen Comparators are easy to use with 4 in an IC. <S> Here's another way. <S> If you never used CMOS before, read about precautions for handling. <S> Minor Fixes to yours <S> Many ways to do it. <S> The main problem of 12s is fixed changing R4 from 1M to 10M assuming 2/3 threshold from 33% hysteresis. <S> In order to not load the comparator pull-up resistor and offset the hysteresis threshold too much, I changed the NPN bias. <S> When cascading saturated high currents to the same voltage use 10 or 20:1 <S> Rb/Rc ratio. <S> 1:1 is ok but not needed. <S> Since the comparator is low Z out when low the NPN base R to gnd is redundant. <S> R for the LED can be increased to 1k or 2k <A> There is nothing fundamentally wrong with your design, in the usual sense. <S> However, you are missing few tricks. <S> Can you change that? <S> Yup. <S> Notice that your output buffer is essentially two inverters in series. <S> Rload is a good deal greater than R1, the LED current limit resistor, and the total required current is on the order of 25 mA. <S> This means that you need about 2.5 mA base drive for the PNP, and a 290x will provide that without blinking. <S> So you don't need the NPN to boost the comparator output, and can safely do away with it. <S> Under the new setup, the + input of the op amp connects to the trigger level voltage divider (R2/R3), so running a large resistor (100k to 1M) from the comparator output to the + input will have no major effects. <S> It will produce unequal trigger levels depending on the state of the comparator output, but that is in the nature of practical hysteresis. <A> The following circuit will do what you are looking for. <S> It debounces the switch and provided a \$1\:\text{s}\$ wide pulse of \$12\:\text{V}\$ to your ERAD4000 Trigger inputs (pins 8 and 9, as I understand it.) <S> simulate this circuit – <S> Schematic created using CircuitLab \$C_1\$ and \$R_1\$ and the threshold voltage required by \$M_1\$ set the timing for you. <S> In this case, I've worked out the approximate values for \$C_1\$ and <S> \$R_1\$ when \$M_1\$ is a BSS145. <S> But you can use other choices for the NFET device. <S> Just be aware that the gate threshold voltage will affect the timing. <S> (For example, a BSS123 would also be fine here but because it has a lower threshold voltage, the timing will be longer than I estimated.) <S> \$R_2\$ determines the base current in \$Q_1\$. <S> As shown, the base current is about \$\frac{12\:\text{V}-V_\text{BE}}{R_2}\approx 1.6\:\text{mA}\$. <S> So the collector of \$Q_1\$ will provide substantially more current compliance than the ERAD4000 requires.
| Adding hysteresis will affect your timing (capacitor waveform) since that is connected to the + input of the comparator. Swap + and - inputs at the op amp, and get rid of the NPN section of the buffer, and the circuit will work just the same. Regardless, it shouldn't be hard to make adjustments to get the desired timing. This needs to be about one tenth of the required current compliance for the ERAD4000 trigger. If you need to increase time duration increasing R2 up to 1M may be done. The circuit will automatically shut itself off after the time period.
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Is a 5V DC Voltage @ 30mA applied, applied to water, unsafe? As my question states, I have 5V DC, with an input current of 30mA, being applied across two exposed metal electrodes which are inserted into water. I am trying to measure the conductivity of water for a scientific experiment. Would there be any concerns about the safety of this system, in case someone touches the water? <Q> It is definitely safe. <S> Also earth your power supply properly. <A> If you are using a certified power supply mounted in a location where it is safe from splashes or ingress of water <S> then that is no more dangerous than touching the metal parts of the connector on a USB phone charger. <S> That is, it is safe. <A> No - that is generally safe as long as the voltage is not going to inncrease beyond that. <S> 5V cannot drive enough current through the body to cause damage even if the skin is broken. <S> It is even safe if someone sticks their head between the two electrodes. <S> I have experimented with tDCS at 9V and saline electrodes attached to the head. <S> That has trouble pushing more than 2mA. <S> However, make sure that the PSU cannot fail to a much higher voltage eg mains.
| But make sure that your power supply doesn't have a risk of increasing the voltage to abnormal level or risk of electrical isolation failings (between the output and mains).
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How do I really calculate these resistors? I am learning about it. I would like to focus on this circuit I am trying to calculate R10 and R11. I want Ic = 10 mA. Transistor is 2N2222A. This is an amplifier. If I understood how amplifiers work, I want Vc to be in the middle of the rail, so the amplifier will have maximum output swing. The circuit is powered by a 9V battery, so the middle of the rail = Vc = 4.5V. So, I have calculated Rc like: \$ R_C = \frac{V_{CC} - V_{CE}}{I_C} \$ \$ R_C = \frac{9 - 4,5}{10 \times 10^{-3}} \$ \$ R_C = 450 \$ ohms. I have calculated Ib like this: \$ I_C = \beta I_B \$ \$ \beta = 225 \$ for 10 mA so, \$ I_{B} = 44,44 \thinspace \mu A \$ Hence, \$ R_B = \frac{V_{CC} - V_{BE}}{I_B} \$ \$ R_B = \frac{9 - 0,7}{44,4444 \times 10^{-6}} \$ \$ R_B = 186,750 \$ ohms. The problem is: when I put that on the simulator, it gives me a Vc equal to 4.24V, almost in the middle of the rail but not exactly and the worst part is that the simulator gives me a Vbe = 0.562V. As far as I know, this silicon transistor will need 0.7V to work and the value from the simulator is not showing the whole thing will work as expected. How are these calculations really done, taking into consideration differences from theory to the real world? <Q> You've done the calculations correctly, but the entire basis for the calculations is just a rough approximation of transistor behavior. <S> As you see, \$V_{BE}\$ isn't really exactly 0.7V (most of the time) and the \$\beta\$ isn't exactly 225 (most of the time). <S> This kind of biasing is very sensitive to variations in transistor parameters so it isn't used much in practice. <S> A better biasing scheme uses four resistors, with a voltage divider for the base bias voltage and a resistor from the emitter to ground (for a little negative feedback). <A> However, your biassing scheme will only work well with one set of transistor parameters, it's very sensitive to variations in beta. <S> A better biassing scheme will maintain a good bias point even with transistor variations. <S> A more stable one is shown below. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The 450 ohm resistor is calculated as you have done, as is the base current. <S> I've then taken roughly 10x Ib, a nice round 500uA, and calculated resistors to give me 700mV and 3.8v drop at that current. <S> That standing current swamps variations in Ib with beta. <S> As the transistor will want to try to keep around 0.7v on the base, it will act as an amplifier to try to keep the collector at 4.5v. <S> R3 can be broken into 2 parts with the midpoint decoupled to ground to restore the AC gain. <S> You'll notice I've not attempted to allow for the base current being sourced from the R2/3 divider. <S> This is deliberate. <S> It shows that when you have a more stable biassing scheme, you can get away with significant errors, and still have a working amplifier. <S> As transistor beta increases, the base current drops. <S> If you have a scheme that will work well with zero base current, it means your design can be changed to use a really good transistor, and it will still work. <S> An even more stable circuit is shown in Dan's answer. <S> You might well bypass R5, or a portion of it, with a capacitor, if you want more gain than Rc/Re. <A> You don't use that circuit in the real world.... <S> Beta is horribly badly specified in real bipolar transistors (For example a random BC548 datasheet <S> I just looked at <S> give Hfe as 110 (min), 800(max), and it will vary with temperature and from device to device, <S> so biasing the way you are trying to do it will give nothing good as a result. <S> Far more common is to use a scheme like the following: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here the emitter resistance provides negative feedback <S> (Its IR drop is subtracted from the base voltage) <S> so Beta becomes (provided it is sufficiently large) mostly irrelevant, generally real circuits make extensive use of negative feedback to remove the dependency on poorly controlled things like device Beta. <S> The other nice thing about this circuit is that the gain is (to a first approximation) <S> the ratio of Rc to Re.
| You have done the calculations correctly.
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Where to find proven design patterns? When developing software, it's recommended not to reinvent the wheel but to use battle tested libraries e.g. for sorting or proven design patterns like the GoF. However when it comes to electrical engineering I'm unaware of such libraries. I have bought Circuit pattern trading cards however a friend of mine told me those implementations are not ideal. Therefore I'm asking: where to find proven circuit design patterns and higher level libraries for electrical engineering tasks? <Q> A good place to start would be the Reference Design Library at AllAboutCircuits.com <S> It has a large searchable database of common circuits. <S> Another option is more vendor specific. <S> For instance Texas Instruments has nice tools for helping design typical circuits. <S> Even Digi-Key has a good reference library. <A> Most (all?) <S> How well these designs are tested varies from manufacturer to manufacturer and probably even depending on which applications engineer wrote the note. <S> Historically there were books that collected these designs, mostly copied directly out of the app notes, with titles like "1001 Electronic Circuits" or "501 Measurement Circuits". <S> But those are hardly needed in the era of Google. <A> Art of Electronics is good, the other sources mentioned in other answers probably as well, but I just have a nagging feeling that there's an underlying assumption at play here, which is false. <S> At the risk of being blunt, I dare suggest that there is no "smarter" method of developing electronic circuits, and no way to cut corners. <S> Trying to save design time and effort will cause even more delays at a later stage in design <S> The analogies between software development and hardware development are not exactly one to one. <S> For example, in software development it may not be necessary to know the inner workings of something like a MPEG video decoding library, you just use the library. <S> Similarly, in electronics, you don't need to design an MPEG decoder if you can pick that up as a component. <S> But there the analogies end. <S> In electronic design, the division beween what you design (your own circuits and printed circuit boards) and what you use (ICs that other people have designed) is quite clear. <S> You cannot tinker with the IC design, unlike that MPEG decoder library, whose source code you can probably look. <S> In electronics, you design what you design and you must know completely how it works. <S> You cannot, generally, just take one circuit from one book and another circuit from another book, throw them both on the same PCB and hope that they work together. <S> I think you must have a complete understanding of how each of them works so that you can evaluate how they work together. <S> No way to cut corners. <S> No way to work smarter, not harder. <S> That's my point of view anyway. <S> You don't need to reinvent the wheel - in fact, you absolutely must not. <S> You better be well familiar with the design of the wheel even before you start designing your product, and then measure and re-measure it so that you're pretty darn <S> sure that it's round before you ship it to your customers.
| If you are looking for more basic/fundamental circuits, I think you would be best off with a book such as The Art of Electronics . chip manufacturers publish application notes showing how their chips can be used in a circuit, and these usually contain reference designs that you can use.
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Does a Crystal Radio coil require enameled wire or insulated turns? When creating a coil for a Crystal Radio, why do I have to have insulated windings? Why does it matter if the individual turns touch each other? When it "shorts out" in this context, what does that mean, as in where did the charge go? <Q> More turns = more field and more inductance. <S> If you use bare wire, the current will take the least resistance path, which would be the shortest path from one end to the other, and there will be virtually no magnetic field at all. <A> Yes it matters if the wires are insulated. <S> Using regular insulated wire rather than enamel will result in a much larger coil. <S> If the individual turns touch each other <S> then the current can flow along the length of the coil rather than going around all the loops. <S> This will change the inductance of the coil so it won't tune to the frequencies you want. <A> If the individual turns are bare and touch each other, then you don't have a coil but just a piece of wire. <S> The effective inductance would be very low.
| If you use insulated wire, the current is forced to go all the way around each loop, and this helps add to the magnetic field that is crucial for the operation of an inductor.
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Connect 90w charger in a laptop in which the charger was 130w Is there any permanent damage on the battery or another part if i used a 90w charger on a 130w laptop? I need to do this, because my new 130w charger arrives only day 9 and I must continue using the laptop. 130W INPUT: 100-240 V - 2.5 A OUTPUT: 19.5 V - 6.7 A 90W INPUT: 100-240 V - 1.5 A OUTPUT: 19.5 V - 4.62 A <Q> it shouldn't cause any damage if it is only used to charge your battery. <A> First, it is not CHARGER, it is an AC-DC adapter. <S> Then, if the 90-W adapter is of the same brand as the original 130-W adapter, and all are for the same laptop brand and have interchangeable barrel plug, the laptop must be able to detect the adapter properties, and scale down its consumption, and likely limit internal charging rate and limit any turbo modes the laptop can execute. <S> This is the normal case scenario. <S> In worst case your "a laptop" will determine that the 90-W power of this adapter is insufficient and won't accept any charge. <S> Your laptop will operate on internal battery and drain it totally, eventually. <A> The laptop might throw you an error and go in an super energy saving mode. <S> That is, lowest cpu and gpu clock possible, which is very slow. <S> If it doesn't. <S> you might go over the rating of the adapter when using the laptop to it's full potential, while charging the battery. <S> Charging when turned off is not a problem. <A> If the laptop power supply doesn't get too hot and is rated for the same voltage (which it is) then its ok use. <S> If its starts to get too hot you may wish to remove it for a while and perhaps allow the charger to cool down to avoid potential damages. <S> Perhaps best to charge for smaller periods.
| However, if you plug it in and it is supplying power to your laptop and charging it at the same time, depending on the power consumption of your laptop, it might overload the charger and damage it if there isn't any protection in the charger. Usually a normal laptop will report what kind of power adapter is connected, somewhere in general properties of "ACPI battery", or through some laptop-specific applet. From your comment it looks like you have the normal scenario, so you should be just fine.
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Is the USB B connector still standard? Nowadays, There are several kinds of USB connectors for USB connection. A type B type micro-B mini-B USB C Etc. The connector sizes are gradually smaller than before. It is a big advantage because it reduces board size. But the B type connector is more suitable for a big size product because it is very easy to connect the plug. But I rarely see the B type connectors now. Is the B type connector is removed from the standard? Can I make a product by using the B type connector? Thank you very much. <Q> Yes - USB B connector is still part of the standard. <S> Yes - you can design a device with USB B connector if you have enough space. <S> You can also design a device with any other connector carrying USB signals. <S> If you use non-standard connector you probably can't claim that the device is fully USB-compliant, certify it easily or can't put the USB logos on it. <A> USB B connector is still part of the USB standard, "filo" is correct. <S> All USB revisions (including latest revision of USB standard, USB 3.2) include backward compatibility with USB 2.0, the general packet-based framework remains the same, and all legacy cable assemblies (which includes USB 2.0 Type-B connector) are fully specified. <S> The new revision of USB standard only splits the legacy cable drawings and definitions into a separate document (from USB 3.2 specs, Section 5 page 50): <S> The electro-mechanical definition and requirements for USB connectors and cables have been removed from this specification and are now located in the USB 3.1 Legacy Cable and Connector specification . <S> The exception is the set of mini connectors, mini-A. mini-B, and mini-AB receptacle, which was retired from USB 2.0 specifications, and superseded by flimsy micro-A-B set of connectors. <S> So using mini-B won't lead to USB-IF certification logo, but the old-style USB 2.0 Type-B receptacle is still a valid design option. <S> CLARIFICATION: <S> USB 3.2 Specifications state, Section 3, page 15 <S> USB 3.2 is a dual-bus architecture that provides backward compatibility with USB 2.0. <S> One bus is a USB 2.0 bus (see Universal Serial Bus Specification, Revision 2.0) <S> Formally it means that all USB 2.0 provisions, with all ECNs are still in effect, including all connector arrangements. <S> AMPLIFICATION: <S> More formally, USB electro-mechanical connectivity nowadays is defined in the USB Type-C specifications . <S> The Type-C specs define backward compatibility by specifying "legacy cable assemblies" like "USB 2.0 Standard-B to Type-C", which implies that there must be USB 2.0 Standard-B receptacles to work with. <A> Is the B type connector is removed from the standard? <S> The old Type B receptacle is superseded by "USB 3.1 Standard-B Receptacle" and "USB 3.1 Powered-B Receptacle". <S> The new receptacles are backwards-compatible with old Type B plugs, so technically Type B is not "removed", since its plug is still supported somewhat. <S> However neither of two new USB 3.1 type B plugs can be inserted into old Type B receptacle. <S> Furthermore, no cable assemblies with old Type B plug are allowed by 3.1 standard, which means Type B socket is deprecated de facto. <S> Can I make a product by using the B type connector? <S> Now, this is different. <S> I did not find any 3.0/3.1 documents on Type B certification. <S> To me this looks like any new device with Type B receptacle will not be certified as USB 3.0 compliant. <S> I don't know whether or not it can still be certified as USB 2.0 compliant device. <A> It's still used. <S> A lot of USB audio equipment uses it, even today, even in new products, because the impression that the cables are better quality than usb mini or micro cables. <S> Some audiophiles even believe that the B cables sound better, although I doubt there's much actual evidence to support that belief.
| A USB device can be made with Standard USB 2 B receptacle, the only downside is that it can't claim "USB 3.x compatibility", it is a "USB 2.0 device", with all corresponding USB 2.0 certifications/implications.
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Why are the SRAM data and address pins numbered? As far as understand, it would make no difference to the operation of an SRAM if you mixed up the order of the address or data pins. E.g. http://www.farnell.com/datasheets/1911297.pdf?_ga=2.220805788.545458806.1537605448-1970519925.1482350159 Removable non-volatile chips, such as PROMs and EEPROMs, have address and data pin numberings defined, as they need to be consistent across programmers and application circuits. Why do SRAM chips have numbered data and address pins? Is there any operation impact if I swap address pins around to make a PCB easier to route? <Q> You are quite correct. <S> I would strongly advise putting a note on the board that you've done this, to avoid freaking out any subsequent engineer who comes along to debug or modify it. <S> Why are they labelled? <S> If the pin functions were labelled 'pool of address pins' and 'pool of data pins', then at some point your CAD tool would insist on identifying them, and you would need to make a choice, even if it was only 'ADR_line_pin_23' sort of identifiers. <S> So they do it for you. <S> It's convenient, and recognisable, each pin has a physical identifier and a function identifier. <S> When the manufacturer does it, and is consistent with standards like JEDEC, it eases the task of engineers debugging stuff. <S> If I've learned the pinout of a particular package style, then I can expect to find those signals on the same pins without an extra translation step through the schematic, the less mental work I have to do when debugging, the better. ' <S> Let's see, this is word addressable, so <S> the LSB adress shouldn't be changing at all as it reads memory, now where is the LSB'. <S> Exactly the same goes for multiple functions in a single package. <S> While a quad opamp in a single package tends not to be labelled amp0, amp1 etc, a triple analogue switch does tend to be labelled s0, s1 and s2. <A> Some reasons : Pinouts are often standardized between static RAM and EPROM/FLASH: "JEDEC" RAMs are usually arranged in matrices with lines and columns. <S> Using consecutive addresses may require a little bit less power. <S> For events that affect the matrix, such as hardware defects or a SEU (high energy particle hitting atoms), knowing which memory cells are adjacent may be useful for finding errors or designing error correction codes. <S> Yes, you can swap address and data pins on static RAM. <A> Is there any operation impact if I swap address pins around to make a PCB easier to route? <S> If your design uses different chips to transfer data stored in a memory chip then the transfer of that data relies on both accessing chips knowing the data and address bus the same way.
| With an SRAM, where all address pins are equivalent, you can indeed juggle the address pins around within their set on your board to ease layout, and also for the data pins.
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MAX485 Receive Output pin current sufficient for optocoupler LED? I want to recreate a project found at this link where an optocoupler's LED is a driven by the receiver output of a MAX485. Now, my question is, the LED series resistor (R5) value is 470ohms, if my calculation is right the led is taking around 10.6mA however I cant find in MAX485's datasheet , about how much can RO pin sink. The most I got is Vol where it says Io max is 4mA. Does that mean it can only sink 4mA? In that case, how is that shield working? because 6N137 needs at least 5mA to turn ON. <Q> The voltage drop is only guaranteed to be less than 1.5V (or <S> maybe 1.25V <S> depending on whether they include Vcc tolerance) when driving high, but the design is not using it in that mode. <S> Chances are good it will be okay at at 6-8mA since the Vds drop is most likely well under the threshold voltage of the MOSFETs where the I-V curve flattens, but it's not explicitly guaranteed. <S> You can see the typical (ie. not guaranteed) receiver output voltage at 8mA here: <S> So it's about 0.42V at 8mA/125°C. <S> That means it would be about 0.21 typically at 4mA <S> and they've allowed a bit less than 2:1 margin for variations. <S> If we assume a 0.8V drop at 125°C then your current would be about 6.4mA nominally. <S> The opto is only good to 100°C <S> so we have even more margin. <S> Generally you should check the opto data carefully to make sure you allow enough forward current for all temperature and unit-to-unit variations and also to allow for aging of the LED (which is accelerated at high temperature). <A> The RO pin characteristics are shown on page 5 of the datasheet: <S> The datasheet does not give an absolute maximum <S> I out rating for RO, but the graphs imply that it can sink up to 45mA and source up to 18mA, but the output impedance is significant, which means that it the output voltage will pull away from the rail as the output is loaded down. <A> If you look at the equivalent driver ESR = RdsOn for Vol/ Iol at any current and supply voltage and temperature, you can include this in the required series R to define the IR diode current limit. <S> Then consider the VI= <S> Pd dissipation which may be small even at 20mA. Typically 5V logic is 50 ohm nominal + <S> /-25% Vol levels then do not matter as you are not looking for logic output levels but rather photodiode current levels. <S> This applies to all drivers, 3.3V logic max tends to be 25 Ohms with the same design/process tolerances of 25% or so.
| The datasheet does not tell you the maximum current is 4mA, it tells you that AT 4mA the maximum voltage drop is 400mV (low) over the full temperature range .
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Toy-like plastic strip for allowing turn-on I'm currently working on an electronic design who need to be turned on just once and continue to run until the battery die.I was searching for a way to do that and the most obvious was to use a plastic strip who act like a "remove before flight" lock.A lot of toys and small consumer products use that but I didn't found any specific component which can allow that. I know that on most of these devices, a plastic strip in inserted before the battery to prevent electrical connection between the circuit and the anode or cathode. But I would like to know if there is a component which could allow that without putting it on the battery. Since my design could make the initial launch very unpractical due to the position of the battery. I will reply the first three comments: the product is shipped with the battery already inserted. Due to confidentiality reason, the final custommer don't have to access the device. Nice idea. Unfortunately, my geometrical limitations will probably not allow that. But I will study it. It's the only way for me to control that the system is launched only when really used by the final custommer. Long time stocking of the full product could impact the battery life during normal use. <Q> You could use two PCB-mounted spring battery terminals mounted next to each other, so that the spring terminals are pressing together. <S> They would probably have to be hand-soldered as they will try to push each other out of the board during soldering. <S> Then add the plastic strip during manufacture, after testing but before assembly of the case. <S> For example, the Keystone 590 . <A> Once you slide it to the off position, plastic of the casing is used to prevent the user from sliding it back without damaging force or deconstruction. <A> How about just a simple slide switch?Put a pulltab on the slide so it can only be pulled to the on side. <S> https://www.digikey.com/products/en/switches/slide-switches/213?FV=ffe000d5%2C1f140000&quantity=0&ColumnSort=1000011&page=1&stock=1&k=slide+switch&pageSize=25&pkeyword=slide+switch <S> I guess that it's similar to the PCB jumper onto pins idea, but perhaps a little more mechanically secure. <A> A strip of paper between two neodymium magnets would achieve this - connections can be soldered to the other sides and the magnets would not need to be very big if weight is an issue.
| A stiff strip and magnets with beveled edges would allow the strip to be re-inserted easily This is coupled with a switch. Some smoke detectors with built in batteries have a one way catch, latch or clip that is used for a disable button.
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Best battery to power a Microcontroller outdoors I live in a place where the temperatures can range from -20°C to +40°C in different seasons. I need to power a small micro-controller that uses very low energy with a solar panel and a battery. Which kind of battery is more suitable for this work? It must have the following characteristics: It should resist to temperatures from -20°C to +40°C. It's not important to store a lot of energy, 500mAh will be fine. It should output at least 5V. If less I will have to put two or more batteries in series. It should not require maintenance. It should live for a decade or more. It will be pretty always charged since the solar panel will be directly connected to it (or maybe, via software, I can make some cycles of charge/discharge). which battery do you recommend? Explanation of the project I'm following an IoT course at my university, and I'm free to propose a custom project that will be part of the final exam. I thought about a series of small boxes that will allow us to check if a parking slot is free or not. The box will be build with my 3D-printer and it will include: a small solar panel with a maximum power of 9V * 50mA in the best case (size 60x80mm); a battery of at least 500mA that will be charged by the solar panel; a proximity/distance waterproof sensor that will check if a car is parked on the slot; a MCU with WiFi to send data to an other sensor or to a master with a raspberry pi. I will take inspiration from internet for the routing of packets through this "network" of boxes. Not all boxes will be directly connected to the master box.... We are already experimenting plastics to build a box strong enough to resist to the wheels of a car. The solar panel will be protected with a non-scratchable piece of plexiglass. Sensors will be cleaned from leaves and earth by rain, by road-cleaners and by the wind. Our roads are very clean. The major problem we are facing is to choose the best battery to be used in this "extreme" application. In winter, at night, the temperatures can drop up to -20°C while in summer, during sunny days, it's possible to reach +40°C.Finally, we will develop an App that will give you the nearest free parking slot. <Q> Run your micro off solar when the sun shines and off the batteries at night. <S> 10 years might be pushing the life of the batteries but not by much. <S> You can isolate the batteries with a Schottkey diode. <A> Without knowing your exact requirements, I would recommend you Ni-Mh AA batteries with nominal voltage of 1.2V (e.g. Eneloop brand) <S> The standard ones have operating range down to -20deg C and they have very high energy density. <S> But attention: At -20deg C you need to fine control the charging voltage and current. <S> See this site for details: batteryuniversity.com <A> Your combined requirements <S> It should resist to temperatures from -20°C (to +40°C.) <S> It should live for a decade or more. <S> sort out any standard accumulator but lead. <S> Lithium may reach this <S> but I personally won't count on the quality of the cells <S> I could get my hands on. <S> Car batteries of high quality are in contrary available in the next car parts shop. <S> As an alternative, you may want to oversize your accumulator and put it into an electrical heated, insulated compartment. <S> Also good for the electronics. <S> Condensation is hell.
| Sounds like you need a couple of alkaline D cells.
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How do I get 5V 10A output from a 3.7V li-ion battery for 64x16 led matrix display module? Please I am new to electronics and new here. I'm doing a project which involves the use of a 18650 li-ion battery (or two as may be the case) to power a 64x16 led matrix display module which (according to seller info and datasheets) takes up a large 10A and 5V. I've read posts which exposed me to BMS modules. I have also researched on the bms modules but seem to find only those of 5A (considering the 18650 works with 3.7V).Can I find a bms module that fits into this specification as a single module?Does connecting two modules of 5A bms in parallel result in both working as one unit of 10A?Will I still need a li-ion charger module to work with the BMS module(s)? (considering I'll have to recharge the batteries) <Q> One way or another you want 5 V at 10 A. <S> That's 50 W. <S> Your batteries have to cough up that, plus some extra for the loss in the converter. <S> That's basic physics. <S> No amount of clever electronics can fix that. <S> Let's say you find a boost converter that is 85% efficient. <S> It will then need 59 W as input. <S> 50 W of that will go to the output, and the remaining 9 W will heat the converter. <S> 59 W is a lot to ask of a 18650 cell. <S> At 3.7 V, that would be 16 A. <S> And, the current goes up as the battery voltage runs down. <S> That's either past the reasonable limit, or high enough that you will get significantly reduced energy per charge, and probably reduced lifetime. <S> Pop up a couple of levels and think about the larger problem. <S> You probably want a different way to address that altogether. <S> Or, realize that what you want to do is not feasible. <A> For a newbee like yourself: forget it <S> Why? <S> You're asking too much from those 18650 cells. <S> That LED panel wants 5 V <S> * 10 <S> A = 50 Watt. <S> That's a ridiculous panel BTW <S> but I digress. <S> A reasonable current that a decent 18650 cell can deliver is about 2 A at an average voltage of around 3.6 V: 2 A * 3.6 V = 7.2 Watt. <S> At that current an 18650 cell of 2200 mAh will then be empty in about an hour. <S> For 50 Watt you need to use in the order of about (50 W / 7W ) = 7 cells. <S> But since there will be conversion losses from converting 3.6 V to 5 V <S> You might even need 8 or 9 cells to have some margin. <S> That's for example how it is implemented in many laptops. <S> Charging cells in series is more complex and definitely not something for a beginner. <S> I would suggest that you use a ready-made battery pack which outputs 12 V for example. <S> The reason that upconverters that can output 5 V at 10 A is that the losses will be huge at such a current. <S> If the upconverter was 100% efficient (which it is NOT) then at 3.6 V around 50 <S> W/3.6 V = 14 <S> A would need to flow. <S> That's a lot of current, any extra series resistance will cause a massive voltage drop. <S> I would urge you to consider using a LED panel with a much lower power consumption as that will make things so much easier. <A> What about the min and max voltage ratings on the matrix display, whould it by any means be safe to run it on either 3.7v (single cell) or 7.4v (dual cell). <S> If you're using standard LiPo cells the current draw is not going to be your problem when you do not need to scale the voltage (it's only 3C for the cell). <S> (I'm drawing 120Amps out of a 4s LiPo pack everyday ;-) @80C). <S> One thing you have to keep in mind though is that LiPo batteries are unstable, so either too low or too high voltage can make them dangerous explosives. <S> So use an existing protection circuit, or come up with some protection yourself.
| Also it is more efficient to use the cells not in parallel but in series and then use a buck converter to convert down the higher voltage to 5 V. Check the datasheet of your battery. What you want to achieve will be very challenging, even for someone with plenty of experience in electronics.
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Punch a hole into a RFID card (125khz) So I have this 125khz RFID card and I would like to punch a hole in it so I can have a thin string go through it, making my card to be able to hang from my neck. Though I have no experience in punching hole to any sort of card, not mentioning an RFID one. Is it possible to punch a hole through the RFID card and still have it work properly? How should I punch a hole through the RFID card safely, easily and using some cheap accessible tools (list of alternative ways to punch a hole into RFID card is welcomed) In the picture below, you can see my RFID card in front of a light source, showing the antenna . On the picture, I put a green circle to the place where I need the hole to be (though I can make the hole a little bit smaller as long as a double-string can pass through). Is that a safe distance? Is the whole too big? Is there a chance that it will break the card? Picture: Note: please make your answers to be understandable for an average person like me (who is not deep into this field yet) RESULT Thanks to @Passerby for the answer. I suggest using the heated needle method if you have a needle lying around somewhere and don't have any special tools. I can confirm that it works perfectly (see image below) Also special thanks to @DarkCygnus for his help in finding this StackExchange . <Q> They punch holes in RFID cards all the time. <S> Using a hand or desk hole punch. <S> The smarter companies order them with the holes pre punched. <S> A hand hole punch is fine <S> if you are no where near the coil or IC, otherwise you risk breaking the coil. <S> You may also risk opening the inner layers of the rfid card to the elements if you open it in the wrong place. <S> The spot you illustrate with green, is just asking for trouble. <S> A millimeter in and you just cut the coil. <S> Alternative, glue something to the back that would let you attach a rope, like a small piece of fabric or plastic. <S> Or use/make an pouch or card holder. <A> Use a "slot hole punch" like this https://www.amazon.com/Badge-Punch-Puncher-Luggage-Credentials/dp/B0006M648E to keep away from the antenna element. <A> The best way is to put your RFID card in a card holder, then attach the cardholder to your lanyard. <S> Your employer should provide one, otherwise the card will get lost.
| Also check the card for some dots or other marking to help with identifying where to make the slot.
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Can a USB Oscilloscope be used reliably and accurately for RF projects? Are USB based oscilloscopes "sensitive" or able to be used with RF circuits? In particular, are they able to see the waveforms, carrier wave, etc. of something like a crystal set? I saw something like these (I picked these a random, first ones I saw), DigiTech QC-1929 (Hantek 6022BE) USB Oscilloscope Hantek 100MHz PC Based USB Oscilloscope The first is much cheaper, but the second is $200 (where I saw it), and that is getting close to a new or used scope. For context, this is for hobby/enthusiast. It is also mostly for RF based projects like a Crystal Radio and seeing AM, Shortwave, and the like. For example, I want to see the carrier wave and the AM modulation and what ends up getting sent to the earpiece, the "information." That may not be the right terminology. Hopefully you understand what I want to see. <Q> What does one thing have to do with another? <S> You look at a big expensive scope <S> it is a windows computer with some boards in it for the measurement interfaces as well as the knobs and buttons. <S> This day and age do we really need that? <S> Knobs and buttons are superior to touch screens <S> yes, but there is no technical reason why they cant build the same quality scope as far as measurement capability goes. <S> Being wrapped by a computer display and knobs does not affect the quality of the measurement for a properly designed scope. <S> No reason other than profit selling is a computer and buttons at an inflated price, that they cant provide high speed high quality scopes that we connect to via ethernet or usb and run software on a computer we provide. <S> Being a non-screen, non-button scope does not have anything to do with the potential quality of the measurement. <S> Nor does having buttons and a screen make a scopes ability to measure and display a signal any better. <S> A bad scope is bad scope in whatever form factor. <S> As is a good one. <S> If your real question is, are these specific products I named any good, I would say that falls under recommending a product and the question should be closed. <A> Scope quality and features has little to do with the interfaces it has. <S> What matters is the specs and the reputation of the brand that allows you to trust these specs. <S> RF is a little vague, but if you want to clearly see e.g. 100MHz signals, you need a scope with at least 100MHz bandwidth (which will translate to 0,5 to 1 GHz sampling rate). <S> As an example, Hantek 6022BE is advertised as having 48MHz sampling rate, and 20MHz bandwidth, and I'm pretty skeptical about the bandwidth. <A> Benchtops tend to be built better with lower noise and larger voltage thresholds, but that's not always true. <S> USB tends to have the advantage on software, analysis, and storage. <S> As long as the USB scope is cheaper only because it doesn't need a full hardware interface, it should be just as capable. <S> The only feature a benchtop will always have over USB is grounding. <S> Not in terms of signal integrity. <S> More for if you screw up. <S> Unless you get a floating scope or exclusively use active differential probes, the passive probe ground lead is coupled right to earth. <S> With USB, that's right through your computer's motherboard and power supply. <S> The one time you slip or aren't thinking could fry both your scope and your computer. <S> Other than that <S> , it's purely a question of specs, features, and user interface. <S> Don't forget to include your computer's specs into that equation if you go USB.
| A USB scope can theoretically be just as good or better than a benchtop with one exception for safety.
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Powering a PC monitor with batteries So, this is where I am at currently( no pun intended). I am assuming that with an inverter for both components and 3 - 4-18650 cells in series - in parallel rated at 4800mAh would be good enough?Ill have to look at the controller again but I think the data and power are delivered in the same ribbon. <Q> 24 W for 8 hours is 8 x 24 <S> = 200 <S> Wh (watt-hours). <S> (Don't worry about the calculation error. <S> It is nothing compared with the ill-defined battery specification.) <S> Since \$ <S> P = VI \$ <S> we can calculate the amp-hours required for a particular battery voltage. <S> With a 12 V battery <S> \$ <S> Ah = \frac {P}{V} = \frac {200}{12} = <S> 17 \ \text {Ah} \$ . <S> Ah is a popular measurement but it is only useful when comparing systems with the same voltage as, for example, in auto batteries. <S> Wh is a true measure of the energy stored in the battery and allows much simpler comparison. <S> From the comments: <S> First I have a monitor that has uses 12 V and a micro controller that uses 5 V. <S> I am assuming since these are in parallel they will equate to 17 V draw on the source. <S> No. <S> Imagine you have two 12 V lamps in your car. <S> Does that "equate to a 24 V draw" on the 12 V battery? <S> Obviously not. <S> Parallel connection increases current , not the voltage. <S> The panel is drawing <S> ~24 <S> W and the controller 4.5 - 5.2 W under load <S> (~32 <S> W max for system). <S> If I made cells consisting of 5 - 4.2 V 18650 batteries in series equaling 21 V cell, would this give me 256 <S> Whr / 21 <S> V <S> = 12.2 <S> Ah? <S> You need a 12 V to 5 V converter for the controller. <S> The monitor and converter are wired in parallel. <S> You need 200 <S> Wh for the monitor. <S> You need 5 W x 8h = <S> 40 <S> Wh for the controller. <S> Where you use a voltage converter you need to include a factor for efficiency. <S> If the 12 V to 5 V converter has an efficiency of 80% then the input power required = <S> \$ \frac{P}{eff} = \frac {5}{0.8} = <S> 6.25 \ <S> \text A \$ . <A> for an 8 hour run time, I would need 20*8-24*8 Ah power source. <S> You are calculating watt-hours not amp-hours. <S> To get amp-hours you need to divide by your battery voltage. <S> However you should also allow some margin for inefficiency. <S> So if you are using a 12V battery I would guess you want to be somewhere around 30AH. <S> how do laptops power the same screen <S> laptop screens tend to be both smaller, dimmer and have narrower viewing angles than standalone monitors. <S> Plus the power systems in laptops are likely to be better optimised than something you can cobble together yourself. <S> And while 8 hours on a laptop is certainly not impossible it tends to require running the laptop with a relatively dim screen and without any heavy GPU/CPU activity. <A> Marginal as to on-topic-ness, but: <S> Don't guess, measure - connect a voltmeter and ammeter to the circuit under load and see what is actually happening - <S> be sure to check what the load looks like as it turns on, as well as while running. <S> This may require cutting a cord, or getting a spare cord you can cut, or some such means (clever clipping, but be careful not to short things) to allow measuring. <S> As pointed out in comments, your calculation does not deal with voltage at all, and is thus watt-hours, not amp-hours <S> (well, unless you are using a one volt battery, which has its own challenges for most applications) and will look more reasonable when you have done that part. <S> As an aside, a Laptop display may not be identical to a similar flat-panel display intended for plug-in use (where the designer does not care that much about power consumption.)
| You need a 12 V supply for the monitor.
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Bad effects of coiling electrical cables I have a desktop computer and a lot of external hard disk cases. I coiled all cables to make them organized and easy to clean. But an electrician told me; if you are using a cable, coiling (wrapping) cables is wrong. It could damage your computer, electronic devices, etc. You should un-wrap cables of devices if you want to use them. So; Are there any bad effects of coiling electrical cables? Note: I found this on a Stack Exchange site, but still I'm not sure enough: Physics of Coiled Cables A quick answer that I learned ; If cable is long and power usage will be high (up to 200 W), don't use a coiled cable. It melt and start a fire. <Q> In the article that you linked to, it says in one of the answers, <S> The most notorious feature of loaded coiled cables is that they potentially generate a lot of heat in a tight space. <S> In most cases it's not an issue, but at high load with little cooling such a coil could be a fire hazard. <S> And I actually have first hand experience with this. <S> There was this extension cord (50 or 100 feet) that was wrapped around a spool, which I needed to power a small room heater. <S> Still it was a 1500W heater which draws over half of what the outlet could deliver. <S> Which means that it easily qualifies as a high load. <S> Lucky ending to the story: I could smell the insulation melting which led me to discover the danger and pulled the plug before it caught fire! <S> (I would have posted as a comment, but my rep < 50 Thought the OP really should have this info) <A> Is there any bad effects of coiling electrical cables ? <S> That depends on the cirumstances. <S> The electrician is correct for the part of electrical engineering that he (probably) deals with. <S> And that is high-power devices (more than say 200 Watt) running on mains (AC) voltage. <S> That's OK if the cable has enough "breathing space". <S> If you coil up the cable that might not be the case. <S> Especially in devices like these: the heat cannot escape if you do not completely unwind all the cable. <S> If you would only be using this for powering one low power device like a radio or a phone charger, the heat generated will be of no concern , you can leave the cable rolled up. <S> Fortunately for you USB and nearly all computer cables cables are quite low power <S> so coiling them up is no issue at all . <S> As winny commented: feel the cables wjhen they have been in use for a while. <S> If you cannot feel that they are getting warm, then there is no issue . <A> I don't think the other answers are correct in stating that the heat is the only issue, though it is the biggest. <S> I came into my office one Monday morning and found that the door had been forced open by the fire department who had put out a fire in the server room where someone had used a coiled cable to power a server - it had been like that for weeks but updates running on the server caused the power use to be higher for a sustained period of time. <S> However, power going through a coiled cable will also create a more concentrated electromagnetic field which can cause other problems. <S> The most frequent problem I've experienced caused by this is a distortion in a CRT monitor caused by the power cable being coiled at the back of it, but others would be possible. <S> For example a long USB cable coiled with a long power cable could result in a significant voltage being induced in the USB cable, potentially destroying the USB port and/or peripheral.
| With high power devices, large currents flow and these large currents through mains wires cause the cable to warm up . But if you use it to power a 300 W floodlight and/or a lawnmower and/or a toaster then you do need to unroll the cable completely so that any generated heat can escape into the air.
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Triac Dimmer Supplying a Purely Inductive load I was asked a question in my lab viva today, on a triac switch supplying a fixed value inductor of 0.2 henry. How will the current and voltage of the inductor vary, if the triac's firing angle was varied between 90 to 120 degrees?Explain the nature of inductors voltage under different firing angles. My answer was that the voltage will follow the relationship of $$V_{load}=V_{peak} \cdot \sqrt{\frac{2\pi-2\phi+\sin 2\phi}{4\pi}}$$ which I derived there only and the inductor current will be out of the phase lagging by 90 degrees to voltage. Vpeak is the peak voltage of the ac mains supply and \$\phi\$ is the firing angle of the triac. Apparently, this wasn't the correct answer maybe. So, what might be the correct answer to this or perhaps the correct generalized way to answer this question? <Q> Considering the integral as being the area under the curve, we can see that, ignoring diode drops and series resistances, for a sine wave conduction will be the area from alpha to 2Pi - alpha by symmetry. <S> For alpha in 0 <S> -> <S> Pi, current is zero outside the interval <S> (alpha to 2Pi - alpha).While <S> the RMS voltage or current is easy to evaluate given this, it is not particularly interesting, far more fun to evaluate how the current varies moment to moment while the thyristor is conducting, which is of course the integral of the applied voltage during conduction. <S> There will also at some angles be an unpleasantly high dV/dt which can be an issue for triacs in real systems. <A> Not even close. <S> The triac is a nonlinear, non-time-invariant device, which means you need to consider separately what happens when it is conducting and when it is not conducting, as well as exactly when it switches from nonconducting to conducting and vice-versa. <S> The key thing to keep in mind is that the triac turns on when the trigger occurs, but it turns off when the current through it drops to zero. <S> The fact that the inductor modifies the phase relationship between voltage and current is VERY significant. <A> Your formula is actually a load RMS voltage vs. triac firing angle for resistive load vs. mains peak volatge. <S> There is nothing useful in finding voltage relation for inductor. <S> The triac can't operate with arbitrary angle (0-180) for inductive load. <S> You would need a current and voltage feedback to properly turn on the triac. <S> These are so called phase angle controllers, for example motor speed controller. <S> The firing angle looses its meaning, it is increased or decreased depending on motor speed. <S> Makes no sense to derive formulas, if at certain angles is totally impredictible, it would conduct on every second period, maybe,...
| The key is that the current starts at zero and goes as the integral of the voltage WRT time over the interval starting at alpha, ending when this integral is again zero. You may find that for some range of trigger phase angles, you get one kind of general behavior from the circuit, while for another range of angles, you get a very different behavior.
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Makeshift SMT Station. Any suggestions? I am trying to do some surface mount soldering for my project and have hit a few snags. My hot air gun works perfectly fine for most of the PCBs I am making but one type in particular is problematic. Due to the fine pitch of the chips I am using, and the number of very small components on the board I cannot seem to get a good board out of my current process as the components have a bad habit of shifting due to the air flow. Due to the fact that the board is 35cm long it will not fit in any desktop ovens I have found and I just don't have the money nor the long term need to invest in any industrial equipment. I've tried ordering two different induction hobs off of amazon in the hope that I can run them at somewhat accurate temperatures but I have hit the same problem in both cases. My tray started smoking and when I checked the temperature with a laser thermometer both heated up in excess of 440 C when I had them set around 160. I think this may be due to the size of the tray I have to use, rather than a fault with the hobs themselves? Anyone able to point me in the direction of an oven that is large enough for what I need but is under £250? Or perhaps someone can recommend a hot plate that will do what I am expecting / an entirely different process? <Q> I've soldered down QFN components using one, you have to be careful with airflow and direction, and/or use a pick to hold the component in place as it reflows, but I've hand soldered all the remaining 0603 package components. <S> Any chip that isn't leadless can be hand soldered, even with fairly fine pitch, you have to develop a technique of dragging a solder ball across the leads, and the surface tension of the solder/solder resist between the pins usually results in no bridges. <S> To reflow an entire board, there are a few projects that can be found on the web about converting toaster ovens, I bought one, with a convection fan for $40, it's fairly large, it does 12" pizzas with a fair gap at the sides, so a 35cm board should fit. <S> the hacks only involve replacing the thermostat with a controller, but for a one-off it'd be possible to manually switch the heaters on and off to approximate the profile needed. <A> Consider the following conclusion: You must find someone locally who will do the soldering job for you. <S> The individual must have proper equipment and skills to perform the job well. <S> You can of course continue yourself, but I am afraid by searching in the dark spots of your experience <S> you risk to invest more than you expect, to say gently. <S> Let me highlight some thigs for your further thought <S> My hot air gun works perfectly <S> If you'd tell us model name of the gun, we may be able to make some useful suggestions; at this point I can only say that you, most probably, can adjust temperature of your gun and speed of airflow. <S> Why not adjusting them so that small caps and other parts would not blow off? <S> I checked the temperature with a laser thermometer both heated up in excess of 440 C when I had them set around 160. <S> They are either faulty, or read the manual. <S> No information on their models - unfortunately no further insight. <S> I'm currently looking at the T-962A+ <S> Did you read manual for it? <S> The quality of manual and the phrase "The warranty for mainbody is one year, for parts is three months, provide lifetime technical guide <S> " should give you an idea that it is use-it-once device. <S> High probability that UI will be in Chinese. <A> At school we had a hot air tool that descends by adjusting a screw. <S> Sort of like a drill press in motion. <S> You can take your time lining up the components under the heat gun, then turn the heat on while it's far away and slowly lower it to the PCB. <S> If you apply symmetric airflow in all directions, you should have some luck in getting the components to stay put. <S> I'll try to see if I can determine the name of the tool <S> I'm thinking of and post an image for reference...
| The hot air guns are good for reworking a single component or two, but not a good choice for entire boards. As a budget, DIY solution you could affix your heat gun to some sort of vertical mount, and raise and lower the heat gun down to the surface of your PCB.
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Is this communication I²C? I need to decode communication between two devices, but I have no information about these devices. All I know is that four wires are needed (GND, VCC and two wires of communication). I suspect that it is I²C communication. I'm trying to decode it with the oscilloscope decoding tool, but I'm not quite sure about it. I can not identify elements of I²C communication appropriately when I visually check the waveforms. Looking at the waveforms I made the following assumptions, and maybe someone can help. These were my assumptions: Everything leads to believe that the clock is the blue signal and the data is the red signal. The clock seems to be inverted because its idle state is not at high level. I'm not sure if the data signal is also inverted, but it seems to be. Are my assumptions correct? In the last figure, the figure with the number 5 indicated in a circle, and there is a part of the signal. I can not identify the start, ack and stop bits. Can anyone identify these elements just looking at the figure? [Edited] Some people asked me about the devices that are in the communication. The communication is between a car key and a tool that I'm not allowed to say, but I'm trying to do a reverse engineering on it. <Q> Given that there are only 8 clocks per byte (I2C requires a 9th clock for the ACK/NAK bit) and the clock idle state seems to be low, I would say that this is more likely a SPI (or SPI-like) interface. <S> Not sure about the extra clock width on the first bit of each byte, however. <A> My guess is that it's some company's homegrown "I2C-like" protocol. <S> There were some of those back in the day when using I2C meant having to give money to Philips. <S> looks a lot like the data line getting passed from master to slave). <S> Oddly, it appears to transmit 7 bits at a time. <A> I'll toss my hat into the ring... <S> If these are old devices you could be looking at some "bare minimum" 7-bit synchronous RS-232 variant: That longer pulse in the beginning of each frame could be a start bit, and The plateau in the clock signal at the very beginning could be return to 0 before going to negative "mark". <S> (You did not provide voltage on screenshots, so I am guessing here).
| It appears to have an ACK (the short pulse on the data line prior to the clock stretch
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AC coupling capacitor in the ground to break the ground loop? I'm implementing the following audio single-ended to differential converter for an ADC: This is the reference design for the PCM1804 that I intend to use. What would happen if I do this: Would it still work? The reason I'm interested in doing this is demonstrated in the following diagram: simulate this circuit – Schematic created using CircuitLab Even though I'm using an isolated power supply for my DSP, my amplifier internally connects the signal ground to the chassis. There's clearly a huge ground loop, consisting of the ground shield in the RCA cables and the car chassis that is a big source of noise. I thought inserting a capacitor in the way of the ground as shown above would break this ground loop. But apparently, this is not a good idea. Any other suggestions are much appreciated. I am aware of ground loop isolator transformers, but that's a band-aid fix, and they create a large amount of distortion. <Q> The whole circuit you provided is a single ended AC coupled to the differential output. <S> 1st stage buffers (x -1) then the output is AC coupled on its output and AC coupled input for the noninverted on the bottom. <S> The output is a differential analog signal. <S> But the input impedance is unbalanced and does not convert stray noise current or ground shift to a differential noise voltage such that the Common Mode noise cancels out. <S> Maybe what you want for better immunity is a differential amplifier with precision matched resistor ratios or an Instrument Amp (INA). <S> Your suggestions won't work at all as others have explained. <S> The cct that works has an input impedance of 4.7k//{3.3k+Z(10uF)} <S> There are other ways to do this <S> but that was not your question. <A> Your circuit will not work because the (+) input of the opamp is missing a DC bias voltage that is at half the supply voltage. <S> A simulation guesses that the opamp has no input bias current and the capacitor on the (+) input is already charged to half the supply voltage. <S> Your first circuit had its opamp (+) input floating without the COM voltage you show in your added full circuit. <S> An opamp input must NEVER be floating. <A> It's common to solve this problem using a transformer in the audio signal line. <S> Most places that sell car audio equipment will also sell the audio isolating transformers.usually a device than handles two channels for use with sterio equipment. <S> the amplifiers are still referenced to ground.
| What you propose will not work because all it does is defeat the shield of the signal cable.
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Find Voltage between two terminals I have the circuit below and I need to find the voltage between the terminals A and B, which steps do I need to follow? And if there are several methods to do so, what are they? <Q> Not many "tricks" for this one. <S> Best to choose a reference point from where you measure all the voltages. <S> By habit, most of us would choose the bottom-most node (I suspect that @Transistor has chosen this point when he asks "What the voltage at 'a'? ). <S> One trick for this particular circuit is to recognize the symmetry. <S> One branch is the inverse of the other branch. <S> So you need only solve either 'Va' or solve 'Vb' voltage. <S> Then compare with half the supply voltage. <S> Double the difference to get the solution of 'V(ab)'. <S> This 'trick' may seem a bit round-about. <S> A different way of viewing the symmetry trick takes the voltage 'Va', and applies Kirchhoff's voltage law: 'Va' <S> + <S> V(ba) <S> + 'Va' = <S> 12V. Be wary of the signs of these voltages...is the required solution "voltage on A with-respect-to-B", or is it "voltage on B with-respect-to-A" ?. <S> It may be best to state your solution: voltage on B is more positive than voltage on A by X volts ... <S> this is unambiguous. <S> As a sanity-check, if the magnitude of your solution of V(ab) is larger than 12V, you've made a sign-error. <S> Are tricks worth the trouble? <S> A non-symmetrical choice of resistors forces the long solution: solve voltage 'a' & 'b' in each branch separately, then subtract to find the difference. <S> This smells somewhat like superposition, where signs of voltages or currents can drive you crazy. <A> My simple "no calculation" (well, hardly any) trick is to see that you have two voltage dividers. <S> The voltage at point A will be 12/(12 + 24) <S> * 12 = 4 volts. <S> The voltages at B and between A and B are left as exercises for the student. :-) <A> I'm going to answer the question just as you asked it. <S> Methods you might use include the node-voltage method, the mesh-current method, Ohm's Law, Kirchhoff's Voltage Law, and Kirchhoff's Current Law. <S> Superposition would probably not be a good choice for this problem. <S> Also, Watt's Law won't help you much.
| You need to solve for the element voltages in the circuit, then use those to determine the voltage difference in question.
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What is this electric thing, hanging on an electric pole? Recently I found this plastic thing on a pillar in the village: Looks like it is not finally connected yet, just attached with tape to the pillar. What is it's purpose? 4 cables go down to this thing: <Q> This is a fiber optic splicebox . <S> It is a weatherproof enclosure where two (or more) ends of fiber optic cables are connected. <S> Yes - electricity poles also carry optical fiber for control of the network and often internet access. <A> Here you can buy a similar one. <A> We've given your photo to the forensics lab <S> and they suspect that it is a fibre-optic data cable using the electrical power distribution poles as the cable route. <S> Photo 1. <S> The enhanced image. <S> The incoming cable. <S> The cable clamp. <S> The down-feed to the junction box where tap-off and through connections are made. <S> The cable out to the next stage. <S> The outgoing cables. <S> Photo 2. <S> A fibre-optic junction box. <S> Source . <S> The optical fibres can be run with power cables as they are insulating and there is no danger of fault currents going to ground. <A>
| It's a junction box for fiber optic cables. That is a junction box which is weather proof, and is probably temporary and will be moved before it gets played with.
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Only getting data when 'rubbing' wires I'm trying to use cansniffer and candump to read the CAN bus data in my car with a Raspberry Pi and the PiCAN 2 board , but I have run into a weird issue I can't explain. Situation: I've located the CAN-H and CAN-L wires in the car. I stripped part of the wires and used wires with an alligator clips on both ends to clamp to the stripped part. On the PiCAN board I attached wires to the CAN-H and CAN-L ports, each wire is stripped at the end, and also connected with alligator clamps. The problem: I only see data from candump / cansniffer when I take off one of the clamps and 'rub' the wire against the clamp. Which seems really weird, and I have never experienced anything like this. Are there some tips that I can try? I have tried using just wires, different wire gauge, but the problem stays. <Q> You need to connect three wires: <S> CAN-H, CAN-L, and ground. <S> The CAN lines are differential, but they are still ground-referenced. <S> That's another way of saying that your CAN receiver probably has a limited common mode input range. <S> If you do have ground connected, then you are not making good connections to the CAN lines. <S> Make sure the clips are digging into the bare wires properly. <S> Make sure that the teeth of the clips are on the wires, then manually squeeze the clips together more. <S> The best way to identify the lines is to look at their voltages on a scope. <S> When the bus is idle, both lines will be at the same voltage, usually around 2.5 V. When the bus is in the dominant state, then CAN-H should be about 1.8 V higher than CAN-L. <S> Keep the total wire length from the car wires to your receiver short. <S> CAN is not meant to be in a star configuration, which is what you are creating. <S> CAN will still work if the stubs are short enough. <S> The CAN bus will already have a terminator at each end, so make sure your receiver does not have any terminator enabled. <S> Remember that in CAN, it is the bus that is terminated, not individual devices. <A> The cables might be aluminum. <S> It is used in cars and power lines as they are very good conductor, lighter and cheaper than copper. <S> The problem with aluminum is that it's very difficult to make connections because of the oxide layer that forms rapidly exposed to the oxygen of the air. <A> I suggest you use STP wire or separately twisted pair signal + ground. <S> AWG24 to 30 magnet wire or thin insulated wire. <S> This may reduce signal wire impedance to prevent ringing on short cables. <S> Examine the wire contact resistance for insulation issues, but also note that your fingers absorb stray coupling to ground. <S> By lowering the common mode impedance while rubbing which is improved by twisted pairs to ground for each signal. <S> It also adds capacitance to the differential impedance too, so CAN bus interfacing requires knowledge of impedance matching and stray ground noise caused by your SMPS which is then improved by bonding of receiver 0 VDC to chassis (like an earth ground).
| Differential logic voltage readings must have good conductance (or contact) and low inductive (short GND wire) and shield of paired signals to 0 V. Otherwise, poor signal integrity will occur. Also make sure you don't have the CAN lines flipped. If indeed they are aluminum, you can try to crimp the cable, or putting some oil on the wire and try to scratch through it while connecting the cables. Some receivers have the option for this.
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Current flow from pull-up resistor in MCU According to the circuit diagram here, when the switch is on, the current directly flows to the ground, but shouldn’t it split at the junction and some of it go to the MCU, as well as the ground, if so then I don’t think MCU will give LOW as output? I may have some mistake in understanding the concept of pull-up resistors. <Q> The digital inputs on most microcontrollers and logic ICs sense the voltage, and pass very little current. <S> In your drawing, if the switch is closed, the MCU will see Zero volts, so will sense a logic Low. <A> I don’t think MCU will give LOW as output? <S> I believe you mean "MCU will read LOW on the input". <S> To get an answer <S> I suggest calculating voltage divider output using 0.1 Ohm as switch resistance (R2) and comparing resulting voltage to Vil (maximum input voltage considered LOW) from your MCU datasheet <A> Your reasoning is right (except that this is no MCU output configuration). <S> There might be some current flowing into the MCU pin. <S> But if the MCU internal resistor is very high (like >100kOhms), the current flowing into the MCU pin is very little compared to the switch and R1 current. <S> In fact, it is so little that you may disregard it.
| Some very small current may flow out of the MCU pin, through the switch, to Ground.
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Car battery (lead acid) discharges much, much faster than it charges? Let's assume a lead acid car battery (12V, 50Ah, 250A output). According to BatteryUniversity article BU-403: The charge time is 12–16 hours and up to 36–48 hours for large stationary batteries. With higher charge currents and multi-stage charge methods, the charge time can be reduced to 8–10 hours; however, without full topping charge. Lead acid is sluggish and cannot be charged as quickly as other battery systems. So the charging rate is no more than C/12. But car batteries (usually 6 cells) can discharge at often beyond 200A. 250A in my example. For a 50 Ah battery, this would mean discharging at a rate of 5C. Does that mean, that a lead-acid battery can be discharged at least 60 times faster than it can be charged? Or have I misunderstood something? <Q> You are entirely correct. <S> The issue comes down to the fact that it's pretty easy to overcharge a lead-acid, but normally you try not to "over-discharge". <S> So you have to be careful about the last stages of charging. <S> This is not (exactly) true for discharging, but that is only true because you don't want to completely discharge lead-acid anyways. <S> Unless you have a deep-cycle battery, you don't want to pull more than about 50% of the available charge out when discharging. <S> If you do, you'll severely reduce the battery life. <A> You said '..a lead-acid battery can be discharged at least 60 times faster than it can be charged?" <S> In general that may be true of a particular charging system, but it varies depending on the charging system and the charge profiles used. <S> Your premise is far too simplistic to cover all situations. <S> Battery University presents quite reasonable information, but it does not usually provide highly accurate technical details that cover all situations. <S> The main premise of the statement in BU-403 <S> you referred is to "Learn how to optimize charging conditions to extend service life". <S> But the charging profile presented is only one of many possible options. <S> For example the design of many car alternator charging system is typically a simple current limited (not CC) and CV profile. <S> You discharge the battery at perhaps 200-800A when starting the vehicle but charge the battery at perhaps 70-100A once it's running. <S> The current profile drops when the battery terminal voltage rises but the profile is simple. <S> Here the discharge/charge ratio may only be 10:1, at least for a short time. <S> If you read relevant information on smart charging profiles you may get a better picture of the situation. <S> Start with something like this from TI. <S> If you want in depth detail for extending battery life you may find this paper on VRLA EV use of use. <S> This uses a ZDV profile to ensure the minimum of overcharging, but maintaining fully charged terminal voltage. <A> For use in boats and RVs, I've seen C/5 recommended as a maximum charge rate, but lower rates are probably kinder to the battery. <S> Engine starting batteries are made to deliver very large currents for a short time. <S> Look at the Cold Cranking Amps (CCA) rating of a starting battery for examples. <S> The diesel engine on my boat requires a starting battery with a 900 A or greater CCA rating, if I recall correctly.
| For marine and RV use, you want to charge the battery as fast as practical (without damage), but for stationary use (UPS and similar applications), you usually have lots of time between discharges, so a slower charge rate is practical.
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I need the name of this one "component" I have been wondering what this blue thing in the top is called. I want to make the same kind of buttons for a project but i dont know what they are called and how to make them. Please help. <Q> These are typically custom parts, and the minimum order quantity (and design costs) are likely to be out of your price range. <S> Depending on your requirements, you have a couple of options: <S> Some rubber dome switches are available in standard PCB-mountable packages (containing both the rubber dome and the switch contacts). <S> The sizes and shapes will be limited, though. <S> Reuse parts from an existing rubber dome switch. <S> For instance, if you needed one just like the one in the picture, you could cannibalize the switches from the original controller. <S> If you needed to position the switches differently, you could cut apart the rubber between the domes and glue the resulting parts in place. <S> Consider using a tactile dome switch instead. <S> These switches use a buckling metal dome instead of a rubber part. <S> The metal dome is a reasonably standard part, and can be positioned on the board with adhesive tape. <S> One caveat is that these switches are much "clickier" than rubber domes. <S> If you just need to detect a finger touch without pressure, you could use capacitive touch sensing. <S> This will require a conductive surface on the switch. <A> I have seen (in fact I bought a bag of 1,000 of them) <S> individual (none-bespoke) <S> rubber switches that include conductive rubber and dome for a single switch. <S> Like most other electronics bits, they should be easily available in China. <S> Photo from a random Alibaba supplier. <A> It appears to be a moulded conductive rubber button. <S> The two black strips are conductive rubber that bridge contacts on the PCB it fits over. <S> These are typically custom devices, although Brady or someone might make some simple strip and matrix devices.
| What you are looking at is a rubber dome switch .
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What's the first pin of OPA2365? Usually I recognize the first pin of an IC by observing the small circle and then comparing it with the layout in the datasheet, but in the OPA2365 there is no circle. In the image, is the red arrow pointing to pin 1? <Q> Indeed that probably is pin 1. <S> But to be 100% sure, consult the datasheet: <S> Note how the pin 1 side has a different edge, noted by the circles. <A> On that part, the white bar marks the pin 1 end. <S> On many parts, there is a notch on the pin 1 end. <S> Also, if you hold the part so the text is readable, pin 1 will usually be the bottom left pin. <S> Many packages have a bevel along the pin one edge (look at the end <S> o the package to see this). <A> It's difficult to tell, because these op-amps have had multiple production runs with different styles silk screened on them. <S> They were originally Brown Burr parts before TI purchased them, and factory changes from one production facility to another. <S> They are old designs, and highly counterfeited, to were its hard to tell if legit unless you get them straight from TI. <S> There has been history of even legit distributors ending up with fakes. <S> Significantly enough to the other side, which is almost squared off/90 degrees parallel to the face. <S> * <S> * <S> In the picture above, you can see it's almost 45 degrees from the face. <S> Your picture is also a bit unclear, damaged or a really bad counterfeit because there is a dimple next to the first line, on the left of the picture. <S> That would typically indicate that that's pin 1, but the text is upside down, which is very very rare for a DIP or SOIC package.
| A legit TI part should have the pin 1 side of it's package beveled.
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LED not achieving max current I have a project that is powered by a 12v power supply. I need as bright of an LED as possible and decided to use this 9v LED . According to the datasheet, its max current is 1A, which is what I would like to get it to. In order to do this, I am using this 3 Ohms resistor According to my calculations, (12v-9v)/1A = 3 Ohms. This should get my LED to the correct current. To test the current, I placed a multimeter in series with the 12v power supply, LED, and resistor. However, I am only reading 0.639A. I tried several power supplies and meters and it's all the same. What am I doing wrong? Is my test setup providing too much resistance, thus decreasing the current? <Q> According to the chart on page 17 of the datasheet , the terminal voltage of a nominal-9V LED rises to between 10.25 and 10.5 V (depending on temperature) at 1 A. <S> You need to size your resistor accordingly. <S> But it would be far better to use an active current regulator to feed this kind of LED. <S> Then, the current wouldn't depend on temperature. <S> Or on your actual source voltage, for that matter. <A> From the data sheet: Look at these characteristics. <S> Measure the temperature as well. <S> For example if the case temperature (that means right under the diode) has a temperature of 85˚C, then you should get a forward voltage 10.25V at 1000 mA. <S> That would need a resistor of approx. <S> 1.7 ohm. <A> It could well be stray resistance. <S> If you are only getting 640mA, then the resistance you are seeing is (12-9)/0.64 = <S> 4.69 Ohms. <S> That is an extra 1.7 Ohms. <S> Your resistor is 3 Ohms. <S> It looks like it is a 5% tolerance, which means it could be as high as 3.15. <S> You have the resistance in the multimeter leads, you have resistance in those white wires, and you also have resistance in your terminal blocks, where the wires connect to the breadboard, and the breadboard itself also has resistance. <S> That will have some tolerance as well, so you may find it is not exactly as the datasheet says. <S> This will also have some significance. <S> Check all of this and recalculate. <A> Page 17 of the datasheet shows the IV curve for the 9V version. <S> For 1 amp, you are looking at a Vf of 10.4 volts or so, depending on the temperature of the led. <S> Note that 1 amp is the ABSOLUTE maximum, and without proper cooling of the led, you are almost guaranteeing a dead led, or multiple internal failures. <S> Note, if your getting 640 mA with 3 ohm resistor, you may want to measure both the voltage and resistor in use. <S> Your source voltage may be higher than 12V, and the resistor may be on the lower range of its tolerance. <S> 640 mA is <S> the If at 9.6V. Actual measurement of values should always be taken over theoretical values, especially when small variations lead to large changes. <A> You have to actually READ THE DATASHEET: <S> Depending on temperature, the LED will drop about 10.2 to 10.5 V with 1 A thru it. <S> Conversely, with 9 V across it, the current will be about 300 to 450 mA. <S> The datasheet is really quite clear. <S> I can't see how anyone would think this LED would drop 9 V at 1 A. <A> Short answer: Use 1.3 Ohm instead of 3 and verify temp and current. <S> It should not instantly burn your finger. <S> You must measure LED voltage rise with current to optimize this R limit value. <S> Or see my design. <S> What you need to learn: All diodes above nominal current have a **constant differential resistance, but those may be +/-50% tolerance <S> Ri = ΔVf/ΔIf {min:max} [Ω] ( also applies to LED's, transistors, Zener's(Zzt), MOV's) <S> Here is a simple adjustable driver using an NFET RdsOn=0.5 Ohm Vgs=4V 2W heatsink warm only 0 to 1 A adj range. <S> Fixed Rs may be adjusted for tuning range to match LED. <S> the VI curves are collected with thermal constant T heatsink 25'C and others. <S> Diodes have a threshold voltage that drops with rising temperature What the datasheet tells me: <S> The graph on p17 measures a slope of 4.25 Ohms @ <S> Tc rising to 85’C. <S> When you added 3 ohms is added in series your current in theory , reduces to 3V/(3Ω + 4.25Ω) = <S> 0.41A <S> Since you measured 0.64A <S> this means ... <S> Req was (12-9V=3V@ 0.35A) <S> so 3V/0.64A=4.7 = <S> (3Ω+1.7Ω) <S> so your sample had 40% Ri of the graph's slope of Ri = ΔVf/ΔIf ( due to wide tolerances on Ri) now recomputing series R , you need 3Ω-1.7Ω= 1.3 Ohm or so. <S> Keep in mind if your heatsink compound or slow fan is inadequate, and Tc rises, then the current also rises from the Shockley Effect, due to table worst case Vf@Tc computed as -1.6V/+60’C case rise. <S> There are many ways to regulate an LED power, linear or SMPS. <S> CC or variable CC. <S> Here using the BJT current Vbe= <S> 560mV <S> so I used Re=560 mOhm which could be made from AWG28 magnet wire with <1W dissipation kept cool by an engineered solution. <S> For best performance use a CPU 1W muffin fan with a Resistor drop for reduced speed with silver paste on LED array.
| The next thing to do is measure the actual voltage drop across the LED.
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How can I troubleshoot an active short circuit? What are some effective methods of diagnosing an active short circuit? By this I mean a short circuit that presents itself only after a PCB is powered on. tl;dr I have a design in the prototype phase. 17 of my 20 boards work great. The other 3 all have a short circuit on a 3.3V rail. This only shows up after the board is powered on. After removing most of the components on the rail, I tracked it down to an Ethernet PHY. If I lift the IC off, my rail is rock-solid at 3.3V. When I put it back on (also tried 2 new ICs), my rail is overloaded again and it drops out. I have thoroughly visually inspected, and probed the board for shorts, but cannot find one. I have lifted off pretty much everything around the IC (crystal, series resistors, ferrite beads, etc.) but still get the same behaviour. I've also tried holding the chip in reset, but that doesn't help. I lifted individual pins on the IC (VDDIO) and that fixed it, but doesn't offer a real diagnosis. I'm starting to wonder if there's an issue with the PCB fab, but not totally sure how it would cause this. They claim to do 100% E-test. Any advice will be appreciated! <Q> It won't always work, but sometimes you can track down a short with a thermal camera... <S> of course you have to have a thermal camera to do that. <A> While I wish you best of luck with your thermal camera endeavour, I would actually expect that the camera will show you the IC affected by the short (the Ethernet PHY?), not the short itself. <S> You'd have to be pretty lucky for the actual faulty spot to have higher resistance than the internals of an IC. <S> If you find nothing with the camera, I would then check continuity to GND / 3.3V rail on any of the IC pins which are not actually connected to GND / 3.3V. <S> Do it with a diode tested, since a forward-biased junction is close enough to a short. <S> If that comes out negative, I would power the PCB with the IC removed and check the voltage on all pins (ideally, a power-up waveform but it may be tedious for a large number of pins). <S> Any voltage outside of 0.. <S> VCC range could potentially cause the IC to latch up, a condition which typically looks like a short. <S> Finally, I would check if all pads corresponding to the output pins can be actively driven. <S> This can be done by connecting a scope and a signal generator together outputting a 0-3.3V square wave (so that you see the square wave on the scope), then connecting the probe to the pads. <S> A disappearing square wave would mean that something else is trying to drive pad that the IC will want to drive as well. <S> This can be justified for open-drain and bidir pins, but not for pure outputs. <A> Or you can use the "Louis Rossmann Thermal Camera without a thermal camera" technique - cover the board with IPA, run power through the short - the thing that's shorted heats up, and the alcohol will evaporate very fast. <S> It makes it really obvious what's shorted. <S> See demo here: https://youtu.be/gRV0cmIj5Ks?t=236 <S> - yeah, it's kind of a bad example, because he finds the short with the thermal camera in this case, but he usually uses the alcohol method, and it works great. <A> If you cannot put enough current into the rails to be visible as heat, you may try to use a sensitive voltage meter. <S> Put the meter in microvolt (µV) range, touch one probe to where the current is coming from, and other probe to where you suspect it might be going. <S> The more current is going through that route, the higher difference the meter will show. <S> This point is then closest to the short-circuit. <S> PCB traces have a resistance of around 1 milliohm per millimeter, depending on width of course (you can check with a calculator ). <S> Thus if the short-circuit is drawing e.g. 100mA, you should see 1mV difference per every 10 millimeters. <S> You can also try to do this from the GND side, which could help if the short-circuit is from an IC output pin to ground. <S> But if the board has a good GND plane, you probably won't see a voltage with small currents. <A> The good options are already mentioned, but I'd add this. <S> Disclaimer : it's more suitable for shorts on signal traces. <S> Stuck node tracer <S> The Art of Electronics, pg. 276, discusses the "stuck node tracer" - a sensitive microvoltmeter, which you use while the circuit is powered, and can point you to where the short is. <S> You keep one of the probes on a fixed location (e.g. the 3V3 rail), the other one you move along the PCB traces. <S> The traces which carry the short exhibit a measurable voltage drop, so the closer you get to the short, the larger the display on the voltmeter will be. <S> I've never built this, but it looks reasonable (and Tony EE also mentions it). <S> But it's really more useful when the PCB traces in question are thin. <S> Fixing a short <S> In the case when the short is within an IC, it's usually a defective IC <S> and I'd just change it. <S> It may have been killed by ESD or some sort of abuse. <S> If it's on the PCB (and you didn't have electrical test), I enjoy fixing those by placing a large supercap (2kF@2.7V) across the short to vaporize the small whiskers that caused it. <S> Once we had a batch of PCBs littered with problems like this (due to bad conversion to gerbers, which made them noncompliant to the fab's DRC). <S> We fixed a bunch of boards that had whiskers like this, it was a thoroughly fun (albeit smelly <S> ) experience :) <S> If your electronics won't get killed by 2.7V, you can use this method on a populated board as well. <S> Just be careful with polarity, otherwise there'll be smoke everywhere :)
| Just power the board up and watch very closely through the thermal camera to see if one area of the board gets really hot, it could help you narrow down the area at least. My usual technique is to keep one probe always on power connector or voltage regulator output pin, and move the other probe until I find the highest difference.
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DSO oscilloscopes with CRT screens To my ignorance I used to think all big CRT scopes were analogue oscilloscopes. But I found out that there are digital storage oscilloscopes which were built by using CRT screens as well. Below is an example: I have two questions: How come text like a number can be displayed on a CRT? I can understand how a signal leaves traces on a phosphor screen through electron beam hits on the CRT screen by deflecting rays. But how is text displayed? Why isn't there any FFT function for such old scopes which are DSOs with CRT screens? <Q> Not all scopes that can display digital information on their screens are DSOs. <S> A Digital Storage Oscilloscope, by definition, digitizes an analog signal, stores those digitized samples, and then displays it. <S> DSOs could be constructed with LCDs, CRTs, or any other display technology--there are even DSOs that don't have any display at all, and rely on being connected to a PC to display captured data. <S> So if there's no digitizing, and no storage, it's not a DSO. <S> It is this storage that provides the DSO the ability to perform arbitrary computations on the captured waveform, whether FFT, or integration, or whatever. <S> But just displaying text on the CRT does not mean you have a DSO on your hands. <S> The sort of thing you see in the picture in the OP is actually not uncommon in analog 'scopes, especially later models. <S> Essentially the CRT can be driven by the analog front end, where the horizontal position is controlled by a ramp generator (the horizontal timebase) and <S> the vertical position is controlled by the channel amplifier, or it can be controlled by perhaps a small microprocessor to draw vector shapes, as you would see in a vector monitor. <S> Essentially, the scope alternates between drawing waveforms and drawing informational text or other information on the screen, presumably doing the vector drawings in between triggers. <S> Some analog CRT scopes could even take digital measurements of voltage or time, with cursor position readouts and text menus that could be displayed and navigated through. <S> However, it wasn't very long after that when DSOs became viable, and quickly took over the market. <A> How come a text like a number can be displayed on a CRT? <S> I can understand how a signal leaves traces on a phosphor screen through electron beam hits on the CRT screen by deflecting rays. <S> But how is a text displayed? <S> You're saying "CRT" when I think you actually mean "vector display". <S> These displayed text the same way any other CRT computer monitor did. <S> On a vector display, you can still display text. <S> You just need a drawing routine that produces the text by routing the beam around the display and turning it on and off as required. <S> You see this kind of text on things like radar displays going back probably to the 1950's or 1960's. <S> Why is there no FFT on such old scopes which are DSO with CRT screen? <S> An FFT takes a fair amount of processing power to perform quickly enough for the display to be responsive to user inputs. <S> This was probably not possible with the microprocessors available at the price point the manufacturer and user wanted before maybe the mid-1990's. <A> You can just paint the text by steering the beam. <S> The X coordinate of the beam is not hardwired to a sawtooth, but also connected to a DAC, like the Y coordinate. <S> The DACs read from sampling RAM, which contains ascending values for the X coordinate next to the sampled values, so the waveform is painted left-to-right. <S> A microprocessor then writes a few extra positions at the end of the sampling RAM, which move the beam to the text area, and paint numbers. <S> E.g.: <S> 0 0 <S> on1 42 63 74 <S> 65 <S> 46 07 <S> -48 <S> -6... <S> 999 -4 <S> off22 -200 on21 <S> -20020 <S> -20020 <S> -20120 <S> -20221 <S> -20222 <S> -20222 -20322 <S> -20421 <S> -20420 <S> -204 <S> The first 1000 positions in memory represent the sampled waveform, as X,Y pairs. <S> Then, we turn off the beam, go back to the left, turn the beam back on and paint a 5 to the bottom of the screen. <S> FFT is a more complex operation that requires computing power, while simple sampling and reproduction of a waveform only require efficient streaming data paths to and from memory.
| Digital scopes had CRT's with bitmapped or raster displays were quite common from the first days of digital scopes until the price of LCDs dropped in the early 2000's.
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CANopen transmission to multiple IDs at the same time I am asking myself, whether it is possible that I simultaneously send multiple signals to my robot via CANopen. Edit: I am trying to build a CANopen connection with my robot. I can turn the motor up by sending a message (8A 00 0A 00 00 00 00 FB) to ID 181h. Now, I am trying to control the movement of the robot via CANopen. However, it does not repond at all. Here is what I did: I first move the robot with my control module. Since I have to press down two buttons at the same time to move the robot, I checked how the data values of each IDs changed for each button press. For example, when I press down the button A, the data value of ID 181h changed like 00 00 00 00 00 00 00 FB -> 00 00 00 FE 00 00 00 FB. When I press down the button B, the data value of ID 301h changed like 00 00 00 00 00 00 00 FB -> 00 00 00 00 32 00 00 FB. So now I tried to move the robot via the CAN connection as the one in the picture. I first wrote "00 00 00 FE 00 00 00 FB" to ID 181h and after that I wrote "00 00 00 00 32 00 00 FB" to ID 301h. However, there isn't any response at whatsoever. Am I missing something? EDIT: Pressing button A changes the value of 181h to 0A 00 09 00 00 00 00 FB and pressing down button B changes 281h to 00 00 00 FA 6F BD 00 FB. So I sent those messages to the TPDOs 181h and 281h, respectively. However, still no response. <Q> In order to control any CANOpen device you should first study the object dictionary of the device. <S> In the dictionary you'll find Service and Process Data objects, which are essentially addressable locations for configuration data, status, control commands etc. <S> So, instead of trying to figure out what CAN messages to send and how to send them, you should learn how CANOpen works first, find communication objects that you need to access and then send messages in CANOpen format to perform the functions that you need. <S> Most likely these won't be messages "button <S> A pressed" or "button B pressed", but rather <S> something like "Node NNN Change object 0x6063 <S> to value XXXX", which will be interpreted as "set new speed for left motor". <S> There are many communication profiles used in robotics, like CiA 402 <S> (CANopen drives and motion control profile). <S> You robot might support one of them. <S> If you do not know the object dictionary of your specific CANOpen device and do not even know if it supports one of the standard device profiles than there is very little you can do. <S> From your question it seems you have some kind of remote control device and you attempting to reverse-engineer the communication. <S> This could be tedious process, since some CANOpen commands require several back-and-forth CAN messages to do something. <S> My only advice in this situation would be to try and get the datasheet from the manufacturer. <A> CAN (and I think CANopen as well) can only transmit one ID at a time. <S> I think this is what you need. <S> If you need to act on a response which is a combination of different devices (where the buttons come from), than use separate CAN message (ID)s to set a mode, and if all modes (buttons) are active, act on it by the robot part(s) <S> that need to respond. <S> In case 2, e.g. if you have button A, B and C that needs to be pressed to perform an action D, you can define: BUTTON_A_PRESSED_ID, sent by the part that has button ABUTTON_B_PRESSED_ID, sent by the part that has button BBUTTON_C_PRESSED_ID, sent by the part that has button C <S> The part that should act on it, receives these messages, sets 3 flags, and if all 3 are valid/received, action D is performed and the flags are reset. <S> Update See comment Maple below... <S> the above only works if the receiver code can be changed, which seems not to be so. <A> This is, in a sense, trivial and can be done with (I think) any CANBus implementation, but it's not done at the CANopen level. <S> In general, any CANBus controller on the market, such as the venerable SJA1000, will allow the implementation of an "acceptance mask". <S> This will essentially cause the interface to treat some bits of the address field as don't-cares. <S> Setting, for instance, the last 3 bits of the mask to don't-care will cause the device to respond to a block of 8 addresses. <S> Just as a reference point, the SJA1000 has four independent acceptance masks. <S> So you can set up as many as four address blocks of varying sizes for each unit to respond to. <S> See, for instance, the SJA1000 app note . <S> So, the work you need to do is configure your units to take advantage of the interface capabilities. <S> Once that's done, you simply transmit an ID to one address block for individual messages, and to another for group messages. <A> In order to succeed with this project, you need to know both how CANopen works and how the specific device works. <S> For example, if you are able to read Object Dictionary item 1000h, it will reveal which device profile the robot is using. <S> All data in CANopen is sent through PDOs (process data objects). <S> Messages coming out of a CANopen node are called TPDO (transmit PDO) and messages coming in to a node are called RPDO (receive PDO). <S> This does of course mean that one node's TPDO must be another node's RPDO. <S> Some insights on reverse-engineering CANopen: <S> 181h is the standard <S> CAN id for the first TPDO of node 1. <S> 301h is the standard <S> CAN id for the second RPDO of node 1. <S> This is in accordance to for example the most common device profile DS401 "generic I/O module", but may match other device profiles too. <S> Notably though, it is unlikely that 181 and 301 have similar functionality. <S> I would expect one of them to be incoming data and the other to be outgoing data. <S> If this is device profile DS401, then 181h will be digital data going in one direction, while 301h will be analog data going in the other direction. <S> It is unlikely that these 2 correspond to 2 different buttons (digital). <S> In device profiles like DS401, it is specified how data should be stored. <S> In a generic CANopen application, this isn't specified, but data could be mapped into PDOs in custom ways. <S> Thus what you describe doesn't seem to make any sense. <S> It is also unclear why these two PDOs have value 0xFB in the end. <S> This is also some kind of data - checksums are handled by hardware.
| But there are some solutions based on what you need: Send a single ID that is received by multiple CAN devices (robot parts).
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Why is my oscilloscope reading jittering, and how might I fix it? I've made a simple inverter circuit and I've read the voltage across the load resistor using a RIGOL DS1104 oscilloscope. Here is the link to the screen capture. I'm using a full-bridge inverter topology, I'm using an Arduino Uno for control, and I have no feedback control system yet in place. I am filtering with a capacitor in parallel with the load resistor as well as an inductor in series with the combination of those two.I'd like to know why my oscilloscope measurement jitters from side to side and how much of a problem that presents to the quality of my sine wave? Thank you! <Q> There is not a lot of information here, but when I see students with jittery sine waves on their oscilloscope screens it is almost always because they haven't set their trigger level well. <S> If your scope has an "autosetup" try that for now (and learn how to set up an oscilloscope), but in general set the trigger near the zero-level of the scope. <S> At the zero level the rate-of-change of the sine is maximum and the jitter in the display will be minimum. <S> Edit: Chris Stratton's comment is quite right. <A> Per sigmadelta, start by setting the trigger properly, first by trying the AUTOSETUP and see if that improves the display. <S> My guess is that you'll still see the signal amplitude vary from cycle to cycle, but at least it will trigger at the same point for each sweep. <S> Once you've achieved that, you can determine why the sine wave isn't a constant amplitude. <S> Hope this helps. <S> Send more info if you can. <A> I'd guess that "I'm using an Arduino Uno for control" is your culprit. <S> I notice that the amplitude of the output is varying noticeably. <S> Instead of looking at the output, try looking at the bridge drive waveform. <S> If you see horizontal jitter there, that's your problem. <S> And if that's true, then, "how to fix it" is straightforward: learn to do a better job of programming.
| Since you say that there is no feedback, and assuming that the raw bridge voltage is remaining constant, I'd suggest that the period of the drive voltage is not constant, and varying the drive waveform is causing the output to vary.
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Can I use 1N4148 diodes instead of a 1N4001 diode for reducing the speed of a 5V 1.8W fan? Looking at this 1N4148 datasheet , it seems that the forward continuous current is 0.3A, thus 300 mA. So if I add two in series, I will get 600 mA. I want to use diodes to reduce a 5V 1.8W fan speed. The current through the fan is P = V * I <=> I = P / V = 1.8 / 5 = 360 mA. Can I use two 1N4148 diodes in series, resulting in a voltage of 5 - 0.7 * 2 = 3.6 V (having thus about 72% fan speed), because 360 mA < 600 mA? (note: I know a 1N4001 is better, having a 1A limit, but I don't have them yet). Or did I make a mistake in my calculation or is there another effect why two 1N4148's cannot be used? <Q> The 1N4148 is intended as a small signal diode. <A> Components in series have the same current through them — it doesn't divide. <S> All of the current goes through all of the devices. <S> If one diode has a limit of 300 mA, then two of them in series still have a limit of 300 mA. <A> If you place two components in series, current flowing through first diode and second diode is equal. <S> It means that through both of your diodes will flow 0.6 A (actually 0.5 A -> <S> you have to use new 3.6 V if you want to compute current to fan) <S> -> they are going to be burned. <S> You should choose other diode with bigger forward current. <S> Also, when you want to compute current to fan, you should use 3.6 V instead of 5 V, because of voltage drops across these diodes.
| As others have said, for diodes are in series, EACH diode carries the full current. The 300 mA rating is in the Absolute Maximum Ratings table - you don't normally want to go near those ratings. If you look at the Electrical Characteristics table, you will see that most specifications are given with a 10 mA test current, so you should only use a 1N4148 with currents in that range.
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Shield grounding for analog sensor I have an analog sensor, a resistance temperature thermometer, which is essentially just a resistor (~50 Ω) which I apply a constant current to (10 mA) and measure the voltage of in order to detect temperature changes. The temperature changes of interest are extremely small, and so it is extremely sensitive to electrical interference, and nearby there are many pumps and stepper motors. I have been told conflicting information about how the cable should be shielded. The two options are: (diagram attached also) A: COAXIAL CABLE I have up to now been using a coaxial cable. The outer of the cable is connected to ground at the oscilloscope I use to record data. The outer is carrying the return current, which I have been told is bad. B: SHIELDED TWISTED PAIR The constant current is carried in the internal twisted pair with a shield connected to ground at the oscilloscope end. <Q> RTD connection <S> RTDs often need a 3-wire or a 4-wire connection to make a precise measurement, because the connecting wires have a resistance which varies with temperature. <S> [@Phil had mentioned this in his comment .] <S> In the 2-wire connection, the variable parasitic resistance of the connecting wires appears in series with the bonafide RTD. <S> You apply the current through the two wires and the RTD, and measure the total voltage drop through the same two wires. <S> You can't separate the contribution from the connecting wires from the actual RTD. <S> The 4-wire RTD connection solves the problem by applying the excitation current through one pair of wires, sensing the voltage drop through the second pair of wires. <S> Kelvin connection. <S> The current in the sensing pair of wires is negligible, and so is the voltage drop. <S> The 3-wire RTD connection solves the problem by measuring the parasitic voltage drop in the excitation wire. <S> Somewhat less accurate than the 4-wire connection. <S> Somewhat more complex front end <S> (it has to subtract the parasitic voltage from the total). <S> Fewer wires, which can be important if the connections are long. <S> (If curious about sensing front end for 3-wire RTD, here's an application note: Microchip AN687, Precision temperature sensing with RTD .) <S> Cable type <S> For a geographically small setup and precise measurements, a 4-wire connection would be a good idea. <S> That calls for twisted pair cables. <S> Where to connect the shield? <S> Which end to connect? <S> A good practice is to connect the shield on the end where noise is generated. <S> My first choice would be to connect the shield to the same ground as pump and stepper motors. <A> Lets model the twisted-pair in flux of 10 amps / 10 microseconds switching/commutator current changes. <S> Assume 10 meters of twisted-pair <S> , assume 2mm spacing between the +wire and the -wire. <S> This gives an AREA we need for the equation below. <S> The twisting we will consider later, because the twisting is never perfect and flux is never totally cancelled. <S> Let spacing between the twisted-pair and the HOT wire model of flux generation be 1 meter. <S> Notice I've placed the Return wire of the motor at infinity. <S> Use this formula Vinduce = <S> [MU0 * MUr * Area/(2*pi*Distance)] <S> * dI/dT <S> which is combination of Biot-Savart and Faraday Law of Induction. <S> For MU0 = 4*pi*10 <S> ^ <S> -7 henry/meter and MUr = <S> 1 <S> (air, copper), thisbecomes Vinduce <S> = <S> 2e-7 <S> * Area/Distance <S> * dI/dT Vinduce = <S> 2e-7 <S> * (10 meters * 2mm) / <S> 1meter <S> * 10^+6 amp/sec Vinduce <S> = <S> 2e-7 <S> * 10meter/meter <S> * 2mm <S> * 1e^+6 <S> Vinduce = <S> 2e-7 <S> * 10 <S> * 0.002 <S> * 1,000,000 <S> Vinduce = <S> 2 <S> * 0.002 = 4 <S> * 0.001 <S> = 4,000 microVolts Is 4,000 microVolts, from the magnetic field of a motor at 1meter, a serious error? <S> If so, then use twisted-pair, but do not expect more than 20dB attenuation. <S> Build it, and measure the noise floor from motors and pumps. <A> There is no general answer for that question. <S> The coaxial cable prevents magnetic field coupling but is susceptible to electric field coupling, whereas with the shielded twisted pair, its exactly the opposite.
| The shield should have direct connection on exactly one end (the other end either unconnected or connected through a small ceramic capacitor).
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What's the formal process to determine the IC by manufacturer code? I've curious if there's an industry standard method used to determine the IC on a board. Specifically, I've been wondering as I have an IC that I want to replace. It seems to have failed, but there's not manufacturer mark. Is there a database somewhere where you can enter the code and get possible hits? In my case, my bad IC is a "DOT 201 139". The 201 is the manufacturer identifier, but no manufacturer has a "DOT" to my knowledge. Also, the SOIC-8 is really tall. This is enough information if there was central database to probably narrow it down, but I've been unable to find anything through web searches. I'm sure this problem isn't unique to me, and there must be some way that the IC industry agrees on how to find replacements, no? update The circuit drives the gates of IGBTs that then run a solenoid. There's two of these drivers where one turns on and one turns off so that the coil field is bidirectional. It's weird that it's an H-bridge where the "center" is GND, so it has V+ and V- and then common, which is GND in this case. The best guess on the schematic is follows (I left off the connection to the 222 resistors): <Q> There is no formal process, unfortunately. <S> Manufacturers may use numbers, letters, punctuation, and arbitrary shapes as meaningful symbols in top marks, I've seen all of these. <S> Some manufacturers include datecodes, manufacturing location, lead-free status, etc, in addition to the part number in their top marking. <S> There are a few websites that provide lists of top marks with manufacturer/part number, however these are invariably incomplete. <S> Keeping such lists up-to-date would be a massive effort. <S> However, they can be worth checking, as you may get lucky. <S> They are especially useful with parts that use numeric markings, such as yours, since numbers are unlikely to provide useful web search results. <S> Speaking of web searches, that's another option. <S> I have had reasonable success in the past using search strings that include "IC top mark" and the marking in question. <S> It may help to include the package as well, although there are usually three or more different designations for a given package that the manufacturer may use, which makes that a bit more difficult. <S> This method works best when the manufacturer includes the top mark in their datasheets in a way that search engines can index. <S> Some manufacturers are better about this than others. <S> If you can narrow down the function of the part, you may be able to use distributors' catalogs to narrow the search based on functional information (for instance, knowing it must have a voltage rating of at least X) as well as package to some number of manufacturers and part numbers, and then it's a matter of sifting through the datasheets and seeing if any of them have the right top mark. <S> The final option, is to simply not worry about what the specific part is, and focus on finding a functional replacement. <S> This depends on a sufficiently thorough understanding of the circuit in question, but may give you a lot more flexibility in sourcing a replacement. <S> You may wind up needing to go this route anyway if it turns out the part is obsolete. <A> A component datasheet often will have a description of the markers that they put on their IC for users to easily identify the variation of the product that they have, the batch number, manufacturing origin, etc. <S> Some smaller chips have marking on the underside too, so you can try to desolder it and check the bottom part for clues. <S> If you don't know what IC it is then you'll have to try your luck with the search engine. <A> The dot is not a logo; it's a pin identifier. <S> 201 <S> could be the manufacturer number and 139 could be a date code or the part number could be 201139. <S> Generally if you can't figure out the part, look at the use of the part on the board and what's it's connected to. <S> If you can find anything on the other ICs around it, you may find an application note with the part referenced. <S> I like this site for as close to a database that I know of anyway: http://www.s-manuals.com/smd
| Some manufacturers do provide an option to look up a part by its top mark, so if you can narrow the part down to a few manufacturers, it may be worth seeing if they offer such a tool.
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Can an EEPROM be one-time-programmable? I'm confused, and I hope you can help me figuring out, what I misunderstood. I am not an electronics engineer, so bear with me :) I have a chip, from the engine control unit of a Toyota Yaris Verso from year 2000, with the following print: ATMEL01025020NB9D3203C I have not been able to find a datasheet. And that really surprises me. I did find one for At25020B, but not for one with the NB suffix. Why would Atmel make a chip and not publish the datasheet? First of all, from what I can find, the Atmel AT25 series are all EEPROM ( source ). So this chip should be an EEPROM. Am I right? I looked up, what the suffices mean and found the following (source: Explanation of Atmel’s Part Number Code ). B means some kind of military compliance. The N says something about the package: N = Leadless Chip Carrier, One Time Programmable I think the N in some cases mean "Leadless Chip Carrier" and in some cases "One Time Programmable", since they are not related. Do you agree? Looking at the picture, it obviously has 8 leads, so it is not leadless, and thus N must mean one-time-programmable. Do you agree? As far as I understand the EEPROM technology it is by definition erasable/rewritable. Is this really a one-time-programmable (and therefore not erasable/rewritable) EEPROM? Or what have I misunderstood? :) The background for even asking is because I have not had luck with communicating with the chip. A friend, who is an experienced tingler tried to with Bus Pirate and command line tool flashrom, without luck. <Q> I did find one for At25020B, but not for one with the NB suffix. <S> Why would Atmel make a chip and not publish the datasheet? <S> Why wouldn't they? <S> Chips are often manufactured for specific customers with specific needs, especially if the numbers are large enough. <S> Probably, though, your IC is an AT25020 of some kind, which pretty much sets the way it communicates. <S> I think the N in some cases mean "Leadless Chip Carrier" and in some cases "One Time Programmable", since they are not related. <S> Do you agree? <S> No. <S> The wording strongly suggest it's a leadless, one time programmable part. <S> Not either. <S> Both. <S> Looking at the picture, it obviously has 8 leads, so it is not leadless, and thus N must mean one-time-programmable. <S> Do you agree? <S> No, it seems that naming scheme simply doesn't apply to the IC you have. <S> As far as I understand the EEPROM technology it is by definition erasable/rewritable. <S> Wikipedia: <S> That functionality is controlled by an on-chip erase controller. <S> If you simply disable that after the first write (e.g. by automatically burning a fuse), then it becomes one-time programmable. <S> Remember, you do not directly interface the memory cells with the pins of your IC; you talk to some kind of controller that takes the messages ("hey, can you give me data from address 0xDEAD?") and interacts with the actual memory based on its own discretion;. <S> The background for even asking is because I have not had luck with communicating with the chip. <S> A friend, who is an experienced tingler tried to with Bus Pirate and command line tool flashrom, without luck. <S> What's the chance of that IC simply being broken? <A> OTP is useful for anything that can't or shouldn't be re-programmed. <S> Manufacturers might want to use it to prevent tampering or misuse by the public or for other reasons. <S> You might want to use it for a bootloader so that a device cannot be "bricked" by a bad firmware update (assuming the firmware resides on eraseable memory). <S> Any application where true mask ROM might be used is a good candidate for OTP. <S> Compared to mask ROM, OTP allows the manufacturer to optimize the product development schedule by finalizing firmware AFTER the hardware design is complete, and it allows for firmware changes to be rolled into production with less lead time than mask ROM. <S> OTP is typically less expensive than similar forms of memory which can be reprogrammed many times. <S> Sometimes chip manufacturers make custom parts for large volume customers. <S> The large volume customer does not want the datasheet for the part made public. <S> This is somewhat common. <A> The precise features available depend on the specific chip, and will be detailed in the datasheet. <S> Many EEPROMs also have protection against accidental erasure or reprogramming, usually according to a JEDEC standard which requires a particular command sequence before the actual erase/write command. <S> Erasure must occur before reprogramming in any case, due to the way flash memories work. <S> Some small EEPROMs include an erase cycle in their programming instruction, so don't need a separate erase instruction. <S> In the specific case of the AT25020B , there's a pair of nonvolatile write-protection bits in the status register, an external write-protect pin and a write-enable prefix instruction. <S> Use the "read status register" instruction to determine the state of the write-protection bits. <S> To change the write-protection bits, you must hold the /WP pin high, issue the "write enable" prefix instruction, then issue the "write status register" instruction; a zero byte should clear the protection on the main array. <A> I have not been able to find a datasheet. <S> Try this one, an older rev data sheet. <S> http://ww1.microchip.com/downloads/en/DeviceDoc/doc0606.pdf <A> Others have answered thoroughly about the overall question; if an EEPROM can be OTP. <S> But I want to add, what I found out about this particular chip. <S> I wrote Atmel and asked for the datasheet, and they were very helpful. <S> This is the response I got: <S> The complete part number of the device is AT25020N-10SI and it is made in 2000. <S> The 9D3203 wafer was made in 1999. <S> That device contains Pb (lead) and therefore is not RoHS 6/6 compliant. <S> f <S> This device is the first generation AT25020 device and was EOL'd by PCN SC030403A that introduced AT25020A. <S> The AT25020A was EOL'd by PCN CC084205D that introduced AT25020B. <S> AT25020 has 10ms maximum write cycle time. <S> AT25020A and AT25020B are 5ms maximum write cycle time. <S> AT25020 may not operate in AT25020A or At25020B sockets due to this difference. <S> I don't know why they would produce this chip with a print other than "At25020", which they call "the full part number".
| EEPROM stands for Electrically Erasable Programmable Read-Only Memory ; so yes, by definition, it's electrically erasable . To answer your question more generally, most modern EEPROMs have OTP (one-time-programmable) areas and/or the ability to "lock" the data within the general area.
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Why some Arduino LEDs are not grounded Looking at the schematics of the Pro Micro, the LED in PD5 (bottom-right) is connected to VCC instead of GND, same for PB0 . Why? Shouldn't they connect to GND? <Q> It works perfectly fine as is. <S> The output of a microcontoller can be set to high or low. <S> High is just connecting the output to Vcc through a transistor inside the processor. <S> Low is just connecting the output to ground through a transistor inside the processor. <S> All it changes is what value you have to write to the output pin. <S> If the LED is connected to ground, then you write a high to the output to make the LED light up. <S> If the LED is connected to Vcc, then you write a low to the output to make the LED light up. <S> No difference in function, you just have to use a different output value. <S> That said, some chips are better at sourcing current (better for output high) and some are better at sinking current (better for output low.) <S> If you need to get maximum brightness without damaging the chip and without using external transistors, then it matters which is which. <S> When in doubt, use an external transistor. <A> Instead of having the switch on the "high" (VCC) side, these LEDs have their switch on the GND side: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The left circuit shows the more "traditional" solution where the LED is grounded. <S> In both cases the switch is a transistor inside the microcontroller. <S> In case of "high side switching" (left) <S> a PMOS will do the work. <S> In case of "low side" switching (right), an NMOS will do the work. <S> The LED doesn't care what you do. <S> The LED cannot tell the difference. <S> If you apply enough voltage across it, it will light up. <A> Figure 1. <S> Image source: 1 GPIO - multiple LEDs . <S> Note the direction of the arrow in the diode symbol. <S> This is the required direction of current (from positive to negative). <S> With the circuit shown in Figure 1: <S> If the GPIO is switched high (+5 V) there will be no voltage across L1 but there will be 5 V across R2, L2 so current will flow through L2 and it will light. <S> If the GPIO is switched low (0 V) there will be no voltage across L2 <S> so it will turn off. <S> Instead, current will flow from the +5 V through R1 and L1 to ground and it will light. <S> Your circuit is the same as Figure 1 but with R2 and L2 left out. <A> There is no real reason that those LEDs need to connect to GND. <S> It is a design choice as to how to connect them. <S> Another consideration may have been related to the default operational state of GPIO pins so as to manage power up and power down behavior.
| In the case shown the design choice may have been to be able to light the LEDs when a low logic level was commanded to the output pin. LEDs can be connected to positive supply or to ground.
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How will an audio power amplifier output voltage behave when loaded? If an audio power amplifier would power a 8 Ohm speaker through its proper(?) output and if the volume knob is set such that the amplifier outputs 8Vrms 440Hz sine wave across the speaker leads the power delivered to the speaker can be found roughly as if Im not wrong: P = V^2/R = 64/8 = 8 Watt But now if we add a parallel speaker the equivalent resistance will be 4 Ohm. In this case will the amplifier act as a current source which will still output 8Vrms but double the current it sources? How about the typical effect of the mosfet power amplifier’s output impedance? <Q> Your scenario should separate linear model from a model that includes limiting values of voltage and/or current . <S> The amplifier is more voltage source than a current source within its linear range... <S> If the amplifier pumps twice the current into a 4 ohm load compared to an 8 ohm load, then it is a voltage source. <S> A high-quality amplifier will have a much lower output resistance, compared to the resistance of the load (speaker). <S> When you turn up the volume, you run into limits: a voltage limit (mostly set by the DC supply for the power amp) and a current limit. <S> A voltage source has difficulty supplying enough peak current into a small load: when you add speakers in parallel, more current must flow. <S> An example audio power amplifier: An amplifier has a damping factor of 180, which basically means that its output resistance is 180 times smaller than its 8-ohm load. <S> Such tiny output resistance is achieved with a great deal of negative feedback in the amplifier. <S> Within the voltage and current limits, the amplifier runs linearly and the model is a simple voltage source, with 44.4 milli-ohms Thevenin output resistance. <S> But the power supply likely sags when more current flows - DC supplies are often not regulated. <S> And the high-power output stage is not perfectly efficient when more current flows, so the peak current limit decreases for small load resistors (like 4 ohms or 2 ohms). <S> Feedback cannot correct these voltage and current limits . <S> When voltage or current limits are reached, distortion increases greatly. <S> So a power amplifier is specified for maximum power at some (small) distortion. <S> The example power amp is tested for maximum power where distortion has increased to 1% distortion (THD): 250W... 8 ohm 410W... <S> 4 ohm <S> 600W <S> ... 2 ohm Note that halving the load from 8 ohms to 4 ohms doesn't double the power from 250W to 500W, and similarly at 2 ohm load, power is less than 1000W. <S> This is a result of limiting of voltage and/or current. <S> When operated within its linear range , you can assume the simple voltage-source model: simulate this circuit – Schematic created using CircuitLab <A> "If an audio power amplifier" Modern Audio Power Amplifiers are designed to have very low output impedance. <S> What you are asking depends on the specifications of that particular power amplifier. <S> If the power amp can supply the amount of current that the load requires, it doesn't matter what the actual load impedance is. <S> In other words, if the amp can supply 100 Watts into an 8 Ohm load (about 3.5 Amps) and you connect a 4 Ohm load at 8 Vrms output, the total load current is 2 Amps. <S> This is well within the current and power ratings of that particular amplifier. <S> The changes in output voltage as the load is varied occur for several reasons: resistance between where the negative-feedback point is in the amplifier to the output terminals, how much negative-feedback is actually used inside the amplifier, some other (minor) factors. <S> My experience with a large variety of Audio Power Amplifiers is that the change in output voltage as the load is a very low percentage. <S> This is assuming, of course, that the input to the amplifier is constant and that the amplifier is not into current limit or power limit. <A> An audio amplifier is not a constant current or constant power device. <S> The Volume control sets the voltage output, and you can see this by simply disconnecting your 8 Ohm load ... <S> the output drive voltage should remain constant. <S> However, all amplifiers have an output impedance (of more than just the FET) and the output voltage will droop slightly with increasing load (lower impedance). <S> How much load you can apply and how much droop you experience depends on the amplifier. <S> As you lower the load impedance, you increase the current requirements for the amplifier. <S> At some point it cannot supply the required current and the amplifier will become non-linear, and eventually potentially blow a fuse or shutdown due to overtemperature. <S> The output impedance of the amplifier should be constant (and not load dependent) as long as the amplifier is still linear. <S> Once it becomes non-linear, it would be design dependent what happens to the output impedance. <S> You would expect that non-linear includes an increase in distortion, severe clipping and loss of gain (Vout). <A> In this case will the amplifier act as a current source which will still output 8Vrms but <S> double the current it sources? <S> Only if the amplifier is designed for that load. <S> Usually, amplifiers are designed for a specific load, and feeding twice the load with it will not lead to twice the power delivered. <S> So, no. <S> See: <S> Impedance matching for maximum power transfer. <S> How about the typical effect of the mosfet power amplifier’s output impedance? <S> These impedances are typically very low with given minimum load impedance, and within operational bounds . <S> Amplifiers aren't perfect nor linear: They might be very linear (i.e. work like a low-impedance voltage or current source) up to a specific current (or voltage), but then simply break down or run into saturation. <S> So, modelling an amplifier as an ideal current or voltage source plus a source impedance usually doesn't cut it.
| In general, the output voltage of an Audio Power Amplifier remains fairly constant as the load changes.
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How to connect LED strip to be powered by charger and batteries I plan on using it mostly plugged in, batteries are there just for extra mobility and won't be used most of the time. Don't know lot about batteries so in first scheme I added a switch (pos. 1 charging, pos. 2 normal usage when plugged in) which would be on pos. 2 most of the time and I don't know if it is necessary?Anyway, I'm not sure any of those two schemes is good so please correct me or suggest a better way to do this.Thanks. Link to batteries. Link to charge protection board. <Q> Looks mostly fine. <S> I'm not overly familiar with this specific charging module <S> so correct me if I'm wrong, but it looks like there is an input to charge the batteries, but <S> no output pins, so you would have to directly draw current from the Li-ions. <S> It should be fine for just a 2A draw, but any higher <S> and I would make sure I have output current protection. <S> When your batteries drop below 4V, you wont be able to directly supply 12V any more. <S> I would suggest a 12V regulator or some kind of LED driver to maintain constant and adequate power. <A> I'm not sure any of those two schemes is good <S> so please correct me or suggest a better way to do this. <S> Neither scheme is acceptable. <S> 20 <S> Amp charging current is way too much. <S> See: Charging Lithium-ion <S> These batteries should be charged at 1625 mA. <S> Source: <S> Panasonic <S> NCR18650B Datasheet Li-ion batteries are nominally 3.6V and three in series will not supply 12V very long if at all. <S> You could use four series batteries and a 16.8V charger rather than <S> 12.6V then add a 12V buck step down regulator. <S> NCR18650B <S> Li-ion Discharge Curve Source: Panasonic <S> NCR18650B Datasheet <S> If you were to use this charger the input power and output to battery would be connected to the input and output of the PCB. <S> The input voltage is not specified by the vendor but it would be more like 19-20V input. <A> I agree with most of what @misunderstood said except the specs. <S> Here are my own opinions. <S> Charging voltage: <S> 12.6V - 13.0V is specified by 3S,10A Protection board on vendor detail link with 10A max if ambient is cool. <S> This assumes you have a smart charger that cuts off when charge current reduces to 10% at 4.2V/cell and not a 12V 10A supply <S> but can be made to work if adjustable with pot or drop 0.7V with one 15A power diode. <S> Thus 11.3V/3=3.77V/cell to 3.8V max. <S> But since you have a steady load, that wont work with a smart charger. <S> I suggest a small cooling fan for both PS and protection board with an air plenum retrofited to flow thru both rather than over the top so nothing is burning hot to touch. <S> Do you understand that most LEDstrips for white have a dim threshold just below 9V and can be operated from car voltage of 14.2 V from built in current limiting resistors , you won’t draw maximum current or radiate maximum brightness of LEDs. <S> But if your strip is rated at 4.5A at 12V , two strips will exceed your 10A supply and rating if the supply is 14.2V. <S> So if you don’t mind a reduction of 0.7V reduction in nominal brightness at 12V, the apparent brightness will be ok but 2.3/3V rise above 9V or 77% of nominal brightness @ <S> 12V. <S> if you can specify the LEDstrip’s actual current vs voltage then perhaps a better solution is possible like 15V @ <S> tbd A with 4 cells. <S> Perhaps then drop 15V by 0.7V to achieve 14.3V (marginal hot LEDs) with one 15A to 20A power diode to reduce both the battery (4S) and 15V float charger voltage to 14.2~14.3. <S> There is some temporary 20% or so reduction in cell mA capacity by using 3.8V float and not charging to 4.2V/ std Li Ion cell but at least you wont accelerate the aging rate keeping it overcharged above 3.8V for long durations. <S> These are just my opinions.. <S> Your mileage may vary.
| Better would be to use a single cell (or parallel) and charge with a 5V input (e.g. USB) and boost the the battery output to 12v. This charger is not recommended for your application. You must have a circuit to limit the discharge of the battery so the voltage does not drop below 3V, regulate the charge current, and maintain a full charge while charger is powered and batteries are in use.
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Is it ever bad to run an IC VCC trace next to a ground plane with 1mm clearance? Since my last board flopped, I looked at it again and noticed a ground loop (because the DB9 casing completed the loop). Now I adjusted my board so that there is no loop of any kind. Instead, I'm running a VCC track next to a GND plane with 1mm clearance right underneath the 40-pin DIP AT89S52 microcontroller (on opposite side of course since I'm doing single-sided PCB). I'm also planning to add a couple of 0.047uF decoupling capacitors. For clarity, I added a picture of part of my circuit. I highlighted the ground wire in red and will convert it to a plane after so I don't waste etchant. The circled green items are the 0.047uF ceramic decoupling capacitors. They say ground loops are bad, but is vcc next to ground this close bad as well? and would modifying my clearance between the two planes affect microcontroller operation? and no I won't use 0 clearance or I'll blow the batteries up. <Q> In general, it's good to run Vcc as close as possible to the ground copper that will carry its return current. <S> This reduces the size of the current return loop, which minimizes radiated emissions and improves radiated susceptibility. <S> If the potential on VCC might be over 50 V, then you need to start thinking about creepage and clearance distances. <S> But I doubt this is the case here, since you're talking about powering a microcontroller. <A> To achieve even lower inductance, widen either or both of VDD and GND traces (neither traces are a large region of copper metal, or foil as used in fabricating the PCB, so neither is a "plane"). <S> By using that 1mm minimum separation all along that region, you will better exploit those two capacitors in supplying transient currents to the microcontroller. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> They say ground loops are bad, but is vcc next to ground this close bad as well? <S> Nope. <S> However, you already are adding capacitors between them, so if anything it's to the good. <S> This principle does not hold for two signal conductors, as the coupling can cause cross-talk, especially if one line is digital with a lot of sharp transitions and the other is low-level analog, but it's fine for VCC/ground. <S> There are other situations where coupling can get you in trouble, but there is no indication they apply here. <S> The thing to look out for is leakage between the two. <S> In the pcb world, the usual rule of thumb is 1 mil (1/1000 inch) of separation per volt of difference.
| So, as long as VCC is less than about 40 volts, 1 mm separation is just fine. Consider: the coupling between the two is largely capacitive, and by reducing clearance you increase the capacitance.
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Can you determine RGB pinouts with a multi meter? I have a Swiftech Apogee XL2 that has an addressable LED inside it. I want to connect it to my other RGB setups but I need to know the pin signals and voltage of the LED. I have a multi-meter at the moment but I can buy anything else that's needed. What do I need to do? Can anybody help? I only have very basic electrical knowledge but I'm ready to learn. I've included photos of the block and the connector in the link below. Thanks! https://photos.app.goo.gl/wtKxNEYCA6qHwcP3A Based on my research, I'm certain it's one of these two variations below. I'm just not sure about the voltage and how to identify the pin order. https://rog.asus.com/forum/attachment.php?s=6ebdc99c23d18e50b41c1d879d3513ac&attachmentid=71018&d=1517278100&thumb=1 <Q> This may not work on adressable LEDs. <S> Other than this you can try to identify the LEDs and Google them to find the spec sheets Fv at the least, but bear in mind that this is usually fairly inaccurate but it could be your only option. <S> You could also check the output of the PSU for the LEDs and go from there about a correct supply voltage. <S> If I needed to test the pinout I would: Try to find a schematic in the manual Tap each contact with a lesser supply voltage till you get lights Attempt to find the schematic and specs of the IC that is being used to make the LEDs adressable throughout the strip. <S> This might also reveal the reccomended input as I would hope the ICs had a constant current driver (better running conditions for an LED than contant voltage). <S> Hopefully this is what your looking for! <A> Popular addressable LEDs are typically powered by +5v. <S> The page you link to has a link to a controller that cautions the user not to insert the ALED into a fan connection as the voltage (+12v) is higher than what is allowed (+5v). <S> So it is somewhat safe to say you are dealing with +5v. <S> If those attachment arms are made out of metal, perhaps they are grounded as well. <S> Try probing the pins of the connector and the metal arms. <S> If you measure a short circuit it is very likely that you will have found ground. <S> More often than not the power rail is opposite the ground pin / wire and the data lines are in the middle. <S> So if you feel confident with 'somewhat', 'likely' and 'often' you could give it a try.. <A> This identification depends on parasitic diodes within the package. <S> Assuming you have a digital multimeter, usually the red lead is positive with respect to the black lead when doing a diode test (confirm by reading the manual or by testing with a known diode). <S> If you test the leads relative to each other you should be able to identify one which tests as a diode (about 550-700mV reading on the meter) to each of the other two leads when forward biased and overrange when reverse biased). <S> That will be the data lead. <S> Connect <S> one of the other two leads to +5 through a 1K resistor and the remaining one to ground. <S> Measure the voltage across the chip (ignore the data line). <S> If it reads about 0.7V the GND and +5 are reversed. <S> Swap the leads and measure again. <S> If it reads more than about 1V that should be the correct +5 and Gnd connections. <S> Do not use much less than 1K or the chip could be damaged. <S> You can then try connecting 5V and hope the smoke does not come out. <S> Use a separate 5V supply, not your motherboard. <S> When you connect it to your motherboard use the same polarity and at least temporarily put the 1K in series with the data line and preferably a polyswitch resettable fuse in series with the supply line to protect the motherboard. <S> No promises here, you may burn it out, or you could conceivably damage something else (such as the motherboard). simulate this circuit – <S> Schematic created using CircuitLab
| Some multimeters have a diode tester so you could use that on any number of LEDs to find their forward voltage.
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Why does the op amp output in this circuit have low slew rate? I have the following circuit on a bread board (with the exception of using AD820 instead of AD822 - the main difference based on the datasheets seemed to be one versus two op amps in one package). simulate this circuit – Schematic created using CircuitLab Simulating the above circuit shows a decent square wave both at input and output. However the actual signal I'm seeing on an oscilloscope is below. Here the slew rate seems to be around 1V/40us, not the 3V/us the datasheet describes. Things I've tried: Add the LM358. (The original circuit didn't have this, which caused problems when the square wave went from 0V to 2V. The AD820 was very slow to react to this. Using the LM358 to offset the signal out of saturation helped with that delay.) Bypass caps on the power rails and LM358 output. Currently I've got a bypass cap on the LM358 output to get rid of high frequency noise (not shown in the schematic). No effect on the actual slew rate. Replace the AD820 with an LM358. This had an immediate effect and the issue was gone. However I'd prefer the AD820 due to its lower input offset voltage. The final use case here is current sensing (and I happened to have few AD820s around). This was more of an attempt to demonstrate that the breadboard reactance isn't a problem here. Various loads on the output ranging from 1k to 100k. Play with the op amp power rail voltage and signal voltage. Most of the experimenting has been done with 5V power supply, +/-1V square wave. I did push the power supply all the way to 10V with no difference in the output (other than it being offset by +5V instead of +2.5V from the ground). Non-unity gain on the AD820 (10x and 100x non-inverting). Inverting buffer (-1x gain). I've also probed the power rails near the AD820 and they were very stable. The only remaining reason I can come up with myself is that the AD820s I've got are faulty. They are cheap AliExpress specimens after all. If this ends up being the case, is this a common thing with the cheap Chinese chips? Scope detail on 6.5 V input voltage (probing the AD820 power pins). Note the 500ns scale - The 2 V transition should be fully visible here at 3 V/us. <Q> Yes, it is common with cheap Chinese chips. <S> You should also try to measure quiescent current and see how it differs from one that is stated in the datasheet. <S> Chinese manufacturers usually tries to adopt chip design to poorer technology and it results in larger capacitances and higher quiescent currents. <A> The very first thing I would try would be to use a known-genuine Analog Devices AD82x in place of your mystery chip. <S> What you have could be a much cheaper part remarked or it could be a bad copy. <S> I would lean towards the former. <S> There are distributors such as Mouser and Digikey that will ship almost anywhere. <S> Aliexpress definitely has some bad sellers and they know it is expensive for customers to return the parts so they seldom have to make good. <S> I have had much better luck with domestic China purchases, believe it or not. <A> I have seen cheap open collector opamps reliabled as more expensive jfet parts. <S> I have also seen factory seconds that were supposed to be junked being resold from China. <S> Do yourself a favor and buy from Digikey. <S> Your time is more valuable than the few bucks you saved.
| Since your op-amp is JFET-buffered it could be the case of a higher Cgs than it should be.
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RFID: Why is inductive coupling used instead of EM waves for LF and HF systems? Is it because of antenna sizes? At 900 MHz a 1/2 dipole has a size of around 15 cm. That would be far higher for 125 kHz, so instead we use coils to power the transponder and use load modulation for the reverse-link. Does the low frequency have any effect on reflection? Reflection is dependent on frequency, low frequency waves tend to "go through" materials easier so we would have a harder time reflecting those through the transponder antenna? Are those the two main reasons? <Q> Inductive coupling is a much better way of transferring energy over a short distance. <S> For (really) long distances, EM waves win the race over inductive coupling. <S> So, it has nothing to do with antenna sizes or reflection. <A> When we speak of EM waves, we generally mean planar waves. <S> This requires us to be in a region known as the far field . <S> The far field is a certain distance away from the antenna, and thus gives us a minimum distance. <S> If we are talking about reactive (inductive or capacitive) coupling, we are talking about the near field . <S> In order to be in the far field at low frequencies, we need very large distances. <S> The wikipedia article on this can be found here: Near and far field <S> In addition, we can extract more energy with reactive coupling (in essence, transformers operate this way, and they go to 99+ percent efficiency). <S> Think of it this way <S> - if EM is propagating, it spreads out as we move away. <S> Our antenna can only cover a small area, and this area is a fraction of the total 'sphere' our EM wave is propagating away from. <S> As a result, we can only capture a very limited part of the wave. <S> This is seen in Friis transmission equation. <A> When speaking about inductive coupling in RFID. <S> The reader emits magnetic field and when tag enters the fild the chip varies its antenna resposnse which will result in a perturbation of magnetic field which can be detected by the reader. <S> The strength of the magnectic field drops sharply with distance from the emitter. <S> Hence inductive coupling is used in a situation where there is a involvement of short range operation. <S> Also electromagnetic waves causes a lot of interference and other multipath problems. <S> Inductive coupling works better for high and ultra high frequency operation.
| No, the reason for inductive coupling wins over EM waves for RFID systems is better efficiency at nearby distances.
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What does nm mean in FPGA When I read articles, FPGA comes with a specification-nm.For example, Xilinx 28nm Virtex 7 or 20nm UltraScale FPGAs.What does -nm means in FPGA? <Q> Smaller gates mean more transistors inside a chip. <S> See here for more details: <S> what do the nm or µm for a process technology refer to e.g 22nm process technology for device fabrication <A> It means the structure width of the manufacturing process (I.e. the width of the transistor structures within the FPGA). <A> As pointed out by RoyC <S> I'll expand the comment to a full answer, giving some more insights. <S> Any kind of semiconductor needs to pass through a production process. <S> The process technology was historically associated to the length of the silicon channel between source and drain in field effect transistors. <S> This has varied a lot in the last 5 decades, passing from a 10 µm in 1971 to the current 10 nm in 2017 and it is prospected to have a 5 nm technology by 2020. <S> Just to remind the length magnitudes <S> : 1 <S> mm (milli-meter) = <S> 10E-3 <S> m <S> 1 µm (micro-meter) = <S> 10E-6 <S> m <S> 1 nm (nano-meter) = <S> 10E-9 <S> m 1 pm (pico-meter <S> ) = 10E-12 m <S> Since technology has evolved in various aspects, also the transistors have. <S> So how it is pointed out by Dan M , nowadays we continue with naming the process with a shrinking size, but this is not referred to the channel's length, but to the smallest element in the transistor or the size of the gate. <S> For more info you can check out these references: Wikipedia general explanation on production process <S> Extra details on naming the process <S> I would also like to point out that the process doesn't continue linearly decreasing, since with smaller technologies more problems arise. <S> There is a theoretical limit to this shrinking, since to create a transistor at least three elements (Source-Drain-Gate for FET transistors or Collector-Emitter-Base for BJT transistors). <S> I also report one interesting research on super small transistors: <S> In 2018, researchers at Karlsruhe Institute of Technology created a transistor with a working single atom gate <S> So actual research is already pretty close to physical limitations in transistors. <S> If we need to push more, other technologies need to be discovered, making transistors obsolete. <S> You can find more research level technologies on this Wikipedia page . <S> Naturally this answer of process technology is not applied only to FPGA manufacturing, but to the whole integrated circuits manufacturing (FPGA, ASICS, CPU, GPU, etc.).
| It's the size of the gate of transistors inside the chip and so a name for the production process.
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Does a dimmed LED lamp draw less power than one at full brightness? Does a dimmable 6W LED lamp, dimmed to 50%, draw c. 3W? Or does it still draw the full 6W? <Q> Usually, dimming on LED lamps is done by PWM (Pulse Width Modulation). <S> The LED will actually turn on and off faster than the eye can see. <S> By modulating the ratio of the time the led is on and off, it simulates the dimming to the human eye. <S> This method is widely used because it is the cheapest and easiest method to dim an led <S> but it can also cause some eye fatigue. <S> With this method the power will reduce with the light intensity (or perceived intensity). <S> Another method of controlling the light intensity of an LED is to control the current flow through the LED, and there are mostly 2 ways of doing it: Through DC/DC conversion like a current controlled buck converter or similar circuit. <S> in this case the current will reduce with the intensity of the LED. <S> With this method, there will be little flicker, which will be roughly equivalent to the DC/DC converter ripple. <S> Using an analog ballast (adjustable current source), like a transistor, mosfet or specialized IC with required regulation. <S> In this situation, the current will diminish but this circuit will likely have more losses. <S> The latest has the advantage of providing a very stable light intensity (if needed for some specific application perhaps like photography), so we can imagine this exists on the market. <S> The power will be reduced when the light intensity is reduced, but it's not a linear factor. <S> F.I. 50% dimming might not have a 50% decrease on the power consumption. <S> The reason is that the control electronic will have some non-linearity and won't have the same efficiency at different operating condition and the second reason is that the eye intensity response is not linear to the actual light. <A> In general yes, the LED itself will use less power when "dimmed". <S> But we have to be a bit more specific what "dimmed" means. <S> So the answer depends a bit on how the dimming is done. <S> If a pulse width modulation is used to control the LED's brightness, then the dissipated power scales directly with the pulswidth. <S> This means full brightness is reached with max pulswidth what equals max power dissipation whereas as a reduced brightness requires less pulswidth and therefore dissipates less power. <S> Back in the days when incandescent light bulbs where used, dimming was done by so called phase angle control, comparable to something like AC pulse width modulation. <S> And before that, you would just add a potentiometer as a pre-resistor. <S> In that case, the bulb itself would dissipate less power when dimmed, and the resistor would burn the rest. <S> But long story short, when you use PWM, the power will scale with the pulse width. <S> But consider that brightness doesn't scale linearly with dissipated power. <A> I will use about half (except for the controller using a bit). <S> Probably PWM is used, this means a duty cycle of 50%, means the light is on only half of the time, many times per second though.
| It depends on the LED driver, but unless it's of a bad (or very specific need) design, a dimmed LED lamp will draw less power.
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How to initiate a ring oscillator So assuming it's a simple oscillator like so : How can I start this oscillating with an initial value then make sure this value doesn't affect the operation of the circuit? <Q> Not only can you do this, you can synchronize the start of oscillations quite precisely. <S> EDIT - Not only can you start the oscillator precisely, this topology will produce a complete last pulse. <S> Simpler on-off configurations may produce narrow last pulses if the off condition occurs during a pulse, but not this one. <S> END EDIT <S> Start with this structure simulate this circuit – <S> Schematic created using CircuitLab <S> A low input will inhibit oscillation, and when the input goes high oscillation will start. <S> As for the time delay, assuming you're using discrete logic, you can take advantage of the common occurrence of quad NAND gate ICs by implementing simulate this circuit <S> Of course, you can also use a couple (or 4 or 6 or ...) of inverters to do the job. <S> Alternatively, you can simply brute-force the issue. <S> Using an open-collector output or a separate transistor, you can simulate this circuit <S> This takes into account that every (discrete, at least) logic family on the market will tolerate a short circuit to ground indefinitely. <S> If it could not, you'd get any number of dead ICs due to inadvertant shorts during test, assembly or use. <S> This approach will cause a slight change in the output frequency when oscillation first starts, due to the fact that the chip would probably get warm and this would affect the propagation delay of the gates. <S> But this would quickly disappear as the IC reached operating temperature. <A> If you are open to use different types of gates, here is what you can do simulate this circuit – <S> Schematic created using CircuitLab <S> The RC delay will keep NOR gate output at 1 from the beginning, then switch to NOT function as required for oscillation. <S> Having said that, there is still "twilight" zone on the power up where gate output might be undefined. <S> Adding pull-up on the output might help with that, but it will affect the oscillation frequency somewhat. <A> How can I start this oscillating with an initial value then make sure this value doesn't affect the operation of the circuit? <S> Your question is unclear whether you mean this circuit being in a circuit simulator or a physical form using real components. <S> In the simulator <S> indeed this circuit will not start oscillating as the circuit solver can find an operating point where all stages basically output the same voltage that is at their input. <S> Usually this voltage is near half the supply voltage. <S> To make it start you have several option. <S> All you need to do is to "kick" it into action. <S> I usually use a current pulse (ipulse component) to make a for example 10 uA current pulse of a few micro seconds. <S> As the current returns to zero after the pulse, it does not influence the circuit anymore. <S> You can inject the current into any internal node of the ring oscillator. <S> Sometimes a pulse on the supply will also work. <S> In the real world this circuit starts up (given that it has enough loopgain) due to noise and disturbances. <S> It will never be on the "dead" point that the simulator finds as that state has a very high gain so all noise and disturbances are amplified kicking the oscillator into action.
| You can also use a vpulse to make a voltage pulse, put that source in between two inverters.
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Detect ring signal using a low pass filter I created 2 boards communicating each other like an intercom system. One of them has a button to send a ring signal to the other. The ring signal consists of PWM signals created by an MCU. I want to detect the ringtone by the MCU of the receiver side. My ringtone contains 4 outer pulses with a period of 300 ms: 200ms of this signal contains the inner pulses with a period of 1.25ms: So, I want to add a filter circuit to the receiver side of the ringtone and convert the outer signal to a straight PWM signal. Below is an example output : So far, I have tried adding an RC low pass filter with a cut of frequency of the outer signal (1/300ms = 3.33Hz ). But I couldn't even get close to the example output. Is there anything I misunderstand about the concept since I'm not very experienced in circuit designs? Regarding my circuit: I just added a series resistor and a parallel capacitor to the node: Edit: I didn't want to talk about the circuit that handles converting the audio line signal back to the PWM since this is not the subject on this question. But I suspect that my RC filter does not work properly because of that part. I tried the solutions on the answer but they didn't work as expected. Below is the schematic of my optocoupler side, line input contains the audio signal of the PWM. The audio signal converted back to the PWM by the optocoupler then I get the PWM signal which I gave the screenshots above. But when I add the RC parts to the OPTO_OUTPUT node, PWM signal changes in terms of volts too. Do you think I have a mistake on adding the RC parts? <Q> If you want the 300ms signal to be received, while filtering the 1.25ms one, then it's the high frequency signal you need to filter out, not the other. <S> Which means your chosen time constant is too large and affects the 300ms one. <S> The high frequency signal has a period of 1.25ms, so choose a time constant that is more than 10x larger, say 25ms, which is also more than 10x smaller than the 300ms. <S> Here's a quick test in LTspice: <S> V(a) shows the modulated output, V(b) shows the filtered output with a time constant of 22ms, and V(c) <S> shows the recovered signal with a minor hysteresis of 10mV to counter the non-ideal filtering of the RC. <S> Notice that the filtered signal has a somewhat thicker trace <S> , that's because of the residual. <S> You could use a Bessel or Gaussian filter for better results, but that would only add to the complexity and, besides, you'll still need the recovery of the signal, so that means you can simplify things. <S> If your Vcc 's value is 5V, then the collector resistor is too much, remember the optoisolators have a strong Ic dependency. <S> If so, then try a 2k2 resistor, which will only use Ic~2.2mA, and you could give up the extra R and place C directly across the collector (10uF, adapted value for time constant). <S> This will get Ic higher when C discharges, though. <S> Otherwise make <S> R=2k7 and C=3u9 (for example). <S> Don't forget that the signal is inverted now, so I've used an inverting hysteresis comparator (the Schmitt trigger). <S> I don't have a TLP opto, so I've just used whatever you see, adapt to your needs. <S> I'm not home right now, but here's an attempt at exemplifying ( V(n004) is the 300ms signal): <A> Depending on how tightly you want to detect the ring signal, you may need to band-pass filter the received signal (at F = 800 Hz) and then envelope detect it. <S> Given your ringing profile, this signal will disappear for about 100 ms every 300 ms. <S> This detects the "envelope" of your ringing signal. <S> You would then need some logic to determine that the envelope shape was approximately correct. <S> There are other things to look for. <S> There are tone decoder ICs that spring to mind - the LM567 has been used in applications like this many times. <S> Or you could change your basic 800 Hz signal incorporating two tones and use a DTMF decoder chip. <A> I would use an envelope detector: <S> If you pick the RC time constant somewhere between the carrier frequency and the signal frequency, you'll get the output looking like this: <A> You can achieve this directly by the micro-controller. <S> Doing so has some advantages: <S> No need of external components. <S> No response time / delay caused by the filtering. <S> Higher accuracy. <S> Basically, I understand you want to know the duty cycle of the PWM signal. <S> To do that with a micro controller is fairly simple, given your frequency is relatively low, it shouldn't be an issue for the uC to handle it. <S> You only need a timer and an interrupt. <S> Connect the signal directly to your uC input. <S> Setup a pin driven interrupt in both low and high edge. <S> Setup a timer with a known time period. <S> When the interrupt is called, first read the value of the timer, then read the state of the pin (low or high). <S> On the next interrupt, do the same, you then know how long the pin was high or low, do the same for the other edge. <S> After 4 edge changes happened, you will know precisely how long it was high and low, then you simply need to compute the ratio. <S> You can repeat this for the whole 200ms signal, recording all the timing and do an average to further improve the stability. <S> Considerations: <S> The interrupt will introduce some delay, it is why it is important to read the timer first, so you have minimum delay. <S> You can use high priority interrupt to avoid having other interrupt blocking. <S> We consider the interrupt delay (with high priority) will always be relatively similar and small differences averaged over the 200ms signal.
| Alternatively (and this may be preferable) you could use a retriggerable monostable circuit that will produce a constant high output when the "inner signal" is activated.
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Problems sizing series input resistor for this optocoupler Under the electrical characteristics, the optocoupler 4N26 typical "forward voltage" is given as 1.2V. But for sizing the input resistor Im have hesitations. In my application as a control signal I will use an Arduino 5V digital output for the opto input. The optocpupler will then switch this SSR which requires 16mA to operate independent of input voltage. Here is the basic diagram: In a net tutorial I found this information: "The CTR depends on the LED input current (IF) and the CTR decreases from a maximum point when the input current is both increased and decreased. If you look at the below diagram, the top of the curve is around the 6mA point. This is where the chip is most efficient." In my case I have the black package and below is for the CTR vs Forward current: Here my questions: 1-) So in my case by looking at the plot above I choose the forward current IF as 4mA and from datasheet the forward voltage is typical 1.2V. So I calculate: R = (5 - 1.2) / 0.004 = 950 Ohm. So I conclude 1k. But in many examples I see much smaller resistor value. Am I doing something wrong here? 2-) In some examples they place the resistor in return path. Does that have any significance? Her an example: edit: <Q> The 4N26 is a really cheap crummy optoisolator and has a guaranteed CTR of only 20% at 25°C. <S> It will be worse at temperature extremes and with LED aging. <S> So you would need a minimum of 80mA, but more like 120mA to be safe on CTR, <S> but that's too much for the optocoupler LED. <S> Note that the curve you show is normalized <S> so CTR drops to more like 12% at 20mA. <S> 4N36 has a more reasonable 100% CTR guaranteed <S> so maybe you can choose 25mA drive <S> but that's still a bit high for the Arduino and for good opto life. <S> Or put a transistor on the SSR side to drive the SSR. <S> Note that the SSR already includes proper isolation so you may not require anything more than a BJT or MOSFET to drive the SSR if you don't need to keep the 12V power supply isolated. <S> In such a case, I would use something like a 2N4401. <A> It is because the maximum input current of the LED is 100mA (60mA for M version). <S> You rather drive the LED around 20-40mA to make sure the opto-transistor is conducting fully. <S> 5-1.2/0.02 <S> = 190 ohm <S> so you can use 180 ohm. <A> The phototranstor browns out over time, so the CTR ratio falls. <S> You need to supply a consitent current. <S> There are other optos with smaller current also. <S> But your SSR device already has opto isolation, so an open collector BJT can solve this without problem. <S> Open collector output can be used for relativelly long distances, if this is an issue. <S> EDIT: <S> You can eliminate the worry about false triggering. <S> The noise could come from Vcc and not from signal. <S> The easiest is to use a BJT. <S> Or viceversa, you add this filter to the input of MCU power supply and left unfiltered for external <S> I/O signals. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> But most likely you won't have any issues even if the 12V Vcc goes 5m away, since you have also a low pass filter before and after LDO which supplies the MCU. <S> simulate this circuit Professional PLC sytem using same power supply for MCU and IO with choke and opto coupled IO signals.
| If noise is a concern, then you might install a differential choke and cap, MOV where you suppy the output Vcc.
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I2C Issues - SCL is stuck high I am working on a custom carrier board using a TI microcontroller as the master and an ADC IC as the slave. It appears that sometimes SCL will stay high when it shouldn't, almost like the master releases it. We only get this fault condition on a reset of the MCU (or total system power cycle) but it does not happen every time, maybe one out of ten on average. Data transfer will be fine most of the time, until one of the unlucky resets causes the issues. You can see in the image below that after the address byte is sent SCL just goes high after a few clock pulses. Any ideas? <Q> From recent comments: the microprocessor does return an arbitration lost flag Thanks, that is vital new info and deserves to be added to the question. <S> So does this info: <S> I discovered this problem happens when the reset is pressed during activity on SDA. <S> If the reset is pressed while SDA is high, the bus does not get messed up. <S> Based on the new info, I believe your problem is a duplicate of previous question " I2C bus occupied ". <S> Your expectation that " I would think the MCU would come back up in a known good state with I2C initialized " is not true, because the MCU does not control the state of all devices on the I2C bus. <S> The I2C Master only knows and controls its own state. <S> It relies on other devices following the I2C protocol, to infer their state. <S> Resetting only some devices on the bus can result in a mismatch between the internal states of the various devices on the bus, and lead to problems like yours. <S> When they restart, I2C Masters can be affected when they see that an I2C Slave is pulling SDA low, waiting for clock pulses from the I2C Master (which the Master won't send, as it was just reset) so the Slave waits, still pulling SDA low. <S> For example: After restarting, some I2C Masters (e.g. MCU), before they start the I2C SCL clock, will check the SDA signal. <S> In your case, it sees that SDA is already low . <S> This state can be interpreted as there being another I2C Master device driving the bus. <S> Actually the cause of SDA being low is the I2C Slave, which was in mid-transmission before the MCU <S> was reset. <S> However now the Slave isn't receiving SCL pulses from the I2C Master and so it cannot complete its transmission. <S> Most I2C Slaves don't timeout and release SDA in that situation, although a few do. <S> The fixes in the answer to the above linked question would apply here e.g. pulsing SCL manually, until the I2C Slave releases SDA, or resetting the I2C Slave (via a reset pin or local power switch to the I2C Slave(s)). <S> Then restart I2C on the Master. <S> The I2C master will see that SDA is no longer being held low, and so it will start to drive SCL again, as it won't be confused that there is another I2C Master on the bus. <A> Look closer: SDA is low . <S> This is the "arbitration lost" case, where the master wants to put a "1" on the bus but reads back a zero. <S> In I²C terms, this would mean there is another master on the bus, an thus the original master ceases toggeling the SCL line. <S> Now you will have to look at <S> why the ADC (or something else in on your I²C bus) exibits this behaviour - but that is a different and much more interesting question. <A> The slave ADC has no idea the MCU was reset and still thinks they are communicating and it is waiting for the MCU to respond with more clock pulses. <S> However since the MCU lost power it does not know where it left off, meanwhile its seeing the SDA line low and thinks there is arbitration. <S> So it is an issue of the slave waiting to send all of its data, but the MCU doesn't know that, <S> so actually it is the mcu's fault. <S> This white paper describes this completely. <S> Link
| This is a problem of the MCU being reset during a bus transaction and essentially forgetting what it was doing.
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Can a 24VDC 14amps rated power supply provide more current than it's rated? I am having some problems with understanding power supply. Can a 24VDC 14A rated power supply provide more current than it's rated? I thought the maximum current the power supply can provide is 14A. Imax= V/(Ri+RL), when RL = 0. (Ri=internal resistance, RL=load resistance) Even if I short the power supply with a wire, the maximum current that will flow through the wire is 14A? If not, what's the maximum current? Can you explain why? <Q> It depends Basically, there are several different possible outcomes, depending on the decisions the designer made when designing the power supply. <S> Some possibilities: <S> The power supply fails. <S> In which case, the current and voltage drop to zero, and possibly the power supply catches fire. <S> This is the cheapest and easiest option for the designer/manufacturer. <S> The power supply has an output fuse, which blows. <S> In which case the the current and voltage drop to zero, and the power supply can be repaired by replacing the fuse. <S> This is only marginally more expensive than the above, so should be the bare minimum in most power supplies you can buy. <S> The power supply has overcurrent protection. <S> In which case the power supply detects the short circuit, and turns the output off, or turns it down to a very low level. <S> It may then automatically reset after some time, or it may need the short disconnected, or a reset button pressed. <S> The power supply has a current limit. <S> In which case the current stays at or near the rating, and the voltage drops to virtually nothing, in keeping with ohms law. <S> This is the most complex to design and most expensive to build option, and a power supply which does this will probably draw attention to it in the manual/datasheet. <S> In all cases, the power supply will probably go a bit over the rating before the above kicks in. <S> This is done because the manufacturer wants to make sure it never ends up less than rated, so they build in a little bit of margin. <S> But you should't normally use a power supply in the narrow gap between the rating and the point it fails/shuts down, as it will probably be unreliable and/or wear out quickly. <A> Although the supply might provided more than its rated current, you can not assume that it will. <S> The manufacturer is guaranteeing that you will get <S> at least 14A. <S> The current might limit at 14A and the voltage drop to near zero. <S> The current might "fold back" to a smaller value. <S> The supply might catch fire. <A> It's a choice of the power supply designer. <S> The possibilities for a supply rated at 14A are - a) <S> 14A - if it's been designed to limit at a constant current. <S> b) much less than 14A - <S> if it's been designed to 'fold back' on overload. <S> This provides a much lower current when overloaded to avoid turning a load fault into a fire. <S> c) much more than 14A - <S> if it's an old transformer + rectifier type supply. <S> The current will typically stop within one second if a fuse blows, or in a few minutes when the supply catches fire. <S> d) <S> roughly zero - a more or less full shutdown into an overload. <S> It might pulse occasionally to test whether the short circuit is still present.
| The amount of current that can flow is dependent on how the power supply has been designed. No one can say, based on the information you give, what will happen if you short the outputs.
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