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Do I need individual connections to +5 V for this circuit? I have the following circuit: See that +5V in the top left corner (ignore the bottom one for now)? It is connected to J, S, J, and S. I know that I can simply grab 4 wires and connect each on to my +5V on my protoboard, but I was wondering - instead of using 4 long wires, can I just use 1 long wire, connect it from my voltage source to some spot on my breadboard, and then connect 4 wires in series (same row) to it to grab +5V? Or would this reduce the voltage each port receives? <Q> The schematic is drawn that way to avoid crossing wires and improve clarity. <S> When digital chips change output state (low to high or high to low) there is often a sudden very short spkike in current demand. <S> For this instant resistance and inductance of the PCB or wiring can cause a drop in voltage to the chip and this can cause intermittent or incorrect operation. <S> To solve this we add a small decoupling capacitor - typically <S> 100 nF - as close to between the chip V+ and GND as we can. <S> This supplies the current for that instant. <S> The problem becomes more severe at high clock speeds. <S> For your application it may not matter but always add the decoupling capacitors to avoid trouble. <S> From the comments: I was just wondering if I can connect 5 wires (1 from the +5V and the other 4 from the chip) in series to draw the same voltage to each pin through each wire. <S> Sorry, I understood the question but didn't answer it well. <S> Since the currents are miniscule no significant voltage drop will result. <S> In any case, the logic thresholds are generally well towards Vcc/2 which gives quite a bit of noise immunity. <A> They are electronically almost the same thing. <S> The wire would act like a resistor of almost 0 ohms. <S> Negligible. <S> This happens in parallel led setups like 12V led strips. <S> But not here. <S> Your breadboard does this. <S> Each spot in a row is a slightly different distance from the source, but the resistance difference is so small it's negligible. <S> FYI. <S> This type of connection is called a Daisy Chain. <A> Most breadboards have two rails along each long edge specifically for power distribution. <S> Typically, one rail would be used for the positive supply, while the other rail is used for Ground.
| As the wire size decreases and the current increase across each node, then the resistance is more significant and would need to be accounted for. At the current this would take, there is no significant difference. All 5 V pins will be fed from the same power supply. Yes, you can loop from one to the next et cetera .
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How can a DC18V ~ 36V LED driver work with DC3.6V LEDs without burning them out? I bought some slightly more powerful leds to build a hobby project and grow some herbs. The kit has 10 LEDs that have the following spec: Power: 3WVoltage: DC3.6Vcurrent::700mA It comes with a driver: Output Voltage: DC18V ~ 36VOutput Current: 600mA The idea is put 10 in series and drive them from AC with that driver. My questions is how can it work without burning them out? I know about the voltage drop, but the first LED still gets a high voltage? They are only rated ~3V. I "tested" one LED with 9V battery and it lighted up very bright and then it was dead. Was it because of 9V or was it because it drew that much current? What am I missing here, it looks like there is some very basic thing I don't understand. I have studied the OHM law and have built some simple things, but I'm out of ideas here. <Q> Your driver will output a fixed current through a string of LEDs. <S> So you can attach 18/3.6 to 36/3.6 or 5 to 10 LEDs to this driver. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> With 10 LEDS the output from the driver will be around 36V. That voltage is divided across each LED, so no individual diode has more than their rated voltage. <S> Or more accurately, when driven with 600mA each LED will generate a voltage across it, close to 3.6V, the total of those voltages will be what you measure at the output of the driver. <S> When you hooked up an LED directly to the 9V battery you vastly exceeded it's rated voltage <S> and it went out with a flash of light. <S> You basically fused it. <A> 1) NEVER connect a led to any voltage source (power supply or battery) without resistors. <S> It could ruin the led, the power source, or both. <S> 2) <S> It seems that this "driver" is a constant current source of 600 mA, with minimum and maximum voltages of 18 V and 36 V. Since the led's specs are 700 mA (max), you can connect the ten leds in series, and the serie of leds can be directly powered by this driver. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> There is an advantage of running the 700 mA leds with 600 mA: they will not bright as their max, but their life spanning will be higher. <A> Others have already covered the driver situation. <S> If you want to test with your 9V battery, try it with a 8.2 or 9.1 ohm resistor in series with the LED. <S> This will limit the current to something that LED can handle. <S> The resistor should be rated for at least 4 or 5 watts though. <S> You probably don't want to do this for very long though-- <S> you're <S> right at (or maybe even beyond) <S> the current a typical 9V battery is intended to supply, so don't expect the battery to last more than an hour or so (at best) if you use it for this. <S> The battery will also probably get pretty warm if you pull this much current from it for very long at all. <S> I'm pretty sure <S> the single largest use for 9V batteries is to power smoke alarms, so they're designed primarily for long shelf life, while providing fairly minimal current (on the order of 10s of milliamps at most).
| How many LEDs you can put in series on a driver depends on the drivers voltage range, in this case 18 to 36V. Since your LEDs are rated at 700mA @ 3.6V, this driver will indeed supply adequate current for the LEDs without burning them out.
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How to turn an SPST momentary push button into a momentary DPDT? How to turn an SPST momentary push button into a DPDT? I would like to close 2 circuits at the same time. (From comments: I want to ground the reset circuit while getting a pin to HIGH on an Arduino like board.) SPST DPDT Image source . <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Use a relay to provide DPDT action from a single push-to-make pushbutton. <S> The relay coil voltage should be the same as the supply, V+. <S> The contacts should be rated for > switching voltage and > switching current. <S> In your case it is small signal so it may be more important to get gold-plated contacts which won't corrode and degrade the small signals. <A> If you have to keep the existing SPST momentary switch then here is how to convert it to a DPDT switching circuit. <A> Based on the info in your comments. <S> simulate this circuit – <S> Schematic created using CircuitLab
| You do not want a DPDT switch, but a resistor and an invertor.
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Silk screen layer over pad resist cutouts on pcb I am designing a SMD PCB and have noticed that some component silk screen outlines extend over the pad resist cutouts. Is this acceptable? If the resist cutouts are made after the silkscreen process I can see this working. <Q> The silk screen is the last step in PCB production. <S> The solder mask is the penultimate step. <S> If your silk screen is over (or too near) <S> a pad, that should throw an error from your DRC. <S> Most likely the silk-pad spacing in your DRC settings is greater than the solder mask expansion so you should not have the silk screen going over the edge of the solder mask unless it is too close to the pad (or unless you are doing something special on the solder mask layer such as removing it over a large area for some reason). <S> Keep in mind that the registration of the silk screen layer(s) may not be as good as the other layers so you may wish to set conservative DRC rules for the spacing, lest it end up on a pad and affect the soldering. <A> This is usually fine, but it's good practice not to do it. <S> I've never had a problem with this, and it should be an obvious check from the board house. <S> To be totally safe, just move your silkscreen. <A> Most board houses will fix this for you, but when you make your gerbers you can fix it yourself by intersecting the appropriate solder mask layer with each silkscreen layer
| If the solder mask cutout is unrelated to a pad, I don't think running it over the edge of the cutout will cause serious problems but it might affect readability if your designators are very small.
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Is it possible to draw circuits through code? Is there a better way not by writing netlist files. something like: Define Battery1 As a BatteryDefine Resistor1 As a ResistorConnect Battery1 First Terminal to Resistor1 Second TerminalConnect Resistor1 First Terminal to Battery1 Second Terminal EDIT: I find the answers pretty helpful. It will be great if there is a way to simulate the circuit produced by the code. <Q> Example . <S> More Examples <A> Check out SKiDL ( https://github.com/xesscorp/skidl ), this is along the lines you're thinking. <S> Edit (as req'd): SKiDL allows procedural description of all circuits (rather than just digital), instead of graphically entering your schematic. <S> The netlist output can then be imported into layout software. <S> It will also perform ERC checks, and is extensible. <S> This means, for e.g., that you could write a filter once and then reuse it in different projects rather than drawing each time. <S> Written in Python, with all the support that comes along with that. <A> Many programs can draw a schematic. <S> None that I know of can draw a good schematic: One that emphasizes the most important information, and arranges the circuit in a clear and easy to understand way. <S> If you're just looking for a hardware description language (no graphical schematic output), VHDL and Verilog are widely used to define (digital) circuits to be implemented in ICs, and can also be used for board-level design. <A> Your example looks a lot like modelica , an object oriented language for simulation based on creating blocks and connecting ports between the blocks. <S> An example using the electrical components library (from maplesoft.com ), encapsulated model ChuaCircuit "Chua's circuit, ns, V, A" import Modelica. <S> Electrical. <S> Analog. <S> Basic; import Modelica. <S> Electrical. <S> Analog. <S> Examples. <S> Utilities; import Modelica. <S> Icons; extends Icons. <S> Example; Basic. <S> Inductor L(L=18); Basic. <S> Resistor Ro(R=12.5e-3); Basic. <S> Conductor G(G=0.565); Basic. <S> Capacitor C1(C=10, v(start=4)); Basic. <S> Capacitor C2(C=100); Utilities. <S> NonlinearResistor Nr( Ga(min=-1) = <S> -0.757576, Gb(min=-1) = <S> -0.409091, Ve=1); Basic. <S> Ground Gnd;equation <S> connect(L.p, G.p); connect(G.n, Nr.p); connect(Nr.n, Gnd.p); connect(C1.p, G.n); connect(L.n, Ro.p); connect(G.p, C2.p); connect(C1.n, Gnd.p); connect(C2.n, Gnd.p); connect(Ro.n, Gnd.p);end ChuaCircuit; Although you can generate a schematic from the model, normally this is done in a GUI that annotates the components with position and orientation information. <A> Cirkuit is an editor to convert a simple text description to a circuit diagram. <S> It provides a set of M4 macros for electric symbols. <S> It can be used together with circuitikz which was suggested by nidhin . <S> circuitikz uses the more modern pgf/TikZ graphic system. <S> The stackexchange community has very active users of cirkuitikz , but there are more solutions in TeX . <S> Picture from https://ece.uwaterloo.ca/~aplevich/Circuit_macros/html/examples.html . <A> PSTricks is another library for TeX users. <S> It can even do complicated mathematical calculation such as differential equations. <S> \documentclass[pstricks,border=12pt,12pt]{standalone}\usepackage{pst-eucl, <S> pst-circ}\psset{ dipolestyle=zigzag, labelangle=0, labeloffset=-.9, intensitylabeloffset=-.4, tensionstyle= <S> pm, tensionoffset=.9, tensionlabeloffset=.9, <S> %tensioncolor=red, %tensionlabelcolor=blue,} \begin{document}\begin{pspicture}[showgrid= <S> none](12,-12 <S> ) \pstGeonode[PosAngle={135,90,45,0,-45,-90,-135,180,45}] <S> (2,-2){A} (6,-2){B} (10,-2){C} (10,-6){D} <S> (10,-10){E} (6,-10){F} (2,-10){G} (2,-6){H} (6,-6){I} % \resistor[intensitylabel=$i_1$,tensionlabel=$V_{HA}$](H)(A){$R_1 <S> $} \resistor[tensionlabel=$V_{AB}$](A)(B){$R_2 <S> $} \vdc[tensionlabel=$V_{BC}$](B)(C){$E_1$} \resistor[tensionlabel=$V_{CB}$](C)(D){$R_3$} % \resistor[intensitylabel=$i_2$,tensionlabel=$V_{HI}$](H)(I){$R_4 <S> $} \vdc[tensionlabel=$V_{ID}$](I)(D){$E_2 <S> $} % \resistor[intensitylabel=$i_3$,tensionlabel=$V_{HG}$](H)(G){$R_5 <S> $} \newSwitch[ison= <S> true,tensionlabel=$V_{GF}$](G)(F){$S_2 <S> $} \wire(F)(E) \resistor[tensionlabel=$V_{DE}$,dipoleconvention <S> =generator](E)(D){$R_6$} % \vdc[tensionlabel=$V_{FI}$,dipoleconvention=generator](I)(F){$E_3$} \newSwitch[intensitylabel=$i_4$,tensionlabel=$V_{BI}$,ison=false](B)(I){$S_1$} \end{pspicture}\end{document} <A> Yes. <S> You can use HDLs to describe your circuit through code. <S> You can use verilator,Xilinx or any other softwares or you can use the https://www.edaplayground.com/ (which works online without having to install anything on your PC).
| If you are familiar with \$\LaTeX\$, you can use circuitikz to draw nice circuits by writing code.
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How is the 220V line in US different from other parts of world? Does a US 220V line need two switches for each wire? Today my A/C compressor had given up its ghost and when I tried to disconnect it from the wall switch after turning OFF the wall circuit-breaker,I assumed like most 110V lines where only one line (Live-L) lights with neon tester. When the circuit breaker was turned OFF the neon tester on A/C switch line turned OFF and I assumed the power was out. But while disconnecting I accidentally touched the other wire and I received a shock. On testing with neon tester I noticed the light was glowing for the other line also and I later discovered it had in fact two breakers(no handle tie) for the A/C circuit. When I checked on my normal 110V lines the neon test light glows only on one wire i.e., Live wire. Even most of my google searches showed a 220V line with just L and N. So,doesn't the US 220V have a neutral? <Q> US residential (and most commercial) uses split-phase electricity , therefore both wires of a 220V outlet are live; the neutral is used to provide two 110V connections, each 180 degrees out of phase. <A> Here in the US we have a split phase, that's true. <S> The "Neutral" is a center point of ONE phase of a transformer, so both ends are not "180 degrees out of phase", they are the same phase, just opposite ends. <S> I know, it's semantics, but it's important to be correct. <S> If you have a device that needs <S> 240V (220, 230, 240V is all nominally the same), the device generally will not care if that is derived as 1 phase and a neutral as you find elsewhere in the world, or 2 ends of the same phase as you find here, just so long as the voltage measured between the two lines reads 240V . <S> But here in the US, if you are using 240V, you are REQUIRED to have over Current Protective Devices (OCPDs, i.e. fuses or a circuit breaker) on EACH of the ungrounded conductors. <S> You can however CONTROL a 240V device by switching only one leg. <S> Many people get this confused and think that they can use a single pole breaker to feed a 240V device, because they see a single pole switch controlling it. <S> But that's an incorrect assumption. <S> 2 poles of protection, regardless of how it's controlled. <S> So your Bosch power tool is fine with just the 2 hot wires going to it, plus a safety ground (unless it is "double insulted, in which case it will have a 2 pin plug on it)). <S> GROUND <S> however is not the same as Neutral , even though they are usually at the same potential. <S> Neutral is considered a "Current Carrying Conductor" and must be insulated, Ground is a SAFETY conductor and must NOT carry any current unless these is an accident. <S> You cannot use the Ground wire as a Neutral connection. <S> People do it all the time, but every one of them is illegal... <A> Most of the world is supplied with 220-240V single-phase. <S> Neutral (and ground) are on one end, and the other end is 220-240V. <S> Neutral and ground are in the center, with 120V to either "hot" pole. <S> That means if you want 240V-only, in Europe only one of the two conductors is at a dangerous voltage, the other is near earth potential. <S> In North America, both conductors have a dangerous voltage.
| In North America, houses are supplied with 240V split-phase. If the device in question has no need for 120V inside of it, you do not need the Neutral conductor brought out to it.
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Parallel access to memory with multiple DDR4 chips Some development boards have multiple DDR4 chips. Does it imply that such boards have multiple DDR controllers and memory can be accessed in parallel. For example this boards: https://www.xilinx.com/products/boards-and-kits/kcu105.html#hardware has 4 DDR4 chips. Thanks <Q> Unlikely. <S> The spec for that shows 64bit wide chunk DDR4. <S> Each of those four chips will be 16bit wide, and all four wired in parallel gives 64bit. <S> From the specifications on the page you link to: Memory 2GB DDR4 component memory (four [256 Mb x 16] devices) at 1200MHz / 2400Mbpsps <S> Note the "four [256 <S> Mb x 16]" - four 16bit devices. <S> Then from the diagram it shows 64bit DDR4. <S> Thus they must be wired in parallel to get the full data width. <A> Yes and no. <S> The four chips (256Mb 16 bit) are wired in a way so that they look like one single 64 bit wide memory chip and the board uses a single memory controller for all four chips. <S> You can only access one address at a time, but you can transfer up to 64 bits of data per access. <S> You can find this info in the board's user guide , chapter "DDR4 Component Memory". <S> In the pin list for the memory chips, you'll find 64 data pins and only one set of address pins that's wired to all four memory chips. <A> Generally speaking, the number of chips does not imply the number of controllers. <S> As you say, the chips on board are 256Mx16, meaning they have a 24 bits address bus, 16 bits data in, 16 bits data out. <S> Plus of course all the signals necessary for a DRAM chip to work. <S> Now, if you want to use them in parallel, i.e. as a single, 256x(16x4) bank, you will need the data lines to arrive separately to the FPGA, and then on the FPGA you instantiate an appropriate controller. <S> If in parallel means that you want to be able to access the four banks independently, then also the address bus must be separated. <S> This adds up to 24x4 (address) + 16x4 <S> (data in) <S> + 16x4 <S> (data out) = <S> 224 pins. <S> It seems quite a lot, I suspect that the board designers did not allow for such a huge usage of the FPGA pins. <S> The answer to the question is then: if the signals are routed accordingly, you can instantiate up to four controllers, if you want and have room for it, and use the chips as you wish.
| In all likelihood (you can check with schematics), each of the four DDR4 chips will share the same physical address lines, thus preventing using them individually.
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Why use an opto-coupler? A side note: please be forgiving of me because I'm just starting out in Electronics so I'm still confused about certain terms. Why are optocouplers used? Yes I know they link a circuit together since there is an LED inside and a photoresistor. I also read that it prevents high voltages. But why use an optocoupler over another component? For example, to reduce high voltages we can use step-down transformers but why, in a certain scenario, we would use an optocoupler over a step-down transformer? In what scenarios are optocouplers appropriate? Hope my question is clear enough for anyone who can help. <Q> Both transformers and opto-couplers can provide isolation between 'hot' and 'safe' regions of a circuit. <S> The difference is that opto-couplers are very small, cheap, and can work on simple DC, so can shift a logic signal from one side to the other with no fuss and no or few other components. <S> Transformers are big, expensive, and need AC to work, they cannot simply be inserted into a logic line. <S> When you need the thing that a transformer does well, moving power from one side to the other, then you have to use a transformer. <S> Otherwise, you use something smaller, cheaper and easier. <A> The main reason for using an optocoupler is to realise a galvanic separation between circuits. <S> The use of a transformer much more expensive and only possible in AC circuits. <S> For safety reasons: separating mains connected electronics from the low voltage electronics section. <S> Low cost <S> : An optocoupler is very small and has a low cost. <A> Others have mentioned the isolation aspects of using an optocoupler. <S> One benefit from this is that you can switch different voltage busses that are otherwise unrelated to each other. <S> Some examples: <S> On a circuit board, you can switch 12V signals using your 3.3V microcontroller output. <S> Or the other way around . <S> In an industrial panel, you can run a power wire from an external system, switch it with an optocoupler, and return the voltage back to the external system. <S> This could help if you don't know the other system's working voltage. <S> It could also help if the external system doesn't share a common ground with yours. <A> I don't know why the other answers didn't mention the term noise . <S> Maybe I have a misconception about the opto-couplers. <S> -which <S> does the processing jobs- from its low frequency and power supply part <S> (and I know about the role of capacitors. <S> That's not the point here). <S> This separation/isolation is partly because high voltage results in higher amounts of noise which particularly affects A2Ds and measurement precision. <S> Again, I am talking from experience and don't have any source for my claims. <S> Please correct me if I am wrong.
| But from my experience in dealing with electronic circuits, one of the main reasons for using an opto-coupler is to separate the high-frequency part of the circuit
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How can we persist the leds blinking rather than swithching from one led to other while using a demultiplexer Demultiplexers are used for many outputs. In the case of led blinking, but its difficult to identify the leds that blinked. Since its output get switched between many leds. How can I digitally switch between many LEDS but have the output persist? Without switching between leds using a demultiplexer? <Q> ... <S> its difficult to identify the LEDs that blinked. <S> If you mean that it is difficult to read them <S> then no, it's not. <S> Multiplexing is the standard technique used for LED seven-segment displays. <S> Provided the blink rate is fast enough and intense enough the light will appear continuous to a (relatively) static observer. <S> Figure 1. <S> A typical 50 mA LED (1) can be pulsed at 200 mA (2) for short duration (3). <S> The note says 100 µs per 1000 µs. <S> Figure 2. <S> Timing diagram for high-current driving of the LED. <S> Using the timing of Figure 2 the flicker will not be visible to a human eye. <S> Figure 3. <S> Source: LEDnique . <S> For further reading see the link above to a short article I wrote on the subject. <A> Multiplexing is used to reduce the number of individual DIO <S> pins you need to turn on each LED and uses scanning frequencies above the human threshold of persistence. <S> There are other ways to reduce the number of DIO pins per LED: simulate this circuit – Schematic created using CircuitLab <S> Here you can have D1 on or D2 permanently on, both off ... <S> but not both on non-multiplexed. <S> Note ... <S> the combined Vf for the LEDS D1, D2 must exceed the VCC voltage or they would both conduct. <A> Persistence can be achieved by using a latch (or memory circuit), such as the 74HCT164. <S> There are also other latches that can be used, a good thing to search for is open drain outputs if you want a low side switch or open collector if you want a high side switch. <S> Source: <S> LED circuits
| For flicker-free displays a latch can be used. If you want an LED to be on all the time your logic says it should be on (so you could measure the voltage with a multimeter for example), then you need a bit (DIO pin) for every LED.
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Is it okay to connect a European 240V transformer to a US 240V outlet? I've an old Bosch German made automotive charger transformer with ratings 14V/9V, 220-250V @ 50/60Hz 132VA . Can I use it on the 220 to 240V outlet here in the US which uses a split - phase system as the transformer was intent to be used on Europe style 240V with a Live and Neutral system?Will there be any difference in its performance? Note: I'd initially asked this question in the comments section of " How is the 220V line in US different from other parts of world? Does a US 220V line need two switches for each wire? ". But has been removed by moderators and was asked to make separate thread so this info is seen by everyone.Thanks. <Q> Yes, you can connect that to a US 220-240 V outlet. <S> Just connect to the two hot lines and connect nothing to the neutral. <S> However, there are normally only 30 to 60 amp 240 volt outlets available in US residences. <S> They are dedicated to clothes dryer cooktop and oven use. <S> To add a 240 volt circuit requires two adjacent 120 volt circuit breaker spaces in the distribution box. <S> You will need a 15 or 20 amp double breaker whichever is the smallest available. <S> You can connect additional 120 volt circuits to the same breaker, but they must not be circuits like bathroom and kitchen outlets that can not be mixed with outlets elsewhere. <S> If you want to do that you should have smaller fuses near the outlet. <S> You probably will not find a suitable converter for a 240 volt US receptacle. <S> There could be a difference in performance related to the actual voltage in the previous vs. present locations. <S> However the transformer will be operating within the specified voltage and frequency range, so that is nothing to be concerned about. <S> If the new voltage is higher, the charging current may be a little higher and the charging time a little quicker and the opposite if the voltage is lower. <S> I assume an old charger may not have any voltage regulation. <S> The operating temperature may be lower or higher. <S> A 15 or 20 amp 240 volt receptacle installed for a window air conditioner is an excellent choice for this application. <A> The plug is unpolarized anyway, so the device has to be able to accept live voltage on any of the two pins. <S> The mains power switches in Schuko devices always switch both live and neutral for this reason. <S> You should just make sure to use an adapter with an earthing pin if the charger uses a Schuko plug. <S> Operating a Schuko device without earthing might be quite dangerous. <A> Sure, there's no problem. <S> You see common 120V receptacles all over houses in North America. <S> You can do exactly the same thing with 240V circuits . <S> The normal 120V plugs are called NEMA 5-15 plugs, and they go into either NEMA 5-15 or 5-20 sockets. <S> The Electrical Code says the breaker must be 15A or 20A, and the socket ampacity must match the breaker, except 5-15 sockets are OK on a 20A circuit if there's more than one socket. <S> The same exact rules apply to 240V circuits (no neutral) except you use NEMA 6-15 and 6-20. <S> The sockets are almost identical but are keyed not to fit NEMA 5. <S> You could take a 120V circuit and just convert it to 240V by changing all the sockets to NEMA 6 and landing its 2 conductors on a 240V breaker. <S> However, Code requires certain mandatory sockets in homes, and those mandatory ones must be 120V. Other than that, fit all the NEMA 6-15 or 6-20 sockets you desire. <S> It's no more difficult than installing a 120V circuit. <S> You cable it with the normal /2 <S> cable (twin and earth) and you tape the white wire to indicate it is a hot. <S> Now, one person talked about making a 240V circuit that serves both 120V and 240V loads. <S> That's allowed but is a little awkward. <S> It needs the more special /3 <S> cable with two hots and a neutral. <S> From the 120v circuit's perspective, it is a multi-wire branch circuit with some funny rules. <S> GFCI/AFCI, if needed, wont be any harder than on a straight 240V circuit, but will probably end up needing to be in the breaker panel.
| You should use a US receptacle and plug rated for 240 volts and the fuse or circuit breaker current rating. It is not a good idea to connect to a higher current circuit because the higher rated circuit breaker might not prevent a fire if you have a short-circuit in the charger or in the charger cord. I don't know if that is permitted by code. If the device has a Schuko plug or an Europlug , you should be able to use it on a split-phase system.
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Why are lithium ion cells mostly round? After watching some tear-down videos on YouTube with various lithium battery products (portable chargers, laptop battery, power tools) they all (apart from mobile phones / tablet battery) seem to feature cylindrical battery cells. Is there a technical reason for this, other than it's a uniform size that tiles relatively well to give manufacturers flexibility in incorporating the cells? <Q> 1) <S> The cylindrical shape is very easy to manufacture - You basically fold the layers around a round object. <S> 2) <S> For the same reason as drinks cans - this shape is something between the Sphere (high strength and damage resistant) and Cuboid (Easy to stack and low space is wasted). <S> It is the best trade-off between strength and volume. <S> 3) <S> This way You can easily stack the batteries - (Parallel and Series connections). <A> Among other reasons already stated, in 1991 Sony commercialized the first lithium-ion cell and later used it to power their 8mm camcorder. <S> They needed a way to make these new batteries quickly and on a huge scale. <S> At the same time CDs were killing tape sales (remember cassette tapes?). <S> The same equipment that coated magnetic tape with slurry could be used to make batteries in the same way. <S> So basically they had a bunch of equipment and factories that were slowing down <S> and it all kind of lined up. <S> That's why it started that way anyway. <S> That's a pretty well told story in the industry and if you google you'll probably find a few articles about it. <S> Everyone peddling some new version of the Lion battery or some new battery technology likes to lead with it to show you how old and outdated the batteries you're currently using are. <A> I'd say the opposite is the case. <S> Nearly all alkaline cells (and zinc carbon etc.) are round. <S> The square-ish packs are made up of round cells ( e.g. 9V ; the larger sizes such as lantern batteries often use C or D cells internally). <S> The main exceptions are lead-acid and lithium , such as phone batteries. <S> Many small rechargable devices like MP3 players and even my current laptop use these "prismatic" or "pouch" cells as well, which minimise the packaging bulk and weight. <S> Note that some round Li cell sizes are the same as the common alkalines <S> (14500 is the same as AA but ~3x the voltage which leads to useful tricks to free up space) <A> A cylinder (or even better, a sphere, but a spherical battery isn't a very convenient shape to use inside a device) is a much more efficient use of material for resisting internal pressure than a box with flat sides, in case the battery fails or overheats. <S> With a box shape, the sides can bulge outwards and then break off where they join at the edges. <S> A cylinder doesn't have any "edges" except where the end pieces join the cylindrical wall. <S> For the same volums, a square section container has about 27% more surface area than a cylindrical one - and that means 27% less "dead weight" and material cost. <A> The simplest reason there are a lot of round cells is the reason that tek states in his comment. <S> The batteries consist of two electrodes seperated by an electrolyte, one set gives the normal voltage. <S> To increase capacity and current capability, the electrodes and seperator are typically very thin and are wound up into a "jelly roll". <S> The easiest way to do this is simply wind it up into a coil and seal it in a can. <S> Hence the round cells. <S> Prismatic cells, the rectangular ones used in phones and similar electronics, are made in a similar way, they are simply flatter and oblong. <S> This is why a lot of prismatics will have a curved edge.
| Aside from the stress analysis considerations, the shape with the biggest ratio of enclosed area to circumference is also a circle - so even for an unpressurized container, a cylinder uses the least material to hold a given volume.
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Current Draw with Stepped Down Voltage If I have a 24V source, and then step down the voltage to 5V. Then at this 5V level I draw 1 Amp, what current draw will this be on the 24V source. My gut tells me it's 5V/V24 * 1 Amp mathematically but is this true in real life?Edit:DC/DC converter <Q> It depends on what you use to step down the voltage. <S> On the other hand, if you use a linear regulator, you’ll see a 1 amp (plus inefficiencies) draw on the 24 volts. <S> In either case, the losses will show up mostly as heat. <S> With the linear, the regulator will put out 19 watts of heat. <A> You can't break conservation of energy, and you have losses, so basically: Power <S> In = Power Out + losses <S> So, if power out = 5V * 1A = 5W, then: Power <S> In = 5W + losses <S> If a regular DC/DC converter has ~80% efficiency, then losses = <S> 5W <S> * ~0.2 = <S> ~1.25W <S> So, Power in = <S> ~6.25W <S> ~6.25W = <S> 24V <S> * I <S> → I = ~6.25W/24V = <S> ~260mA <S> In summary: 24V, 260mA in (6.25W) <S> 5V, 1A out (5W) <S> (~1.25W in losses) <A> That is true if you have a 100% efficient step-down converter. <S> Probably the best you can get is 85% efficient - and that's when running at rated load. <S> At a lighter load the efficiency will fall off. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> A linear regulator. <S> Your worst option would be to use a linear regulator. <S> It will act as resistance in the current path and waste 19/24 of the power. <S> Figure 2. <S> A plug-in 7805 switching regulator. <S> You should be able to find a switching regulator or ready-made board to do the job. <S> Have a look for 7805 switching regulator replacment and you should find something like that shown in Figure 2. <A> It's always best to go to a datasheet for these sort of questions. <S> Not all manufacturers list input current. <S> But from Traco Power DC/DC Datasheet : Input Specifications <S> Maximum input current (at Vin min. <S> and 1 A output current) 1 <S> A Model <S> : TSR 1-2450 Input voltage range: <S> 6.5 – 36 VDC Output voltage: 5.0 VDC Output <S> current max: <S> 1.0A Efficiency typ. <S> @Vin min. <S> : 94% Efficiency typ. <S> @Vin max. <S> : 84% <S> So at Vin = 6.5V, efficiency is 94% when Iout = <S> 1.0A <S> then Iin = 1.0A. <S> It's probably slightly larger. <S> As Vin increases with Vout and Iout constant, efficiency decreases and Iin <S> > Iout. <S> Most convertors have efficiencies between 75% and 98%. <S> In your case, Vin = 24V and Vout = 5V, so you'd be towards the bottom. <S> I'd guess Iin < 1.25A for Iout = <S> 1A.
| If you use a switching regulator, you are mostly correct, less inefficiencies in the regulator.
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5A connector that fits through 1/4" hole? I'm powering a load that draws 5A at 18V from a brick power supply. The power supply and load are separated by a thin metal wall that has 1/4" (6.35mm) holes in it. So I'm trying to find connectors that can fit through 0.25" diameter holes. The best I've found is small JST plugs , but they're more like 0.29" so we'd have to file them down to fit through the holes. I'd appreciate any suggestions, either for a 2-circuit connector or two 1-circuit connectors. <Q> Why not just put the wire through with a grommet (you could even use a gland) and then put a connector on one or both ends? <A> A bit old fashioned by modern standards but 2mm plugs and sockets would do the job. <S> This example is a line socket that is rated at 10A and measures 5.7mm in diameter, plugs would probably be thinner. <S> The only snag is lack of polarisation <S> so you'd have to rely on colour coding. <A> Two options (both crimping, so requiring specialized tools but not plugging in a soldering iron) that come to mind: Wire ferrules (also known as end sleeves or bootlace ferrules) can be crimped on the ends of stranded wire and not be very much thicker than the insulation. <S> They would be inserted into screw terminals (the kind with a hole and a screw-driven clamp, not the kind where you wrap the wire around the screw). <S> image by Wikimedia Commons user Simon A. Eugster <S> The advantage of these over bare (stranded) wire is that they do not fray <S> and they are easier to insert and clamp securely in a screw terminal. <S> They do ideally use a specialized crimping tool to form the high-grip pattern of indentations you can see in this picture, though anything might do well enough with a good screw terminal. <S> Anderson PowerPole connectors (most famous for being popular in amateur radio applications) consist of any number of crimped terminals inserted into plastic housings. <S> The crimped pins can also be removed from the housing again, though with some fatigue to the metal. <S> image by Wikimedia Commons user Cqdx <S> Crimping these terminals requires a specific crimping tool, and there is a specific removal tool, but the insertion can be done by hand when the wire is of heavy gauge, or with the aid of a small screwdriver or similar blunt-corner-ended tool otherwise.
| The terminals themselves are very narrow, so if you want a proper connector that is easy to connect and disconnect but is semi-permanently installed through the hole, all you need to do is crimp on the terminals to your 2-conductor cable, insert that through the hole, and then add the housing.
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I painted myself in a corner. How to layout? I'm a hobbyist working mostly in DC projects. For development and testing, I use breadboards, and later a perforated board. My projects are normally low count: an Arduino Nano/Pro Mini + a few ICs. All my works are one-of-a-kind, and making a PCB is out of question for me. I use the ICs and physical board itself to test the layout of components on board. I have Fritzing, but it's not up to the task for planning the physical layout; it's more like an ideal vision of the circuit where wires have no thickness and every thing fits neatly in the matrix. My problem is that sometimes I paint myself in a corner: after soldering a few things, I found the next pin buried under other wires and components; no way to solder anything. I'd been searching for tips, found none. How is your flow work? How do you do these things? Here some questions: Solder all component first and wire them later? Wire each component directly to Vcc, Gnd and shared signal or use rails ? How to layout for changes? Maybe I need to add something in the future. <Q> Your suggestions are spot on. <S> Solder your components first, but think about what you could use as a wire, e.g. a resistor can be used as a jumper. <S> ICs are the main thing to put in first, as they are set and will have a lot of dense connections. <S> Try and work in a matrix, with all wires/components running along rows and columns. <S> Rails are best - you'll have lots of things going up to V+ and down to ground so fit around that. <S> Leave yourself space in the prototyping phase. <S> You can scale down later if need be, but once you've started small <S> it's hard make it larger. <S> A bit of planning will go a long way :) <S> You could also try some simple stripboard layout software. <S> A number of them are paid (and surprisingly expensive), but PEBBLE is free and is a good starting point at least. <A> Even though you are using perf-board to build your projects, there is nothing stopping you from using schematic capture and PCB layout software to lay out the placement of your components. <S> Doing this would benefit you in several ways: You would learn a useful skill doing this. <S> The software can generate and check your netlist to verify you've actually accounted for each connection. <S> While performing the layout, you may discover that you don't have enough space between parts to route, or that the parts need to be moved. <S> It is easier to move things around in the layout software than it is on a perf board. <S> I've had good success with the following method: Match the PCB layout on a grid spacing that matches your perf board. <S> This is key, as it allows you to verify on the computer that you will physically be able to place and wire to each part. <S> Keeping on the grid pattern will keep your layout neat and easier to follow later. <S> Route your traces as needed, keeping in mind that tracks on the bottom layer will be implemented as solder bridges/buss wire and tracks on the top layer(s) will be run as insulated jumper wire. <S> If you follow the grid and make an effort to avoid running multiple jumpers on top of each other, any needed troubleshooting will be much simpler. <S> Print <S> both sides of the PCB on a 1:1 scale, with the bottom layer as a mirror image. <S> On the top layer print only the component pads and silkscreen and any top-layer traces. <S> Place the printout of top layer on the top (non-copper) side of the perf board, aligning with the grid of the board. <S> Place components through the paper and solder them into place on the bottom side of the board. <S> Any jumper wires needed to implement top layer traces are installed next, soldering them into place. <S> Flip the board over to run any bottom layer traces. <S> Use the mirror-image bottom print as reference to locate and solder those. <A> As @mguima and @spuck said, use PCB software to do your layouts. <S> They will help you a lot to figure things out. <S> Also, don't be afraid to bridge your perfboard contacts to make tracings. <S> This might be a pain for longer stretches, so you can still use a wire, but just fine for shorter runs and helps avoid weaving wires over your components or the pads you need to get to for your crucial/longer wiring. <S> (I'm not assuming that you already do it, but I hope whoever taught you how to do electronics also taught you this "trick".) <S> (If you don't know where the nearest maker space is, there are online maps to research them.) <S> Some of them have the capability to build PCBs using CNC machines, laser cutters, and other techniques to even make them near professional quality. <S> Quite often, you'll find other hobby level or, possibly, professional level electronics people that can help you even further. <S> As far as the actual professional services go, several of them can take a file online, then simply ship you the board. <S> Some can even pre-populate the board for you. <S> I was going to suggest Seeed Studio, but it looks like their min quantity for most PCB orders is 5. <S> You say making your own board is out of the question, but CNC machines that have accuracy good enough to make PCBs can be as small as about 12" cubed. <S> They can cut the traces as well as the holes, then you don't even need the smelly acids. <S> I've seen them on Amazon and eBay for less than $500. <S> If you don't care too much about tiny traces, even a less than $300 CNC machine will work. <S> Noise might be a factor, though, since they can be about as noisy as a drill, Dremel, or other small rotary tool. <S> Some of them even use a Dremel as the cutting head. <S> Depending on the software that comes with the CNC machine. <S> you might be able to use your layout file directly in the CNC machine. <S> Depending on your layout software, you should still be able to at least export the file into something the CNC machine can use.
| If you have PCB software for layouts, you might be able to take the file to your local maker space to make a legit custom board. Using different colors for different top layers, maybe to match the colors of the jumper wire you have, would be helpful.
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Are AVR chips open source? I am working on a new product, hoping to start up my company with it. I'm an electronics noob. I want to know whether AVR microcontrollers are open source, and if I use them, should my product be open source as well? Because Arduino also uses an AVR chip and its schematics are open. <Q> No, AVR chips are definitely NOT open-source. <S> However, that doesn't imply anything at all about whether a product you build using them can or cannot be open-source. <S> Arduino is open-source, but there are thousands, if not millions of commercial products that use the same chips that are completely proprietary. <A> AVR is proprietary. <S> It can be programmed using open-source tools like avr-gcc and avrdude (both work great) but using these tools does not mean you have to opensource your code. <S> arduino also uses an avr chip and its schematics are open. <S> This is a choice made by the arduino creators, you don't have to do the same choice. <S> What would force your code to be open-source would be using code that is under a licence like GPL. <S> So please check the licensing requirements of code and libraries you use. <A> "should my product be open source as well because arduino also uses an avr chip" <S> No. <S> Unless you use designs or codes, in part or in whole, that are covered by a license demanding so. <S> If you use your own design, and don't steal anyone's licensed code that prohibits use, then you own it <S> and you can do as you please. <S> There are always lawyer-ly details. <S> If you're concerned about that type of thing, then hire a lawyer, and don't trust strangers on the internet. <A> You are allowed to distribute your commercial product using their code by Atmel's license. <S> No open sourcing is required. <S> Arduino code is open sourced, but using open source does not always compel you to open source your product. <S> There are licenses which do this (they're nicknamed "copyleft" licenses), But the Arduino does not use these licenses. <S> From their FAQ : Open-source hardware shares much of the principles and approach of free and open-source software. <S> In particular, we believe that people should be able to study our hardware to understand how it works, make changes to it, and share those changes. <S> To facilitate this, we release all of the original design files (Eagle CAD) for the Arduino hardware. <S> These files are licensed under a Creative Commons Attribution Share-Alike license, which allows for both personal and commercial derivative works , as long as they credit Arduino and release their designs under the same license. <S> The Arduino software is also open-source. <S> The source code for the Java environment is released under the GPL and the C/C++ microcontroller libraries are under the LGPL. <S> (emphasis mine) <S> They state explicitly that their original design files are licensed in a way that permits commercial works. <S> This means you could take their schematics, modify them, and produce a commercial work from them. <S> However, in your case, you are almost certainly not selling schematics. <S> You're using existing AVR chips made by a foundry. <S> This part of the license won't be important for you. <S> What will be important is that the C/C++ libraries are LGPL-ed. <S> The LGPL license is written explicitly to permit you to use these libraries inside a commercial closed-source application. <S> So thus you can be confident that you do not need to open source your product.
| AVR's code is proprietary, but are licensed for your use by Atmel.
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SMPS output voltage too low - what to look for? I have an old switching power supply for HP 712/60 workstation (+5V, +3V3, +12v). It was working about ten years ago when i've last tried to boot the system, but now one of the output voltages is at half of the nominal value (12V output is at 5.6V). I have replaced all of the electrolytic capacitors in the SMPS, and resoldered all points on low voltage section - still 12V output is at 5.6V. What to look for next? General question - SMPS has output voltage lower than nominal - what troubleshooting strategy should i take to find the culprit? <Q> Chapter 7 of this document MAY have useful information: https://parisc.wiki.kernel.org/images-parisc/a/a8/712_technical_manual.pdf <S> I can't read German well enough to tell if this document contains useful information: <S> https://computermuseum.wordpress.com/2006/05/09/model_712_60/ <S> This website MIGHT offer some replacement parts: <S> https://www.partsit.com/product/hp-712-60 <S> This website offers some manuals: https://www.manualslib.com/products/Hp-Model-712-60-Workstation-3127355.html <A> Apart from the obvoius things like checking electrolytics and for dry solder joints and for broken windings in the transformer, you should also check your feedback circuit. <A> One of the most important troubleshooting procedures I do, is to measure the resistance between output rails to the ground. <S> If the resistance is too low, then one of semiconductor parts on the secondary side; e.g. diodes or MOSFETs might be broken. <S> Otherwise, which I'm not so practical with is the capacitors also might be faulty. <S> Good luck, <A> A simple answer is: measure as much as you can :-) <S> If you have a clamp amp-meter and the PSU output wires are exposed enough, check the currents out the 12V and 5V rail. <S> What about power consumption at the primary side? <S> Is it possible to disconnect the PSU from the motherboard?If yes <S> , can you start the PSU disconnected from <S> the mobo?If yes, can you still see the disbalance in voltages, 5V vs. 12V? <S> There's a single feedback loop, sensing the two output rails via a simple resistive divider (hence the "average" regulation). <S> I don't have a schematic of your particular PSU, but take a look at R25/R26 in the schematic here , near the label saying "FEEDBACK". <S> I would expect the feedback to pass through an opto-coupler, maybe preceded by a TL431 reference (also working as a comparator) <S> but in general outlines it's a fairly basic PC PSU arrangement. <S> The disbalance that you observe likely means that the +12V rail is heavily loaded and the +5V rail is not loaded at all. <S> Look for a short to ground in the 12V section. <S> Hopefully it's not in a secondary winding in the main transformer. <S> Or it could indeed be a rectifier diode, conducting both ways. <S> Some do this only under RF load (not if you check the diode alone with a tester). <S> If there was indeed a short, I'd expect something to smell and burn. <S> If there's no "Ampere odeur", that's weird... <S> BTW, if your 12V rail is actually lower than your 5V rail, it might be a clue that each rail has a dedicated secondary winding. <S> I.e. the windings are not connected in series. <S> Hmm... have you tried measuring voltage between the 5V and 12V outputs? <S> Could there be a short between those two rails? <S> After rectifiers, or before rectification? <S> What if the secondary windings are in series, and the section "between 5V and 12V" is shorted? <S> And the difference after rectification could merely reflect a difference in load (in the diodes' forward voltages). <S> Are you saying that you've replaced all the elyts? <S> With what family / model? <S> Do you have an oscilloscope? <S> That's a key tool when debugging SMPS... <S> Unfortunately the resistance in secondary windings is so small that checking it with an ohm-meter yields <S> no useful data :-(
| If something is wrong with the resistor voltage divider or optocoupler in the feedback circuit, this could have the effect of a lower output voltage than expected.
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Does an ideal wire dissipate energy? In this circuit, I calculate the current through R1 to be 10mA. What happens if I take into account the electromagnetic radiation in the ideal wire? simulate this circuit – Schematic created using CircuitLab edit: This is a pedagogical question, trying to establish that no, an ideal wire does not dissipate energy because that's not a useful definition of "ideal". <Q> If you accounted for the wires your circuit would look like the diagram below. <S> It would have inductance from the magnetic field around the wire and resistance from the conductor. <S> The magnetic field effects are represented by the inductor and only apply if the current through the inductor is changing. <S> Which in this circuit, once it is turned on the current is constant and the inductance does not matter. <S> If you had an ideal wire with electromagnetic effects, you could nix the resistors and just model the inductance. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Well, it depends on the definition you give of "ideal wire" and the exact context you are involved into. <S> Usually in lumped element circuit theory (CT), a connection between two terminals is considered a piece of ideal conductor of zero length . <S> An ideal conductor is a conductor with zero resistivity, hence that connection can be considered like a zero ohm resistor. <S> In basic lumped element CT we assume the EM fields vary very slowly, so slowly to be almost stationary (i.e. constant with time). <S> "Almost" here means that we can neglect all the terms of Maxwell's equations that involve time variations. <S> This also implies that any element in the circuit has physical dimensions that are much less than the wavelength of any signal component in the circuit. <S> Moreover, it also implies that the whole circuit is much smaller than that wavelength (because we neglect the propagation delay of the signals in the circuit due to the finiteness of speed of light). <S> In other words, in basic CT we assume that signals travel in the circuits with no delay other than those introduced by the lumped elements. <S> That assumption, together with the fact that those connections have zero length (well, to say things more precisely <S> , they have dimensions much smaller than the lumped elements they connect), also imply that there are no parasitic effect. <S> All this is a really drastic simplification. <S> When you begin to relax your definition of "ideal", e.g. you assume connections are made with a conductor having some finite dimensions and non zero resistivity, you get what other have already said in this thread: parasitics. <S> In particular, some residual series resistance (due to non-zero resistivity of the material. <S> Conductors having length also give you some stray inductance, because of them forming loops in your circuit and being coupled magnetically. <S> Their proximity makes them develop stray capacitance between different parts of the circuit. <S> Moreover, if you begin to take into account also the dielectric that separates the wires (e.g. FR4 circuit board substrate) <S> this both affects stray capacitance and introduces some parallel (leakage) resistance. <S> Another problem with "ideal wires" having non-zero physical dimensions is that the resistance of the conductor varies with frequency (no, I'm not talking about the equivalent impedance or its real part, just the raw resistance of the wire) because of the skin effect and the proximity effect (ouch!). <A> All the radiation emitted by the power supply will travel parallel to the wires, then dive into the resistor, bending its path so it strikes the resistor surface at 90deg. <S> For zero-ohm wires and any low frequency, the electrical energy travels like this: <S> A 2-plate battery with electrolyte on the left, is powering a grey resistive element on the right. <S> The vector-field, showing the power-density and direction, is always perpendicular to both the e-field (shown as grey lines) and the b-field (not shown.) <S> Wires don't leak any radiation unless they're hundreds of miles long, or unless the power supply is RF rather than DC. <S> The model for radiation-leakage is the model for empty space: a vast array of capacitors and inductors chosen to produce c-velocity wave propagation. <S> The array is connected at many points to your two wires. <S> With pf and nH array elements, there won't be much leakage until you get far above 1MHz, or build an array hundreds of KM long. <S> Note that the e-field and b-field of this simple circuit actually looks more like the one below. <S> Fill in all the planes, including distortions caused by the elbow-bends and the battery and resistor. <S> Then sketch in the poynting-vector field perpendicular to those planes, so the energy-flow vectors connect all the junctions where the flux-lines of e cross the b. <S> Also, more stuff.
| With ideal wires, none is absorbed, but also none escapes.
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Do I need to solder every header pin to the board? When soldering MCU (Arduino and like) to perforated board I apply solder only to the pins in use, leaving the unused ones just sitting in the hole. I do this to minimize soldering accidents. The headers themselves had all their pins soldered to the chip. It's this a good practice? <Q> No, it's not good practice. <S> For this kind of task, I'd expect very little soldering mistakes to happen, so go ahead and do it. <S> Furthermore, this is confusion waiting to happen if you decide later on that you want to use a pin <S> you haven't soldered yet. <A> That's fine for prototyping, the only downside is the reduced mechanical strength. <A> You can do it, but you have to take mechanical resistance into account. <S> What I do is solder all the pins I use, plus pins on both ends. <S> on a 2.54mm pitch (like the arduino headers) <S> soldering accident should not happen (assuming decent equipment and careful operator), but not soldering every single pin makes rework/desoldering easier. <S> Since an Arduino is prototyping work, mechanical resistance does not have to be perfect anyway, and you can always solder the rest once the project is finished.
| Obviously, the header will be better fixated mechanically if you solder all pins.
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Pull-up vs Pull-down for contact switch? I am designing a simple contact switch(tactile, which conducts when you pushes, and springs back to not-conducting when released) interface with a microcontroller. It's GPIO has option for both internal pull-up and pull-down. 1) Which one should I use? For 2 layered board, pull-up makes me easier to do PCB artwork, since one-end of switch connects to GND. Other than that, is there any difference or reason for me to use one configuration over another? 2) Is there any possiblity that the noise from physical contact make MCU malfunction? <Q> 1) <S> Which one should I use? <S> Whatever suits your overall design best 1) <S> For 2 layered board, pull-up makes me easier to do PCB artwork, since one-end of switch connects to GND. <S> That is exactly the reason pull-ups are used more often than pull-downs (and <S> some chips only have pull-ups): a ground line is often more conveniently available for the other side of the switch (and open-colledror/drain outputs are more common than their top-side counterparts). <S> 1) Other than that, is there any difference or reason for me to use one configuration over another? <S> 2) Is there any possiblity that the noise from physical contact make MCU malfunction? <S> I would guess that (in a reasonably well designed system) the power is more noisy that the ground, so an active-high switch would potenetially induce more noise. <S> But even that should be no problem on a digital input with sufficient margin. <S> So by all means, go for pull-up and an active-low switch. <A> Standard practice is to enable internal pull-up of ~100k with a debounce cap across switch such that RC= <S> T > <S> bounce time or use software debounce. <S> Depending of ESD risk the contact may also have a series current limiting R to allow the IC diodes rated for 5mA to do their job or better , add Transil or TVS protection. <A> Either will work, a pull up with the switch shorting to ground, or a pull down with the switch shorting to Vcc. <S> Noise from switch bounce shouldn't upset your microcontroller, especially if you've followed the manufacturer's recommended decoupling scheme.
| If layout is easier for you using a pull up then that's fine.
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Why are old digital ICs families still so "popular"? Why is it so? The basic 74 series, the 4000, are such poor ICs compared to the newest families (by the 74: AUC, AUP, ALVC...), which only have advantages over the oldest one: lower power consumption, lower propagation delay... <Q> Why do you think TTL 74/54 series are popular? <S> They are a relic of the past now (unless you have to maintain old hardware). <S> Metal gate CMOS <S> 4000 series is also somewhat obsolete, but it has some advantage in some specialized digital application where its wide supply range (up to 15V at least) is convenient. <S> Of course it's slow and power hungry, compared to newer CMOS families. <S> 74HC/HCT is mainstream because it is mature and stable (but it's losing terrain). <S> The other, faster, families are used only if you really need their speed. <S> A faster digital device is not necessarily an advantage in a circuit that doesn't need that speed. <S> If you are a student, I suspect that your intuition is skewed because many high-school labs (and even university labs) purchased tons of TTL (and CMOS 4000) chips and many teachers love beating a dead horse (sadly). <A> 5 V logic was "standard" for many years, and there are still many systems and devices out there that need 5 V levels for their interfaces. <S> If you're working with one of them, you need a logic family compatible with 5 V. <S> Maybe not 7400 or 4000 series, but something like ALS or ACT. <S> Some systems even need higher voltage levels, such as 12 V. <S> In these cases you might still want to use 4000 series logic. <A> Cost, availability, multiple manufacturers, simplicity.. are all factors, but I'd argue they are not really that popular any more for new designs. <S> They are still readily available though because they are still used everywhere in older equipment that is still manufactured. <S> But even today, sometimes it's just quicker and cheaper to plop on a simple NAND gate into a circuit rather than the alternative. <S> There is something to be said for the basic families though. <S> They are after all foundational to all the higher order parts. <S> You need a decent knowledge of them to really understand the rest so they definitely have their place. <A> Those "fancy" SSI/MSI technologies were developed at the tail end of the heyday of the wide use of SSI/MSI in high-volume commercial products, and were never manufactured in any great volumes. <S> They were not generally incorporated into new equipment designs at the time, because of proprietary and single-source issues. <S> Microcontrollers, CPLDs, FPGAs and ASICs are the mainstream technologies used commercially today. <S> Only hobbyists are still using SSI/MSI for design, and it is only the older technologies that were manufactured in high volumes (and are still manufactured for older equipment that is still being manufactured) that are generally available to them. <S> There are a few niche products that are used in modern design, such as the "single gate" logic that's available in SOT-23 packages, but again, because of the packaging, these are not popular with hobbyists.
| Older families also have the advantage of being low cost and multi-sourced.
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How to detect if a nearby parked car has started engine? We have a project with a machine servicing a car parked next to it. It's comparable to a car washning machine. Before the machine is started, we need to be 100% sure that the engine is off. Also, if we detect that the engine is turned on during the process (which takes severa minutes), the machine should withdraw its operation immediall, to avoid damages. The machine is supposed to refuel the car without any manual action by the driver. Electrical cars are not a part of our current plan. But it certainly is relevant that some cars stop the engine when the brake is down. Our problem is to detect whether the car parked at the machine is running or not. We have already been thinking of these solutions: Microphone listening for engine sound. Problem: In a noisy environment, we might have too many false alerts, when a big truck is parking close nearby. IR camera looking for hot exhaust gas. Problem: We have tried that and can't see any significant changes whether the engine is on or not Camera detecting whether the engine lights are on. Problem: Not all cars have lights on when engine is running. Not all cars require the driver to push the brake to start the engine. What other options do you see for us? <Q> This sounds like an XY problem. <S> Your problem is you need to prevent damage to your equipment. <S> You are myopically focused on detecting when the vehicle begins moving. <S> Instead, I propose you prevent the vehicle from moving. <S> An electric or hydraulic bollard is cheap ($2000-3000) and designed from both a mechanical and visual perspective that a car needs to stay put. <S> Give the user a flashy screen, like a car wash. <S> Show them the status, gallons filled, estimated time remaining. <S> Retract the bollard after your robot's sensitive bits have tucked themselves away, then change the screen to a big friendly "you're done!". <A> Most Petrol Cars can be heard on a sensitive AM radio .Most <S> of the electrical noise comes from the spark plugs .You <S> can try using a regular MW AM radio to start experimenting with .If <S> you are close to the car like feet away <S> you should hear something .Better <S> would be a SW radio if you can get your hands on one .It <S> has been said that car ignition noise has a very broad peak around 20 to 40 MHz .If <S> you can hear something from your experiments then you could process the Audio with a DSP .Electric <S> cars do make engine noise too .I <S> can hear my Nissan leaf on my experimental MW AM car radios when I use an external test Antenna and plug them into the Standard 12V Cig lighter plug .The sensitivity of these test car radios is good but not excellent <S> .This means that you wont need some really expensive comms radio to achieve this . <S> I was told in 1974 by my father who was a Civil Engineer who worked for the city council that the traffic light signalling in Dunedin where I grew up detected cars by antenna wires buried under the road .Diesel cars were rare then but large trucks were diesel .I <S> do not know how well thier system worked . <A> Definitely sounds like an XY problem, you don't need to detect whether the engine is running, you need the car not to move. <S> Bryan's idea made me think of one that I think is even better, as there is no risk of damage to the car, if someone were to run into the bollard when it is up. <S> Mount rollers in the ground (parallel to the direction of travel) that are normally locked (preventing them from rolling), allowing them to be driven on. <S> When your apparatus is engaged, unlock the rollers, thus preventing the car from driving away even if the engine is started and the driver attempts to drive. <S> When done, and the device is retracted, lock the rollers again. <A> Can be done invisibly with IR lasers. <S> All you need to do is point it at a piece of bodywork and look for vibrations in the range of engine RPM with suitable magnitude. <S> You'll need to calibrate it so the vibrations of passengers moving around or loud sound systems don't trigger it. <S> You can also detect vibration at a distance with doppler ultrasound, but I think this is harder. <S> http://ieeexplore.ieee.org/document/1293292/ <S> With electric cars I think this may be impossible; but there may also be ways of wirelessly contacting the car's control electronics to monitor this. <S> Note that existing fuelling stations handle the risk of people driving off with the hose attached - <S> if this does happen it breaks off in a safe-ish manner and you charge the driver's insurance. <A> Exhaust pipe detection requires a Flir or IR THERMAL camera, not an IR security camera. <S> Thermal cameras are far more pricey, use exotic lenses (germanium, chalcogenide, etc.) <S> and they see IR longwave, not the shorter near-visible wavelengths. <S> There are even thermal cameras with narrow filters tuned to CO2, and can see the gas itself rather than the high-temp emissions. <S> " Gas FindIR " product from Flir corp. <S> Experiment with those $12 <S> Harbor Freight keychain thermal IR detectors. <A> OBD <S> You should be able to find ready to use hardware and possibly software libraries too, or you could write your own. <S> Only when you have established that Vehicle engine is dead, should you start remainder of service process. <S> OBD will also allow you to sense if/when engine comes to life. <S> And it will cover a vast majority of current vehicles. <A> I would go with a nitrogen and carbon dioxide detector on the exhaust pipe. <S> The gas will start coming out the pipe the moment the engine starts. <S> Don't count on heat as it might take a couple of second for the whole area to heat up especially on a cold day, same goes for the hat of the gas.
| Your rabbit-hole solution is to assume that you need to retract your equipment when the vehicle begins to drive. If you have a system that finds the fuel opening you wont have any difficulties finding the exhaust pipe. Laser "microphone" vibration detection. Make it a prerequisite to have your OBD hardware plugged into your Vehicle. Cheaper might be a handful of IR thermometers, pointed to various locations where you'd expect to find the exhaust outlet.
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Can I place a relay at the output pin of a 555-timer I recently remember running across a bunch of articles on web about the 555-timer, it’s my first IC. I’m getting more familiar with it, and I want to know run it in a-stable mode, with a relay between pin-3(Output) and ground. I will be using the relay to switch between two different LED arrays. What concerns me is whether placing a relay directly between pin-3 and ground would cause a high current situation in the 555-timer. Would it be necessary, and or possible, for me to put a resistor in series with the coil to limit current? I’m just now getting experience with relays to, I know that they require a certain applied voltage to magnetize the coil to a point where it can effectively move the armature. I’m just not sure how this relates to KVL? At the high state of pin-3 12V would be applied to the coil, I’m not quite sure, but I believe without a resistor, that would probably mean a lot of current would be sinked to pin-3. If I were to put a resistor in series with the coil of the relay, At high of pin-3, would the coil maintain the required voltage across it? <Q> To elaborate on Peter Bennett's answer, here is the standard relay driver circuit I use. <S> It removes the requirement that the 555 can supply the full current of the relay. <S> simulate this circuit – <S> Schematic created using CircuitLab Notes: <S> Relay power doesn't have to be the same power supply as the rest of the logic, but it can be if that is adequate voltage for the relay. <S> Ensure the relay power supply voltage is adequate to close the relay. <S> Select Q1 for adequate relay drive current. <S> For large relays, the 3904 may not be appropriate. <S> Also ensure that the device will saturate. <S> A darlington connection may be required for very large currents. <S> D1 is a flyback diode required to protect Q1 from voltage spikes caused by the relay coil. <A> If you are running the 555 on 12 volts, you should use a 12 volt relay, and ensure that it does not require more current than the 555 can supply. <S> You might use a resistor in series with the relay coil if the relay requires a lower voltage than the 555 output provides. <A> Relay coils have internal resistance so if you know this number, can measure it, or can infer it from the relay data sheet, you can then consult the 555 datasheet of the 555 variant you have chosen to determine if the current the relay needs will be available from the 555 output. <S> NE555 can source 100mA from the output pin, so a1 at 12V is should be capable of driving a relay with a resistance of 120 ohms or higher. <S> this would include most 12V "sugar cube" relays and common automotive horn and lighting relays. <S> that said the the output can sink 200mA, so if the 555 logic can be inverted connecting the relay between 12V and output gives stronger drive. <S> UA555 can source only 10mA bunt can sink 100mA, so can still drive a relay connected to +12 but is too weak to directly drive a ground-connected relay
| If the 555 cannot supply enough current to operate the relay, or if you must operate the relay and 555 from different voltages, you may require an additional transistor to drive the relay.
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How can I distinguish between LTE and WiFi antennas? I've a LTE router with WiFi, it switches automatically between LTE bands 3, 5 and 40, depending on the respective signal quality. I want to improve the band 40 reception. So I think the easiest option is to remove the band 3 and 5 antennas. As the propitiatory firmware provides no interface to switch or lock a specific band. The main problem is identifying the correct antennas to remove. The device has 4 patch antennas, printed on yellow mylar films. All of them looks similar to me! So how can I identify the band 3/5 antennas ? The WiFi chip is RTL8192EU , which is capable of 2T2R MIMO, so is it possible there's 2 antenna for WiFi? I can provide internal pictures if needed. Device is made by Pegasus Telecom, subsidiary of Haier. Pictures, main board. Antenna on top Antanna on left and right , seems like inverted F monopole. Antenna on bottom <Q> Based on the wavelength-to-antenna ratio, it can be assumed that the largest antenna is for the 5 band (850 MHz), the left antenna is for wifi (2,400 MHz), right is for the 40 band (2300 MHz) and the lower is for band 3 (1800 <S> MHz).UPD: as written on the antenna, abbreviation RJ-PRA and RJ-DIV mean that the antenna is intended for mobile communication (main antennas). <S> And with the reduction of RJ-WFA (maybe left right side antennas, you need to check) and RJ-WFB, it's a wifi antenna. <S> (here is the info about antennas https://www.eximpulse.com/import-product-I-phone-6.htm?tpages=35585&page=3596 ) <A> The LTE Band 40 is 2300 and Band 3, 5 at 1800, 850. <S> This means that the shortest features on the LTE antenna/s will be what you want to keep. <S> You may be able to cut the longer 850 MHz dipole parts off without much affecting the 1800/2300 sections if they are on a single foil section. <S> One LTE antenna may share band 40 with one of the longer antennas. <S> There are likely two WiFi antennas for diversity, they are also likely dual band at 2400 and 5000 MHz on each antenna <S> so these are more likely the two smaller side antennas. <S> To maintain a similar band coverage a wider (like the flat conductors they have used) or larger diameter element (screen connection of a thin co-ax cable). <S> Trimming it to length without network analyser gear will be challenging. <S> Here is an interesting paper with a lot of theory and some history on LTE antennas. <A> You really cannot distinguish them most of the time as many antennas can be use for both LTE and 2.4 GHz Wlan, and they look similar. <S> The 5 GHz wlan antenna could be shorter. <S> If you look at the datasheet, some LTE and Wlan antenna has similar frequency response in the 800 - 3000 MHz band. <S> That is because a lot of antenna vendor choose to design it such that they can be use for cellular and wlan.
| You may be able to replace the LTE antennas with a simple 1/4 wavelength element selected for 2300 MHz.
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Should I check battery current by ammeter? I had the concept that in order to check the maximum current a battery can supply, it is fine to connect an ammeter in series with battery because ammeter has low resistance in series and this will yield the maximum current a battery can supply. Many people have said it is wrong, but I can't understand why. <Q> "Many people" are correct in this case. <S> The problem is that an ammeter has a very low internal resistance. <S> (It is designed to measure the current with minimal loading effect on the load.) <S> If you connect it across the terminals of a battery a large current will flow, limited only by the internal resistance of the battery and the meter - both of which will be low. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> The wrong way and the right way. <A> If you put a voltmeter in parallel with the ammeter you'll see that the voltage is very depressed. <S> What you need to do in order to measure maximum current available is to measure it through a variable resistor while also measuring voltage across the battery. <S> The resistor should be adjusted to the point where the voltage is at the desired minimum operating voltage and then the current read from the ammeter. <A> Drawing short-circuit current from a battery can damage or even explode the battery even if it is done for just a few seconds. <S> The only way to know the current that a battery can handle safely is to find the information in the manufacturer's specifications for the battery. <S> The reason for that is the internal battery parts have resistance. <S> That resistance causes internal power loss in the battery that heats the components. <S> The components can deform or melt if the temperature is too high. <S> Designing those components to carry a certain maximum current is part of designing the battery. <S> Testing the voltage drop at various current levels can provide a good indication of safe operating current, but only the manufacturer can determine that with a reasonable amount of certainty. <A> When you connect an ammeter into an automotive circuit there will be a current draw well in excess of 50amps which will blow up multimeters. <S> This is due to all of the modules on the vehicle starting up! <A> Testing a battery's current supply capability by shorting it with an ammeter is a very bad idea in many cases, and an effective but informal method in selected cases. <S> Where it works: For Alkaline and carbon zinc batteries in the AA size, short circuit current capability is usually inder 10 amps even when new. <S> A short duration short circuit with a multimeter on its 10A or 20A range will give a good indication of the battery's state of charge. <S> The short circuit should be for less than 1 second. <S> I have used this method for many years for testing only this class of batteries . <S> It provides a rapid and effective means of assessment of battery state. <S> For AA NimH cells this will usually produce less than 10A but <S> the energy available from the cells makes this a marginal act. <S> Where it is a very bad idea: <S> For most other batteries - larger Alkaline etc (C, D) or LiIon, lead, acid and other higher energy cells applying a short circuit will be liable to damage or destroy the meter and may damage the battery - possibly dangerously. <S> High quality meters costing typically 100's of dollars <S> US (or equivalent) may have over current protection on their high current range. <S> This may be via a formal fuse or perhaps will utilise electronic circuitry to prevent damage. <S> Most low or medium cost multimeters do not have formal protection against overcurrent on their high current range - usually rated at 10A or sometimes 20A max. <S> significant overcurrent will usually damage or destroy the meter. <S> The 10A arrangements for two meters are shown below. <S> The first meter has no over-current protection and is typical os most meters that cost say under $US100 - and possibly rather more. <S> The second meter has a wire link acting as an informal fuse. <S> This is a higher rated and priced meter than most non-main-brand meters. <S> Typical 10A current shunt in low to medium cost DMM. <S> There is no fusing and excessive current will probably fuse the track on the pCB between the shunt and meter jack. <S> 10A current shunt and informal fuse on a somewhat more upmarket meter. <S> Ends of 10A shunt ringed in green. " <S> Fuse" consisting of wire strands circled in red.
| Instead, figure out what the battery should be able to supply, connect up a suitable load resistor or lamp which would draw that amount of current and measure the result.
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What is benefit of diode in the RC circuit This picture for reset circuit of the 8284A clock generator can you tell me what is the benefit of diode in the circuit? Second question why has the resistor been added between crystal an ground? <Q> It looks non-functional doesn't it! <S> However, the designer has cleverly anticipated what happens when you cycle the power on this circuit. <S> When the 5V goes away, without that diode the capacitor would remain charged for some time. <S> Reapplying power too soon might prevent the reset signal being generated. <S> It may however be superfluous if the input of the device has internal diodes to protect it from over-voltages on the pins. <S> But even then, those diodes might not like 10uF charged to Vcc over Vcc <S> hung on them. <S> BTW <S> it is a common mistake to forget to think about, and designing for, what happens when the power is removed. <S> Your second question... <S> See Transistors answer. <S> Generally devices like this indicate how the crystal should be wired in their data sheet. <S> It is important to accurately follow their recommendations. <S> Even then, once the PCB stage is reached, some tweaking can be required. <A> Can you tell me what is the benefit of diode in the circuit? <S> C1 is the reset time delay capacitor. <S> On power up it should be 0 V and after \$ \tau = <S> RC = 10k <S> \cdot <S> 10 <S> \mu = <S> 0.1 \ \mathrm <S> s \$ it will release the reset. <S> D1 ensures that when power is switched off that the capacitor will rapidly discharge into the positive rail. <S> This ensures that the proper reset will take place on power-up. <S> Second question why he adding the resistor between crystal an ground? <S> It looks like its part of a common base crystal oscillator. <S> Figure 1. <S> A common base crystal oscillator. <S> Source: Bipolar Transistor Cookbook . <S> Finally, Figure 1 shows an exceptionally useful two-transistor oscillator that can be used with any 50 kHz to 10 MHz series-resonant crystal. <S> Q1 is wired as a common-base amplifier and Q2 as an emitter follower, and the output signal (from Q2 emitter) is fed back to the input (Q1 emitter) via C2 and the series-resonant crystal. <S> This excellent circuit will oscillate with any crystal that shows the slightest sign of life. <S> R4 and R5 are R1 and R2 of your schematic. <S> C2 may not be critical. <A> Second question why has the resistor been added between crystal an ground? <S> Data sheet suggests that another crystal arrangement is possible, provided that stray capacitance from pins X1 and X2 to ground is small. <S> A small-value variable capacitor in series with the crystal provides a method to adjust oscillator frequency: <S> simulate this circuit – <S> Schematic created using <S> CircuitLab Where a "series-resonant" crystal is specified, the oscillator pins (X1 and X2) likely work best at low-impedance. <S> The extra 510 ohm resistors may ensure that this is the case, and they would tend to swamp out the effect of stray capacitance to ground. <S> This is a guess, since no internal circuit description (of any data sheet that I could find) shows internal amplifier configuration.
| The diode provides a short path to discharge the capacitor when the rail goes to ground.
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Design of 7400 Series IC Why is pin number 7 GND and 14 VCC in 7400 ICs ( logic gate ICs) ? Could the designer have put VCC and GND in some other pin number? Is there a constraint for that specific number? (Except a few cases, like 7490, Please consider gate ICs for example) <Q> There are no significant constraints and early ICs i DIP packages did try a variety of pinouts. <S> After a few years the corner power pins dominated and most 7400 series ICs standardized with power on the corners. <S> It is fairly convenient for routing power on few layers on a PCB by using a grid arrangement <S> but it is not ideal for power system integrity as it requires large loops for the decoupling network. <S> An arrangement where the power pins were adjacent would be better from that point of view as the decoupling capacitor could be placed with very short leads. <S> These power pinouts only really apply to 7400 series digital ICs in DIP packages or SMT variants of the same device. <S> Analog ICS, ASICs and others such as microcontrollers invariably use different power connections to improve power integrity. <S> As WhatRoughBeast commented ground connections in particular became more important as the device speeds increased. <A> It did/does also make it really easy to place decoupler caps right above the DIP packages, tight to the Vcc pin in nice neat rows/columns of ICs. <S> When chips were in rows, the left side of the decoupler was nicely right next to the ground on the IC above it. <S> On a densely populated board that was important. <A> This made pcb layout easier, as ground and power connections could be bussed along the rows and columns of a grid array of ICs. <S> However, Kevin is incorrect that there are no significant constraints, at least not always. <S> In the 1990s, the technology had progressed to the point that logic families which maintained compatibility with 7400 started to get so fast that switching transients interacting with physical issues started producing ground bounce <S> https://www.fairchildsemi.com/application-notes/AN/AN-640.pdf <S> issues. <S> There were even a number of buffers constructed specifically to minimize ground bounce, such as https://assets.nexperia.com/documents/data-sheet/74ALVC_ALVCH16244.pdf <S> This class of parts was largely superseded as design steered away from the physically large DIP package and toward much smaller SMT packages.
| These ICs featured multiple ground pins, and the pins tended to be located in the middle pins of a row, so as to minimize the length (and inductance) of the internal lead frame and chip bonding wires. As Kevin White has answered, the earliest 7400 chips showed a variety of ground/Vcc pins, with pin 7 and 14 (or 8 and 16 for 16-pin packages) quickly becoming the standard.
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5v relay causing short circuit when turned off I am using an Atmega 1284 microcontroller and four 5v electromechanical relays to turn on/off appliances based on some sensors, time, etc. The ATmega has its own 5v power supply in parallel with an 8v supply that I am using for the relays. Both of these power supplies are getting their DC from the same transformer after its been rectified. I am having an issue where the appliance that I am connecting to the relay shorts the entire circuit when turned off. It turns off the LCD screen and resets all the variables in the program on the Atmega. After about 2 seconds the circuit reboots to the my home screen but all the variables have been reset. I have measured the current coming from one leg of the transformer and before the appliance is turned on the current is at a steady 30ma. After the appliance is turned on it jumps to 70ma. After I turn the appliance off the current drops to about 15-20ma and shorts the Atmega. I have tried to solve this issue by using an optocoupler/optoisolater in between the signal pin of the Atmega and the transistor just before the relay to try to isolate the branches as much as possible. That hasn't solved the issue. I have read all of the place that isolating the circuits is the best way to go to try to save the Atmega from damage. But how can I isolate the circuits when they technically share the same ground from the transformer? Or is that even the issue here? Here is the circuit: [ ] <Q> Your problem is lack of awareness to EMI solutions <S> ( hundreds on this site alone) <S> Noise glitches may be conducted or radiated by E field (Voltage) or H field (current) when there are high impedance lines close to untwisted inductive back EMF surge voltages. <S> Since you have tried conductive isolation, that leaves radiated noise isolation. <S> The solution to this depends on your schematic and layout. <S> Check for : inductive load diode clamp for DC , RC snubber for inductive AC loads use ground from power source not shared power ground cable for pulsed loads twisted pair for inductive loads and coil driver <S> shielded cable helps as well to suppress the emission over twisted pair or STP - cable proximity and non-parallel orientation add ferrite sleeves to suppress CM noise. <S> Opinion (based on experience) <S> Opto-isolation was unnecessary. <S> solenoid wiring needs to be twisted pair <S> solenoid needs a snubber to reduce bandwidth of voltage spike and thus crosstalk voltage to signal/return lines. <S> DC signal cables ( if any) need to be twisted pair. <S> plastic film X-rated snubber cap avail from avail from Digikey etc. <A> The only thing that makes sense about resetting the circuit when turning the relay off is that you're not handling the stored current from the solenoid part of the relay. <S> You do this with a flyback diode, which it appears that you have one, but it isn't clear how you have wired it. <S> Make sure <S> thAt when the relay is energized, the diode is not. <S> Also, follow the direction of the current through the coil, and when you deenergize the relay, the resulting current should find a recirculating path through the diode. <S> If you have a scope, put a probe on the switching node of the relay coil. <S> When you shut it off, if you don't have it hooked up right, you'll see it spike up to 10s or 100s of volts. <A> The indication from the information that you have provided is that you are not actually getting a "short" happening. <S> Instead it would seem that instead you are getting latchup happening inside your microcontroller. <S> Latchup is a phenomenon that happens on complex IC chips when illegal signal conditions are presented at the pins of the part and internal stray circuit paths act like an silicon controlled rectifier and turn on and conduct current through parts of the silicon that are not valid for current flow. <S> Latchup can vary from annoying and requiring a simple Power Cycle to reset the problem to a more serious type where the latch current critically destroys all or part of the IC chip. <S> Latchup will typically occur when some IC pin is taken to a voltage way over the VCC rail or below the GND rail. <S> Voltage spikes coupled into the circuit are the usual latchup trigger and can come from static discharge or switching circuits. <S> In your case it is most probable that the AC mains voltage switching is causing spikes in your circuit. <S> Switching AC mains with an electromechanical relay can cause noise and voltage spikes up to 100s of volts. <S> The noise can come from contact bounce, contact arcing and switching inductive loads. <S> There are a number of things you may need to consider for investigating your problem. <S> Take a look in your circuit with an oscilloscope to determine if you have high voltage spikes. <S> Use the scope with proper probing techniques to see if there are serious ground rail surges going on. <S> Learn about AC switch snubber circuits and how these can be used to reduce spiking and noise on relay contacts. <S> Evaluate <S> carefully how you have your circuit constructed. <S> Make sure to isolate and keep all the AC power wires keep completely away from any of the low voltage portions of the circuit. <S> Evaluate having a fully separate transformer for the 8V supply and use that only on the relay side of the optocoupler. <S> This would allow full isolation of the grounds between the MCU subsystem and the switching subsystem. <S> Check into the actual use of real solid state relays that incorporate zero crossing switching. <S> These can significantly reduce AC switching noise and spikes. <A> Assuming it only does this when the appliance is connected, the problem is EMI from the contacts scrambling the operation of the MCU. <S> Your optoisolator does nothing of value since you are using the same supply. <S> Try using a different supply just for the relay coil and try to clean up your layout around the micro, and the wiring to the load. <S> In particular, avoid having any more connections than those you have shown (such as to a debugger or programming pod). <S> You can try an MOV across the contacts but typically the long term solution is a better relay and/or isolation and better layout (a 4-layer board with ground plane for example).
| The problem is not that Relay shorts MCU , rather that a reset condition from ground shift or voltage noise on lines i.e. EMC issue.
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Discharging one side of a capacitor? What I'm asking about is if it's possible to discharge one side of a capacitor. I've found a question about that, but I also found the answers to be a little unclear or questionable. I imagine that if you were to do so, you would need to put energy into it, because you would be removing charge from one side of the capacitor and creating electrical potential energy. Could this be done if you connected the positive and negative ends of a battery to a negative and a positive side of two capacitors? Could doing that cause capacitors to break down? <Q> It is physically possible for there to be more electrons on one side of a capacitor without there being a corresponding number of holes (absences of electrons) on the other side. <S> In fact, your proposed configuration of two capacitors and a battery would do that — but by a very, very small amount — about the same amount as if you cut a single capacitor in half and spread the plates apart to the same locations, then connected the battery. <S> This effect, which applies to any conductor, not just capacitor plates, is called self-capacitance , as opposed to mutual capacitance . <S> It is defined in the same way as capacitance, $$C = <S> \frac{q}{V}$$ — <S> but it is immensely smaller for a given physical size. <S> The amount of charge 1.5 volts — or 9 volts or 240 volts — can push into such a capacitor is so small that it has a negligible effect in typical circuits <S> — we do not bother to think about it. <S> (It is also true that there is some amount of (mutual) capacitance between the unconnected ends of the two capacitors. <S> Every pair of conductors is a capacitor, but they're usually bad ones with small area and large plate separation! <S> Both self-capacitance and mutual capacitance contribute to how much charge you can stuff into a conductor for a given voltage.) <S> In electrostatic systems, working with kilovolts and up, the effects of self-capacitance can become significant. <S> If you walk across a carpet and touch a CMOS IC, destroying it, what was the immediate source of the energy at the discharge? <S> It was your body having a net positive or negative charge. <S> The opposing charge was left behind on the carpet. <S> The self-capacitance is the ratio between that amount of carried charge and the voltage between you and the carpet. <S> (Where did the large voltage come from? <S> Separating the “plates”. <S> Where did the initial charge transfer come from? <S> The triboelectric effect. ) <S> A physical example of essentially an “only one side charged” capacitor is a Van de Graaff electrostatic generator. <S> The sphere on top is one plate; the entire surroundings including the Earth (assuming the generator is grounded, as it usually would be) is the other, but the Earth is so much bigger that the charge imbalance is insignificant for it but very significant for the sphere. <A> No. <S> The charge on a capacitor is defined by the voltage difference between the two plates, the geometry of the plates, and the chemical properties of the dielectric. <S> That is.. <S> the charge is between the plates, across the dielectric, not on the plates. <S> You need to understand it is the presence, or absence of electrons on one plate that drive away or attract electrons on the other plate. <S> You can't change one without changing the other. <S> As such, the concept of removing charge from one plate is incorrect. <S> If you remove electrons from the negatively side of the capacitor, the voltage across the plates would drop, as would the charge in the entire capacitor, not just that side of the capacitor. <S> So we just went round in a nice circle. <S> This is of course what we do all the time when we discharge a capacitor, we apply zero volts across it. <S> EDIT <S> There is one way you could achieve what you suggest and that is to use actual plates in a capacitor configuration. <S> Charge them up then disconnect them from the source and then separate the plates. <S> Both plates would still be "charged". <S> You could then discharge one of them to ground and then put them back together. <S> You would then have an unbalanced capacitor. <S> Of course, as soon as you hooked it up to anything, it would immediately try to rebalance itself. <A> Capacitor discharge is a process of reducing the stored charge in the capacitor. <S> That is going to be a relative operation of the component itself and not something that you do to just one lead of the capacitor.
| In fact, the only way to remove the electrons is to change the applied voltage across the capacitor.
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Finding the current dissipated on each resitor I'm given a question and part (1 asks me to find the total resistance of a 3.0 and 6.0 ohm resistor when I) in series and ii) In parallel. The answers to those are 9 ohms and 2 ohms respectively. Now, part B) says that a 4.0 ohm and 6.0v battery of negligible internal resistance are connected in series with the combination In (a) Now it wants me to find the current, p.d., and power dissipated on each resitor but I'm not sure how. I tried several attempt but they've not been successful. Thanks. <Q> The only voltage source in the system is the 6V battery, so use \$\frac{V}{R_1+R_2+R_3}\$ to find the current \$I\$ going through each resistor. <S> Now <S> \$V= <S> IR\$ and <S> \$P = <S> IV\$. You can substitute to get <S> \$P = <S> I^2 × R\$. <S> You can now find the power dissipated by each resistor. <A> The resistors are in series, so the current in each resistor is the same. <S> Now you know the current and resistance of each resistor, you can use ohm's law to calculate the voltage drop across each resistor <S> (add them all up and make sure the total is 6V).You <S> now know the voltage and current in each resistor, multiply the two to get power. <S> You can also use one the form of ohm's law that relates power to resistance and current and do the last two steps in a single formula. <S> If you make an attempt at the procedure above and show your work, I'll help you verify your answer and correct any mistakes. <A> This will be answered with Ohm's law and Watt's law. <S> I'm sure your textbook has one chapter devoted to each law. <S> Hint
| First find the equivalent resistance of the three resistors (in series, so add the resistances).Next, find the circuit current, you now know the load resistance and the battery voltage, so you can use ohm's law to calculate the current.
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What is the output voltage in this case? What happens when the output voltage of an op-amp (say in a voltage follower/inverting configuration) should be higher than a certain amount (say 10V) and on the output I place a 5V zener diode in reverse, followed by GND? simulate this circuit – Schematic created using CircuitLab On the one side, it should be Vin, right? But I think it must also be less or equal than 5V because of the Diode <Q> If the inverting input is low, the op amp <S> it will try raising the output voltage, so via the feedback it will try to match the +15V from the noninverting. <S> As hard as possible. <S> However, at the same time the zener sinks as hard as possible. <S> In the end you get the equivalent of a short circuit and either the opamp, or the zener, will eventually become very hot and might get destroyed. <S> Realistically, the TL081 is short-circuit protected and the output current capability is not incredibly high, so both parts will just get hot and the output voltage will be somewhere in-between 5 and 15 volts, probably close to 5, because, as said, the current capability is not incredibly high. <A> This is a impossible situation for a ideal opamp and a ideal zener diode. <S> The opamp will source or sink whatever current is required to maintain its desired output voltage. <S> Real components, however, have some maximum current they can source or sink and still adhere to the other specs they promise. <S> If the opamp can only put out 10 mA and the zener can sink 20, then the zener wins and the voltage stays at 5 V. <S> If the opamp can source more current than the zener can handle, then the zener is destroyed and the voltage goes to 15 V. <S> Once you run parts past one spec, you can no longer rely on operation within any of the other specs. <S> In addition, permanent damage can occur. <A> The op-amp will current-limit trying to drive the output voltage to 15V (assuming the power supply voltage on the op-amp is greater than about 18V, otherwise it may not be able to drive the output that high). <S> The TL081 has a maximum 60mA short-circuit output current. <S> A small axial-lead Zener (eg. <S> the 500mW 1N751A ) is rated for 70mA maximum current, so it will survive- <S> the output voltage is going to be higher than 5.1V most likely, but not a lot higher (probably less than 6V). <S> This is outside of even the typical curves and the zener impedance listed is only valid near the test current <S> so it's a guess, not something you can hang your hat on, and therefore not good engineering to depend upon.
| The zener will sink whatever current is required to limit the voltage across it to 5 V. For ideal components, the current goes infinite and the voltage underterminable.
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Measuring negative pressure with a raspberry pi and I2C Pressure Sensor I want to measure negative pressure which is generated by a diaphragm / membrane pump with 0,5 Bar (according specification). I am not sure if such a sensor as this https://sensing.honeywell.com/HSCDAND001BASA3-amplified-board-mount-pressure-sensors would work? Measuring range is from 0 to 1 bar. But my pump is generating negative pressure which I want to measure. Is this possible? Can anybody help me please... <Q> The real question is, the pressure range in the sensor's datasheet is absolute or relative pressure. <S> As I see, it is absolute. <S> The relative scale is what we use mostly, the 0 is the pressure measured at sea level. <S> It is 1ATM (~1bar) on the absolute scale, where 0 is full vacuum without any particles. <S> The answer is yes, <S> you can measure vacuum with that sensor. <A> Ostensibly, no, it won't work as it's not rated for vacuum, if itmeasures gage pressure. <S> That said, the pressure on one side is always greater than the pressure on the other, so if you measure atmospheric with respect to your vacuum, you're measuring positive pressure, and get the info you need. <S> If you can't do that, shop for a sensor that can go in either direction. <S> This works if you're sensor measures gage pressure, and not absolute. <S> If it measures absolute, you're fine. <A> The linked sensor is an absolute pressure sensor. <S> Which means it is fine to use it, but the value you get might not be quite what you expect it to be. <S> As it measures the absolute pressure, it will measure against a close to total vacuum, your pump is working against air pressure. <S> So to get the pressure of the pump, you have to store the pressure value before turning the pump on. <S> Then turn the pump on and measure again. <S> The difference between the two values is the pressure (or rather vacuum) <S> your pump is generating. <S> For example, the current air pressure is 960 mbar. <S> You will get a reading of 960 mbar from the sensor. <S> Then you turn your pump and it will create some vacuum. <S> The reading you get is then say 500 mbar. <S> The vacuum your pump generates is then 960 mbar - 500 mbar = 460 mbar. <S> They are sometimes sold as vacuum sensors. <S> A range would read like -1 to 1 bar or something like that. <S> There you can use just the reading and don't have to calculate something.
| There are relative pressure sensors available which are rated for vacuum, which would make the task a bit easier.
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Why do most battery protection circuits have 2.5V discharge cap? I was looking for a battery charge- discharge circuit from ebay, and most if not all, state to have overdischarge protection voltage at ~2,5V. Isn't this way too low? By my knowledge Lithium batteries (LiPos in this case) can only be brought to around 3.5V before riskin damage to the cell? Certainly no lower than 3.0V. Here are some examples of the boards I found: https://www.ebay.com/itm/5-10-20Pcs-TP4056-5V-1A-Micro-USB-Lithium-Battery-Charging-Module-Charger-Board/172856541571?hash=item283f0d8183:m:m1YzA8TerE_cywHj69klcVA https://www.ebay.com/itm/10x-3A-Protection-Board-for-1S-3-7V-18650-Li-ion-lithium-Battery-W-Solder-Belt/282474147244?hash=item41c4c56dac:g:hu4AAOSwY3BZFV3A Am I missing something? Would it be safe to connect these to a bare lithium polymer cell? <Q> This may not be a definitive answer, but a couple of points to note when choosing the threshold voltage. <S> The output voltage of a cell while discharging is lower than its quiescent voltage and the difference is higher for higher current. <S> That's why discharge controllers should have hysteresis or memory to prevent them from turning on again without a charge cycle. <S> Another thing is if you look at a discharge curve of a Li-Ion battery, you will see that the voltage drops very quickly at the end of the discharge cycle. <S> There is just little difference in terms of remaining state of charge between 2.5V and 3.0V. <S> So the actual implication of choosing 2.5V vs 3.0V threshold voltage may not be as dramatic as it looks. <S> UPDATED. <S> This document includes a discharge diagram: https://engineering.tamu.edu/media/4247819/ds-battery-panasonic-18650ncr.pdf . <S> At low current (0.2C), the difference in charge between 2.5V and 3V is ~3%, the cell is practically empty. <S> At higher currents (2C) the difference is higher, however, this is partially due to the discharging battery being "unable" to yield all stored energy quickly. <S> In this case, voltage recovery will likely be more significant. <S> UPD2. <S> Summarizing, to make my points a bit more clear: When talking about voltage of a battery, it is important to distinguish quiescent voltage and voltage under load. <S> 3.5 V is certainly too high. <S> UPD3. <S> By the way, all diagrams on the datasheet above are specified under V_cutoff = 2.5V, including the particularly interesting "Cycle Life Characteristics" one. <A> Yes, you are correct. <S> Li-ion batteries can be discharged to a minimum of 2.5V, but it is recommended that the lowest voltage that a Li-ion or LiPo battery be discharged be 3.0V only. <S> This will increase the longevity of the battery. <S> Draining it to 2.5V causes changes to the chemistry like increased internal resistance etc. <S> I've accidentally discharged below 2.5V and then the battery goes to short internally <S> (I've had experience). <S> If you take a look at some ESCs from RC electric cars they usually shut-off at 3.0V. <A> The usual protection IC on most batteries I have seen which incorporate protection is the DW01A which cuts of at 2.4 - 2.5V. <S> If you want to cut off at 3.0V then you need a different protection IC. <S> The Ablic IC for 3.0V is the S8261ABTMD G3TT2. <S> https://www.ablic.com/en/doc/datasheet/battery_protection/S8261_E.pdf
| 2.5V to 3.0V is a reasonable interval for shutoff voltage under load. If you disconnect the cell after reaching the low voltage limit, its voltage will recover a bit. So, I guess it depends on the brand and quality of the protection circuit used.
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Is there a gap in an LED between the plates (post & anvil) or a wire (or are there different kinds of each)? I previously believed there was a gap between the plates inside an LED (between the post and the anvil). However, I recently saw the following drawing on wikipedia ( https://en.wikipedia.org/wiki/Light-emitting_diode#/media/File:LED,_5mm, green (en).svg ) : I noticed that it shows the "Wire bond" which seems to indicate a wire from the post to the anvil. Question 1 Is there actually a physical wire there? I always thought the electrons jumped the gap and that there was no wire and that was part of why LEDs lasted so long. Question 2 - for clarity If there is a wire there, is it the part that lights up? Or, do the electrons still jump the gap? (I'm assuming they are conducted on the wire, but just attempting to clarify.) Question(s) 3 If the wire is there, is this a (more) recent design change to LEDs? Did LEDs of the past have or not have this wire? Is this a special type of LED with the wire and there are some that don't have it? <Q> Answer 1: <S> Yes, there is actually a physical wire there. <S> However, the wire does not short out the anode and cathode, it connects from the anode onto part of the semiconductor die. <S> Answer 2: <S> The wire is just a wire, it does not contribute to the light output of the LED. <S> (Other than providing a necessary electrical connection) <S> Answer 3: <S> This is not a new LED design. <S> Further comments: <S> All of the light emitting function happens on the small part labeled "Semiconductor Die" in the figure. <S> In fact, that is the only part necessary. <S> You could produce an LED with just the die, and it would work just fine. <S> Although it would probably be quite difficult to use, and fragile. <S> LEDs are a special type of diode that emit a significant amount of light under forward bias. <S> There is no "gap" that electrons jump across. <S> 1 <S> The "Wire Bond" in the diagram connects the anode wire on the LED package to the anode connection on the semiconductor die. <S> The purpose of everything else in the diagram is mechanical stability, or light focusing/defocusing. <S> The band gap is a part of the energy band model of semiconductors. <A> To show a good image of what's going on, here is an LED in cross-section (taken from this nice article on LED failure analysis ) <S> Here you can see the semiconducor device sitting in the cone <S> well of the cathode. <S> This well is there to work just like the reflector in a normal flashlight. <S> The light emission from the device comes from the edge of the chip at the junction between the p-type and n-type semiconductors and the reflector deflects the light to exit the top of the LED device. <A> Here is an example of a different style LED package. <S> The active part of the LED is the die which one side is the lead, and the bond wire attaches to the other connection. <A> LEDs create light (photons) by electroluminescence. <S> Light sources that create light from a heat glow wire are called incandescent. <S> LEDs are fabricated with a diode p-n junction where the p-type and n-type material is separated by a bandgap material. <S> When a diode is forward biased, donor free electrons on the n-type side diffuse over to the p-type side where they encounter many p-type holes with which they recombine. <S> This recombination can be either non-radiative (diodes) or radiative (LEDs). <S> When a diode is forward biased, donor free electrons on the n-type side diffuse over to the p-type side where they encounter many p-type holes with which they recombine. <S> This recombination can be either heat generating non-radiative (diodes) or radiative, light generating (LEDs). <S> In a radiative recombination event, one photon with energy equal to the bandgap energy of the semiconductor is emitted. <S> The bandgap is commonly called a multi-quantum well (MQW) Textured Surface <S> The light emitting surface is no longer flat. <S> In current LEDs the light emitting surfaces are structured in columns. <S> Columns can be horizontal or vertical.
| It is called a "wire bond" because "wire bonding" is the method used to connect these wires. 1 Electrons do transition across the energy band gap, which is what causes them to emit photons, but this is not a physical gap, it is all within a single block of semiconductor material.
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What is the Chip-on-board ic used in calculator, how to obtain one and how to use it? In any calculator there is a black dot blob such as this : After searching on what it is, I find that it is a Chip-on-board controller but i can't find what is the type of or how to buy or how to program it.can any one supporting me with any information about it? <Q> @Ahmed, if you want make a calculator, of course is just for fun, right? <S> If you were NEEDing a calculator, you would buy a chinese one VERY cheap, and it would be better than anything you could build yourself. <S> But, as a hobby project, it can be really fun build one for yourself. <S> The starting point SHOULD NOT BE a commercial product. <S> Forget about it. <S> The chip for this calculator isnt available to be sold retail, and even if you can get some samples, you wouldn't be able to apply this chip to a circuit, because it requires expensive industrial machinery. <S> I think that the approach should be different. <S> An easier form would be to buy a complet kit, like this or this . <S> If the links don't works, try search for KIT DIY CALCULATOR on Aliexpress or Banggood. <S> Even if you don't want to do it so easy, I think that kits like these are a good starting point to you decide what you want to do by yourself and what you'll buy for your project. <S> You should think about the hardware and the software. <S> you could just use an Arduino, attaching a keyboard and a 16x2 display to it... <S> you could buy a simple keyboard to attach to Arduino (and there libraries to Arduino work with those keyboards), or could build your own with your own PCB and switches... <S> You could use a LCD display, or 7-segments led displays, and the hardware and software shall be able to use them. <S> You could chose a microcontroler and write its software yourself, and load it in the microcontroler, even you could need a chip programmer;you should learn a lot about software, because the calculator use numbers of many digits, sometimes more than the microcontrolers registers use. <S> you can use the PCB from a kit, you can build your PCB from an online project, or even, you can project and build your own PCB... or have your PCB project made in a factory... <S> I'll stop here, since you started only asking how to get the calculators chips. <S> I fell that this question doesn't exactly address what you were needing, since the correct approach to build a calculator as hobby project is not try to reproduce or copy a commercial product. <S> Good luck! <A> The pcb looks like it used to have a more traditional square IC. <S> Those pads can be used as a breakout or test contacts for the chip while installed. <S> The COBs are not meant for hobby work. <S> If you can find out what chip it is you may be able to find (a clone of) it in a more manageable form factor. <A> This isn't really an answer, just some information. <S> You might be able to find info on it on the internet, either because its a super common chip that all calculators use, or because the specs for the exact calculator model youre using are online (but you would need to tell us what model it is), however you will never be able to tell for certain. <S> That black blob is just some epoxy, with the real chip underneath. <S> This is an incredibly common technique, used in hundreds of thousands of different products. <S> Because the actual chip is obscured, and there is no visible manufacturer or chip information, all you can do is guess about what it might be. <S> You need a pretty powerful microscope for this, some very strong acid, and lots of patience, so I would avoid that if I were you.
| The only way to fully identify the chip would be to remove the epoxy and decap the chip, and compare the silicon die inside it to some other chips until you found the right one.
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Electronics Newbie - How can I mute speakers when 3.5mm is plugged into female jack? (with volume control) So I'm a recent newbie to the world of electronics and have been working on a personal project using the Pi Zero W but I've reached a snag when it comes to sorting out audio. I have 2 problems I'm trying to figure out: 1) Mute speakers when 3.5mm male is in female jack and vice versa.2) Control the volume for both speakers and 3.5mm. Now I've seen tutorials online on how to create an amp op circuit using a 3.5mm plug for amplifying a internal speaker but seen nothing so far online in regards to using a 3.5mm jack. I was told by an electronics buddy I could use an audio switching MOSFET and to look at differential OP amps as circuit boards but still alot of it confuses me. At current my project is using a piezo speaker with the goal of simply playing single notes. Although for other parts of my project I will be using sound files and want to also eliminate any loss of sound quality when transferred between speaker/3.5mm. Any help will be appreciated if able to help. Remember I'm newbie to this so apologies if I don't pick up answer straight away and know exactly what you're talking about. Edit: I see answers I can understand the 5pin jack and the pins shorting out stuff. What I'm more concerned about is forming the circuit itself. As in are any helpful videos or images that you could point me? Cause there's 1 thing saying it but to be actually able to see it then makes all the better sense to myself. Edit 2: I think I've managed to understand it all correctly and following google searches as well got the circuit wired for potentiometer and speaker but could do with a checking with to see if I've got it all right (See image below). <Q> This question is pretty old, but I can maybe elaborate on what the other answers are saying. <S> The socket has five pins - You might think this is odd, because your stereo plug only has three connections: <S> Left, Right, Ground. <S> What are these strange extra connections? <S> The symbol you have shows two connections with weird angled parts at the end. <S> Those are the bits that actually interface with the plug. <S> The top one is the tip, and the bottom one is the ring. <S> The sleeve is connected to gnd. <S> The other two connections are the ones that act "odd". <S> Each of these is normally connected to one of the angled contacts, except when a plug is inserted. <S> When the plug is inserted, they are not connected to anything. <S> This means that you can use these connections to detect if a plug is inserted or not. <S> Going even further, you can use these to directly solve your problem. <S> If you connect your audio source to the two angled pins, when nothing is inserted, audio will come out of the two non-angled pins, so you should connect those to your speakers. <S> You shouldn't need any of those other components, except the potentiometer, which will need to be far smaller, around 20 ohms as Peter says. <S> If you want to control the speaker and the jack with the same potentiometer, it should be in series with the audio, and you'll likely need two of them, or a dual one (to control both channels). <S> Dont be tempted to put a single pot between the jack and ground, because you'll get strange sounding music as the common signals start to attenuate. <A> The 5-pin jack should have 2 pairs of contacts that are shorted together when there is nothing plugged in and are disconnected when something is plugged in. <S> On of each pair will connect to the plug and the other will be left floating. <S> Connect the speaker across to the common pin and the pins that end up floating (using a few resistors to create a rudimentary mono mixer) and the audio output to the pins that will connect to the plug. <A> You don't need a fancy circuit. <S> The extra terminals in the 3.5mm socket do it for you. <S> With no plug in, the "left" and "right" inputs just connect through to the "output" pins opposite - wire <S> those output pins to the speaker(s). <S> With a plug in, the outputs are disconnected, and the inputs connect to the plug instead. <S> It's that simple.
| When the plug is inserted, it will disconnect those two pins, muting the signal to your speakers.
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Does a circuit of 3 LEDs connected in series to a 9v battery work? Excuse my noobishness.... I'm buying these 10mm white LEDs on Amazon, and I want to connect 3 of them in series to a 9v battery. Is the 9v battery enough to power all 3 of these LEDs? Is this a circuit that I could add a resistor to? If so, how many ohms should the resistor be? Thanks!! <Q> With a series connection, the same amount of current flows through all devices. <S> You'll want to look closely at this circuit and ask yourself "what will keep these 3 LEDs from flowing more than their rating of 20ma? <S> The number on the data-sheet won't do it. <S> That only admonishes you not to design a circuit which would allow them to draw more than 20ma. <S> If you're thinking " <S> well it says 20ma draw <S> therefore it must surely self-regulate to 20ma", nope. <S> Components don't work that way. <S> Consumer appliances do; a light fixture that claims 120ma draw on AC power will, because it's engineered as a system (and approved and listed by UL) to do so. <S> The constant-voltage AC power system forces appliances to work in a linear way (self regulate their draw). <S> Components do whatever is their nature to do. <S> LEDs are highly non-linear and a small change in voltage will result in a high change in current. <S> So if you engineer for worst case, a 9v battery outputting 9.7V and the LEDs dropping you say 3V each, that's 0.7V to drop in the resistor, and you wnt 20ma flow. <S> E=IR is Ohm's Law. <S> 0.7 <S> = 0.02 <S> x R 50 x 0.07 = R 35 = R 35 ohms is a place to start. <S> You'll want to test it real-world to see what it actually does. <A> These are very cold white (bluish) LEDs that I expect are 3.0 to 3.1 at 20 mA but have an ESR of about 16 Ohms each. <S> {ESR~1/60 mW) <S> Meanwhile 9V alk. <S> batteries range around the same or less ESR, so you dont need any series R. <S> but it will end up being a battery voltage meter than a flashlight. <S> But you will likely get <20mA except with fresh Panasonic Alkaline batteries. <S> If the LEDs measure 3.1V each then with 9V across 3 in series, you get <20mA <S> .e.g. <S> the Vf best case is {3.0V- (20-I(mA)] <S> * 16R} * 3 LEDs = <S> Vf for diode string. <S> If the 9V cell has pulsed short circuit current of 1A or an ESR of 9 Ohms then <S> 10 mA <S> then voltage as much as 90 mV . <S> You will get far better results with CR2 battery cells x3 (cheap online) <A> Battery powered LEDs should not use a current limiting resistor. <S> You'd have to get them and measure the actual forward voltage. <S> Many white LEDs are less than 3 volts, more like 2.8v typical. <S> If the actual Vf is below 3V then the current needs to be regulated. <S> Battery powered LEDs are best driven with a CCR rather than a current limiting resistor. <S> Example CCR: <S> On-Semi <S> NSI45020 <S> 3 white LEDs and a 9V battery is not recommended. <S> The LEDs will be constantly getting dimmer. <S> Better would be to power them with a 3.6V battery and use a 3 output CCR such as the ON-Semi <S> CAT4003B or Microchip CL320 <A> A 9V battery, or any battery really, is a non-perfect voltage source which is better represented as a voltage source and a resistor in series. <S> For an alkaline 9V battery these have a high Equivilant Series Resistance compared to say AA batteries. <S> Leds will draw a forward current at a given forward voltage. <S> If the voltage is lower than your target current, then it van be connected without any other type of current restriction. <S> In this case, the 9V battery also acts like a resistor, not that it is needed. <S> You can put 3x 3V @ <S> 20mA leds across an alkaline 9v battery without any issues. <S> I regularly put 2x AA batteries without a resistor with 3.2V blue leds.
| The LEDs you chose have a forward voltage specified at 3.0V-3.2V, so a 9V battery would be borderline.
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Is it okay to make a loop of LED strips? I have 10 m of RGB LED strips (common anode) and want to put it around my room. I realized on testing that the "far of" LED s were very dim compared to those that were "electrically" near the driver circuit. Would it be okay to loop back the common positive and ground to the starting point. Basically loop back the power and ground rails? Is there any problem if I am driving them with a PWM signal of around 1 KHz?edit 1: I asked the last part because a loop would might create some EMI due to the 1KHz PWM signal running through it. <Q> A antenna for a 1 KHz signal must at least be 75km long <S> so i would not expect any problems in terms of radiated emissions. <S> Since you are using a square wave you must consider the slope of the signal though, which you can control by controlling how fast your switch turns on (eg. <S> with a gate resistor when using a MOSFET). <A> Making it a ring circuit would not be that bad, assuming you can do it neatly ;) <S> As for emissions, although 1kHz is not that high a frequency, if the edges are sharp that 1kHz will be producing some very high frequency harmonics. <S> As such, it may be prudent to limit the edges using some form of high frequency filter on the feed output, if it does not do that already. <A> Ideally, you should be injecting power at 2.5m, but in a pinch you can simply provide power at both ends, like you describe. <S> This reduces the effect of the high resistance of the FPC. <S> As to EMI from a 1khz pwm, it's unlikely to affect anything.
| If you connect your positive supply cable to one side of the LED Strip and the GND cable on the far end of the LED Strip the LEDs will be the same brightness.
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Difference between maximum amps for chassis wiring and power transmission I am putting together a test system that'll be carrying over 50Amps. I need to rate the wires that will be powering the device under test and also querying voltages and signals on the device. Looking at the AWG table hosted by powerstream there are two current guidelines, one gives the maximum amps for power transmission and another gives the maximum power for chassis wiring. There is a huge difference between the two e.g. for AWG24 3.5A versus 0.577A I would hazard a guess that chassis wiring is less than 1 metre and power transmission is greater than 100m. Or maybe chassis wiring is something that carries a signal or is only on for a limited amount of time? What exactly is meant by chassis wiring and power transmission, what is the difference between the chassis and power transmission wiring? <Q> Chassis wiring assumes each wire is routed separately, and power wiring assumes they are wired in a bundle. <S> In chassis wiring the cooling of the conductors is better, because they are all exposed directly to air. <S> In a bundle, some of the wires are not in direct contact withe the air. <S> This is why maximum current for chassis wiring will be more compared to an equivalent power transmission. <A> There's a statement above the table: <S> To me, it does look extremely conservative. <S> Maybe it assumes cables tightly bundled inside a cabinet full of hot equipment. <S> For cables in free air, in a cooler environment, you could increase the current considerably, provided the volt drop is acceptable. <A> Use the 60C column of the table below. <S> This is generally closest to what most people use without getting into the nitty gritty of what type of wire is being used and what type of load is being fed. <S> https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes <S> Beyond that, you haven't specified your voltage. <S> If you're using 50V and above, you really need to pay attention to the voltage rating of the wires insulation.
| The Maximum Amps for Power Transmission uses the 700 circular mils per amp rule, which is very very conservative
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What is the use of SENSE + and - in a PSU? I was in search of a good power supply for my radio. Apart from the batteries I have tried all the AC wall adapter or the SMPS based supplies seem to introduce noise in the radio. The SMPS being the worst. I have an old PSU as in below pic unused in my collection. As per the data sheet its a high quality voltage source which made me think of using it for this purpose. It has pins A,1,2,3,4,5,6,7 which corresponds to Adjust, - Sense, B-, +Sense, B+, Not Used, AC, AC. AC is rated for 115V. Its max current is 150mA, which is well within my current draw. What is the use of the _ and + SENSE pins? I've seen this on an old PSU unit I had with me back from the 1960s. Will it improve my output in any manner? Datasheet here . Update: So, I unpacked the PSU today and did some adjustment and checking now.I connected the (-SENSE & -B) and (+SENSE & +B) together and adjusted the voltage to 8.8V . Now, I tried removing the SENSE lines from both +B and -B terminals and measured the voltages across the each terminals: -SENSE & -B: 13.6V,+SENSE & +B: 8.1V,+B & -B: 22.4V,-SENSE & +SENSE: 0V .So, as our members have explained the sense lines seem to be very important in this case. <Q> Cables have some resistance, and when you pass current through them, they have a voltage drop. <S> This causes the load to "Receive" less voltage than what is actually set on the power supply. <S> Sense wires measure the voltage at the load (sense wires pass very little current <S> so they don't suffer that much voltage drop), and the PSU corrects the voltage based on the feedback they get. <S> Hopefully, this schematic will explain it better: simulate this circuit – Schematic created using CircuitLab <A> Sense pins are for remote voltage feedback. <S> Without connecting Sense wires to corresponding B wires the PSU won't work properly, if at all. <S> If you are using a very long cable between the PSU and your load, the main power wires (B+ and B- in this case) can suffer a substantial voltage drop. <S> When you connect the Sense- to B- and Sense+ to B+ at the POINT OF LOAD, it closes the feedback loop, and difference amplifier inside the regulated PSU will compensate the drop of voltage along main feed wires. <S> As result, the PSU will maintain its rated voltage level right at the destination point. <S> NOTE: <S> the sense wires don't carry much current, and can be made of a fairy fairly thin wires. <S> NOTE2 <S> : If the connection to load is not that long and/or voltage drop of little concern, the Sense-/+ can be connected to B-/+ right at the output strip connector of the PSU. <S> Many PSU came with this kind of jumpers out of the box. <S> And the local jumpers can be removed if necessary to make the 4-wire connection to load. <S> Alternatively, to be safe out of the box, the Sense pins might have a weak internal connections to corresponding B pins. <S> In case of using the external 4-wire hook-up, the feedback wires simply override the internal connections, and the point-of-load regulation will take place. <A> The sense lines tell the power supply what the voltage is at the point of load, so if you have a 20ft cable that connects the power supply to the load for some reason, you can run separate sense lines and hook them up right at the load. <S> This will remove the voltage drop caused by the wiring resistance, and the supply will margin up the voltage at the output until the voltage at the load is at the commanded point. <S> The key here is to run separate lines, as no current will flow through the sense lines. <S> If you're not going to this trouble, then you can just hook the sense lines up directly to the output of the supply. <A> Another use is if you want to put a multimeter in series to measure the load current - most have fairly high resistance on the current range and the voltage drop can stop the device operating. <S> This is particularly common on devices that operate on a low voltage. <S> Some PSU with sensing I have used in the past have an internal diode between the sense terminal and the output terminal. <S> That way the voltage only increases by 0.6V if the sense wire becomes disconnected. <S> Of course this also limits the voltage drop on each output lead to 0.6V.
| The danger is that if the sense wires become disconnected the PSU will go to it's maximum voltage probably destroying the load.
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Do we need any kind of protection for AC relay coil? I have seen and dealt with mostly DC relays until now I have a circuit that involves AC relays. In most of the DC relays we use a diode(like in the pic below) to protect the components like transistors, MOSFETs etc from the high voltage back e.m.f spikes. My question is do the AC relays also need some kind of protection across its relay terminals to prevent arcing when its turned off? The relays I'm working on are 120VAC ones. <Q> It's not unusual to switch one relay with the contacts of another with no "protection". <S> Such a snubber can also be useful across an inductive load. <S> Or across the coil to prevent high dv/dt at turn-off from preventing commutation of a thyristor (if you are using an SSR, it will often be built into the SSR). <S> Sometimes MOVs (Metal Oxide Varistors) are used, but I would suggest not using them for such repetitive transients as they actually wear out over time and still allow rather high voltages and don't reduce dv <S> /dt signficantly. <S> If you must use them, take into account their eventual failure mode (usually shorted, followed by a certain amount of flames and smoke and then opening up, if they are across the power). <A> You don't need one if you turn off the AC relay with a SCR or triac, because those turn off at current zero automatically. <S> In that case, there isn't any voltage induced at turnoff. <S> If you turn off the AC relay by another contact, you place an RC network, a snubber, over that contact, so the current isn't turned off immediately, but slowly reduced to the much lower value of the RCL series connection. <S> That way the induced voltage is low and will not arc over the opening contact. <A> Instead of using a snubber you can also use a varistor. <S> As long as the relay is energized the varistor has a very high resistance. <S> If you switch of the relay the voltage created by the self-induction of the coil rises to a higher level than the supply voltage and the varistor becomes conductive. <S> The varistor must be selected by the supply voltage of your coil and the energy stored in your coil. <S> You can calculate the energy of your coil by:\$ E=\frac{1}{2}LI^2 \$
| You can add an RC snubber (typically something like 100 ohms in series with a few tenths of a microfarad) across the contact or across the coil to reduce EMI and sparking.
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What software was used for drawing this schematic? I was reading a book on digital electronics and the gates with more than two input are drawn differently. I have seen many books that draw schematic in this style. I was wondering what software did they use. I am specifically talking about the way the OR gate is done here. <Q> As others have mentioned, it looks like that was drawn the old fashioned way, with one of these: along with some of these: <S> Tools that once were in every electrical engineer's arsenal, along with one of these: <S> Battery life was great, and they seldom crashed! <A> That doesn't look like any software was used, but a good old-fashioned drawing board, maybe a few symbol templates/stencils/curve templates used by someone who probably is a trained technical draughtman . <S> Making such drawings is a job where you actually needed quite some expertise, so technischer <S> Zeichner <S> (at least in Germany) is a proper Ausbildungsberuf (a recognised occupation requiring formal training). <S> Nowadays, you'll find a lot of circuit drawing software, but my guess is that you'd need to extend them quite a bit to make it easy to draw such legacy diagrams. <S> Other than that, standard vector graphics software can be used to draw anything that primarily consists of geometric elements. <A> As others have said, they probably didn’t have any particular software available at the time of publication. <S> If you are interested in a modern solution, however, check out the Circuit macros package for LaTeX. <S> It has the wide gate in its library. <S> From the manual: <A> It appears to me to be a reproduction from a late-80s/early-90s era databook from before the days of online/digital datasheets. <S> Go to ti.com and look up some of the CD4000-series logic datasheets that are reminiscent of this period. <S> Dual 2-Input NAND <S> As to the "wings" on the OR, this is typically done to allow the user to better follow the interconnect of the wires by spacing them out.
| I don't believe this was drawn with any modern software tool.
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Pull-up resistor - why input pin is pulled to ground when the switch is closed? Let's say 1 is an input pin. So when the switch is "OFF" (open circuit), the output is almost equal to Vcc since there's no current. 1 is pulled to Vcc . When the switch is "ON" (closed circuit), 1 is pulled to the ground and there's a current between Vcc and the ground. But why 1 is not pulled to Vcc when the circuit is closed? Assuming Vcc is 5V , how much voltage goes to 1 when the circuit is closed? If 1 is pulled to ground when the circuit is closed, does it mean that the ground has higher voltage than Vcc in this case? UPDATE : All answers are appreciated and all of them helped. Thanks a lot for the effort. <Q> Actually, to confuse you more.. <S> One, is current going into the pin, and Zero is current coming out of the pin. <S> For one you have to supply the correct current to maintain the input at a high enough voltage. <S> For a low you have to extract enough current to keep the input voltage low enough. <S> If the resistor is not too large, the voltage drop across the resistor will be small and the voltage at the pin will be near Vcc, or more accurately, high enough above the high level logic threshold. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> When the switch is closed, obviously current is flowing through the resistor to ground, but there is another simultaneous current path from Vcc through the device, out the pin, through the switch to ground. <S> Since, with a switch, the pin is hard connected to ground, the voltage at the pin will be zero volts with respect to that ground. <S> simulate this circuit Notice when the switch is closed the resistor does nothing useful for the power it is dissipating. <S> For a standard (non CMOS) <S> TTL device the zero logic level current coming out of the pin is MUCH larger than the high level current that enters the pin. <S> For CMOS parts the high level and low level currents are almost equivalent and very small since all they do is maintain the charge, or lack of charge, on the input capacitance of the device. <A> As @MD says, you can think of it like a potential divider: When the switch is off, the resistance of the switch will be very high (probably much larger than 100Mohm). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If we use the divider equation, we can find the voltage at the input to the inverter: V=Vin*(R2)/(R1+R2)=5 <S> *100,000,000/(10,000+100,000,000)=4.9995V, which is basically 5V. <S> When the switch is closed: simulate this circuit <S> We can again use the divider formula:V=Vin*(R2)/(R1+R2)=5 <S> *0.01/(10,000+0.01)=0.00000499V, <S> which is basically 0V. <A> But why 1 is not pulled to Vcc when the circuit is closed? <S> If you like, consider the closed switch is a resistor with a very low value, probably 1 ohm or lower. <S> You can look in the datasheet to get the maximum resistance for the switch you actually plan to use in your circuit. <S> Now you have a voltage divider, and $$V_1 = <S> V_{cc}\frac{R_{sw}}{R_{sw}+R_1}$$ <S> Since \$R_{sw}\$ is (we're assuming) <S> about 1 ohm and \$R_1\$ is 10 kilohms, if \$V_{cc}\$ is 5 V we get about 0.5 mV at 1 , which is well below the the threshold for a logical low. <A> So let's replace the electricity in your circuit with water: <S> Your wires are now hoses that the water flows through. <S> Vcc is a tap connected to your house plumbing. <S> Ground is a drain. <S> The resistor is a coil of really narrow tubing that only lets the water slowly trickle through. <S> The switch is a valve that you can open or close. <S> The NOT gate, which basically acts as a voltage sensor, is now a water pressure gauge. <S> The important part to remember about the water analogy is this: Current is flow. <S> Voltage is pressure. <S> Now, let's first see what happens when we open the tap and close the valve. <S> Of course, the water from the tap will flow into the hose until it reaches the narrow tubing. <S> The tubing will slow down the water flow, and cause pressure to build up in the hose feeding into it, but some water will still slowly trickle through. <S> This trickle of water will next reach the drain valve (and the pressure gauge, which is connected to the same hose), but since the valve is closed, it cannot get any further. <S> So the pressure in the hose between the tubing and the valve will slowly build up, until it becomes equal to the pressure on the other side of the tubing (and to the pressure in the water mains feeding the tap). <S> At that point, all the water will be at the same pressure, there will be no more flow, and the pressure gauge will register a high pressure equal to that inside your house plumbing. <S> Next, let's open the drain valve. <S> Now all the water in the hose between the tubing and the valve can flow out, and the pressure will drop. <S> The water in the hose between the tap and the tubing will still be at high pressure, though, and so will start to flow out through the tubing. <S> But since the tubing is narrow, the water can't flow through it very fast, and only a trickle of water will come out. <S> And that trickling water will immediately flow out through the open valve and down the drain, as fast as it comes out of the tubing. <S> Thus, despite the trickle of water coming out of the tubing, the pressure gauge will read approximately zero.
| When the switch is open, a small current flows into pin 1 and eventually ends at ground through the device. For an intuitive understanding, it might be useful to pull out the old water pipe analogy .
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Why are my solar cell currents not adding up? I'm outside in full sunlight trying out some new solar cells I soldered together. I have a string of cells lined up like this: My DMM reads 3.14V open circuit and 0.8A short circuit. I take two more strings and connect them like this: My DMM should read something like 3.14V and 2.4A. Instead, it reads 3.14V and 1.61A. I thought I may need diodes for each string, so I did this: Diode: https://www.mouser.com/ProductDetail/Vishay-Semiconductors/BAW76-TAP/?qs=sGAEpiMZZMtoHjESLttvkqndoGbqj5C%2FaqMW7rnfy0M%3D I expected a voltage drop and sure enough my DMM read 3V and 0.52A. Next, I made this: I was expecting 3V and ~1.5A. Instead, I got 3V and 0.82A. I've never experienced parallel solar cells having a current drop like that in my other solar panels. Not only is the current lower after I add a diode, it's even lower when I connect them in parallel. Outside of possible bad workmanship on my part, is anyone aware of a possible cause? EDIT!!! Here are pictures of the (HOW-IN-THE-WORLD-IS-HAPPENING) phenomenon. Each string of solar cells are color coded. The cells having this current reduction issue are the bigger ones. Each picture shows short circuit current with my DMM (sorry Kevin White haha). I used a lamp, so yes the current is gonna be tiny, but it suffices. 1st string of malfunctioning cells: 0.51mA 2nd string of malfunctioning cells: 0.36mA 1st and 2nd strings of malfunctioning cells in parallel: Should be 0.51 + 0.36 = ~0.87mA. Instead, it's 0.43mA And for the sake of a control group, 1st string of normally functioning cells: 0.3mA 2nd string of normally functioning cells: 0.28mA 1st and 2nd string of functioning cells in parallel: Should be around 0.3 + 0.28 = ~0.58mA. It's 0.6mA :) And finally, to validate my point, 1st and 2nd string of normally functioning cells and 1st string of malfunctioning cells in parallel: Should be 0.6mA + 0.51mA = ~1.1mA. Instead, it's 0.53mA. It's almost as if the bigger cells are siphoning off current. I tested for shorts. Looked for corrosion and bad construction. Nothing stands out. WHAT IS GOING ON?! <Q> It is necessary that the illumination of solar cell elements was uniform at the time of measurement. <S> Damage (or shading) of one element leads to loss of efficiency of the whole chain. <S> The rated open-circuit voltage of the serial-connected elements must be the same (the number of serial-connected elements must be the same). <A> You probably have a wiring error - possibly one array is not connecting properly. <S> Try disconnecting them each in turn, probably disconnecting one of them will not affect the current. <S> The currents should add up The short circuit current with three panels will be slightly less than the sum of the individual panels as the voltage across the meter will be higher since it is passing more current but it will not be much in error. <S> If you look at the V-I curve diagram: <S> You can see that even if the voltage across the panel is not quite zero the current remains almost the same - they act as constant current sources. <S> The diodes are not needed as under these conditions there will not be any current flowing back into the other panels. <S> The main situation where diodes are needed is where you have a solar panel connected to a battery - in this case current can flow back from the battery when the panel is not illuminated. <S> The diode avoids this. <S> You realize that in general you should not measure the current capability of voltage sources by using a DMM in that way. <S> You would normally put the DMM in series with the load. <S> For example if you measured the current from a battery that way you would either blow the fuse in the meter or possibly damage it. <S> Solar cells are a special case in that when you load them heavily their voltage drops and they become constant current sources. <S> alternative-energy-tutorials <A> You are assuming that the Superposition Principle applies to non-linear PV sources just as it does with linear circuits. <S> It does not apply to non-linear circuits. <S> Similarily This does not apply to superposition. <S> edit <S> Your test report is disorganized and not concise and have no test specs and less than ideal lighting conditions. <S> First you report 0.8A for 1 string and 1.6A for 3 identical parallel strings without results for each one. <S> Then you report Should be 0.6mA + 0.51mA = <S> ~1.1mA. <S> How, which ones dropped from A to mA? <S> Your test report is impossible to understand. <S> It should be a concise table of results with clear item numbers. <S> Compare specs , estimate light power and then report results to validate this. <S> You may be experiencing a leakage resistance under low current conditions. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> PV's are almost ideal constant current sources below 1/2 of the Voc, except where the input light power is so low that internal leakage affects the current source. <S> Why are you only getting 1mA? <A> The cells are not matched, this could have been for a few reasons: the factory could have screwed up, they could have been damaged. <S> Regardless it is unwise to use mismatched cells (or string of cells) in the same module because they can dissipate power in the cells and run down your efficiency. <S> What can you do about it? <S> I would just leave the offending string out. <S> You could find a replacement and match it. <S> Ideally the I-V curve of the module or cell needs to be matched as closely as possible to the other cells. <S> Mismatch in PV modules occurs when the electrical parameters of one solar cell are significantly altered from those of the remaining devices. <S> The impact and power loss due to mismatch depend on: the operating point of the PV module; the circuit configuration; and the parameter (or parameters) which are different from the remainder of the solar cells. <S> Differences in any part of the IV curve between one solar cell and another may lead to mismatch losses at some operating point. <S> A non-ideal IV curve and the operating regime of the solar cell is shown below. <S> The impact of the mismatch depends on both the circuit configuration and on the type of mismatch, and is demonstrated in more detail in the following pages. <S> Image and text source: PV education <A> If you see, your measurement of short circuit current is giving large disparity between two strings of larger cells(0.51 & 0.36) than your two strings of smaller cells(0.3 & 0.28). <S> I am also suspecting Multimeter problem. <S> It might have internal non-linearity for some portion of measurement range. <S> Try changing the current measurement method. <S> Put 1 Ohm resistor across output of the solar cell string output and measure the voltage across the resistor. <S> This might give you the confirmation that your Multimeter is the problem or not. <S> Also, try to measure the current directly using your current method of amp meter across panel but with 20A current range. <S> You might get low resolution reading but will give you confirmation that your problem is not due to your measurement method.
| Although mismatch may occur in any of the cell parameters shown below, large mismatches are most commonly caused by differences in either the short-circuit current or open-circuit voltage. Elements connected in series must be of the same type. It has different shunt for measurement than your mA range.
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Help on turning starlight in to voltage Here’s my issue. I would like to turn bright starlight into voltage. Here’s what I have done. as you can see I have placed a photodiode in a telescope eyepiece tube. I have it below the lens because this is where prime focus is, the place where the scope’s light comes to a point. Here it is on the scope My problem Is that I am not able to generate voltage. The diode I am using is this one. The scope is an 8 inch f-10. Would a zero bias diode be better than a reverse bias. If your suggestion involves a diode with more than three pins please advise how wire. <Q> I think that in order to maximize sensitivity you are going to need to use an amplifier. <S> There are two basic modes of operation: photovoltaic mode and photoconductive mode. <S> In photovoltaic mode, when using a transimpedance amplifier, the voltage across the diode is held at zero volts, and the amplifier converts the diode output current to voltage. <S> Zero light means zero current and zero voltage. <S> There should be no dark current. <S> In photoconductive mode, the diode is reverse biased. <S> This mode also uses a transimpedance amplifier to convert current to voltage. <S> In this mode, a certain amount of reverse leakage current will be present, and thus must be calibrated out of the measurement. <S> Unless you can find a good reference, you may need to try both modes. <S> Here is an image from thorlabs (my main source of information): <S> Note in the diagram that the basic connection is the same for the two modes. <S> It is just a question of whether the anode is connected to GND or to a negative bias voltage. <S> I found the image here: https://www.thorlabs.com/tutorials.cfm?tabID=31760 <A> You are expecting the diode to act like a solar cell. <S> You may simply not have enough light for it to be detectable that way. <S> Try it this way instead. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You may have to play with the resistor value to get what you want. <S> But again, if there is not enough light on the sensor you may need something much more sensitive. <A> For power? <S> There is not a lot of energy in starlight, even if focused. <S> There are just not that many photons striking Earth from a distant star... let alone the aperture of your telescope! <S> Silicon devices are not sensitive enough to create any usable power this way (power generated less than leakage). <S> For signal? <S> Just trying to detect the light or measure it? <S> If you can't get enough signal with silicon diodes or transistors, you have to upgrade to scientist level ;-) <S> This is a Photomultiplier Tube (PMT): Wikipedia's description is correct, but not very helpful: Photomultiplier Tubes. <S> A photomultiplier tube, useful for light detection of very weak signals, is a photoemissive device in which the absorption of a photon results in the emission of an electron. <S> These detectors work by amplifying the electrons generated by a photocathode exposed to a photon flux. <S> When a photon enters the tube it is accelerated rapidly by the strong field gaining a lot of energy in the process. <S> It then collides with a metal plate at high speed (energy). <S> When this happens it knocks some electrons off the metal plate. <S> These electrons are also accelerated rapidly by the field. <S> By arranging the internals of the tube so that multiple collisions occur you can make the signal very strong as each collision creates more free electrons, which after colliding with the next plate, will each create more free electrons... <S> and so on until you reach the end where you measure the significant number of collisions as voltage. <A> Diode leakage current (or dark current) will be the most significant competitor of tiny starlight current. <S> Diode leakage current is highly dependent on diode temperature - a cold diode has less leakage. <S> To see if your photodiode is starlight-capable, a circuit is required that yields an output affected by diode dark current. <S> The digital voltmeter (VM1) in the following circuit should have 10 Megohm input resistance, and a sensitive scale of 200 mV, with a display that would show 199 mV as 199.9 . <S> That last digit increments every 100 microvolts. <S> When this voltmeter is placed in series with a reverse-biased photodiode, it converts to \$ 100/10 \$ picoamps resolution. <S> That may be sensitive enough to be affected by dark current. <S> Such digital voltmeters are quite common. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Such a feeble light source as starlight may not be measurable by a photodiode. The way they work is that you create a very strong electric field inside the tube.
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Properties of an Ideal Transformer and the reasons behind them Why should the magnetising reactance of an ideal transformer be very large i.e. approach infinity? What practicality does this imply? <Q> Suppose you have connected a transformer primary to an AC voltage source. <S> There is no load connected in secondary. <S> In this scenario, you would ideally want no power to be consumed from the source. <S> However, with a finite magnetising inductance this is not the case. <S> According to faradays law, when a voltage is applied across a coil, a finite flux is established in the core. <S> The establishment of flux requires energy in a non ideal material and that is manifested in the form of a current flowing from the source. <S> In short, in the non ideal world, the current flowing from the source has two components, one flowing to the load and the other flowing to establish flux in the core. <S> The former component is captured in the circuit model of transformer as a magnetising inductance, although it is a non linear inductance which depends on the material properties. <A> The magnetizing reactance is what you see looking into the transformer primary when the secondary winding is open circuited. <S> So this means that without a load on the secondary you see a high impedance looking into the primary. <S> The magnetizing reactance appears in parallel to the transformed impedance of the load you are going to place across the secondary, so you want it to be large. <A> An Ideal transformer has only one property, the turns ratio. <S> All other properties are ideal, that is either zero, or infinity. <S> No resistive or core losses, no magnetising current, no leakage inductance, able to sustain an arbitrary core flux. <S> When we come to model an ideal transformer with SPICE, then the way we communicate the turns ratio to the program is as the ratio of primary to secondary inductance, so we need to put finite figures in for those. <S> If they are very large, then the magnetising current will be very small, which will usually do for a first approximation. <S> When we come to make a transformer form real materials, then all of these ideals have to yield to what's practical, and what we can afford. <S> Things that were zero in the ideal transformer <S> , we make as low as we can afford, things that were infinity we make as high as we can afford. <S> Apart from silver (too expensive for general use) and superconductor (too difficult for general use) <S> copper has the lowest resistance of all conductors, so we're stuck with using that, and accepting whatever resistive losses we get. <S> Depending on the frequency, we use carefully engineered iron or ferrites, for high permeability, low hysteresis losses, and low eddy current losses. <S> It's in quadrature with the load current, so even if it's as high as 20% of the full load current, the increase in heating when the transformer is on full load is usually negligible. <S> When a transformer is designed to run at light load for a long time, it's worth reducing it further for efficiency. <S> Core saturation will limit the maximum voltage we can use, and copper cross section the maximum current.
| We generally want the magnetising current to be small, but there's no need to make it very small.
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Why place capacitors in front of the line to a (headphone) audio speaker? I'm using this module based on the popular bluetooth module CSR8635 https://www.tinyosshop.com/index.php?route=product/product&product_id=875 I was wondering why, in the right-hand side of the circuit, there are two capacitors before the speaker. In building this circuit, I have found that leaving out these capacitors result in no sound whatsoever. Tinier capacitors give less sound while higher valued capacitors give more sound. In fact, it seems the audio quality also rises with the size of the capacitor? Why is this? Why does the audio signal become more "intelligible" after the capacitor? How does the capacitor value affect the frequencies that resonate inside the speaker? Shouldn't the capacitor act as a LPF and filter out any high frequencies? <Q> The two large capacitors are acting as AC coupling capacitors. <S> A capacitor in series with a signal acts as a high pass filter (if you're curious about that, look into the impedance model of capacitors). <S> Basically, the smaller the value of the capacitor, the higher the cutoff frequency and so more of the low frequency signal (in this case, the audio bass) is going to be filtered out. <S> The DC bias would just drop across the speakers and dissipate extra power. <A> The capacitors are there to block the DC voltage from the Bluetooth module. <S> The BT module is unipolar, that is it does not have positive and negative supplies. <S> If you allow this DC voltage to reach your headphones, it will pull them to one side and is likely to cause damage. <S> The capacitor forms a high pass filter, not a low pass filter. <S> The headphones act as a resistor to ground, and so high frequencies pass through the capacitor but low frequencies are blocked. <S> The sound changes because of the capacitor's affect on the filter. <S> Low values of C will move the frequency roll off to higher frequencies and so you'll lose the bass tones. <A> Just to pass (filter out) only the high frequency components i.e., AC signals and to block the DC component. <S> For a given frequency of the incoming AC, capacitive reactance decreases with increase in capacitance ie., increase in size. <S> Hence you obtain better sound output. <A> The lower the capacity of the capacitor, the higher the frequencies allowed to pass while blocking lower frequencies. <S> A large value capacitor would pass most audible frequencies, so you would indeed hear 'better sound quality'. <S> The capacitor in line would be to block the DC bias. <S> If there is enough DC bias on there, it would lock the headphone driver in place and prevent much sound from being produced. <A> Do the math. <S> 10uF into a 32 ohm earphone <S> cuts frequencies below 500Hz. <S> 100uF cuts frequencies below 50hz. <S> 220uF cuts frequencies below 23Hz.
| The reason why the capacitors are so large is so that the cutoff frequency is lower, filtering out the DC bias while also keeping more of the low frequency signal.
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How are function name abbreviations of a chip conceived? The microcontroller on the Arduino has functions such as RXD, XTAL1, ADC4, AVCC, MISO. These abbreviations are quite commonly used, and often make sense on first sight. One thing I noticed, that it's not just "first letters of each word": XTAL1: Crystal 1 input MISO: Master Input, Slave Output of the Serial Peripheral Interface AVCC: Voltage supply for analog digital converter RXD: UART receive input How are these names conceived? Is there a guideline somewhere that I can follow for my own designs? <Q> If you just do first letter of each word, you usually end up with a) non-unique identifiers and b) non-informative names. <S> From your examples: <S> XTAL1: Difficult to use C for "crystal" as it's more widely used for capacitor. <S> X is possibly from the Greek abbreviation of Christ (X), so ChrisTAL. <S> [see comments] MISO: <S> Describes the direction for each device, which can be ambiguous otherwise. <S> AVCC: <S> Analog, VCC. <S> VCC comes from V-Collector-Collector from (older) bipolar devices where the collector is usually at the highest voltage. <S> RXD: RX is a very common abbreviation for receive. <S> RXD is then Receive Data. <S> In your own designs, as long as it's documented and provides some guidance as to what the pin does, it's fine. <S> If it's been done before in another design, keep using it. <S> It'll make your designs easier to understand for you and others. <S> As the comments say, there's nothing particularly special <S> but I think it helps to have consistency and some guidance for a knowledgeable reader. <A> They are mostly historical/traditional, rather than strictly logical. <S> Except for MISO, these date from long before the development of integrated circuits. <S> XTAL - This is just my guess, but in Britain, "X" is frequently pronounced "cross". <S> So, "XTAL" becomes "cross-tal", a close approximation to "crystal". <S> AVCC - V CC is the conventional way to denote the collector supply voltage in a BJT circuit, and is frequently also used in MOSFET circuits instead of the slightly more correct V DD . <S> So this is just short for "analog supply voltage". <S> Note that depending on context, this could be either an input or an output of a particular circuit. <S> Sometimes the abbreviations become a little more obscure when details such as lower case and subscripts are lost to the limitations of ASCII strings. <A> On addition to the other great answers given here, you have to understand that devices are developed on the shoulders of the previous art. <S> Having the RS232 line called RXD and naming your pin SIP ("Serial input pin") would just confuse people. <S> So you go with what folks know. <S> As such, over the years, an informal standard has evolved. <S> Other pins that are specific to your device, or schematic come to that, you generally name with as few characters as you can while still portraying the function of the pin as best you can. <S> The fact that you need to be able to fit all these names onto a symbol on your schematic dictates that long names are out. <S> It is also important to be careful not to use names that can be easily confused, for example. <S> A pin named "Drive_A" could be an output or an input. <S> If it's an input and there is a complimentary pin named "A_Drive <S> " you can see that remembering which is which might cause a bit of confusion.
| If you are developing a new micro, and are including the functionality of a UART or other well developed technology, you give the pins the same name as what it connects to or whatever previous state of the art used. RXD - Rx and Tx are traditional abbreviations for receive and transmit, so this is just "receive data".
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Which thing/component does increase the current on same volt in the charger What causes a mobile phone charger or any adapter to increase the current on same voltage. I saw that chargers are rated with different amp on same volt ( 5v.1A 5v,500ma 5v,250ma). Ohm law says (V=IR)I am confused....!!! <Q> To translate ohm's law to weight for easier understanding. <S> \$V <S> = <S> I×R\$ = <S> > <S> \$g = <S> cm³×(g <S> /cm³)\$, one is weight, one is volume, and one is density. <S> You can't choose to have 1000 cm³ water weighing 30 kg. <S> If you choose to have 1000 cm³ then it will weigh 1 kg, if you want 30 kg water then you will have 30000 cm³. <S> This is assuming water weighs 1 kg with a volume of 1000 cm³. <S> I know this question will be closed within the hour, but meh, thought I'd at least elaborate on ohm's law. <A> Ohm's Law applies to things that draw current, not things that supply voltage. <S> If you connect a load to a 5V power supply and the current drawn is 1A at 5V then Ohm's Law says the load has a resistance of 5Ω. <S> Your 'chargers' are actually power supplies rated to deliver a maximum of 500mA or 1A. <S> Maximum current output of a power supply is determined by several components (transformer, rectifier diodes, filter capacitors, transistors etc.) <S> all of which have to be matched to the current rating. <S> Switch-mode power supplies often have a current-limiting circuit in them which prevents drawing more current than the supply can safely deliver. <S> If the current limit is exceeded the power supply may shut down until the load is removed. <A> Ohm's Law applies to the unknown equivalent load resistance, Req. <S> I= <S> V/Req Although charging devices are dynamic, the current they draw, depends on hard limits imposed in the design and the battery condition and if the device display is also on as both increase demand current. <S> Since the target device usually has a smart battery charger, the current is limited by the target before it is limited by a voltage source, capable of more current. <S> I recently bought a 6 port hub 2.4A 5V adapter for about $35Cdn at Walmart. <S> It avoids the "octupus outlets". <S> It supports 2.4A per port but only 12A max = <S> 60W. <S> Devices such as Ipads and iPhone X are capable of accepting 12W charging or 2.4A when needed. <S> Smaller devices with smaller batteries must use less current to reduce heat & ageing on the battery, so the "smart" load determines current. <S> Here is just an example of a cheap 60W USB hub. <S> ( 12A/6port ) <S> Not a testimonial but $0.50 <S> /W is typical for USB hubs and this will come down with volume. <S> Cheapest is not the best. <S> https://www.ebay.com/itm/Multi-Port-60W-USB-Charger-12A-Rapid-Charging-Station-Desktop-Travel-Hub-iPhone-/162596367150?_trksid=p2349526.m4383.l4275.c10&var=461598838749 <S> If you just have a passive heater resistor, then it is linear and again <S> I= <S> V/R for example <S> 5V/2.4A=2.08 ohms for 12 watts.
| The charger in the phone etc. will draw whatever current it wants, up to the maximum the power supply can deliver.
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Powering my Arduino Nano with 24 V DC This will be my first time trying to power my Arduino from an external source rather than an USB port. I will need to power my Arduino Nano (Rev 3.0) with an external 24V DC power supply (from a 24V, 36Ah lead acid battery), so I was wondering what I could do. I have three possible solutions: Connect the 24V power supply to a 24V-to-12V DC buck converter, and supply the 12V voltage to the Vin pin on Arduino Nano to power my Arduino Nano. Connect the 24V power supply to a 24V-to-12V DC buck converter, and supply the 12V voltage to a voltage regulator (LM7805), and supply the 5V voltage to the 5V pin on Arduino Nano to power my Arduino Nano. Connect the 24V power supply to a voltage regulator (LM7805) directly, and supply the 5V output voltage to the 5V pin on Arduino Nano to power my Arduino Nano. Do all of the 3 solutions work theoretically? Will all of them successfully power the Arduino Nano? Which one is better? EDIT I will get my 24V DC supply from a 24V, 36Ah lead acid battery, and I will use my Arduino Nano to power a DC motor via L293D and buttons. I am not sure how much output current I will need, but I think it wouldn't be a lot. Will the 24V, 36Ah lead acid battery affect my choice? <Q> The most efficient solution would be to use a 24V->5V DC->DC converter. <S> Adding a 7805 at any point will simply waste power - the 7805 (or any other linear regulator) just acts as a resistor. <S> The current drawn from the 24 volt source will equal the current required by the Arduino (plus a bit for the regulator operation). <S> With a DC-DC converter, the power drawn (not the current) from the 24V source will equal the power required by the Arduino (plus a bit for the DC-DC converter operation and inefficency). <A> *7805 has a pretty bad effectiveness <S> --12->5 will really hot <S> --24- <S> > <S> 5 don't ever think... <S> maybe u can try LM2596 to solve it <A> Your first choice is the best one imo. <S> The 5 V pin on Arduino nano is intended to be a low current local output supply for projects so not recommended to supply the board via the 5 V pin. <S> It is possible if you have a regulated supply which won't exceed 5.5V. <S> See: <S> https://www.rugged-circuits.com/10-ways-to-destroy-an-arduino/ And https://forum.arduino.cc/index.php?topic=271158.0 <S> Use the 24 V to 12 V convertor and apply it to Vin on Arduino. <S> The Arduino has a built in power regulator that provides the local 5 V on-board voltages. <S> Ie the second part of your choice 2 is already implemented on board the nano. <S> This way you will have 24 V, 12 V, 5 V and 3v3 sources available for your project. <S> 5 V and 3v3 via the Arduino itself <S> but those are current limited. <S> Eg Entire current for arduino nano is 500 mA max. <S> Cant recall off hand <S> what max current is via 5v or 3v3 pins. <S> A 12 V 2956 converter would give you up to 3 A. <S> The 24 V supply whatever your battery could provide. <S> A well designed 2956 circuit would stabilise the 12 V input to the arduino if your project suddenly drew a large parallel current from the 24 V supply making the voltage drop due to internal resistance on battery. <S> Similarly if you used the 12 V output from 2956 to power the arduino and the project in parallel. <S> Vin is 7 V to 12 V. <S> So a 5V supply wouldn't work on Vin. <S> But I like to use that for serial comms with PC. <S> Don't like powering projects via computer usb as that can cause power surges on pc especially if there's an unexpected fault during development (accidental shorts can happen!). <S> If you power via USB and Vin simultaneously then i believe the on board regulator selects highest voltage source. <S> So 12 V will win and current isn't drawn via PC.
| You could also supply 5 V via the usb connector.
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Use of very large bulk capacitor for microcontroller power stability I have a PIC32 microcontroller (MCU) system that experienced intermittent (though regularly) resets, most likely due to ground bounce caused by the switching of inductive loads (solenoids). All the decoupling capacitors and bulk capacitor are present as per the datasheet's recommendation. The MCU is contained on a breakout board that plugs in to the rest of the circuitry. For modularity and ease of replacement if necessary. One solution that nearly eliminated the resets was to put a very large value capacitor across the main MCU power pins of the breakout board. The MCU is powered by 3.3 V. In this particular case, it was a 1000 uF electrolytic capacitor with a 25 V rating (which is a rather large capacitor). However, the odd intermittent resets still occurs now and then, but not nearly as much as it did without the electrolytic capacitor. I did some testing with a power supply and LED where I removed the power to the breakout board, and the LED still remained powered on for a couple of seconds, as expected. However, it seems that there is a saturation point at which more capacitance doesn't help the situation anymore. Herewith follows a few questions: Is there a saturation point at which any larger capacitors will not help the situation (i.e. that remains charged for longer)? Does the fact that the capacitor is rated for 25 V make it a bad choice for such a bulk capacitor for use at only 3.3 V? Is there a minimum voltage at which a particular capacitor will reliably provide adequate stability as a temporary power supply in the event of a ground bounce occurrence? Would a super capacitor be a better choice in this case (particularly the ones with low rated voltage and high capacitance in the Farad-range)? Taking all the above questions into consideration, what is the best way to provide an MCU with a reliable, stable power supply to eliminate susceptibility to ground bounce (other than using external power supplies like switch mode supplies). <Q> You want to attempt to create a star ground where the power supply ground connection is in between the power and digital sides of the circuit. <S> Ground bounce is caused by resistive losses when a large current flows through the ground traces. <S> By putting the power circuitry on the "other" side of a star ground, you put the resistor that is causing the drop/bounce in series with the power load and not in series with the digital load. <S> Given your construction <S> this might or might not be practical. <A> Smaller or larger capacitors are not the question. <S> The question is what is the frequency content of the transient, and do your capacitors help at those frequencies. <S> All capacitors have parasitic inductance and don't respond to frequencies if they are high enough. <S> Larger capacitors are usually worse at this. <S> 25 v capacitors might have higher parasitic inductance than 3.3V capacitors. <S> The type of capacitor and package construction matters a lot. <S> Capacitor performance is not usually affected by voltage unless it is so high that you blow it up. <S> The best way is to solve the transient problem. <S> Do you have transient suppression diodes on the coils for the inductive kick when you turn the coils off? <S> Here is a test <S> -- Turn all the solenoids on and then turn them all off at the same time. <S> Does this consistently cause the problem? <S> Off is usually the problem with transients. <A> Insert a large inductor in the power lead between the main PCB and the breakout board; this provides some additional transient suppression for your MCU circuit; what you have now, with the 1,000uF cap being in parallel with the main PCB, leaves the MCU exposed to the voltage sag caused by the sudden current demands. <S> Or you can simply try 10 Ohms or 33 Ohms or 100 Ohms. <S> The purpose is to isolatethe MCU board from the main board.
| I would guess that a super capacitor would not help because you likely have high frequency transients and super capacitors aren't great at high frequencies.
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Using differential line drivers for digital signals I need to send a slow (steady state most of the time) TTL signals over 10m - 1km (max) distance. Shielded CAT6 wire will be used and I can't use radio based solutions. Wire is only used for that purpose (eg. not shared with anything else, all 4 pairs are available) Electrical isolation is needed because sender and receiver side may have different ground potentials. I want to avoid protocols/standards that needs additional complexity at receiver side. A receiver IC that outputs the signal is OK but I don't want to have a uC for decoding packets over something like RS485/CAN etc. Since I have twisted pair cabling, using differential line drivers seems to be most appropriate solution for these requirements. Here is my confusion begins.. Most of the line drivers I found labeled as "RS485 transceivers". The thing is, I don't intend to use RS485 protocol. Is there any side effect for using these in my application? I can't see any but I want to be sure. Again, many line drivers works with single 5V supply. That means it can only generate ~+-2V differential signals. With long wires like I'm using, this can be a problem. Is there alternatives that can generate/receive higher Vpp signals (for example +-12V for improving signal), if any what is the keyword for finding them? All these solutions will also need a shared ground between sender and receiver. Since I'm going to have multiple receivers, even if I isolate ground from sender side (seperate logic / line driver supplies with optocouplers in between for signals), the sender board will need to connect all receiver grounds to a common point. How can I improve this situation and electrical isolation? Thanks. Note: In an other question , I stated to want control a relay with that and got related answers. This question focused on transmitting signal over the wire. <Q> RS485 <S> /RS422 hardware layer is based on differential line transceivers. <S> The 5V is chosen, because of high speed transfer. <S> Note that a 12V level signal with same slew rate needs more time to "reach" the HI or LO signal level, so small voltage signal is needed for high data frequency. <S> The protocol layer is something totally different story. <S> Typically a RS485 communication can go up to 1.2km @ 9600 bps, twisted pair, with a conductors diameter > <S> =0.64 mm (AWG 22)- courtesy of Siemens. <S> You won't find differential signals at high voltage, except some current loop, also used for data communication over distance. <S> This is now obsolete in industry and replaced with RS485, because at time there were no such RS485 transceivers, I guess. <S> The current loop was made with an optocoupler and a current mirror. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> With an isolated RS485 comm. <S> you would need to supply the RS485 transceivers with a galvanically isolated, floating PSU, like DC/DC converter 5V/5V. <S> You have to bias the signal pairs with Vcc and GND potential through bias resistors, and of course the termination resistor at both ends. <S> In such way, the PSUs are equalized to common potential voltage. <S> There is no need to connect the earth on them, and of course it has to be avoided to connect the GNDs of both PSUs throuh a third conductor like shield. <S> The shield is earthed on one end to sink the environment interference. <S> At each side you put a bias and termination resistors. <S> simulate this circuit <S> Link <S> to ready made transceiver with bias and terminating resistors. <A> Use the MIDI physical layer! <S> Opto coupler at the receiver (with a few resistors and a diode for protection) and a drop dead simple 5mA current loop transmitter, easy, simple, isolated and utterly reliable for this sort of thing. <S> There will be loads of examples on line showing how to do it. <S> Regards, Dan. <A> Please find below answers to your questions. <S> Q1: <S> Most of the line drivers I found labeled as "RS485 transceivers". <S> The thing is, I don't intend to use RS485 protocol. <S> Is there any side effect for using these in my application? <S> I can't see any <S> but I want to be sure. <S> Ans1: These are called transceivers because they follow the physical layer specification of RS485. <S> Mean the voltage level output of the ICs will be of RS485 standard. <S> But if you can send the data on variable speed in the form of 1 and 0, then why can't you send very low speed control signal of 1 and 0. <S> Also, these transceivers are just voltage converters and they do not contain any intelligence of making any kind of logic operations on your inputs. <S> Q2: Output +/-2V, <S> With long wires like I'm using, this can be a problem. <S> Ans2 <S> : This can be a problem when you are making ground of transmitter and receiver. <S> The output signal you will get will be differential signal and it is designed to overcome the technology drawback for such long distance application. <S> Q3: All these solutions will also need a shared ground between sender and receiver. <S> Ans3: No, you will not need to connect GND of any of the transmitter or any of the receiver. <S> Also, you need to provide isolation using opto-isolators on the input and output signals.
| If you want to isolate the communication part, you need to power these transceiver ICs with isolated power supply on both transmitter and receiver side.
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Calculate the electric bill of a phone charger (AC adapter input or output?) I need to calculate the amount of money I pay for charging my phone with this charger I know that I need to calculate the watts then kWh, so I multiplied 5 V by 1.55 A to get 7.75 watts and then use that to calculate kWh. I'm so confused though because whenever I asked anyone in my school they would say, "you don't pay for the output, you pay for the input, so what you did is wrong". Are they right? Why or why not?What is the correct method? Note : I'm not looking for a very accurate number because I don't wanna use a Kill-a-Watt. In addition, I watched a video of someone using it on their charger and it gave a number around 4 or 5, that makes me think my teachers are wrong. Phone battery I'm adding info about the battery since some of you asked for it.It's a Samsung secondary Li-on battery. Nominal voltage 3.85 V / 11.55 Wh. Charge voltage 4.4 V / 3000 mAh. <Q> Now the question is, "How do I calculate the input when I only know the output?" <S> We need one more piece of information: the efficiency of the PSU. <S> For this type of device it should be about 85%. <S> $$ P_{OUT} = <S> P_{IN}Eff <S> $$ where Eff is efficiency. <S> You can rewrite this as $$ P_{IN} = <S> \frac {P_{OUT}} {Eff} <S> $$ <S> From this you can calculate your watts and multiply this by your hours of use per year. <S> Note that your result will be on the high side because the charger will taper off the current as the battery becomes charged and eventually drop to a trickle. <A> Yes they are correct, you need to consider the efficiency of the supply. <S> Also the nameplate will not necessarily reflect what is happening under the hood, as it were. <S> The phone will draw a great deal of current while charging and the adapter will waste a certain amount getting that power from the mains plug. <S> When the charging is finished the current will drop, but not to zero since most phones will continue to operate from the adapter while plugged in, leaving the battery at 100%. <S> Finally, when you unplug the phone but leave the adapter in the wall, the power supply will waste power just sitting there (efficiency drops to zero since there is no useful output current). <S> The minimum (active) efficiency of that Level V supply is 0.0750 <S> * ln(5V*1.55A) <S> + 0.561 = 71.4%. <S> Assuming your supply takes full current for 2.5 hours to charge it will use no more than 5 <S> * 1.55/0.714 <S> = 10.8W for 2.5 hours or 0.027kWh. <S> After the battery is charged it will drop considerably. <S> With the phone unplugged the draw will be no more than 0.3W, so if we assume the phone is drawing no more than 0.5W with efficiency in the 70% range, we get 1W consumption while the phone is just sitting there after having charged. <S> If you leave it plugged in 24/7 <S> the total consumption is less than 0.05kWh/day. <S> If your electricity costs you $.15 <S> /kWh <S> it will cost around 50 cents a year to keep your phone charged. <S> In fact the charger is bound to be a bit better than the minimum required, so probably somewhat less. <S> Also it will never discharge if you leave it plugged in all the time. <S> Note that the charger can use a significant part of the total consumption just plugged in and idling with no phone connected. <S> This "vampire power" adds up with millions of adapters so connected, and governments have been tightening the regulations on quiescent consumption. <A> This is like nailing jello to a wall. <S> First, the cell phone itself will not have a constant draw down the USB side. <S> Obviously it will throttle back to almost nil when fully charged, but it will also vary during charging based on the battery charging curve and other phone load (e.g. Plugged-in usage). <S> Second, the charger is not 100% efficient, and even that efficiency is not linear. <S> The 0.3A figure is the nameplate rating and is useless. <S> Its purpose is to tell electricians how much power to provision for the device; i.e. If she is wiring a cellphone factory or cellphone store, can they wire all the kiosks with one 20A circuit or will they need 3. <S> This figure accounts for worst case use x power factor <S> ** x whatever fudge factors UL wants to see in order to approve the item. <S> On your charger the difference between rated input and output is 28 watts, that can't be real, something that small would get too hot! <S> Back to the Jello-nailing, I would guesstimate the rating of the phone's charger (typically 5w) and then double that to account for inefficiency (10W) , for the runtime of the recharge (e.g. 2 hour). <S> Watts x hours = watt-hours. <S> Divide by 1000 for kilowatt hours. <S> Americans pay typically 10 to 20 cents per kilowatt-hour of power. <S> If you pay 12 cents a kwh for power, 1 watt continuous 24x7 costs $1/year. <S> ** <S> Power factor is the ratio between the power you actually use (cherrypicking parts of the AC sinewave), and what the transformer/wiring must carry to deliver the entire AC sinewave. <S> For instance if you use a half wave rectifier, your PF is 50%. <A> Those numbers tell you the current that the supply can provide, not how much your phone needs. <S> Your phone will only draw as much as it's designed for, and so you don't know how much power is coming from the mains. <S> So, they are correct. <S> You can't get anything meaningful from the adaptor text itself. <S> There are two ways to get this figure. <S> You can measure the current going into the adaptor itself, or you can measure the current going to your phone. <S> The latter means you also need to know the efficiency of the adaptor as there is some loss in the adaptor when it converts mains voltage to 5 V. <S> You could potentially estimate it from the mains current input (0.3 A), but that doesn't say what voltage or frequency it's at. <S> Adding calculations for the second method <S> Ok <S> , so your phone is probably charged over USB which used to be limited to 500 mA. Also <S> , that 1.55 A is the maximum the supply can provide, you don't want to have that all the time. <S> So, let's assume (crudely) <S> that \$ I = <S> 500\ <S> \mathrm{mA} <S> \$. Power to your phone is \$P_{out} = <S> IV = 0.5 \times 5 = <S> 2.5\ W\$. <S> Then, let's assume the adaptor is 80% efficient, so \$P_{in} = <S> \frac{P_{out}}{0.80} = <S> 3\ W\$. <S> The multiply by the time to completely charge, and you're there. <A> Assumptions: 120V, 2 h to charge, 12¢/kWh P = <S> VI = <S> 120V <S> × <S> 0.3A = <S> 36W <S> Energy <S> = P t = 0.036kW <S> × 2h = <S> 0.072 <S> kWh <S> Cost = <S> Energy × Cost/Energy = 0.072 <S> kWh × <S> $0.12/kWh = <S> $0.00864 Or 0.864 of a cent. <S> You should be able to follow this.
| They are correct that you pay for the input.
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Small lathe induction motor - off-grid alternatives to capacitor start or modifying SP PH for capacitor (UPDATE 19/09/18)I have experimented retrospectively adding capacitors in series with the (centrifugally switched and originally capacitorless) start winding of the split-phase induction motor fitted to the small lathe. Testing it on the noisy 5kVA diesel genny revealed about 8A instantaneous start current with no capacitor. This reduces to 5.6A with 14uF and 3.9A with 47uF. There is no appreciable increase in time taken to reach operating speed; it is still quite instantaneous so it's not like it's lingering on start for too long causing mischief. The running current is 2.6A. This now runs on the smaller generator as required although it still causes a bit of strain on starting. It would be nice to tweak it one step further with a higher capacitor. Tests with no pulley belt fitted. I have available a 100uF and 200uF capacitor to take the test further but we are approaching the run current anyway. Does anyone forsee any adverse effects on machine longevity with such a modification? Thank You. (UPDATE 15/09/18: I was mistaken - Lathe motor is SP PH and looks like no capacitor) I have removed the lathe motor and thought I'd check for the correct function of the centrifugal switch of the original motor now that's easy to get to. The centrifugal switch works fine and is used to cut out a second parallel winding once the RPM is high enough. The diagram in the cover describes an alternative connection for the start winding for opposite rotation. I do not find a capacitor anywhere obvious. The plate is stamped (WDG) SP PH so do I presume split phase and not capacitor? A very similar (but higher output) capacitor version from the same Hoover motor family is shown in https://www.youtube.com/watch?v=esRAVdc1drI but that is stamped something like CAP ST and the type stamp is different. Does this change the state of play? Can one add a capacitor to my SP PH to achieve softer starting? [230V (UK), 2.6 Amp, 1425 RPM, 1/4 HP, Rating: Continuous, 50Hz. This is manufactured at Cambuslang, Scotland.] Thank You [230V ac, single phase, off-grid, UK] I have a pillar drill and small lathe both with 1/4HP motors which are plated at 2.6 amp. They are both capacitor type - one has the capacitor piggy back on the outside and the other is inside the free end bearing cover. I am running a 1KW inverter generator which is nice and clean. (The light-dimmer switch on the cylinder vacuum cleaner for a forge blower can crawl at a snails pace if I need it to and the variable speed jigsaw works fine.) When I use the Pillar drill (belt-driving the chuck spindle) it starts fine after the expected momentary surge. When I try to start the lathe (with all belts disconnected so only motor inertia) it overloads the genny. So I am trying to work out why they are the same capacity but different electrical loads and how to reduce the start requirement of the lathe. I made a simple experimental arrangement of a single gang and double gang mains sockets in series. The lathe is plugged into the single. The double carries a (fused!) "shorting" plug in one socket and any resistive load in the other socket. The current for the lathe therefore has to also pass through whichever of the other two plugs is switched on. The plan was to start on the resistor then short the resistor to run. So I took a few resistance measurements and tried a few different things like a hair drier and electric kettle for a resistor but had no luck. The hair dryer at 66ohm holds it back so it only flinches and the 26ohm kettle is not enough. Maybe my steps for splitting the voltage are too coarse. Is this a correct approach? I don't mind the inconvenience of the extra switch. Might a light-dimmer switch (I'd need to source a higher current one) work on an induction motor? The requirement is only for soft starting and NOT run speed control. There are many budget motor speed controllers. What are the differences between these and a light dimmer switch? I can get the motor plate details and try to find out any spec differences if that helps. Thank you. <Q> The vacuum cleaner and jigsaw probably have universal motors. <S> They are essentially the same as series DC motors and can be expected to work ok with light-dimmers and even better with cheap speed controllers. <S> A cheap speed controller will likely have a variable DC output. <S> A light dimmer could have either an AC or DC output. <S> It is likely that the motor with a capacitor in the end cover is a permanent-split-capacitor type of motor. <S> The capacitor has less capacitance than the motor with the external capacitor and it is connected all the time. <S> The capacitor is disconnected by a centrifugal switch or some other device once the motor reaches full speed. <S> The starting torque capability is higher, but it takes more current to start. <S> A cheap speed controller or dimmer that has an AC output might be able to get the motor started without overloading the generator. <S> The right value resistor might also. <S> If it works, it may take so long to get the lathe up to speed that the capacitor or the winding it is connected to could overheat. <S> There is a VFD that claims to be able to do the job, but I believe the price may be higher than a new motor and controller combined and it may not work. <S> There may be only the one manufacturer making something like that. <S> You might be able to find a 1/4 <S> Hp universal motor. <S> That would probably work without a controller and you could add one if needed. <A> No, a light dimmer will not work. <S> You control AC motor speed by frequency, not voltage. <S> And since it is capacitor start, you will probably damage the capacitor. <S> Motors are not purely resistive loads (like your hair dryer or kettle). <S> Change the motor out to a DC motor, and get a little DC Drive controller. <S> Done. <S> And with variable speed. <A> If you remove the capacitor then you can drive the 2 coils separately using a variable frequency drive. <S> Though I'm not sure a VFD exists that will do this job. <S> Though instead you can kickstart the lathe using a smaller motor that doesn't have the massive inrush current and then switch over to the main motor. <S> A one-way clutch will then let the starter motor shut off again.
| The motor with the external capacitor is likely a capacitor-start motor.
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LEDs in Parallel, each with its own resistor I am currently planning a project to light up a model of a cruise ship. I've done this before with some success. To keep things simple I want to wire in Parallel. (Assuming I've understood the difference correctly; Each LED will have a direct connection to the power source) I will be using a mix of colours; White, Warm White, Blue, Red, Green. Every LED is pre-wired or will be fitted with a resistor (470) to allow them to be run from a 12v source. Every online simulation I run and my own previous experience says this will work perfectly fine, but every article I read seems to scream "no don't do it" and even then, opinions seem to differ. Above is an example of my wiring diagram, forgive the crudity. There will be a lot more LEDs used than this, but this gives you an idea of my plans. Any thoughts? Thanks {{{}}} EDIT: This is the overall schematic that I have planned: Sadly I cannot provide the direct link for this circuit as it is too long for browsers apparently. <Q> Your schematic is not ok, but your description is. <S> If you are using one LED and then the proper resistor the connection is fine. <S> For instance (led values taken as example): simulate this circuit – <S> Schematic created using CircuitLab <S> Anyway, you will be wasting at least 3/4 of your power on the resistors. <S> My suggestion, whenever possible, put the maximum amount of leds in series: <S> simulate this circuit Just keep in mind that you should not exceed 9-10V of drop on the leds <S> (so for instance max number of 3V leds is 3) <S> all the leds in the same line will sense the same current <S> you will not be able to individually control the leds, just the lines (not a problem if you plan to turn them on for every branch, you will need to have at least one current-limiting device <S> (a resistor is the cheapest option, but if you want higher quality control you can also use a constant current circuit made by two transistors and two resistors) <S> Note : as Olin said in the comments, many times you may want different currents for different LEDs. <S> For instance, low brightness red LEDs may need a higher current than the standard blue ones, so it needs some balancing. <S> So putting the leds in series should be the last step; please test the various currents before, in order to choose the appropriate value so that the effect is ok for you, then you can group them by current like shown before. <A> but every article I read seems to scream " <S> no don't do it" <S> LEDs should not be wired in parallel when driven with a constant current source. <S> Ideal is where each LED or string of series LEDs are driven with its own constant current source or regulator. <S> The drawback to resistors is the forward voltage will vary depending on current, and temperature and therefore the current and luminous output will vary. <S> Not too applicable in your project. <S> Resistors should be fine for your application unless it is battery powered. <S> It appears you are going for 20mA (12V / 470Ω). <S> Keep in mind the forward voltage varies especially between red and the other colors. <S> 12V is kind of high for single LEDs. <S> With 150 LEDs this project will draw about 35 watts where 25 of the watts is due to the resistors. <S> The model will get very warm. <S> If possible 5V would be much better and 3.3V would be ideal. <S> Keep in mind <S> the luminous output at the same current can vary significantly. <S> You may want to adjust the currents to match luminous output rather than forward current. <S> For example within the same Cree XPE2 product line the luminous flux for Red Blue Green White at 350mA varies from 33 lumens (blue) to 126 lumens (green and white). <S> To choose the resistance value use a calculator such as: LED Series Resistor Calculator <A> As long as you have a resistor per LED then this should work fine. <S> At 12V: say worst case scenario there is 0v drop across the LED, then current is: $$ <S> I = <S> \frac{12\ <S> \mathrm{V}}{470\ \Omega} <S> = <S> 26\ <S> \mathrm{mA}$$ Power through resistor:$$P= <S> I^2R = <S> 0.026 <S> ^ <S> 2\times470 = 0.32\ \mathrm{W <S> } $$ <S> The only suggestion I would give is to use 1/2 watt resistors just to be safe <A> I like your circuit. <S> Keep it simple. <S> If one branch goes out you only loose one LED. <S> Make sure your supply can handle the current. <S> Assuming each branch is designed to 20mA per led. <S> With 20 LEDs in parallel you're up to almost 1/2 amp. <S> Good luck with the display!
| When wired in parallel with a current limiting resistor and powered with a constant voltage source will work okay, but not ideal.
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why three phase induction motor draws current with spike at low frequency by using VFD? I have built Variable frequency drive (VFD) for a three-phase induction motor using V/F constant control. while I was testing 15HP three-phase induction motor without load with my VFD, the motor draws current with a spike at low frequency. Low frequency means (10 to 20 HZ), and the current value is about to 1A to 3A. This only happens when I start the motor from 0 HZ to 20 Hz. when the motor reaches to above 20 HZ, the current value becomes low (1A) and smooth. My question is to you what will be the factor that draws the current with variation at low speed or at low frequency. ? what should I have to notice? is there any winding arrange of the three-phase induction motor for starting torque that is drawing current more also at low frequency? or I should not run motor below 20HZ? <Q> You have to understand an induction motor is basically a transformer with an odd rotating secondary. <S> It has all the properties of the transformer. <S> What beats you here is magnetic saturation. <S> The voltage you may apply to a given transformer primary depends linear on frequency (within the bounds of the used magnetic material). <S> Saturation does exactly this, during one half of the AC cycle the magnetic material cannot "eat" more voltage-time and so the current increases as it wasn't there. <S> That's why you see those spikes. <S> The lower your frequency drops, the longer the time per AC cycle in which the magnetic material is saturated. <A> I suspect the PWM is not behaving like <S> is should or <S> it is providing boosted V/Hz below 20 Hz. <S> If you provide constant V/Hz programmed to close to zero V at zero Hz, the motor should run fine without a load from 1 or 2 Hz and above. <S> It should also work well with a fan or centrifugal pump load. <S> To get enough torque to start a constant torque load, you need some increase in V/Hz below 10 Hz <S> or so. <S> You need to find out by trial and error how much the motor will tolerate without drawing too much current. <A> Assuming you are using a commercially available VFD, most of them provide a "Torque Boost" function at low speeds when using V/F control, because of the reasons stated by the previous responders. <S> Most manufacturers ship their drives with this as an optional feature that you must program to turn on, some ship with it turned on as the default setting and you must turn it off in programming. <S> It could also be that someone else before you turned it on. <S> Look through your programming parameters.
| For your motor/transformer, as soon you go below 20Hz, you have to reduce the primary voltage to avoid saturation of the core.
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Cheapest way to limit 4xAA to 5v or less I need to limit the voltage from AA batteries to 5.25V or less. Essentially, my circuit can operate from (minimum) 3.5V up to 5.25V, and I want to maximize run time by using 4xAA. Should I just run it off 3xAA batteries (giving 4.5V when full) or is there a way this can be done (without the expense of a switching regulator)? I have read that using zener diodes is not great since it wastes power and limits current, which is critical (I need to draw around 1.5A from the batteries). EDIT:Sounds like a boost converter with 2S2P batteries will be best. For reverse polarity protection, should I just include a regular diode in series? Wouldn't that lose 0.7v? <Q> What is it? <S> Is cost THAT much of an issue? <S> A linear regulator will burn excess voltage as heat, so if you don't wish to waste power, then its not a great option, but better then a zener IMO. <S> 1.5A is way too much for a typical joule thief I would think, but the idea is OK so... <S> So I am afraid the best option IS a switching reg, you can have 3A units off ebay for like a dollar, or even less. <S> For something that will cope with the low voltage I would suggest adapting the AA supply to 2 sets of 2xAA in parallel, to provide 3V. <S> If you do this you can get away with just a boost circuit (to save cost over a buck/boost), and will reduce the current draw on each AA - enabling better runtime and using the boost converter to draw every last drop out of the batteries. <S> If you're willing to risk a dollar, you would get much more use out of the batteries and have a better item with the latter option (boost conv) - think of the environment ;-) <A> The really inexpensive 5V USB power supply that I purchased many years ago was simply 4- "AA" cells in a battery holder with a couple of 5.1V 1 Watt Zener Diodes soldered right across the battery terminals. <S> Fresh cells discharged very quickly down to just below the Zener voltage and then stayed at that voltage for many months. <S> Completely cheesy <S> but it did work. <S> I still use that supply to this day. <A> You could run your application on only 3*AA 1.5 V cells, but I'd suggest you need to use Li primary cells. <S> Any other solution will introduce losses and you can use almost all the energy in the battery by just using 3 batteries. <S> Here's a datasheet for a AA Alkaline ...which I would not recommend you use. <S> Notice here that the current you want to draw is way outside the battery capability. <S> This will severely reduce the capacity you can access and using 4 with a buck regulator <S> will still not get you into it's current rating. <S> If you use Lithium primary AA cells , these have a no load voltage of about 1.8V, so with new batteries you have to ensure your application can sustain the voltage (5.4 V) when not operating. <S> As soon as you load the battery the voltage drops about 100 mV so brings you well within spec. <S> The datasheet shows a comparison of Alkaline vs Li as below:
| If you set the boost to somewhere near your circuit minimum (3.5V as you suggest), then the boost will be working at the highest possible efficiency (close to the 2-3V VCC from your AAs in parallel), and potentially reducing losses elsewhere in your circuit. Using a zener will work, but to work well with such a large load, 1.5A, your going to waste people's batteries IMO, and to be fair 1.5A is not trivial for AA batteries, though good alkaline batteries will be OK.
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How to make a self-recognition socket for diferent connectors (sensors) I am working on an electronic device, to which will be a lot of sensors on wires connected to a panel with sockets. Is there any option that will recognise which sensor is connected to a which socket? On the device-box will be a lot of same sockets. I never made something like this, but I think, that it will be possible with resistivity between some wires and the computer recognise trough this which sensor is it. I will have also Arduino as a slave, so i will be able to use analog inputs/outputs. May there is an option where three wire sensor is connected to a more pins connector and some pins will be connected together and some logic behind the socket panel recognize what is connected. Sensors are all digital some of them are using 1-Wire (DS18B20) and some I2C (BH1750) Datasheets DHT22 . DS18B20 (1-wire). MH-Z19 . BH1750 (I2C). simulate this circuit – Schematic created using CircuitLab <Q> You could use an analog method of resistance or voltage to determine the sensor, but that requires an extra pin per sensor, plus a way to communicate that back to the RPI. <S> Instead, look at i2c sensors. <S> Each sensor would have a unique address and the RPI can scan for these sensors using the LM-sensor or i2c-tools package. <S> Of course, you want to mix and match how these sensors are wired so a custom solution by you is require. <S> You already have the basic idea down. <A> Details of your sensors are not well understood (from your question - if you fix this you may get better answers). <S> For example, a VGA monitor self identifies using pins 12 (ID1/SDA) and 15 (ID3/SCL) of the VGA cable . <S> Using this technique, it would be trivial to employ the correct calculations necessary to deal with a T-type verses a K-type thermocouple . <S> Likely one of this simplest (and durable) sensors possible. <A> The I2C ideas already mentioned by others are good, but they do suffer from the complexity needed to be added to the sensor or cable and, since the devices are active, powering them can be an issue if you do not add sufficient pins to the connector to do so. <S> A simpler approach is to simply add a resistor to each device. <S> Then include circuitry on your board to scan each socket for the presence of said resistor using a ADC circuit with a MUX and pull-up. <S> The latter could all be hooked up directly to your micro, or a cheap scanner micro could be added to monitor the connector and communicate with your main micro via I2C or serial, or whatever. <S> Again though, without more information on what your idea of "sensors" is, this question is just too broad.
| If we can assume nothing about the type of sensors, consider adding a small form factor I2C or SPI EEPROM in the sensor's cable to identify the type of sensor.
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OLED 5V-tolerant SCL and SDA Looking online for small SSD1306 OLED screens to use with Arduino over I2C, I find products that look like this: The sellers don't offer datasheets - and the product descriptions don't say anything about whether the SCL/SDA lines are 5V tolerant. Based on what is visible in the pictures, there appears to be a regulator for the VCC line - so this will indeed be taken down to 3.3V or whatever that screen needs to work. But the rest of the board is just resistors and capacitors - no level-shifting in sight. I worry that these sellers simply play on the tolerances of these screens - and will either fry immediately with 5V SDA/SCL signals, or will age quickly and fry a bit later. What do you guys think? <Q> I2C is properly an open collector (well, today, open drain ) bus, which is to say that the active devices only ever drive low, and the pullup resistors are relied on to raise the bus to the supply rail otherwise. <S> So for a starting point, you'd want to be sure that your only I2C pullup resistors connect to a 3.3v rail, and that you have none on the Arduino board or elsewhere which would connect to 5v. <S> inputs when running off a 5v supply, <S> so reading the signal should be fine. <S> The greater challenge is that you would need to be sure that at no time does the Arduino ever actively drive the I2C lines high, for example, due to software error or loading of a different program. <S> (Arguably you also want to make sure it never activates the internal pullups, but those are fairly weak and so less likely to be a problem). <S> If you can't make this guarantee, I2C is also fairly compatible with MOSFET-based level shifting, and such modules are sold by many sources. <S> Here's an excerpt from one channel of the the schematic of Sparkfun's version . <S> You'd probably want to replace (or simply parallel) <S> those 10K resistor with resistors of a more typical I2C bus pullup value. <S> And of course you'd also want to measure the actual operating voltage of the board rather than assume that it is 3.3v. <S> With the MOSFET level translator, you have wide latitude even for something like a 1.8v I2C on the low side. <S> But without the level translator, you need to be sure that the I2C bus voltage is high enough to be reliably read as a 1 by your Arduino, which it may not be if it is substantially lower than 3.3v. <A> The vast majority of these displays are based on the SSD1306 which is indeed a 3.3 V maximum I2C device. <S> Connecting to an AVR type MCU can be problematic and there is a good description of the problem here in Arduino land ....and a simple cure using separate pullups. <S> Depending on the display you buy you may find there are onboard pullups to 3.3 V rail or level shifters such as Adafruit use in their displays. <S> Adding the two 4k7 resistors also acts as a divider with any internal pullup on the MCU, but the voltage is very close to maximum and can exceed it depending on your MCU. <A> The controller is not 5V tolerant. <S> From what I've read this board needs external pull ups. <S> There are plenty of simple transistor plus resistor voltage translators for i2c around. <S> Or use a 3.3V arduino or other MCU.
| The ATmega chips used in typical Arduino's have specifications that indicate they work fairly well with 3.3v Unless there is a resistor voltage divider, which can be used for 5v to 3.3v translation, you should provide your own.
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Voltage measurement at different adc ground I'm trying to measure a voltage drop across a resistor divider with a supply from ldo. The single ended adc is connected as follows. Will the adc readings give the expected readings if the gnd1 and gnd2 are not connected? Update added November 27: Thanks a lot from your inputs. So I've managed to make this work by having GND1 = GND2. <Q> No it will not work. <S> The microcontroller ADC will measure the input voltage relative to its local ground. <S> If this is not connected to the other ground, then this measurement will not accurately reflect the difference between the LDO output and Gnd 1. <S> It's even possible that the to ADC signal is outside the power supply rails of the microcontroller, and could damage the micro. <S> If the ADC has a differential input option, and you can arrange that the 'To ADC' signal and 'Gnd 1' voltage will always be within the recommended input voltage range of the ADC, then you could measure this voltage using two connections and the differential mode of the ADC. <S> Otherwise, your options include Use an external ADC grounded to 'Gnd 1' (and possibly some isolated data transfer mechanism like opto-couplers). <S> Use a "high-linearity optocoupler" (for example, HCNR-201) with an appropriate feedback circuit to transfer the analog voltage between the two ground domains. <A> Will the ADC readings give the expected readings if the gnd1 and gnd2 are not connected? <S> Only if you expect wildly varying readings. <S> The problem is that you don't have a "circuit". <S> You are expecting a small current to flow into the ADC input <S> but you have provided no return path. <S> The "circuit" is open so no current can flow. <S> As drawn your ADC input just has a large antenna connected to it so it will be prone to picking up electro-magnetic noise and this will cause random readings from the ADC. <S> Edit your question to explain what the real problem is and what you have tried to get around it. <S> Someone will help. <A> You can use a DPDT relay to sample the voltage on the left, then flip the relay to place the (1uF? <S> 0.1uF? <S> 0.01uF) capacitor across the ADC input/gnd. <S> ADCs, if a successive approximation ADC with moderate # of bits, usually have about 10pF of Cin (that provides, with sqrt(K*T/C) <S> a 20 microvolt RMS floor).If your relay charges <S> 1,000X <S> that 10pF, or 0.01uF cap, and then places that 0.01uF across the ADC input, you'll have a slight gain error but a fine and isolated voltage acquisition system. <A> To know the maximum error in your measurement, you can check the LDO and your micro-controller's data sheet to find the worst case ground bounce and/or ground dc value shift. <S> Now the LDO's ground will shift based on many parameters like the current drawn or sinked, amount of de-caps present to ripple out ground noise injected by power switching circuits etc. <S> Depending on the LDO you buy, you may have a way to ensure the ground voltage by connecting the thermal power pad to a ground plane (if you plan on using it in a Pcb).
| Typically if you would like to find the voltage across the resistor divider, you may have to connect the node above R2 to the ADC input of the uC. Two different grounds can be at different voltages given that you've not connected them.
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Will a LED strip be less bright when I drive my mosfets with 3.3V instead of 5V? I was originally planning to drive my 12V RGB LED strips (5m + 1m) from an Arduino Uno. It's not an addressable strip so I'll use a mosfet per channel. Since I am now looking to connect it to my Domoticz, I want to drive it from a NodeMCU (ESP8266) which only has 3.3V on the GPIOs. Will this influence the brightness of my strip? I've got some IRLZ44N N-channel mosfets laying around that should work with the 3.3V GPIOs (Vgs(th) = 2V, if I recall correctly.) EDIT: forgot to mention that I'll be drawing about 2.5A per channel (i.e. 2.5A per MOSFET) for the complete 6 meters. <Q> The key number is the guarantee of Rds(on) at the lowest gate voltage shown in the "Specifications" section. <S> Vgs(th) is the guarantee that the MOSFET will be almost entirely off (250uA). <S> There is a region between Vgs(th) and your operating point and <S> you want the voltage across the MOSFET to be small or it may get too hot and your LEDs may not be at full brightness. <S> In this case, the Rds(on) is guaranteed to be less than 0.039 ohms at 25 degees C Tj with 4.0V Vgs. <S> The transfer characteristics curve is only a typical characteristic and is not guaranteed so it is bad engineering to use it for your particular purpose. <S> It may be appropriate for a hobby project to give it a try, of course. <A> Try to find the datasheet for your IRLZ44N mosfet. <S> From there you should look at the transfer characteristics shown in the Vgs-Id diagram. <S> If you look at that diagram than you can see the Vgs required for the drain current needed. <S> The drain current is the current needed by your RGB leds <A> That would depend on how much current your LEDs take. <S> Since you failed to mention that it's pretty hard to answer this question. <S> The graphs on the data sheet indicate the currents you can expect to drive at specific gate voltages. <S> At 3.3V the current capability is about 20% of a five volt gate. <S> Whether that is enough, and whether your device will need extra heat-sinking in that state is what you need to calculate. <S> At2.5A it will be fine. <S> Mosfet will be a little warmer than optimum, but ok. <S> However, since you appear to have 12V available, you might want to consider augmenting your circuit to have the GPIO switch that to drive the gate, while keeping the gate under the Vgs max (10V). <S> Or just buy a more appropriate MOSFET.
| They probably will, in my opinion, but there is no guaranteed with such a MOSFET. I'd guess you will be ok if your current requirement is not huge.
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Question about the resulting real voltage of pwm (pulse-width modulation) I understand that pwm simply emits a voltage in either a LOW or HIGH state. But I read that for practical reasons, one can simply take the average voltage. If I would measure the voltage in practice, would that apply? <Q> No, not in general: If you measure fast enough, you'll see the high and lows. <S> Now, if you add a low-pass filter after the PWM unit, you will indeed filter out the changes, and "smoothen" the PWM to an average value. <S> So, whenever "I read somewhere <S> that one can simply...", it's always worth asking which simplifications were made. <A> Not the same. <S> Imagine <S> you have a 1 ohm resistive heater, designed to run on 3 volts. <S> You send 3 volts, E= <S> IR 3V=3A*1ohm, <S> EI=W 3V*3A=9 watts, and it is rated for 9 watts. <S> Happy dappy. <S> Instead, you send it 6 volts at 50% PWM. <S> Half the time, <S> E= <S> IR 6V=6A*1ohm, EI= <S> W 6V*6A=36W. <S> An 18W average, which exceeds the heater's 9W rating. <S> Ditto ditto: <S> LED good for 3A@3V (assuming magic LED). <S> Except LEDs are non-linear with a much sharper V-I curve. <S> As it happens 6V gives 20A. EI=W, 6V*20A=120W, half the time , or 60W average. <S> Whoops! <S> OK, so you put a capacitor on there to smooth the rattle and nominally average <S> 3V. <S> Except during the "on" time, the capacitor gulps up lots of current, and sustains the load at near 6V the rest of the time, effectively defeating the PWM. <S> You have to put some more stuff in there - limiting resistor, choke, something - to "tune" it <S> so the smoothing keeps things near spec 3V@3A. <S> That assumes the load has a fixed characteristic like the heater. <S> If the load is dynamic, you'll need more active circuitry, and the best way is to dynamically tune the PWM to other than 50%, and now we're talking a switching constant-voltage or constant-current supply. <S> if the goal is to divide the input voltage by exactly 50%, it could certainly use the input voltage as a pilot signal for the correct output voltage. <A> PWM works by varying the duty-cycle <S> (let's call it D) of a square wave (with 0 as low and some max value as high), as you know. <S> It can be shown fairly easily that the average value of that square wave is proportional to D. ... for practical reasons, one can simply take the average voltage. <S> They are also theoretical reasons, because the average value is what the system being controlled is sensible to. <S> That's the whole point of PWM: controlling an analog system using a digital signal without a complex DAC scheme. <S> A digital system like a microcontroller can easily generate PWM signals with adjustable duty cycle, therefore that's an easy way to control a system. <S> To be more specific: when you design a PWM control system correctly, you set the PWM frequency so high that the system being controlled cannot follow the rapid variation of the PWM signal, hence it acts, by its own inertia, as a low-pass filter. <S> Therefore the system being controlled turns out to react only to the average value of the PWM, which varies much more slowly than the instantaneous value of the PWM signal itself (that's a requirement of PWM: <S> duty cycle variations should be slow compared to the period of the PWM signal). <S> In other words, the intrinsic inertia of the system behaves as a low-pass filter, which, in turn, extracts only the DC component of the PWM, i.e. its average value. <S> If I would measure the voltage in practice, would that apply? <S> It depends how you measure the voltage. <S> If you use an oscilloscope, you'll see the exact PWM waveform (but a fairly modern, cheap, digital scope can calculate its average value and display it on screen). <S> Bottom line: for a correctly designed PWM control scheme, what really matters is the average value of the PWM signal, both from a practical and from a theoretical POV.
| If you use a simple voltmeter it depends on the PWM frequency and its measurement method.
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How to measure positive, negative and inbetween currents I am making a variable +12V / -12V bench power supply, where I will have 3 output pins: 0- -12VDC | GND | 0- +12VDC, where I want to measure all possible voltages and currents.I will use voltmeters to measure potentials between: -12V and GND, GND and +12V, -12V and +12V; but I got stuck here, how to measure same currents. I want to be shown at all time the current between GND and -12V, -12V and +12V & GND and +12V, but don't know how to wire the ammeters here. Red resistor is between 0 and +12V;Blue resistor is between 0 and -12V;Black resistor is between -12V and +12V. <Q> For the voltmeters you need to connect them between +12V and GND and -12V and GND.The ampere meters you should connect them in series with the outgoing +12V and the connector and the same for the -12V and the outgoing connector. <S> If the power supplies are separated ( missing information!) <S> then you could also place the ampere meters in both GND lines and combine the output on the connector. <S> In the situation that you are only interested in the differential current in the GND connection then you connect the two GND connections before the outgoing terminal. <S> The ampere meter should in that case come between the two ground connections and the outgoing terminal. <S> Depending on the load the current could be sourcing or sinking. <S> If you are using digital volt/amperemeters you should be carefull. <S> For most units there is no galvanic separation between the supply and the measuring connection. <S> In that case you might end up with separate supplies for each meter or at least for two of them. <S> For the amperemeter in a common GND situation you need to select a type that can handle both positive and negative current. <S> Not all types can. <A> Your connection of the ammeters and shunts is incorrect. <S> Figure 1. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 2. <S> Ammeter and voltmeter layout. <S> Since you want to measure the current in the common rail you have no option but to add the ammeter in series with this. <S> This has the disadvantage that your ground voltage will shift with current. <S> You will have to determine whether or not this is acceptable. <S> The layout gives an interesting design decision requirement. <S> Which current flow direction should be reported as positive? <S> I suggest that you connect ammeter positive terminals to the PSU outputs. <S> This will allow you to clearly monitor the current out of the terminal. <S> The -12 ammeter will always display a negative value as it is only capable of sinking current. <S> The common ammeter will indicate positive when sourcing current for the negative supply and negative when sinking current from the positive. <S> When +12 and -12 loads are balanced the common ammeter will read zero. <A> Then take the difference of these for the GND current. <S> The type of error you'd see in the GND current indication differs between a three sense resistor and my two sense resistor scheme, but both schemes will have errors, and the error will improve with matching components, so that's not really a significant differentiator between the methods.
| As +12v, -12v and GND currents all sum to zero, use a sense resistor in the two 12v lines, sense these with differential input amplifiers, and use these voltages to represent your line currents. Correct wiring of ammeter shunt in series with load - not across the supply rails.
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Relay to control electrical plug and non-electrical wire I'm not sure what I'm looking for, but what I need is a relay switch that will control two things, or maybe two different relays are needed, I'm not sure. Basically, from my Raspberry PI I want to control two things; An electrical outlet and it's power, with NC side wired, so that my plug is on unless I turn it off from the PI. Part 1 can I do on my 2 channel relay: https://www.amazon.com/SainSmart-101-70-100-2-Channel-Relay-Module/dp/B0057OC6D8 I'm having trouble with part 2. Have another wire interrupted also at the same time the plug is turned off. This wire is it's own power source and will not be powered by the wall electrical line like the plug, I simply want to interrupt the line when off. Not sure if I need a second relay or something else for this purpose, as I don't want the power from the wall routed to this secondary line. <Q> As Trevor sugggests, you could use a double or triple pole relay, but the two single pole relays on your board will work just as well. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Two options. <S> Figure 1 shows the wiring of the relay options suggested in the other answers. <S> Since you already have two single-pole relays (b) is the simple option for you. <S> You just need to switch the coils of both relays at the same time. <S> RLY2 is wired as an NC, normally closed, circuit. <S> It will disconnect when the relay is energised. <S> RLY3 is wired as a NO, normally opened, circuit. <S> It will connect when the relay is energised. <S> (b) also has the advantage that there is better isolation and no risk of mains on the DC circuit in the event of catastrophic failure of the relay. <A> You can use a multiple pole relay, but chose one with sufficient isolation between the poles. <S> Maybe a three pole with the centre pole grounded for some added protection. <S> If you must use a patch board, the one you mentioned should also work assuming it brings out the normally closed connection. <S> Hard to tell from the amazon information.
| Use one of the relays on that board to switch the AC outlet, and the other relay to switch the second circuit.
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Packet loss on LAN point-to-point I would like to ask if packet loss in p2p LAN is a normal behavior in LAN. And now in more details: I am connecting PC to target board with TI's chip and LAN9221 controller.I am doing a ping-pong test, with udp protocol: PC is sending the first packet. target wait for packet, and on receive, it send the same packet back to PC. PC wait, and on receiving send a new packet (increment data in the UDP packet), and so on.... Running the ping-pong test, I see that once in a while the target did not receive the packet correctly. It is easy to notice the failure because the ping-pong test stops whenever one of the side did not receive a packet.When the failure happens, I see that the first bytes the target detected is actually the the 5th byte that the PC send (I see in WIRESHARK). I also observed that these failures always happens when a another packet is sent about the same time from PC as the packet from the test , as if a collision happened. If I add firewall rules, which take care that only the ping-pong packets shall be delivered in the LAN, then there is no failure at all. So, the question is if the collision failures is a normal behavior in LAN ? EDIT: seems the problem was related to (sw) driver error, not to collision, thank you for the solutions, which emphasized that in p2p there are no collisions. Thank you! <Q> Collision failures <S> yes, but you haven't any of those in a p2p or switched setup. <S> I test with flood pings usually and any connection with a single dropped ping within a minute <S> or so at least isn't good. <S> It would work, though, as dropped packets should be resent by upper layers. <A> No. <S> There is no collision or packet loss in p2p Ethernet network. <S> If you are connected point to point Tx/RX with no switch in-between <S> ...there simply cannot be packet loss on the wire (it's typically a software problem). <S> If you are connected through a switch, then you have a store and forward system, again with no potential for collisions or packet loss. <S> If the store RAM gets close to filling then the switch sends a signal to the sender to stop sending. <S> If you are connected to a larger network there can be packet loss, but this is a decision making process in the network routers to drop packets (typically due to route congestion). <S> Your software needs to react appropriately to this and handle it (usually with timeouts). <S> If you are doing ping-pong, then you should simply never see an Rx failure. <S> You may have a poor hardware implementation (perhaps unmatched magnetics or noise). <A> It is normal behavior. <S> In fact any LAN technology which doesn't account for lost or dropped packets is basically useless. <S> Consider sending duplicates after a certain time period with no reply, or use a transport protocol like TCP which establishes connections via acknowledgement packets.
| If you are dealing with partial packets (runts), this is typically a software problem in your stack. We have not had collisions/packet loss since the introduction of Ethernet switches.
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Using Raspberry Pi to read open collector outputs from tally connector Please forgive my ignorance, I'm a software programmer who is new to the electrical and hardware side. Our Panasonic HS300G video switcher has a Tally connector. Below is the pin diagram and circuit diagram. Pin 1-6 are for telling me which Camera input is active (Camera 1-6). Using a Raspberry Pi 3, I need to be able to detect which pin is "active", and my software will then use that to run a software-based tally system. I've been trying to research as much as possible myself to understand how "open collector outputs" work, but I'm still confused, and I don't want to fry anything on this video mixer by sending 3.3V to a wrong pin. Would this design work as follows? Attach Pin 9 to GROUND on the Raspberry Pi. Attached Pin 1-6 to 6 individual GPIO pins set as "input" on Raspberry Pi Send 3.3V from a Raspberry Pi GPIO to Pin 7. Result: My GPIO "input" pins (1-6) will receive 3.3V whenever the corresponding camera input is selected. This will give the GPIO a "high" reading and my software will know that camera is active. I would like to test this theory but I can't afford to send too high voltage and mess something up on this mixer! Please help me understand what is going on here, really appreciate it. EDIT: How do I wire up the 3.3V into the circuit properly? What am I missing? <Q> Send 3.3V from a Rasberry pi GPIO to Pin 7. <S> No. <S> Pin 7 should simply be left unconnected. <S> The outputs are enabled when pin 7 is left open. <S> Other than that, <S> a simple 1k to 10k resistor as a pull up to 3.3V on each output is all you need. <S> You have the general idea. <S> , except as another answer mentioned, these are active low outputs. <S> Open collectors are pulled up to VCC and actively pulled low to turn on. <S> So your RPi code requires you to look for a high to low change, a 0 <S> For on instead of 1. <S> This is inverted logic. <S> This is the typical connection scheme. <S> The 3.3V is connected to the resistors, which pull the lines up (hence, pull-ups). <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You can connect them directly, but you must configure the internal pullup resistor on the relevant GPIOs. <S> Since the pullup is around 50K it won't affect the signals much. <S> The internal pullups are rather high value so if you have a long cable it may pick up noise- <S> in which case each output could have a resistor such as 4.7K to the 3.3V supply of the Raspberry Pi. <S> Edit: <S> In your diagram. <S> Option 1: <S> Configure GPIO16/20/21 to be inputs with internal pullups in your RPi code. <S> Replace 10K resistors with something between 0 and a few hundred ohms (eg. 470 ohms). <S> Don't connect the 3.3 anywhere (it's supplied internally by switches in the Broadcom SOC with series 50K-ish pullups). <S> Option 2: <S> Leave internal pullups off. <S> Replace 10K resistors with (say) 470 ohms. <S> Connect a separate 10K resistor between each open collector output (D-sub pins 1-4) and RPi <S> pin 17 <S> (+3.3V). <S> You can use shorts rather than 470 ohms. <S> Schematic- <S> two outputs shown. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You've already got good answers for some of your points from Spehro Pefhany and Passerby <S> so I won't repeat those parts. <S> I'll just answer (and correct) <S> one of your points which hasn't been mentioned so far: <S> Would this design work as follows? <S> [...] <S> 4. <S> Result: <S> My GPIO "input" pins (1-6) will receive 3.3V whenever the corresponding camera input is selected. <S> This will give the GPIO a "high" reading and my software will know that camera is active. <S> No. <S> Look at this (partial) schematic which you kindly supplied: <S> Assuming that the example LED shown in that schematic, turns on when that camera is active, <S> the transistor shown will conduct to switch on the relevant LED. <S> That means the truth table will be: Output OFF <S> (Camera not active) <S> Transistor not conducting = <S> voltage on connector pin is same as voltage used for the pull-up resistor (internal or external to the Raspberry Pi) <S> - should be 3.3V (logic <S> High detected on Raspberry Pi input connected to that pin) <S> Output ON <S> (Camera active) <S> Transistor conducting = <S> voltage on connector pin is 0V (logic Low <S> detected on Raspberry Pi input connected to that pin)
| Therefore your point 4 is exactly the wrong way around i.e. your GPIO inputs (assuming pull-ups to 3.3V) will be logic Low when the device output is selected and that camera is active, and logic High otherwise. The grounds must also be connected together and there should be nothing else (other than the Raspberry Pi) connected to those pins. If you want to be ultra-cautious, put a 470 ohm resistor in series with each of the outputs.
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Why not produce household electricity with natural gas generator? I don't have an engineering background so I apologize in advance if this seems like a silly question. So I noticed natural gas in my area is a lot cheaper than electricity. So I was asking myself why not get a natural gas generator to produce electricity? My utility company charges .17 kwh for electricity and .35 per Therm/100 cubic feet for NG. A 10kw generator uses about 200 cubic feet NG per hour. So that's .35x2 = .70 for 10kwh. Or .07 per kwh. Of coarse there's the initial costs for the generator too. Am I doing this correct or am I missing something? <Q> This means using the waste heat from the generator to heat your home in the winter. <S> In this case, even if the electrical generator efficiency is only 20%, global efficiency will be close to 100% since you'd use the heat. <S> However this isn't very practical as the generator would only run when heating is required, which may not be at the same moment than you need electricity. <S> Note that the gas use of the generator you quote is most likely specified at full output power. <S> If you only use a small part of the 10kW power, efficiency will drop a lot. <S> If you don't use any power, the engine will idle and consume gas for nothing. <S> Also, an internal combustion engine doesn't last that many hours... <S> If your air conditioning bill hurts, first thing is to add insulation and shade the windows, then you can use solar panels to power your A/C for example. <S> And it doesn't use any gas... <A> Some reasons: Noise. <S> Maintenance of generator, e.g. replacing oil. <S> Repair of generator and not having power when it is broken. <S> Most generators use up natural gas even when you don't need more than a little electricity. <A> Gas generators are the expensive for the use. <S> The energy companies have spare gas turbines ready to be used at any time, this is for immediate start to service in case of sudden rise of power demand. <S> They are intended to work occasionally because of their costs and maintenance, but they do have this property to be launched in operation very fast.
| Considering the investment for the generator, its maintenance, etc, the only way this would be viable would be to use cogeneration.
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Altium Designer - How to Remove Green X's (Error Markers) I often find myself clicking on a PCB component in Altium and accidentally dragging it ever so slightly. Altium then draws a bunch of green circles with X's in them, which I believe means there is an overlap problem (not enough room for the move). See image below: My issue is that even after I press CTRL + Z (undo), the moved item is returned, but the green X's are not removed. This is a problem for reading text on nets and pins, and making screenshots. The only solution I have found is to close the PcbDoc and re-open it, which is in no way convenient. How can the error markers be removed without reopening the file? Bonus question: Is there a way to disable the ability to move the component in the first place? <Q> The direct solution is to go to the top menu: Tools <S> > <A> It looks like there is a polygon pour in your picture. <S> In this case, right-click on the pour and select "repour polygon" which will redraw it and should get rid of the violations. <S> You can also activate auto-repour, but it can slow things down if you have large polygons. <S> To disable moving the component, open its property page and click the "Locked" checkbox. <S> This is quite useful, as you will now be able select stuff like tracks and vias under the components without Altium asking you every time exactly what you want to select... <A> Bonus question: <S> Yes. <S> You can lock the component. <S> For instance from PCB list, with the checkbox column "Locked"
| Reset Error Markers ...Or simply press T then M Running a Design Rule Check will also remove the error markers, but seems to disable them completely until the file is re-opened.
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Why are input logic levels are more tolerant than output logic levels? An input 1 is defined from 2V to 5V and an input 0 from 0V to 0,8V while an output 1 is defined from 2,4V to 5V and output 0 from 0V to 0,4V. Why are output logic levels defined to be more tolerant? Why not the same? <Q> <A> NOTHING is EVER the same! <S> If the levels were the same for both, the slightest change in tolerance would cause some levels to be incorrectly interpreted. <S> The point is the outputs are defined to always be within the range of the inputs to guarantee any receiver (of a compatible type) will understand whatever level is being sent by the sender. <S> Having that tolerance band also provides you with some built in noise immunity, up to 0.4V, on the signal. <A> Why are output logic levels defined to be more tolerant? <S> Why not the same? <S> Rearranging the data from your question: Output Low Input Low Output High <S> Input <S> HighMin 2,4 <S> V 2,0 <S> VMax <S> 0,4 <S> V 0,8 V Figure 1. <S> The problem is "fan-out". <S> Source: Electronics4U . <S> The point is that an output has to be able to drive multiple inputs reliably. <S> Each logic family has a maximum "fan-out" which is determined by the worse of the output high fan-out and output low fan-out. <S> (An output may be able to pull 16 inputs low but only able to drive 11 inputs high. <S> The fan-out specification would be 11 in this case.) <S> In Figure 1a we can see that the output (on the left) is supplying current to each of the three inputs. <S> The effect of this is to cause the voltage on the output to droop a little. <S> The inputs must work at the lowest value and a safety margin. <S> Similarly in (b) <S> the more inputs we try to pull low the more the output voltage will rise. <S> Again we need some safety margin to ensure that the transistors are definitely switched.
| Having a closer tolerance on the output levels than on the input levels ensures that an input will correctly recognize the output of another circuit as being the intended level.
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How can I light up 3 LEDs in sequence but only one at a time? I have 3 LEDs in a row on a breadboard. I want to simulate a visual depiction of a current pulse by lighting up LED #1, then turning that off and lighting up LED #2, then turning that off and lighting up LED #3 then turning that off. How can I achieve this effect? To keep things simple, assume that it is irrelevant whether this process repeats endlessly or if it is activated based on some event. Also assume that only simple components are available, i.e. I probably can't use any component that a PhD in Electrical Engineering would consider "the most optimal component to use in this situation", as that would probably be too complex. The "most complex" components I have available are things along the lines of op amp, transistors (bipolar, FET), comparators, MUX's, and DAC/ADC. Summary of requirements: LEDs must only light up one at a time in sequence Must have a way to adjust the speed at which they light up Cannot do this by mechanical methods <Q> here is a mechanical way <S> just drag the lead across, contacting each LED in turn <S> not meant as an answer (it is the only way to include a schematic) <S> simulate this circuit – <S> Schematic created using CircuitLab simulate this circuit <S> A variation on the them. <S> Use a ball-bearing running on two wire rails - <S> one segmented with each LED connected to a segment - to switch the LEDs in sequence. <A> What you want is an LED chaser . <S> There are a few basic ways this is done. <S> Mechanical. <S> That doesn't lend itself to repeating or automation. <S> Build a circuit of flip flops. <S> Just add a 555 timer. <S> Use a 50 cent micro controller that you can get started programming for under 5 bucks. <S> MSP430, Arduinos, etc. <S> This is how many people would do it today, not just PhDs . <S> How you physically arrange the LEDs has nothing to do with how you control them electronically btw. <A> Even the tiny and cheap PIC 10F200 can do this: <S> This processor has a internal oscillator, so needs nothing more than power applied to run the code. <S> You can program it to light the LEDs in any combination and sequence. <S> For a more tedious, klunky, and less advanced approach, you can use a shift register as the basis for sequencing thru the LEDs. <S> Wire up the shift register so that it is circular, and initialize it with a single 1 bit. <S> Each clock pulse, the 1 output will shift to the next bit. <S> The speed of running thru the pattern is dependent on the clock speed. <A> Use a capacitor to integrate the current to create a linear voltage ramp. <S> Use a set of comparators to turn on the LEDs when the ramp output is between certain voltages. <S> 1) <S> First LED 0V to 1.66V. 2) <S> Second LED 1.66V to 3.33V 3) <S> Third LED 3.33V to 5.0V <S> Connect the negative side of the first LED between to the output of the 1.66V through a 1K resistor. <S> Connect the other side to ground. <S> The positive end of the second LED to the output of the second comparator through a 1K resistor. <S> Connect the other side to the output of the 3.33V comparator. <S> Connect positive side of the third LED to the output of the 3.33V comparator through a 1K resistor. <S> Connect the other side of the third LED to ground. <S> The LEDs will then turn on/off in sequence when you apply power. <S> You can adjust the timing either by changing the 100K resistor or the 100uF capacitor. <S> The basic pattern can be extended to any number of LEDs by just using more comparators. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You don't need amps, transistors, comparators, muxes or dac/adc, just use a (simple) microprocessor. <S> When using a microprocessor: Just connect with 3 different GPIO (generic) pins, one led to a separate GPIO, with a resistor (like 470 ohm) before or after each led. <S> By setting the output of a GPIO pin to either High or Low per pin, you can decide whether to light the LED or not. <S> Use something like this <S> : If you don't have experience with microprocessors, buy an Arduino Uno, the first example is to blink a light ... if you can do that, duplicate the circuit for each LED instead of one LED and you get the idea. <S> There are also tons of moving lights (like the Knight Rider led) which is sort of what you want (but even simpler). <A> Have the 1 light hit a phototransistor, which in turn lights the 2nd LED. <S> This one would shine on the next phototransistor and so on. <S> Having a small capacitor on each leg would look like consecutive pulses.
| Use two PNP transistors to create a current mirror. Use a CD4017 Decade counter, looping the 3rd output to the reset. The obvious way to do this is with a microcontroller. Basically a pre-built flip-flop circuit.
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Powering a router with a USB power bank I have a usb powerbank(w 1.1 & 2A outputs) and I would like to power a router since I have no access to a power source. My issue is the least I've seen routers rated for are 9v min. usually 12v. Im assuming I don't need to worry about amperage because that is built into USB to self-regulate. However I'm worried about voltage. I'm wondering if there is a way around this? I've searched the internet and come up with nothing. <Q> Yes. <S> But check the current required by your router first. <S> Power is voltage x current. <S> You have at most 5 V (USB standard) <S> x 2 <S> A = 10 W. <S> At 12 V, that's ~0.8 <S> A maximum. <S> Probably less with switching losses. <A> What you need is a boost converter, which will raise your voltage from 5V to 9 or even 12V, as you need. <S> The schematics is quite simple, such as shown on picture below. <S> For this project, I recommend you to use an already built one with regulated output voltage, which you will pre-set for the desired voltage, as your router requires. <S> You can get them cheap on largest retail store on the web. <A> One possibility is to save your existing power bank for other projects, and get a power bank with Quick Charge 2 or QC 3. <S> These can be tricked into delivering 9V or even 12V for some models, by using a so-called Quick Charge trigger, a cheap device that you can buy online. <S> You will need to fashion your own power adapter from USB to whatever power plug the router has. <S> And of course you need to do some tests to ensure that: the trigger is doing what you want it to do (namely tricking the power bank into delivering whatever voltage you want) <S> the power bank is actually able to deliver the voltage you need at the current you need, without significant ripple - power banks vary in this regard and ultimately that the router is comfortable with the whole situation <S> When all this is done, you might still want/need a USB voltage meter in order to verify that you are selecting the correct voltage at start up, unless the QC trigger has integrated indicators that can reliably provide you with this information.
| You need a "switching boost regulator".
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How does a DC clamp meter measure current? I understand that for AC clamp meters, a changing magnetic field induces a voltage/current in a wire loop and therefore it is possible to measure AC currents by induction. But how does a DC clamp meter work? In essence a DC current means a static magnetic field, and by Faraday's Law, no voltage/current is induced. <Q> This is one way of doing it. <S> Ref <S> A normal transformer can't deal with DC currents. <S> Therefore the operating principle of DC current probes is rather different from AC probes. <S> Here also is the current carrying conductor the primary winding and is inserted through the core opening. <S> There is also a secondary winding, but now it functions as a compensation coil. <S> The core is provided with an air gap that holds a sensor, e.g. a hall-sensor, which measures the magnetic flux in the core. <S> The current in the primary wire will magnetize the core. <S> This magnetic field is measured with the sensor and as a result of this, the control circuit runs a current through the compensation winding in a way that the magnetic flux in the core is kept zero. <S> As a result of this the core will never be magnetized. <S> The advantage is that the non-linear properties and hysteresis of both the core and the magnetic sensor have little influence on the measurement results. <S> Beside the described measurement circuit there is also demagnetization circuit. <S> Before using the current probe the core must be degaussed without wires inserted in the core. <S> Gain can be increased with compensation turns ratio at the expense of bandwidth. <S> Side Note <S> True RMS RF power meters work something like above, except instead of magnetic loop matching, they use a thermal resistor and using a bridge to compare RF heat with a DC heater for accurate measurements and obviously much lower bandwidth from thermal time constant, but very accurate. <A> Static magnetic fields can be measured with an Hall effect sensor . <A> Feeding that pulse to an integrator gives the current because the pulse voltage is equal with the flux growth velocity through the winding and the final flux is proportional to the current.
| If an AC clamp is put around a wire with DC current, a DC pulse occurs at the output of the winding of the clamp.
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What is a good PCB that you can fit with 8AWG wires? Me and a couple of other students are working on a school project that is using voltage up to 100v and 40amps. So I figured the that 8AWG wires are the ones to to use for this. They are pretty thick wires compared to the wires I would use for an Arduino. So since they are pretty thick, it does not look like they can be fitted onto regular PCB. Anyone know where I can find PCBs to that can fit them? <Q> Using your favorite Internet search engine perform a search using the keywords "pcb wire-to-board connector". <S> You'll discover multiple companies that manufacture (e.g., Molex, AVX, TE Connectivity, etc.) and distribute/sell ( <S> Newark, Mouser, Digi-Key, etc.) <S> wire-to-board connector products. <S> Plan B would be to visit an electronics parts distributor website (Newark, Digi-Key, Mouser, etc.) and use the site's search tools to find the connector products they sell. <S> You can also request a printed catalog from a parts distributor (Newark, Mouser, Digi-Key, etc.). <S> Sometimes it's easier to find the parts you want by perusing a distributor's printed catalog versus searching the distributor's website. <S> For what it's worth, if wire flexibility is a concern, have you considered using multiple smaller gauge wires, rather than using a single 8 AWG wire, to deliver the 100V/40A power, where each wire carries only a portion of the 40A maximum current? <S> For example, if you create an eight-wire harness where each wire is the same wire gauge, then each wire carries roughly 40A/8 = <S> 5A. <S> Each of these eight wires can be much smaller <S> , the mounting holes in the PCB can be smaller, the wire-to-board connectors can be smaller, etc. <S> If you do this you'll need to be careful about how the individual wires are bundled together, encased, etc., to manage and prevent unwanted heating. <A> If I had to build something that required #8 wire, I would do my best to keep the #8 wires away from the PC board. <S> To safely handle 40 Amps on a PC board, you would need very wide traces, and very thick copper on the board <A>
| I would use a relay, or contactor, which would enable you to use a smaller wire on the PCB and route that wire to the relay, mounted remotely.
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Power connectors: 2.1mm and 2.5mm, 1.35 and 1.7mm I refer to DC power connectors used for a wide variety of common consumer devices. I notice that I have to specify between 2.1mm and 2.5mm when I want to purchase these online. Are the two interchangeable or is the tolerance so precise that 0.4mm makes a difference? Separately, I have an existing 5V adapter with the smaller connector. I can't tell whether it is 1.35mm or 1.7mm. I am trying to purchase an adapter to convert it to 2.1mm/2.5mm. Same question: does it matter whether it's 1.35mm or 1.7mm, or is there a good way to find out what my connector is? <Q> Yes, it matters. <S> In the 5.5 mm O.D. connectors, if you use the 2.5 mm I.D. plug in the 2.1 mm jack it will not make good contact because the outer housing of the jack may not allow the plug to be displaced far enough for the center to make contact, and in any case the spring will not be exerting the proper amount of pressure since the plug has less thickness. <S> It may work but it will not be reliable. <S> In the reverse case, the plug will not fit over the center pin of the jack at all. <S> I am not familiar with the smaller connectors but I expect that you would find similar results. <S> If you are choosing a power jack for use in your design, and you are not looking for a specifically unusual one, I would recommend choosing the 5.5 mm O.D., 2.1 mm I.D. size. <S> In my experience it is the most common in hobbyist electronics and related areas. <S> Due to their use in the CCTV camera field, you can readily obtain both male and female adapters to screw terminals, which are handy for making temporary connections either to a power supply or device where the other end does not have a barrel plug/jack at all. <A> DC barrel connectors are so tight that the outer diameter and inner diameter must both match between plug and jack. <S> Otherwise they won't fit at all (2.1mm plug, 2.5mm jack) or they swim loosely ( <S> 2.5mm plug in 2.1mm jack). <S> The only spec that is slightly interchangeable is depth. <S> But again, you can get really short ones that won't fit in a jack designed for longer ones. <S> As to a size adapter, don't bother. <S> You won't likely find one for arbitrary sizes. <S> Instead, cut off the old connector and replace it with screw or spring or push in type connector. <S> Cheap, easy, convienent. <A> https://forum.digikey.com/t/measuring-power-supply-barrel-plug-id-2-1mm-vs-2-5mm/401 <A> Yes, it matters <S> The dimension you refer to does matter -- 400 microns of center pin diameter is the difference between "fits" and "doesn't fit at all". <S> A pair of decent calipers is the correct tool for the job when it comes to measuring such small dimensions, by the way.
| “Barrel” connectors work by having exactly one spring (the outer contact of the jack) which pushes the plug sideways so that the center makes good contact as well as the spring. It's pretty hard to measure the inner diameter (ID) using a pair of calipers given the relative size difference. Here is a wonderful post by Digi-Key about using a toothpick, a pen, or multimeter test lead to quickly determine if it's 2.1mm or 2.5mm.
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How does LED matrix circuit work in the following example? The circuit below shows how to build an LED matrix with shift register multiplexing. From what I understand, R1-R8 is the column vector setting the LEDs high or low and the shift register will pick one row at a time. What is the output of the shift register Q0-Q7? In order to light up bottom left LED, I Would need to set R1 3.3V and Q0 grounded (or 0V I guess?) whilst Q1 - Q7 must be floating. However, in this circuit, I don't see how to float the Q1-Q7. Or are Q1-Q7 actually 3.3V so that all other LED in the first row have zero potential with the Q1-Q7? In the example below, it's more understandable as the shift register is switching the transistors, therefore, multiplexing the rows to the ground. A - How does the first circuit work? B - What are advantages and disadvantages of the first and the second circuit? <Q> ... <S> whilst Q1 - Q7 must be floating. <S> No they don't. <S> They simply must not provide a voltage that will light the LEDs to any real extent. <S> Pulling them low, and therefore the anodes to 0V, suffices. <A> If you examine the first circuit closely you can see that to turn on an LED you need a low from the Arduino GPOI line and a high from a single output of the shift register. <S> So to light the bottom left LED you would output a 0 (low on pin(13) and a 1 (high) on Q0, all the other Qx outputs would be 0(low). <S> Effectively the shift register allows you to select a single column from the 8 in the display, and the Arduino pin(x) selects an individual LED in the selected column. <S> The second circuit you show is incomplete, but operates in the same manner, this time using a FET to sink the current for an LED to ground. <A> The data sheet for the SN74HC494N shows the part's timing diagram. <S> Basically the Q0 - Q7 outputs are a "marching one" sequence: 0 <S> 7 clk10000000 001000000 <S> 100100000 <S> 2... <S> So with Q0 logic HIGH, you would need to program pin 13 on the Arduino as logic LOW in order to turn on the LED that's located on the bottom left corner of the matrix: <S> +5V ->- <S> > 74HC595N VCC 74HC595N Q0(HIGH) <S> ->- <S> > <S> LED Anode <S> LED Cathode ->- <S> > <S> 220 ohm resistor R1 ->- <S> > <S> Arduino pin 13 <S> (LOW) <S> Arduino GND ->- <S> > <S> GROUND <A> In this type of circuit you need to think of the shift register as a power-router. <S> Lets redraw it a bit to make it clearer. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Perhaps you can see in the above that the two parallel circuits are driven by two common data lines at the bottom. <S> Now, obviously, which set of LEDs light up will depend on which circuit it powered. <S> However, in this instance "powered" actually means. " <S> At a voltage greater than the Low level signal level applied to the data lines PLUS the diode voltage drop." <S> As such, the HC595 outputs, when high, provide that power to the appropriate circuit. <S> When the outputs are low, the voltage level is too low to turn on the LEDs in that circuit. <S> This type of circuit is used to multiplex many LEDS into a few IO lines from whatever is driving it. <S> Each row or column will only be on for 1/(row or column count) of the time. <S> As such, you lose some brightness from the LEDs.
| What you really have is a number of circuits hooked up in parallel.
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why is a GROUND electrode needed with a referential amplifier? Physiological amplifiers seem to often (always?) have both ground (G) and reference (R) electrodes along with one or more active (A) electrodes.Typical voltage at an active site is determined AFAIK as [(A-G) - (R-G)] = A-R.I haven't managed to find out why the ground electrode is needed at all and why the amplifier cannot simply do the A-R calculation directly without the need for the ground electrode? Is it because when there are multiple active electrodes that A1-R and A2-R couldn't be done without them interfering with each other? <Q> Differential amplifiers such as instrumentation amplifiers, with or without unity gain buffers, have a certain working range for the common-mode voltage. <S> In general they will stop working if either input voltage gets too close to the power supply rails. <S> For example, in the case of the INA117 , both input voltages must be within the range of V+-4V to V-+4V, so if you have +/-10V supplies, each input must be within the range of <S> +/-6V. <S> There are also internal nodes in instrumentation amplifiers which can saturate. <S> Here are some curves for the INA118 which show the permissible input voltages: <S> The reference ground makes sure that the inputs remain within the common mode range for proper operation and minimizes the common mode AC voltage. <S> The latter will show up in the output even if the inputs remain within the common mode range, attenuated by the common mode rejection ratio (CMRR). <A> After looking at some schematics: (Google: "physiological amplifier circuit" and select Images tab), this is one example: <S> Many amplifiers do use A-R directly similar to instrumentation amplifiers. <S> safety: since the amplifier must be mains isolated (to prevent electrocuting the patient) any charge build-up cannot escape without a ground connection. <S> If a charge were present it could escape via the patient, not harmful but still unpleasant. <S> safety: if the mains isolation fails on the amplifier then the ground connection prevents the mains voltage from reaching the patient. <A> The reason is actually because of the buffers. <S> A simple differential amplifier does not care about the reference point of the input. <S> Without any other common reference, all it cares about is the difference between the positive and negative inputs. <S> The difference itself of course has to be within the tolerable range of the inputs. <S> Notice in the circuit below, the output is correctly -1V. simulate this circuit – <S> Schematic created using CircuitLab <S> Why is that? <S> Generically, the inputs stage of all op-amps basically looks something like this... <S> That is the input is a balanced circuit. <S> That balance is disturbed by any applied voltage/current difference between the two inputs. <S> The reference point for those inputs is, in reality, the junction between those two emitters. <S> As such the applied voltage need not be referenced to ground as long as the two inputs are referenced to each other. <S> You should be able to see, how attaching a battery between the plus and minus pins in this circuit should be acceptable. <S> The buffer on the other hand needs a reference. <S> Obviously the circuit below, attaching a voltage with no reference, to a unity buffer makes no sense. <S> simulate this circuit <S> But how about this.. <S> simulate this circuit <S> In fact what you end up with here will depend on the op-amp. <S> Some may get you what you expect +0.5V on one -0.5V on the other, some may rail out to either or both sides. <S> All will be very sensitive to any noise. <S> So <S> what does all that mean. <S> simulate this circuit <S> It means a standard differential amplifier measures the difference voltage <S> \$Vr\$. <S> An instrumentation amplifier measures \$V_a - V_b\$. <S> Mathematically that may be the same, electrically, it is not. <A> Touch the tip of a 10MOhm scope probe, and you'll likely see 100 or 200 volts at 50 or 60Hz; you are the antenna, the 2nd plate of the capacitor gathering charge from all the power line wiring around you. <S> A physiological amplifier has that same input --- 100 or 200 volts--- plus the tiny signal of interest. <S> That GND (to your ankle) greatly reduces the undesired voltage.
| The ground connection might be needed for: shielding around the A and R signals to prevent coupling to external noise like other equipment.
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How to remove turn on ringing/overshoot in MOSFET output for 1 MHz and 70V Switching? I have to make a pulse generator square wave for ultrasonic transducer with frequency 1 MHz and 70V for the amplitude, the input is function signal generator with 10V VPP output and 1 MHz. I have made it and it works, but there is a problem. There is always a ringing or overshoot in turn on or rise time but not for the fall. It happens for all frequency, not just in 1 MHz, in 100 KHz for an example, the overshoot is so tight or narrow. this is the circuit that i made and this is the result of the output what i have done1. try to add resistor 1M or 10K in gate of MOSFET, the result : no output signal2. Add Decoupling capacitor, the result : no effect UPDATE how i add up the resistor So, how to fix the overshoot or the ringing of the output? Thank you <Q> what i have done 1. <S> try to add resistor 1M or 10K in gate of MOSFET, <S> the result : <S> no output signal <S> You've done that what you shouldn't do! <S> If you check the datasheet then you'll see that the input capacitance, \$C_{iss}\$, is quite high: 2.16nF. A 1M gate stopper resistor plus this capacitor will form a nice low-pass filter having a cut-off frequency of \$f_C=1000/(2\pi <S> \cdot 2.16 <S> \cdot 1) = 73.7Hz\$, so the input pulses will be totally chopped off thus the MOSFET will never turn on. <S> That's why you get no output. <S> Cure: <S> Remove 100k and change 1M to a resistance so that it forms a LPF having a cut-off frequency of at least \$3 <S> \cdot f_{SW} = <S> 3MHz\$. <S> Add Decoupling capacitor, the result : <S> no effect Of course! <S> Any inductance causes ringing. <S> Where do you have inductances? <S> Answer: Cables and internal drain inductance of the MOSFET. <S> Cure: <S> A snubber network across Drain and Source/GND. <S> You can find a lot of info about snubber design on Internet. <S> PS: Your MOSFET does not seem to be suitable for switching at 1 MHz. <S> You should use another like this one <S> (I picked this one randomly. <S> You select according to your voltage and current needs). <A> Layout is very important to reducing ringing in gate voltage. <S> So think short traces thicker traces can also help with this a bit. <S> The other part of reducing gate ringing is slowing the transition time down which is what the other comments/answers are trying to accomplish with the resistors. <S> It could also be a measurement artifact if you are not using the ground spring clip on your scope <A> I just wanted to add that there will be some inductance between the power source and the drain of your MOSFET (and also on the return side). <S> Remember that the faster you switch the load and the larger your current, the larger the voltage spike you'll see. <S> $$V = <S> L\frac{di}{dt}$$ <S> Where V here is the voltage across your parasitic inductance. <S> Wide traces should be used with heavy weight copper if you're switching large loads quickly. <S> This voltage will appear in addition to your source voltage (70V). <S> Less resistance in series with the gate will cause faster switching and more resistance will cause slower switching. <S> Keep in mind the gate of the MOSFET will need a gate driver if you want to switch loads very quickly and reduce power dissipation in the FET. <S> In addition to reducing the inductance at the MOSFET drain, you should also try and keep the loop distance small connecting you power sources positive and return terminals.
| You want to optimize for low inductance on the gate drive trace.
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Why is the neutral pin hotter than the live pin? I frequently use an Immersion Rod Water Heater (shown below) and I have noticed that the after unplugging the heater, the neutral pin of the 3 pin plug is quite hot (about 5 degrees more - but not very hot) compared to the live pin. How is this temperature difference possible and why? <Q> I just had one of those last week. <S> The plug was hot, and the socket was hotter. <S> I'm the electrical guy <S> so I popped it off. <S> #12 stranded wire shoved into a smaller #14 backstab hole. <S> Two thirds of the strands had missed the hole! <S> Dumb things like that happen all the time, and you have to nip 'em in the bud the moment you see them. <S> Don't use this outlet until it's fixed. <S> It probably feeds thru to other outlets on the same circuit, so check this outlet while using those. <S> Get it fixed ASAP. <S> If I had used the next outlet over, I would've never found it. <S> It can start a fire. <S> (We use steel junction boxes, that helps.) <A> The voltage on each pin, live or neutral, is irrelevant. <S> Also the current flowing through each pin will be the same. <S> The difference is the resistance of the connection the current is flowing through. <S> It's not unusual for heavy current users like kettles to make the pins slightly warm. <S> However a cup heater like this takes very little current, so a warm plug pin is warning that the socket could need replacing. <S> With a heavier current using appliance, it could get hot enough to cause a fire. <A> To put all that was said in a simpler form, in case a layman stumbles upon this question: <S> This is an indication that the plug/cord, the socket or both are defective and need to be checked ASAP, since there is a potential fire hazard. <A> The high resistance of the plug connection is often due to oxidation on the surface of one of the mating pins/receptacles. <S> Regardless of which one it is, the heat will often cause oxidation on the other surface which also causes undue heating. <S> If this is now plugged in elsewhere, this heated surface encourages oxidation of the new mating surface and so on and so on. <S> It is the same epidemiology as zoological venereal diseases. <S> Terminology is similar, mating, male connector, female ....
| This will be caused by the springs in the socket not making good contact with the plug pin.
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How do generate a negative supply for op amps ICs? I know if you reverse + and - on a voltage supply you can measure a negative voltage using a multimeter, but for an op amp they only have one negative pin so I don’t think reversing + and - would work because you would be connecting the - voltage terminal to ground, I’m a little bit confused about how to achieve this, could you please explain how to get a - supply railThanks, Be <Q> Some type of switching supply is the easiest approach if you only have a positive rail available and actually require a negative rail. <S> There are easily available DC-DC converters that will accept a single DC voltage input and produce (say) <S> +/-15V rails that are galvanically isolated from the input. <S> It's easier for your meter because it has an inherently isolated supply in the form of the battery. <A> You may find this thread helpful. <S> Negative voltage from Arduino? <S> I would suggest you to do it with two batteries as shown below. <S> However, your question lacks detail, so I can't recommend what would be the best possible solution for you. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If you are going oldschool you have to have two transformers or one with 3 output leads of which one is in the smack middle of secondary winding. <S> Usually op amps require +/- <S> 15 V DC. <S> Now all you need is 7815 and 7915 ICs.
| Use a DC-DC converter, a charge pump capacitive inverter or an inverting buck converter.
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Does a voice coil vibrate at the same frequency as the source sound file? As a disclaimer I know very little about this so I'm probably describing this in a very imprecise way. Say you generated a sine wave on your computer at 800Hz and played it through a speaker. Would the speaker also vibrate at 800Hz? <Q> I would expect so, assuming the speaker is capable of vibrating with the same speed. <S> Mostly bigger speakers are better suited for vibrating at lower speeds (and having more 'pressure', more air that is moved), small speakers can resonate faster (thus higher frequencies). <S> That is why for lower frequencies you need bigger speakers, and for higher frequencies smaller speakers will do. <A> An ideal speaker will vibrate at whatever frequency of signal you drive it with. <S> However, in reality, there are a few constrainsts. <S> The amplifier output you are driving it with has a frequency range it can amplify. <S> Extremely low frequencies, including DC, and high frequencies will be severely attenuated in the amplifier. <S> The physical construction of a speaker also has a frequency characteristic. <S> It will transfer some frequencies, and harmonics, much better than others. <S> The speaker can resonate. <S> That means the vibrations you introduce with the signal can set up a resonance with the natural vibration frequency of the speaker making that nasty rattle sound you sometimes hear. <S> This resonance can also be induced by harmonics in the sound output. <S> The voice-coil also has an inductance and capacitance, which again, produces a different amount of movement depending on the frequency of the signal. <S> The physical size of the speaker matters. <S> The speaker not only has to vibrate, which means inertia is important, but it also has to "push" air, which also has inertia. <S> That inertia makes it harder to push the faster you try to, which is one of the reasons speakers get smaller for higher frequencies. <S> Because of the inductance and the mechanical inertias, the movement of the voice-coil and speaker cone, and ultimately the wave-front in the air, lags the drive signal. <S> It is at the same frequency, but out of phase with it. <S> An 800Hz sine-wave should make it through to the speaker pretty much unmolested. <S> An 800Hz, square wave, will have a far more interesting output. <S> A square wave is actually a sine wave summed with an infinite series of increasingly smaller harmonics. <S> As such, each frequency is attenuated differently as it passes through the amplifier and speaker. <S> The higher frequencies are "lost" entirely, and some of the lesser harmonics can, and do, cause the speaker to resonate. <S> Which is why square waves sound so "tinny". <A> It depends how you define "vibrate at 800Hz". <S> If (for example) the signal is strong enough so the voice coil moves beyond the constant region of the magnetic field (nonlinear response), then the motion will not be sinusoidal. <S> It will still be a periodic motion with period 1/800Hz but, as Trevor describes, this will be a sum of sinusoids of various frequencies.
| As far as I know each speaker has a range in which it can vibrate (usefully).
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Differential to single ended I have a differential signal ranging from 0 to 2 MHz and need to transform it into a single-ended signal. I found out that there are 3 methods: Transformer Amplifier simply ground the negative part with a matching load How can I decide which method to use? I mean I would just take the load-to-ground method but I guess it's not that simple to decide. Edit: The transformer method would fall out, because DC cannot pass, although this would be no problem. <Q> 1.Transformer <S> If your signal is a single, or narrow band, frequency this method is fine. <S> However, anything else and you introduce the transformers frequency response into your signal. <S> 3.simply ground the negative part with matching a load <S> This method is ok, IF the signal source is close to the receiver. <S> Since you are are throwing away the common mode noise reduction that is the reason the signal was sent differential in the first, this method should be limited to part to part connection locally on a PCB. <S> 2.Amplifier <S> A differential amplifier is ultimately your best choice. <S> Not only will it be easier to control the frequency response of the receiver, but you also take full advantage of the common mode noise removal. <S> Further, this method decouples the input signal from wherever the signal is headed to giving you the appropriate output impedance to feed the next stage. <A> DC to 2Mhz is differential amplifier territory, they can have excellent phase and gain flatness over that sort of range combined with reasonable CMRR (This tends to fall with frequency). <S> Have a look at the high CMRR parts from THAT Corp, they are aimed at audio but IIRC have plenty of GBP and usefully bootstrapped common mode impedance. <S> 300KHz <S> to 2MHz is easy (and compact) for a transformer to handle, excellent CMRR and common mode range, and over a mere decade of bandwidth phase and gain should be flat. <S> These are however best in fairly low impedance circuits, so not what you want if trying to measure a high Z source. <S> Minicircuits have what you need among others. <S> The resistor to ground is usually something you see at the line driving end rather then the receiver as it buys you no CMRR at all in the receiver while helping with the impedance balance in the driver. <A> To add to existing answers, a typical differential signal has the same signal with opposite phase on both signal lines. <S> Your solution 3 has the disadvantage of giving you half the signal amplitude which a differential signal would see, because you only have one signal line. <S> Professional audio has a workaround for this, because audio applications frequently need to drive single-ended sources such as guitar amplifiers. <S> One half of the differential signal has the actual signal, and the other half is simply driven at 0V - that's driven by the same output impedance (typically the same op-amp) as the "signal" half, not simply connected to 0V. <S> The screen of course is connected directly to 0V. <S> A differential input will subtract the 0V reference signal line from the signal line, which removes common-mode noise from the signal but does not otherwise affect the signal amplitude. <S> A single-ended input will simply see the signal line and the 0V, which does not allow noise to be removed of course, but does not change the signal level. <S> This is fine for your signal. <S> It does have a downside though (as does everything). <S> Equal-and-opposite signals on a twisted pair radiate very little, which reduces their effect on equipment around them. <S> A single "hot" signal line does not have this advantage, and this could be an issue as you head up into MHz territory.
| What to do depends on your design parameters (Such things as source impedance, common mode range, required CMRR...).
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Length measurement of a moving hot deformed bar with a sensor In a hot rolling mill moving deformed bars are cut according to the desired lengths, which sensor is suitable to use for measuring the length of the moving hot bar and sending the instruction to the shear cut to apply the cut, e.g. 40 feet length is desired from a 240 feet long bar, keeping in view the temperature factor. <Q> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> Use a through-beam sensor positioned 40 feet from the cutter. <S> When the infeed reaches the sensor then actuate the cutters. <A> Feeding a known material is best done with a drive roller in conjunction with a metering roller. <S> The two are arranged to "pinch" the material as it passes such that when the material is present the drive roller/material turns the metering roller. <S> Metering encoder can be an actual shaft encoder or some sort of hall effect sensor detecting details on the metering roller. <S> Slipping can be minimized by using an appropriate spring force and if need me adding knurling or other surface "teeth" to the roller. <S> I'd also use an additional sensor on the tension arm to detect when the stock is in the mechanism. <S> Note 1: <S> Some tweaking can be required to get this to work consistently especially if the uniformity and speed of the material can cause the tension arm to bounce. <S> Your vague indication that the material is somehow "deformed" may make this a non-starter. <S> Note 2: <S> A small change in diameter over 40' in length will add up quickly if the idler is not very large. <A> I would suggest that the rotation of one of the rollers can be used : avoids any temperature issue , but slip may be a factor...
| Using a through-beam sensor to detect cut-point of steel stock. Depending on how hot.. hot is, some adjustment or compensation may be required to compensate for thermal expansion in the metering roller which will take heat from the material.
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Prevent corrosion of circuit board fallen in water I apologize in advance if this question is off the topics answered on this forum. Some hours back my drone fell into some dirty water. I was able to rescue the device, unplug its battery, open up its circuit board, dry it thoroughly using a hairdryer and finally putting it in place. It worked fine afterwards.But my question is how I could prevent or rather reduce the chances of the circuit board been corroded soon? <Q> You need to wash it. <S> Seriously, what will cause corrosion is any salts that were dissolved in the dirty water that have remained behind when you air dried it. <S> You need to buy some distilled water and repeatedly wash the board a few times in fresh distilled water to dissolve and remove as many of those salts as possible. <S> Flushing under components may require the use of something like an ear syringe, or as Peufeu suggests, a dental water jet. <S> Then, once the thing has had sufficient time to dry, and if there is a risk of this happening again, you should coat the board in a urethane, or better.. <S> epoxy, so this will not happen again. <S> TIP: <S> It is an old trick to stick your board inside a bag with a desiccant for a few days to completely remove any moisture trapped inside hidden places on your board, wires and other components. <S> Rice apparently, despite the rumours, is not a great choice and there are better desiccants just don't use a salt, and keep the desiccant itself from directly touching the parts. <S> NOTE: <S> Salts + <S> humidity = electrolyte. <S> Not only will that corrode stuff, but it also conducts which can cause shorts, arcing, and function/part failure. <A> If there are still ions (salts) on your board, any humidity which condenses on your board will become conductive, and corrode the copper away through electrolysis. <S> My favorite tool for washing a board is a dental water jet. <S> It's great for getting junk out from below SMD chips. <S> If you don't own one, you can also use a spray with the nozzle set to make a jet, but it is a bit more exercise. <S> If you use tap water, there will still be a few ions on the board, though. <S> So it would be best to use distilled or demineralized water. <S> As an alternative, you can use compressed air (or canned air) to blow most of the water off. <S> You can also spray some isopropyl alcohol to blow the water off, but this wastes a lot of it. <S> Make sure you also look at the connectors and motors in your drone, unplug all connectors so the contacts dry, etc. <S> The hair dryer is nice because it will blow the water off too, and dry the hard to reach places, but you don't need to dry things ASAP. <S> So rushing to unplug the battery probably saved your board. <S> You can use spray-on conformal coating on the board to protect it from water. <S> This is easiest to remove compared to epoxy, if need be. <S> However, be careful about the connectors... <S> the stuff is insulating, so if it gets into contacts, potentiometers and the like, you're in trouble. <A> It is so called conformal coating. <S> High end electronics for industrial, automotive, ...use are coated with a varnish. <S> Could be based on epoxy, silicone and other based upon the final use. <A> If you don't want to coat your stuff in a hard potting compound (epoxy) or conformal coating (acrylic) <S> then you can smear/immerse it in dielectric grease. <S> Amazon has something like this readily available, <S> But really, any "waterproof marine electronics grease" will work.
| You should wash the circuit in isopropyl alcohol and then dry it, as the water can contain salts. A circuit can stay wet for a while without anything bad happening if it is not powered .
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Control 5 V load with transistor 3.3 V - Raspberry Pi I'm currently having this project where I put a Raspberry Pi, batteries and a touch screen inside a book that can fit my pocket. This screen is constantly on, and I want to be able to turn it on and off with a transistor. I'll be controlling the 5 V instead of ground because it is also getting ground over HDMI. The base will have 3.3 V, and the output should be 5 V. I've tried many things with both NPN and PNP. I'm running a Python script that outputs 0.02 V as LOW and 3.3 V as HIGH. I'm ending up with the screen being either on/blinking, grey/grey, grey/off or off/off. It stays totally off with 5 V with PNP transistor. This is curcuit that has almost worked (grey/off): When GPIO is HIGH (3.33 V): Monitor: Grey Base: 4.32 V Collector: 2.21 V Emitter: 5 V When GPIO is LOW (0.02 V) Monitor: off (no back-light) Base: 4.38 V Collector: 1.6 V Emitter: 5 V I tried with two 2N3906 transistors to make sure it wasn't broken. I'm kind of confused over this circuit and it seems that my knowledge doesn't match. What can I do to make this work? What am I missing? <Q> Figure 1. <S> In this example Vss is greater than the 5 V supply of the micro-controller. <S> The protection diodes keep the transistor always on. <S> Figure 1 shows the internal schematic of a 5 V powered GPIO in "output" mode. <S> A pair of transistor switches pulls the output high or low. <S> (Only one can be turned on at a time.) <S> Note the internal protection diodes. <S> This will keep the PNP transistor permanently turned on and may damage the chip. <S> In your case your micro is powered from +3.3 V and Vss <S> is +5 V. <S> The result is the same, as you have discovered. <S> Figure 2. <S> To drive a high-side transistor from a GPIO pin we need a level translator. <S> An NPN transistor does the job nicely. <S> Note that Q2 inverts the logic so you may need to modify your code to suit. <S> Links: The images are mine and more on the topic can be found in the article GPIO high-side driver fail . <A> There is a standard way to do this with a P-channel MOS transistor (PMOS). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This is the standard way of doing it. <S> You may or may not need to add C1. <S> If the load has capacitance, then sometimes when you turn on M1, the 5V rail will suddenly dip, and that can cause problems for anything powered from 5V. C1 can help with that just by being a bigger capacitor. <S> Q1 does not have to be an NPN. <S> You could use a logic-level N-channel FET such as a BSS138. <S> There are lots of choices for M1. <A> <A> As others have already noted, the 3.3 V from the digital output is not high enough to turn off a PNP transistor with emitter connected to 5 V. <S> There are some simple ways around this. <S> Here is something really simple that would work in a pinch: <S> The two resistors form a voltage divider so that the E-B voltage is held at less than 400 mV when the digital signal is at 3.3 V. <S> When the digital signal is low, the transistor is driven with about 2.3 mA base current, and the digital output has to sink 5.2 mA. <S> This is rather inefficient in the use of the possibly limited current sink capability of the digital output, and it leaves some current thru the protection diodes when the transistor is off. <S> Here is a much better approach, but slightly more complicated: <S> This is similar to what others have suggested, but simpler and uses the digital output source current more optimally. <S> Q2 acts as a switchable current sink. <S> Figure about 700 mV for the B-E drop of Q2, <S> so 2.6 V across R1, which results in 9.6 mA current thru R1. <S> If the transistors have a gain of 50, then that results in 9.4 mA base current from Q1, which would support up to 470 mA of load current. <S> The current sourced by the digital output is only 190 µA, which any remotely normal digital output can do easily.
| The protection diodes on most logic chips creates a sneak-path to positive supply. I also got stuck on this until I had a moment of inspiration, instead of controlling the power to the screen with a transistor circuit I now control the video signal as my control board for video shuts down when there is no signal present. A BSS84 might work if the load current is low.
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capacitive reactance in a stranded wire Can a stranded wire exhibit a capacitive reactance? I measured the real and imaginary part of a piece stranded wire using an old LRC meter( HP 4284a). Applied an open loop and a closed loop compensation for the parasitic impedance of the lead wires.Measured reactance is negative. This left me puzzled, is this phisically possible? I know that a real conductor will display an inductive behaviour because of self inductance, with positive reactance, increasing with frequency.Could a stranded geometry cause a negative reactance to be measured? Or is this more likely a wrong calibration issue? <Q> All wires have a distributed inductance and capacitance and a resonant frequency which becomes a short circuit at 1/4 wave and beyond this inverts the wire reactance and impedance rises to the free space impedance at 1/2 wave length and then repeats. <S> This is the characteristic of a whip antenna. <S> So it depends on the length of wire and what f was used in the test. <S> So a single wire is fairly low pF /m <S> compared to loose twisted pair which is ~ <S> 50pF/m. <S> Conclusion: measurement error or at least ill-defined measurement conditions. <S> Although proximity of wire to hand or any dielectric that is grounded by stray a capacitance will affect pF/m. <A> No, a piece of wire can't exhibit a capacitive reactance, a guess. <S> I think there is a problem with meter/connection. <S> What results are measured while closed-loop compensation?What output meter gives at 'short connection' after 'closed loop compenstaion'?What output meter gives at small coil with known inductance?What output meter gives at the stranded wire? <A> This is a 'small difference between two big numbers is meaningless' issue. <S> Your meter's open and short calibration will be slightly contaminated by noise, as will your measurement. <S> Your corrected reading is therefore dominated by measurement noise, and is approximately zero, plus or minus some noise.
| The wire is inductive based on ratio of length to diameter and capacitive based area/gap ratio with the ground signal.
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Why wouldn't the diode in this transistor scheme turn on? In the schematic above, the diode should not light, but why? (Ignore the pencil drawings on the schematic). <Q> The transistor is off since it's <S> base-emitter junction is reverse-biased. <S> That allows no current to flow thru the collector, which keeps the LED off. <S> The B-E junction looks like a diode to the external circuit. <S> The special thing that a BJT (bipolar junction transistor) does is allow C-E current some multiple of the B-E current. <S> That multiple is the gain. <S> Since you have the B-E "diode" reverse biased, no current flow thru it. <S> That, in turn, does not allow for any collector current. <S> Being a silicon diode, the B-E junction drops about 600-750 mV when conducting useful current. <S> Depending on the value of the base resistor, you could start to see the LED come on with the input being about 1 V or more. <S> Let's work thru some numbers to pick good values for the resistors. <S> Let's say this is a common green LED rated for 20 mA, and that it drops 2.1 when fully on. <S> The C-E drop of the transistor in saturation is about 200 mV. <S> After the LED and the C-E drop, that leaves 6.7 V across the LED resistor. <S> We want 20 mA to flow, so use Ohm's law: ( <S> 6.7 V)/(20 mA) = 335 Ω, which is the minimum resistance to not over-drive the LED. <S> I'd round that up to at least the next common value of 360 Ω to give some margin. <S> Working backwards, the LED current would be 18.6 mA. <S> You wouldn't notice the difference in brightness between that and 20 mA, even in a side by side comparison. <S> Let's require the transistor to have at least a gain of 30, which is very easy to find in a small signal NPN. <S> You want 20 mA collector current, so you need at least (20 mA)/30 = 670 µA base current. <S> Let's assume you want to drive this LED from a typical digital output that is either 0 V for off and 3.3 V for on. <S> Figure <S> the B-E junction will drop 700 mV, so that leaves 2.6 V across the base resistor. <S> Again using Ohm's law, the base resistor needs to be no more than 3.9 kΩ. Round down to the next common value of 3.6 kΩ. <A> The transistor is working in cut off region here. <S> So no current is flowing in the collector circuit. <A> A mental model for the semantics of an (NPN) BJT is an ideal current source from the collector to the emitter, where the actual current is some gain times the current from the base to the emitter. <S> You see inside the BJT a diode between the base and emitter <S> and you should immediately think "diode drop" : as long as the voltage bias from base to emitter is over the drop voltage, a lot of current can flow. <S> To actually get the gain and drop you would consult the manufacturer's data sheets for your BJT in question. <A> No current will flow since the transistor is "closed", because Vbe = -2V <S> and it should be at least positive and to be fully open above ~0.7V.
| Your diode is reverse-biased causing no current to flow from base to emitter, and therefore the current source is set to 0.
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3.6V Circuit and 2.8V LED I'm trying to figure out a good LED + resitor pairing for my circuit. However, I'm provided with a challenge and limited knowledge. Here's the situation:I'm creating a minimized version of THIS sensor board. The board breaks out THIS sensor, which requires a 3.3V supply voltage. To minimize the size, I took out all voltage regulating components. Now I need to power a white LED with 3.3V and add a current limiting resistor. Getting a sufficient voltage drop is the challenge for which I could use some help. Now, my solution was to increase the supply voltage to 3.6V instead of 3.3V, which shouldn't be a problem for the color sensor according to the datasheet. I found THIS LED with a supposed forward voltage of 2.8V and current of 20mA. With this situation I CALCULATED that I would need a resistor of at least 47 ohm. Would my suggested solution work or did I miss something?? <Q> I found THIS LED with a supposed forward voltage of 2.8V and current of 20mA. <S> It's worse than you thought! <S> Figure 1. <S> The ASMT-UWB1-Nxxxx datasheet LED could \$ V_f \$ could vary between 2.8 and 3.6 V at 20 mA. Figure 2. <S> The blue shaded region represents the possible values of I versus V for your LED. <S> The 47 Ω load-line is superimposed on the chart. <S> Data is for the LTST-C170TBKT but looks similar to OP's LED. <S> Source: Variations in Vf and “binning” . <S> Depending on the actual value of \$ V_f \$ for your LED <S> the current will be 12.5 mA at lowest \$ <S> V_f \$ to about 1 mA at the highest. <S> Figure 3. <S> The LED is powered from 3.3 V. <S> The easiest solution is to disconnect R5 from the 3.3 V supply and power the LED from the unregulated voltage. <S> Figure 4. <S> The load-line for an 82 Ω resistor on a 5 V supply. <S> With a 5 V supply and an 82 Ω resistor the current would be 20 mA for the mid-range \$ V_f \$ but could vary by up to 5 mA each way depending on production spread. <S> This is the difficult world of electrical engineering. <S> The load-line will be less steep and the resultant currents will vary less for a given range of \$ V_f \$. <A> An alternative is to use a simple current limiter circuit instead of a series resistor. <S> That will minimise the volt drop in cases where current is low. <S> The following is about as simple as you'll get. <S> Pick Rsense so that it produces about 0.6V at your max current. <S> So for a 20mA LED, choose Rsense = <S> 30Ω. When the current limit is hit, T2 will turn on, drawing the base current away from T1 and therefore inhibiting the current into the load. <A> The four options you have power the LED separately from a higher voltage. <S> Use 3.6V with the original 3.3V led and a larger resistor. <S> Use 3.3V with the original led and no/a smaller resistor and/or a lower current. <S> You don't have to run it at the 20mA <S> current and associated Vf. <S> Use what you calculated, which is 100% fine. <S> You could use a constant current led tester set at 20mA to find out the actual forward voltage at Vf and adjust your led calculation based on that. <A> If you have the room a simple solution is to add a small capacitive booster regulator. <S> There are many available, but here is one, the LM2750-ADJ from T.I. <S> They have nicely given you the schematics to drive LEDS in the application examples. <S> (I know you only need one LED, so ignore the extra five shown.) <S> Notice <S> it regulates the current through the LED, you simply have to chose a value for R1 that sets up your If at 1.23V. <S> It also has a simple logic level shutdown pin you can use to turn the LED on and off. <S> If you had to drive more than one LED, I'd suggest using one of the fixed output boosters to power all of them, and a traditional transistor/resistor switch for each individually.
| If you can increase the supply voltage you can also increase the series resistor which will make it more like a constant current source.
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Wiring a Crompton Parkinson single phase motor I have acquired an Elliott Progress No.1 pillar drill with a Crompton Parkinson motor. It must date from the 1960s.The motor worked fine when I first acquired it. The rest of the drill was in a bad state, so I stripped and restored it and gave the motor a good clean. It's single phase with two windings, a centrifugal start switch, and a start capacitor. Once reassembled, I wired it back up the way it was when I dismantled it, and switched it on. The motor emitted a loud buzzing sound but did not move. The bearings are good, there are no obstructions, and the spindle turns smoothly by hand. One thing I did notice was that with the power on the spindle locked – I couldn’t turn it in either direction until I cut the power. I've spent weeks trying to get it to work, and have scoured the web for ideas. I found a few posts on various forums that described similar motors, but none quite the same. I've run out of options and hope someone with more knowledge may be able to help me. I'll describe the configuration as clearly as I can. There are notches for six terminals on the connection plate inside the motor marked, from top to bottom, K, Z, AZ, S, A, T. K and S are empty. Mains live was connected to T, neutral to A, earth to the earthterminal. A wire runs from T to what I believe is an overload switch,and from the switch to AZ. From AZ, one wire runs to the centrifugal switch. On the other sideof the switch a wire runs into one of the windings. Another wire runsfrom AZ into another of the windings. Apart from mains neutral, two wires were connected to A, one runningto a capacitor, the other into one of the windings. The other side of the capacitor was connected to Z. Another wire ledfrom Z to one of the windings. I’ve tested the resistance of the windings, one is quite high (around 19 ohms), the other much lower (around 4 ohms). I presume the first is the start winding, the second the run winding. The centrifugal switch seems ok, no missing, damaged or corroded parts or obstructions. I replaced the original start capacitor with one of the same specification, but this makes no difference. I’m 99.9% sure this is exactly how it was wired when I dismantled it for cleaning. I can’t work out what the problem is. Although my electrical knowledge is basic, I got the impression that the start capacitor is supposed to be placed before the centrifugal switch in the circuit to provide the start winding with additional torque. As it is, it seems to be placed between the two neutral terminals, A and Z. Much Googling later, I am none the wiser as to why this should be. How do I get this motor working again? UPDATE : one of the winding wires snapped while I was working on the motor and I lost the end. It's now beyond recovery unless I replace the windings, which I suspect would cost more than a new motor. Thanks everyone who helped! <Q> One thing I did notice was that with the power on the spindle locked – I couldn’t turn it in either direction until I cut the power. <S> One possible problem is that the rotor is moving axially. <S> If you removed the rotor, you may have made an error in reassembly. <S> There were probably several washers of different types at each end. <S> In addition, the centrifugal switch must engage with a mechanism on the shaft. <S> I can't think of a wiring error that would make the shaft lock. <A> The centrifugal switch contacts must be CLOSED when the motor is at rest. <S> The Run winding has no relative phase rotation, so it just vibrates a little back and forth, what you perceive as being locked up. <S> Most likely you installed the centrifugal switch backward and the contacts are OPEN when at rest, so no power is getting to the Start winding and the cap is not in the circuit. <S> Here is a wiring diagram of a Capacitor Start single phase motor. <S> Note that in this, and most diagrams available on the Internet, the centrifugal switch is shown in the OPEN position, leading people to think that's how it is supposed to be wired, but it's kind of a mistake in the depictions. <S> That switch is only open AFTER the motor gets to about 80% speed. <A> The wiring resistance seems different to this site... <S> (* different model) yet similar. <S> Z to T resistance = <S> 10.2 ohmsA to AZ resistance = <S> 7.4 <S> ohmsWhen wired up for forward rotation <S> the resistance between the Live and Neutral wires is 4.2 ohms <S> http://www.instructables.com/id/Dismantling-a-Brook-Crompton-AC-motor-from-a-Myfo/ <S> I believe the speed switch goes to AZ. <S> This is just the DC resistance or DCR and this impedance determines the start current but the current drops back <S> EMF cancelling the applied voltage as the speed rises. <S> Check your windings. <S> A to AZ must be the switched winding to produce >5x start current. <S> If this motor does not turn and does not hum then the Line to neutral resistance is wrong which should be less than the A to AZ since it is in parallel with the run windings Z to ZT
| There may be something not right about those items either because of a reassembly error or a mechanical failure. The description of what it's doing sounds as though the Start winding is not energized.
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How to check for quick oscillations without an oscilloscope I'm working on a 555 timer oscillator circuit, and I'm trying to get an LED to blink. The LED turns on, but I'm not sure if it's blinking. I do not own an oscilloscope I've tried a few things, such as recording with a slow motion camera. Is there a good way to test it? I was thinking I might be able to hook up a transformer to it and check for a constant voltage. <Q> Use a RC low-pass filter and a multimeter. <S> Here's how: Connect a 10 kΩ resistor to the point you believe is oscillating. <S> Connect <S> a 10 µF capacitor between the resistor and ground Measure the voltage across the capacitor, let's call this voltage <S> \$V_c\$ <S> If \$V_c\$ is around 20-80% of the max value of the signal you believe is oscillating, then it's most likely oscillating. <S> If \$V_c\$ is <S> around +99% of the max value of the signal you believe is oscillating, then it is most likely not oscillating. <S> Extra info: <S> Using a low pass filter in this configuration will give you the average value of the oscillating signal. <S> So if it is 50% high and 50% low and it is a square wave signal that is between 0 V and 5 V. <S> Then in an ideal world you will read 2.5 V across the capacitor. <S> If it is high 80% of the time then you will in an ideal world read \$0.8×5=4\$ V across the capacitor. <S> If it is high 100% of the time, then it's not oscillating, and you will therefor read 5 V. <A> They are reasonably cheap and a quick test tool. <A> You could try to take a movie of your LED at a high rate. <S> I think most smartphones can film at 60 Hz. <S> Plot <S> the amplitude of the pixel associated with the LED and see if there is a distinct frequency. <S> You might be able to say if your LED is blinking but might not be sure about the frequency due to aliasing.
| If you can't afford a scope a decent logic probe will tell you if a logic signal is pulsing.
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Takes days to charge phone from 10W solar panel I bought a small 10 W solar panel which is rated to deliver 570 mA at its peak, according to the specifications: Power: 10 Watt Max. Voltage: 18 Volt Max. Current (Imp) : 570 mA So I bought a step-down converter that drops the voltage from 18V to 5V. Then I connected a USB port to it to charge the phone. Now, the problem is that it takes days to charge the phone battery, which is rated at 7.22 Wh. So with a maximum power of 7 W from the panel (let's say that it will never reach 10 W), it should charge the phone in roughly one hour. Well, this is not happening and I would like to know why... I initially thought it is the converter, so I have tried also with a 5V Voltage Regulator (the well known LM7805 ), and it has the same problem. Moreover, when I use an LM7805 and connect the phone, the voltage drops to 3V, so I suspect that there is not enough current from the panel. The Step-down converter is this one , which has the specifications: Rectification: non-synchronous rectification Input voltage: 7V-35V Output voltage: 1.25V-30V Output current: adjustable maximum 3A Conversion efficiency: 92% (the highest) So I have these two voltage regulators, and it takes days to charge a phone. Now, the question is: am I doing/understanding something wrong or can the solar panel not handle this? Is there any way to check this out? EDIT:After seeing the comments and the answer, I tried to measure the current in full sun. I get this: This mean that the current is 0.7 mA, right? It's the first time I'm doing this and I followed this sketch to measure the current with a multimeter: In my case, the battery is the solar panel and the light bulb is the phone. I hope the measurement is correct. <Q> First make sure everything is working as intended. <S> Point <S> the solar panel at full sun and see what kind of current you can get at 5 V. <S> Try a 5 Ω power resistor. <S> That should draw 1 A, which takes 5 W of power at 5 V. <S> You should be able to get at least that according to your specs. <S> If that works, it may be that the phone is expecting a "smart" charger. <S> It uses the USB charging protocol to communicate with the charger to find how much current it can draw. <S> If it can't talk to the charger, the phone falls back to a rather low current. <S> Measure the current with the phone connected <S> , first making sure the output really is at 5 V. <A> You're hooked up wrong. <S> The meter's big rotary switch has a legend around the outside indicating range. <S> Obviously, there are as many range markings as there are switch positions. <S> Only two of them have the "20A" icon, which indicates you should use the 20A socket instead of the V/ma/ohm socket. <S> You have not selected one of those positions, yet you are using the 20A socket, That's not going to work. <S> You're sending the power through the ammeter shunt, but the meter doesn't know to rescale. <S> Try turning the meter knob one notch clockwise. <S> If that yields a measurement that is too small or out of range, then select an appropriate range and use the other socket. <A> Please note that some USB connectors in cars show the same performance even though the battery of car rated 120AH. <S> Phones require special charging circuits. <S> Please check: <S> How do I design a 2A or more power supply for my consumer USB devices?
| You may need to solder a few resistors between power lines and D+/D- lines.
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Does the formula v=Ldi/dt apply to materials with zero resistence, such as superconductors? I've been trying to find information about the formula \$ v = L \cdot \frac{di}{dt} \$, it seem to have something to do with voltage drop, and I know voltage drop is caused by resistance, so would a material with zero resistance still have weaker levels of induction with higher frequencies of changing current? <Q> Your formula is the basic law on how inductors behave in electric circuits. <S> Actually it's derived from a relation between magnetic and electric field, more specificly from the one that states how strong electric field is caused if magnetic field changes. <S> That law is as valid in all materials and in the empty space. <S> Conclusion: <A> Given that superconductors have the lowest resistance, this does not affect their inductance. <S> Inductance is controlled by the geometric ratio of the conductor Length/Width and the relative permeability of any surrounding magnetic material. <S> The voltage drop you may have heard about could be due to the reactive impedance at some f which has nothing to due with real loss, from IR drop. <S> \$Z_L=\omega <S> L\$ <S> Then if a load R, is a applied to a superconducting source with some voltage and frequency , the load will see \$V_o= <S> V_i\dfrac{R}{\omega L+R}\$ <S> The basic transient behavior of inductors is as you stated.\$V_L= <S> L dI/dt\$ <S> However the voltage change at a switch, relative to 0V, when released, will be a pulse in the opposite polarity of the voltage when current is applied. <S> i.e. A high side switch opening results in a sudden drop in current or a negative ΔI thus a large negative spike is produced. <S> So a Low-side DC switch releases a +ve spike and High-side switch to L <S> releases a large -ve spike. <S> The other important formula is how much energy is stored in an inductor depends on the current. <S> This is the amount of energy dissipated in the arc of a switch or FET-internal clamp diode when turning off an Inductive load. <S> \$E=\frac{1}{2}LI^2\$ <S> [J=Joules = <S> Watt-seconds] <A> Yes, it does. <S> Actually, for an inductor, \$ v = L \frac{di}{dt} + Ri \$ R is the resistance of the wire in your inductor. <S> If you use a superconductor, it would be zero. <S> The first part of this equation models a theoretically perfect inductor (ie, superconducting, like all inductors in spice simulation unless you specify its resistance in the spice model). <S> The second part models its resistance. <S> In an equivalent circuit, the inductance and its resistive part can be put in series, this does not change the behavior.
| Your formula is as valid in all circuits, no matter if some parts of it are superconductive.
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3-phase Motor RPM depends on power line frequency? I have a 3-phase motor but my knowledge on the topic is almost null.But, you know, I have also curiosity to satisfy. The motor's nameplate reads: 50 Hz, 7.5kW, 400V, 13.8A / 690V, 8.0Acos phi 0.892910/min60 Hz, 8.6kW,460V, 13.5Acos phi 0.893492/min It looks like the RPM is proportional to power line frequency. Is this true? Applying i.e. 340V 50Hz, may I assume that the RPM remains 2910/min? <Q> But given that the ratio of RPM to grid frequency is the same in both cases, yes, it is very likely that the speed will stay very similar at 340V. <S> This assumes that the load is low enough for the motor to run at the reduced voltage, of course. <A> You have a squirrel cage induction motor. <S> The synchronous speed <S> \$n_S\$ (speed of the magnetic field pulling the rotor) is: $$n_S = \frac { <S> 120 f}{P}$$where f is frequency and P is the number of poles. <S> You have a 2 pole machine, which gives a \$n_S\$ of 3000rpm at 50Hz and 3600rpm at 60Hz. <S> It's an induction machine, because rotor speed n is 97% of \$n_S\$. <S> Slip is 3% at full-load. <S> To answer the question. <S> At 340V, 50Hz, the motor will spin at 2910rpm dependent on the load torque. <S> $$ T \ <S> \alpha \ <S> V^2$$ <S> Torque is proportional to the Voltage squared. <S> 340V is 85% of 400V, so load torque would be approximately 72.3% [\$(85\%)^2\$] of full-load torque at 400V. <S> If the actual load torque was greater than this, the motor speed would slow down to drive the load. <S> Slip would increase. <S> Edit... <S> Torque Reference <S> Red line is load torque. <S> Motor will slow down until it is capable of driving load. <A> S = f <S> * 120 <S> / N <S> Where: S = Synchronous speed in RPM f = Frequency in Hz <S> 120 = mathematical constant N = Number of poles in the motor design Actual speed = <S> S <S> * (1-Slip %) <S> So yes, speed of a 3 phase motor is based on the frequency applied to it.
| There are different designs of electric motor which can run on a 3-phase supply, and we don't know for sure which you have.
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How to reduce the number of pins for three 7 led segment? In my project I need to use 7x tact switches 6x LEDs 3x 7 led segments 1 arduino uno board my prof says we can use any ICsbut arduino uno's pins cannot accomodate all those needed values. I need to manipulate all the tact switches and the 6 LEDs so the only thing that can be reduced is the issue with 7 led segment. Can you suggest a way that I can reduce the number of pins needed to be used in arduino uno? <Q> You can probably reduce the GPIOs usage as follow : <S> Instead of using 7 GPIOs for 7 x tact switches, you can use a 3x3 matrix for 9 switches : 6 GPIOs (you can add 2 more tact switches) Instead of using 6 GPIOs for <S> 6 x leds you can use a 2x3 matrix for 6 leds : 5 GPIOs Instead of using 21 GPIOS for 3 x 7segment you can put it on a 7 bits bus, and use 3 more pin to control the 7-segment : 10 GPIOs <S> For switches you need to scan the matrix rows and to know what tact is pushed, and for the led and 7segment you need to refresh quickly, because of eyes afterglow (remanence) <S> the user could not see the flickering/blinking, user think light is fixed on this different leds. <S> Before : 34 GPIOs After : 21 GPIOs (With 2 more switches) <S> You can also consider the 6 leds as a "fake 7segment" ans reuse your 7 bits bus to control them, you only need 1 GPIO to select this "fake 7 segment". <S> Before : 34 GPIOs After : 17 GPIOs (With 2 more switches) <A> 3 x 7 segments and 6 LEDs can be multiplexed with 7 + 4 = 11 pins, in a 4 x 7 arrangement (wire the LEDs as if they were a 4th digit). <S> So a total of 12 pins will be sufficient for a simple (no extra chips required) approach. <S> Using shift registers or I/ <S> O extenders you need even less pins. <S> An MPC23017 for instance requires 2 (I2C) pins and gives you 16 GPIO pins, which is more than sufficient to do everything you asked for. <S> But some programming is required.... <A> Use ' Charlieplexing ': - 6 LEDs can be driven by 3 PINs - see link here - 7 (up to 12) <S> switches with 4 PINs - see this link , chapter named "Charlieplexing also to interface buttons" - <S> For 7-segment display I would recommend I2C driven 4-digit LED display (2 PINs - SDA and SCL) - to see available options drop a "7-segment display i2c" line into google. <S> Total: 3 + 4 + 2 = 9 PINs. <A> For 7 switches , MM74HC164M SHIFT REG 8BIT PISO 14-SOIC <S> For 6 LEDs and 2 LED digits , (3) <S> SIPO's MM74HCT164MX SHIFT REG 8BIT <S> 14SOIC <S> For (3) LED digits choose LEDs with BCD to 7seg decoder integrated. <S> The clk & data lines are shared but a chip selector is needed,.
| If you use 6 diodes with the 6 switches you can share the segment pins of the LED segments, and use one shared input pin (with pull-up) to read the switches.
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Convention on pin shape association for headers on a PCB? I want to use such headers below as terminals on PCB to solder power or signal wires to the PCB. Is there a convention for square or circle pins association? I mean should square always be the GND or? <Q> Basically a polarity indicator, so you can tell which pin is which from the underside of the board. <S> The latter normally does not have silk-screened details to indicate such information. <A> There is no hard and fast rule, but generally designers use round pads for most through hole pins and square pads are often used to indicate pin 1. <S> You can also run a trace closer to the curved edge of the circular pad than you can diagonal to the corner of a square pad if you are running trace clearances close to the process limits. <A> There's not a formal convention as far as I'm aware, it's up to you. <S> What I have seen and used is that square (1) is the more positive connection. <S> You'll see this often in library parts for polarised caps, diodes etc... <S> Whatever you choose, label it!
| As far as I am aware, or have seen used, the square pin is only used to indicate Pin #1.
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IRF530 getting very hot I have a IRF530 connected to a 5V micro. The trace is on a PCB and very short, with only a pull down resistor on the gate: simulate this circuit – Schematic created using CircuitLab The MOSFET is switching a long length of 24V LED strips, which draws 6A. There is a 5m 2 core cable from the MOSFET to the strip. When I set the pin to high, the mosfet gets very hot. On shorter lengths of LEDs, it only gets marginally warm. When I PWM the mosfet (f=500Hz), it still gets hot but takes a little longer to warm up. How come this is happening? I am above the threshold voltage of 4V, so I assume the FET is not in a linear state, and the power dissipation according to Rds(on) is only 0.96W, which the un-heatsinked (un-heatsunk?) package should handle. <Q> How come this is happening? <S> I am above the threshold voltage of 4V, so I assume the FET is not in a linear state <S> The FET datasheet specifies RdsON for Vgs=10V. <S> It is not specified for 5V. <S> Therefore it will not conduct fully and act as a resistor. <S> Threshold voltage is the Vgs at which is begins to conduct just a little bit, you need way more to have good switching. <S> Solution: use a FET with RdsON specified for Vgs=5V, and suitably low RdsON. <S> For 6A and without heatsink, aim for 15 mOhms or lower. <S> There are plenty of such FETs. <A> Your problem is insufficient Vgs for the device chosen. <S> Hence not the rated RdsOn. <S> MOSFET's come a many different part numbers, with distributor search filters to help selection. <S> How can you choose one to keep cool? <S> How much does a heatsink cost vs a better FET? <S> Copper area? <S> Alum. <S> sub. <S> ? <S> How many do you need? <S> 1? <S> 1k? <S> +? <S> e.g. <S> >= <S> 30A <S> Ensure RdsOn is rated at your available Vgs or in other words Vgs > <S> > <S> 2x <S> Vgs(th) <S> pref. <S> 3x. <S> prefer SMD over THT for better heat transfer to board <S> if suitable always do a thermal resistance analysis of temp rise including case and ambient cooling or lack of. <S> This change will: reduce heat loss, reduce cost of thermal design and only slightly increase FET cost but multiply input capacitance to gate Example: IRFH5301 <S> $1.25 (1pc) 1.85 mOhm @ <S> 50A, 10V Alternatives <S> Use 12V to drive the gate 1k pullup, 1k down divider from 24V and shutoff with an additional NPN inverter to shut off gate. <S> simulate this circuit – <S> Schematic created using CircuitLab R values may be increased slightly to reduce Pd. <S> with Ciss affecting risetime. <A> While others have adequately answered the question, I thought I'd approach it from another perspective, in order to give you some insight as to how to read data sheets. <S> Go back to your IRF530 data sheet, and look at the Vgs(th) rating of 2 to 4 volts. <S> Now look closely. <S> Even at 5 volts gate drive <S> , you are not guaranteed to get a whole lot more than this, and this explains your high Rds. <S> Now, it's true that 4 volts is worst-case, and you are surely doing better than this, but it's the heart of the problem. <S> Now look at figure 3. <S> This deals with "normal" behavior, rather than worst-case. <S> At 5 volts gate voltage, you can only "expect" 3 to 4 amps of current, and you want 6, so this also should give you warning. <S> Finally, look at figure 4, Rds vs temperature, and notice how it rises. <S> This allows the temperature to stabilize, rather than running away. <S> This is in contrast with BJTs (bipolar junction transistors) whose effective resistance drops with rising temperature, which can lead to runaway thermal failure.
| Generally for slow PWM, high current, low voltage; consider devices rated for > 5x the current you plan on using. It's very useful for connecting FETs in parallel, since if one is hogging the current, its resistance will rise and it will draw less current, allowing equal current sharing without too much fuss on your part. Notice that at the rated voltage, the drain current will only be 250 uA.
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Cockroach or insects triggering conductive ink sensor I want to modify a cockroach to get it to trigger a conductive ink sensor or Arduino with foil capacitive touch setup. At the moment there is no reaction as far as I can see (in the serial monitor) I was wondering about spraying it with water to make it wetter?Or modifying its feet somehow?Maybe some little shoes? :) Alternatively, suggestions of better insects? <Q> I doubt you'll have much luck with resistive sensing - the chitin shell and legs of an insect will have a very high resistance. <S> Any voltage high enough to give reasonable current is likely to cook the little buggers. <S> Not that that's a bad thing, of course. <S> By the same token, I'm dubious about the capacitance, as well. <S> I'd suggest a different approach, such as optical sensing. <A> If it were me, I would try increasing the sensitivity of the capacitive sensor. <S> Actually increasing the sensitivity of a conductive ink sensor might work too. <A> Try one of these... maybe with aluminum foil booties.
| I would expect insects to have a relatively high resistance, so you'd have to really send a high voltage through the sensor to see a response.
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Does a BJT controlled relay need protection diodes? I am fairly new to electronics, so please pardon my mistakes. I am attempting to build a Wi-Fi controlled relay. The problem, however, is that the relay runs on 5 V, but my ESP-12E can only supply 3.3 V. I figured to connect the (+) end of the relay to the VCC (5 V input) and then the GND end to a transistor which also connects to ground. The base of the transistor is connected to my ESP-12E's 3.3 V GPIO output. Here is what my circuit looks like: <Q> You are pretty close, except your transistor is upside down. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Something else I noticed... <S> It looks like you are trying to power your ESP through a resistor divider? <S> That is a bad idea, the current through the divider depends on the draw of the circuit which can alter the amount of voltage fed into the ESP. <S> Generally resistor-dividers are OK for measurement purposes, but do not use them to adjust power or as a step-down power supply. <A> You are close. <S> Your transistor is upside down and you need a base resistor. <S> You should also add a fly-back diode to protect the transistor. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As Ron spotted, you can not divide down the voltage like that to power your module, You should use an appropriate 3.3V LDO Linear voltage regulator. <A> You can google it . <S> You are missing a resistor between your ESP and the base of 2N2222 <S> The 2N2222 is also upside down <S> The relay is missing a fly-back diode <A> Your 2N2222 is oriented with the emitter connected to VCC, so the PN junction from base to emitter will not conduct with GPIO16 in either state. <S> Think carefully about the structure of the NPN transistor and you'll see how you can protect the GPIO pin and switch the relay.
| You should also add a diode across the relay coil contacts to avoid back EMF and a resistor on the base of the transistor.
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Need a buck circuit to go from 48V down to 12V-9V less than 30mA I need a circuit to go from 48V down to 12V-9V less than 30mA, it doesn't need to be a buck converter. I don't want to use transformers as simplicity is good (and adding components to the library is bad) and I probably already have inductors that I could use on my shelf. Most of the stuff out there in IC's only goes up to 36V on bucks or regulators. How do I make a circuit to do as I described? Points for being more than 60% efficient as I don't want to burn up 60% of my power on regulation. <Q> Use a rough pre-regulator to halve the supply. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> ADDITION: <S> SInce you added >60% efficiency as a new requirement... you need something switching... <S> How about one of these ... <A> Quick & dirty HV regulator <S> : everyone has a TL431 stashed somewhere, right? <S> I just pulled this one off the net, so you'll need to adjust resistor values. <S> It is not short circuit protected, unless you count Rb limiting the current via pass transistor hFe. <S> It will be stable with output capacitors depending on ESR, can be confirmed via a quick sim with TL431 model. <S> I'd use a general purpose aluminium cap like 100µF on the output. <S> Transient response can be made faster with a cap in parallel with the upper feedback resistor. <S> For example at LT . <S> I remember seeing lots of such chips on DigiKey when I did a search a while ago, there are some with integrated switch in small packages. <S> example <S> example <A> With a new spec for 60% efficiency and a voltage ratio of 12/(12+48) =20% for any linear design it is impossible to meet this with a linear design. <S> Your spec could be defined some other way. <S> You could use a small SMT transformer, a bipolar input clock and rectifier output without using a SMPS design . <S> Also this may cost more than a SMPS. <S> Suggestion <S> This design took 1 minute and may be exported into 5 standard S/W packages.
| If you have a 36V input switcher then dropping the input voltage with a bunch of diodes or a zener to 30V then running the switcher from that should give you the 60% you want... If you want a switcher , look into the ones designed for Telecom 48V standard, or Power Over Ethernet.
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Why would you put multiple resistors in line vs one resistor of the same value? Why would you put multiple resistors in line instead of one resistor that is the same value? In this case, I have a PCB with three 1k ohm resistors. What could be the reason for this instead of just one 3k ohm resistor? Thanks in advance! <Q> Three reasons I can think of: to allow for a higher power dissipation than one resistor can handle. <S> to tolerate a higher voltage than one resistor can safely handle. <S> (Higher power or higher voltage resistors would be larger, which may not be convenient for the auto-place machine) to minimize the number of different resistor values required on the board (if I use lots of 1K resistors elsewhere, it may be cheaper to use three here, than to buy a few 3K resistors to use in one place)("cheaper" may include loading and programming the auto-place machine as well as the actual cost of the resistors). <A> It may be a reason of power dissipation. <S> Each resistor package of that size can support only a certain amount of power on the resistor so putting multiple in series spreads the power across multiple packages. <S> Another reason may be voltage rating. <S> An SMT device like that will have a MAX voltage rating that it can handle without insulation break down. <S> Multiple resistors in series will split the voltage drop across the various devices. <A> In this case, reduction of production cost. <S> Using a common value three times (1k) instead of a less common value once (3k) reduces the total number of different parts required by your design. <S> Loading new reels into the feeders can be a significantly greater cost than using a couple more of the same part. <S> Other reasons to make a chain like this can include increasing the effective voltage rating of the part, such as in the high voltage frontend of a meter or probe, but that doesn't apply here. <A> I wonder if the board has three resistors in a row because that's what the circuit needs: they aren't just substituting for a single resistor with three times the resistance. <S> Look to the left between each of the triple resistors, e.g. just above the "3" in the R23 label. <S> Aren't those vias ? <S> It looks like there's a via next to each of the junctions in all of the triple-resistor groups. <S> (Someone please let me know if I'm misinterpreting the photo.) <S> So my guess is <S> the three resistors in each group could be wired in series, thus the placement, but there's also a tap at each of those junctions running to some other part of the circuit. <S> Or maybe the vias are there for some other reason of convenient layout - but it does seem like an interesting coincidence. <S> (Also, thanks everyone for the other answers! <S> Always fun to learn some things about PCB manufacturing.)
| A last reason that this is sometimes done is that the circuit board design is done to it is easy to swap out certain resistors to be able to select a combination of different valued resistors that total up to certain calibrated requirement that may not be available in a standard valued single resistor. In the pick and place process used in manufacturing, you are looking to minimize the number of different parts used, within reason.
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Bi color led with tri-state pin (high->D1, tri-state->D2, low->off) I have a tri-state pin and I would like to use a bi-color led: Pin Low High Hi-ZD1 off on offD2 off off on There are a lot of examples around but none in the combination of the above table. How can I do that with minimum components? <Q> Assuming a 3-pin LED common anode it can be done as follows: simulate this circuit – Schematic created using CircuitLab . <S> If it's acceptable to have both LEDs on for a high input (which would give youred + yellow or green + yellow for a red/green LED) you can reduce it to just two components total (plus the LED) by using a prebiased array for Q1/Q2, a resistor array for R3/R4 and eliminate Q3/R5. <S> simulate this circuit <A> Specification: IN L1 L21. <S> L 0 02. <S> H 1 03. <S> X 0 1 Figure 1. <S> A totally bonkers method of obtaining the required LED logic. <S> How it works: <S> With IN low U1 turns on and L1 is shorted out turning off. <S> U2 is off <S> so L2 is off too. <S> With IN high, U1 turns off so L1 lights and L2 is off. <S> With IN tri-stated, both U1 and U2 turn on. <S> L1 is shorted out and L2 turns on. <S> Notes: <S> R1 and 2 have to be low enough to turn both U1 and U2 on in tri-state. <S> The current in the opto-LEDs has to be fairly high to get enough through the photo-transistors. <S> Read up on "current transfer ratio" to learn more. <S> R3 will pass current all the time. <S> I haven't tried it and it isn't efficient <S> but I think it meets the user requirement specifications. <A> Splitting rails for fun and for profit <S> The basic approach we take here is similar to Cleary/Sheridan/Thomson's work in US Pat. <S> 6133753 <S> -- we use a bias generator to weakly pull the output to the half-supply point, then evaluate the voltages that result. <S> The main difference here (and why the claims of the patent cited wouldn't read on what we're doing here) is that we don't care about power consumption nearly as deeply as the patentees do, so we can dispense with the sampling complexity they claimed and employed. <S> Note that this requires a 4-pin bi-color LED (i.e. one where the LED chips have no pins in common). <S> This configuration isn't seen in through-hole parts, but is reasonably common in SMT bicolor LEDs -- for prototyping purposes, simply use two separate LEDs instead. <S> Also note that this is intended for a 5V supply -- for a higher supply voltage, change R3 and R4 as appropriate to keep the LED currents reasonable, and change D2 to a Zener to set the threshold for Q3 turning on above the half supply mark. <S> (This approach won't work well with lower supplies though -- you'll want to use comparator hijinks instead of Q3's arrangement in that case, as we run out of supply margin there.) <S> simulate this circuit – <S> Schematic created using CircuitLab
| A simple voltage divider suffices to bias the output weakly to the half-supply mark, while from there, we can then use a few bog-standard BJTs and some diode trickery to handle the voltage sensing.
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Is there any standard for the input impedance of an audio amplifier Ebay and other stores are full of items like this . These are audio amplifiers for various uses and output impedance. But for almost none of them, I was able to find any specification regarding the input impedance. I suspect there is something to know about that: is there any standard input impedance for audio amplifiers? at least sufficiently probable for taking the risk to buy such an item? <Q> This standard is 600 Ω. All "line level" audio in such a setup is expected to drive 600 Ω. <S> This is also the impedance of the microphones. <S> However, many many amps don't follow this. <S> In fact, most amps try to be "high", which means to not appreciably load the output stage of whatever they think is producing the signal. <S> Of course everyone's idea of "high" is different. <S> The most common is probably around 10 kΩ. <S> That's what I've done with audio gear aimed at consumers. <S> Anything that produces a consumer audio signal won't be damaged, loaded to the point of distortion, or even appreciably attenuated by 10 kΩ. <S> Otherwise, you want the input impedance to be as low as possible avoid picking up stray noise when nothing is connected. <A> Generally the input impedance to an audio amplifier is 'high'. <S> That is, high enough to be driven comfortably by any audio source output. <S> But of course, without a proper specification, you are taking something of a gamble. <S> 10k <S> to 100k is the usual sort of range we find, though some can be less. <S> Even something as low as 1k will be driven easily by most op-amps outputs, whether they're designed to drive headphones (many are) or not. <A> Buying one is a gamble. <S> No spec = <S> no data beyond reverse engineering the photos, which can be fake. <S> Your example seem to be based on IC type <S> HXJ8002. <S> Input impedance is dependent on used resistors, in this case there's a resistor just like in the input of inverting amplifier made of an opamp. <S> That resistor = the input impedance. <S> The gain also depends on it. <S> With extreme good luck the input impedance can be 20kOhm. <S> (=component value in the example application circuit). <S> But as well it can be only 2kOhm. <S> See the datasheet of the used IC: http://www.shenzhensum.com/products/datasheet/8002(2.0W).pdf <S> If you can do SMD soldering, you can easily change the resistors and even it's possible to change the configuration to non-inverting. <S> But then you should as well be able to build exactly, what you need. <S> Consider to do it.
| There is a sortof standard for a certain class of professional audio gear.
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