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Creating an Assembly Drawing for Fabrication Houses I have a design for a PCB that I have been thinking about producing. One thing that I have found is that I need is something called an assembly drawing. So far, all I have been able to come up with is that they are used to describe which components go where for PCB assembly. But I think they are also used for turnkey supplying as well. What are the key things that fabrications houses look for in a good assembly drawing? If I am using turnkey manufacturing, how do I describe what parts I want, and how specific do I need to be for things such as capacitors, resistors, and IC's? What does a good assembly drawing look like? And is there an easy way to create them with Eagle? If I order, say 500 PCB's, it is common practice for a supplier to create the first 10, ship them, make sure they work as required, and produce the rest? <Q> Assembly drawings are the master prints to completely assemble a unit. <S> They should contain (at a minimum): <S> The quality standard to be used for assembly and inspection (usually IPC-A-610 class 2; class 3 is difficult to achieve unless you are willing to pay a premium). <S> A reference to the schematic with current revision (which you should also supply). <S> A reference to the Bill of Materials with current revision which, once more, you should supply. <S> The Bill of Materials should specify the bare board and revision information. <S> Any special instructions (not everything required to completely assemble the board will be in the normal tool outputs): <S> Bonding of components perhaps for high vibration environments where you should also specify the material type <S> (at least the standard to which they must be made). <S> I usually specify a particular bonding agent. <S> You should clearly identify on this drawing which components are to have instructions applied to (using a callout ) Any modifications to be performed - we do not spin a PCB for every modification which may require a wire or two. <S> Any conformal coat requirements, including masking information (you can do this on a fabrication layer, but that layer information needs to be called out). <S> Any test requirements (you may need to provide test equipment or work with the assembler to define the tests). <S> Any other information: perhaps some Kapton (r) tape needs to be applied to avoid short circuits. <S> A view of any surface with components with reference designators clearly visible. <S> An isometric view is useful for some information. <S> The applicable standard is IPC-D-325. <S> I usually require a certificate of conformance to the instructions provided and that assembly has been carried out to the quality standard referenced. <S> Note that I also generate fabrication master prints in addition to gerbers / ODB++ and the other various outputs <S> ; that is a subject in its own right. <S> There are other things that may end up on the drawing, but this should get you going; I do not use Eagle, so I do not know if there is a simple way of doing this (I use Altium with the drafstman tool) <S> Edited for comments. <S> Most passives are commodity parts; it is your responsibility to identify the parts that can be eaaily changed. <A> I never heard of the need for an "assembly drawing" for producing circuit boards. <S> Assembly drawings are required when you want to produce a fully assembled device (including plastic shells, screws, spacers, brackets, free-hanging wires and connectors, whatever...). <S> They are usually mechanical drawings. <S> Now, if you want to order assembled PCBs (just the circuit board with all components soldered), you simply have to provide: <S> The gerber files and drill files for producing the PCB (including the silk-screen masks where the component designators are given). <S> These files are produced by your cad tool, using a standard format. <S> A Bill of Materials sheet, identifying each and every component with their designator, and giving their values (for resistors/capacitors), the manufacturer from which they are purchased and the manufacturer part number (especially for chips), and eventually the package (e.g. the size for passive components, standard package code for chips, ...). <S> Basically all information that is relevant for purchasing and identifying the components. <S> There is no standard format for this information. <S> Usually a excel spreadsheet or a CSV file is fine. <S> See this other question for details and an example of BoM . <S> Now, with just this, the fab house should be able to produce boards. <S> The BOM allows them to source the components, and the silk screen gerber files let them know where to place them on the board (be sure to clearly indicate the component orientation: <S> pin 1 mark for the chips, anode/cathode clearly shown for diodes, polarity for capacitors, ...). <S> If the fab house is using automated SMT pick-and-place for assembling the board, you can also give them a pick-and-place file (generated by your CAD tool). <S> It gives the position and orientation for each component on the board. <S> This will ease the job at the fab house and avoid mistakes. <S> I am not sure there is an actual standard format for these files, however. <S> The CAD tool usually let you configure the format. <A> Parts: if the brand is important specify it, or specify device propertieselse <S> you'll get whatever was easiest for the manufacturer. <S> Drawing depends on the product - it must have sufficient detail that it describes an acceptable product and does not describe an unacceptable product.all dimensions need tolerances. <S> 500 units is a small order. <S> If you order 500 units you will pay the cost of starting a second run if you insert a stoppage after 10 parts. <A> It's usual to have a couple of prototype boards made if you are contemplating a run of 500. <S> You can then sort out any problems at relatively little cost. <S> Board manufacturers cannot interrupt a large production run so that you can check the first few boards.	If there are some particular instructions (non-standard way of mounting a specific component, additional manual assembly steps, ...), be sure to clearly indicate this somewhere, too.
Is the NAND logic gate perfectly symmetrical? In other words: if we swap A and B, will Q behave exactly the same in DC and transient analysis? <Q> Depends on the environment. <S> Maybe in your circuit above and in an FPGA they are the same but in an ASIC library you find differences between the various inputs. <A> There will be a very small difference in that circuit because of the differences in VGS in the N stack while the circuit is sinking current during switching. <S> M1 will be marginally slower than M2 under some conditions. <S> There are however likely to be other factors, say in how the circuit is laid out, that will have an equally large effect. <S> Define perfect. <S> Much of what we do in EE is about modelling. <S> If we let very small differences in a circuit that typically would include tens of these gates effect us <S> we will never get anything done. <A> As the M1 and M2 devices are in a different configuration, there will be a difference between the A and B inputs. <S> However, you may have to look very hard and carefully to see the timing or threshold effects of that difference. <S> When you design a logic gate into a system, you work on the maximum specifications, but expect it to behave nearer to typical. <S> There's often a 2:1 or even 3:1 variation between max and typical specs. <S> It's likely that any difference in performance between the A and B inputs will be much much smaller than the difference between the max and typical timings. <A> If you care about precision pulse processing, as in building the FlipFlops of a low-jitter PFD, phase-frequency-detector, you should understand <S> all <S> the various ways charges will battle inside the circuit and remain lodged to upset the next pulse, to cause inter-pulse-delay-variations and thus deterministic jitter. <A> I once made a chip with purposely asymmetric NAND gates, for a ripple-carry adder in which the speed from one input needed to be optimized, and the other not so much. <S> So no, not necessarily symmetric. <S> But usually very nearly so.	The model is never perfect and at most levels of abstraction the behaviour of this circuit would be considered to be symmetrical.
Are the Offset Voltages in a Dual/Quad Op Amp Correlated? On a single piece of silicon on which there is more then one op amp, are the input offset voltages at all correlated, i.e. would they be expected to be same direction and similar magnitude? <Q> No. <S> The offset voltage comes from the difference between the two input transistors in the same opamp. <S> The input transistors in this TL072 are interleaved so that they have the same center, so that if the transistor parameters are varying linearly across the die, the average parameters are the same. <S> Despite this the transistors are still slightly different because the variation is not exactly linear. <S> So if the two transistors right next to each other are mismatched, why would the ones on the other half of the die have the same mismatch? <A> I don't remember ever seeing a datasheet say anything about offset voltages of opamps on a chip relative to each other. <S> Think about it: offset voltages are due to the slight mismatches between transistors in a chip. <S> However, the input transistors of an opamp are already near each other, and likely much closer to each other than the input transistors of other opamps on the same chip. <S> This doesn't leave much mechanism for the offset voltages between nearby opamps to be correlated somehow. <A> For absolute values-NO. <S> For drift over time, they will have there independent offsets and gain tracking issues as well. <S> For dual and quad op-amps, precision means independent gain and offset for each channel. <S> With independent gain and offset, there should be some correlation over time, but no datasheet would ever state that. <S> Reason is that the user ambient temperature and voltage and loading of outputs is unknown. <S> If one of four channels has a heavy load, any hint of correlation is gone. <A> No guaranteed correlation assured. <S> Best you can expect is that they should be within data sheet specs as long as you correctly apply the part within rated operational ranges.	No, you can't assume anything about correlation between opamps on the same chip unless the datasheet explicitly says so. Some transistor parameters are random, but others may correlate with where they are on the wafer.
High voltage, low current and vice versa Can someone please explain to me in simple terms how is it possible to have high voltage and low current and low voltage and high current and what actually does harm to human body. Here is what I don't understand: 12V / 1000 Ohm = 0.012A (12mA) 300V / 1000 Ohm = 0.3A (300mA) So if I go and touch 12V battery terminals, my body is 1000 Ohms 12mA of current flow through me and it is safe. Then if I go and touch 300V battery terminals and 300mA of current flow through me it is fatal. Yet 12V battery vehicles have fuses rated for 30A and more. Well, that means if there is a short etc. 30A will flow through a wire and break the fuse. 30A in a short circuit wire with a 12V is safe and 0.3A is a 300V is not? How is this possible? <Q> First of all, 30A in a short circuit wire with 12V is not safe . <S> Chances are the thing will go bang or get extremely hot before you even have the chance to touch it. <S> As for your actual question ' how is it possible to have high voltage and low current and low voltage and high current? '. <S> That question you have sort of answered yourself. <S> It's Ohm's Law. <S> you have used the equations to work out the current in those batteries. <S> Now, change the resistance in that circuit and you will see the voltage changes, hence that is how it is possible to have high voltage and low current, and also low voltage and high current. <S> As for what damage can these currents do to a human.... <S> Well, this link <S> HERE will tell you all you need to know about fatal currents and electrical safety. <S> There is also a discussion on what damage can be done in a question asked on this site HERE if you wanted a bit more of a read! <A> The voltage is the "pressure" needed to overcome the resistance to the flow. <S> Having said that, if a high voltage is needed to drive a moderate or otherwise "safe" current though the body it can cause burns from the dissipated energy. <S> In short, what will kill you <S> is fairly complex given the variability of skin resistance, the current paths involved etc. <S> Rule of thumb - anything over 30mA is seriously bad no matter what the voltage. <A> If you have a short circuit on a 12V battery you have a short circuit and these 30 amps will take the path with the lowest resistance (namely the short circuit itself), this has nothing to do with you touching it. <A> As far as I understood its not the voltage nor the current (to a specif limit of course) which harms the body. <S> Its the energy or lets say power over time which can effect serious issues to the body. <S> I think it is a more medical question. <S> Maybe a PhD can answer this question in a serious scientific way. <S> I just believe (how should someone be tested?) <S> the physical body is a varistor thus a non constant resistor.	In terms of the Human body, what causes harm is the current, or number of electrons flowing per second.
Implementing a good connection for pogo pins in eagle I'm currently working on my very first SMT PCB, and would like to connect a number of PCBs to the main PCB using spring loaded pogo pins like these: http://uk.farnell.com/harwin/p19-2221/probe-convex-1-90mm-pitch/dp/9960171 My question is in 2 parts: 1) What kind of connection would you use for the 'female'? I wanted to use some small pads like the ones shown here: https://www.harwin.com/product-highlights/pogo-pins-pads/ Somebody suggested leaving plated through holes where the pins would connect but I am skeptical about the long term durability of this method. 2) How would I best go about implementing that type of connection on an eagle schematic / PCB? Must add that I do not literally mean those exact products, just included the links as an example <Q> That kind of pogo pin is designed to contact the inside of a plated-through hole. <S> So a hole that takes you about halfway down the cone would make sense. <S> The cone is 1.5mm diameter so a 1.0 to 1.2mm hole, with a pad that at least meets the minimum annular ring specification of your PCB maker. <S> There are other variants of probe heads. <S> The serrated head is designed to contact a pad. <S> The concave head goes over a pin or clipped (but not cinched) <S> lead wire. <S> This should be fine for testing or programming, however the pressure may not be high enough to ensure long-term reliability if you intend it as a "permanent" connection. <A> The serrated and sharp variants are used to pierce flux residue. <S> As for the receiving side, a pad usually does the trick, however, if your PCB surface plating is prone to oxidation, you can use selective gold plating on those pads. <A> For temporary connection (test points for signal checking or flashing microcontroler) <S> I am using similar connection to this one... <S> but in my case pins are on the right angle related to PCB and they are going going inside PCB. <S> However, for permanent connection I am using low profile female (individual) pin sockets.	There are pogo pins for more permanent usage, usually, they have either flat or domed tips.
How should I connect my PCBs? I will need to connect several PCBs within a single enclosure for the project I am working on (They are split to reduce cost should there be a manufacturing error). I need to connect an I2C Bus, 5V , an analogue and 3 digital signals (and importantly, the ground plane) from each 'slave' board to the master. Is there a universal connector that would be appropriate for all of these signals?I am also under the impression that I want as 'wide' a connection for the ground planes as possible (Multiple connection points?) to reduce RF emissions? to clarify: all of the digital signals are 5V logic, either from a PSOC, arduino or a CMOS chip. The analogue signal is 0-5V. Signal integrity of the analogue line is a priority as I cannot connect the external reference signal to the line due to pin restrictions and am relying on an internal reference. The PCBs are 35mm Wide (Though I will take them down to an inch if I can manage), will be placed basically touching each other, and arranged like this: . _____________________ \|######MASTER#### \| \|____________________ \| # S # \|# S # \| # S # \| \| # L # \|# L # \| # L # \| \| # A # \|# A # \| # A # \| \| # V # \|# V # \| # V # \| \| # E # \|# E # \| # E # \| \| # # # \|# # # \| # # # \| \|______\|______\|______\| Sorry for the bad ascii art, I made bad paint art but couldnt attach it. <Q> I would use standard 2.56 pitched pin rows. <S> Preferably gold coated. <S> Either vertical or at 90 degree, according to the most rational set up geometricaly. <S> For even better contact, look for gold fingers . <A> Given this reasoning: (They are split to reduce cost should there be a manufacturing error) <S> Solder wires between them. <S> If you decide after assembly that there was an error and something needs to be replaced, clip the wires, remove the board, unsolder the wire stubs, and wire in a new board. <A> If you are not doing anything high speed (e.g. I2C is the fastest relevant signal) <S> I suggest something like short 2mm ribbon cables. <S> Pre-made IDC ribbon cables are durable and don't place a lot of strain on the connector when you have them hanging around without a bespoke mechanical enclosure. <A> This is quite an open-ended question but something to consider is whether you want a rigid or flexible connection. <S> I usually opt for a flexible connection using wiring terminated by a JST connector of some type. <S> JST homepage listing their various options	You don't really need a connector, if you are not really making something with subsystems that are removed and replaced, and connectors are both expensive and prone to issues with connection quality.
what causes voltage harmonics People keep telling me that Nonlinear loads causes current harmonics and when i ask them about what causes voltage harmonics, they tell me wherever there is current harmonics there is voltage harmonics. So i did the experiment myself, powered a rectifier and DC motor using AC, and monitored using the Fluke PQ analyzer, the current THD was 32% but the voltage THD was almost 0% ! Could you please explain to me how harmonic voltages are created ?and Does it happen in the case of inverters only ? <Q> A true voltage source cannot be altered by load currents of any type (harmonic, transient or otherwise). <S> This is because an ideal voltage source has zero output impedance. <S> If a voltage source has non -zero output impedance (and harmonic or transient currents are drawn from the source), then there will be remnants of these disturbances seen on the output terminals of the voltage source. <S> This is due to the non-zero impedance of the source. <S> The current passes through that series impedance and creates a volt drop that adds/subtracts to the original immutable voltage source. <A> Just add a current-to-voltage-translator (somethimes called "resistor") in series with the non-linear load and you will see also voltage harmonics across the load (and also across the resistor). <S> By definition the voltage across a (ideal) voltage source cannot be changed (as also noted in another answer). <S> So if the voltage source provides pure sinusoidal voltage (i.e. no harmonics) <S> no matter what load you have (linear or non-linear) <S> the voltage will be pure sinusoidal. <S> This changes as soon as the voltage source has a series resistance. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Left circuit: <S> current throuh non-linear element (diode) will have harmonics. <S> Voltage across non-linear element will not (cannot) have harmonics because V1 is an ideal voltage source. <S> Right circuit: <S> current throuh non-linear element (diode) will have harmonics. <S> Voltage across resistor will be proportional to non-linear current through resistor, i.e. will have harmonics. <S> Also voltage across non-linear element (diode) will be pure sinusoidal minus voltage with harmonics, i.e. will also have harmonics. <A> Voltage harmonics result from voltage drops across the source impedance due to current harmonics. <S> If the current harmonics are low in comparison to the short circuit current capacity of the source, the voltage harmonics will be low. <S> Any nonlinear load will have current harmonics. <S> Nonlinear loads include rectifiers, controlled rectifiers, voltage converters etc. <S> Saturation of magnetic circuits also causes harmonic currents. <S> Inverters that are powered by grid power and supply powers for variable speed motor control can cause harmonic currents in the motors. <S> That type of inverter receives grid power through a rectifier, so the rectifier can cause harmonic currents. <S> All loads that cause harmonic currents can have the harmonic currents reduced by mitigation measures in the load device. <S> Harmonic currents are also reduced by adding source impedance. <A> Harmonics are a result of our mathematics, where correlation transfers energy into narrow-band filters or basis-functions of the Fourier math. <S> Fast rise and fall times will correlate with anything; a reoccurrence of the fast edge, of the same polarity, but exactly one period (of the narrow band filter) later will produce an interesting behavior. <S> Here is 0.31MHZ squarewave into a 1.000MHz filter with Q = <S> 100	Inverters that supply power to the grid can cause harmonic currents.
What does "normalized" mean in this datasheet? I came across the graph below in ONSemi's 2N3904 datasheet that depicts the DC current gain. I don't understand what "normalized" means. I've read other explanations by Googling, but found all of them to be more vague than helpful. Please explain what "normalized" means in this context, and why they would use it in place of the DC gain numbers? <Q> I think there can be a rather large variation in the gain between individual parts. <S> A range of a 100 <S> ..300 is given elsewhere in the datasheet. <S> (page 2 in this datasheet from ONsemi ) <S> For a particular part, the actual gain at 10 mA/25 °C might be 120 and for another part it might be 150. <S> The graph seems to be fixed so that the 25 °C curve is 1.0 at 10 mA, so it would appear the other values are given in relation to that point. <S> The absolute values aren't useful here since they might be different from part to part. <S> In other words, whatever the gain at 10 mA/25 °C for a particular transistor, the graph tells us the gain at 1 mA should be around 0.75 times that. <S> (Not that this is mentioned explicitly in the datasheet, so I might be mistaken.) <A> In this case, "normalized" means "relative to the values stated elsewhere" . <S> In the example you show, note that 1.0 is about in the middle on the Y axis. <S> That Y value of 1.0 tells you that you get the gain stated elsewhere in the datasheet. <S> At 125 °C and 5 mA collector current, you get about 1.5 times more gain than the numbers in the table otherwise indicate. <S> The way the datasheet is organized, the tables give you the expected gain at a few points. <S> Many things effect gain of transistors. <S> The purpose of the graph is to give you some guidance on how the gain changes as a function of collector current and temperature, relative to the numbers stated in the table . <A> The data sheet normally gives a clue: - The table doesn't explicitly state that hFE is normalized at 100 to 300 (Ic = 10 mA) <S> but this is what they actually imply. <S> So a normalized 1.0 can cover transistors with a range of hFEs from 100 (minimum) to 300 (maximum).	A normalized value of 1 therefore means "the same as the value stated elsewhere" .
What is the voltage potential of ground during a short During a short, is it possible for the voltage potential of the ground to change? Say for example, we short a 9V battery, how does the ground stay at 0V and not become 9V? <Q> Ground is a reference point so it is always regarded as 0 volts. <S> Of course in a lightning strike the localized area around the strike can create massive potentials per metre so if you value your life (and the lives of any future children you wish to bear), keep your feet close together if there's a risk of lightning. <S> Picture source <A> Circuit ground is just some point from which we measure or reference voltages. <S> If you like, it's the point where you connect the black lead of your multimeter. <S> All other voltages can then be measured with respect to that point. <S> It should be clear from above that if you connect the red lead of your multimeter to the circuit ground that it will always read 0 V no matter what is happening to the battery. <S> (In real life it can be a little more complex if there is a current flowing in the ground circuit and there is some resistance between the two points.) <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Both VM1 and VM2 will read zero volts. <S> Battery internal resistance limits the current flow. <A> The resistance of standard PCB foil is 0.0005 ohms (500 microOhms) per square of foil ---- for any size square of foil. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Each square is 0.0005 ohms. <S> To exit the entry point, there are 8 surrounding squares, thus R of that first ring of square is 0.0005/8 = <S> 0.00006 ohms. <S> Then the next (3x larger) ring adds another 0.00006 ohms. <S> Etc. <S> But in the larger rings, the current is mostly heading toward the DESTINATION output node, and not all squares get equal current density. <A> Ground is just a reference point, it does not need to be an actual connection to the earth. <S> When you short a 9 V battery, things which you can normally ignore (like the battery's series resistance) come into play. <S> This prevents the ground potential from changing depending on what you choose as your ground reference point. <S> In the left circuit, the 1 ohm series resistance of the battery can be ignored as the resistance of the load is 100 times higher. <S> R1 doesn't do much really. <S> In the right circuit R3 does come into play, it limits the current which will flow. <S> The full 9V will be across R1. <S> V by definition . <S> Theoretically if you had an ideal battery with zero series resistance, an infinite current would flow if you shorted that ideal battery. <S> But also then ground remains 0 V. simulate this circuit – Schematic created using CircuitLab <A> The term "Ground" has historically meant ground literally: an electrical connection to the earth itself, typically via a rod driven into the earth. <S> This was used in power and telecom systems to obtain a voltage reference, and sometimes as a current return path. <S> In electronic circuits, the term has persisted and is typically used to denote the negative side of the power supply, even in cases where the circuit is isolated from the actual earth ground. <S> In circuit analysis, "Ground" is often used to specify a reference point from which other circuit voltages are measured; therefore it is by definition at zero volts. <S> In the 9V battery example, suppose you specify the negative terminal as Ground. <S> If you short the positive terminal to Ground, a current flows through the short circuit, and the positive terminal will be (nearly) at zero volts, since a physical battery, unlike an ideal voltage source, has an internal resistance. <A> Ground and Earth are often confused. <S> Personally I only refer to Earth as a connection to to the local electricity supply earth which is literally connected to the planets ground as pointed out by others. <S> Ground on the other hand means a circuit reference and may, but is not required to, be connected to earth. <S> Consider the circuit below <S> we have two volt meters what do they measure? <S> VM1 if sensitive enough should measure a positive voltage because there is current flowing in the connection between its inputs. <S> Theoretically we have zero resistance <S> so no matter how many amps are flowing out of the shorted battery it should be zero <S> but we never, except possibly with super-conductors, really have zero resistance. <S> VM2 does not see the drop from the battery current so sees 0V as expected. <S> We talk about 'local grounds' for this reason each bit of circuit cares about the point it is referenced too which may not be the same everywhere if significant currents are involved or at 'high' frequencies due to inductive effects. <S> simulate this circuit – <S> Schematic created using CircuitLab	The ground will still be 0 V, it cannot even change as ground means that it is 0
Voltage divider ratio limitation applied to potentiometer I'm trying to find the general formula or more like understanding of how to choose resistor values to limit the voltage ratio when using a potentiometer. A potentiometer is equal to a voltage resistor divider where the ratio depends on the resistor value. Okay, so you can tune from 0% to 100% which means for a Vin of 5V for example, from 0 to 5V. Now let's say I want to limit this ratio to a certain value as from 50% to 70% to obtain a Vout ranged from 2.5V to 3.5V. So then, I'm sure when I will change the potentiometer value the voltage will be limited between this voltage range. I understand by adding a grounded series resistor the ratio will then change to constrained value since when the potentiometer will be at min/max value, there will still be this resistor to take into account: Which will give a new ratio of 50% to 100% (so if VCC = 5V, 2.5V to 5V) due to the 2k series resistance added. (2k/4k for min potentiometer state and 4k/4k for max potentiometer state) But I can't get my head around how to now get arbitrary ratio as 50% to 70%.What I am not seening ?Thank you <Q> 1) <S> Choose R total for the string. <S> 2) Choose Range difference , thus 70-50% = <S> 20% which is your RV ( variable resistor) <S> 20% of R total 3) <S> Calculate the remaining fixed R's using Min and Total-Max, i.e. 50% and 30% of R total. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If your 2k pot is to give you a span of 20% from 50% to 70% <S> then you need \$ \frac {2k}{20} = <S> 100 \ \Omega \$ per %. <S> So the bottom, 50%, resistor will be 50 x 100 = 5k. <S> The top, 30%, resistor will be 30 x 100 = 3k. <S> Pick one that suits and calculate the other resistors afterwards. <S> I'm trying to find the general formula or more like understanding of how to choose resistor values to limit the voltage ratio when using a potentiometer. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Using min and max in percentage form: <S> $$ R_1 = <S> \frac { <S> R_2 \cdot <S> (100 - max)}{max - min} $$ <S> $$ R_3 <S> = \frac {R_2 \cdot min}{max - min} $$ Testing for your example with a 2k pot and a 50% to 70% adjustment range: \$ R_1 <S> = \frac {R_2 \cdot (100 - max)}{max - min} <S> = \frac {2k \cdot (100 - 70)}{70 - 50} = <S> 3k \$ <S> \$ <S> R_3 <S> = \frac {R_2 \cdot <S> min}{max - min}= \frac {2k \cdot 50}{70 - 50} = <S> 5k \$ <S> Obviously, min can be set as low as 0% and max to 100% which will result in R3 or R1 being 0 Ω. <A> simulate this circuit – Schematic created using CircuitLab <S> Consider the circuit above. <S> The pot can be adjusted from 0 to 100% so first label the node between R1 and R2 with the higher percentage voltage, 70% of 5V is 3.5V. <S> Now label the node between R2 and R3 with the lower percentage voltage, 50% of 5V is 2.5V. <S> Now since you know the value of your pot and the voltage across it you calculate the current in all 3 resistors using ohms law. <S> Since you know the voltage across R1 and R3 together with the current we can calcu1ate the value of R1 and R3 again using ohms law. <S> Simple. <S> My answer above assumes there is no current being taken from the wiper of the pot as do all the other answers posted here so far which may be a reasonable assumption but may not be the case if the current taken from Vout is not insignifiant. <S> So lets add a load resistor RL between Vout and 0V <S> We now have to solve two simultaneous equations <S> assume you know RL and R2 $$\begin{align}\\BottomPercent & = 100 <S> \times <S> \dfrac{R3 \|\| RL}{R1 + R2 + R3 \|\| RL} \\ & = 100 <S> \times \dfrac{R3 \cdot RL}{(R1 + R2) <S> \cdot (R3+RL) <S> + R3 \cdot RL}\end{align} <S> $$ <S> $$\begin{align}\\TopPercent & = 100 <S> \times \dfrac{(R2+R3) <S> \|\| <S> RL}{R1 <S> + (R2 + R3) \|\| RL} \\& = 100 \times \dfrac{(R2 + R3) <S> \cdot RL}{R1 <S> \cdot <S> (R2+R3+RL) <S> + (R2+R3) \cdot <S> RL}\end{align} <S> $$ <S> Note <S> if the output is loaded then output will not adjust in a linear fashion. <A> Work it backwards.... <S> A simple trick is to figure out the percentage range you want from the pot. <S> In your example you want 50 to 70% of Vcc or 20% range. <S> Not pick a pot which is a nice multiple of 20. <S> Perhaps 200R, but lets say 2K. <S> Now it is a simple matter to deduce that 50% would need 5K and 30% would need 3K Hey presto. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> But say you wanted 60 to 75%, 15% range Pick a 1.5K pot, and the bottom resistor R2 is now 6K and R1 is 2.5K. simulate this circuit <S> Of course, you can not always get pots of the right value so you may need to scale it to a different total value.	Generally you will be constrained by the available potentiometer values.
Calibrating a capacitive moisture sensor I am doing a project which require to build a moisture sensor to measure wood chip moisture content. I built the the cylinder and put all the capacitor plate around it and I was able to measure the capacitance of the wood chip. However, I do not know how to convert capacitance that i got to the moisture content. I did a lot of research online but they do not talk briefly about it. Is there any specific equation or way to convert capacitance to moisture content? <Q> There's not going to be an equation online for you to just plug in your values, as it's going to depend on your specific device, <S> how you're measuring the capacitance, and probably even the wood you're using. <S> Instead, you're going to need to calibrate it yourself, so you'd collect a bunch of points at known moisture values (ex. <S> 0%, 10%, 20%, ...) <S> and measure the capacitance for each. <S> Assuming they form a smooth curve, you can then interpolate between capacitance/moisture values to get your reading <A> Prepare samples of wood-chip with, if possible, a range of moisture values that cover the range you expect to see in practice. <S> This may involve drying some and steaming some or just leaving them outside, etc. <S> Bag and label each sample. <S> Take a moisture content measurement of each sample using your sensor. <S> Record the reading and the batch used. <S> Immediately weigh the sample. <S> A digital kitchen scales should suffice. <S> Record the readings, (\$ m_1 \$). <S> Dry <S> the samples in the oven at a gentle heat until you think the are completely dry. <S> Weigh each sample again. <S> Record the readings, (\$ m_2 \$). <S> Plot the results on graph paper. <S> If you are a good lab technician your data points will align closely with the best-fit curve between the points. <S> Moisture content can be calculated as \$ \frac {m_1 - m_2}{m_2}\cdot 100 \$ (%). <S> Figure 1. <S> Microwave Oven Drying, by Richard Jones makes it look easy. <S> Moisture Content by the Oven Dry Method may be of interest. <A> To get the best accuracy, you'll need to do the tests at a stable ambient temperature then repeat the tests at different ambients. <S> You will probably find a variation in your readings which you can also plot on your graph. <S> If you wanted to take it further, you could then try different materials and repeat all the tests again. <S> Plugging your values into excel will give you (hopefully) a nice visual representation of your results which you can then use as a means to calibrate your device. <S> Be sure to double check the results, don't just do it once. <S> You need to make sure your results are repeatable if you want to use the data for calibration! <A> You need to buy/rent an appropriate sensor to calibrate your own. <S> You soak your wood chips, then measure the moisture content using your calibrated sensor and the capacitance with your device. <S> That gives you the first point. <S> By example 1.000nF => 95% moisture. <S> Then you dry your wood chip and measure with the sensor and your sensor. <S> You get second point, by example 0.990nF => 5% moisture. <S> You plot the line and you get moisture level as a function of capacitance. <S> In fact, it will be much more complicated: you may have non-linearity in your device forcing you tu use either a non-linear model or to take a lot of measurements <S> (maybe 1% moisture apart). <S> The temperature may offset your curve and so on. <S> This is yours to try...	The best way to do this would be to use some sort of reference meter and several objects of known moisture levels in 5 or 10% steps and plot the capacitance read on each of them.
Multiplexer not switching any audio input I have designed this audio switch. 1 out of 4 TRRS inputs is selected. I am testing it but it does nothing. I press the switch to switch the next channel and nothing works. The idea here is: the 555 debounces the key and a pulse is injected on the flipflop when the button is pressed. the flipflop is a 2 bit counter, producing a sequence from 00 to 11 as the button is pressed repeatedly. the two bits from the flip flop selects one of the four channels of the MAX359E switch. The leds display 00, 01, 10, 11, depending on the input selected, obviously... I am not sure what to do with the mic input in relation to the input/output capacitor. This is because some TRRS inputs, like iPhone's, send a voltage on the mic line, to supply the mic. If I put a capacitor it will remove that voltage... Any ideas why it is not working? I may have connected something wrong on the test board. I have double checked the connections several times but at this point I may be tired and not seeing the problem or the circuit is bad designed... Any ideas? Thanks EDIT: I have modified the circuit according to WhatRoughBeast suggestions. <Q> The most obvious thing I see is that your mux inputs are not ground-referenced. <S> That is, you should have a largish resistor to ground on each analog mux input. <S> If you don't, there is no way to control the "resting" level of the inputs. <S> Try something like 100k. <S> Second, tie your floating R and S inputs on IC6 to ground. <S> NEVER leave a CMOS input floating <S> - you won't believe the weird things that can happen. <S> EDIT - <S> In response to a question in the OP comments, you stated that "doesn't work" means that you get no sound from headphones. <S> Well, there you go. <S> You're not going to, either, with the MAX359. <S> I'm going to assume that you understand impedance/resistance. <S> If you look at the datasheet characteristics (page 2), the very first line is ON resistance, and you should note that typical will be about 1.2 k. <S> A headphone (earbud) will typically have an impedance of about 32 ohms. <S> This means that the signal power at the headphone will be about (32/1200)^2 of normal, or something like 1/1000 of what you expect. <S> You might try something like this: start by cranking your iphone volume all the way up. <S> Now, make sure that your area is really, really quiet - and I do mean quiet. <S> No background music. <S> Wait a few minutes listening to dead silence. <S> Now put on your head phones and listen, switching between channels and see if you can't hear a faint signal on your desired channel. <S> The MAX359, and virtually all analog mux ICs, are simply not intended for power switching. <S> They are intended to drive high-impedance loads, such as the input to a subsequent amplifier stage. <S> So, I'd suggest you get an amplifier stage to connect your output to. <S> And a simple op amp won't do very well either, since their outputs are usually not real happy driving loads of less than 500 - 1k or so. <S> Plugging into a headphone amp would be close to ideal. <S> END EDIT <A> I would venture to say that it’s not working because you aren’t generating the -5V rail. <S> The input pin to the negative regulator is floating. <S> You can test it by removing the power jack and adding to 9V batteries as shown below <A> You should not be switching signal ground, only the left and right signals. <S> Follow your signal grounds closely <S> and you will see the output is not wired the same as your inputs.	You have miss-wired your ground connections, and should only need one MAX359 IC to do all the work.
Good equipments to measure current consumption for IoT devices I was asked to bought some equipment to measure the current consumption for the IoT devices. The company is a startup and has a limited budget for this (let's say less than 5k$). I'm looking for an instrument that can: Capture the sleep current (around a few uA) Capture the active current (around a few mA) Capture the transmiting current (around a few hundred mA) Can store the recorded samples in file (about 100 millions samples or more). Testing time can last for 1-2 days to estimate the battery life. Sample rate >= 1 KSamples/s Would you please suggest me some names (DMM or oscilloscope)? Do I need a current probe for this purpose? <Q> I think a keithley SMU or an equivalent would be your best bet. <S> It is within your budget and does exactly what you need. <S> https://www.tek.com/keithley-source-measure-units/keithley-smu-2400-series-sourcemeter <S> there is also a battery simulator, but I think it lacks some of the features you need. <A> Not a complete answer but a few pointers: <S> Nordic Power Profiler kit: <S> https://www.nordicsemi.com/Software-and-Tools/Development-Kits/Power-Profiler-Kit <S> Works great with Nordic dev boards, easily usable for any other microamps to milliamps range circuit. <S> Specs are 77kHz, 0.2uA resolution. <S> Can record the millisecond long high current spikes when the nRF51 radio is on. <S> EEVBlog <S> uCurrent: <S> http://www.eevblog.com/projects/ucurrent <S> Solves the problem of voltage drop through a sense resistor quite nicely. <S> Good analog bandwidth at 300kHz (better than Nordic), converts <S> uA/ <S> nA <S> current to mV voltage (so can measure much lower current than Nordic), but you still need a DSO or fast data logger to record the output. <S> Hours to days at 100ksps works. <S> I've used both of the above for fairly complex projects. <S> Never had access to the big rigs (Keysight, HP etc) <S> so I can't compare. <S> All of the above together is well under $1k. <A> Disclaimer : I am the designer for ZS-2102-A. <S> There is one from qoitech which retails on Digikey and other distributors. <S> Another good one is "ZS-2102-A" from ZSCircuits. <S> It is available for sale at www.anglercircuits.com. <S> This tool can measure from few uA to 1A with 2us response time. <S> It can record practically unlimited record time. <S> Available for $699 with added discounts and free shipping.	A power profiler is most suited for this application. uCurrent+Saleae Logic is a good combo for very long recordings (really, just limited by your hard drive/SSD).
Analysing a band pass sallen key filter I'm trying to understand the transfer function of the band pass sallen key filter , which looks like: With the following circuit: How can I analyse it to get the transfer function? Thanks for any help. <Q> Once you get the node voltage, you can find the transfer function. <S> For the analysis, I denote node and current as in the picture below. <S> Now you can write KCL for every node and a constraint by OpAmp. <S> You can get 7 equations: <S> Then you can solve 7 equations to get all unknown voltages and unknown currents. <S> Finally transfer function is just V5/V1. <A> General hints (works not only for this particular configuration): <S> Do following steps in \$s\$-domain: <S> set up equation vor \$v_+\$ <S> (pos. <S> input of OpAmp) as expression of \$v_{in}\$ and \$v_{out}\$. set up equation vor \$v_-\$ (neg. <S> input of OpAmp) as expression of \$v_{out}\$. solve the equation \$v_+ <S> = v_-\$ for \$v_{out}\$ (e.g. by nodal analysis or by network transformations) <S> plug in the resulting expression in definition of \$H(s)\$: <S> \$H(s) = \frac{v_{out}}{v_{in}}\$ <S> The resulting expression won't contain \$v_{in}\$ (or \$v_{out}\$ or \$v_{aux}\$) any more but will be just an expression of the parameters \$R_1\$, \$R_2\$, \$R_f\$, .., \$C_1\$, .. and the independent variable \$s\$. <S> You can use symbolical math tools, e.g. sympy package for Python, Maple, Mathematica... <S> Here is a python script doing the algebra (using sympy ); not sure though if correct: # deriving the transfer function of a Sallen-Key band pass filterfrom sympy import Symbol, symbols, solve, collects = Symbol('s')def Xc(C): global s; return 1 / <S> (s <S> * C)Vin <S> , Vout, Vaux = symbols('Vin Vout Vaux')R1, R2, C1, C2, Rf, Ra <S> , Rb = symbols('R1 R2 C1 C2 <S> Rf Ra Rb')X1, X2 = Xc(C1), Xc(C2) <S> # get expression for Vaux by solving KCL in node_aux: <S> Vaux_expression = solve( (Vin - Vaux) <S> / R1 <S> + (Vout - Vaux) / <S> Rf - Vaux / X1 - Vaux / (X2 + R2), <S> Vaux)[0]Vpos = Vaux_expression * R2 / (X2 + R2) <S> # voltage at pos. <S> input of OpAmpVneg = <S> Vout <S> * Ra / (Ra + Rb) <S> # voltage at neg. <S> input of OpAmp# get expression for Vout by solving equation Vneg = <S> Vpos for VoutVout_expression = <S> solve(Vpos - Vneg, Vout)[0] <S> # get transfer function H(s) by defining formula:H = Vout_expression / <S> VinH = collect(H, s)print H Result: <S> C2R2Rfs(Ra + Rb)/(C1C2R1R2RaRfs*2 + R1Ra + RaRf + s(C1R1RaRf - C2R1R2Rb <S> + <S> C2R1RaRf <S> + C2R2RaRf)) <A> This filter can be analyzed using the fast analytical circuits techniques or FACTs . <S> The principle is to determine the time constants of the circuit in two different configurations: when the excitation is reduced to 0 V and with a nulled output when the excitation is back. <S> Reducing the excitation to 0 V, means replacing the input source \$V_{in}\$ by a short circuit. <S> Then, "look" at the resistance \$R\$ offered by the energy-storing elements (the caps) to form time constants, \$\tau_1= <S> RC_1\$ <S> and \$\tau_2= <S> RC_2\$. <S> The below drawings illustrate this principle: Using these two drawings, you determine the following time constants: \$\tau_1= <S> C_1(R_1\|\|R_f)\$ <S> \$\tau_2 <S> =C_2(\frac{k_1(R_1R_2+R_1R_f+R_2R_f)-R_1R_2}{k_1(R_1+R_f)})\$ <S> \$b_1=\tau_1+\tau_2\$ <S> Then, you set capacitor 1 in its high-frequency state (a short circuit) and determine the resistance \$R\$ looking into \$C_2\$'s terminals: <S> You have \$\tau_{12}=C_2R_2\$ and \$b_2=\tau_1\tau_{12}\$ <S> Finally, you determine the gain \$H^2\$ when capacitor \$C_2\$ is a short circuit: <S> The complete transfer function is determined according to \$H(s)=\frac{sH^2\tau_2}{1+sb_1+s^2b_2}\$ <S> However, even if this expression is mathematically correct, you have zero insight on the plateau <S> gain \$H_{MB}\$ and the tuning frequency which are truly the design goals. <S> You need to rework the formula according to the following low-entropy format: \$H(s)=H_{MB}\frac{1}{1+(\frac{\omega_0}{s}+\frac{s}{\omega_0})Q}\$. <S> This is what the following Mathcad sheets show and compare the various responses: <S> So the point is not just write a transfer function linking \$V_{out}\$ to \$V_{in}\$ but more rearranging the result in a meaningful form from which you gain insight and can design for a certain goal in tuning frequency and quality factor: this is all what FACTs are all about.	You can use modified nodal analysis to solve for all unknown node voltages and unknown currents.
Practical way to generate 1500V? Designing the power supply for an radiation detector, need to supply up to 1500V at 2 mA from mains (170 Vdc). I've largely considered a flyback converter, but cannot find a flyback transformer with a gain of N=~16 (for a gain of 8 at 50% duty cycle) rated for 1500V at the output winding. Looked into charge pumps/voltage multipliers some but not in depth yet. Also just recently learned about CCFL inverter transformers, but still need some time to understand them better. This project requires a small size and weight, so using a large transformer is not preferred. What are other ways to step up voltage which I may look into? <Q> Use a cascade. <S> There are two main choices a) <S> Villiard aka Cockcroft-Walton <S> aka Greinacher multiplier <S> Each stage is stacked on the previous stage, so capacitors and diodes only need to be rated at the input voltage, not the output voltage. <S> However the output impedance increases as the number of stages squared. <S> b) Dickson multiplier <S> Each stage returns its pump and storage capacitors to ground, so capacitors need to be rated for the output voltage. <S> However the output impedance increases only as the number of stages, so it can use smaller values of capacitance for the same output impedance as the other cascade. <S> Given a specification for output voltage, output current, and output impedance, you would need to fully design one of each type to determine which will give you the best cost or volume. <S> Very high voltages would naturally favour type (a), but with only 1500V, both types are contenders. <A> Try researching Geiger counter power supplies. <S> They don't produce 1500 volts but they do lend themselves for modification such as this design: - Picture from MAXIM website . <S> As you can see the output stage is a cockcroft walton multiplier so you can add more stages and get more output voltage. <S> Alternatively you build two of these (with fewer added CW stages) and make a bipolar supply that spans +/-750 volts. <S> The circuit above runs from 5 volts but the principal is the same for any DC supply voltage; you make a (circa) 50 kHz oscillator and amplify it to produce a large peak-to-peak voltage swing then use the CW multiplier to make a larger DC voltage. <S> More Geiger Muller tube power supply images <A> There are many companies which will build transformers to your specifications. <S> Books such as Switching Power Supply Design by Abraham Pressman (ISBN: 978-0071482721 ) and Switchmode Power Supply Handbook by Keith Billings <S> (ISBN: 978-0070053304 ) go into great depth on how to design custom transformers for switching power supplies. <S> To prototype your design you can use ferrite cores from companies such as Magnetics, Inc. and Ferroxcube <S> As for other methods of producing high voltage I would consider a simple boost converter . <S> Properly designed and provisioned for safety, I believe it would also be the smallest and cheapest. <S> (Considering that you can easily buy fast diodes rated >2kV these days) <S> [Edit] <S> The reason I would not recommend a voltage multiplier is that the requirement is 1500v * <S> 2mA <S> = 3W. <S> It is unlikely that a voltage multiplier will practically work.	When building switching power supplies it is more common to use a custom transformer than it is to use an off-the-shelf component.
How does this radio transmitter circuit oscillate? Hello. I am trying to understand how this circuit operates. I understand how the circuit works on the right side of the transistor, but the oscillation stage with the crystal confuses me. It appears that the crystal has no feedback from the output of the oscillator. I researched this and found out that the collector-base capacitance of the transistor provides a feedback path, but wouldn’t that only give a 90° phase shift instead of the 180° phase shift required for positive feedback? I have seen similar circuits where a variable capacitor is included with the crystal to adjust frequency. Would that give the phase shift for the remaining 90°? Thanks, your help is appreciated. <Q> Yes, it could oscillate, but in a SPICE simulator, it didn't. <S> Not quite. <S> A few component changes did start oscillations. <S> The 7MHz crystal equivalent circuit is a guess (C1, L2, R5, C2): The base-to-emitter capacitance of the 2N2222 is large enough that this is a Colpitts-type oscillator. <A> This question has quite interesting answer history - at least for +10k <S> rep members who can see the whole history. <S> But there's done some reductions = <S> > <S> I think that now here's also room for my answer: <S> At first: The crystal can be any reactive impedance from nearly zero ohms to very high number of ohms. <S> The reactance can be as well inductive as capacitive and <S> the losses are extremely low when compared to practical LC-circuits. <S> And all those reactance values are found from very narrow frequency band around the stamped frequency of the crystal. <S> = <S> > <S> It's well possible that at some frequency the CB capacitance of the transistor and the crystal form together a phase inverting voltage divider which attenuates less than the amplifier amplifies => oscillation. <S> In practice also the input impedance of the transistor must be taken into the account => exact full 180 degrees phase shift in the feedback route doesn't happen. <S> But the amp also doesn't cause exact 180 degrees phase shift, because the loading is partially reactive = <S> > <S> It's still well possible that oscillation happens. <S> There's no need to try to classify this oscillator "is it hartley or colpitts or or clapp or some other well known type". <S> Those well-known LC oscillators were designed to make oscillations possible and controllable with low gain triode electron tubes. <S> We have here a high gain transistor and the crystal. <S> But if someone forced me to name one old electron tube oscillator that can be considered to be the grandma of this circuit, I would write TGTP (=tuned grid, tuned plate). <S> ADD: <S> Radio circuit engineers do amplifier stability calculations. <S> It's not uncommon to find that amplifier is unstable due the reactances of input signal source, load reactance and the internal feedback of the transistor. <S> Microwave oscillators are often constructed as unstable amplifiers. <S> In place of the crystal there's a high-Q microwave resonator. <A> Draw the circuit like this. <S> The inverting amplifier inverts between base and collector. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The missing knowledge is that:A cap's current leads <S> it's voltage by 90 degrees. <S> An inductor's current lags <S> it's voltage by 90 degrees. <S> When they are in series, the current is the same for both, so the junction voltage is 180 degrees at resonance. <S> Thats also why a series resonant circuit appears as a short. <S> Now reason out the parallel resonance circuit, where the two elements both have the same voltage. <S> As mentioned above a crystal is a series or parallel resonance circuit. <S> Yes the transistor's collector-base capacitance provides driving energy. <S> BTW: <S> Often at a frequency so high, it is Only noticed as a DC shift when you wave your hand over it. <A> If you temporarily remove the crystal, you should see that the circuit will oscillate at a frequency primarily determined by RFC1 and C1. <S> The only thing the crystal does, is stabilize <S> the oscillation frequency!	Many FETs oscillate due to gate inductance and drain to gate capacitance.
Transistor to turn 3 Amp motor on/off I have 3Amp (max) 12v DC brushed motor and power supply. I'm thinking of turning this motor on/off and (maybe) control its speed with a transistor. I'm not an expert in electronics at all (I'm a programmer). Another variant is using relay, but I see transistor as a better solution. To control transistor I want to use Arduino signal. So I tried to search for a transistor that can maintain 12v and 3amp and use signal of 0-5v, but it seems there are no such transistors. So my question to you - is my way of solving this problem is right? Or should I use relay for it (but obviously I can not control the speed of the motor with the relay). <Q> Use a MOSFET as a switch. <S> MOSFETs have high switching frequency and can pass high current through them. <S> Use a flyback diode to prevent the back emf of the motor from damaging the Arduino. <A> Consider that DC motors can have start currents up to 10x rated currents, you either must have a soft start or drivers rated those currents and Pd heat loss from voltage drop. <S> Consider a 3A <S> * 12V motor uses 36W at full load but 360W (peak) at start if full voltage is applied. <S> Losses of <2% are desirable for thermal design but also affect the cost of the drivers, so tradeoffs are required. <S> A full bridge is used for bi-directional control. <S> Depending how you control the soft start (acceleration) with coasting or braking and you can decide what power you need. <S> When coasting the motor turns into a generator. <S> e.g of a high power driver. <S> https://www.pololu.com/product/1451 <S> Cheap and hot driver <A> Here is the common MOSFET circuit for Motor Control.	You could try PWM concept in Arduino to control the Motor speed and it can be done via MOSFET.
10k Resistor Pull Up/Down Standard for 74 series chips I've been having some issues lately with the usage of 10k ohm resistors as pull down resistors for 74ls series chips. Most people seem to choose 10k resistors as their default pull up/down value but I've found that the pins seem to float a little anyway when trying to pull down on the TTL 74ls series chips. My solution was to switch to a smaller value that I had on hand such as 1.1k ohm. My question is why 10k is the default assumed by many making electronics projects, and why might I be having a problem with this value that is eliminated by using a lower resistance value? I haven't been playing with hc series much lately so I am unaware if the problem is particular to the 74ls series. Edit: I also experienced this problem with a 28C16E EEPROM chip and solved it with 1.1k instead of 10k. Could this be related to noise in my power supply setup (c.M.T-305D set to ~5V DC)? Edit: I notice people seem to have interpreted the pull down resistors as being used when the input is unused (could be tied directly to ground for 74ls or 74hc). This was not what I meant. The case for these pull down resistors is with the use of dip switches or I/O with a 5V bus. <Q> Unused high inputs for 74S and 74 series logic with multiple emitter inputs should be connected to Vcc through a resistor ( for protection of the input). <S> The resistor can be calculated as follows (from these ancient scriptures): <S> If you run the numbers you will see that 1K to 10K or even higher is fine for a pullup (but you'll get less EMI immunity with the higher values). <S> For other logic families (74HC and other CMOS types, 74LS) <S> the inputs can be connected directly to Vcc. <S> (74LS is really DTL logic, there are no multiple emitter inputs). <S> There is no reason not to connect any unused 74x input you want <S> low directly to GND (use a 0 ohm resistor if you want to make it removable for some reason). <S> Anything higher will reduce the noise immunity. <S> Of course you can also connect the unused input to an output that has the desired logic state. <S> For example, grounding the input of an unused 74S04 inverter and using the output to drive up to the maximum fanout of inputs high <S> If it isn't really unused and you actually need a pull-down- a 250 ohm resistor will drop about 400mV at 1.6mA (numbers that mean something with the old 7400 series logic- 400mV is the maximum output voltage for 400mV noise immunity and 1.6mA is the maximum input current). <S> The input current for 74LS logic is 1/4 as much (400uA), <S> so 1K is the maximum pull-down for 74LS logic that still guarantees 400mV noise immunity. <S> If you were doing some kind of funky diode logic this might be useful. <S> Keep in mind that the input current is much less for a high input (40uA) vs a low input (-0.4/-1.6mA). <S> Internal pullup/pulldown resistors are necessarily a trade-off and tend to be on the high side for high-EMI environments, so we often add lower value resistors in parallel. <S> On the low side, drive capability comes into play. <S> For example, for I2C bus pullups the bus capacitance vs. pullup resistor value can come into play. <A> They never used pull down R's for TTL since the currents and threshold voltages were asymmetrical with 0.25mA for 74LSxx. <S> CMOS on the other hand <S> almost no static current. <S> This is a typical 74LS TTL schematic. <S> The input threshold is 2 diode drops compatible with all TTL families. <S> Added. <S> —- If a pull down on true74LS were used 0.8V/0.25mA= <S> 320 Ohms the driver has insufficient Ioh current to reach 2V/320=6mA <A> 10k resistors as [..] <S> default pull up/down value <S> You can't assume that this is in general a good choice. <S> For CMOS logic (e.g. 74HCxx) <S> this is ok (and you can use even much larger values for pull-up as well as for pull-down) but for the 74-series 10k as pull- down is not ok; only for pull-up. <S> Note: an open input acts as H for the 74xx- <S> (TTL) or 74LSxx-series. <S> So you can image that you need a smaller resistor value for pull-down. <S> See also Tony's answer and henros comment. <S> Inputs with pull-down resistors are rather unusual for TTL/LS logic. <A> In TTL 74LS serie, a input signal between 0 and <S> 0.8V is considered “LOW”, and a input signal between 2.0 and 5.0V is considered “HIGH”. <S> Any voltage between 0.8 and 2.0 volts is undefined. <S> Therefore, you have to guarantee in your design that you will never enter in the uncertainty zone. <S> Pull-up Resistor Value $$Rmax = <S> \frac{V_{CC}-V_{IH(MIN)}}{I_{IH(MAX)}}$$ <S> where: $$V_{IH(MIN)}$$ is the minimum input voltage guaranteed to be recognized as a logic “1” (2V approx. <S> , but see in datasheet).$$I_{IH(MAX)}$$ is the max current flows into the TTL input when the input is a 'HIGH' (see in datasheet). <S> If you use a higher resistor, you will have a voltage within the undefined zone. <S> For the pull-down resistor, the analysis is similar. <S> Pull-down Resistor $$Rmax = <S> \frac{V_{IL(MAX)}-V_{IL(MIN)}}{I_{IL(MAX)}}$$ <S> where: $$V_{IL(MAX)}$$ is the maximum input voltage guaranteed to be recognized as a logic '0'.(0.8V approx., but see in datasheet).where <S> : $$V_{IL(MIN)}$$ is the minimum input voltage guaranteed to be recognized as a logic '0'. <S> (0V approx., but see in datasheet).$$I_{IL(MAX)}$$ is the max current flows into the TTL input when the input is a 'LOW' (see datasheet). <S> Again, if you use a higher resistor, you will have a voltage within the undefined zone. <S> A much more detailed explanation, you can find it here . <S> In order to know why the 28C16E EEPROM did not work with a 10k pull-up, it would be useful to see the schematic circuit.	On modern CMOS logic, the choice of pullup or pulldown resistor is largely a matter of noise immunity, speed (to charge input and stray capacitance) or in extreme low power situations, of leakage.
Converting a push on/push off SPDT switch into a toggle I'm working on a small electronics project: trying to convert a guitar footswitch into a push-to-talk foot pedal. At my disposal, I've got a Teensy acting as the actual HID device (to be connected to a computer via USB), a single-pole double-throw footswitch button identical to this one , and the footswitch case itself. My issue is that I'd like this footswitch to be a toggle: at a high level, putting your foot down should enable speech capture, while releasing it mutes a connected microphone. Is it possible to reliably implement this kind of logic in the Teensy given that the switch I've got on hand is a push on/push off? <Q> <A> It's simple. <S> Treat it like a spst, leaving one side disconnected. <S> The side you use, wire it like any momentary push button, with a pull up resistor. <S> Your microcontroller code handles the rest as you describe. <S> When pressed, an interrupt enables the speech control, then waits until the switch is pressed again, to send the mute command. <S> The physical state, the mechanical action of the switch doesn't affect it's electrical behavior, and the rest you handle in your code. <A> Is it possible to reliably implement this kind of logic in the Teensy given that the switch I've got on hand is a push on/push off? <S> No of course not - when you release your foot the switch stays latched until you press the switch again. <S> Implementing a solution using all the processing power in the world doesn't overcome the physical obstacle that a push_on push_off switch doesn't do what you want.	If you want a push-to-talk switch, replace the button you have with a momentary pushbutton - no microcontroller or logic required.
How to convert a metal work table surface for electronic work? I have a 23" x 60" metal work table...the kind that restaurants use for prep work. I'm using it as my computer table & electronics workstation. I want to make the surface non-conductive in the cheapest way. I've seen cheap neoprene mats and more expensive mats with grounding wires. An expensive example out there: Static Control Work surface mat What are some cheap alternatives to cover a large area? Can I get something from the hardware store? Rubber shelf liners? <Q> While it is risky, if done properly, can provide an ESD safe place to work. <S> Since you live on the beach near LA, this may be of no importance. <S> Humidity above 60% or so makes ESD protection somewhat redundant. <S> I'd get a sheet of 3 or 6mm plywood for the top (full sheet, not pieces which will let the table show through). <S> An overhang of 2-3 inches on all sides is best, but will require a thicker sheet. <S> Make sure no live wires are below the table. <S> A simple stand to hold the power points can be made using wooden batons (1"x1") and fixed to the sides or rear, so all live wires are well above the table top. <S> My wooden shelf constructed with 1"x0.5" batons is behind the table and holds all my gear, and a wooden stand with power sockets and wire spools, tape rolls, multimeter probes, is attached to the table side. <S> Makes it all very handy, and safe. <S> I work at a metal table with 3/4" blockboard top. <S> edit: pics you requested <S> The curtains are a hazard, but it's simply too dusty in India. <S> Definitely not recommended. <S> [Almost all are brick houses in India, so hopefully only me and my room will burn in case of fire] <S> The full setup. <S> table is 3'x2', always wish it was at least a foot wider. <S> Rear shelf is 40" wide, 10" deep (fitted inside unused doorway). <S> Bottom steel shelves hold multimeters, magnifying glasses soldering accessories, pliers wire cutters etc. <S> There is a fuse on the left, and another fuse where this strip attaches to the wall socket. <S> Wiring is very heavy gauge (4 square mm). <S> Shelves lined with cardboard, but no cardboard under laptop. <S> Keeps it very cool. <S> The batons are 1"x0.5" shorea robusta... extremely strong, and dirt cheap. <S> As you can see, I do not throw away any packaging... <S> ever :( <S> two 12W LED bulbs fitted on the top of the near edge of the table, so no shadows under my fingers while working. <S> Also, 2' away from the curtains. <S> Can there be a fire? <S> yes. <S> Do not sleep at work. <A> You can obtain it from the big box home improvement stores in 4' x 8' sheets. <S> It's available unfinished as well as melamine coated. <S> I prefer the white melamine coating because it's easy to see small surface mount parts and it cleans well. <S> Use an ESD mat as needed when working with vulnerable components. <A> Linoleum has nice anti-static properties and can be put on top of any table. <A> I've done some research on self-healing cutting mats such as this one...because my wife happen to have a couple of these lying around. <S> These are supposed to be static free, too. <S> They're not the cheapest thing out there, but b/c <S> I had some lying around, they cost $0 for me. <S> These silicone pastry mats seem good too.	The sides of the table top can be covered with heavy duty wide insulation tape. I've used tempered masonite to cover a work table. Do not skimp on safety. Never leave the workplace unattended. Always switch off everything when not in the room.
How does a maximum power point charge controller regulate the charging current, when it always draws the maximum current from the panel? I want to understand exactly how an MPPT charge controller works and all I could find is this: An MPPT charger is a DC-DC transformer, that when it lowers the input voltage, it raises the output current as well, therefore, except for a small loss, the input and output power is the same. But I also know that a battery bank needs to be charged in different steps having different voltages and currents (bulk, floating, ..etc) So my question is that, if the MPPT charger always tries to get the most current out of the panel, how does it achieve those different charging steps? Also, most references say that the output voltage is constant, so the control is done on the current, i.e. the output current is constantly compared to the maximum value and it alternates around a maximum reference value. How is the voltage constant - doesn't the battery voltage also change when the bank is charged and discharged? <Q> Good question. <S> You are right that there is a contradiction there. <S> How can you do maximum power point tracking if the load is not able to accept the maximum power? <S> For a grid tie inverter, it is no problem because the grid can accept all the power that the solar array can produce. <S> But what about a battery? <S> What if the battery is charged already, or if the maximum power would result in too high of a charge current? <S> The way this contradiction is resolved is that the MPPT charge controller for an off-grid system will do MPPT only when the battery can accept the maximum power. <S> At other times, it will abandon the MPPT protocol and function as a normal charge controller. <S> In a typical off-grid system, the battery will be large enough to accept the solar array's maximum power output during bulk-charging, but once the charge controller transitions to constant voltage, the maximum power will not be used. <A> MPPT stands for Maximum Power Point Tracker. <S> It tries to draw the Maximum Power from the panel at all times, and Tracks the panel's operating point as the amount of sun arriving changes. <S> In order to stay operating at Maximum Power, the MPPT needs to have a load connected to it that can absorb safely all the power it produces. <S> This means a sufficiently big, uncharged battery, or a good grid connection. <S> If the load cannot absorb the power, perhaps the battery is not big enough or is nearing full charge, then the MPPT has to drop back from maximum, and must only draw what is permitted to charge the battery safely. <S> An MPPT works by drawing a certain amount of current from the panel, converting to a the voltage/current the load needs at that moment, and measuring the power that represents. <S> For a load like a battery, it can simply measure the charging current as a proxy for the power. <S> It then varies the amount of current it's drawing from the panel by a small amount, and does the sums again. <S> It keeps this, or the last, current setting, whichever produced more power, and varies again. <S> Rinse and repeat. <A> This is done by varying the effective resistance of the load typically by varying the output current of a switching power converter (by varying the switch duty cycle) charging a battery. <S> Loads where the current cannot be controlled usually cannot be used with MPPT schemes, it would be possible to use a mosfet/bjt or other programmable series resistance to reduce the current, but the losses associated with these schemes offset any optimization associated with MPPT.	MPPT controllers vary the current draw in order to maintain the optimum voltage.
how to design circuit of high side MOSFET switching with bootstraping and mosfet driver ic? Hey im starting to learn about MOSFETs at my university classes, but Im having trouble trying to design a circuit for a buck boost converter for a MPPT solar charge controller. The part pushing me down is the high side switching, I kinda understand the boostrap operation with the capacitor and the diode, but dont know how to integrated with my MOSFET driver ( TC4420 ). I know that boostraping is needed because I will be using an N channel MOSFET, but I dont have a clue how the circuit might look I know is probably really simple but im kinda stuck. If someone can show me how the circuit can be I would really appreciate. Im using TC4420 mosfet ic. <Q> You wont get 100% duty cycle if you are using bootstrap circuit. <S> The design process will become easy if you can use DC-DC converter for high side power supply <A> The simplest would be to use an intergated driver with bootstrap: http://www.farnell.com/datasheets/9094.pdf <S> https://www.infineon.com/dgdl/ir2110.pdf?fileId=5546d462533600a4015355c80333167e <S> This is a very common problem, so you will find lots of chips on the market to do this. <S> The search terms to use are "half-bridge driver" (because a pair of FETs, one high and one low side make one half of a H-bridge). <S> A H-bridge driver would also work, but chips intended for this application will usually include extra features like current monitoring that you probably don't need here. <S> If you're more interested in learning how it works, the datasheets should provide useful information. <S> There is a gotcha: some of these drivers are unable to start if the bootstrap cap is not charged, which means they will not start if the output voltage is present. <S> If you want to make a buck converter to charge a battery <S> then you need a driver which is able to turn the low side FET ON which then charges the bootstrap cap and allows it to start, even if output voltage is present. <A> A bootstrap circuit is just a capacitive voltage doubler in disguise. <S> The current requirements are rather minor, so this type of DC-DC converter is all that is needed. <S> You start from a low-impedance AC/switching signal ( <S> in a bootstrap this would be the output node of the driver). <S> You attach a relatively small capacitor to it (no need to load the signal too much) <S> You connect a diode from the power supply to the capacitor. <S> You connect another diode from the capacitor to another similar or larger capacitor (the other node of that capacitor can be gnd or vdd). <S> This capacitor holds the bootstrap voltage. <S> (Sorry, I have no access to the schematic editor right now). <S> Of course, the actual capacitor sizes would depend on switching frequency and current consumption of the gate drive circuitry. <S> The charge transfer has to be enough to account for the current consumed (Vpp C f <S> > Igate) <S> Once you see this idea, I hope you would realize that the bootstrap voltage is only needed when the NFET gate has to be turned on, so with careful design of the gate drive the second diode and capacitor can be removed altogether.	Instead of bootstrap circuit you can use isolated dc dc converter for highside mosfet.
Choosing the correct Gate Driver for a MOSFET in a DC-to-DC Converter I am building my first simple buck converter, stepping down 12 V to 5 V and driving a small load of 5 Ω - 10 Ω. Something like the following: or: The above images both show an N-channel MOSFET. I am confused as to what would be the best option to drive the N-channel MOSFET at a 10 kHz switching frequency, keeping in mind the above parameters. I have read here , that for an N-channel MOSFET a special drive circuit may be required to turn the transistor ON, so a P-channel MOSFET may be easier to implement. As for the gate-driver I have seen plenty, and a popular, cost-effective circuit seems to incorporate a non-inverting totem-pole driver as shown. I was thinking whether the above circuit would be good in my case, and simply give the PWM pulse at Rb . - What would be a suitable gate driving IC for an N-channel MOSFET, and why? - IF a P-channel MOSFET is more suited for this application, then I would go for the least 'complex' solution and use a P-channel MOSFET. What would be a suitable P-channel drive circuit/IC? I have never designed something like this before, so please keep this in mind :) Any tips and/or suggestions would be appreciated! Edit: Just for clarification. Cost is not an issue in this case. Simpler design (fewer parts) is better. <Q> I'd simply use a driver like ADP3120 or one of its cousins. <S> This one drives 2 MOSFETs in synchronous rectification mode, which will be a bit more efficient than using a FET and a diode. <S> Considering the low price of this driver, there is really no reason to build a complex circuit for this. <S> If you want to keep the diode, there are also high-side driver chips. <A> There are many possible solutions. <S> What is the best way to travel from London to Paris? <S> By car, train or plane? <S> They all work but the cost and performance will be different. <S> The same applies to your problem. <S> If you are set on a high side switch, there are a number of different options. <S> I will outline a few common approaches: Use an isolated DC-DC power supply plus a standard low side gate driver, e.g. MCP1416, whose reference point is the source of the MOSFET. <S> This solution is expensive and has a high parts count. <S> Use a pulse transformer. <S> Simple. <S> The duty cycle here is limited to < 50%. <S> Use a bootstrapped high-side buck driver or a half bridge driver, e.g. LM5109. <S> This is one of the simplest means of driving a high side switch. <S> The following diagram illustrates this approach. <S> You can read more here . <S> Just be aware that you need to charge the bootstrap capacitor at startup. <S> For a buck converter, you can also shift the MOSFET to the ground rail. <S> This is known as a low side buck. <S> However, measuring the output voltage becomes more complicated. <A> For the low power application your describing (~5W output power) and your demand for a simple design i would recommend just using a simple buck controller IC with integrated switch (some even have a integrated inductor/diode aswell). <S> The only external components would be input and output caps, (diode and inductor) and in some cases (if output voltage is adjustable) <S> a voltage divider in the feedback path. <S> These ICs normally have comprehensive application notes which makes the choice of external componens more or less foolproof.	With a low-side buck you can use a standard low side MOSFET driver, e.g. MCP1416, which is very simple.
Can anyone tell me what an 0R-Jumper Varient C is? I found this component in a schematic I am looking at and have searched the internet far and wide to find out what it is, alas to no avail. If anyone could explain this component and why one would use it that would be awesome. Here is the link to it: https://www.diymodules.org/eagle-show-object?type=std&file=jump-0r-smd.lbr&package=C0R-JMP <Q> for 2 position jumpers with 3 pads use Variant 1 or 2. <S> for 1 position jumpers with 2 pads use Variant 3. <A> An example could be one of these components being the sole connection between two local ground planes on a board. <S> Here's an example datasheet for \$0\Omega\$ chip resistors: https://www.vishay.com/docs/31017/rcwp99.pdf <A> It is nothing but a zero ohm resistor, also called a jumper. <S> "Variant C" probably refers to the physical shape. <S> You can probably find that out by checking the properties of the part when you place it in Eagle. <S> I can't tell from the linked image what it is supposed to look like. <S> It is most likely a surface mount part - I don't see any holes for through hole connections.	0R-Jumper is a "zero-ohm" resistor, typically used on PCBs to allow for different configurations based on how these components are populated.
AWG for connecting a USB What is the recommended AWG size for connecting a USB male connector? The link from Instructables.com below uses only a spare LAN cable and no mention of the size: http://www.instructables.com/id/Creating-you-own-USB-cables/ <Q> Note that 28AWG is rather thin for supplying 500mA (the maximum current that USB 2.0 specifies for power), and will result in a high voltage drop for long cables. <S> Most USB cables use 26 or 24AWG for the power pair, or even bigger (typically 22AWG) for the longer ones (~3 meters). <S> As Passerby said, Ethernet is usually between 22 to 26 AWG. <S> But it is not only about the AWG. <S> USB cables are supposed to have 90Ohm impedance, and ethernet 100. <S> The impedance mismatch could be a problem for long cables, especially if you use it as an extension to another regular 90Ohm USB cable. <S> USB High Speed will be more sensitive to this than Full Speed. <A> It's mainly dependent on how you will use your USB port. <S> In the USB 1.0 and 2.0 specs, a standard downstream port is capable of delivering up to 500mA (0.5A); with USB 3.0, it moves up to 900mA (0.9A). <S> The charging downstream and dedicated charging ports provide up to 1,500mA (1.5A). <S> AWG 24 can handle upto 2,000mA (2A) of current safely. <S> So it's applicable on all situations. <A> USB can use between 20 AWG to 28 AWG wires. <S> Ethernet tends to be 22 to 26 AWG. <S> The lower the number, the more current it can carry.	USB specifies 28 AWG for the signal pair, and 28 to 20 AWG for the power pair. In short, it will most likely work, except if you're making a very long cable, but you'll definitely be violating USB specs.
What is the name for this Encapsulated SMPS I found this and I'm a little confused. It appears to be a simple component in to which I can put mains voltage and out of which get 3.3V DC without needing any other components - precisely what I want. Maybe I'm missing something, but from various googling this doesn't seem to be common, or used in circuits as a component of its own. Could I, for example, use this single component to run an Arduino from mains? (there is a 12V version here )? Or to charge a mobile phone? Or, in my specific case, to connect a mains supply to a 3V relay 24/7 using minimal standby power. Assuming I've understood what this component does, is there a better name for it? Googling "encapsulated electronic transformer" only returns this MYRRA product, but surely this is a generic thing... or is it? <Q> This component does what you expect. <S> It takes AC mains as input, and outputs the specified DC voltage (at the specified power). <S> It is actually made of a switched-mode controller and transformer, all integrated in a small plastic cube. <S> It seems perfectly appropriate to use them for the usages you describe. <S> Of course, for high-volume products, designers usually go through the hassle of designing their own, because it can make the product cheaper. <S> This is why it is not usually found in consumer products. <S> There is a wide choice for these components available from distributors, and is typically called "AC/DC power module" . <S> See those available from Mouser , for example. <S> Note <S> : this component is galvanically isolated, so you need to be careful in your layout to maintain the isolation between primary and secondary. <S> Just don't route the AC mains side and secondary side traces close to each other. <A> There isn't a single standard name, but most people would expect a PC-mount power supply to be something like what you show. <S> At least one distributor lists such things under On-Board Power Supplies . <S> Other possible interpretations <S> I thought the salient part you cared about was that it is one simple device that you can mount right on your PC board. <S> If you just want to convert line power to DC, then that's any power supply . <S> If you want it to be a switching power supply (you might care about size, efficiency, and heat produced, but how those specs are met shouldn't really matter), then switching power supply , SMPS , or <S> even just switcher would do. <S> If the point is to be able to solder it directly onto a PC board, then see my original answer above. <S> You question is like showing a picture of a orange, then asking what it is. <S> Fruit , citreous fruit , orange , and spheroid are all correct answers without you specifying what aspects you care about or what level of detail you want to know about. <A> The data sheet shows a block diagram and describes it: - <S> Could I, for example, use this single component to run an Arduino from mains? <S> (there is a 12V version here)? <S> Or to charge a mobile phone? <S> Or, in my specific case, to connect a mains supply to a 3V relay 24/7 using minimal standby power. <S> The only caveat is ensuring it provides enough load supply current and that some (older) phone chargers might have a more constant current output meaning this device won't be suitable across the board. <A> You should read the web page properly. <S> The description tells you that it is a switched-mode power supply, SMPS or "switcher".	Generically it is a switch mode power supply (SMPS). It is a rather common component used mainly for simple appliances requiring relatively low power, where the designers don't want to waste time designing a specific AC-DC supply. If the point is to have a single block like you show, then use encapsulated power supply, or less formally power brick .
uln2003a: Switching ground of device works, however, optional USB function of device doesn't work correctly I use the uln2003a to switch a device (with use of a MCU) that is capable doing several things, like fm-radio, bluetooth, playing MP3's etc. It also provides an USB function when it is connected to a computer, it provides an usb soundcard and a SD-card interface. The function is only enabled/started when selected on the device. The switch works okay, however, when connected to a computer the device is recognized but the computer is unable to communicate with the devices. In Windows there is an exclamation mark present with the comment "Device doesn't start, code 10" . When I bypass the uln2003a switch with a wire, connect ground directly so the uln2003a has no function, all of the devices operates just fine. I tried the following experiment: Start the device, Connect point A and B with a wire (see schematic below) to bypass switch, Select USB mode and working fine. As soon as I disconnect this wire (ground still connected via uln2003a), devices starts to malfunction in Windows. Question: The uln2003a switch the same ground and I don't understand why USB is not working. Can somebody explain what's going on? Simple schematic to give some idea what's going on: ULN2003A datasheet: http://www.ti.com/lit/ds/symlink/uln2003a.pdf <Q> Take a look at the graph "6.14 Typical Characteristics" in the ULN2003 datasheet. <S> Even with only a small amount of current the output is still at least 0.7V above ground. <S> That means that your device's USB signalling is all going to be (at best) shifted 0.7V higher than it's supposed to be (could be even worse) - making the USB host in your PC very unhappy. <A> The ULN2003 datasheet says Vce(sat) can be as high as 1.6V which is pretty high for a "low" output. <S> When you shorted the connection with the wire you got 0V which is why it worked. <S> The ULN2003 isn't really useful for this application because it can never pull the output low enough. <S> What you are doing is "low side switching" -- for USB <S> I would recommend high side switching and put a Texas Instruments TPS2051B power switch between +5V from USB and your device's +5V input. <S> Then connect the device to ground like normal. <S> http://www.ti.com/product/TPS2051B <S> The TPS2051 has a simple interface; there's an enable pin and a fault status indicator. <S> It's easy to use and can handle up to 500mA for your device you are switching. <A> When you have a Darlington output stage you can expect significant volt drops: - To turn on Q2 you need about 0.7 volts applied to its base with respect to the emitter. <S> Given that the base voltage is supplied via Q1 and Q1 sources this voltage from Q2's collector you cannot realistically have the output falling below 0.7 volts unless you have a very light load on the output. <A> Good answers here already, but I feel it important to add ground side switching of devices is generally a bad idea. <S> We spend a lot of time ensuring the integrity of the grounding system to minimize noise in systems, ensure proper return current management, and to maintain as close to a constant reference voltage across the circuit/system as we can achieve. <S> (The latter being the issue that stopped you!) <S> As such, it should be avoided wherever possible. <S> Furthermore, when you switch the ground side you leave the device hot. <S> Internally it is still connected to whatever rail it is attached to. <S> That means the output pins will tend to pull or drive whatever they are attached to up to the rail. <S> Even input pins can do some strange and unpredicted things with the ground disconnected. <S> Having the device not directly connected to a power source is safer. <S> Of course, having a device in circuit with Vcc disconnected can also be problematic. <S> The device can actually take power from the input pins via the protection devices if whatever is connected to them can supply enough current. <S> As such, depowering devices takes some careful design and is often not a trivial matter.	Adding a switch in the ground side negates all that and adds all kinds of hard to debug issues.
How to get 20 V from a USB-C charger? I want to power my DIY stuff with a MacBook Pro USB C charger. The label on a charger says that it can provide three different power profiles: 20.2V - 4.3A 9V - 3A 5.2V - 2.4A There is a ton of information about what USB-C power distribution is capable of, but I can't find any examples of how exactly to do it. Is there an easy way or workaround to request one of those power profiles without using a microcontroller? For example, I got apple MacBook Pro charger, apple USB-C cable, and a breakout board like this one: <Q> In a word, no. <S> You need to implement the USB Power Delivery protocol through the CC line of the Type-C connector and that's a two way communication at 300kbps complete with preamble, <S> CRC and so on that is pretty much impossible to do without a microcontroller. <S> The PD message format looks like this: <S> For full details you'll need the USB Power Delivery specification , but there's a useful introduction here <S> which is where the diagram above came from. <A> As Finbarr noted, you need to implement Power Delivery communication protocol to negotiate the 20 V output profile. <S> The PD specification was designed by a community of more than 300 fine engineers from top semiconductor and software companies. <S> It took 3 major revisions and about 5 years of work to come up with current functional standard. <S> Trying to implement the protocol from scratch (as Finbarr seems to suggest) is an obviously losing proposition. <S> However, there is a solution at DIY level. <S> Due to horrible protocol complexity, several semiconductor companies offer a set of ICs that embed the PD protocol, a turnkey solution. <S> To start, look at overview of PD solutions at TI , and select proper cluster. <S> You already have the functional PD source. <S> So you need to select "Device UFP" and "Sink". <S> You will have about 5 variants of ICs performing the function of PD controllers, something like TUSB422 or TPS65981. <S> Unfortunately, the IC will need some control over I2C interface to perform the actual negotiation, so you will need some microprocessor with some software support for PD. <S> They should offer development kits with full examples how to do this. <S> So, good luck. <S> EDIT: <S> Other companies as NXP, Maxim, Linear, Cypress, ON Semi, STMicro, might offer controllers with pin straps (with no microcontrollers), but you need to search for this. <A> It's complicated but doable. <S> There's someone who has implemented exactly that and offers the PD buddy sink on tindie: https://www.tindie.com/products/clayghobbs/pd-buddy-sink/ Code: <S> https://git.clayhobbs.com/pd-buddy/pd-buddy-firmware/src/branch/master/docs/console_config.md <S> Hackaday project site: https://hackaday.io/project/20424-pd-buddy-sink/details <A> Is there an easy way or workaround to request one of those power profiles without using a microcontroller? <S> This will be a very complex circuit though and will require some reverse engineering and detailed study of the protocol. <S> Since this is an Apple product, it might not strictly adhere to the public standards either. <S> In practice: no as a microcontroller will offer much more flexibility than a bunch of logic chips. <S> Also precise timing will be much easier to achieve. <S> But for both cases: I do not think it is worth the effort to do this in the first place. <S> It will be much cheaper and much less effort to buy a power adapter which will output the voltage you require. <S> Also, there is the risk that you somehow break or damage the adapter. <S> Then you have no way of charging your MacBook until you buy a new adapter. <S> Since this is an Apple product, that might not be cheap. <S> So why risk that? <A> I've just completed a board based on STUSB4500 which is super easy to use requiring only a few passive externals (no custom firmware needed). <S> Will be open sourcing the design soon as well. <S> You only need an Arduino to flash the chip once with your voltage/current requirement and it will negotiate with the charger automatically. <S> EDIT: <S> The new version of the board can be configured via web browser.	In theory: yes, you could make a circuit using standard logic chips to emulate the protocol and "fool" the charger to output more than 5 V.
Receive GPS Antenna and cable simple confirmation of operation I am in need of testing a GPS antenna and cable assembly after installation at a reasonable cost. I have a GPS antenna installed on an airplane, then the coax cable from the antenna goes down to the receiver. The system is already assembled and validated technically and in production, but we find that after some shipments around the country something goes wrong in the cable assembly. We are testing out a complete system inside of a metal hanger so there is no available sky view for real satellite reception. Besides we are trying to control the test and provide a go/no-go simple automated test. We really want to know that we have not pinched the coax cable and confirm that all the connections are still good from outside the antenna through the aircraft and all the way down the cable. I am thinking it should not be necessary to pass a GPS capture and location measurements. I am thinking that I can find a simple method to transmit an RF signal in the GPS range and then receive it on the other end of the cable to confirm the antenna and all the connections and the cable are still good. I am thinking that I could attach an instrument on the receive end of the coax cable to detect what is transmitted from a non-contact transmitting antenna held a few inches above the GPS antenna. Any ideas out there? Any simple equipment that does this already that can be connected to a Windows or Linux PC to initiate the test and read back the response signal if not to say go or no-go? ----- edit of March 25 -----At this point, I believe the best way to go on this test effort is to provide a signal generator driving an antenna, then provide a receiver / detector at the "bottom" end. Now I need to find hopefully a single PCIe card that will give me an L-Band range signal generator and a receiver with an API to control and read from s/w. I think the SDR version is more complex than it needs to be but that still may work. And since the quantities of those are high, that may get those prices down to a good range anyway. Any ideas out there? I am looking at Pentek, Agilent, NI and any others that may be appropriate. I haven't yet figured out what to call it to do my search for what I need, other than a Signal Generator and Receiver. <Q> It will tell you if there are any shorts or opens (and where they are) and whether the end is terminated in a proper load (antenna connected). <S> Its a relatively pricy piece of equipment at about $1500-$2000, but its a huge labor saver. <A> My intuition would be that you could use a network analyzer for this. <S> MiniVNA <S> Tiny: <S> http://miniradiosolutions.com/54-2/ Software to use with MiniVNA <S> : http://vnaj.dl2sba.com <S> I've used a very cheap USB network analyzer ($550) for simple impedance matching. <S> This seems like the cheapest and most direct approach. <S> You can even get one of these for $330 from China on AliExpress (not sure if they ripped off the Italian guys who originally created this or not...). <S> I've never worked with SDR radio, but there seems like a bit of a learning curve compared to just using a VNA and getting the measurement right away. <S> I am with Brian Drummond <S> using an SDR will be a learning project. <S> If you guys have a GPS system on an airplane <S> I'm guessing $550 is not going to a deal breaker? <S> A tracking generator is not crazy either (check the manufacturer Rigol for "cheap" spectrum analyzer -- I'm guessing it is $1800 for a spectrum analzyer + tracking generator). <S> Make sure to order a set of calibrations from the vendor you choose <S> so you don't have to hand solder them. <S> (I ordered from Wimo in Germany). <S> You are just going to calibrate your unit, get an SMA to Coax adapter because the MiniVNA has SMA ports. <S> (Technically, if you calibrate with the SMA standards from MiniVNA <S> but then add a coax adapter you are adding some inaccuracy, but for a cable good or cable bad approach, I think you should be okay). <S> NXP MiniVNA Setup Document: https://www.nxp.com/docs/en/application-note/AN11535.pdf <S> This seems like the easiest path to get your measurement. <S> Check <S> if MiniVNA also has a bluetooth version of sorts, you may be able to setup the VNA at the antenna end and then remotely scan if the antenna is hard to get at. <S> If you decide to go with this approach, just comment in the answer if you need help later on. <S> I can step you through some stuff if you need help. <A> Well, if you just need to test the antenna + coax, then sure: Unplug your GPS receiver, plug in (if necessary, but probably is) <S> a bias-tee to power the LNA in the active antenna, and connect something that can receive at least L1 frequencies. <S> You'll need something very weak transmitting in the GPS range – <S> not only because otherwise you'd risk saturating the amplifier in the antenna, but also because you don't want to get a visit from the friendly neighbors of your radio regulation agency (FCC if this is the US). <S> Then, you'll need your receiver to look for that very weak signal in the receiver noise - that's exactly what GPS does, correlating until it finds signal, only for GPS signals. <S> Both ends, Transmit and Receive <S> , can relatively cheaply be implemented using an SDR (Software Defined Radio). <S> A system that transmits a known sequence on one end of your hall, and another one that looks for it would probably be two separate cheap SDR devices. <S> Transmitting the known signal would probably be easiest with GNU Radio as signal-generating software (darn easy, tutorial ) and a COTS SDR like the Ettus USRP B2xx family (or other manufacturer's products, but no experience with those) with... a paperclip instead of a proper antenna and low transmit gain. <S> Reception: again, GNU Radio, correlating with the same sequence, and as receiver another one of these SDRs (could've used the same one, but you really want good TX/RX isolation), or, as stupid as that sounds, a $30 RTL-SDR dongle, as your quantization and noise don't matter, due to the immense correlation gain you'll be getting. <A> I recommend a signal generator with tracking generator. <S> The tracking generator produces a signal at the same frequency that the spectrum analyser is looking, making it into a simple one-path scalar network analyser. <S> With a length of cable, a bias T, an attenuator and antenna, you will be able to generate traceable records of performance of the GPS antenna system. <S> You can define a pass/fail limit line and test against that, store and print the result. <S> You'll need to make a small antenna jig that sits atop the GPS antenna on the aircraft roof. <S> I'd use a plywood frame or a solid block of polystyrene foam, some padding for the aircraft, supporting a transmitting antenna about 400 mm above the GPS antenna. <S> I mention the attenuator because a) you don't want to transmit a whole milliwatt of RF at GPS frequencies, and b) you'll saturate the GPS antenna if you transmit much more than -30 dBm. <S> Used, a R&S FSH-3 or similar from Agilent/Keysight, Anritsu site master, etc, will cost only a few thousand dollars. <S> The nice thing about GPS is that it fits into the budget price range for test equipment, it's between the cellular bands. <S> It's battery powered and fairly rugged. <S> A spectrum analyser is a very handy tool for other radio purposes too. <S> Check that it has the tracking generator enabled, sometimes its a software option you pay extra for. <S> There are also a range of low cost USB network analysers, copper mountain make some for about $10,000.	A TDR (Time Domain Reflectometer) would seem to be the answer here - we used these in the Navy to ring out long RF cable runs/transmission lines.
Detect if coil is near steel piece using MCU How to detect inductance change using MCU (lowest power methods preferred). I want to detect if the coil touches steel piece (higher inductance) and when it does not (lower inductance). Basically I need to compare inductances at two positions and compare the values. Then I will derive a threshold value to detect when the coil is away from the steel piece. <Q> I used a PIC12F509 to make this kind of sensor, years ago. <S> It used 100uH leaded-resistor type inductor L=6mm D=2.5mm. <S> The actual capacitor values I can't recall. <S> It was used to make a waterproof switch that reliably worked in salt water. <S> In my case a spring (ID=8mm L=4mm) slid over the coil forming a shorted turn. <S> simulate this circuit – <S> Schematic created using CircuitLab C1,L1 form a resonant circuit at 150kHz (for values shown). <S> One half cycle is approx <S> 3usThe output is enabled, then a single pulse energises the circuit (hi for 2us), then you switch back to input and sample the input. <S> The LC tank will now resonate. <S> i.e <S> it will going - + - <S> + at the resonant frequency for a couple of cycles. <S> Immediately after energising it goes negative, then a half cycle later the voltage will peak +ve, and you will sample as logic 1 . <S> Code is very simple - probably less than 10 opcodes. <S> You need to choose L and C to suit how fast your micro works (PIC was 1us instruction cycles). <S> You need to experiment with coil types to suit your situation. <S> You need to empirically test the best excitation pulse (how wide, single +ve pulse vs 1/0 pulse) and explore what the resonance looks like and how it moves, using an oscilloscope. <S> Note that there are two detection modes possible. <S> I used change in L and therefore change in position of resonant pulse. <S> It is also possible to look at the decay-time of the ringing, by counting the number of ring cycles. <S> In this case the presence of metal will increase the loss, and the ringing dies faster. <S> e.g ring count drops from 3 to 1. <A> There are two obvious approaches. <S> Make the coil part of a LC oscillator and look for the frequency changing. <S> This has already been covered in another answer so I won't expand on it here. <S> The second way is to create a fixed frequency oscillator. <S> Apply it across a series LR circuit or LCR and look for a change in the current by measuring the voltage across the resistor. <A> For inductance change detection the conventional scheme is to design the coil into an oscillator circuit. <S> Changes of inductance are then determined indirectly by detecting the change of frequency of the oscillator. <S> An MCU timer can be used to check the frequency by one of two methods. <S> 1) Measure the time period of cycles of the oscillator signal. <S> Take several readings and average. <S> 2) Count cycles of the signal from the oscillator over a given time interval. <S> Then only start the oscillator and make your readings on a periodic basis that is compatible with the maximum latency that you can live with in your application. <A> Create an LC (inductor-capacitor) oscillator. <S> The L is the inductance of your sensor, which varies. <S> C is a known capacitance. <S> The frequency of the oscillator will depend on the inductance. <S> There are several methods for determining frequency by a microcontroller. <S> Some MCUs have peripherals which can measure time periods. <S> There are also frequency to voltage converter circuits. <S> Output voltage is DC proportional to frequency. <S> The resulting voltage can be read by an ADC on a microcontroller. <S> edit: Apparently, there are specialized ICs which are inductance to digital converters <S> (also here ). <A> Research EFM32 MCUs with LESENSE peripheral - very low power inductive sensing. <S> Silicon Labs has an application note covering your exact topic.	Some MCUs have counters, so you can count the number of LC circuit oscillations over a known period of time. When there is metal forming a shorted turn the peak happens earlier, because L is effectively reduced, and you will sample 0. To make the solution as low power as possible design the oscillator so that it can be gated off via a separate GPIO from the MCU.
How do phone chargers have variable input voltage with constant output voltage? My basic understanding is that a transformer can step down a voltage by the ratio of the primary and secondary windings, since this is a ratio the output is not constant. Thus my question is, how are chargers like the apple phone charger (a Fly-back Switch mode power supply) able to take an input of 100v-240v ~ 50/60 Hz to create a constant 5v output? Above is a supposed circuit diagram of the apple phone charger. is this constant output voltage an effect of the flyback transformer? (i have little experience in AC to DC power supplies) Any help is appreciated. <Q> Modern AC-DC power supplies do the voltage conversion in three steps. <S> Roughly speaking, the process is as follows. <S> First, they rectify the AC into DC, so 100 V AC gets into about 140 V DC, and 240 V AC results in about 340 V DC. <S> This is a first step. <S> This is the range of voltages that the second stage of converter is dealing with. <S> And this voltage has horrible ripples at 100-120 Hz. <S> The second stage is a "chopper" that modulates the high-voltage DC into high-frequency pulses, 100 kHz or something. <S> The transformer, as you duly noted, has a fixed winding ratio, so the output pulses would have the variable amplitude proportional to the input DC (which is 140 to 340V, not counting ripples from 50/60 Hz primary rectification). <S> However, the chopper also makes these pulses of different width, which is called PWM - Pulse-Width-Modulation. <S> Thus the output of the transformer, when rectified by "half-way" diode rectifier and smoothened with a large output capacitor, on average can have variable amplitude: <S> narrow pulses make lower average amplitude, and vice versa. <S> This is the third stage of AC-DC converter. <S> So, while the transformer has a fixed winding ratio, the PWM still allows to change the output of rectifier in considerable range, thus accommodating the fixed transformer ratio and vast input voltage range, including voltage ripples. <S> The final control and voltage stabilization is done via negative feedback mechanism using linear opto-isolators. <S> If the rectified voltage goes too high, the feedback makes the controller IC to produce narrower pulses, so the voltage goes down, and vice versa. <S> This feedback mechanism not only takes care of the voltage, it also controls the overall power delivered into PSU load. <S> There are some fine details how the transformers tolerate the asymmetic waveforms, there are some fine engineering tricks behind the scenes, but basically that's it. <A> If you want to identify one 'component' that's responsible for the constant output voltage, then it's the 'feedback'. <S> The forward path which includes the flyback transformer pushes a controllable amount of power to the output. <S> The voltage on the output is measured, and the feedback requests a smaller or larger amount of power moment by moment, to keep the voltage constant. <S> The forward path is designed to be able to run from any voltage in the input range, which needs a bit of care with design, but is fairly straightforward. <S> The way a flyback converter works is that its output voltage adjusts to whatever voltage is needed to deliver the power it's been asked to deliver. <S> It can step up or down by a large ratio, to allow it to match the input and output voltage ratio. <A> The phone charger has to do several things in addition to regulating the voltage. <S> It has to convert AC to DC, step down the voltage substantially and provide substantial isolation between input and output. <S> Since we're only concerned with regulation lets instead consider a DC-DC "in car" charger, that accepts DC over a typically wide voltage range possibly up to 28V, and converts it to 5V. <S> The charger probably uses a fast switching transistor and diode to rapidly switch between the input voltage and ground, then an LC filter to smooth out the switching and output the average voltage. <S> The resulting transfer function is Vout=D*Vin, where D is a PWM duty cycle. <S> For reasonable input voltages there will be a "D" value that yields 5v. <S> In its simplest form D is set by a controlling "error amplifier" comparing Vout with a reference voltage. <S> In more refined versions the PWM circuit is modified to cancel out the influence of Vin, two examples of this are "feedforward" and "current mode". <S> In current mode the PWM pulse ends when the current in the inductor reaches a value. <S> If the input voltage is higher the value is reached sooner but the output is relatively unaffected. <S> If this DC-DC design is "upgraded" to include a transformer then it gives the popular "forward" configuration which can be more compact and efficient than flyback as the transformer can use magnetic parts optimized for transformer use (ferrite), and the inductor can use parts for inductor use (iron powder). <A> The "transformer" in a flyback converter is technically not a transformer but two coupled inductors. <S> Unlike a transformer it stores magnetic energy in an air gap. <S> The energy store is charged via a switch (transistor) during the scan and discharged via a diode during flyback. <S> Source and load are never connected simultaneously, and thus the ratio of turns does not apply. <S> Instead, the duty cycle, or the on-off ratio, is what matters, since the average voltage over any inductor must be zero. <S> This ratio is easily varied. <S> The flyback converter generates the high voltage for a CRT display, making use of the fast flyback (or retrace) of the horizontal deflection, hence its name. <S> Edit: the turns ratio matters too, but not as much.	The output voltage is usually actively regulated, i.e. stabilized against load variations, by a regulator with feedback. There is a controller IC that drives a pair of powerful MOSFETs, which are loaded with primary winding of the isolation transformer.
How can I divert current to higher resistance components in parallel? I am working on a circuit and I need to divert the current to higher resistance components in parallel without changing their voltage input. There are 4 thermoelectric devices (T1, T2, T3, T4), a heatsink, and a pump. They all ideally need 12 volts but the heatsink needs 0.09 A, the fan needs 1.1 A, and the thermos can take any current. The battery is 12V and 2.9 Ah. Since the thermoelectric devices have such a relatively low resistance, nearly all the current goes through them and none will go through the pump or heatsink as shown in the schematic below. Is there some component that will help divert the current to the pump and heatsink? We are avoiding using extra batteries due to the weight and ease-of-use. We have run out of ideas since we have limited experience with circuits. Any suggestions would be greatly appreciated. Thank you! <Q> There will be no need to divert current. <S> If the battery cannot maintain 12 volts while delivering the required total current, you need a bigger battery. <A> If your battery is powerful enough to supply all the current those loads need, then it will maintain a nominal 12v across the rails. <S> If your battery is not powerful enough, then your rail will drop, and you're in trouble. <S> At 12v, each parallel load will take whatever current it wants to. <S> The fan will still take its 1.1A, the heatsink still take its 90mA, and the thermos will take whatever they want as well. <S> There is no need to 'divert' current into the fan and heatsink, just because the thermos are taking a large current. <S> There is no way to 'divert' current without changing the voltage on various components. <S> At the high currents involved, a buck converter is usually the best way to do this. <A> I might not understood your question, but as long as you have resistors - or circuits very similar to resistors - their resistance will determine their voltage to current ratio. <S> For example by inserting a buffer with the desired voltage gain. <A> For most components/circuits the current is dependant on the voltage. <S> For resistors there is a simple relationship \$ <S> I = <S> \dfrac{V}{R} \$ <S> this is also almost true for a lot of other devices but lots of things are not resistive. <S> A zener diode for example will take almost no current below its zener voltage and current will rapidly increase if you try to increase the voltge across it until either your power supply can't provide more current or the zener gets too hot and fails. <S> My PC power supply will take less current as you increase the supply voltage as it needs to supply the same power to my laptop whether I have 100V mains in Japan or 230V mains in the UK. <S> You can not divert current between parallel devices. <S> You can add series resistance to divert current <S> but then the devices have different voltages across them <S> so are not in parallel any more.	If you have parallel resistors, and you want to change the current through them, you have to change the voltage on them. If the battery can maintain 12 volts while delivering the total required current, each device will draw the current it requires. You might want to reduce the amount of current your thermos are taking, by regulating their voltage down.
Design of supply for nichrome wire heater I have two Cr20Ni80 nichrome wires having 1m length each, 24 SWG and a resistance of 4.390 Ohm/m. I need to heat up a furnace to 450 degree Celsius. How can I calculate the supply voltage and current for the purpose? I need some formulas to design a power supply to power this nichrome wire. Is there any relation between temperature, voltage and current? The furnace has a base diameter of 54cm and top diameter of 34cm with a height of 40cm. Total volume is 61869cm^3 and is made of mild steel with no insulation. Also I need to know:Can I power it using AC or DC supply? If I give a constant supply for some time, will the nichrome wire temperature increases over time? or will the temperature remain constant? <Q> I have read about people saying about furnace size, heat loss, etc. <S> Please leave all that. <S> Nope. <S> That is the stuff you need to figure out before anything else. <S> Start by defining your interior volume, and then specify your insulation. <S> Assuming 450 C inside the box, and knowing the thermal resistance of your insulation, you can figure the power necessary. <S> Once you know the power, you can make an operating assumption about the wire temperature. <S> Let's say the wire will operate at 900 C, which will allow 50% duty cycle. <S> At this point, you have a known wire length and resistance. <S> Then, given the power you need (remember power?) <S> you can calculate the supply voltage as Warren Hill suggested. <S> Dividing power by voltage will give you the current required. <S> Then you will need to calculate the resistance of the wire at room temperature, and rerun your calculations. <S> The previous numbers gave power and current at operating temperature, and the wire will dissipate a good deal more than that at turn-on when the wire is cold and has less resistance. <S> But first things first. <S> The very first thing you MUST do when designing a system is to determine your requirements. <S> In this case, the most important requirement is the power needed at operation - all else follows from this, and in order to determine it you must know the insulation values of your oven. <S> "Please leave all that" doesn't work. <A> You're going to need to estimate how many watts your furnace needs to get up to 450 degrees. <S> A simple no-math way to do this would be to look up similar devices on the internet and see how many watts of heating they have. <S> Next you're going to calculate the voltage required on your nichrome wires to get that wattage. <S> It's quite simple, WATTAGE = <S> V^2/R. <S> The voltage can either be DC or AC, doesn't matter. <S> Lets say for example you need 750 watts. <S> If you hook up your nichrome wires in series, you will need sqrt(7508.8) = <S> 81V. <S> In parallel you will need sqrt(7504.4) = 57v. <S> The amperage your furnace will draw is V/R. Find a power supply that can supply at least 1.2x this many amps, at the voltage you calculated earlier. <S> You may be able to use the mains voltage directly if you adjust the nichrome wire lengths a bit. <S> Make sure to redo all the math and check it. <S> You may also be able to use mains through a transformer to lower the voltage. <S> This will probably be cheaper than using a burly DC supply unless your furnace is very small. <S> Continuing our example: 57/4.4 <S> = 12 Amps (in the parallel configuration). <S> 81/8.8 = <S> 9.2A in the series config. <S> So we want a supply that's either 57V <S> >15A, or 81V <S> >11A. <S> It looks like you'll probably want to increase the nichrome wire length a bit and get the target voltage up to 110V, <S> so you can run it straight from mains. <S> .. <S> but anyway. <S> Last you're going to need to buy a PID controller to control the temperature. <S> These are cheap and commonly available. <S> For example: https://smile.amazon.com/Inkbird-Temperature-Thermostat-Controllers-Fahrenheit/dp/B01KNXETWS/ Just make sure: The temp sensor can handle 450C <S> The controller (or bundled relay) can handle the voltage and amperage your going to run at. <S> The amperage will be V/R. <S> Make sure the output is AC if you're using AC, and DC if you're using DC. <S> If you bought the product that I linked, you would have to buy a new thermocouple, since the one it comes with is only rated up to 400C. <S> This shouldn't be an issue, as you can hook any old thermocouple up to these controllers. <S> Just make sure you change the termocouple type when you set the controller up. <S> Here's one that's rated to 800C: https://smile.amazon.com/Perfect-Prime-TL1815-HeadProbe-Thermocouple-Temperature/dp/B0142S9J4S/ <A> Also I need to know <S> : Can I power it using AC or DC supply? <S> A heater wire will work equally well on AC or DC. <S> If I give a constant supply for some time, will the nichrome wire temperature increases over time? <S> or will the temperature remain constant? <S> The temperature will rise until it stabilizes at some temperature. <S> It's very difficult to tell what that temperature will be without knowing all the thermodynamics of the system.	Using the date sheet for your nichrome alloy you can find the resistivity of the material, and determine the wire length and diameter.
How does a capacitor resist changes in voltage? I was reading about how capacitors work and two sentences confused me: "A capacitor’s ability to store energy as a function of voltage (potential difference between the two leads) results in a tendency to try to maintain voltage at a constant level." "When voltage across a capacitor is increased or decreased, the capacitor “resists” the change by drawing current from or supplying current to the source of the voltage change, in opposition to the change." If the capacitor draws current from the source when the capacitor's voltage increases, how is this considered a form of resistance by the capacitor? I mean - the way I understood it - for the voltage to remain constant in this case, shouldn't the capacitor not draw any current but instead, limit the flow of current through it? I am really confused in here - I feel like I am missing something very fundamental... Here is the full paragraph: "Because capacitors store the potential energy of accumulated electrons in the form of an electric field, they behave quite differently than resistors (which simply dissipate energy in the form of heat) in a circuit. Energy storage in a capacitor is a function of the voltage between the plates, as well as other factors which we will discuss later in this chapter. A capacitor’s ability to store energy as a function of voltage (potential difference between the two leads) results in a tendency to try to maintain the voltage at a constant level. In other words, capacitors tend to resist changes in voltage drop. When the voltage across a capacitor is increased or decreased, the capacitor “resists” the change by drawing current from or supplying current to the source of the voltage change, in opposition to the change." <Q> "Resists" may be an unfortunate choice of word. <S> If I were to describe a capacitor like that I think "reluctant" or "reticent" would be a better choice. <S> The voltage that develops across a capacitor is the result of charge carriers (electrons typically) building up along the capacitors dielectric. <S> From Wikipedia: The build up of charge carriers takes time , and therefore the change in voltage will also take time. <S> Contrast this with a pure resistance where the voltage that develops is dependent on the current flowing through it at any instant in time. <S> So when the book says the capacitor "resists" changes in voltage, what it is referring to is that any voltage change will take some time depending on how quickly the charge carriers flow in or out of the capacitor. <S> Hope this helps. <A> To see why it's said that a capacitor 'resists', or 'objects to' changes in voltage at its terminals, it's useful to compare its behaviour with a resistor (don't confuse the 'resists', meaning 'tries to stop', with anything to do with the component 'resistor'). <S> If you have 10v across a 1k resistor, then 10mA will flow. <S> If you now try to change the voltage to 20v, ramping it up at 10^6 volts per second, so it takes 10uS to change from 10v to 20v, the current will smoothly increase from 10mA to 20mA in that time. <S> If you have 10v across a 10uF capacitor, and the voltage has been steady for long enough, then no current flows. <S> If you now try to change the voltage to 20v, ramping it up at 10^6 volts per second, so it takes 10uS to change from 10v to 20v, the current will go to 10A for that 10uS, and back to zero when the voltage is steady again. <S> If you try to change the voltage at 10^7 volts/second, the current pulse will be 100A. <S> That sort of current amounts to a fairly violent 'objection' to the voltage being changed. <S> If the power supply cannot supply it, then the voltage will not change as fast as expected. <A> Think of this as water in a pipeline. <S> If the capacitor limits the current i.e. like closing a tap in our analogy. <S> Now if the tap is closed water will start to accumulate in the pipeline and pressure(voltage) will start increasing at the point where water flow is cutoff. <S> On the other hand if the capacitor draws current that is we allow the current to flow freely, there will be no accumulation of water and thus increase(change) in pressure (voltage) will be limited. <S> Think of the capacitor as a water tank that stores water. <S> Now the tank takes time to fill. <S> Also a change in pressure does not instantaneously fill the tank. <S> Let the tank have a input and output pipe. <S> If pressure instantaneously increases the tank cannot fill instantaneously so current will flow through output pipe. <S> If pressure increases slowly the tank has time to fill as well. <S> This explains low impedance of capacitor for low frequency (slow change) and high impedance for high frequency (fast change). <S> This is just a analogy to help visualize the phenomena and is not the exact explanation. <S> Comment if you feel anything is out of place. <A> The word "resists" here is nothing to do with resistors. <S> It's the plain English meaning of the word. <S> A capacitor opposes changes in voltage. <S> In doing so, it will tend to drag down the supply voltage, back towards what it was previously. <S> That's assuming that your voltage source has a non-zero internal resistance. <S> If you drop the voltage across a capacitor, it releases it's stored charge as current. <S> That tends to keep up the voltage for a short while. <S> In the example below, Rs is the resistance of the voltage source, assumed to be low, but not zero. <S> All the component values are arbitrary, just for illustration. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> There is a totally different way of looking at it. <S> A capacitor has an impedance given by 1/2πfC, where C is the capacitance, and f the frequency. <S> Notice that as the frequency increases, the impedance drops. <S> Whereas, at DC, the frequency is zero, and the impedance is infinite. <S> With a bit of hand-waving, we can consider frequency as equivalent to rate-of-change of voltage. <S> So if we put a capacitor across an unstable DC supply, then once it's charged up, it has no further effect on the DC, as it has infinite impedance to DC. <S> But it effectively shorts high-frequency AC components to ground. <A> Capacitor impedance reduces with rising rate of change in voltage or slew rate dV/dt or rising frequency by increasing current. <S> This means it resists the rate of change in voltage by absorbing charges with current being the rate of change of charge flow. <S> So at DC in theory there is open circuit high resistance but for rising AC slew rate, the effective resistance drops inversely with rising dV/dt = <S> I/C	If you increase the voltage across a capacitor, it responds by drawing current as it charges.
Potato battery not powering clock So for a school project where we have to make a lesson plan for kids to learn stuff out of "lame" experiments, I was assigned potato clock. I could not for the life of me find a simple 1.5 volt led clock, and I was not allowed to purchase something that is a kit or part of a kit. I ended up purchasing a simple rotary clock, instead. The potato battery I rigged up puts out 1.60-1.80 volts so says my multimeter, but when I alligator clip it to the clock, nothing. I pulled a double a battery from a controller, and it works, but not the potato batterie (also tested the battery with the multimeter to see if the multimeter is broken, it came up right around 1.5 as well). Why will my potato battery not power the clock? <Q> a simple 1.5 volt led clock <S> Well, there's your problem. <S> You want an LCD clock. <S> LEDs take (relatively speaking) a lot of current, and a potato battery simply won't supply it. <S> As a check on this, go back to your potato/clock combo, and measure the voltage with the clock connected. <S> Notice something? <S> Your voltage is almost zero with a load on the potato. <S> I suggest you go on Youtube and search for potato clock. <S> This will give you videos which provide a source for an appropriate clock. <S> If, as you say, you cannot buy such a clock, you are probably out of luck. <S> About the only alternative is to find a clock, or perhaps a watch, which runs on a single alkaline battery. <S> You can figure out which battery connections to connect to which electrodes you have and go from there. <S> But LED and mechanical clocks simply will not work with a single potato. <A> The motor based clock (the square quartz clock module that takes a 1.5V AA cell and ticks once per second) requires short pulses (like 30ms) of fairly high current (few mA) more than you can get from a potato. <S> You could add a fairly large capacitor in parallel with the potato. <S> It will charge up from the potato and supply those short current pulses to the motor. <S> I think you'll need about 1000 uF. <S> I just measured the coil on one, about 260 ohms <S> so just under 6mA at 1.5V. dQ = <S> I <S> * <S> dT = 6mA <S> * 30ms = 180uC, so 1000 uF would give <S> dV = dQ / C = 180mV <S> ripple from the current pulses; good enough. <A> Try putting more potatos in parralell, or find a clock that uses less current. <A> The battery source series resistance must be much less than the load resistance at any time ( no pun intended ) like pulsing the solenoid wheel otherwise the voltage ratio declines according to the resistance ratio. <S> You can measure resistance or ESR on a potato with short circuit current ( expected being less than max range of meter) <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The Short Circuit current must be >10x than the load current to maintain battery voltage within 10%.	This is probably because the clock requires more current than the potato battery can provide. You don't want an LED clock.
Is it possible to make current flow through a transistor from the emitter to the collector without supplying voltage to the base? I've started studying bipolar junction transistors, and I'm trying to understand how they work. When learning about how a transistor work, you always see it connecting 2 circuits, the smaller one involving only the emitter and the base, and the largest one with the collector too. According to what I understood current flow in the largest, only if it flows in the smallest. If I disconnect the base and the smaller cirucuit, or remove the voltage generator from that circuit, current stop flowing in the largest one too. For this reason a transistor in a circuit if no voltage is supplied to the base (or the base not attached to anything) is said to be "off". As my current understanding (it may be wrong), in this scenario, however you put the voltage generator the current can't flow due to the presence of an electric field inside the transistor ( depletion layer). But what would happen if I increase the voltage? Is there a trashold limit where the depletion layer is overcomed? Or is, more generally, possible to make electricity flow from the emitter to the collector without supplying voltage to the base? <Q> The NPN transistor in reverse direction can be viewed as Emitter-base Zener diode with the breakdown voltage larger than 5V plus "ordinary" PN diode between base and collector. <S> simulate this circuit – <S> Schematic created using CircuitLab And the typical values for Veb is <S> BC337-40 <S> Veb=8.2V at I=5.5mA <S> BC549B Veb=8.3V at I=5.5mA <S> BD139-16 Veb=8.5V at I=5.5mA <S> BC639 <S> Vbe=7.7V at I=500uA <S> BC337 Veb=7.9V at I=500uA <S> 2SC945 Veb=8.1V at I=500uA <S> And the emitter-collector breakdown voltage ( Vec ) is: <S> BC337-40 <S> Vec=6.7V at I=5.5mA <S> BC549B <S> Vec=7.2V; I=5.5mA BD139-16 Vec=6.7V; <S> I=5.5mA <S> BC639 Vec=6.3V; <S> I=500uA <S> BC337 Vec=6.4V; <S> I=500uA <S> 2SC945 Vec=7.5V; <S> I=500uA And <S> the "equivalent" circuit is : simulate this circuit <S> And what is more interesting is that the emitter-collector "equivalent" diode act as a "tunnel diode" hence the negative-resistance region in emitter-collector avalanche breakdown. <S> Look at the exampel: <S> http://www.cappels.org/dproj/simplest_LED_flasher/Simplest_LED_Flasher_Circuit.html <S> http://jlnlabs.online.fr/cnr/negosc.htm <A> There will always be some leakage current, even with the base tied to the emitter. <S> However, that is "small" in practical applications to the point it is generally ignored. <S> At about 1000 V per millimeter, air conducts, for example. <S> Each transistor has a maximum C-E voltage spec for that reason. <S> Up to that voltage, the transistor works like a transistor, and the C-E current with the base open is leakage that should also be specified in the datasheet. <S> If you apply more than the maximum C-E voltage, there is no guarantee what the transistor may do. <S> At some point enough current will flow to irreversibly damage the transistor. <S> The leakage will be higher with the base floating than with it tied to the emitter. <S> One way to look at this is that the little bit of leakage across the reverse biased C-B junction ends up being base current. <S> If you really want a BJT to be off, actively hold its base close to the emitter voltage. <S> A transistor in a circuit with its base floating can also pick up and amplify noise. <A> Is there a trashold limit where the depletion layer is overcomed? <S> Or is, more generally, possible to make electricity flow from the emitter to the collector without supplying voltage to the base? <S> There are three useful circuits that employ a transistor without anyvoltage supply to the base (through external wires, at least). <S> The first is a temperature-compensated Zener diode: base-emitterbreakdown may be about 6V, and a forward-biased collector addsa single diode drop to the breakdown. <S> The temperature coefficientof the Zener breakdown and the forward diode <S> are opposite, andapproximately equal, so that this makes a good voltage reference. <S> This application note AN-71 recommends 2N3252. <S> The second is as a phototransistor; in a transparent package,light impinging on the collector-base junction causes leakagecurrent and can bias the transistor into conduction. <S> This isa very common kind of light sensor element. <S> The third is rather exotic; if, instead of Zener-breakdownin the base-emitter, one breaks down the collector-base junctionwith a high applied voltage, one may find (depending on theinternal structure of the transistor) an avalanche-transistorcapabiity to do very fast switching of high current. <S> Manytransistors do not do this usefully, but a few types areavailable ZTX-415 datasheet	If you apply enough voltage to anything, current will eventually flow.
Capacitor Value in Bass amplifier to speaker circuit Replacing electrolytics in a 30+ year old Ampeg BT-15 bass amplifier. The PS filter caps are \$ 2500 \mu F/80V\$ which are rare to be found; I know that 3300/100 would be okay for the filter. There's also a \$0.1 \mu F\$ in parallel which I asked about earlier and got helpful answers. There is also a \$ 2500 \mu F/80V\$ electrolytic in parallel with a \$0.1 \mu F\$ just in front of the speaker connection, after the power transistor network. I know that it is in poor condition because pushing on it causes all kinds of static and popping. The replacement needs to be a through hole mount, preferably radial leads based on the configuration of PCB. I can't find that specific capacitor: the largest 80V cap is \$1800\mu F\$, and going to 100 V gives me either a \$2200\mu F\$ or \$3300\mu F\$ as options. Does it matter whether the replacement is higher or lower in capacitance? What effect on tone might either of these choichave? simulate this circuit – Schematic created using CircuitLab <Q> In your diagram, the speaker is shown as a simple resistive load which is far from reality. <S> In fact, the speaker is a reactive component (see Thiele-Small equivalent circuit). <S> This implies that the whole output circuit has a (hopefully) well designed resonant characteristics, which in turn depend on the capacitance. <S> I'd recommend choosing the capacitance as closely as possible. <S> The voltage rating should be not lower than 80V of course, but it also should not be much higher as well (like 250V or higher for example), but I don't think you could fit a high voltage cap in there anyway. <A> You don't show the schematics, but if the output is coupled via a capacitor, then I assume the amplifier is single-supply, which means the output voltage before the capacitor idles at half-supply relative to ground. <S> Probably something like +30V or so, and it would swing between "close to 0V" and "close to 60V". <S> Or maybe more volts, I'm assuming the designers used 80V caps for a 60V supply with a bit of a safety margin. <S> This means the cap doesn't need to be bipolar. <S> If the output voltage of the amp is +30V at idle, then it isn't going to change sign, and the cap is always going to have a rather large DC voltage across it, and this voltage will always have the same polarity. <S> So a standard polarized cap is fine. <S> (Please check the amp's output voltage before the cap: if it is is close to mid-supply <S> then I'm right). <S> Since the output cap is the same as the cap used in the power supply, then it is safe to assume the manufacturer simply used the same value for convenience. <S> The value of the cap doesn't matter much* as long as it is large enough. <S> A smaller cap will result in less bass, as its impedance increases at low frequencies. <S> You can substitute with a larger cap. <S> Your 3300µF 100V cap would be a good substitute for the 2500µF original cap. <S> It will have a slightly lower impedance at low frequencies, so you'll get a bit more bass, something like 3dB. <S> If the loudspeaker enclosure was tuned for a specific value of cap, then using a different one may produce a small bump or a dip in the frequency response. <S> It may sound a bit different. <S> Caps aren't that expensive, so you can try and adjust to taste. <S> This is a musical instrument, you should pick value that sounds best to you. <S> That will probably be the largest. <A> Bigger V is better as well as C Must be bipolar type	I believe that 2200uF should be well within the actual manufacturing tolerances (which are usually high for electrolytic caps).
PCB ribbon cable drop out easily I have a PCB board and the ribbon cable keep drop out easily. I'm planing to find a way to solve it. In my mind, I was thinking of: Add in another connector (adaptor) with the locking type on top of the connector on the PCB board. Replace the connector on the PCB. (I can not do it but the pin to pin distance is 2 mm) <Q> <A> Desoldering the connector may be best done with a hot air gun. <S> But there is a risk of damage/accidental removal of other components. <S> This will make the fit tighter, but still leaves you the option to easily unplug it later. <A> I would say change the connector. <S> Unsolder it and solder on a new one. <S> Do the same thing on the cable. <S> It's most likely, that it's worn out or maybe even a bad product, that came like that from the company. <S> Before you change both, you can check the fit of connectors first, so you won't do the Sizif's work.	I would test both the connector in the PCB and that on the cable using another set of connectors of the same type to see which is responsible for easy drop out. A quick fix that doesn't involve heat would be to wrap some thin tape around the connector on the cable end.
How to test the maximum secondary current of a step down transformer? I got a small step down transformer from my local market which has a total of 6 individually isolated winding in it (1 primary of 230V and 5 different secondary wingding with 5 different voltage). My question is that how can I find out the maximum output current for the secondary winding? I was told generally these used to be made for "Small Sound Systems". As this from a local market it has no datasheet. Transformer weights about 500 grams and looks well made. The dimensions of the transformer are near about 6cm X 3cm X 5cm and all the secondary wires are less than 1mm in dia. This is my first question on stackexchange. Thanks in advance for all help and suggestions! The winding voltages and resistance measured are as follows: Primary Winding (Red thick wire) 230V with a resistance of 72.6 ohms Blue Wire 8.8V with a resistance of 1.4 ohms Yellow Wire 11.1V with a resistance of 2.0 ohms Green Wire 12.8V with a resistance of 2.7 ohms White Wire 23.1V with a resistance of 9.1 ohms Red wire 3.7V with a resistance of 1.8 ohms These are all the output voltages when the transformer is given 230VAC as input. I tried measuring the maximum output current by a short circuit test (Using some high watt ceramic resistance in series with my DMM in amp mode).But unfortunately the DMM is not showing anything, I double checked my DMM for a blown fuse inside but it was working properly when I tested a DC load. I want to use this transormer for building up a variable power supply circuit, but without getting sure of it's maximum output current of the secondary windings, I can't proceed further.I am not that much experienced of power electronics hence can't figure out what to do next. Here are some pictures of it Thanks again for helping me, Best Regards, Robbin <Q> I got a small step down transformer from my local market which has a total of 6 individually isolated winding in it (1 primary of 230V and 5 different secondary wingding with 5 different voltage). <S> And As this from a local market it has no datasheet. <S> A transformer that drops AC dangerous voltages to much safer lower voltages is: - A COMPONENT THAT IS RELIED UPON FOR SAFETY <S> So, if it doesn't have a data sheet and you bought it from a market I have to point out that you are risking your life using it. <S> I would not use this device at all. <S> Go buy one with a data sheet from a reputable dealer. <A> First of all, a small transformer like that is going to be limited to tens of watts total power, based on the core size alone. <S> Determining how that power is distributed among 5 separate windings is going to be tricky. <S> You can start by estimating the relative wire gauge by looking at the ratios between voltage and resistance for each winding — from that, you can infer what relative current values the designer had in mind. <S> This basically boils down to calculating \$V^2/R\$ for each winding and then figuring out what fraction of the total power it represents. <S> In this case, we find that the four highest voltage windings each get about 24% and the low voltage winding gets about 3%. <S> If we generously assume that the transformer can handle a total of 50W, we get: 8.8V @ <S> 1.4A <S> 11.1V @ <S> 1.1A <S> 12.8V <S> @ 0.94A <S> 23.1V @ <S> 0.52A <S> 3.7V @ <S> 0.41A <A> To estimate a transformer's throughput, weigh it, find a similar build transformer of the same weight in an online catalogue, and read off its VA rating. <S> The decide how to partition that total VA rating between secondaries, calculate \$\frac{V^2}{R}\$ for each secondary winding, where V is the open circuit voltage, and R is the DC resistance of that winding. <S> This will give you the relative power rating of that secondary winding compared to the other windings. <S> Scale the total of those to the total VA for the transformer. <S> Although it's approximate, it's much better than guesswork, or trying to estimate the wire gauge from the leads. <A> Technically I believe it is the RMS resistive load current at 85'C winding temp. <S> This may help get a rough estimate. <S> Note that for a DC bridge this is not true as Peak to Average drops it another 40% resulting in 50% drop from no load to full load. <S> Max power is usually rated as VA (=W+VAR) output which depends on total losses and rated insulation temperature such as 85'C. <S> But VA = <S> W only for linear loads and must be de-rated for Diode bridge cap loads by some 40% depending if DC ripple is very low <<10% <S> My quick and dirty estimate. <S> For a rough estimate use 10%Voc <S> /DCR = <S> I max rms. <S> Another Rule of Thumb Small power transformers have a total weight <S> is mainly steel core which has certain losses according to grade of core laminates and balance from copper weight and losses. <S> My Rule of thumb is 36VA <S> ~ <S> 60VA per kg. <S> How much does yours weight? <A> This looks suspiciously like those marketed for DVD players. <S> I bought one and it was disappointing. <S> The core heated up with no load. <S> color Vrating Vopen current draw at rated voltage ( <S> mA)red 230white <S> 11.5V <S> 12.0 <S> 150yellow <S> 21.5 <S> 20.8 !!! <S> black <S> 3.5 4.0 150blue <S> 13.5 <S> 13.8 <S> 50green <S> 12 <S> 12 <S> 50purple <S> 10 <S> 10.1 <S> 100 <S> You may or may not have something similar... <S> From experience, there is only one way to figure out maximum secondary current output. <S> Make an electronic load and measure it. <S> Make a plot of VxI, and VxVA. <S> Linear region of the VxVA plot will show the max current that can be safely drawn. <S> The VxI plot will be absolutely linear (one can possibly use this property to test transformer voltage drop at the market, before buying). <S> Double check with a thermometer after the transformer has warmed up under load. <S> Make sure not to overlook airflow from table/ceiling fans, overly hot transformers might require air cooling. <S> Most enamel coatings are rated at 130C, with a safe limit of max 85C. <S> The copper winding will generally feel cooler than the core. <S> Also, plugging in the primary and secondary inductances and resistances into ltspice, gives a ballpark figure of the voltage drop under full load. <S> However, it is not too useful. <S> Also, calculating only with secondary resistance is inaccurate at low currents. <S> This is because the current will be limited by the inductance ratio, and not the wire diameter. <S> PS: Here's my first question on stackexchange <S> How to find step down transformer secondary output current rating? <A> Perhaps by monitoring for the 10% drop (suggested by a previous post) from the open circuit voltage with a varying load, the current measured at the 10% point might be the best ballpark figure.	Measure Voc ( open circuit) and secondary DC resistance (DCR) then use the max current as the point where voltage drops 10%. I measured the current draw on the one I bought....
How to safely measure VCEO or VCES of a transistor? As an example, NPN Silicon Power Transistor TIP31 VCEO (Collector-Emitter Voltage) is rated at 40Vdc Maximum. I have many unknown transistors for which I would need to measure the VCEO. Many are power transistors, many are low power. Is there a safe way to measure VCEO (or VCES) of a transistor? (avoiding to fry the unit) <Q> A curve tracer would do it, but if you had a Tektronix curve tracer you wouldn't be asking (photo from here ). <S> If you guess right and don't put too much current for the type of transistor there will be no damage (assuming a BJT- it will destroy a MOSFET gate if there is no internal gate protection). <S> Test some known transistors to get an idea of the safety margins manufacturers use. <A> Nothing puffs the final smoke out as long as you do not burn anything. <S> You can well exceed some breakthrough limit voltage. <S> Only keep the dissipated energy small enough. <S> That's not true if you test insulators, because the area of the breakthrough isn't under your control - very small energy can cause irreversible heating if it happens in area small enough. <S> But transistors and diodes generally do not get destroyed, if you keep the dissipated power low enough. <S> If you are lucky enough to have an oscilloscope, make an XY plotting circuit which shows Ic vs Uce when a fast and sparse ramp or sine pulse is applied. <S> Oscilloscope is not a must. <S> As well you can construct a circuit which applies growing Vce. <S> A current sense circuit switches Vce off as soon as the pretermined Ic is reached. <S> A hold circuit saves the highest Vce into a capcitor and you can read it from a DVM. <A> There is a very simple way to measure it. <S> You'll need a voltage supply rail that is high enough to reach the voltages you want to test, of course. <S> (\$V_\text{SUPPLY}\gt V_\text{CEO}\$.) <S> If this is a lab supply that allows you to set a current limit, use that feature. <S> If you don't have a current limiting feature to access, then you will need to fabricate something (not hard, but again if you are using BJTs for this you need to be sure their \$V_\text{CEO}\$ is large enough.) <S> Just set the desired current and feed the current source to the collector of the BJT. <S> Leave the base open, of course. <S> Hook the emitter to the other side of the power supply. <S> Attach a voltmeter across the emitter and collector and measure the voltage there. <S> The current source (or sink, depending) should be set to a value you feel comfortable with considering the BJT. <S> Make sure that you test this current source/sink with a few different resistors to be sure that it is doing what you expect and yielding a value you want -- before applying it to your transistor. <S> That's about it. <S> Don't test too long. <S> Just long enough to get your measurement. <S> See Vishay's Measurement Techniques for an example <S> (Figure 3.) <A> The VCEO is similar to operating a Zener 12 Watts is considered safe for this 40W device. <S> ( 0.12A @100V) <S> Simply use a sufficiently high R value to measure V with a DMM. <S> simulate this circuit – <S> Schematic created using CircuitLab Anecdotal <S> My tube TV in late 70's needed a 160V regulated voltage (HSweep) <S> so I looked up the Vceo of my spare parts and found one in that range <S> and it worked same as a Zener. <S> Cheap and quick fix.	If you have a suitable power supply you can manually trace curves by putting a relatively high value resistor in series with the transistor and slowly increasing the voltage until you start to observe breakdown (don't put too much current through the transistor or it may be damaged).
Get Feedback from Internal Potentiomet of Servo Motor Hello! I was trying to control an MG996R Tower Pro servo motor using Matlab and I need to get some feedback from the internal potentiometer of the servo. However, I don't know where to connect the wire so that I can get some feedback from the potentiometer. I have attached a picture of the board of my servo. Thanks in advance! <Q> Generally speaking questions about modification of undocumented devices are not allowed here. <S> That said, it is likely that the servo amplifier wires the variable resistor as a potentiometer, which is to say applies some relatively constant voltage to the outer leads and measures the voltage on the inner lead. <S> If you want to "snoop" on the position you would need to connect a high-impedance ADC input to the middle lead, and perhaps also the the higher voltage outer lead to measure the scale. <S> Do take note that some MCU ADC inputs are not high impedance at all - especially in faster conversion modes, they can draw a fair amount of burst current. <S> If that is the case you may need a buffer amplifier to avoid disturbing the servo amplifier's own reading. <S> You could potentially use one of these and modify it to report position back over a bidirectional interface scheme. <A> Every single servo that I've taken apart or built <S> * has the V+ lead connected to one side of the pot, the GND lead connected to the other side of the pot, and takes output from the middle. <S> There's only a few (two that I know of, or perhaps just one) <S> analog servo amplifier IC designs out there. <S> The new "digital" servos may do it differently <S> *, <S> but probably don't. <S> (So @ <S> Chris Stratton -- now they're documented!!). <S> If you've replacing the servo electronics entirely, then just run three leads to the pot (it's under the electronics board in your picture). <S> If not, then your second-best bet is to grab the center lead and ADC convert it -- this will be noisy. <S> If you really want a good reading, then run all three pot wires back to your data acquisition system, treating them a signals independent of ground. <S> Then measure all three (or apply ground and center tap to a differential amp, and +V and ground to a different differential amp and measure those two). <S> The ratio of (center tap - 'gnd')/(+V - 'gnd') is the number you're looking for. <S> <S> I'm old enough to have built half a dozen Heath and Ace R/C servo kits. <S> ** <S> For best performance you'd want to at least RC filter the incoming power for the purposes of driving the pot, or regulate it down -- both of these would take up physical room and voltage headroom. <A> As you mentioned DC servo motors usually use a potentiometer to return position feedback for their own driver. <S> The only thing you need is to follow potentiometer track on driver PCB. <S> Next solder a wire to it and bring it out. <S> Then you need to do an identification of your motor position feedback. <S> To do this only connect the potentiometer wire having been already brought out to a voltmeter. <S> Finally command your servo motors to some specific positions and write their related voltages. <S> Now you will be able to obtain a linear equation which relates position and voltage. <S> I have already done a similar project. <S> This page helped me a lot. <S> I hope it would be helpful for you too.	There are also various retrofit designs/kits out there which replace the electronics of common servos with an MCU based solution, often offering a more digital communication scheme.
Why is input impedance matching done for amplifiers? I am trying to design a single transistor RF power amplifier. For a single transistor amplifier, the voltage at the base/gate of the transistor determines the current through the transistor and ultimately the output power. Then why is power input to the transistor maximized by impedance matching. Why can't impedance bridging be done to increase the voltage at the gate/base of the transistor and hence increase the collector/drain current? Edit: If your answer is maximum power transfer, I dont understand how impedance matching ensures maximum power transfer. I understand that impedance matching ensures that maximum power enters the transistor but the transistor is not a power amplifier. It is usually a transconductor or a voltage amplifier. How does the output power maximize when the input power is maximized? Edit again: I realized that this question Why do we care about matching the input impedance of receiving RF amps? is very similar to the one I am asking. The accepted answer there is that maximizing the power input to the amplifier through impedance matching increases the SNR of the input signal thereby increasing the SNR of the output signal too. Is the same argument applicable here too? Is maintaining the SNR of the signal the sole reason to maximize the input power to the amplifier by impedance matching? <Q> Impedance bridging is done if you are interested in Voltage rather than Power. <S> It will ensure that maximum voltage is driven to the amplifier input with minimum signal degradation. <S> It consumes less input power, as the current drawn by the series combination of source and load impedances is low. <S> It is relevant in voltage amplifiers. <S> Impedance matching is done when you are interested in Power. <S> It is relevant in Power amplifiers, where output loads are of low resistances. <S> Here the input impedance is equal to the source impedance. <S> The amplifer will now extract maximum Power off the source at its input (MPT-Theorem). <S> Compared to previous case, the input current of the amplifier is higher too. <S> Hence, the amplifier can now drive higher current in the output collector circuit. <S> i.e., More Power can be extracted at the output. <S> Also, it is not just about the output Power. <S> Maximum Power efficiency is obtained only when the impedances are matched. <A> Impedance matching is important in RF circuits to prevent reflections. <S> When the input impedance of the load matches the characteristic impedance of the transmission line there is no reflection at the load. <A> View the transistor as a box that provides transconductance between input/base and output/collector. <S> For maximum output at the collector, the base needs the maximum input voltage. <S> In the presence of various reactances at the base, such as Cemitter_base and Ccollector_base (which will vary as collector resonances change with temperature and the collector VPP changes), the designer still needs to achieve some guaranteed VinPP on the base. <S> You may need to de-Q (include some dampening) if your RF signal must cover a wide range of frequencies. <S> You need to examine various stability factors, and place the various resonances at frequencies, and with adequate dampening, that you the designer can guarantee the RF amplifier will not oscillate nor will become peaky enough to cause unacceptable variations in output VPP. <A> If you look at the base or gate of a transistor, it is both resistive and capacitive. <S> Hence it is called input impedance. <S> You want the input power to be put in the resistive part of the input <S> so you have to eliminate the capacitive part. <S> This is called impedance matching. <S> Assume the input comes from a 50 Ω source, you have to match it to the complex impedance of the transistor. <S> You can use a L matching network. <S> The same applies for the output of a transistor.	It is also about Power efficiency in terms of input power and output power of the amplifier.
Advantages to solder over headers? I'm in the process of assembling an embedded (Raspberry Pi-based) machine that has a lot of sensors wired together via I2C. I find myself adding headers for the I2C lines to all the little tiny breakout boards on which the sensors are mounted, and I have an I2C "bus board" that I've assembled using a prototyping board to simplify connecting all the sensors in parallel, but I'm wondering if there's any advantage to soldering the connecting wires instead of using 0.1" headers and jumper wires. The boards are not very rigorous timing-wise (the bus is set to a low bandwidth.) For the record, what I have now -- everything assembled via headers -- works just fine. I'm concerned about reliability, noise, capacitance, etc. What's best practices when it comes to assembling systems like this? <Q> Advantages of solder: <S> Lower part usage <S> Mechanical strength <S> Part cost Disadvantages: <S> There isn't much to worry about for capacitance and noise for a header + connectors vs soldering. <S> It really boils down to assembly cost/time/effort. <S> Keep in mind commercial products use both in similar environments. <S> The most critical thing here is if your use case involves a lot of vibration. <A> In addition to what @Passerby said, I can think of few other considerations. <S> 1. <S> Space limitations: <S> Fairly obviously, headers and pin connectors require more physical space to accommodate them without pinching or stressing the wires. <S> 2. <S> Fail-safe needs for wire <S> snags/ tension: Depending on where and how the sensor will be mounted, and how the wires will be secured if at all, the relatively insecure connection of headers vs solder could actually be an advantage. <S> Better to have a connector unplug than to damage a PCB. <S> Of course good wire management and proper strain relief is the best option when possible regardless of termination type. <S> 3. <S> Corrosion risks: <S> Solder is inherently less susceptible to corrosion than pin connectors. <S> Granted, unless you've got a fully potted or well coated board, by the time corrosion becomes an issue with header pins you'll probably also start having to worry about any small pin pitch and sensitive components on the board. <S> Gold plated headers and pins can help here too, albeit more expensive. <S> 4. <S> Redesign and re-purposing flexibility: <S> I do a lot of prototyping work. <S> I'll have to change the wire lengths or scrap the project. <S> No clear overall winner. <S> For me, the decision usually ultimately comes down to how I have to mount it and how likely I am to unplug/replace it. <S> Really just a matter of balancing project needs and life cycle expectations with personal preferences. <A> Except soldered wires fatigue where the solder filled strands meet the strands. <S> Support with hotmelt glue <S> so the bending cannot occur where the solder has wicked up the strands. <S> If you are making pcbs, I find it very easy to have oblong pads 0.050 <S> " apart, and solder ribbon down directly without splitting the conductors apart. <S> For I2C we use Micro-match connectors (see i2c connectors ), which are low height mass-terminate ribbon cable, can crimp in a vice -no special tool needed can make daisy chain cables with multiple males on one ribbon for bussing. <S> drawback: <S> zigzag footprint bad for Veroboard	Nothing wrong with solder, until you have to unplug it. Effort to replace if needed Effort to assemble Sometimes I go with headers simply because I know there are good odds Solder would beat a non locking terminal strip.
History of ESD protection These days you can't read anything about electronics without hearing about the dire consequences of omitting ESD protection from your designs. It seems this wasn't always the case. If you look at most older designs (late 90s and earlier), you hardly ever see ESD protection components on I/O ports - even when CMOS ICs are involved. Why is this? One might assume that modern ICs are more vulnerable than older processes. I don't think there's much truth to that though. Looking at the datasheet for a typical 74HC gate, they specify a maximum HBM of 2kV. This is the same as a modern MCU or FPGA. I could understand designs using 74LS gates omitting protection, but I also see lack of protection when 74HC or CMOS LSIs are used in older designs. And having looked at more recent datasheets for 74LS parts, even those are specified only up to 2kV HBM. Is it just that we have a better awareness of ESD now? Or could it be that you can actually still get away without ESD protection just as they did in the 90s? Is there a minimum level of ESD robustness required by law for consumer products, that wasn't enforced in the past? <Q> I can only speak from my personal view... <S> In the 80's, there was a thorough awareness of the need for ESD protection and handling precautions. <S> My then-employer modernised its factory in 1987, taking ESD precautions from only in the electronics assembly areas to every assembly area. <S> The consensus was that we were very late in the day to do so. <S> The understanding of ESD that I see is no more stringent or widespread than then. <S> What has changed considerably since then are the costs of electronic components and of PCB design, manufacture and assembly. <S> I've seen these get dramatically cheaper. <S> So the costs of adding extra parts, such as transorbs, TVS or input filters are much cheaper. <S> Back then, in the commercial industry I worked in, these protection parts would be an unacceptable increase in cost. <S> It's not just the component cost, its the price of the inventory, assembly and testing that goes with having those parts there. <A> As long as some junctions are tied to input pins, and not just MOS-gates, the older CMOS seems tolerant to modest charges such as sliding into ESD_proof mylar envelopes. <S> The outputs of CMOS are by definition "junctions", because FET drains are tied to the output pins, to aggressively pull up or pull down. <S> These drains are not optimized for ESD energy processing. <S> For ESD survival, the charge needs to be taken DEEP into the bulk silicon where lots of mass is available to absorb the ESD energy. <S> Fast CMOS PMOS and <S> NMOS are designed for speed, and are built on the surface. <S> Thus modern ESD structures may use special implants, deep implants, to include mass for energy absorption. <A> Any manufacturing plant worth a nickel pays a great deal of attention to ESD-or else suffer losses while a board is still in process. <S> The 74cxx and 74HCxx have some diode protection, but the CD4000 and outdated 74Fxx series has little to none. <S> Walk on a carpet and touch a lead while it is on a table top and consider it gone. <S> It was easy to test for a blown CMOS part, as I would get a positive voltage reading out of an input pin. <S> Along with IC's we had trouble with the newer generation of LED's blowing from the stock room not handling them with anti-static bags. <S> As part of engineering I had warned them, but it took blown parts and a verbal assault by the company VP to get things in order. <S> You can no longer 'get away' without grounding wrist and ankle straps, and a grounded solder iron. <S> ISO does pay attention to this issue, but only in terms of how a company manages internal ESD issues. <S> Do they have procedures in place and do they follow them when an ESD incident occurs and take corrective action? <S> There is <S> no government issued ESD regulations unless you care to follow military or NASA procedures, but none the less ISO will force a manufacturing plant to work out and implement ESD procedures. <S> No, you can't pay the ISO person to look the other way. <S> They make plenty of money and are often escorted by a senior employee or a trainee. <S> Once ESD protection permeates a manufacturing plant, an ESD based failure can usually be traced back to someone new or who has poor English skills, so it becomes a re-training issue, or a documentation issue. <S> ISO looks into ALL your documents to find a root document, at the base of the document tree. <S> This is a link to a 3M document on ESD and the fine details: LINK	So there is much better awareness across the board in terms of ESD, mostly a stockroom and assembly issue.
Choosing pulldown resistor for 74LS86 I am trying to build a half adder using a 74LS86 XOR gate. Initially I connected both input pins that I am using (9 & 10) on the 74LS86 to the ground rail using 10K ohm resistors. I then connected 5V to pin 9 and expected the LED fed from pin 8 to light. No luck. I then pulled pins 9 and 10 high by connecting the 10K ohm resistors to 5V. When I pulled pin 10 low by connecting it to ground, the LED lit. I substituted 330 ohm resistors for the 10K resistors and reconnected them to ground. Now the circuit works properly when I pull either pin 9 or 10 high. Is 330 ohms an appropriate choice? Are pins 9 and 10 sourcing current when they are not active (pulled high)? I read that low power schottky ICs can only source 8mA. I am calculating that with 5V and 330 ohms that the current draw could be 15mA. I am wondering where I should look on the datasheet to properly size pulldown resistor for TTL ICs. <Q> Inputs on bipolar TTL (plain 74xx, 74LSxx, 74ALSxx, 74 anything without a "C") <S> all source current, and must be pulled below 0.8 volts to be seen as a low. <S> CMOS parts (anything with a "C" in the middle letters) have a very high input impedance, so can be used with a 5 - 10K pull-up resistor with a switch to ground, or 5 - 10 <S> K pull-down resistor, with a switch to +5V. <A> See below clip of this datasheet: <S> The first number Vil is the maximum input voltage that is guaranteed to be interpreted as low. <S> It is 800mV. <S> However we almost always want a lower voltage than that to give some noise immunity. <S> The next number Vol shows the guaranteed output voltage when sinking 4mA, which is 400mV, yielding 400mV of noise immunity. <S> It would be good to at least match this, so input voltage between 0V and 400mV. <S> The final number Iil is the maximum input current when the input is at 400mV, -600uA. <S> So the maximum resistor value on a single input (on this particular chip) for 400mV of noise immunity will be (from Ohm's law) <S> 400mV/0.6mA = <S> 667\$\Omega\$. <S> For example, the 74LS30 has the circuit shown below: There is only a single resistor for all the inputs, so the input current is the same for one input low or 8 inputs low. <A> I will expand on Peter Bennett's answer: Don't use use pulldowns at all. <S> Instead, use 1k pullups to +5, and ground the input for a low. <S> In other words, simulate this circuit – Schematic created using CircuitLab Tieing inputs directly to ground will not cause excessive current to flow in (LS)TTL, although the same is not true for tieing to +5. <S> So use pullups. <S> For CMOS, you can tie inputs directly to ground and Vcc. <S> Also note the connection of the LED. <S> As you have realized, TTL works much better as a current sink rather than a source, so go with it. <S> If the polarity inversion (a low output turns the LED on) bothers you, buffer the LED with an inverter such as an LS04, or use another stage of the '86 to do the job.	Common practice for switch inputs was to put the switch between the input pin and ground, with a 5K or so resistor to +5V to ensure the input would really be seen as a High when the switch was open. If you just want to use the X-OR gate as an inverter, it's better to just tie the input to GND, of course, as that saves a part and gives you 800mV of noise immunity rather than 400. Note that in some cases multiple inputs to a gate will source the same current as a single input.
In a flyback transformer with multiple windings, how are voltages controlled for different current loads? This is the flyback transformer I will be referring to: Wurth 760871543 There are three output windings as follows, 24 V at 0.30 A, 5 V at 1.15 A, 14 V (unspecified but in mA range aux winding). The voltage which I need to control precisely is the 24 V winding using secondary side sensing. Here is my question: If the output I am sensing has a very low current load (1 mA), but the 5 V winding has a very high current load (1 A), won't the voltage at the 5 V winding drop much faster than at the 24 V winding? If this is correct, how would I then go about controlling all the voltages simultaneously? For example, if I instead sense the 5 V output winding for feedback, would switching the flyback transformer more frequently (to compensate for the larger current load) cause the voltage at the 24 V output winding to increase more than the regular 24 V? Just some additional background on my design...I am using a Power Integrations TinySwitch-4 switching controller. The 24 V will get increased to 1 kV through a converter and voltage multiplier. The 5 V will power a Raspberry Pi. The 14 V auxiliary winding will power the switching controller IC. <Q> If you need multiple precisely controlled voltages, you will need multiple independent controllers. <S> That being said, in theory the output voltage of a flyback converter is dependent ONLY on the duty cycle and transformer ratios - NOT on the load current. <S> Unfortunately, this simple relationship breaks down because of losses, but in a well designed converter with fairly low loss components, the other output voltages should be close . <S> Whether they are close enough depends on your allowable voltage range and the detailed design of the converter. <S> Typically, multi-output flybacks make the assumption that they know approximately what the current draw of each of their outputs will be, and set the turns ratios to get close enough to the intended output voltage over the expected range of load currents. <S> However, all output voltages except the master output (whose voltage is being controlled) will vary over load conditions. <A> You simply can't control the output voltages of that transformer separately. <S> If you have an uneven load on its secondary, then the "unregulated" output voltage will go up or down depending on the situation, just like you already mentioned. <S> Do you really need the 24V to be that accurate? <S> A voltage multiplier's output goes down quite a lot under load so you might need additional regulation there anyway (sensing the 1kV high voltage line). <S> If both of your voltages have to be accurate, you can design the converter to deliver 6V instead of 5V and use an additional linear regulator to get regulated 5V. <A> Your load regulation <S> Error and Relaxation voltage time constant <S> Will both be dependent on the load resistance ratio with source impedance or the percentage of maximum current rating. <S> Therefore your control feedback should be from the highest power load which is probably your 24 V boost regulator . <S> Since the 5V load is small, <S> The overvoltage can be regulated with an LDO.	A multiple output flyback converter (with only one controller, i.e. PWM circuit/feedback loop) can only precisely control one voltage.
How much power does a strain gauge use? Is it essentially nothing given how small they are? I see most configurations are given 5V, but does that mean they draw amperage as well? <Q> This means that their current draw will vary depending on the strain. <S> The actual resistance of the gauges will be found on the datasheet, but can vary a lot between different products. <S> For example, this one is 120 ohms, and this one is 10 kohms - clearly a big difference. <S> The overall current depends on your configuration: for example, you could use a Wheatstone bridge or a simple potential divider. <A> As said by the others, they are resistors, so they will consume power depending on how much current runs through them. <S> That said, if you do not need continuous measurement, you can only activate the measurement current/voltage when you need it. <S> For example, if you make a battery-powered scale that wakes up when the user stands on it, you would keep the whole circuit in sleep, and periodically wake up, send current into the strain gauge during a very short time, just enough to take a rough measurement and determine if the user is standing on the scale or not. <S> If the scale wakes up for 100µs every second, then power consumption will be 10000 times less. <A> In strain gauge bridges that I have used (120 Ω and 350 Ω) I've tended to excite them with 2 to 4 mA rather than with a fixed excitation voltage and so 1 or 2 mA would flow through each gauge. <S> Power = \$I^2R\$ hence <S> , for a 350 Ω gauge it is up to 1.4 mW. <A> Bridge/Strain Gage Signal Conditioner Omega <S> CAD530.00 <S> DMD4059 <S> Maximum Output: 10 Vdc @ <S> 120 mA <S> You can operate from 1 to 10V and power is increased with sensitivity accordingly. <A> Since these sensors are essentially resistive networks, the excitation energy could theoretically be arbitrarily small - in practice, noise, interference, leakage and dry contact concerns, and power requirements of the conditioning circuitry, will dictate the lower limit.	Strain gauges change their resistance depending on the strain they're subjected to.
Negative Power Supply for Op Amp I have a 12V (single ended)power supply available for and op amp circuit I am designing. I want it to be very precise around zero volts for an output. I already though about using a virtual ground, but I am putting the output into a Max 10 FPGA, so I do not think that I can accommodate for that. My question is, is it possible to use a second op amp and make the Vcc of it ground and the Vee of it as 12V and use it as an inverting op amp, to then use that as the power supply for the first op amp. The op amps that I have are LM358, but I do not think that it makes a difference for this scenario. I am shifting the input and scaling it with a circuit like the one below, I am just trying to improve its resolution at small outputs. <Q> Opamps don't produce voltage or current. <S> They let more or less current pass from the powersupply to the output. <S> To get a negative voltage out of an opamp, you have give it a powersupply with a negative voltage. <S> The output of an opamp is limited to what is available from the supply rails. <A> Figure 1. <S> Internals of the ancient 741 opamp. <S> Source: Wikipedia . <S> From the internal schematic of the 741 op-amp it should be clear that the output can source current from the \$ V_{S+} \$ rail via Q14 or sink current to the \$ V_{S-} \$ rail via Q20. <S> It is not possible to generate an output voltage higher than the postive supply pin or more negative than the negative supply pin. <A> If you are not using the 12 volt supply for anything else, then you can float it and generate a +/- <S> 6 volt supply by connecting what you think of as a virtual ground to your system ground. <S> This does require that the 12 volt supply is "isolated". <S> simulate this circuit – <S> Schematic created using CircuitLab <S> A few notes: 1) <S> The virtual ground op amp should be significantly faster than your application op amp. <S> 2) <A> No, that will just break the op amp. <S> If you truly need -12V, then you need a second power supply to generate it. <S> Something like this might work, but if you need a precise reference then you will want to use a commercial module. <A> Your question is over 1 year old, so you most likely have a solution already. <S> If not, here is my advice:	The virtual ground op amp should have better current output capability than the application op amp. If you have rails of 0 and 12V, then that's the limits of your output.
How to prevent power overload when loads exceed a generator's capacity How to solve the following practical problem that I currently have in my "food bike" design in a practical manner: Power supply (AC): Generator: 800W Loads (AC, purely resistive, each controlled with its own thermostat): water heater: 750W syrup heater: 517W Rules: The available power may never be exceeded. If so, a breaker trips, which must be avoided. The syrup heater load demand must always be satisfied. The water heater load demand is allowed to be interrupted in order to satisfy the syrup heater load. Edit 1:Additional rules to narrow the solution space: Human intervention is not allowed to achieve the desired result because the food bike operator is too busy serving his customers. Edit 2: I prefer to keep both heaters unmodified so their certification remains valid. Therefore both heaters (including their thermostats) are considered as "black boxes" with only their power cords as the "interfaces" to an external circuit. Additional system details: This is a real-world problem that I try to solve in order to design a food bike for myself in order to have my own small business, in contrast to food-trucks, and in the long term to be powered with PV-charged LFP batteries. For now, I use a noisy and stinky generator. Each thermostat switches its heater on to keep it at a given temperature. Assume, for the sake of conversation, that this happens randomly at intervals of roughly a minute or so. The water heater is expected to demand power predominantly after the food bike operator washes his hands because that heater serves the hand-washing station. The budget to solve this specific problem is in the order of $100. Weight is an issue, so a lightweight solution is desired. Fictitiously, an "XOR" power strip with an configurable socket prioritization (the syrup heater socket would get priority) would solve the problem, by that I mean a pair of AC power sockets that mutually exclusive provide power without temporal overlap that would trip the generator's breaker. You could also see this as a "power strip with load-shedding". Project's geographical location: California, USA (needed for appropriate certifications / regulations). <Q> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> (a) Existing circuit. <S> (b) Syrup priority relay modification. <S> How it works: RLY1 is an AC relay. <S> Coil voltage should match the generator voltage. <S> Contact rating should have a minimum voltage rating to match the generator and current rating of at least that of the load. <S> The SYRUP and WATER thermostats are normally closed and will open when temperature is exceeded. <S> When the SYRUP thermostat closes the relay will energise and power the syrup heater. <S> The water circuit will be disconnected. <S> When the SYRUP reaches temperature its thermostat will open, the relay will drop back to its shown position and the water circuit can work as normal. <S> simulate this circuit Figure 2. <S> Using a thermostat with a changeover contact makes the modification simple. <S> If you could replace the thermostat with one with a changeover contact then the modification becomes very simple. <S> simulate this circuit Figure 3. <S> (a) Existing circuit. <S> (b) Syrup priority current sensing relay module. <S> This circuit does not require modification of the syrup heater. <S> It is similar to the socket monitor in Jeroen3's answer. <S> It requires the use of a current sensing relay with an AC coil rating to match the current drawn by the syrup heater. <S> The problem with this circuit is that if the water heater is on then there is a transient high-current load when the syrup heater first switches on. <S> Both heaters will be on for a fraction of a second until RLY1 moves. <S> This may trip out your generator. <S> simulate this circuit Figure 4. <S> A current limiting lamp to prevent overload on switchover. <S> This circuit uses a lamp to limit current while the relay energises. <S> When it does relay contact 1b bypasses the lamp and applies full power to the heater. <S> Meanwhile the water heater is already disconnected so current is limited. <S> This solution relies on finding a relay that will energise on the low current but survive with the high current running through the coil. <S> Fuses and circuit breakers have been omitted for clarity. <A> Here in Europe we have special extension cords to have two big loads on one breaker. <S> Advertised for washers and dryers. <S> Energy distributor switch, Energieverdeler, Energieverteiler. <S> When the white socket is asking current, the black one disconnects. <S> Link to dutch shop. <S> You can also assemble this yourself with a industrical current monitor relay and a power relay. <S> Or a simple circuit with a LEM module and a comparator with delay that controls a relay. <S> But $100 is very a tigh budget if you do not want to bodge something. <S> (safety!) <A> A device called a "current sensing relay" solves this problem within the financial budget. <S> For example this one from CR Magnetics: it contains a current transformer and a SPDT (== Form C) electromechanical relay. <S> The typical power consumption is typically 80mA @ <S> 120Vac, so 10W of heat needs to escape. <S> Edit: In my experiments, the heat production was negligible. <S> Enclosure in a plastic box without venting holes did not result in a noticeable temperature increase for at least the first 15 minutes. <S> However, more testing is needed to confirm this. <S> Note that this solution still has a small temporal overlap of all loads in the order of (half) <S> the period of the 60Hz (USA) sine wave. <S> It work if the total load does not exceed the generator's "maximum starting Watts" as it is called in my generator's manual. <S> I actually tried this out and it worked as hoped: No tripped breaker! <A> When the switch is one way the water heater is on, when its the other way the syrup heater is on. <S> When the switch is in the center position neither is on. <S> Digikey has two switches that appear to meet your requirement. <S> CW Industries GTS448E101AHR <S> 250V/12A, $4.42 <S> each https://media.digikey.com/pdf/Data%20Sheets/CW%20Industries%20PDFs/GTS448E101AHR.pdf <S> Bulgin C1720H <S> 250V/20A, $5.14 each www.bulgin.com/media/bulgin/data/A_Toggle%20Switch.pdf	You can do what you require with the addition of one relay with a changeover contact and a little rewiring. Use a dual pole switch.
what are the implications of using a 45kW ACS580 ABB VSD on a 22kw induction motor? What are the implications of using a 45kW ACS580 ABB VSD on a 22kW induction motor? <Q> It means the ACS580 will run nice and cool. <S> The motor will only take the power it wants. <S> If it has programmable current limits, you may want to halve the default settings, to help protect the motor if it became overloaded, though the primary protection for this would be a thermal trip on the motor itself. <A> You're spending more money than you need to. <S> If you have one "Lying around" no problem, but if we double rate every component and have to buy the equipment your wasting money. <S> For example, If your motor only needs to do 10kw of work why would you use a 22kw motor? <S> inefficient, also if you have a constant load on the motor just use the VFD for starting i.e. switch the motor DOL once it's running instead of running the VFD constantly, also inefficient. <A> The nominal VFD kW rating is the maximum motor kW rating. <S> The VFD kW rating may need to be higher than the motor kW rating in various situations. <S> The VFD output current rating must also be equal or greater than the motor full-load current rating. <S> Some motors, such as those with more than 4 poles, may have a higher than typical current rating. <S> The nominal kW rating is the rating for light duty use. <S> For heavy duty, refer to the table in the manual. <S> Refer to the derating instructions in the manual. <S> If the drive is to be mounted in an IP55 enclosure, derating is required. <S> Refer to the derating factors in the manual. <S> If the switching frequency must be set above 4 kHz, the drive will need to be derated. <S> Theoretically, mean time before failure (MTBF) for any electronic device is inversely proportional to component operating temperature. <S> It may be desirable to increase the drive kW rating to reduce the component operating temperature and increases MTBF. <S> The VFD is designed to assure the proper performance and provide protection for a range motors of various power, current, frequency, and voltage ratings and number of poles. <S> The drive start-up procedure requires the motor rating plate data to be entered into the drive to make that possible. <A> I don't know specifically about the ABB drive but in some VFDs that use Vector Control, the settings for motor current values will not go below 50% of the rated current of the drive. <S> This is because the current sensing devices inside of the drive have a tolerance for their accuracy and when the motor is too small PLUS lightly loaded, the Vector Control algorithm my not be able to accurately track the extremely low values of current it is seeing, which affects performance and low speed torque accuracy, which is likely why you wanted Vector Control in the first place. <S> If you are using a VFD in simple open loop V/Hz (Scalar) control mode, it makes no difference at all. <S> In fact in that mode the VFD may not even know if the motor is connected or not. <A> See <S> ABB NOTES: Sizing VFDs for HVAC Applications, Based on the National Electrical Code	The drive may need to be derated for altitude or ambient temperature higher than the nominal limits. The answer here is that the input power wiring to the VFD must be sized for the drive’s nameplate input current, even though there will never be a demand for this much current.
Why Electro Mechanical Relay makes controller reset? In my current project, I am controlling Peltier Element Temperature which is current controlled load using Arduino. I am controlling Temperature of Peltier by applying PWM to H-bridge. In addition to this, I have to change the polarity of Peltier Element. For this functionality, I am using two Electro Mechanical Relay. Once Relay starts switching it makes the controller Reset. What could be the possible reason for this error? And How could I solve it? <Q> if the maximum output current of the voltage regulator feeding the circuit is exceeded <S> its output voltage drops causing a drop in the supply voltage of the chip in the arduino, and the chip resets if its supply voltage is under the minimum recommanded value giving in the datasheet. <A> Probably when the relay coil releases, there is a big voltage (flyback) spike from the coil. <S> You need a diode straight across the relay coil pins. <S> Alternatively there isn't enough power to energise the coils, and VCC is falling. <A> In my current project, I am controlling Peltier Element Temperature which is current controlled load using Arduino. <S> I am controlling Temperature of Peltier by applying PWM to H-bridge. <S> If you use a H-bridge then you don't need relays to control current direction (polarity) because by driving the H-bridge appropriately you can get full supply reversal: - Pictures from here . <S> If Q2 and Q3 turns off and Q1 and Q4 turn on you get current reversal: - <S> Once Relay starts switching it makes the controller Reset. <S> What could be the possible reason for this error? <S> And How could I solve it? <S> There could be several reasons: - Bad circuit layout	Bad design Power supply not sufficient to drive all loads Flyback diodes required
Edge-mount SMA connector for ~100mil board thickness I have a footprint on a ~100mil thick PCB for an SMA edge connector: I find millions of SMA edge connectors for 62mil board thickness and I found a single one for 32mil ( https://www.adafruit.com/product/1864 ). However, I found none for ~ 100mil thickness. Do they exist? And if not, is there any hack to still install an SMA connector reliably? (Frequency range is up to 3 GHz, RL as good as possible but not critical; price secondary). It is worth to mention that the PCB is already fabricated with the footprint above. That means I cannot change footprints, holes, thickness etc. <Q> This type was I believe originated by Southwest Microwave, but is now available from a couple other vendors: <S> This type originated with Rosenberger: <S> But either of these types has lots of parts so it will be very high priced for a 3 GHz application. <S> Why not just use the old standard through-hole right-angle type? <S> The pins are typically a bit over 100 mils, so you shouldn't have trouble soldering them. <S> For best signal integrity, mount this type on the opposite side of the board from where your signals are routed. <A> There are some options to accommodate your pre-existing footprint. <S> Like this one from Amphenol, Or several similar. <S> They offer edge-mount connectors for 0.093" and 0.11" board thick, but they are not just like a "fork", and need two holes to mount. <S> However, your footprint doesn't look very "impedance matching" friendly, so doing hacks will definitely cause some serious impedance mismatch. <S> Also, "Adafruit" is not the best place to shop for RF connectors. <S> Try Mouser, Digi-key, or more specialized places. <A> In addition tot he answers given by The Photon, Rosenberger also have these: <S> But these are expensive. <S> Really expensive. <A> Thanks to The Photon < I posted the wrong connector, but you are in luck. <S> this one works as an edge insert to thick boards 0-18GHz <S> $6(1) <S> 116mil depth Best bet for 3GHz for cost and high performance if it can be made to suit application. <S> Maximum VSWR is 1.20 (~21dB) at 11GHz Molex £1.37(1) <S> Farnell <S> Molex 0733910060 <S> $3 (1) <S> 18GHz <S> When you find an RF connector with no VSWR or Return Loss specifications, be very concerned. <S> Consider machine and plating tolerances and the precision of the feed pin , how its soldered and any variations. <S> These all significantly affect Return loss at 1GHz and much more at 10GHz etc... <S> Then decide if 10 dB RL is adequate for a cheap connector. <S> You ought to expect this in cheap mass-produced RF connectors without quality controls. <S> Would rather pay extra for 30dB .. <S> You can't afford 40dB or even need it because the geometric tolerances are < 100ppm and dielectric tolerances for d constant, also very tight. <S> Since you asked about 100 mil boards, I assume rigidity is important to you. <S> In this case consider adding rigid substrate or other stiffener mechanics. <S> From my experience, SMD connectors are just bad news mechanically for any type with forces applied to the cable and jack. <S> But I understand if you have a height restriction or packaging problem. <S> There are alternatives. <S> yours on left, on right Amphenol p.n. 901-10511-3 $2.50 left , $13 right Edge connectors take a lot of abuse and SMA connectors MUST be careful torqued by a calibration tool for optimal performance. <S> $25 <S> (1) 02K249-40ML5 <S> Rosenberger $105 (1) <S> My experience came from setting up Iridium Satphone LNA testset for mobile radio chips with Anritsu VNA calibrated to 0.1dB accuracy and 0.01dB resolution and working with Motorola, until they decided to decommission 1st system from lack of sales. <S> for kicks 110GHz rated >$1k	There are a couple of styles of SMA edge-launch connector that are very flexible about board thickness.
How to calculate 4 MHz crystal trace width on PCB? I have no idea about how to draw crystal trace on pcb. I have stm32f0 chip and 4 MHz external crystal and 15 pF capacitor. Can you help me that how to get suitable trace for crystal trace? Should I use gnd trace around them? <Q> I have designed a lot of boards and 4MHz is a very low frequency and worrying about trace width is not very important for this scenario. <S> Even worrying about the ground layout, while generally important, is not very much of a deal at 4MHz. <S> In the 80s when everyone was only using 1 or 2 layer PCBs they would be huge boards with thick traces. <S> For example if you look inside a pinball game from the 1980s you'll find a CPU board which is almost 18" on a side and the components are all through hole and are placed inches apart from each other. <S> These board were using frequencies up to around 8MHz or so with 6800 series processors and the traces were all laid out by hand by drawing on acetate. <S> None of them even have ground planes. <S> Even now, boards like the Arduino Uno are 2 layer for cost reasons. <S> So yes to proper high frequency design, but don't worry too much. <S> (However for breadboard or 2 layer designs make sure you have a lot of decoupling capacitors to help reduce supply bounce.) <A> The important thing is to get the crystal close to the processor, and try to make both sides as symmetrical as possible. <S> The trace width does not matter very much. <S> A 5 mil (0.13 mm) or 8mil (0.20 mm) trace will be fine. <S> But a 10 mil (0.25 mm) or even 12 mil (0.3 mm) will be possible. <A> Google "crystal oscillator layout". <S> Some guides: <S> ST has prepared such document <S> AN2867 for your MCU <S> Microchip <S> AVR186 <S> (can be applied to other chips too) <A> Agree, ST's note about crystal is worthy reading. <S> And designing RF traces matched to particular impedance, say the 50, is quite long story. <S> needles to say one must take Dk constant which is specific to substrate, various FR4s have different Dk, substrate thickness and transmission line model (track with ground, surrounding ground - so on). <S> For crystals no one does this because there is no impedance match constraint and average crystal impedance at frequency of series resonance is about 10 <S> ..15 <S> +j0 ohms	The most important thing is to get the crystal as close as is possible to the microcontroller chip.
What is this tiny Iphone 4s component C7 What is this tiny square Iphone 4s component with a tiny label of C7 on it? I bought few damage iphone 4s trying to fix them, one of the phones' PCB is in a pretty good condition, but it's missing that c7 chip. I want to know what it does and is it possible to swap it from another pcb. Here is the photo of the pcb for that missing chip. <Q> That would be U10, RP106Z121D8. <S> Looks like a 1.28V regulator. <S> (hint: google the schematics) <A> <A> OK Since posting this I just saw the answer above and if the schematics are indeed available that must be the answer. <S> There are so many small square surface mount ICs. <S> But rather than just delete my comment <S> this is what I wrote without that info (probably wrong) <S> : <S> It's very hard to tell from the photo. <S> What we need to know is how many connections does it have in the land pattern. <S> It has five connections. <S> Four on the corners plus the central thermal pad in one version. <S> It looks more like an integrated circuit to me than a diode or something passive. <S> I would check out the texas instruments <S> SN74LVC1G17 Single Schmitt-Trigger Buffer. <S> Annoyingly they seem to have a number of versions of the same chip which are all about 1mm square, with different connection styles, e.g. ball grid, or whatever. <S> http://www.ti.com/lit/ds/symlink/sn74lvc1g17.pdf <S> Update: <S> Yeah it could be a voltage regulator absolutely - with the same package. <S> Probably you'd have to buy one from Mouser/Digikey to see if it looks the same.	Having looked it up myself I can agree that the first poster is correct about it being a voltage regulator The amount of gunk around it makes it hard to tell what the copper pad pattern looks like. I would carefully look at the datasheet and maybe try to expose the copper land pattern of the board and see which it might be. Some web searching for C7 makes me think it might be a SN74LVC1G17 which has a square package. It makes sense electrically that it would be near a connector because typically a buffer like that is to receive an input from off board which might be noisy.
Identify this [Molex Minifit Jr] low voltage electrical connector for lighting On the other end of the wires from this connector is a MR16/GU5.3 Socket with Retaining Clips for a bulb. My clips need to be replaced so I am looking to replace the entire part with both connectors. But, I cannot identify this clip plug. The male components are exactly 9mm long. <Q> It looks like a Molex Mini-Fit Jr Connector. <S> You can compare the dimensions from the linked drawing and/or the picture below. <A> But since you are replacing the 12V lamp and connector, it might be easier to replace both connectors (Plug and socket) using standard connectors that are easier to locate. <S> e.g. 4 pin Molex HDD plugs used in PC's are easy to shop for in pairs. <S> This part is usually specified by the shell , the crimp pins and the crimping tool and there is not indication you are aware of this. <S> Unless you need 10k pcs, . <S> good luck finding stock. <A> That is a Molex MiniFit-Jr 5566 series plug. <S> It is the mate for my header; the one you see on the PCB view. <S> Also attached is another screen-shot with data for your Molex plug.	It is something like the Molex MLX™ Connectors. Looks like the one I am currently working on.
If single phase supply can not create rotating MMF, how does Induction motor generate torque? I do understand that in a Three phase Motor, a rotating magnetic field is produced in stator and the induced magnetic field in rotor tries to aligns itself with the stator field which makes the rotor rotate. But in case of a single phase supply, there is only a pulsating stator field. So to start the motor, an auxiliary winding is used which has a field that lags the main field by 90 degrees. After the motor starts, the auxiliary winding is disconnected and now we only have the pulsating field in stator which can not generate any torque. Then how does the motor continues to produce torque and rotates? <Q> It's a bit like being on a swing. <S> Once you get it going you only need to "push it" at the right time each cycle to keep it swinging. <S> The same goes for a single phase motor. <S> Once the starter coil gets it going, the "pushes" from the AC power <S> keep it circling in phase with the power supplied with a lag appropriate to the torque taken by the motor. <S> Too much torque <S> and it will cause the lag to reach a point where the push is at the wrong point in the cycle and will stall the motor if it does not include a restart mechanism. <A> The stator magnetic flux can be visualized and mathematically described as two flux waves rotating in opposite directions. <S> Because of the shape of the torque vs. speed curves and the effects of the rotor magnetic flux, once a direction of motion has been established, the forward torque is higher than the reverse torque as shown below. <S> The solid line is the sum of the forward and reverse torques. <S> The above illustration is taken from Fitzgerald, Kingsley Umans, Electric Machinery <S> 4th ed and the paragraph above is a summary of a 1-1/2 page explanation in that text. <S> Also in that text, reference is made to another text for "an extensive treatment of fractional-horsepower motors." <S> Forward vs. Reverse Torque <S> The torque produced by the forward-rotating magnetic fields is higher than for the reverse because of the shape of the curves. <S> The torque vs. speed characteristics can be determined by analyzing the equivalent circuit of the motor. <S> That circuit for a single phase motor that has been started is shown below. <S> A more complex version of the circuit may be required for an accurate determination of the characteristics. <S> Fitzgerald, Kingsley Umans, <S> Electric Machinery 4th ed <S> Split-phase motors have a phase displacement between the currents main and auxiliary windings. <S> That creates an approximation of a two-phase motor. <S> An analysis of a split-phase motor still makes use of the theory of two revolving fields. <S> Another concept that is used is the symmetrical-component concept. <S> Each of the two concepts has advantages and disadvantages. <A> Also, despite what is often depicted in textbooks, if you look at the rotor cage of an induction motor you will see that the orientation of the rotor bars is skewed / slanted. <S> This is so that the magnetic fields around the rotor bars are in constant contact with fields in the stator and the torque pulsations are minimized. <S> In larger 3 phase motors that is generally not necessary because the mass of the rotor carries it through will little pulsation, so the the rotor bars are often straight. <A> After the motor starts, the auxiliary winding is disconnected and now we only have the pulsating field in stator which can not generate any torque. <S> Then how does the motor continues to produce torque and rotates? <S> Figure 1. <S> A single-piston steam engine. <S> Source: Kiddle . <S> This problem is not unique to electric motors. <S> Steam engines had this problem and some locomotives would, if they started the wrong way, have to reversed quickly by the drive to run in the intended direction. <S> (Controls were probably marked "One way" and "The other way" rather than forward and reverse. <S> Bicycles are also, in effect, single phase with a weak forward / reverse thrust available from the rider's legs. <S> Most of the force available is vertical. <S> The trick in both of the above and in the case of the induction motor is to get the cycle started somehow, let the momentum carry you over top-dead-centre and then push hard. <A> Once the motor begins rotating (with the aid of the capacitor and extra winding), it's purely inertia that keeps it rotating under mechanical loads and if the load is too big it will stall the motor and it will remain stalled even if the mechanical loading is removed. <S> If the "starting switch" is automatically reactivated upon speed being below a certain threshold, it will re-start. <S> It's like a bridge rectifier circuit and reservoir capacitor - <S> the capacitor only gets charged for a short period each cycle and in between charging events, it's effectively free-wheeling. <S> The load reduces the capacitor voltage until along comes another charging event. <S> This isn't a problem and it doesn't have to be a problem for the motor under light (ish) mechanical loads.	Split Phase Starting and Running A single-phase motor has two stator windings with different resistances and inductances or with a series capacitor in series with one of them.
What type of oscillator circuits are used in Capactive proximity sensors? Capacitive proximity sensors rely on a change of capacitance to alter the resonant frequency of the oscillator circuit. However I'm unsure as to what type is generally used in industrial type sensors. The types of oscillator I'm most familiar with are LC type such as Colpitts and RC type such as Wein Bridge. Which of these (if at all) are used more commonly in industrial type capacitive proximity sensors? <Q> There are plenty of capacitance probes <S> but I feel compelled to explain the type that I have been involved with. <S> The type of oscillator was based around a common-collector colpitts circuit and used an operating frequency of around 10 MHz. <S> Using this frequency you get a decent signal when there is a high speed situation to be detected such as on the blades of a turbine passing at full speed. <S> The 3 dB bandwidth was up to 70 kHz from memory. <S> The down side is that at 10 MHz, transmission line effects came into play so this was utilized to good effect; a changing capacitance at the end of the probe became a changing inductance at the oscillator terminals and frequency was modulated as metal passed-by the probe end. <S> More often than not an LC oscillator is used (in my types of application) <S> but I can't rule out that an RC oscillator design is used in some other applications. <A> There are many ways to do this. <S> Here I just demonstrate using a diode to change capacitance using an RC Relaxation Oscillator. <A> As there are plenty of RC suggestions I thought I would summarise 2 key difference between using LC and RC oscillators. <S> Voltage and Temperature LC oscillators are primarily dependent of L and C and not the active components or supply voltage. <S> RC oscillators are dependent on voltage thresholds to set the frequency. <S> These are temperature dependent, and often supply voltage dependent. <S> This was a big issue 30 years ago, not so much now if you use stable voltage regulators and precision op-amps. <S> This means that an LC arrangement can resolve much smaller changes in L or C (per unit measurement time) <S> When using an external C, the LC has two further advantages: the tank circuit rejects out of band noise picked up by the open C sensing plate, since it is resonant <S> the frequency is not sensitive to loss R in the sensing C. <S> If there is significant resistive loss this is part of the frequency of many RC oscillators <S> The LC oscillator was a simple transistor circuit (ideal for gooping in an M12 tube) at a time when a decent RC oscillator was complex with DIP IC's	LC oscillators have low phase noise compared to RC oscillators, and much greater supply noise immunity.
Cheap options for high precision distance measurement? I built a rather cheap(~€100) 3 axis CNC router for my workshop for cutting wood and aluminium in all sorts of fancy ways. It works great for most stuff but for things that require high precision, its always a little off. For example, if I make it do 2 identical cuts, one right after another, the difference in any axis can be up to 1mm. Not too bad for wood, considering the total work area is 800x800x400mm and the size of things I make but kind of a problem for aluminium pieces that have to fit together. It can be worse if significant amount of time/work is between the 2 cuts. I've had identical cuts made a few days apart for replacement parts differ by up to ~2.5mm on large pieces. Now I figured I could get better rails, better bearings, engines, whatever to try and make it more consistent but I think the biggest gains could be had by having the ability to calibrate and check it in software. TLDR: So what do people use for measuring distances of up to 1000mm with lets say 0.1mm precision? I looked around chinese estores but all I found was laser sensors for large distances, like 100m with precision +- 0.2m and ultrasonic proximity sensors for shorter distances but pretty terrible precision. They were pretty cheap though (<€10) which gives me hope. I also have the advantage of having complete physical control over both the points that I want to measure the distance between instead of just one. <Q> I think you should get to the bottom of what is causing your errors. <S> Either you are missing steps or your mechanical setup is too flexible or has backlash in the nuts/bearings. <S> Your homing switches may also have poor repeatability if you are not using an edge finder to locate the tool precisely. <S> Or perhaps it's a combination. <S> Things like backlash and flexibility are very difficult to compensate for with software. <S> For example, depending on the direction of cut you may start off with the tool in the right position but as soon as it bites in kerchunk <S> and you've dug into the work as the cutter pulls itself in. <S> Or you may be doing climb milling and the cutter runs well outside of the desired path if the gantry and slides are too flexible. <S> Accuracy over the full scale might be 10-15um over 1m for a cheap one. <S> They typically have a quadrature 5V digital output (incremental), some may have quadrature sinusoidal signals. <S> But each axis will probably cost about as much as you've put into this so far, and there is no guarantee you will be able to do much better in part accuracy. <S> Photo from this page If you win the lottery you can consider Renishaw and Heidenhain encoders, which can reach resolutions orders of magnitude less than a wavelength of light and do it with absolute measurement. <A> Figure 1. <S> A draw-wire sensor. <S> Source: Environmental Engineering . <S> I couldn't find an image of one with a built-in display but someone must be making them. <A> How about Magnetic Linear Encoder Tape ? <S> You can often find it and the associated sensors on eBay. <S> The tape is encoded with opposite pole magnets at a fixed spacing between poles. <S> A separate magnetic head senses the pole positions and interpolates between them for higher precision. <S> A standard pole spacing on the tape is 1mm, and interpolations of 10, 25, and 50 steps between the poles are available. <S> The control electronics measure the movement of the sensor relative to the magnetic tape. <A> I hate to be the bearer of bad news, but you have stumbled into one of the major problems in machining: repeatability. <S> It is not easy to fix in a cost effective manner. <S> What is typically used is optical encoders with diffraction gratings. <S> A linear scale is attached to one part of the machine and the encoder read head is attached to the other. <S> A laser is typically used to make microscopic incisions, called marks, into the scale. <S> As the read head moves between two scale marks it outputs a sine wave due to diffraction. <S> The markings are typically around \$20\mathrm{\mu m}\$ apart. <S> This is what they look like: <S> Furthermore, you will have thermal expansion problems. <S> A one meter aluminum beam will expand by \$23\mathrm{\mu m}\$ per degree Kelvin / Celsius. <S> If the parts of your machine are securely fixed together, the beams will buckle, due to different parts of the machine having different lengths and the expansion being proportional to the length. <S> This is why you use holes that are larger than your bolts with rubber washer so they can move, as shown in US patent 6,058,618: <S> I’ve tried to be brief and pointed only the major issues. <S> I have avoided things like bearings, calibration, tool wear etc. <S> Hopefully this can point you in the right direction. <S> All in all, to solve these problems you need to spend a lot of money. <S> The encoders alone are more expensive than your CNC machine. <A> If you look for cheap digital readouts (DRO) you'll find ones that have data ports. <S> They will not only provide the data over the port, but on an LCD with options for zeroing the axis: https://www.amazon.com/gp/product/B01G5SUZEG/ <S> They can be used as sensors once attached to your machine, and if you do read them you'll get closed loop feedback for your control system which should resolve your repeatability errors with a resolution much greater than your target 0.1mm. <A> Distances this large are generally measured (for calibration) using an indicator against a reference bar. <S> You can get dial indicators accurate to 0.01mm at a reasonable price, and you should be able to get someone to machine a steel or SS bar or rod to 1000mm pretty inexpensively (or you can pay 2K+ for a "real" metrology reference bar). <S> Note that the bar will only be 1000mm at a specific temperature. <S> (attach the dial indicator to the spindle) Google "metrology" and you will find a wealth of material on techniques and products. <S> You could also add inductive sensors to the motion rails at a specific distance apart and verify the distance as part of the calibration process each time you mill something.	Anyway, glass scales are a sort of mid-priced way to measure a few microns down to 1um resolution. Have a look at draw-wire sensors or "yo-yo encoder".
Help me identify a component on a PCB What's the white component? The one besides the green resistor. It has 5 pins, the pictures are below. Here is the circuit. <Q> Due to small slit on top for a screw driver and wires on the sides, I would guess a variable inductor. <S> (Like a potentiometer, but for inductance instead of resistance). <S> Example from coilcraft: Example datasheet (see page 4) Extracted from document: <A> It could be a trimmable transformer. <S> Such things were used often in radios to tune all the circuits to the IF frequency, or whatever. <S> This was a case where the result of 5% components wasn't good enough, before the age of digital trimming. <S> The coil or transformer had a ferrite slug in the middle that was threaded. <S> Turning the slug would move it up and down relative to the coils. <S> That would change the inductance of individual windings, and the coupling between windings. <S> Tranformers that size are only for RF signals. <S> It is too small to handle significant power or to work at low frequencies. <A> The green leaded component is a fixed value inductor (coil / choke) as that is also the schematic symbol for a inductor / coil on the board under neither of it. <S> the opposing side is center tapped primary / secondary.	The white body device is a variable inductor containing a powdered ferrite type tuning slug for minor adjustment of the inductance value and this appears to be a transformer configuration with a fixed primary / secondary and
Ohm's law doesn't seem to be working for this electric motor I'm a beginner in this field so please forgive me if I'm confusing with my question. There is a component that I can't understand with Ohm's law which is a washing machine drain pump. Washing machine drain pumps from most manufacturers have similar specifications. Their winding resistance is usually between 10-20 Ω and it operates under 120 VAC. However the specifications written on the label are quite different.120 VAC, 1.1 A, and 80 W. The actual current draw, 0.9 A, is close to the specification value which is 1.1 A. I really don't understand that according to Ohm's law the resistance value calculated per the specification should be (R = U/I) 133.33 Ω where U is 120 V and I is 1.1 A. But why is the winding giving me 14.8 Ω? Shouldn't it draw 8.11 A as I = U / R = 120 V / 14.8 Ω = 8.11 A? <Q> Have you ever played around with an electric motor connected to something like a light bulb or another motor? <S> If you spin the motor, the motor acts like a generator and spins the other motor or lights the light bulb. <S> The same thing happens when the motor is spinning under electrical power, the motor will behave like a generator, looking something like this: simulate this circuit – Schematic created using CircuitLab Notice how although you see 12V across the motor, the motor resistance only sees 1V, making the current through the motor 100mA instead of 1.2A. <S> This phenomenon is called Back-EMF, and is the reason why motors will draw a huge current on startup, but not much when running normally (when you turn on your vacuum the lights dim for an instant). <A> You're missing the reactance , which is the AC resistance (EDIT: And back-EMF - see comments). <S> When you measure the resistance with a meter you're only measuring DC resistance <S> and you're missing a significant part of the system. <S> Reactance comes from either capacitance, inductance, or a combination of the two. <S> In the case of a motor most of the reactance will be inductive due to the inductor-like nature of the windings. <S> When using Ohm's Law in AC systems you use impedance instead of just resistance. <S> Impedance, usually denoted Z , is a combination of the DC resistance and the AC reactance. <A> Aside from the excellent answers on the differences with AC motors, the thing you need to understand is that what they were wanting from having you check the DC resistance would be to see if it was too LOW, which would indicate it was shorted out, or WAY TOO HIGH, as in an Open Circuit because of a broken conductor. <S> Anything in between just meant it was NOT one of those obvious forms of failure. <A> The DC resistance of the winding conforms to Ohm's law perfectly, and if you actually and directly (without eg a commutator) fed that winding 120V DC, it would perfectly dissipate 80 watts of heat and perfectly go up in smoke, perfectly in accordance with Ohm's law. <S> The actual power draw is dominated by inductance - any power dissipated in the DC winding resistance is actually LOST, all it does is heat up the motor (there is a magnetic field built, but you would get just the same field from a lower voltage if the winding resistance was lower). <S> The inductance of the windings alters with motor load (energy conservation law has something do with it) - an idling motor (if the motor design is safe to idle - some are not!) <S> might actually draw even LESS current than the nameplate says, while a heavily overloaded motor (say if you pumped molasses with that pump) will get closer to the above scenario - very little inductance will be in effect, and DC losses will dominate and eventually overheat the motor. <A> There are two effects at work here: one is that the DC resistance may be \$15\Omega\$, but that does not tell you the impedance of the coil. <S> At 1.1A, 120V, and 80W, the \$\cos\phi\$ is 0.6 which does not actually correspond to \$15\Omega\$, so what gives? <S> If the motor is allowed to turn (and to pump water), additional countervoltage is induced significantly lowering the amount of current that is actually flowing. <S> So you have <S> both the difference between DC and AC impedances here, and the difference between blocked and rotating (though loaded) motor. <A> Ohm's law is not a fundamental law of nature . <S> It's just a law that some very specific components observe; we call those resistors . <S> Now, it so happens that quite a bunch of components that aren't specifically designed as resistors still behave as if they were resistors – but only under specific circumstances . <S> In particular, simple homogeneous metal parts obey a local Ohm's law. <S> That includes also the wire with which the coils of an electric motor are wound, which is the reason you can some sort of reading when using an Ohmmeter with the motor. <S> Nevertheless, the motor as a whole does not actually obey Ohm's law, because the wire is electromagnetically coupled to other stuff: in operation, there's a constantly changing magnetic field inside the motor, and such a field induces voltages in the coils. <S> It is these voltages that dominate the electrical behaviour of the motor in any real use situation, not the voltage from Ohmic resistance. <S> Only if you let a small DC current flow through the coils, nothing actually moves in the motor, the magnetic field is everywhere constant, and since induction only depends on the time-variation of the magnetic field , you then get a much voltage reading that corresponds to the ohmic resistance of the wire alone. <S> That's why your Ohmmeter shows such a small value. <A> The manufacturer is stating the coil resistance so that you as the technician can determine the "health" of the motor winding/s. <S> Each winding should be the same as the others (if 3 phase) and the same as the manufacturers specification. <S> This as well as an insulation resistance test between each phase and earth and between phases should form part of any motor inspection regime to determine the servicability of the motor windings.	It turns out that the DC resistance as measured corresponds to a blocked motor.
When should I use USART instead of UART? I want to make a communication between my PIC18F4550 and my PC but I'm getting stuck whether I should use UART or USART for long distance. When it's more advantageous to use the one instead of the other one? <Q> UART stands for universal asynchronous receiver/transmitter . <S> The additional S in USART stands for synchronous . <S> It's just a little added capability Microchip gave the module to make it more useful in some cases. <S> That extra capability doesn't apply in your case. <S> The PC COM port only takes a UART to talk to. <S> The fact that the USART could have been used in a different way is irrelevant to you, except that maybe you have some additional configuration bits that have to be set the right way. <A> Short Answer: <S> Use whichever you chip has To fully examine the question, it's necessary to decode the acronyms: A U niversal <S> A synchronous R eceiver <S> T ransmitter is your traditional "serial port". <S> It is asynchronous in the sense that only a single signal is involved - no clock is transmitted, and instead the receiver must recover a clock, typically by oversampling. <S> In contrast a U niversal S ynchronous <S> A synchronous R eceiver T ransmitter is a more versatile device with UART-style asynchronous modes, but which can also optionally be configured to operate in synchronous modes where a clock is sent along with the data. <S> Depending on capability, this might include interoperation with well-known synchronous serial formats, for example <S> SPI or I2S. A few MCUs may offer both types of peripherals. <S> For a basic asynchronous serial need you could choose either. <S> However your choice might be influenced by the pins on which a given peripheral can operate, other needs in the system, etc. <S> A factory ROM bootloader might only operate on some peripherals and not others. <S> There may also be differences in buffer support, word lengths, parity support, associated control signals, etc. <S> And the software interface may be completely different between the two. <A> They are basically the same thing for your microcontroller. <S> USART stands for universal asynchronous and synchronous receiver/transmitter. <S> UART stands for universal asynchronous receiver/transmitter. <S> Asynchronous data transmission would most of the time be used in this communication protocol. <S> The Synchronous data transmission is rarely used because you have much better synchronous communication protocol such as SPI & I2C. <A> Your microcontroller (MCU) has a Universal Synchronous/Asynchronous Receiver/Transmitter (USART) in it. <S> This functional unit supports a synchronous communication mode and an asynchronous communication mode. <S> In synchronous mode, transmitter Tx is connected to receiver Rx by a CLOCK wire and a DATA wire. <S> Once per CLOCK period, Tx sends another bit on DATA and Rx takes another bit from DATA. <S> The transfer timing is governed by CLOCK and therefore known to both Tx and Rx. <S> So both Tx and Rx can use higher bit rates than in asynchronous mode. <S> However, two wires/connector pins are needed instead of one and excessive skew or jitter between CLOCK and DATA leads to corrupted data being received. <S> The timing within line drivers and line receivers carrying CLOCK and DATA must be closely matched to reduce this skew. <S> Both the interface and the two-wire factors become more prominent over longer distances, with long cables and/or multiple connection hops. <S> In asynchronous mode, transmitter Tx is connected to receiver Rx by a DATA wire. <S> Once per timed bit period, Tx sends another bit on DATA and Rx takes another bit from DATA. <S> The transfer timing is governed by the frequency separate oscillators in Tx and Rx, each of which is unknown to the other and will be slightly different. <S> Therefore the maximum reliable bit rate is lower than for synchronous mode. <S> At the start of every new byte, Rx uses the starting STOP-START bit transition to resynchronise to the incoming bit sequence timing. <S> This makes the delays from line drivers/receivers, cables and connectors irrelevant to the bit period timing, though not to the bit quality. <S> Only one wire is needed per communication signal instead of two, reducing the cost in cables, connectors and line drivers/receivers. <S> So it depends upon your acceptable costs, the distances you are travelling and the capabilities of Tx and Rx. <S> Your PC is likely to only support asynchronous mode on a standard COM port, requiring a special port (PCIe or USB, likely USB) for synchronous mode.	For your purposes, the UART and USART are the same thing.
Is a fuse the right way to measure current I saw my friend in a workshop use fuses to measure current of a generator and he said that a fuse is more accurate than connecting a light bulb because the light bulb works in current range. My question is : is he right? (that is more accurate to use a fuse instead of a light bulb to measure generator current? ) does fuse consider a load? Is it a proper way to measure the current of generator? is Kill-A-watt meter ( like this device: https://www.amazon.com/gp/product/B00009MDBU/ref=ox_sc_act_title_1?smid=ATVPDKIKX0DER&psc=1 ) accurate way to measure current of generator ? <Q> Both methods are not recommended and will not be an accurate measurement. <S> In order to measure current you should use an ammeter in electric circuits. <S> Fuse and light load will not be considered as a good method. <A> I wouldn't call either the bulb nor a fuse a measurement method. <S> The bulb may give a subjective idea of how much current based on how bright. <S> The fuse will blow, indicating whether the current has reached the fuse's blow conditions (like how fast it'll blow, tolerance or it could just end up being a bad fuse and blow early). <S> The indicated meter looks like it will measure your current, among other things. <S> My suggestion is that when deciding on a measuring tool, you consider the expected usage and then compare that to the specifications of the tool. <A> There two quality of fuses available, the good quality one like Bussmann blow up quick & the cheapies <S> melt and blow with a delay(kinda bad in some sensitive circuits). <S> The major issue is that fuses don't blow exactly at their ratings. <S> A fuse rated for 5A usually never blows at 5A in most cases. <S> It usually takes 1-2A more to blow the fuse.	A fuse isn't a good way to measure current. The best method would be to use a clamp ammeter or using a series ammeter of the desired range depending on the capacity of the source.
Why do high current wires have fabric/nyon sheathes? I saw two wires going into a halogen light bulb whose outer sheathes were not a plastic (like PVC, or polyurethane), but rather a woven fabric (like a nylon shoelace). Why is woven fabric better for this high current application? Sure, PVC may not have as high of a temperature rating, but why specifically does the material transition from solid to woven construction? <Q> I haven't made a study of the matter, but a reasonable hypothesis is that the insulation needs to withstand high temperature, which limits the choice of materials, and the particular material chosen would not be sufficiently flexible if it were made in a solid tube rather than a weave. <S> For example, a quick web search suggests that materials used for high-temperature insulation may be "mineral textile fibre or fibreglass". <S> If fiberglass insulation was not made of strands, it would just be glass, and thus very rigid. <S> Of course, solid glass and ceramic are also used in high-temperature applications, but you think of them as custom-made fittings rather than strange insulation on ordinary wire. <A> The woven fabric is typically not plastic but glass based, for the temperature performance. <S> Solid glass tends to be difficult to bend, woven fibre is more flexible. <A> Typical fibers are glass, but ceramic fibers are also a possibility.	The reason that the fabric is woven is that it is made from what would otherwise be brittle in solid form.
Can I supply a device rated for 1A with 15A power supply? I'm talking about a big difference between output current of the power supply and rated current of the device here. Would it burn up? <Q> Think about it. <S> You have an electrical supply to your house. <S> How many amps can it supply. <S> My main fuse is rated at 63 A. <S> You have a light bulb. <S> It's rated at 20 W. How many amps will it draw. <S> Answer: \$ <S> I = \frac { <S> P}{V} \$ <S> so that's <S> about 0.1 A in 230 V land or 0.2 A in 110 V land. <S> I'm talking about a big difference between output current of the power supply and rated current of the device here. <S> Me too. <S> Would it burn up? <S> No. <S> You may want to consider limiting the current to a safe value in the event of a fault. <S> Add a 2 <S> A fuse in-line <S> and you should be fine. <A> The power supply provides up to 15 amps. <S> This does not mean that it has an output of 15 amps, always. <S> The power supply will give the current requested by the device up to a maximum of 15 A. <A> Short answer: <S> yes. <S> The 15 amp rating shows the maximum current which it will supply. <S> It will also supply lower amounts of current just fine. <S> Think of it as a speed limit (although you shouldn't take the analogy too far). <S> If your car can do 100 mph, that does not mean that it will always go that fast - only that it will if you need to. <S> Likewise, a 15 amp rating only says that it will supply that amount of current if it needs to.	As in the case of the light bulb, the load will draw the current it requires and that's all.
How to convert 9 bit binary to BCD for 3 Seven Segment Displays I have been learning logic gates and binary so I made an 8-bit addition and subtraction calculator. When I add or subtract, the output is 9 LED's in regular binary. I have researched ways to convert Binary into BCD to be displayed to a seven segment display using a 7447 IC. I have checked other posts on Stack exchange which I found one but it didn't really answer my question. So the question is: Is there a simple way or Ic chip that I can use to convert 9 bits into BCD without worrying about the negative numbers? I do not want to use a microcontroller so I can do everything without programming. I have also heard something about using ROM which I don't know how to use or code with Assembly but I am still interested in that. I am not experienced at all with those sort of things. Any suggestions would be very most welcome. <Q> Of course, you'd need one with (nominally) 20 bits of IO - 9 in and 11 out. <S> (hint - what is the maximum value of 9 bits binary). <S> You can get away with fewer IO lines if you add external circuitry: <S> multiplexers on the input and latches on the output. <S> The old-school way is to use dedicated ICs, specifically the 74185 . <S> Of course, you'll need 6 of them, they're no longer made (although you can find them on eBay) and they need a lot of power, but it's doable. <S> The intermediate technique is to use a couple of PROMs. <S> A couple of 2716s will do you, but you'll have to do your own programming. <S> They're cheap enough although they take up some board space. <S> This is the approach I'd use, but you can make your own choice. <A> I would say just do the maths in BCD in the first place, but anyway... <S> Back before micros were the solution to everything, I had two similar problems. <S> One used two separate counters one binary (tuned the radio synthesiser) and one bcd (displayed the frequency). <S> They counted up/down together, so no conversion was ever required. <S> The other had a bcd counter/adder/alu with a display, that had to be punched to paper tape as binary. <S> The conversion was done destructively, by clocking the bcd down and binary counter up, until the bcd counter underflowed (up/down counters). <S> This only required a couple of cmos IC's. <S> [why? <S> I think binary = less paper tape = <S> fewer after hours / weekend trips out to change the tape] BTW this technique remains a compact way to do binary->bcd on some micros, if a little slow. <A> A slight variation on the ROM method will almost halve the hardware. <S> Since it is for display, you can multiplex the display without latches and let the user's eye provide the storage. <S> For example, use an x8 Parallel EEPROM with at least 10 address inputs and feed a clock to the 10th. <S> Connect the 8 data outputs to two 7447s and to three displays. <S> The anodes of the least significant pair get driven by the clock and the anode of the MS digit by the inverted clock (add a bit of dead time if ghosting is an issue). <S> Write a little program in your favorite programming language to spit out a binary or Intel hex file for your EEPROM programmer. <S> You could also multiplex the outputs with a 2 bit counter and 1 of 4 decoder, replace the 7447 with transistors and convert 11 bits with a single 8Kx8, or retain the two 7447s and biplexing and convert 12 bits to 4 digits (0..4095). <S> You will notice that the parts for most of these schemes are obsolete or soon to be.. <S> nobody does wasteful things like this in any quantity, even for very old existing designs, so the market has long since dried up. <S> A micro and/or an FPGA is generally the answer.	The modern technique would be to use a microcontroller to perform the conversion.
What would be the effect of adding small (~<10mm) lengths of conductor between a CPU's contacts and socket [I am interested specifically in the case of state of-the-art consumer processors and motherboards] To clarify what the setup might be in practice, imagine wires running vertically between each pair of contacts on the CPU and socket. I don't have much expertise in this area, but these are the potential issues I am concerned about: The additional length of the conduction path from the processor slows down the speed of processing. The change in length of the conduction path means that the processor won't work at all as the devices are designed to operate with a specific length of conduction path. There is interference of some sort between the conductors running from the CPU to the socket. In each case I am wondering if this is a problem and, if possible, what sort of effects would be observed. <Q> Your CPU has several types of pins/signals: <S> Slow signals like SMBUS, etc. <S> These wouldn't care. <S> Fast signals like PCI-Express, RAM, etc. <S> These all use transmission lines, single ended or differential. <S> This can be done (and it is done ) but it will involve tiny high precision machined parts and obscene prices because this is a very low-volume, specialty custom product. <S> Not to mention how crosstalk between nearby signals would be handled. <S> Perhaps tiny coax pogo pins? <S> Add an extra zero to the price. <S> Power/Ground <S> These require extremely low inductance which is achieved by having literally tons of pins. <S> Here's a random PC CPU pinout from the internets: <S> Enlarge it and look at all the supply and ground pins (anything that starts with "V"). <S> More than half the pins are supply and ground, and this adds to the cost. <S> Adding inductance (ie, length) would degrade power integrity. <S> These are the reasons why motherboard CPU sockets sit as low as possible on the board and keep the connections as short as possible... <S> So i would strongly recommend you find a cooling solution that doesn't require this. <S> It is a lot easier to move heat up by a few mm (a 10 cents copper shim will do the job). <A> This would dramatically increase the inductance associated with each connection from the PWB to the chip. <S> It wouldn't affect (much) <S> the ability of the processor to continue to do computations at high rates. <S> But all interfaces to the outside world <S> (for example to memory and any peripheral bus like PCI) would have to be slowed down considerably to be made to work. <S> For example instead of accessing RAM at 200 million transactions per second you might have to slow it down to 10 million transactions per second or slower. <S> And of course the existing memory chips aren't designed to work with those slow access rates. <A> If you can slow the output edges (output drivers) by 10:1, to perhaps 10nanoseconds, and you solder 0.1uF <S> capacitors across all VDD/RTN pairs on this modern MCU, you might get away with this.	Any impedance discontinuity in the transmission line would create signal reflections and corrupt signal integrity, so you would need to make a riser connector with controlled impedance. It would also increase cross-talk between data signals on those connections.
How do you solder something big? I have got an iron nut with a diameter of about 5 cm. I would like to solder a pin to it so I can connect a wire to it. (I want to make a capacitive sensor.) I tried it already, but it won't attach to the nut at all. How can I do that? <Q> You probably have two issues: getting enough heat, and surface compatibility. <S> Iron (more likely steel, likely plated) may not solder easily, but it can be done with care to get the surfaces extremely clean and flux. <S> Heat wise, you might use a soldering gun, heat gun, hot plate, (non-food!) <S> oven, or even plumber's propane torch to pre-heat it. <S> Be careful however not to get it too hot - zinc based plating is a respiratory hazard if it boils off, solder will not behave properly when a joint is far above the appropriate temperature, and excessive heat will just make the metals oxidize faster, which is your real obstacle to making a good connection.. <S> A far better method of connection might be to cross-drill the large nut and tap it for a small machine screw which can affix your wire directly or better yet with a crimped terminal ring. <S> Using thread cutting taps is a skill, but one worth learning. <S> And you probably have six faces on the nut to work with if you accidentally break a tap in the first hole you try... <S> For anything used outdoors or in a tough environment you might also need to worry about corrosion between dissimilar metals, but this sounds a bit more like a temporary hobby or science fair project. <S> Of course if this is a very temporary initial "does this even work" proof-of-concept, you might also just be able to wrap the wire through the nut a few times and twist tightly - but that may be unreliable, and for good reason looks a bit unprofessional. <A> You can try to drill a small hole such as 2.5 millimeters in it (I assume, it's big enough to do this, if it's 50mm wide) <S> deep about 5mm and tap a M3 thread. <S> It should be enough. <S> Then use a crimping tool to crimp a eye-hook terminal to the wire, use teethed washers and tighten it with a short M3 screw with flat or rounded head (make sure it's not countersink head). <S> Here's the list of pictures: <S> Here's how it should look on the end: <A> Other than brazing, arc welding or ultrasonic welding a threaded ring lug will be the easiest reliable method of attachment. <S> A tapped threaded screw with > 3x surface area more than a smooth surface will engage to achieve a low series resistance in the xx mOhm range. <S> But why use a "nut" to sense capacitance when sheet metal or aluminum foil or copper clad PCB may work better? -- other info <S> If the wire is long it will add inductance 6nH/mm appox for 1:10 diameter/length ratio, shielded or twisted pairs will add 100~30pF/m which then results in a resonant frequency. <S> But for short sensor to IC, 30 mOhms contact is practical for a small screw, wide head to interface to a ring lug and crimp or soldered wire. <A> <A> Preheat the nut to a temperature close to or at the temperature your solder needs to melt, then solder to it. <S> A blowtorch, a kitchen oven, a heat gun would all work for that. <S> Avoid oxidising/tarnishing the nut more than necessary - your flux will have to work harder. <S> Optionally, tin it (if heated above that melting point), but be careful not to get solder or flux in the threads. <S> Tinning only the spot where you want to attach the pin is another option. <S> Coating the intended attachment spot with flux before preheating might be a good idea with SOME fluxes (avoid the kitchen oven then, though!). <S> Avoid fluxes not meant for electrical connections. <S> Make sure your nut is not galvanized, not stainless steel, and not actually made out of pot metal - the first can create fumes that make you feel funny when heated, the second will be VERY hard to solder to due to the chromium content, the third will suddenly melt. <S> Be aware that an iron piece heated to 250°C+ will stay hot for a while, and can give you far more unpleasant burns even on short contact than a flame. <A> I had a similar problem bonding a ground wire to my boat trailer. <S> A spot of braze applied with an oxyacetylene welder provided a clean spot where good contact is achieved. <S> Break the rules! <S> Acid flux is ok but wash it off when the brazing is done. <S> No welding torch? <S> Put the nut on the kitchen stove and use a bernzomatic torch on the spot to be brazed. <S> Solder bonds nicely to the brazed spot. <S> If you eat organic vegan gluten-free non-GMO, you will want to wear a hazmat moon suit with a crash helmet and safety glasses when doing this. <S> Good luck with your nuts!	May I suggest using flux, scoring the surface with fine grit sand-paper (or nail file) and using a large pistol grip style soldering gun >100watt.
Using transistors to replace push buttons I'm looking for a little help designing a transistor circuit. I have a small LCD (I took from an after-market car backup cam system), and I’m trying to control its built-in menu using the buttons of an old handheld game system. The LCD has 3 buttons to navigate its menus which are on a separate PCB with a single 3.3V signal line. The way it works is each button shorts the line to ground through a different resistor. My idea was to just rebuild that circuit and replace the buttons with NPN transistors, then connect the base of each transistor to the output of a logic gate, which would detect when two of the game system’s buttons were pressed simultaneously. The problem I seem to be having is the transistors aren't being triggered, and I don't really know enough about them to know if I've designed this wrong (the logic gate seems to be outputting correctly, so I must have done something wrong with the transistors). The parts I’m using:Transistors:https://www.digikey.com/product-detail/en/on-semiconductor/2N3904BU/2N3904FS-ND/1413Logic Gate:https://www.digikey.com/product-detail/en/texas-instruments/SN74AHC02N/296-4514-5-ND/375964 Thank you very much for the help! simulate this circuit – Schematic created using CircuitLab <Q> Your circuit forces 5-0.6 = 4.4V onto the end of R, regardless of which switch is used. <S> To work you need a high value R in series with the base, and better to have emitter to gnd, and R to collector. <S> An N-fet would be easier, but still has to be S to gnd simulate this circuit – <S> Schematic created using CircuitLab <S> Bipolars will always be adding current/switching voltage, and making the apparent resistance wrong. <S> Cmos transmission gates are best for this job. <S> 74HC4066. <S> As an add on, you can connect them to the circuit directly in place of switches, or across existing switches, without re-wiring the resistors, and without regard to polarity, multiplexing arrangements etc. <S> You can also use them for logic functions e.g in series to get the AND function i.e. the 4th one is in series with LCD_CTRL and controlled by _Mode. <S> simulate this circuit <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> The LCD module appears to use a single analog input with pull-down resistors to create different voltage levels for each button or combination press. <S> simulate this circuit Figure 2. <S> The OP's NPN transistor is wired on the "high side" of the load resistor and acts as an emitter follower. <S> The emitter voltage will be about one diode voltage drop below the base voltage. <S> When the NOR gate output goes high the input to the LCD will be high and will not give the required response. <S> Because \$ V_b \$ <S> > <S> \$ V_c \$ <S> the collector voltage will also be at 4.3 V independent of the value of R1. <S> It will also overdrive the LCD analogue input which is probably 3.3 V max. <S> simulate this circuit Figure 3. <S> A simple diode connection arrangement. <S> Since the LCD module senses buttons by ratiometric voltage division from its internal pull-up and the external pull-down resistors we can generate any voltage in a similar fashion externally. <S> Note that the R2, 3 and 4 are in a (roughly) 1, 2, 5 ratio and so each button or combination can be differentiated by the ADC. <S> This arrangement uses diodes D1, 2, 3 & 4 to prevent feeding +5 to the LCD as this may damage it. <S> This creates a small problem as there will be a voltage drop across it. <S> Setup and calibration: Measure the voltage at CTRL+ for each button press and combination to be simulated: + <S> - MENU <S> V R90 <S> 0 <S> 0 <S> __ __0 <S> 0 <S> 1 __ __0 <S> 1 <S> 0 <S> __ __0 <S> 1 <S> 1 __ __1 <S> 0 0 <S> __ __1 0 <S> 1 <S> __ __1 <S> 1 <S> 0 __ __1 <S> 1 <S> 1 <S> __ __ Connect diode D1 and R9. <S> Press LEFT and adjust R9 to get the required voltage. <S> Power down and measure the resistance of R9 and record on the table. <S> When complete replace R9 and add R10, 11 and 12 with the required values. <A> The outputs of your logic gates already have output transistors to ground. <S> That's how a logic gate produces the low on the output. <S> Use a NOR gate with open drain output, e.g. a 74LS33. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If you can't find a NOR with open drain, you could try placing diodes between the output and the resistors, so current can only flow into the output (to GND).	The OP's hardware seems to consist of a game controller with four buttons, pull-up resistors and connected to four individual controller inputs.
When NOT to tent vias? Tenting vias is great for: If a via is placed over some silk screen (designator, instruction, warning, etc.) and you want the text to be more readable A compact design with high component density and potential shorts nearby More aesthetic look (solder mask across the board, rather than gold/silver dots everywhere) Stop solder being sucked away during reflow if placed under a component/near a pad On the contrary, though, I can't seem to find any reason not to tent vias. Assuming you have appropriate test points and the design works 100% (i.e. no need to probe around and check things), is there any reason not to tent every via on the board? Even if you do need to probe, you can still dig through the solder mask and make a connection if worse comes to worse anyway. Only thing I can think of is increased heat retention if it is a high-current via, but besides that I'm lost. Any ideas? <Q> I use those spots as probe points, and as points where I can solder some botch wires when needed. <A> In production, you will often want to test every product to verify it works. <S> Non-tented vias could double as test points for this. <S> In some extreme precision circuits this might not be acceptable, since current leakage in solder mask can be an issue (if you look at high-sensitivity measurement devices you often see that the low-current tracks might not have soldermask on them, or at least the guard traces around them isolate the soldermask). <A> I tent ALL vias unless I plan to use some of them for probing for the 1st proto pass. <S> Plus, with tented vias you can place them closer to the break-out SM pins without any worry of solder-mask slivers; short traces are also great for keeping the trace inductance low too for de-coupling caps, TVS's and the like. <S> In addition, the less exposed Cu. <S> the better in terms of solder-bridging or under a metallic element of a part. <S> I too have heard about the "volcano" effect but never seen it in real life. <S> There should be no cost adder (to tent vias) either.	For some crazy edgecases: Tenting vias means you have soldermask on your vias.
How can I create a resistor (rheostat) network adjustable from 10kohm to 1Mohm? This is for a part of a system that provides feedback to a voltage reference of 0.6 V . The output voltage has a range of 125 V to 500 V. Using a voltage divider, 414k and 2k will provide the 0.6 V when the output is 125 V. Similarly a 1664k and the 2k will provide the 0.6 V when the output is 500 V. In order to control the feedback, I plan on using a resistor network adjustable between the 414k and 1664k for the top resistor. I have split that resistor into a fixed 390k resistor and a 24k - 1274k variable resistor network. What approach should I take to solve this problem? <Q> First off it appears that your calculations are incorrect. <S> I also think you are using the wrong approach. <S> You will want to come up with an approach that puts your variable resistor in the low voltage side of the divider so that the pot does not get exposed to the high voltage. <S> So assume you have the typical voltage divider configuration: <S> For your two endpoints: Vin = 125VRa = 220KVout = <S> 0.6VRb = <S> 1061.09K Vin = <S> 500VRa = 220KVout = <S> 0.6VRb = <S> 264 <S> It will be a whole lot easier to use a readily available 1K pot with a series resistor of about 60 ohms. <S> The pot can easily adjust down to ~200 ohms to get the control range you need. <A> You really haven't provided enough information for a definitive answer. <S> By the way, your title and your actual requirements don't match: 10 kilohms to 1 megohm vs 414 kilohms to 1.664 megohms vs 24 kilohms to 1.274 megohms. <S> You haven't specified the resolution you need or if the resistance is to be manually adjustable or computer controlled. <S> For low resolution, a single turn potentiometer can be used. <S> For higher resolution, use a 10 turn pot. <A> Following the advice from Michael Karas, I looked into a low side controlled divider, using a MCP41XX series digital potentiometer. <S> The voltage divider is composed of a fixed resistor on the high side, and the digipot and a fixed resistor on the low side (with the fixed resistor between the digipot and ground). <S> Here are factors I need to consider when selecting the resistor values.-The Vout of the divider needed to be less than 5 V, per the digipot limit. <S> Still, I wanted to get a high enough voltage swing to improve resolution.-The low fixed resistor <S> is what prevents the output voltage from shooting sky high.-The bias current should be in the 500 uA range. <S> My final design was the following. <S> A 620 kohm high resistor, a 5 kohm digipot, and a 750 ohm low resistor. <S> With these values, the max voltage is fixed at 500 V per the 750 ohm resistor. <S> The min value is 65 V when the feedback is across the 5750 ohm low combination. <S> The bias current is between 104 uA and 800 uA <S> when 65 V and 500 V, respectively. <S> The maximum voltage on the digipot high side is 4.59 V, close enough to my desired range.	One simple approach is a ordinary variable resistor (potentiometer) in the 1 to 2 megohm range.
How to safely insulate the Power PCB of a display computer monitor? CONTEXT The other day I was checking the heat coming off an aluminium heatsink on the power PCB of an eizo EV line led display monitor. It was okay, not too hot, just warm and tad bit hot. When I touched this it was okay. The display for my custom project was on and still works perfectly. Then I touched the aluminium bezel of the monitor and that was okay too. BUT... when I touched with both handes BOTH the lcd panel's aluminium back (to check if the led backlighting was warm) with one hand AND the aluminium heatsink on the power board of the monitor with my other hand, I got a huuuuge electric continuous shock for a full second. Boy it hurt! I've had a few times in my life touched live wire accedentally while fixing lightbulbs etc, but this time it was long and Im lucky to still be here telling the tale. This seemed or felt more dangerous / higher voltage! How can that be I thought LED was cute and safe!! QUESTION How can I insulate the tiny aluminium heatsink on the power pcb of the monitor? I am planning on covering these PCBs with an aluminium cover as a ground encasing for protectingmyself and children, so first I must insulate this PCB and make sure no heatsinks touch the aluminium plate covering it. HOW can one insulate a Power PCB? Can I pot the power PCB and the monitor PCB completey submerge it with hot glue or with epoxy resin for example? Or could I just cover the pcb with a rubber layer? All suggestions will count as answers as not only I appreciate your input, it (your answer) might in some cases even save lives! So, thanks in advance! <Q> You see that thick white line on the power supply board? <S> Everything to the right of it is to be assumed to be at mains potential, that line marks the safety isolation barrier. <S> You cannot pot a HEATSINK! <S> Think about it, the whole point of the thing is to have surface for air to flow over to remove the heat. <S> I would note that each of those board has a perfectly good set of mounting holes, get some ally, some standoffs and make up a case (With vent holes too small to admit a finger), job done. <S> You will want to make sure the metal case is connected to the power supply earth connection ( <S> The green/yellow wire bottom right of the power board). <A> Before you continue messing with that for your personal safety get a GFCI to protect the outlet from which you are powering your test setup. <S> Plug-in inline devices exist. <S> That way if some current leaks to ground (potentially through you) it will kill the power. <S> To help the GFCI detect that you should connect the aluminum chassis to the ground wire (green and yellow green) coming from the power connection. <S> Everything on the right side of the white line of the bottom PCB should be treated as live at mains voltage. <S> with the only exception being the ground wire. <S> The insulation can be handled using airgap of 1/4 inch (5 mm) using standoffs or a plastic sheet lining the inside of the enclosure where you cannot have that gap. <S> For airflow you want slots in the encasing and matching ones in the plastic sheet. <S> You can skip the plastic sheet if the enclosure is plastic. <A> A metal or plastic box, with plenty of holes or slots, but with the individual openings too small to admit a finger. <S> Potting the PCBs, or anything else that would reduce airflow across the heatsinks, would be a very bad idea. <S> This is precisely why the monitor originally had a case!	Obviously, the electronics need to be in an enclosure that allows plenty of airflow while preventing physical access.
Powering an Arduino with a Supercapacitor For the purpose of a project I wish to power an arduino using a supercapacitor charged to 5V. The supercapacitor will be fed straight into the power Vin and GND terminals on the Arduino. No power plug or USB connected to a computer will be connected, so all power is to be drawn from the supercapacitor. Previously I have powered the board using a 0-12V variable voltage supply, and the arduino draws between 20 and 35 milliamps. My question is this, if I connect the supercapacitor straight to the arduino board, will the arduino govern how much current to draw from the supercapacitor? Or will I need to place a resistor in series with the supercapacitor discharging in order to regulate current and stop the supercapacitor rapidly discharging and frying the board? Lastly is this feasible at all? I am aware that as the 5V from the supercapacitor will drop off the period during which the arduino will run is limited. Any advice is appreciated as I do not wish to fry my Arduino board attempting this. <Q> The voltage on a capacitor is proportional to the charge stored in it. <S> That means that as the device draws current, the voltage will drop. <S> You have to decide how low is acceptable. <S> Do the math. <S> For example, let's say the device can still run from 4.5 V. <S> That means the capacitor voltage can drop 500 mV before the system doesn't work anymore. <S> Let's use a 1 F cap as example. <S> (500 mV)(1 F) / (35 mA <S> ) = 14.3 s <S> That's how long a 1 F cap charged to 5 V <S> can run your device at the worst case current draw. <A> First to answer your question (Although that has been done in the comments already): <S> No you do not need a resistor to limit the current. <S> The Arduino will take just what it needs. <S> Next:Your discharge over a small voltage range 5v down to 4.5V <S> will be ~CV/ <S> I. Olin gave a figure for that. <S> BUT! <S> You can get a lot more time using a smaller and cheaper capacitor if you would use a 5 Volt regulator and e.g. a 12V capacitor. <S> Let's say you want 10 seconds whilst the voltage drops from 5V to 4.5V. <S> You need a capacitor of 0.03510/0.5 = 0.7 Farad. <S> Some assumptions: We use 12V, have 0.5V drop over the regulator and it uses an extra 1ma. <S> The voltage can now drop 6.5 Volts (and we still have 5V on the Arduino).You now need a capacitor of <S> 0.03610/6.5 = 0.055 <S> Farad. <S> Yes, you need a higher voltage supercap but it can be much, much smaller. <A> This can run on an input of 0.9V - 5V. Based on your 15F capacitor and a voltage drop of 4V (for a 5.5V supercapacitor charged to 5V) <S> you could get 4000mV <S> * 15F / 41mA (85% efficient) = <S> 1,463s or about 24 minutes run time.	Another option could be to get a DC-DC step up converter which gives a 5V output.
(logic level) Gating the output of a 900hz (5v p2p) oscillator Seems a simple enough problem this, and perhaps all that is needed is confirmation of approach (although a better solution always appreciated). So the problem: I have a 555 based square wave based oscillator (clocked at roughly 900hz 5v p2p) which i need to gate the output to enable/inhibit to the Power Amp section of the design (audio use only). I can't use a specifically clocked PWM output pin on mcu (all in use for other needs). I don't really have room to push in an AND gate to the output (on the target PCB), so was thinking of a classic emitter follower (npn) with the 'collector' fed by the output of the 555, the 'base' driven by an mcu (arduino 5v) output logic pin, and the output to the power amp taken across the emitter and the emitter resistor. In my mind, on low (from the 555) fed to the collector, the transistor will be off, and the emitter, pulled to low either via the emitter resistor or low oscillator output (depending on Vb logic level from the mcu). Not spectaculary excited about Vbe drop , but can live with it (if its circa the 0.6v expected). Anyway the question(s)a) is the right way to proceed or is there a better solution (bearing in mind the small pcb space available and my reluctance to make the space for either a 8pin or 14pon standard logic gate ? b) is there a different approach for generating a logic enabled 900hz audio tone ? (I found using a nand oscillator with enable fed to the first nand wasn't reliable for fastish logic switching from the mcu - I suspect because of interupting the charge/discharge of the RC network on Gate 2 - but happy to guided). Hope all this made some sort of sense, have a nasty feeling I am missing an obvious solution to this. <Q> I think I'll just answer b): <S> is there a different approach for generating a logic enabled 900hz audio tone ? <S> (…for fastish logic switching from the mcu ) You already have an MCU. <S> Practically all MCUs I know contain at least one, typically multiple independent, PWM units . <S> These are designed to give adjustable-frequency, adjustable-duty-cycle rectangular waves. <S> In other words: Drop the damn 555 and just use either your existing MCU to generate a 900Hz tone, or replace the 555 with a really cheap MCU, and low-pass filter appropriately to convert square wave to sine wave. <S> Note that 900Hz is really slow, from an MCU's perspective. <S> You might, especially if you've got other timers/counters to spare, very simply do this in MCU software. <S> The potential jitter won't matter, as it would be a high-frequency component to be filtered away, anyway. <A> Just use the reset pin on the 555. <A> If you have a timer that is already used by something else and can trigger an interrupt at an even multiple of 900Hz, then use an interrupt to flip a pin to generate your signal. <S> Otherwise, I'd use Andy's solution, but without the FET: since you use an arduino which is 5V tolerant, you can set a pin to output a logic level, or to be high impedance (input). <S> Now connect this pin to the 555's "C" capacitor. <S> Set it to output a logic 0 to disable oscillation, and set it to high-Z to enable it. <S> Free solution with no parts! <A> so was thinking of a classic emitter follower (npn) with the 'collector' fed by the output of the 555, the 'base' driven by an mcu <S> (arduino 5v) <S> output logic pin <S> A 555 used as an astable self-resets via the discharge pin like <S> so: <S> - So, put a small MOSFET across the capacitor (which gets reset via R2 and pin 7) and activate this from the MCU. <S> Or directly wire the MOSFET to pin 7 and ground. <S> Notice the internal transistor inside the 555 connected to pin 7 - using an external transistor will give you the functionality you need without distorting the output voltage. <A> I haven't designed with a 555 for a while, so here goes. <S> In my opinion, the simplest approach is the one given by @RoyC <S> - use the reset pin. <S> There are other approaches; e.g., feed the output to a simple diode/resistor AND gate and use that output to drive whatever you're driving - that's three parts: a pair of switching diodes (e.g., 1N4148s and a pull-up resistor). <S> Some notes on the 555: <S> You'll never get a square wave (with an offset) even if you use steering diodes to control the capacitor's charge/discharge. <S> If you really want something with exactly a 50% duty cycle, you need to output the signal to a flip-flop and divide the frequency by 2. <S> And then, depending on the technology you're using, you may have to bypass the 555's output to ground with a small capacitor <S> (I used disk ceramic); if not, you may end up with spurious flip-flop triggering (school of hard knocks here). <S> On startup, the capacitor has to charge to something like 2/3 Vcc; it then discharges to something like 1/3 Vcc, back to 2/3 Vcc, down, etc. <S> All this is documented in the original data manuals.	Another option is to connect the output of the 555 to an input, and copy this to a microcontroller output with a pin change interrupt, which can be enabled or disabled.
What is this symbol near an AC motor? What is this symbol circled in green in the image? Is related to that induction motor, and I guess it is a french project (if that helps). <Q> The capacitor symbol inside it may be a clue; the block next to it is one way of representing a resistor, and the staggered lines above it suggest there are three. <S> Given this, I suspect it's three R-C snubbers , wired in delta, to reduce noise and switching transients on the 3 phase supply. <S> This IS a guess, but there's not much else it could be. <S> Component values? <S> Resistors in the tens of ohms, capacitors a fraction of a microfarad (e.g. 47R,0.1uF) would be typical. <S> The physical unit is probably a sealed metal can, and may or may not have such useful information printed on it. <A> The graphic is likely a lazy way of depicting it, this is the more complete version: Or <S> the image is a copy of a copy of a copy, and some of the lines have been lost. <A> Boy, that looks like a 3-phase power line filter/conditioner. <A> My guess would be that the box is perhaps referring to another sheet that might have a schematic for that box, referenced with the Z2K that's by it. <S> But, I can't be certain because I don't recall that kind of box and reference scheme including any symbols inside the box, usually.	Given that it is on a 3 phase motor, I'd say it is a set of Power factor Correction capacitors, which typically come with discharge resistors.
What effect does having a constant +12V on the drain have on an N-channel MOSFET? I have a circuit where I have 12VDC source that is connected to the MOSFET (A2SHB) via a timer that applies the voltage to an LED. I have also wired the 12V straight to the LED via a resistor for a constant 8 volts on the LED. The result is that the LED is always on and then the timer comes on and the LED gets brighter making the LED appear to be blinking. My concern is that with a constant 12V on the drain of the MOSFET, I am causing it to fail because sometimes the LED won't blink until I cycle power. I am a complete novice at this. I have included a schematic that I think is how I have it wired. Thanks in advance for any help on this. <Q> Pretty much everything about this circuit is wrong. <S> If your schematic is accurate, you are blinking the LED by shorting out your 12V rail. <S> This is NOT a good idea, for a couple reasons: 1) <S> I don't know what your "timer" is, but I doubt it likes having its power supply shorted every time the LED is turned off. <S> 2) <S> 3) <S> Depending on what kind of supply the 12V is, shorting it out could damage it or at least trigger an overcurrent reset which will mess with your timing. <S> Below is an example of a correct way to switch an LED. <S> I'm assuming the "timer" is "some circuit which is powered from 12V and generates a square wave output at suitable levels for the MOSFET gate limits ". <S> (As Christ Stratton and other mentioned, 12V gate drive is too high for this MOSFET) <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Your drain is fine, but you are operating the gate outside of what your MOSFET is rated for unless you magic timer block somehow outputs a lower voltage. <S> Take a look in the datasheet under Absolute max rating and gate-source voltage. <S> +-10 V. <S> For a short bench test, I wouldn’t worry about it but you will most probably break down the gate oxide layer over time and short the MOSFET. <S> However, your schematic is bogus as the MOSFET shorts out the powersupply. <S> You want your MOSFET in series with your LED+resistor, not parallel to it. <A> When the gate voltage is low, there will be no problem having 12 V on the drain, assuming you've chosen a MOSFET rated for more than 12 V. <S> When the gate voltage is above threshold, this circuit will likely damage either the FET or the 12 V source, because there is nothing to limit the current from the source through the FET when the transistor is "on". <S> As another answer points out, your FET is not rated for 12 V between gate and source, so if your "TIMER" block connects the full 12 V to the gate, that is another problem.	Depending on how much current the 12V supply can source, it could easily destroy the MOSFET because the current is too high (nothing to do with the voltage).
Do switching regulators usually require a lot of external components? The DE2-115 Altera FPGA development board uses "LM3150 Wide-VIN Synchronous Buck Controller" to generate two supply rails:1) 1.2V/5A from a 12V rail2) 3.3V/6A from a 12V rail These switching regulators require quite a lot of external components to be used. Is this common with switching regulators? Why not use an "integrated version" of switching regulators which do not require such a large number of different types of external components. <Q> Switching regulator components are sold in several configurations. <S> A switching regulator controller is an IC contains the control functions, and perhaps a gate-driver boost circuit. <S> But it will require external inductors, external switching components (FETs), external capacitors, feedback resistors, etc. <S> But it will still require an external inductor and capacitors. <S> A switching regulator module is typically a complete circuit with multiple components mounted on a small PCB. <S> It usually includes an IC, inductor, some capacitors, etc. <S> This can be a nearly complete solution in a single orderable part number. <S> Using a modular solution is a good idea when you do not want to devote resources to designing the regulator circuit, and when you can find a existing product that meets your needs. <S> A regulator IC is a good idea when you want more flexibility to tune the design to your requirements and can't find a modular solution that meets your needs. <S> Cost might also be lower than for a modular solution. <S> A controller IC often allows handling higher load currents than an integrated regulator IC because the switches can be larger and not as much thermally coupled to each other and the rest of the circuit. <A> Is this common with switching regulators? <S> Yes <S> Why not use an "integrated version" of switching regulators which do not require such a large number of different types of external components. <S> Not possible for this current and power range with dual outputs. <S> In order to improve cross load regulator noise, separate converters are used. <S> Consider that every part has a design purpose that for better or worse was the most cost effective choice if the design is in high volume and has high quality specs. <S> The design you show is one of the best buck regulators, a Half-Bridge feedback ratio variable voltage buck regulator with fixed parts. <S> The concepts start simple then get more complcated with low error and high power specs. <S> Small lithography analog and digital ASICs cannot do low RdsOn MOSFETs on the same wafer. <S> Nor can they do large reactive parts or large current shunts. <S> Consider : <S> high energy storage in LC reactive parts and high current in semiconductors, input transient OVP, output current limiting , output LC ripple filters. <S> Consider the specs for your DC-DC multiple output converter. <S> 12Vin, Output 1.2V,3.3V, total 27W output <1% Voltage regulation <1% 50% load regulation error, < 2% ripple, crossload regulation error 0.5% , low EMI, etc etc. <S> Can you think of a better way? <S> Look on your MOBO to see how little space it takes to supply all the internal voltages for RAM & CPU , all controlled by BIOS. <S> Locate the coils near the CPU. <S> Then for AC to DC converters we have lightning transients and HIOT isolation requirements. <S> Not every design is scaleable due to I^2R exponential demands. <S> But here is a simple AC-DC converter. <A> To add the the existing answers: <S> some components must be external to chip (and at some distance to other components) to keep ambient temperature as their characteristics may depend on the temp they operate in. <S> If built into a single chip their operations features would suffer from heating; chips in your example assume varying output in terms of voltage, thus some components (like feedback resistors) must be external in order for you to adjust output voltage/amperage to needed level; <S> some components are just too big to fit into the chip's casing. <S> If you would carefully look into the regulator's datasheet, you will most probably find the section explaining the best component layout on the PCB - and also how you should NOT lay them out. <S> These recommendations focus on heat dissipation, EMI within regulator's circuit and crosstalk with other circuits, stability and reliability of the operation, and safety.	A switching regulator IC often integrates the switching components (FETs) with the regulator controller. In order to get the best bang for the buck, it is not feasible to combine all the parts into silicon.
Is it OK to route a trace through a pad? Will I have any trouble routing the traces in this way? (VCC and GND) Is it OK considering that the entire circuit current is under 50mA? <Q> There is no problem routing traces through pads (like you have done). <S> Be aware when routing power/GND of the current that will be travelling through these traces. <S> This will dictate trace thickness. <S> I can see where your confusion may come from. <S> I'm not a fan of how Eagle renders traces connecting or passing through vias/pads. <S> When you do this: This is how the copper will actually look on your PCB: <A> No, it is not a problem routing through a pad. <S> You might wish to consider adding ground and power planes to the design. <A> That will increase the current carrying capacity many times. <S> Additionally, each side of the ring looks about as thick as the trace, so even without soldering, current capacity has been doubled. <S> But what does current carrying capacity really mean anyway? <S> The pad is tiny, there will hardly be any voltage drop across it. <S> And because it has greater surface area compared to volume, it will heat up less than the track. <S> So unless there are a bunch of pads on the track, there is no cause to worry. <S> The real problem of course is if the pad is small, drilled and not soldered. <S> In this case a track might be broken due to a bad drill bit. <S> And, may not be noticed in a complex layout. <S> Much more importantly, an undersized pad may not be mechanically strong, especially when connectors are involved. <S> I would widen the tracks on both sides of the pad just for the mechanical strength alone. <S> Has saved me many times. <S> The epoxy that holds the copper to the board can only take so much. <S> Also make sure the drill holes fit tight. <A> If the connections to those pins are the same VCC and same ground you will have no issues. <S> Physically the copper track will only go as far as the pad it wont actually be left hanging over the hole upon manufacture. <A> In relation to current, your trace is not 24mils (0.61mm) passing across the hole. <S> This is a custom PCB, not one of those cheapo Veroboards. <S> It's actually about 3.81mm (150mils). <S> You need to consider that if your PCB is the standard 1.6mm thickness, and the hole is through plated, the hole has tinning on it's cylindrical perimeter. <S> Like so:- <S> The implication is that whatever your actual trace width is approaching the hole, even if it was a micron and the annular width was a micron, you'd still have 3.2mm of copper along /across the depth of the hole. <S> So it doesn't really matter if the hole is filled or not. <S> It's actually one of the highest current capacity parts of your copper layer unless you have something > 126mils wide. <A> As mentioned earlier, use Ground planes. <S> In Eagle, draw a polygon around the entire board and Name it Gnd. <S> Do that for both the Top and Bottom layers. <S> Rip up any Gnd traces you have. <S> Add Vias in spots around the board and Name them Gnd as well to connect the top & bottom Ground layers. <S> On a 2 layer board, creating a Vcc (5V, or 3.3V)) is harder, those are usually routed as traces.	This should not be a problem if the pad is used, i.e. soldered before use. In addition, search for "power planes", "ground pours" for more information. The thickness of the annular ring needs to be considered to ensure that it can carry the required current.
Can I cut off a thermocouple's metal head? I want to attach a thermocouple to my 40Watt soldering iron in order to make it temperature controlled. The thermocouple I'm going to buy is a K type MAX6675 thermocouple that looks like this: But the thing is it has this bulky metal head attached to it which doesn't seem to be detachable. Since I want to attach the thermocouple to my soldering iron the smaller it is the better for my application. I was wondering if I could just cut off the metal head of the MAX6675, twist the heads of the wires together and turn it into something like this: Will the thermocopule stay intact if I do such thing? And if not, is there any other way I can get rid of that bulky head? (Buying another thermocopule is not an option) <Q> I wouldn't necessarily just twist them up, though. <S> A blob of solder is sort of like a braze. <S> I'd be concerned about oxidation eventually causing intermittent connections. <S> Crimping should work too, in a pinch, and there are gas-tight ways to crimp. <S> https://www.physicsforums.com/threads/okay-to-twist-thermocouple-wires.793193/#post-4982730 <S> Reference <S> https://www.physicsforums.com/threads/okay-to-twist-thermocouple-wires.793193/ <A> Will the thermocopule stay intact if I do such thing? <S> I'd argue that "intactness" is broken the moment you cut something off. <S> Functionally: the contact of the two different metals is indeed what gives the thermovoltage. <S> The thermocouple <S> I'm going to buy <S> is a K type <S> MAX6675 thermocouple that looks like this: … (Buying another thermocopule is not an option) <S> (emphasis by me) <S> I call nonsense on that. <S> You need a differently sized thermocouple head. <S> You can buy hundreds of others out there. <S> So, get a different thermocouple. <A> You can do anything with thermocouple wires as long as you can figure out how to WELD the tip of thermocouple. <S> Soldering is a bad option. <S> You shouldn't worry much about "calibration", especially for the soldering iron application, because the voltage output is determined by metal alloy compositions of the two wires, and by nothing else. <S> If you would require results to a fraction of a degree C, then it is always wise to get your own calibration curve. <A> Yes, theoretically that will work; mechanical issues aside. <S> Marcus Müller's answer warns that metal contamination of the junction will effect the thermocouples accuracy; but that's not correct. <S> If you introduce a dissimilar metal you're basically creating 2 thermocouple junctions at the point: A-B-C. <S> Well it turns out that if both junctions A-B, and B-C are at the same temperature the effect of B cancels out, and it's equivalent to A-C. <S> So no, <S> a little metal from your pliers will not ruin the thermocouple junction. <S> You can check out this document for more in depth information (see page 5): https://www.omega.com/temperature/Z/pdf/z021-032.pdf	Yes, you should be able to cut the wires and reattach them without losing calibration. A weld would be best, and brazing second best ( http://eleceng.dit.ie/gavin/Instrument/Temperature/Laws%20of%20ThermoC.html -- the law of intermediate metals can be used to show this). Even a bad solder is likely good enough. Twisting will give you some life, but it may not be gas-tight.
Power Supply negative voltage as ground and ground I have a PC PSU and can get GND +5V +12V +3.3V and -3.3V -5V -12V out of it. I need to power a device that needs more than 12V. Am I able to take the -5V as GND and connect +12V to VCC and still use the usual GND as ground? So I found a solution. I had an other Mosfet Driver laying around so I will use that. My PSU actually has the -5V wire but not the -3.3V so my bad sorry. Thanky you for your answers. <Q> No, you can't still use the GND as ground if you want to also connect -5V to the device's ground. <S> That would mean shorting -5V and ground rails. <S> The GND from the Pc power supply will be grounded to the mains ground, so it's like it is an "absolute ground". <S> -5V rail from the ATX PSU will be at an "absolute -5V", if you then connect an usb cable which has it's grounding reffered to the mains ground you're basically shorting a -5V generator to ground. <S> That's bad. <S> From now on, let's assume that you aren't going to connect anything else. <S> What you can do is take for example the -5V and make it your "new ground", now all the other voltages from that PSU will be shifted by those 5 volts. <S> So your -3.3 becomes -3.3 + <S> 5 = +1.7V, your GND becomes 0+ <S> 5V = 5V <S> , the 3.3 becomes 8.3V and so on. <S> In this configuration remeber that you can't exceed the -5V rated current <A> I have a PC PSU and can get GND <S> +5V <S> +12V +3.3V and <S> -3.3V -5V <S> -12V <S> out of it. <S> You are mistaken. <S> First: There is no -3.3V rail on an ATX power supply. <S> This has never existed. <S> Second: The -5V power rail was removed from the ATX standard in 2003. <S> Unless you are working with a very old power supply, pin 20 is not connected, and there is no -5V supply available. <S> It's primarily useful for supplying low currents to devices which need a split-rail supply, like operational amplifiers. <S> Do not use it as a substitute for a 24V supply. <S> If you need to provide more than 12V, an ATX power supply is not appropriate for your needs. <S> Use a standalone power supply designed to supply the voltage you need. <A> in terms of basic DC power supplies using +12V as VCC or your positive lead, and using -12V as your VEE or negative lead, that will give you 24 volts at your load... or whatever voltage depending on the combination of 5 ,12, and 3.3 you use for + <S> and -. Use the +12 volt as positive and +5 volt as negative <S> , that's 7 volts to your load. <S> However, on an ATX power supply check the current ratings or DC output specs of the different voltage rails <S> , that's what will either make or break what you are doing. <S> Generally for an ATX PSU <S> the +12v rail has a high amperage rating like 100+ amps to support high end graphics cards, much higher than any of the others rails. <S> the -12v rail may be rated to do less than 1 amp, so for whatever two voltage rails you use to get a desired DC voltage, the minimum current rating from those two is what you have to stay under or blow the PSU. <S> only if you use the GND wire and 1 rail will the current rating of just that 1 rail apply. <S> and also check if the PSU has one 12v rail or multiple. <S> If multiple, make sure you know which rail you are using and <S> what it's current rating is if teh multiple 12v rails are different. <S> at this point if doing the above you use the PSU gnd as system ground on the device (i.e.) load that is fine, that's how every pc motherboard currently works, that DC ground is connected to PC chassis via mothererboard <S> just like the AC earth ground is to the PSU shell connected to the pc chassis... <S> that you handle while powered.	But yes, you can power your device from -5 and 12V rails, assuming that your device haven't and doesn't need another grounding (e.g. a device which has an usb connection or some sort of other connection with the outside world). There is a -12V rail available on an ATX power supply, but its current capacity is minimal -- a typical 600W ATX power supply may have a current capacity of 300 mA on the -12V rail.
Best battery for low power, long term application Im designing a product which needs around 5 to 7v. I need to power it from a rechargeable battery. It spends most of the day in an off state, consuming about 0mA. Sometimes is powered by the user, draws 40mA for 4 seconds and 200mA for 1 second, then it shuts down. The peculiarity of the project is that it needs to run without any charging for 5 months a year. It will be subjected to ambient temperatures up to 50°C. I've excluded, for one reason or another, pretty much every type of battery i know: lithium based batteries are not happy to stay at full charge and almost full discharge for such long time nichel based ones seems to have a big self discharge lead acid batteries seems like the more suitable type, but it also seems that they are very heat sensitive. I understand that there are also some "deep cycle" ones, that would be more suitable for my application, but are pretty rare. I don't have any weight or space constrain (well... As long as they are reasonable) At this point I'm quite lost, i hope for your help. I'm not an expert on batteries, some of my considerations can be wrong. <Q> Total energy per day (Joules= <S> Amps V seconds/efficiencyreps/day) <S> (40x4+ <S> 200x1)5.5 <S> 30/0.85 =70J/day, 25kJ/year,11kJ/5 months. <S> If you only get to recharge after 5 months, why use rechargeable at all if the energy is low? <S> D-Alkaline=75kJ, AA=10kJ (55C max) <S> (alkaline shelf life 5 years) or lithium-irondisulphide duracell AA (1.6V, 10kJ) <S> 60C max ( shelf life 20 years) Alkaline performance is excellent at 50C, (poor when cold) <S> Shelf life will be hugely degraded at 50C of course, but still manages 50% at 5 years <S> see fig 15 <S> , self discharge graph <S> In your case the choice of cell size is about peak current at lowest temperature, not capacity. <S> I think your understanding of lithium rechargables is not quite right: To maximise the energy stored in them, they are recharged to a terminal voltage. <S> The higher that is, the more energy you can get, but the more damage you will do. <S> 4.3V = <S> short life, 4.0V = long life. <S> NiMH batteries have also improved with versions called LSD - low self discharge, which gives shelf life to 2 years. <S> A small micro can run directly from the batteries (2.7-4.5V) without needing the switchmode, to keep track of sleep time etc. <S> A (cheap) stepup, with poor idle current can be turned on for transmissions. <S> See also this answer to a similar question Use a tiny solar panel (1" square, 40mW) to stretch out the alkalines life if you want. <S> 70J/day / (3hoursSunPerDay <S> 3600secs) = <S> 6.5mW. <S> An alkaline may last years with solar float charge <A> Don't use a rechargeable battery. <S> Most of them tend to have short self-discharge times. <S> 40mA for 4s followed by 200mA for 1s <S> sounds like a ZigBee or other similar 2.4GHz wireless sensor application... <S> I built a wireless sensor product with similar energy/duty cycle requirements (compared to yours). <S> We used a lithium thionyl chloride primary battery. <S> Specifically we used the Tadiran TL-5930. <S> http://www.tadiranbat.com/assets/tl-5930.pdf <S> It is a D-size battery with a capacity of 19Ah at 3.6V. <S> It has an operating temperature range of -55 to 85C. <S> To get the 5V to 7V from the 3.6V battery just use one of the many available low power boost converter chips. <S> Texas Instruments has a good selection. <S> Using this battery and a low power boost converter, our testing has shown that the product we designed can last anywhere from a few months to a few years depending on how often the user accesses it. <A> @MarcusMüller i'd say that we can assume 30 as the daily maximum number of cycles. <S> I'm planning to use a mosfet to interrupt the power supply to the board to save more batteri, there will be some leakage <S> but hopefully it will be in the order of tens of microamps at most <S> So that amounts to 0.45 Ah in 5 months. <S> With cell capacities significantly above twice what you need, and mature battery management ICs that prevent deep discharging, I'd also argue that cell aging is not much of a problem over the next decade. <A> High density batteries of consumer grade tend to leak more at high density and high temp, just as capacitors but depends on chemistry and quality. <S> Lower density type chemistry tend to also have lower leakage current. <S> LiPo's tend to have lowest self discharge between 50 to 66% SoC <S> and that's why they are shipped with products, this way for longest shelf life. <S> Low mAh rating but low self discharge rating. <S> Alkalines tend to have 8 yr <S> ~10 , more or less shelf life and I concur with @Henry's suggestion and have done this many times for my own verification using a regulated float voltage on Alkalines but didn't use a PV. <S> So you have many choices actually, including <S> these from Panasonic	You can buy much higher capacity lithium batteries from many distributors (e.g. Mouser ), and as long as you keep them within the specified operating conditions, you'd really don't have to worry about damage to the cells. An arrangment with 3 cells and stepup to 5V is good for this sort of job. @Ali's suggestion of Tadiran TLI LiPo cells fit this category from a very reputable company.
Disconnected laptop's internal mic, but sound-card is still picking up sound, why? I have a HP Mini 5103. A security minded friend wanted me to physically disconnect the internal webcam and stereo mic, no problem as both are individually connected to the motherboard via a wire harness. I unplugged both. After this simple modification, I decided to verify that no sound was being recorded, so I hit record for a few seconds in Audacity. At first it only seemed to be recording white noise, sounds in the room weren't captured, which is what I expected. Then I tried tapping the laptop, and suddenly spikes of activity showed up in the recording with each tap. Playing it back, the taps can be heard relatively clearly. Is it possible that the sound card's circuitry is sensitive enough to be picking up these vibrations without a mic? My initial suspicion was that there was a third mic, maybe just for low frequencies, but I couldn't seem to find anything else on the motherboard. HP only lists the stereo mic on their major parts list: https://support.hp.com/us-en/document/c02492286 <Q> Is it possible that the sound card's circuitry is sensitive enough to be picking up these vibrations without a mic? <S> Yes. <S> MLCC caps are known to be sensitive to vibration, and being both cheap and small there will be some in your signal path. <A> Mechanical vibrations cases voltage to appear across them and circuits which sense in the mV range from the start, such as your microphone amplifier will be subjected to “hearing” these vibrations. <S> https://en.m.wikipedia.org/wiki/Piezoelectricity <A> What if someone (NSA?) hacked your soundcard software and implemented a 24-bit ADC? <S> Would you be happy with that? <S> If the ADC runs at high-sample rates (96,000 per second), but you only need 8,000 sample per second to handle voice, then that ratio of 96,000/8,000 or 12:1may provide log2(12) <S> = <S> 3+ more bits of resolution, lowering the needed signal level by 12:1. <S> If your basic soundcard ADC can run even faster, or if the card's digitized samples can be processed in the CPU fast enough to feedback using the DAC, you may achieve excellent 20 bit results. <S> Or even higher. <S> Using those capacitors as the "mic".	Voice should be well under the noise floor though. It’s called the piezoelectric effect and many materials used in modern electronics exhibits this effect.
Can I use dimmer voltage regulator for single phase multi speed AC motor? I have a swamp cooler motor which is single phase 3 speed motor, 1/7 HP i.e. 106 watts. I wish to reduce its speed (rpm) by 50%. Can I use this dimmer circuit to do so? This dimmer regulator is rated for 1000 watts exhaust fans. I only have school level electrical knowledge. I was thinking of connecting this dimmer regulator in between live (positive) wire, kinda like a ceiling fan regulator. <Q> A 2-wire overhead fan controller would work. <S> You can keep the 3-speed switch and put the speed controller in series with the lowest speed tap. <S> That way you have all the higher speeds, and the LOW becomes a LOW-LOW. <S> In your photo, the heat sink seems small. <S> If it gets too warm, add some metal or put the dimmer into the airstream. <S> Remember, don't go too low on the speed setting. <S> But if your motor is already turning when you switch it to LOW-LOW, then the setting can go a bit lower. <S> It's like trying to drive your car using second gear. <S> It helps if the car is already rolling. <A> It should be possible as long as the power drawn by the motor does not exceed the absolute maximum ratings of the regulators switching device (triac). <A> If the dimmer circuit reduces the AC voltage and does not rectify it to DC, it may reduce the speed of your fan. <S> The fan may not start with the voltage set too low. <S> It may stall if you turn the voltage down too far.	If you connect the motor to something other than the original fan, it may not work at all.
Is this circuit acting as an amplifier and how does it work? I simulate a diode connected BJT circuit as follows: And here are the plot for the currents through Rin I(Rin); the wire then collector-base then to base-emitter junction Ic(Q); and through the base-emitter junction Ib(Q): Almost 99% of the current is flowing through the wire which shorts the collector and the base. It seems the like majority of the electrons flowing from the ground into the emitter-base junction are pulled by the collector and then all the way to the source. Is this circuit acting like an amplifier? How is the collector pulling the electrons even though its terminals are shorted i.e zero potential difference across CB junction? <Q> What you have is called diode connection configuration. <S> Basically, Vb is always a diode voltage drop higher than Ve. <S> Shorting Vc to Vb only helps with biasing the transistor. <S> Your circuit is equivalent to this: simulate this circuit – Schematic created using CircuitLab <S> This kind of configuration can be used in current mirrors when you want a forward voltage drop from a diode to match the drop from another transistor such as in Wilson current source. <S> But it is useful only from transistors on same die. <S> You can also use this to replace a diode in a circuit if you will. <S> At low voltage the voltage drop is usually quite small. <S> Yet, you are better off using a proper diode unless this is part of a much more complex circuit where such an arrangement is needed (eg. <S> wilson current source). <A> VBE is about 0.7v. <S> That means VCE is also 0.7v. <S> For all reasonable transistors, this means it's in the linear region, as it's above VCEsat, which tends to be in the 0.1 to 0.4v region. <S> It's not the zero VCE that matters, but <S> the 0.7v VBE, and the >VCEsat VCE. <S> That means the collector current is beta times the base current. <S> So yes, it is acting as an amplifier. <S> A figure of 100 is often used for beta (when no better figure is to hand) <S> so 99% of the current going into the collector is to be expected. <A> In a diode-connected BJT like this, the BJT is kept right at the edge of active vs saturated mode and so only a small base current is required. <S> Most of the "diode current" flows via the collector to the emitter. <S> The exact value of \$V_\text{BE}\$ isn't immediately obvious. <S> But it will be whatever is required for the collector current. <S> In this case, from your plot, it appears to be \$V_\text{BE}=800\:\text{mV}-(100\:\Omega+330\:\Omega)\cdot 170\:\mu\text{A}=726.9\:\text{mV}\$. <S> Suppose that \$\beta=100\$. Then you'd expect to see \$I_\text{C}=\frac{100}{100+1}\cdot 170\:\mu\text{A}=168.32\:\mu\text{A}\$ and therefore the base current would be the remainder, or \$I_\text{B}=170\:\mu\text{A}-168.32\:\mu\text{A}=1.68\:\mu\text{A}\$. <S> Your plot would appear to be consistent with this idea.	Your transistor acts as a diode.
Relays with NC and NO connections: does high vs low matter? Is there any difference between: An active low relay with the load wired to the NC (normally closed) connector, and An active high relay with the load wired to the NO (normally open) connector? In either case, when the control signal is low, the relay will be open and the circuit will be incomplete. When the control signal goes high, the relay will close the circuit. I ask because I've come across a number of situations in which people build a logic inverter to get active-high behavior from their active-low relay (which seems to be the most common type, at least for the N-channel relay boards commonly available online). If you have both NO and NC connections, is the logic inverter necessary? <Q> The power-off state is usually required to be safe or safe-er. <S> This is a system-level requirement. <S> In most cases that will require the normally open connection to be used. <S> There are cases (for example certain types of heat applications) where hot is safe-er than cold and you want to use the normally closed and depend on some other device as a cutoff. <A> A relay by itself is neither "Active High" or "Active Low". <S> The Normally-Open contacts will close when power is applied to the coil. <S> I assume you are referring to the relay boards sold for use with Arduinos and other microcontrollers. <S> Those boards have some additional ciruitry - Opto-coupler, transistor, etc. <S> - that may provide an "Active Low" or "Active High" input. <S> With an "Active High" interface, the relay should operate when the input is High. <A> In logic both are equivalent but the power offstate needs to be what is needed. <S> There is also the inverted logic case, low side coil drive to NO contacts. <S> The advantage here is a simple SPST-NO or 1P1T is cheaper than SPDT and is compatible with open collector drivers or bipolar or CMOS alike. <S> This latter option which you did not include is the most common. <S> Although off-the-shelf arrays of opto isolated relays permit you to drive either anode or cathode side to choose normal or inverted logic. <S> THe choice is NC vs NO is always determined by your application needs and nothing else. <S> i.e. "What is Normal" is relative to your operation.	With an "Active Low" interface, the relay should operate when the input is low.
Why use an optocoupler with opto-driver instead of isolated amplifier using optocoupler for feedback Why do so many feedback circuits rely on a optocoupler driver instead of using an isolated amplifier to directly transmit the feedback voltage to the switch controller? <Q> Isolation amplifiers can have a high cost. <S> Analog Devices AD202KN is $15, good to 400 HZ at best, and <S> the AD210KN which samples at 500 KHZ is about $30 each. <S> Compare that to the $1.00 USD for many opto-couplers. <S> Also the isolation amplifiers have a 'conversion' time, as they are transformer coupled. <S> Some INA series use capacitive coupling but <S> the delay to integrate the signal back to analog is still there. <S> If you need a low cost and very fast responding isolated loop a fast opto-coupler cannot be beat. <S> They can be a thousand times faster <S> then a isolation amp, because they do not have to convert the signal to PWM, then intergrate the signal to convert it back to analog. <S> The response time of a fast opto-coupler is in the nS range, while a isolation amp may take several uS to 100 uS. <S> Isolation amps are good for measuring accuracy over a wide dynamic range, with at least 1,500 volts isolation, so you can measure line voltage at 600 VAC with no problem, and measure the current as well. <S> What isolation amps are NOT good for is servo-loops. <S> It can make the circuit 'ring' as it settles down or worse yet the circuit may oscillate because the feedback is too slow. <S> Also isolation amps often need a modest low-pass filter at the output to remove any conversion noise, <S> well up in the 100 KHZ to 500 KHZ range. <S> Some need buffers as they may not have a substantial drive current. <S> The AD202KN and AD210KN can only source or sink about 1 mA. <A> I <S> if there is enough gain before before the opto it will find the linear region and the exact position of the linear region will only result in a small offset in the output. <A> Both are linear analog (emitter follower vs OA feedback) isolated optos but the IL300 is a poor choice for 3 reasons ; <S> cost reasons IL300=$5 in volume vs LTI4430=$2 and high BW opto ($0.21) 700kHz worse response time good opto choices offer higher BW than IL300 <S> ~100k~300kHz <S> low (unity) <S> gain <S> adds stability challenges for regulator to improve transient error. <S> LTI adjust gain from 0 vs gain from a Vref <S> so it is not as good. <S> no need for LMV431 Adj zener ref to boost gain or "431" ref type which is included in LTI4430 adds inverting 6x gain to Opto out which improves tracking error of output voltage by almost same amount.	basically it's cheaper to just use the single output optocoupler, it has a region where it's reasionably linear. They greatly increase the time it takes to correct for a change in voltage, current or load.
Trigger 2 triacs in parallel simultaneously I am using a STM TPDV1240RG triac to control an AC load. This triac is rated for 40A. The current in my circuit is up to 80A so I was thinking of putting 2 triacs in parallel to share the load. I am using a MOC3063 with DC pulse to turn on and off the triacs. The problem is that only one of the triacs is conducting (whichever is first to turn on). Is it possible to get them both to fire simultaneously? Note: I will also be using heatsinks on each or perhaps one between the two if suitable. Thanks. <Q> First, the one that latches faster will collapse the voltage, possibly preventing the slightly slower triac from ever turning on. <S> Second, even if both triacs turn on, they won't share the current well. <S> Triacs are sandwiches of effectively bipolar devices and silicon junctions. <S> These have lower forward voltage at higher temperature. <S> The one that takes a bit more current will get warmer, which lowers its voltage, which causes it to take a larger fraction of the current, which makes it even warmer, etc. <A> Use one of the following circuit. <S> Replace SCR with Triac. <S> First method is simple. <S> You need two resistor capable to handle 40Amps. <S> But with that much current, there will be too much waste heat. <S> For Dynamic sharing, you need two magnetically coupled inductor. <S> i.e <S> An isolation transformer (1:1 voltage ratio) capable to handle 40Amps through its winding will do the job. <S> Be careful about the polarity of the two coils though. <S> You want them to be opposite to one another. <S> Let's say, when you apply the gate pulse, traic T2 turns on first. <S> As current flows through the coil and T2 , a voltage of reverse polarity is induced in the other coil connected to T1 . <S> As the voltage is of oppostie polarity, it adds up to the supply voltage. <S> So the voltage across T1 becomes supply voltage + voltage induced in the coil , which is enough to turn it on. <A> To handle 80A with 40A triac you can try period-interleaving control of them. <S> It will spread heat dissipation between them. <S> Better idea is to switch to two 80A SCR BT155W(or BT158W) in inverse parallel configuration. <S> Like this one:	Both Inductor coils are wound on the same core. Find something rated for 80 A to switch the load. Paralleling triacs is a bad idea.
How to latch a signal high at the first rising edge of a pulse train? I have a question about digital circuit design. This is the situation. I have a pulse train. The length of the pulse train is not fixed. So the use of a counter is out of the question. I need to drive a signal high at the first rising edge of the pulse train and need to keep it high as long as I wish. And also I must be able to drive it low when I wish. How can I design a digital circuit to do this? Thanks in advance! <Q> Look up something called a Set/Reset Flip-Flip . <S> In their pure form, these have two inputs. <S> When the SET input is asserted, the output goes high. <S> When the RESET input is asserted, the output goes low. <S> When neither is asserted, the output retains its state. <S> S/R flip-flops can be made with two NAND gates, but you can get them ready made in logic chips, like the 74xxx series. <A> Consider an Set-Reset Latch (or SR Latch : <S> Such a latch can be made of NOR (as above) or NAND gates. <S> And is usually found as a complete circuit in many integrated circuit logic families. <S> In the above example, the Reset and Set lines are normally low. <S> Bringing the S input high forces <S> Q high and Q-Not low. <S> Bringing R input high forces <S> Q-Not high and Q low. <S> Q and Q-Not feed back to the input of the NOR gates to maintain the new state. <S> You can send your pulse train into the S input and force the Q output high upon the first falling edge. <S> Then, momentarily bring the R input high to reset the Q output low. <A> What is a 1 bit edge sensitive memory cell? <S> Async State Sensitive : <S> Two gates with cross feedback form a Latch . <S> This has two inputs Set and Reset, it can be positive or negative logic on input depending on gates if normal or inverting type with NOR, NAND This is now a 1bit register. <S> Your choose the output depend such that Reset overrides Set. <S> Edge sensitive input , async reset: <S> But if both are high or Reset during a pulse train level=1, then you use an edge sensitive FF for pulse and R to clear.	Use an inverter on the pulse train if the first rising edge is desired.
How do I tap into a power supply's standby wire? So I'm working on an Arduino project, I'm going to make a circuit board that's going to be inside my computer. The circuit needs to always have power, after a bit of googling, I found out that the standby (purple) wire seems to be perfect for what I want. I took an old computer and did this. The LEDs barely glow, so it seems I don't draw enough power from the power supply. So I'm wondering, if it's because of the way I tapped into it? or if it simply doesn't provide enough current? <Q> The +5VSB supply should at minimum be capable of delivering 2 amps. <S> Look at the ATX spec 3.3.3, and most modern ATX power supplies exceed this and can deliver 3-4 amps. <S> The Arduino needs a minimum of 6 V to power it from the Vin pin, so you can't use that. <S> You need to be connected to the 5 V regulated line or to the USB connector XUSB line and leave Vin unconnected. <S> The Arduino will never have a current flow over 500 mA <S> (that's the XUSB fuse rating. <S> Please refer to the schematic . <S> I'd suggest that the best way for you to connect the Arduino to your ATX +5SB is to hack a USB cable and use the USB connector on the Arduino. <S> That gives you a fused supply line with a removable connector ... <S> much better than cutting into the ATX cabling the way you have. <A> The picture of your setup clearly shows that the board IS NOT powered via right connector. <S> The red circle shows no wires, maybe to Vin only: <S> You should be connecting the +5VSB to "power connector" +5V pins, More, you also seem to have an extra diode in series with power supply, which likely adds another 0.7 V of voltage drop. <S> Just connect the +5VSB wire directly to "5V" pin, and everything will be fine. <A> The computer MOBO may use a significant amount of the available current on the 5VSB output of the power supply. <S> You should check the voltage from 5VSB to GND when you have your load hooked up using a digital multimeter (DMM). <S> If the load is causing the voltage to sag down to less than 4.75 volts you are overloading the power supply. <S> In that case remove your load. <S> BTW. <S> You should really arrange for heat shrink tubing or electrical tape to insulate those connections you made.	I'd suggest you are either connected to the wrong spot in the wiring harness or (much more likely) are using the Arduino Vin pin instead of powering it from the 5 V pin.
Does the contact pad need to match the footprint shape of the SMD If the contact pad is larger than the footprint, or has a different shape but is still large enough to make full contact with the SMD, will this lead to problems during reflow soldering? I am designing a metal core PCB on which I will be mounting LM561c samsung diodes. I am striving for a high thermal dissipation factor, to preserve the effeciency of the diodes. <Q> In statistics as well in Contract Manufacturing DFM design for Manufacturability the probability of defects in soldering increases with the degree of mismatch to IPC footprints for the desired soldering method < RElow vs Wave solder has different footprints and layout orientation recommendations. <S> In solder reflow, and contract manufacturing, 99% of the defects are related to poor solder and many or most of these are design related to board layout. <S> Without details on your actual part, and pad design , it is impossible to assess. <S> I advise you to read about IPC solder pads and Mfg's recommendations and/or consult with EMS builder for DFM analysis or use your mentor and software to do the same. <S> DFM= Design for Manufacturability. <S> DFT= design for Testability DFC= Design for Cost DFX = <S> Design for Excellence. <S> ( all of above) <A> Solder creates surface tension during reflow between the part and the pad. <S> This surface tension can be controlled and apply different amounts of force to the part. <S> The footprint also needs to contain all of the solder in the solder paste that will flow onto the part. <S> If your only doing short runs and low volume, then none of this really matters as you can fix the parts after reflow if they are tomb-stoning or becoming misaligned during reflow. <S> If your doing larger runs then the manufacturer can help you tune your footprints to allow for manufacturability, because they don't want to fix parts. <S> I usually do low volume (under 100) <S> and I have done a similar thing with a voltage regulator. <S> You can always put solder mask and expose the same footprint while keeping the same area for the recommended footprint. <S> Or you could create a very large trace around the footprint. <S> If additional heat 'spreading' is needed, you can also create vias to a different layer also ( if there is room). <A> Check the datasheet for a manufacturers recommended footprint, copy that into your layout software and you shouldn't have any issues. <S> You can connect the footprint to as much copper as you want so it doesn't really effect the thermal properties unless there's an optional pad and you omit it.	Usually the footprints are larger to allow for solder to be applied to the part as the solder is applied via a stencil in a thin layer, and the volume of solder around the part may need to be larger. Just keep in mind that if the pads have uneven thermal dissipation then it might cause re-flow issues.
Placing TVS or MOV or Transzorbs for a very basic protection This is a question for a general scenario I encounter. So I will not be specifying much details besides the voltage level, cable length/type and the power supply type. Imagine a sensor placed outdoors is measuring some air parameter and it is powered from a receiving unit which is far away and indoors. The sensor output(in the diagram signal SG and signal ground SG_GND) is received inside the receiving unit and measured. When I see such units they always have sort of protection such as MOV or TVS in the receiver box for over voltage protection. simulate this circuit – Schematic created using CircuitLab In general I need to use a single dual power supply(which in the diagram is represented with V1 and V2 above) and also in general power ground and the signal ground are connected inside the sensor. Does anybody have experience with such setup and its TVS or MOV protection? Between which terminals should one place TVS diodes? I have no experience and couldn't find such practical info. Where should I place the TVSs or MOVs for a very basic protection? If we focus on the receiving unit, I named the terminals as A, B, C,D and E for simplicity. <Q> At these low voltages, a TVS diode rather than a MOV is the standard choice. <S> Typically, each diode is connected from a signal or power line to ground, ideally chassis ground. <S> You indicated that the cable is shielded, but didn't show any connection between the shield and ground. <S> The shield, any metal enclosures, and the ground connection of the TVS diodes should all connect to chassis ground, which should connect to your main power ground at one point. <S> For ratings, you want the working (reverse) voltage to be higher than the highest voltage you'll normally see, and <S> the clamping (peak) voltage to be lower than the maximum voltage you can tolerate. <S> If you're already running close to tolerances, this can be difficult or impossible: you may need some scaling on the ADC so that it can tolerate more than ±10 V, if your signal can really reach ±9 V. <S> Another note is that by default, TVS diodes are unidirectional, so if the voltage can go both above and below ground, make sure to get a bidirectional one. <S> Based on your new schematic, it looks like you're close. <S> You shouldn't need bidirectional TVS diodes on the power supply, since each rail stays on one side of ground (nothing wrong with them, they just tend to cost a little more). <S> Where you do need <S> a (bidirectional) TVS is on the signal line, ideally on both the sensor and receiver. <S> If this is a high-speed signal, make sure to use a sufficiently low-capacitance diode. <A> I just retired from working in the TVSS/SPD industry. <S> There are a few hard set rules about surge protection. <S> For power lines AC or DC MOV's are the better choice, especially the 40mm size. <S> Their high capacitance has no effect on power. <S> They have a soft clamp at 50% above your working voltage but a hard clamp at double the working voltage. <S> You normally buy them rated just above your working voltage AC/DC.150 volt MOV's are for 120 VAC circuits, 275 volt MOV's are for 240 volt circuits, and 320 volt MOV's are for 277 volt AC circuits. <S> They go up to 1,200 VAC but those get expensive. <S> Also available are ones with a thermal fuse built in for about $7.00 USD each. <S> Do not use them on data lines as they will choke down the signal. <S> They can be wired phase to ground and phase to neutral or across DC power lines and to Earth ground as well. <S> To protect most data lines TVS diodes have a sharper clamp, much like back-to-back zener diodes, and low capacitance so they will not bog down digital data lines. <S> For super-fast data line SMD versions can be had with 1.5pF of capacitance, good to about 2 GHZ. <S> They have taken the place of gas discharge tubes as they are much smaller in size. <S> Protect data line in differential mode (across the signal pairs) and line to Earth ground. <S> Buy them with a clamp voltage 50% above the expected highest voltage, so small voltage swells do not get clamped. <S> There is no such thing as perfect protection on a wide scale, so you buy what works for this and that and protect both ends of power and data feeds. <S> A close lightning strike will induce substantial current in all nearby wiring. <S> You do not need a direct hit to get lightning damage. <S> By the way we called TVS Transzorbs, or we would have drowned in semantics. <S> Most circuits built to last have Zener diodes and back-emf protection diodes built in. <A> Typically, in the same case size, an MOV can withstand more current than a TVS. <S> But it is degraded by every activation (and can catch on fire). <S> That means that a TVS will tend to be favored when a size-acceptable part can address your external threat. <S> You will need the TVS to be located between each line and chassis ground; the other thing to be careful about is leakage induced by the protective devices. <S> Low voltage devices (uni or bi directional) especially have high amounts of reverse leakage. <S> If you can accept a 10+ ohm impedance on a signal, a resistor (anti-surge 2512 or large wire-wound) can be added between the external connection and TVS to accept much larger faults compared to the TVS rating. <S> If you add a ferrite bead between the TVS and <S> whatever IC you are using, it can also help ensure energy transfers mostly to the TVS. <S> When you get to layout, place the TVS as close to the input as possible and route through the pad (30-mil+ track, ensure TVS fails-short before trace vaporizes) before going to downstream circuitry. <A> If you aren't terribly concerned with power consumption and your sensor is not high current consumption <S> then it is a good idea to provide some impedance ahead of your supply TVS on the sensor side to limit clamp current and therefore limit clamp voltage. <S> Also an ESD rated nF range capacitor in parallel with the TVS helps limit slew rate especially with the aforementioned resistor. <S> Below is the internal schematic for an ESD protection IC. <S> Change the values to suit. <S> Also if possible it makes more sense to have your IA drive the cable instead of sending the low voltage signal before amplification over the long cable. <S> Bandwidth limit the signal cable to whatever range is acceptable with an RC filter and provide differetial TVS.	For low speed data or DC power feeds series resistors of a few ohms combined with Zener diodes or TVS will offer the best protection.
Driving a 12v LED, substituting a PWM chip for a linear regulator? I'm trying to reverse engineer a broken control box for some dual intensity motorcycle LED spotlights , and I note on the circuit board there is an AX 3007-50 12V input, 5V output PWM chip. I assumed the 5V PWD output was to drive the low beam of the LED spotlight, and in a bid to simplify the new circuit I wanted to replace with a 5V linear regulator. In my tests, this hasn't worked. I've been trying to understand why, and I'm thinking that the LED needs 12v to run, the 5V output from the PWM is 12V, but just cut up so it's only on for ~40% of the time. My constant 5v output is not enough to drive the LED, so it simply won't turn on. Would this line of thinking be correct? I know the LED works with 12V, should I be looking to replace my linear actuator for a PWM setup like I think it used before? Some pictures of the LED, and the PWM chip on the control board: <Q> You need to sketch out the circuit to find out how it works. <S> The circuit board is simple enough <S> it's a single-sided board that you can trace. <S> That said, the LED is a simple 3.3 Volt or so multiple amp package. <S> With the right voltage and current limiting device, AKA a resistor, it should turn on if you solder directly to it. <S> If it's not turning on then it's due to your power supply not giving enough current. <S> That said, a switching regulator is more efficient than a linear regulator. <S> If you're dropping 7 volts at 2 amps that's 14 watts of power it's wasting in heat. <A> simulate this circuit – Schematic created using CircuitLab <A> As well as asking this question, I had also asked the manufacturer for their input also. <S> I wasn't expecting an answer <S> but they did get back to me with information on what I need to provide the LED unit to operate the dual intensity functionality. <S> Red: 12 Volt Postive Black: 12 Volt Negative <S> Yellow: <S> Duty Cycle of 10% to 90%, 500Hz, 0-5 Volts 10% duty cycle will be 90% light output, 20% duty cycle will be 80% light output and so on. <S> (100% duty cycle is off, ignoring the input will provide 100% light output )	Your regulator may be shutting off in thermal protection mode. The circuit is more complex than just a dc-to-dc switching regulator..
When current flows from a battery, does voltage decrease? I understand voltage to be a potential for electrons to be pushed through a circuit. However, in a battery, you have an electron build-up that creates the voltage. Once current begins to flow, electrons are now moving through the circuit. Does this mean that the voltage actually begins to decrease as a direct result of current flow? Specifically are electrons "used up" or do they simply lose energy (dissipated as heat in circuit) which leads to a lower voltage potential?Or another scenario, do the electrons, upon returning to the battery, still act as voltage potential to continue to push a consistent amount of current through a circuit despite the loss of energy? Or is there something else I'm missing entirely? Thanks. <Q> For ex, a Lithium-Polymer cell has a nominal voltage of 3.7V and that of a lead-acid cell is 2V. <S> For cells belonging to a particular chemistry, the voltage depends on many factors, the prominent one being the concentrations of the electrolyte, electrodes etc. <S> The increase and decrease of cell voltages while charging and discharging is due to the changes in the these values. <S> However, the number of electrons inside of a battery doesn't change in normal operation <S> no matter what. <S> Applying Kirchhoff's current law, you can check it for yourselves. <S> No matter your circuit and its operating conditions, the current going out of the battery should be equal to the current going in. <S> The voltage only changes because the chemicals inside the cell are changed slightly and not because of a change in the number of electrons. <S> Coming to the heat part, the heat generated in the circuit is compensated by the loss in potential energy of the battery. <A> In general, the answer to your your question is, "yes". <S> As other answers have pointed out, a battery has an effective resistance. <S> This is, in part, do to the fact that the internal structure has an intrinsic resistance. <S> However, it also reflects the fact that the ions in the electrolyte, which are involved in the production of energy, have limited mobility, and this limits the current available and reduces battery voltage under load. <S> However, just to make your life difficult, it is possible for a battery voltage to rise with increasing load. <S> I've seen it. <S> It's a long-term effect, and I observed it while doing capacity testing on a NiMH battery pack which was going to be used on a missile shot (suborbital). <S> Running the battery with a constant current load, I observed the output voltage gradually rise over time. <S> The cause was fact that the internal power dissipation produced a temperature rise in the pack, and the output voltage rises (all else being equal) with temperature. <S> After running for a while (the test duration was designed to deplete the battery in about 45 minutes), the output voltage under heavy load was actually higher than the battery produced under light load at the beginning of the test. <S> It was weird, but made sense when I figured out what was going on. <A> Strictly speaking, batteries are a means to push electrons. <S> They don't use them up. <S> However, a battery also has an effective internal resistance. <S> This resistance is dependent on a number of things (cell chemistry, temperature, cell age). <S> The higher the internal resistance, the more voltage will be dropped internally, and the less force the battery has to push electrons. <S> This reference is an excellent read on the subject. <A> Voltage drop starts with a flow of current thru a resistive loss. <S> This is the equivalent circuit. <S> It can also be an exchange of charge between multiple internal capacitors Q= <S> CV each with different ESR. <S> This is why shorting a battery momentarily returns to some charged voltage level by the exchange of charge Q=CV between multiple layers of dielectric charge. <S> Current is simply the rate of change of charges per second. <S> I= <S> dQ <S> /dt simulate this circuit – <S> Schematic created using CircuitLab <S> Yes all batteries have a series resistance which causes voltage drop and ESR is somewhat inverse to mAh capacity and voltage. <S> e.g. 3.7V <S> 2600mAh <S> Lipo can have 5 to 50mOhm ESR 50Ah car battery can have 700CCA or < <S> 7 mOhm ESR @ <S> 5V drop with 700A <A> Electrons aren't used up they just stop migrating from one pole to the other because the battery is depleted. <S> Imagine a double ended recipient with a diaphragm : <S> The pressure is the voltage (Diaphragm pushing). <S> As soon as you open the valve you get a current. <S> Opening the valve will also lower the pressure. <S> The more you open the valve the higher the current gets. <S> When the current rises the pressure will fall. <S> Everything that goes out comes back in, nothing is lost (closed circuit). <A> A real physical battery, in particular a real physical Lead-Acid battery, has significant capacitance because of the arrangement, orientation, structure of the chemical elements. <S> The capacitance means that electrons do "build up" and "run down" in the battery. <S> But... this is not what a chemical battery "is": it is a feature of what a particular implementation does. <S> For a discussion, read up on capacitors first, then batteries. <S> Real lead-acid batteries also exhibit concentration gradients. <S> This too is a "build up" and "run down" phenomena. <S> Again, this isn't "how a battery works", it's "how actual batteries that people make and sell" work.	The voltage of the battery depends on the chemistry of the cell it is based on.
What does this symbol (three connected triangles) mean? What do the symbols \$GD_{SN}\$ and \$GD_{SI}\$ mean? What is the component’s purpose? The second picture shows its schematic. <Q> What do the symbol ,GDSN and GDSI , mean? <S> It looks like you have answered your own question. <S> Your second image shows what the three triangles are. <S> It's simply three cascaded logical inverters with an enable. <S> The driving force of each logical inverter gets stronger as you go to the right. <S> In other words, the input may be very weak and the output will be very strong. <S> and what is its purpose? <S> The triple inverters are driving high capacitive loads, namely the gates of the two MOSFETs. <S> If there wouldn't have been any triple inverter, then it would take too long to charge and discharge the gates of the MOSFETs. <S> So this is clearly a design meant for high speed, or just low power during the time when the MOSFETs are opening/closing. <S> The gate capacitance of the input to the triple inverter will be considerably smaller than the gate capacitance of the MOSFET pair. <A> To me looks like a three stage inverting buffer. <S> But it's not an IEC 60617 standard symbol as far as i know <A> that symbol is called inverter chain	It's an inverting bufferer.
How can I tell if a relay coil is rated for 240VAC I want to monitor a 240VAC line from an Raspberry Pi, my thought was to hook the 240VAC line to an appropriate relay then just sample the contacts from the Pi. I have a relay which I think would work but I don't know how to be sure the coil is rated for 240VAC. It's about 1 inch in each dimension, the coil resistance is 670k Ohms. I've attached a photo, I did google around for specs on this but I can't find any clear info. <Q> The coil voltage should be shown somewhere on the relay, though I don't see it on yours. <S> In any case, \$640\Omega\$ is way too low for 240V, as it will be dissipating 90W, so instead of a relay, you'd have a lightbulb and a house fire. <S> After looking at the data sheet (see TonyM's answer), and looking in the chart on page 2 for 'AZ-2110-1A-24D', we see that the coil voltage is 24V, and the winding resistance is ~\$660\Omega\$, which matches up nicely. <S> As for solving your problem, I'd recommend using an optocoupler. <S> Something like this: simulate this circuit – Schematic created using CircuitLab <S> Basically, it's a full bridge rectifier feeding the optocoupler (the led and transistor, the schematic editor didn't have a real optocoupler) with a current of ~5.5mA. <S> The two 22k resistors will need to be 1W or larger to remain within temperature tolerances, and need to be rated for greater than 120V. <A> According to this , its a Zettler AZ2110, contacts 1 Form A, coil 24D. <S> The specs on that webpage list the above variant as having a 24 VDC coil. <A> I also confirmed it is a 24Vdc relay <S> SPST <S> AZ2110–1A–24D <S> 24 <S> as in voltage for coil <S> 3/4 <S> Hp rated As @C_Elegans posted, With 240Vrms or + <S> /-340Vp <S> +/-??% the IR LED current will be <S> 7.7mA <S> Peak is to have a logic level pulse for each half cycle for instant detection by IRQ or sustain the output by some routine that samples the input, more than 1 cycle intervals then change the output from 1K pullup to 100K with RC= 100ms ( 10 cycles) , using C = 1uF using hysteresis or a Schmitt trigger on the input.	Your solution is unlikely to work, as the relay is going to buzz a lot, and probably start a fire.
Reading a potentiometer on MCU I want to read a voltage of 3.3V on my MCU's ADC input.I have a potentiometer that is to provide some feedback. I am wondering if it is necessary to add the R1 resistor, I added it since my MCU has a current limit of 8mA for GPIO, although I think this pin won't draw current but only read the voltage? simulate this circuit – Schematic created using CircuitLab <Q> No, you don't need R1. <S> In fact, you can't have R1 if you want the output to go all the way to 3.3 V. 100 <S> Ω is way too little. <S> That will draw 33 mA just sitting there doing nothing. <S> That could well be more than the processor. <S> The 8 mA figure applies to how much the pin can source or sink when configured as a output. <S> This has nothing to do with using it as a A/D input. <S> Use a 10 kΩ or so pot directly across 3.3 V to ground. <S> This is assuming that the A/D also uses the 3.3 V supply as its high end reference. <A> I have done this before with a 5 volt PIC MPU. <S> 1K potentiometers work the best. <S> They keep the impedance low and you can connect directly to the 3.3 volt source. <S> To filter out wiper noise I would add a 100nF capacitor to the analog input of the MPU. <S> If you insert a 100 ohm resistor in series with the input pin then have the filter capacitor you also limit noise transients from the Vcc line. <S> You will have to take care of 'scaling' the 0 to 3.3 volt range internally as software. <S> If you have a 10 bit ADC you have a range of 000 to 999 (1024 rounded to math.floor). <A> Assuming that you have a 10-bit ADC with a 0-3.3V range, your ADC has a 3mV resolution. <S> The range of possible values for your wiper voltage is 1.65V, which would span about 512 values on your ADC. <S> Allowing a bit for noise, say, you'd probably be able to effectively differentiate about 256 values. <S> It's about 1/3 worse on a 5V ADC. <S> If this is good enough performance for you, you're good to go. <S> If you do, just use a potentiometer with a higher resistance. <S> You can read up on what impendances your particular ADC can deal with. <S> In any case, it shouldn't suck much current. <S> I believe charge time of your sample and hold starts to kick in, though. <A> There are a few things you may want to consider: <S> Current draw <S> ADC inputs are usually high impedance. <S> So you dont need to worry about the input current on this pin and can therefore drop R1 as a current limiter. <S> However, current will constantly flow through your potentiometer. <S> If you use 100 Ohms only, the current will be I= U/R = <S> 33mA. <S> Depending on the circuit this is quite a lot. <S> You could go with a 10k potentiometer instead. <S> Input voltage range <S> Usually, the ADC value is scaled. <S> The input voltage is compared to some reference voltage (that may only be 1.0V for your MCU, please check the datasheet), and the result you get is basically Vin/Vref*2^NumberOfBits. <S> So make sure Vin is smaller than Vref (to both get useful results and not damage the ADC). <S> In order to achieve this, your R1 may be replaced by another resistor that is meant as a voltage devider instead of current limiting. <S> Make sure that the votlage drop over the potentiometer you select will not exceed reference/max input voltage of your ADC.	If it's not good enough, you can toss R1, if you don't care about the 33mA of current you need for the circuit
Controlling a 3-pin 12V 3A fan using a Raspberry Pi I have a Highfine FFC1212DE 120mm fan that operates at DC 12V 3.0A max and want to control it variably using a RPi. The third pin on the fan is a tach reading, not pwm control. I'm looking to change the speed of the fan based on temperature input so using a relay isn't a good solution. What is the best way to do this? Can I use a higher rated digital potentiometer? <Q> You can simply use a MOSFET: simulate this circuit – <S> Schematic created using CircuitLab <S> Based on the temperature, you have to generate a PWM wave at the digital out pin. <S> The MOSFET chosen should be able to withstand the current requirement of the fan and switch on at 3.3 V input. <A> Please keep in mind that you need a freewheeling diode across the terminals of the motor. <A> The pi's output is 3.3v and only a handful of milliamps. <S> You'll need to use something that can control 12v and can handle the amperage too- that is most easily done with a FET. <S> I'm partial to the NTD5867NL , which has a low enough gate threshold for 3.3v. <S> If you are doing on/off only, not PWM, you could use a relay. <A> PWM voltage tends to interfere with the Hall sensors commutation to make some audible aliasing noise, which you can reduce with a diode and big cap but why not a tiny series inductor and reverse clamp diode with a low ESR load cap at some high f (100kHz) and variable duty cycle , you now have a SMPS buck regulator. <S> Congrats but you can buy these for < $10 or less or even with remote <A> These fans won't work well for PWM control, it will interfere with the onboard commutation control. <S> You need to control the DC voltage fed to them, and will only get about 50-60% range in speed. <S> You could use a controller like this to get temp control without an MCU involved. <S> The board itself is MCU controlled, it uses a STM8S003F3P6 <S> so is quite capable, monitoring both temp and rpm feedback. <S> If you really want an R'Pi to control the set point/range <S> then you could potentially fake out the NTC thermistor on this board with an I2C driven digital pot. <S> You DO NOT need to put a diode across the fan (or indeed any of the PC fans produced) ... <S> there is no back EMF voltage generated by these fans, they are clamped internally in the commutation driver/controller	One way would be using PWM and controlling FAN speed with the duty cycle.
Why do we need a resistor here? In this circuit Why is there a resistor if the ground already is 0V? <Q> This circuit is supposed to act like an OR gate. <S> You need the resistor for the diodes to work correctly, especially in the off state. <S> $$\begin{array}{\|c\|c\|c\|c\|}\hline\rm A & \rm B & \rm D1 & \rm D2 & \rm Out \\ \hline0 & 0 & \rm off & \rm off & 0 \\ \hline0 & 1 & \rm off & \rm on & 1 \\ \hline1 & 0 & \rm <S> on & \rm off & 1 <S> \\ \hline1 & 1 & \rm on & \rm on & 1 \\ \hline\end{array}$$ <S> When A and B are both zero, D1 and D2 are both off. <S> Without the resistor, the output node would float, meaning it would be affected by nearby electric fields and static electricity. <S> In general, you should always have a DC path to ground for every node in your circuit. <A> The resistor is not intended to change the ground (which is our 0 V reference). <S> It is intended to ensure that OUT is pulled to ground when both the input signals are low. <S> Don't forget that the charge can't dissipate back through D1 or D2. <S> Update for clarification: simulate this circuit – Schematic created using CircuitLab Figure 1. <S> (a) <S> With one diode forward biased the output is a definite +5 V (less the voltage drop across the diode). <S> (b) With neither diode forward biased the output voltage is undefined. <S> They are effectively out of circuit. <S> Vout can float to any voltage between 0 and the reverse breakdown voltage of the diodes. <S> simulate this circuit Figure 2. <S> If the following stage presents a path to ground then the problem is solved. <A> This is also known as a pull down resistor. <S> The circuit is essentially an OR gate (as pointed out in the answer by Adam Haun). <S> When one of the inputs becomes a logic '1', the output becomes logic '1'. <S> If both inputs are at logic '0', the output is '0'. <S> Now, if you imagine the output is connected to something. <S> Without a pull down resistor, the output is 'floating'. <S> If some excess noise or static or some other type of electrical interference occurs, then it could cause the logic level of that output to become undefined, so it is 'floating' between a logic '1' and '0', and could have undesired results. <S> In an ideal circuit, the resistor would not be needed, but, alas, in the real world, things are not ideal. <S> The resistor ensures any stray interference has a path to GND, which keeps the output at 0V, or a logic '0'. <A> Another thing to mention, is don't assume that ground will always be 0V. <S> If that circuit was hooked directly to a battery <S> and there was no other circuitry, then it most likely would be 0V. <S> However, once you are adding other circuits that share a common ground and factor in noise (from a spinning tire on a car or factory equipment for example) that voltage may rise slightly and it could be enough to make your circuit read a logic 1 voltage. <S> Look up resources for Raspberry Pi and Arduino circuits. <S> All of their documentation is geared towards beginners. <S> They won't get as detailed as the college textbooks, but they'll give you enough info to understand what is going on without overwhelming you with a bunch of other info. <S> Pull Up/Down Resister Info: <S> https://playground.arduino.cc/CommonTopics/PullUpDownResistor <A> The simple way to explain it is this. <S> If OUT is directly connected to GND, then IN-A and IN-B don't matter. <S> OUT will always be low. <S> So, OUT cannot be directly connected to GND. <S> With the resistor, OUT will be low whenever IN-A and IN-B are both low. <S> But it is possible for OUT to be high when IN-A or IN-B is high. <S> If the resistor is removed, then OUT may tend to stay high even when IN-A and IN-B are both low. <S> The diodes prevent IN-A and IN-B from pulling OUT low. <A> If you don't add the resistor, the Output and ground will be in the same node therefore the output will always be zero volts	The reason the resistor is in place, is to ensure the output stays at 0V when the output is logic '0'. Another way of saying it is that R is to ensure that any stray charge on OUT is discharged to ground and that OUT falls to 0 V rather than float at an undefined level when IN-A and IN-B are low (0 V).
What is the most accurate way to measure internal resistance of a consumer battery? The typical way to measure the internal resistance of a battery, that I've found through research, is by connecting the battery in a circuit with a resistor, measuring voltage through the battery, calculate current, measure voltage through the resistor, find the voltage drop and use kirchoff laws to calculate the remaining resistance, which would be internal resistance. A professor mentioned that this is too inaccurate for batteries with very low internal resistances, so is there another, more accurate way to test the internal resistance of a standard battery? <Q> The easiest is to use an AC current, just like a network analyzer. <S> You can use a switchable current sink of known value, or a resistor and a FET. <S> If you keep repeating this on/off cycle, this is equivalent to using an AC current. <S> An advantage of AC measurement is that you can use a capacitor to get rid of the DC, and only handle the small AC voltage. <S> It also ignores any DC offset in the signal chain, from opamps etc. <S> You can use something like this as an experiment: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here the voltage source on the left drives an AC signal through R1 and C1 into the battery. <S> This is AC coupled via C1. <S> C2 extracts the AC voltage on the battery. <S> Knowing AC voltage V1, once you measure the AC voltage on "OUT", battery internal resistance is easily calculated, as it forms a resistive divider with R1. <S> You can do this with a soundcard, or with a function generator and a multimeter. <S> Make sure you calibrate and verify your test rig by replacing the battery with a resistor of known value, which should be of the same order of magnitude than the expected resistance to be measured. <A> The objection by your professor to the method you described is wrong. <S> If a battery has low impedance, you just increase the load, so the difference can be better measurable. <S> The problem, however, is quite deeper. <S> Batteries usually don't have a steady DC load, and often people need to know the battery reaction to impulse load. <S> The issue is that the battery doesn't have a single "impedance" that can be used for estimation of its response to load. <S> Every battery have a spectrum of impedances, different impedance for different frequency of loading. <S> This topic of engineering is called "Electrochemical Impedance Spectroscopy". <S> An example of presentation . <S> See also this article from BatteryUniversity , with an example of "electrochemical spectrum". <S> So, the most accurate way to characterize internal impedance of a battery is to measure its spectrum, using, for example, the circuit suggested by peufeu . <S> However, this is not all. <S> The EI-spectrum is usually collected using "small signal" applied. <S> As Lorenzo Donati commented below, the value of impedance is not only dependent on frequency, but its is also nonlinear with amplitude of applied signal/load, which adds another dimension into complexity of the problem. <A> Internal resistance is an approximation of a complex V(I) function of a battery, so there is no "true" internal resistance value you could accurately measure in the first place. <S> The most representative measurement you can do is the actual V(I) curve for relevant current values. <S> If, by chance, you obtain a straight line - bingo, your battery can be accurately described by a single internal resistance value. <S> Otherwise, you'll have to use the curve you've got to determine current and voltage values for any given load.	Measure voltage with current on, then with current off, substract, divide by current, you get internal resistance. AC measurement is the most accurate method.
Why is a current signal preferred to a voltage signal for long analog transmission? Some sensors act like current sources, and I have seen it several times, especially for very long wires even at outdoors like wind vanes. 4-20 mA current loops are used instead of 0-10 V voltage for instance. What can be the physical explanation for this? How is current more advantageous? (I'm also wondering in terms of EMI interference whether a current loop signal is more immune and why.) Please explain this concept by using circuit diagrams, voltage current sources with some components. How common mode interference is coupled in both cases, etc. and why a current loop is immune to noise. EDIT: After reading the answers, here is what I understand(click to see the simulation diagrams and corresponding plots): I apply common mode Vcm interference in all scenarios. In the first top figure a current source with 1Giga Ohm impedance is transmitted via an unbalanced/inbalanced cable and even the receiver is single ended the output is immune to noise. (1G Ohm makes the noise small, the lesser this Rcur the more the noise at receiver) In the middle figure a voltage source is transmitted via an unbalanced cable and the receiver is single ended , the output is very noisy. In the bottom figure a voltage source is transmitted via a balanced cable and the receiver is differential-ended , and common mode noise is eliminated. Is my conclusion/simulation correct to represent this question? <Q> Actually what matters for immunity against noise is the power that is needed to disturb the singal. <S> I.e. a current signal at an input with nearly zero impedance is just as bad asa voltage signal at an input with nearly infinite impedance. <S> What is needed is a receiver with non-zero as well as non-infinite impedance so that the signal involves some power . <S> I.e. if the information is coded as voltage, there still should be some current flowing into the receiver and if the information is coded as current, there still should be some voltage across the receiver. <S> So both cases are similar, but you just have decide whether is is better to code <S> the signalas voltage or as current (another alternative would be coded as power).For measurement purposes voltage or current signals are most appropriate. <S> A good wire for a current signal just needs to ensure that no current is lost (or inserted), i.e. ideally no leakage, i.e. perfect isolation. <S> This can be accomplished in practice quite well. <S> A good wire for a voltage signal needs to ensure that no voltage is lost, i.e. ideally no voltage drop, perfect conductance along the wire. <S> Unless you are using a superconductor this is almost impossible to accomplish in practice. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In any case receiver resistance should be well above 0 and well below infinity. <S> It's easy to have the isolation resistance practically infinite. <S> It's practically impossible to have the series resistance 0. <S> Therefore if the signal has to be sent down some distance along a wire it is better to use a current signal than a voltage signal. <A> Current is great in that it is equal at all parts of a conductor. <S> I.e. if you are pushing in 15 mA from one side, the other side is seeing 15 mA even if it is 200 m away. <S> This is very easy to sense and makes data transmission reliable. <S> The same is not true for voltage. <S> If your conductor has a high impedance and has electrical interference, then your input voltage signal will degrade and a valid voltage may not reach the other side. <S> See here why this matters: <S> Why are high impedance circuits more sensitive to noise? <A> Current signalling has different advantages in different situations, so there are several different answers. <S> In the case of low frequency signalling. <S> A constant current source (sender) has a very high impedance (and a CV one has a very low impedance). <S> So when you put fairly high series resistance in, it has no effect: the CC source is already super high, what effect is a few hundred/thousand extra ohms going to make? <S> Likewise when you couple noise into the cable (C1,2) <S> the high source R means that both wires go up and down together - it is common mode noise and has no effect on the current. <S> Meanwhile the receive end has a low R. <S> This damps down any capacitively coupled noise, and is robust. <S> A voltage system is the opposite. <S> The source should have very low impedance. <S> Series R is going to matter. <S> The rx needs to be very high input impedance or you get a voltage divider. <S> It will capacitively pickup noise, and will be prone to damage. <S> Capacitively injected noise flows through RSource, and you get differential voltages at the receiver. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In the case of high frequency signalling (e.g video) <S> The current loop has basically constant voltage on both sides of the cable. <S> Therefore capacitance across the cable does not pass any current, and does not have any effect. <S> The signal is immune to cable C, and is immune to extra C added to protect from noise and emi. <S> Much less power is used as C does not have to be driven. <A> As far as I'm concerned, these are the two main reasons for choosing current loops in several cases: <S> You don't care about the length/resistance of your wires. <S> You can change a 3m wire to a 50m one, changing its resistance, the signal will be the same (as long as the source can deliver enough voltage/power, of course). <S> You can detect damage and failure. <S> If you get 0mA, either you sensor or your wire is broken. <S> With voltage loops is not that easy to figure out. <S> About EMI, it won't affect most of the times. <S> EMI usually comes at (very) high frequencies, way faster than your signal changes, so you can filter it. <S> Also, it seems this is related to the old pneumatic controls systems, where 3-15psi range was used. <A> Something else to remember regarding analogue signals is the ability to then integrate the HART communication protocol. <S> HART (Highway Addressable Remote Transmitter) is a digital signal which is overlaid on top of the analogue signal allowing for additional information to be sent via the same wiring. <S> Most smart industrial instruments nowadays operate with HART capability. <S> So the benefits are far greater than just voltage drop and EMI.	The noise immunity comes from the fact that current loops are a low impedance system.
How can voltage of a battery be calculated? If voltage is calculated using the formula V=IR, how can the voltage be calculated of a battery if it is not connected to anything (i.e. no current)? I am not 100% sure about this, but neither do I think there is a resistance. Thus, how can V be calculated when both I (current) and R (resistance) are unknown? <Q> The theoretical voltage is determined by the Nernst Equation . <S> You can look up the half-cell potentials. <S> There is a temperature dependence which the Nernst Equation will predict. <S> Here is a simple undergrad lab where the students are asked to work this out for an ordinary lead-acid battery. <S> In practice electrical engineers are usually going to refer to the battery or cell datasheet for the range of voltages and the (usually more important) characteristics under various discharge conditions (and charge conditions, where applicable). <A> If the battery is not connected to anything, no current is flowing but voltage still exists across its positive and negative terminals. <S> It is called EMF of the battery. <S> It's related by standard reduction potentials of its electrodes in chemistry as:$$ <S> E <S> = E_{cathode} <S> - E_{anode} <S> $$ <S> When the battery is connected to a resistor R, some current I will flow through R, as well as through a small series resistance r, which is the internal resistance of the battery. <S> So effectively, we will get a terminal voltage, V across R, which is lesser than the EMF of the battery. <S> Relations will look like: $$ V = <S> E - Ir = IR$$ <A> Voltage of a battery cannot be "calculated", it is a given, inherent property of the electrical element. <S> The potential V exists regardless of load or no load. <S> But if you connect a load R, then the current across the R will be I = <S> V/R. <A> If voltage is calculated using the formula V=IR, how can the voltage be calculated of a battery if it is not connected to anything (i.e. no current)? <S> We don't calculate voltage, we measure it. <S> To do so generally requires that we draw some current from the battery but only a tiny amount. <S> Most digital multimeters have an input impedance of about 10 MΩ and hooking this up to a 9 V battery would cause a current draw of \$ \frac {V}{R} = \frac {9}{10M} = 0.9\ <S> µA <S> \$. <S> This is unlikely to cause a problem in most cases. <S> Figure 1. <S> The ICE Supertester 680R analogue multimeter showing the DC and AC loading. <S> The author had one of these for many years. <S> The old analog multimeters were typically 20 kΩ/V and the user had to take this into account when making a reading. <S> For example, when on the 10 V DC range the load on the circuit would be 20k * 10 = 200 kΩ and this could load a high impedance circuit significantly. <A> It is for the special case where you have a known resistance (R) and a known current thru that resistance (I). <S> Neither of these apply to "calculating the voltage of a battery" , whatever that is supposed to mean. <S> To a first approximation, you can model a battery as a voltage source in series with a resistance. <S> Put another way, you model it like a Thevenin source. <S> To find the value of the voltage source, simply measure the battery with nothing connected. <S> Since there is no current (the current drawn by the voltmeter is so small that it can be ignored in this application) thru the resistor, the voltage across it is 0. <S> The battery's open circuit voltage is therefore the voltage of the internal voltage source. <S> To find the series resistance, apply a modest load on the battery, and measure the external voltage drop. <S> You assume that the internal voltage source remains the same, and that the drop in external voltage is due to the drop across the resistance. <S> You find the resistance by Ohm's law: <S> Ω = V / A, where V is the voltage drop across the internal resistance and A the battery current in Amps.	The open-circuit voltage of a battery is based on the electrochemical potentials of the constituent materials. V=IR is not a universal way to calculate voltage.

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