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Temperature Dependent Frequency of Relaxation Oscillator I built a relaxation oscillator to convert a capacitance into a variable frequency square wave. The passive components have a very small temperature coefficient, but when we change the temperature of the LM2903HYPT comparator the frequency shifts by around 8kHz. How do changes in the comparator cause the frequency of the circuit to change? It normally operates at around 150kHz or so. How can I remove these temperature dependent changes in future designs? Thanks!-Hunter <Q> The comparator will also have some offset voltage shift with temperature, the bias current will change, and the response time will change. <S> If I was going to make a wild-a <S> ** guess I would say it's probably the response time , which will be probably several microseconds with a ramping input <S> (note that the time is specified as 1.3usec typical , only at 25°C, and with an overdrive of 100mV) -- so a change of only maybe +/-100ns will account for your measured change. <S> The output voltage could too, but only with a rather large change in temperature (which you did not deign to <S> mention- <S> it's kind of important). <S> What you can do to make it better next time- <S> control the voltage that sets the thresholds and the integration current better, pick a comparator that is much faster than the required switching time and that has very good DC stability. <S> As an out-of-the box suggestion, you can get much better performance even with a crap circuit by making two circuits or switching inputs so that you measure your unknown capacitance and a reference NP0 capacitor. <S> This will also cancel out some other sources of error such as resistor drift (use 5% resistors if you want), and even a certain amount of power supply noise. <S> That's what I would consider if I was making a whole bunch of them. <A> The attraction of that type of relaxation oscillator is that the output voltage goes back to both the timing capacitor and <S> the threshhold setting resistors, so to first order, the actual value of output voltage does not matter. <S> Comparators are not precision parts. <S> This one has typical propagation delays in 0.3 to 1uS, depending on overdrive. <S> Your half period of 3uS is not much more than that, so any small changes in delay will cause proportionately large variations. <S> A second order effect of the change in output voltage will be a change in slew rate, which would change the propagation delay. <S> If you increase the value of R11 by an order of magnitude, it will increase the period of the oscillator, which should reduce the magnitude of variations in comparator delay. <S> Unfortunately I would expect it to increase delay variation (but not as much as a factor of 10) as it reduces the slew rate. <S> Check out this answer which suggests using one or more amplifiers to increase the slew rate before a comparator. <S> Reducing the frequency, then increasing the slew rate with an amplifier, could well give you much better temperature sensitivity. <S> An amplifier with adequate GBW will have its performance almost completely controlled by the gain setting resistors, rather than thrashing around open loop as a comparator does. <S> Note this is not using the amplifier as a comparator, merely as a slope amplifier. <A> Thanks so much! <S> Both of your suggestions were very helpful. <S> I tried replacing R11 by a 118K resistor. <S> This brought the oscillating frequency down to about 20-30kHz and seemed to greatly reduce the temperature dependence. <S> Thanks so much!-Hunter
| The saturation voltage of the comparator is one obvious contributor.
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Are slew rate frequency and cut off frequency same? I am bit confused about the cut off frequency and the max frequency above which the output response starts to distort which is slew rate frequency. Are these two things same? Cutoff frequency (f) = GBP / Av Slew Rate Frequency (f) = SR/(2*pi*Vp) Is cutoff frequency = slew rate frequency? Thanks <Q> They are not at all the same thing. <S> The slew rate frequency is a non-linear large signal phenomenon. <S> It has to do with how fast the output can change in response to a large input signal. <S> It's due to the finite amount of current available to charge <S> /discharge the dominant pole compensation capacitance. <S> Slew rate limitations will make a sine wave with high enough amplitude and frequency distort to start to look like a triangle wave. <A> There is a third cause of distortion. <S> And it also is frequency sensitive. <S> The Input DiffPair also affects distortion. <S> Bipolar diffpairs, without emitter resistors to linearize, have predictable IP2 and IP3. <S> These numbers, somewhat different for IP2 versus IP3, are near 0.1voltsPP across the Pin+ to Pin-. <S> That means 0.1volt at 1KHz would produce 0.1volt at 2KHz. <S> This is IP2. <S> Dropping down by 10:1, to 0.01volt across Pin+ to Pin-, would produce 0.001 volt at 2KHz. <S> In dB, the 2nd order distortion would drop dB_for_dB. <S> For the 3rd order, given 1KHz and 1,100Hz of level 0.1 volt each, the behavior of 2*F1 +- 1*F2, and 1*F1 +- 2*F2 will produce "shoulders" at 900Hz and 1,200Hz of level 0.1volt also. <S> Dropping the level of 1Khz and 1,100Hz by 10:1, thus 0.01vpp for each, will produce "shoulders" at 900 and 1,200 at 0.0001vpp. <S> In db, the 3rd order distortion would drop 2dB_for_db. <S> Why is this important? <S> At low frequencies, the huge open loop gain is your friend, and causes voltage of Pin- to Pin+ to be tiny, implementing "virtual Ground". <S> At high frequencies, the dropoff in open loop gain causes the voltage of Pin- to Pin+ to grow 10:1 as the frequency increase 10:1. <S> And large voltages between input pins is what causes the diffpair (bipolars or fets) to distort. <S> Example: opamp UGBW is 100MHz, your circuit is unity-gain, your frequency is 10MHz, and your input voltage is 1voltPP. <S> The gain at 10MHz is only 10, thus the input error (how much the virtual Ground differs from Zero Volts) is Vout/Gain = <S> 1vpp/10x <S> = 0.1voltpp. <S> Thus the distortion behavior we predicted in our first paragraphs is our distortion situation here. <S> And as I cautioned, IP2 level is not same as the IP3 level, but they are only a few dB (4:1?) apart. <A> Short answer: <S> Primarily, the slew rate plays a role in high-gain amplifiers (opamps) with feedback. <S> The amplifying capability of such amplifiers has frequency limitations which are characterized by two different bandwidth definitions: (1) Small-signal bandwidth - determined by the small-signal open-loop gain response (gain-bandwidth product) (2) <S> Large-signal bandwidth - determined by the slewing capabilities of the amplifier with feedback: B=SR/(2*Pi*Vmax). <S> Within this bandwidth sinusoidal input signals will be amplified without any remarkable distortions for amplitudes not larger than Vmax.. <S> Larger frequencies will cause gain reduction, triangular distortions and additional phase shifts. <A> Is cutoff frequency = slew rate frequency? <S> Just look at the formula for the slew rate frequency: - Slew Rate Frequency = <S> \$\dfrac{S.R.}{2\pi\cdot <S> Vp}\$ <S> It contains an amplitude value <S> Vp. <S> This means it is dependent on the amplitude of the output signal. <S> The cut-off frequency is not dependent on anything other than how the device is designed/built. <A> They are different. <S> For example, if an op amp is outputting a small waveform within the gain-bandwidth requirements, all is well. <S> Then, you try to output a bigger signal that exceeds the slew rate, and the output gets very distorted.
| Assuming we are talking about an op-amp amplifier, the cutoff frequency is the point where the small-signal gain is down by 3dB.
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How can I slow down a 240V AC motor used for an exhaust fan? I've got a fan that is way too powerful and would like to find out how to slow it down. Variable speed would be nice but I would be just as happy if I could say reduce it to a fixed speed i.e half speed. It currently only has an on/off switch. The specs on the fan are: 240 volt 50Hz 1500 watt Single speed: 2900rpm My ability level is enough to pull things apart and solder but I have no education in electronics. Any insight into this would be great. I'm just going to live with the fan as is and reduce air flow by other means to suit my needs. Thanks everyone for educating me a little more about fans and electric motors. <Q> The fan most likely has an asynchronous AC motor and this means the speed of the arrangement is bound to be near the synchronous speed (of 3000rpm at 50Hz). <S> You cannot lower it a substantial amount without having substantial losses within the rotor. <S> Which means much additional heat. <S> As the fan motor is cooling itself by rotating the fan, this is double bad. <S> So the only reasonable way to change the speed of the motor (and the power of the fan arrangement) is lowering the frequency. <S> You would need a VFD for that. <S> For 1.5kVA, it would be quite expensive. <S> Better use a smaller motor/fan arrangement. <A> In addition to what Janka said, if your fan's motor isn't directly connected to the blades, but uses a belt instead, you could alter the size of one of the pulleys to change the speed that it spins at. <S> The motor will still be spinning at the same speed but less stress would be put on it, as it won't be moving as much air. <A> I wonder if there is such thing as XY- answer <S> ;) <S> If the goal is to reduce air flow then it can be easily done without slowing the motor by using smaller blades. <S> I would look for ready-made fan blades with different diameter. <S> If that is not feasible here are some DIY options: <S> if you have even number of blades you can cut two opposite blades off; if the blade material is not too thick at the hub you can twist them to reduce pitch; as the last resort you can shorten the blades, but be careful to cut exactly same length, otherwise you'll get nasty vibration. <S> BTW, if by "reduce air flow by other means" you meant to restrict it somehow (e.g. by covering part of the opening) <S> then it is not a good idea, most likely result will be motor overheating. <A> I'm going to assume this is a single phase motor. <S> So simple speed control CAN be done on CERTAIN TYPES of single phase AC induction motors, called Shaded Pole (SP) motors and as it so happens, that is a common type used on fans because in general, SP motors are not very powerful. <S> Shaded Pole motors are inherently current limiting so in slowing them down, the motors WOULD overload, but can't because the current cannot increase so they just lose torque. <S> If your motor is Shaded Pole, it should say so on the nameplate, or it should at least say "SP". <S> There are some types of fans that use what is called a Permanent Split Capacitor (PSC) motor, they too can be controlled by simple voltage reduction, but are at greater risk for long term damage if they over heat. <S> Many overhead ceiling paddle fans are PSC type and can have their speed reduced with a dimmer because the case is designed to dissipate that extra heat. <S> If you are not talking about a ceiling paddle fan, I would not attempt it. <S> Basically if your motor has a big "bump" on the side, like it is about to give birth, that's an indicator that it is not SP or PSC <S> and you cannot change the speed.
| With a Shaded Pole motor, you can simply reduce the voltage, say with a light dimmer and the motor produces less torque, then with less torque the air friction on the fan blades causes it to slow down, which in turn moves less air.
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How does voltage know How is it possible that in scenario 1 resistor R1 converts 5V of energy to heat and in scenario 2 the same resistor R1 converts 2.5V of energy to heat? The resistor R1 in both scenarios are exacly the same, I don't seem to grasp what happens on the physical level to the electrons how they "know" when to dump all their energy at once in resistor R1 in scenario 1 or to divide the energy they have proportionally over the resistors R1 and R2 in scenario 2. I have read an explenation here , but it's still not clear. How is it possible that the voltage is different at point A going through the same amount of resistance? I would expect a linear relation between voltage and resistance. Could someone explain what physically happens that causes this? simulate this circuit – Schematic created using CircuitLab Edit Is my understanding right? According to here The resistance drops liniar along the path. In scenario 1 at the start of R1 the resistance the current encounters is 100\$\Omega\$ and at the end of R1 it encounters 0\$\Omega\$. In scenario 2 at the start of R1 the resistance the current encounters is 200\$\Omega\$ and at the end of R1 it encounters 100\$\Omega\$. If this is true, why does point A in scenario 2 have more volt left if it has encountered a higher resistance of 200\$\Omega\$ at the beginning opposed to scenario 1 where it encountered 100\$\Omega\$ at the beginning? <Q> As you may anticipate, the electrons don't actually know anything. <S> You may think it as its hydraulic counterpart: pressure. <S> Such that pressure drives the water through tight and wide pipes at a certain flow rate, voltage drives electrons through resistance with a certain current. <S> If you are interested in a more "scientific" explanation, this is what actually happens: <S> You connect the battery, complete the circuit <S> The electric field starts to propagate inside the wires, exerting force on electrons while doing so <S> Some materials show more resistance to electrons and thus slow them down <S> The slowed down electrons apply their own electric field behind them, taking away some of their potential to move on. <A> How is it possible that the voltage is different at point A going through the same amount of resistance? <S> In scenario 1 it is 0 volts (as you have labelled it). <S> In scenario 2 is is half the voltage applied i.e. 2.5 volts. <S> I would expect a linear relation between voltage and resistance. <S> Not when potential dividers are used unless both resistors change value proportionally. <S> Ohms law, I = V/R <S> so, in scenario 1 the current is 5/100 <S> = 50 mA. <S> In scenario 2, I = <S> 5/(100 + 100) <S> = <S> 25 <S> mA. <S> Ohms law again: <S> V = IR so the voltage across R2 is 25 mA x 100 ohms = 2.5 volts. <A> The electric fields do the "knowing", do the "exploring", do the "urging". <S> Electrons explore all possible paths, to get back home to the battery's other terminal. <S> Electrons will go meters off to the side, just because that is another path back from whence they came, even if only 1 electron/second explores that path. <S> Why do electronics do this exploring? <S> because the electric fields urge/push/drive that exploration. <S> Our circuits often include huge volumes of air, and we think our circuit is just the copper wiring and the resistors and capacitors and ICs and PCBs. <S> But the air is there, extremely high resistance, so we ignore the air. <S> In your two circuits, the electric fields encounter different length paths, and the volts/meter is different. <S> Most importantly, the volts/resistor is different and we use Ohms Law to handle that situation, while we ignore the air. <A> I have found an explanation what I was looking for here , here and here .I think the main thing I needed to realize is as @OlinLathrop and @GregoryKornblum mentioned in a comment, that voltage is not energy itself, voltage is just a difference in "height" so to speak and it tells nothing about the electron itself. <S> It only tells how much energy the electron would have gained if it rolled down the hill. <S> In the following illustration the voltage is not the ball at the top of the hill, but the voltage is the top of the hill and the ball can be used to transform the energy from potential energy to heat/light or in this example to kinetic energy if it hits the ground. <S> And as to the diagram here, the 2.5V is there because its not at the ground level yet (0V) and the real energy can be seen in the current because that is the energy carrier and if the current is slowed it loses potential energy. <S> I think the resistance in this situation can be seen as if you drop the ball from the hill and blowing wind <S> agains it upwards (creating drag) <S> it slows down and when it arrives at the ground it has less kenetic energy because its speed was reduced by drag and thus lost gravitational potential energy. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Correct me if my understanding is still incorrect.
| Voltage is actually a battery's difference of potentials of its poles to drive electrons.
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There is Hardware, Software, Firmware and ...FPGAware? I think by this day and age we all probably have a pretty good understanding of the distinction between hardware and software. In addition it is widely accepted that firmware is software that is injected into an embedded system in a way that it is not readily changeable for each time you power up and run such system. Many designs of systems use FPGA components that are "programmable" in the sense that they load an image from some non-volatile FLASH memory at power up that configures the FPGA to perform some intended functionality in the system. My question here is what is the generally accepted name for the FPGA data collateral that defines its functionality? As you can see in the title I have made up the loose term FPGAware but I rather doubt that it is the term really in use. For purposes of this question let us consider that the FPGA consists of basic logic that includes gates, flip flops, registers, sequencers and state machines. The FPGA would NOT include a fixed or synthesized processor core that runs some "software". <Q> In a project, engineers typically call a functional unit by its function. <S> If they have a baseband processor in programmable logic, they will call it "the baseband". <S> Likewise for modems, rf frontends, crypto-cores, network processors etc. <S> Firmware is by no way "firm" in the development process. <S> Exactly this is what makes programmable logic so attractive for development. <S> The software vs firmware vs hardware distinction is more about how easily a unit can be replaced in the field. <S> Since FPGA-bitstreams are often stored together with other firmware in non-volatile memory, its OK to also call it firmware ... since the process to update it is the same from a user perspective. <A> Software is binaries that are "soft" they live in ram and are lost when the power goes out, you load then when you use them, discard them when you are done. <S> Firmware is a bit more "firm" than software, it tends to be programmed once and stays there, is always used, not loaded and discarded, not temporary, it is more firm than software. <S> The bits that are used in a flash next to the fpga to make it work are no different than the bits in the flash next to some other chip, both are firmware. <S> Because they are firm, not soft. <S> The programming languages are programming languages <S> I dont have a problem calling vhdl nor verilog a programming language because it is. <S> Some folks want to say hdl or rtl instead <S> and that is just fine too to describe the source code just like saying python or C or high level programming language. <S> Nothing wrong with saying FPGA code or FPGA source code either. <A> I'm not quite sure if you are asking about what to call it prior to synthesis or post synthesis, so I'll add my 2p worth for both. <S> Prior to synthesis, you have either have code based or schematic based hardware descriptions or a mixture depending on which you prefer. <S> These are what I would simply call "source code" , "source files" or just "code" . <S> Once synthesised you have what I would call simply "firmware" . <S> The compiled bitstreams or object files are no different from the machine code files that would be used for an embedded processor. <S> It really doesn't make much difference to the definition that one contains machine instructions while the other contains initialisation information for the various LUTs and memories in the FPGA. <S> I would also go one step further though and distinguish fully compiled complete firmware (that you can upload as is to the FPGA) from its building blocks. <S> The building blocks of the firmware are typically referred to as "IP cores" . <S> These can be pure source code, or fully synthesized "netlists" which can be integrated as part of a larger design. <S> This is pretty much akin to libraries in the software world. <S> As a side note, I've personally never heard of the term FPGAware, nor would I ever use it - even the thought of its existence makes me cringe. <A> "Gateware" is one term I have seen used. <S> But to me, it's all firm/software - it can be changed. <S> It happens to configure some hardware elements, but key to me is that it provides functionality and that functionality is not fixed. <S> (Having said that, I see others' point that calling it "software" implies some things that are not valid to people less familiar with FPGAs...) <A> Just like Tom Carpenter, in the pre-compiled state, the description is simply "code", "source code". <S> For the/synthesized thing that you load onto your FPGA, I'd say "Configuration", because configure is the verb commonly used for the process of loading the "FPGA image" (another term that describes the configuration in their stored state) onto the FPGA. <S> We often speak about designs , which is kind of nice, because it highlights the fact that, just like a schematic and board layout leading to a PCB design, HDL / visual design approaches lead to an FPGA configuration.
| They would both be considered firmware in the sense that they are permanently compiled and stored into what is essentially a read-only state.
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Modifying USB Battery (Portable Power Banks) to charge while charging (to use as ups) I've been checking all the questions and can find many similar ones, but unfortunately none that addresses my problem. I ended up reading many interesting questions and answers (including e.g. this one Can i connect Power source (like usb charger) parallel connection with power bank output for keep Uninterruptible power supply (UPS) ) I however already have a few PowerBanks mainly something like this . I would like to use one of them as a ups, but my problem is it does not seem to charge when it is charging. Is that a "feature" of the battery, or is it to "protect" me from wearing the battery down? How difficult/dangerous would it be to disable that feature? Could i just go ahead and get the power directly from the end-poles of the batteries bypassing all circuitry, or that would be dangerous? Last, I have a second brand of powerbank, this one. Here things are even more complecated, as the battery requires me to press a button for it to start charging... Is this standard? I have seen it be done, since I have found that there are similar power banks out there that can fact provide while charging e.g., with other power banks, e.g. http://raspi-ups.appspot.com/en/index.jsp . It also seems to be an "ok"-ish thing to do ( https://www.quora.com/Is-it-okay-to-charge-your-phone-through-a-power-bank-while-the-power-bank-itself-is-being-charged ) <Q> The power bank will not power the targe while charing. <S> The reason is simple: <S> The wall wart (power supply) is not able to deliver enough current to do so. <S> Most will also use standard USB cables which do not allow much current above 2A. <A> Power banks are a class of devices to serve as emergency power sources in stand-alone mode. <S> You charge it, and keep in your briefcase just in case. <S> As such, most power banks are not designed as UPS (uninterruptible) supply. <S> Some power banks even provide a warning label, saying "Do Not Charge & Discharge At Same Time!!!" <S> The powerbank can not be modified as UPS without replacing its active electronic circuit. <S> The new circuit must contain the charger for battery with direct path to VBUS , AND an additional automatic power switch from direct feed to the battery booster to OUTPUT if external power is lost. <S> Texas Instruments has a sizable portfolio of PMICs (Power management ICs) where this function is integrated into controller IC, as BQ24075 for example. <S> Other manufacturers as Linear Technology, Maxim Integrated, Richtek and others have similar ICs. <S> This board from SparkFun can likely do the job. <S> ADDITION: Here is a solution for 1-cell UPS offered by Maxim. <A> You can get bypass enabled POWER BANKS. <S> Otherwise, it's not possible. <S> Only few power banks support it, we can charge while its in charging mode. <S> MI power bank supports it, and it also supports the Quick charge3 for Qualcomm Processor phones. <S> Example: <S> MI Powerbank2i Link: https://www.mi.com/in/10000mah-mi-power-bank-2i/
| While some banks can do this by design (as the OP has discovered this himself), it does not mean that all power banks would charge its internal battery and provide power to output port.
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Using two power supplies to power microcontroller I have a condition where I need to power MCU with the second source when first is not working. MCU I am using is ATMEGA328p . The circuit is solar charge controller. I am using a battery to turn on the MCU. Battery voltage is going into LM7805 and its output (5v) is going into the VCC of MCU. Now I also want to use the solar voltage (whenever available) to give power to MCU, so that if battery is faulty or is not working, then solar voltage will be used to power the MCU, thus at least load will on during day time. I have the below schematic in my mind. Now I want to know if battery is healthy and is giving voltage ~12v and during sunlight solar panel is also giving ~20v, then what will happen in this case.? Is the above design safe to use. If yes, then what type of diodes should I use. During day time both (solar + battery) voltage will be going to the regulator, will it create any trouble.? Please help. Thanks. <Q> Your circuit will work as expected. <S> Diode: You should use diodes with low forward voltage drop, to reduce power dissipation. <S> Also you need some caps around the 7805 to stabalize <S> it's regulatoin. <S> However in this case I would suggest the usage of an DC/DC converter like the TSR 1-2450 . <S> It's a plug and play replacement for the 7805. <A> The highest input voltage will win, and it won't create any trouble (basically, it means solar panel will have priority if its voltage is >12V and will be unused if <12V). <S> Given the input voltages seem much higher than what you need at the 7805 input, and current requirements are probably low, any diode will do. <S> If you had voltages much closer to the limit of what is required for 7805 (5V + <S> ~1.5V dropout at regulator), you would want to use schottky diodes to reduce the drop. <S> Now, this is all fine if the current requirements are not more than about a hundreds of mA. <S> If it is more, the 7805 will heat too much and waste a lot of power. <S> In this case, use a DC-DC converter. <A> Your basic concept is OK, but there are some issues: <S> You really need to add input and output caps close to the 7805. <S> Read the datasheet. <S> This circuit will work with 20 V in, but consider the power dissipation. <S> Let's say the 5 V circuitry draws 100 mA. With 20 V into the 7805, it drops 15 V and dissipates 1.5 W. <S> A 7805 can handle that, but will need a heat sink. <A> So pay attention the PV panel is large enough and remember that this power will drop earlier in the evening and be available later in the morning.
| One thing i don't see they mentioned yet is that the power itself may not be a problem as long as you can actually supply it.
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protect PC against noise I have a PC that controls some equipment, such as, cameras, solenoids, and DACs. There is also a 45KW nitrogen laser. The laser has its own service installed, but, when it turns on, there is a voltage spike that disrupts the USB devices on the PC. The PC is on a surge protector, but the spike still affects the PC in this way. When this happens, shutters open, and the general operation is disrupted. One option is to avoid the issue, and shut down the PC, power on the laser, and then turn the PC back on. This works well, but is tedious. I have talked with the building manager, and he will not take any measures to improve the quality of the electrical services available (such as hire an electrician to install a Whole-House Suppressor). I am looking into putting ferrite beads on the existing USB cables, but those have not arrived yet. Are there other options for noise suppression of this nature? <Q> Obviously your overall setup (with all USB devices and associated USB cables) was never tested to IEC 61000-4-x standards, especially to 4-4 EFT ( fast transient ) section. <S> The position of your building manager is understandable. <S> Before installing some extra equipment, you need to find the root cause of your issue. <S> If the interference is coming from AC power mains, a battery-powered UPS is an easy solution. <S> But if your PC doesn't reboot, the power is not likely the cause. <S> If USB is disrupted, you might want to look into the scale of communication disruption, whether any individual USB devices are disrupted, or the entire PC root hub gets reset and/or massive disconnects are happening. <S> However, most regular PCs have the shielding routed in completely wrong manner (shield is usually directly connected to signal ground plane on most mainboards at USB connectors), the ferrite beads might be not enough. <S> In this case a good total (optical) isolation will solve the problem, see USB3.0 optical cables from Corning Communication . <S> This would be the best, although a bit pricey. <S> [ CORRECTION : <S> The Corning cable does not provide galvanic isolation between grounds on connectors, as some audiophiles have discovered, and tolerance of data signals to common-mode is not specified or known. <S> Corning didn't return the request for clarification ] <S> USB has a limited range of how much of common-mode shift it can tolerate (officially about 1-2V, or less for HS). <S> If the disruption comes from excessive bouncing of common ground across multiple power sources supplying your USB devices, you may want to reduce the ground loop area by consolidating all power bricks into one area if possible, and use another local UPS to power them. <S> Overall, fighting Electromagnetic Interference is always a challenge, so good luck to you. <A> Your PC is probably suffering from a brown-out, a surge protector will not help with this issue. <S> If you do not need steep edges on your laser (a 45 kW laser is probably not modulated for communications?), then limit the inrush current to the laser. <S> There are some thermistor based limiters available for the mains connection. <S> If you cannot change the mains connection (45kW are not plugged to the wall outlet, right?) <S> , you should consider an online ("continous conversion") UPS for the PC. <S> If it turns out that your PC really resets due to radiated electromagnetic interference, I would think the whole setup is not FCC compliant. <S> That would be a bigger problem unless you are in a shielded lab environment. <A> Scenario 1: Brownout Hypothesis: switching on the laser lowers mains voltage for a short time. <S> You don't say if your USB devices are bus-powered or self-powered. <S> Even if the PC does not crash from a brownout, any USB device powered from mains could also experience a transient loss of power. <S> Its internal microcontroller could go haywire or reboot. <S> One misbehaving USB device can make an USB bus go into la-la-land. <S> Solutions: <S> Does it fix the problem?... <S> inquire with laser manufacturer about soft-start Scenario 2: <S> EMC <S> This is difficult to debug. <S> I won't go into details... <S> Adding ferrites could be beneficial, but I would rather suggest USB isolators. <S> For USB1.1 devices, these are available and cheap. <S> Recently USB2.0 high speed have become available, they are more expensive of course. <S> For example, Intona makes a USB2 isolator. <S> Icron makes an USB to fiber optics bridge, which can span long distances. <S> The famous Corning cable is useless as it does not isolate.
| If the interference (fast transient) is coming in a radiated way (coupling into USB cables), the simple ferrite blobs on USB cable coming to PC might help. Apparently your system is suffering from susceptibility to EMI and low immunity to transients. borrow UPS from the IT guys and run everything (including all USB devices) from it.
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Design help for switching a 5V-supply: FET or BJT, relay or not? Background: I've used BJTs before, but never FETs. Been reading the short intro at https://oscarliang.com/how-to-use-mosfet-beginner-tutorial/ and that's about everything I know about FETs (i.e, not a thing really). My scenario now is that I have a voltage monitor for a 5V-rail. The monitor, well, monitors the rail and if/when it hits 5.5V I get a signal for that, and I will react (cut the supply) upon it. The "cut the supply" part is where I am now. The easy way - i.e. the way I know how to do - is a BJT controlling a relay that cuts the rail. This works, but my concern is that 1. The BJT is slow on reacting, and 2. The relay is slow on reacting. Finding a real fast BJT will solve 1, but finding a relay with reaction time not within milliseconds is harder, so 2 is unsolved. Here's where I thought of a FET. A real world scenario is a better incentive for me than just playing with numbers. I could use a rapid FET + relay, but that will still keep the bottleneck of the responsetime for the relay. So, here's the idea that just popped my head: what if I crap the "transistor + relay" idea and simply let a power-FET act as a rail-switch? No more than 2A will flow through the rail. Questions: Will a single FET be able to make or break the power rail? I'm thinking that a BJT has the Uce_sat drop of about ~0.4V, would a FET have this drop as well? (0.4V-drop from 5V is very bad in this scenario, that's why kept trying with the relay, to avoid the drop) Similiar to the above question, how would an AC-rail work through a FET? Let's say I want to switch 12VAC, will there be a voltage drop / loss, or will the full 12VAC come out clean? <Q> The tutorial you've linked is way too basic. <S> I'd recommend you to take a look at this one, which has deeper detail without becoming excessively complex: MOSFET as a switch - Tutorial Q1: <S> If my understanding is correct, your application requires cutting off a rail that supplies 5V to a load (or more than one) connected to ground. <S> If you want to do that, you'll need a high side switch. <S> Check Vishay Siliconix, International Rectifier or any other manufacturer MOSFET Selection Guide for a suitable device. <S> This one <S> Si4477DY could be good your application due to its very low RdsON (< 6 mOhm i.e., 12 mV drop at 2 Amps) and because it will be deep driven when grounding its gate (you'll have Vgs = -5V). <S> You could also trade off some of its capabilities for a lower priced or more conveniently packaged device, it's up to you. <S> If you need to an extremely low RdsON, you could opt for a N-channel MOSFET, but then you'll need additional circuitry to drive its gate (for example, <S> LTC1154) <S> and complexity can escalate pretty quickly. <S> I'll recommend against it. <S> Q2 : <S> You can switch an AC load through two MOSFETs, but be aware that their body diodes will be conducting alternatively thus dropping some voltage. <S> For a 12V AC load you may be better off just using a relay. <A> Yes, FETs are very well suited for switching DC power. <S> Actually FETs are better than BJTs in this respect. <S> There is no problem to have FET that can handle 2A and much more than that. <S> If you want to switch AC power, I would recommend you to look at triacs and optotriacs. <A> I think the device you are looking for is e.g. the LTC4360 in case of the 5V protection. <S> It's a easy solution for over voltage protection and turns your output off within \$1\mu s\$ of (according to the FET).Attached the matching schematic from LT.
| A P-channel enhancement mode power MOSFET may be your best option, as there are plenty of devices available with logic-level compatible gate threshold voltage (a very convenient feature).
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Is 5V 20A power supply output safe? I have a 5 Volt 20 Amp DC power supply. I was wondering if it is safe in terms of electricity shock due to high current output? Do I need to seal the connection to the device for protection? Edit: Thanks for your answers. Basically, I am going to connect the power supply to a string of LEDs which are all parallel (they have a series resistor to control the current of each branch). I want to know whether I need to seal the connection of the string and the power supply? <Q> That power supply can, if it's demanded of it, put out enough current to start a fire in the wires or a resistive load. <S> The wires coming from it always need to be big enough to carry 20A safely. <S> If they are thinner, then they need to be protected by a fuse at the power supply. <S> 5v will not give you a shock. <S> If the supply is mains powered, rather than battery powered, then a fault in it could give you a shock. <S> One way is that supply is grounded. <S> If this is the case, then you must make sure the ground wire is always connected using a 3 pin socket. <S> If the supply only has two leads, then it is 'double insulated', and does not need earthing. <A> Good thing you ask. <S> If you are sure this is a 5V DC power supply then it is safe. <S> Please ask a adult to do so. <S> Please take a look at Ohms law . <S> $ <S> $V = <S> I <S> \cdot R <S> $$ <S> You will get hurt if: your body resistance (R) is too low, for instance if you are in the bathroom and your skin is wet. <S> Or when your skin is damaged. <S> When the voltage (V) is high. <S> This will result in a high current stopping your heart or will burn you. <A> But a short circuit is surely not a wanted thing. <S> The sealing helps. <S> You did not specify, is your power supply galvanically isolated from the mains voltage. <S> If it's not, then a proper insulative sealing is a must. <S> MUST ADD: <S> Get an experienced electrician to check your system!!!!!!!!!!!!
| It is always wise to measure the voltage. 5V does not generate harmful current when touched by hands. Commercial electronic equipment is designed so that a single fault anywhere should not give you a shock, two or more faults are required.
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If light is an EM wave, why don't antennas pick them up? Or do they get picked up but don't show on spectrum analyzers? Although the photons in light are greater in energy than radio waves! Wouldn't there be constant noise from light? Even if it's small. <Q> Yes, you can do this. <S> The antennas have to be very small <S> so you need a lot of them, but using antennas + rectifiers ( optical rectennas ) is a plausible way of converting light energy to electrical current. <S> The structures have to be made with a process that has a resolution finer than a wavelength of light, which is tricky since light is the photo in photolithography, but we are getting down there in resolution by various methods such as using shorter wavelengths and more sophisticated gambits. <S> Similar (but cruder) methods have been proposed to convert microwave energy transmitted from space-based solar power stations to electrical energy on earth. <S> The energy density would be kept low enough that it would not be as much of a death ray as you might think. <A> Effective antennas must have about the size of a quarter of the wavelength (rule of thumb). <S> For visible light wavelength is about 400-800nm. <S> So you can imagine how small those antennas for visible light must be: They have the size of molecules. <S> They are molecules. <S> And in fact if substances are colored the molecules are antennas "tuned" to particular frequencies of EM radiation in the visible range. <S> Although in most cases absorbed energy is converted to mechanical energy (i.e. rotation and/or vibration of the molecules eventually heating up the substance) not to electical energy. <S> The energy doesn't show up on converntional spectrum analyzers because of their limited badnwidth ;-) <A> Your antenna surely picks the light until it's a mirror. <S> Unfortunately ordinary circuits made of wire, capacitors, transistors etc. are too big to notice the resulted current. <S> They are heavily out of the right tuning. <S> Try a photodiode as an integrated antenna and detector. <S> You can increase the effectiveness of your antenna by adding a lens or concave reflecting mirror. <S> Need some frequency selectivity as in radio receivers? <S> Buy a colored filter. <S> Too weak signal - need an amplifier BEFORE the detector? <S> No problem. <S> Insert a laser in it's original form without the resonating mirrors that make it an oscillator that do not amplify, but generate light. <S> Addition due the comment that asked "can antenna convert the frequency?" <S> An antenna is not considered to have an ability to convert the frequency- it only catches the wave and feeds it to some unit that converts the frequency. <S> The frequency converter is a mixer with another input from an oscillator. <S> About 25 years ago the general talking changed. <S> When one went to a shop to buy a satellite antenna for his tv, he assumed to get also the frequency converter from microwave to normal tv channel. <S> To minimize the signal loss it's best to do the frequency downshift for the tv as soon as possible. <S> Thus the mixer+oscillator is in the antenna Not asked, but maybe nice to know : There exist materials that convert the frequency. <S> Some fluorescent materials absorb light or ultraviolet. <S> Absobrtion causes electron orbit exitations that reverse, but do not collapse exactly to the same. <S> They produce radiation that has a different wavelength. <S> The color in safety vests has a wide input band, but the output band is narrow which causes a high color contrast. <A> Wouldn't there be constant noise from light? <S> Even if it's small. <S> Infra red light creates heat and that in turn creates noise in any circuit that has resistance. <S> See this wiki article . <S> The formula is: - Or do they get picked up but don't show on spectrum analyzers? <S> Visible light is in the range 400,000-800,000 GHz - I don't think there are many conventional "radio" spectrum analysers that will work up to this range: -
| Special spectrum analyzers for EM radiation in the range of visible light are called optical spectrometers .
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Feed a speaker only with positive voltages I am using the timer of a microcontroller to create square wave signals from 10KHz to 40KHz, but my microcontroller generates the signal only with positive values (0 - 3.3V). I have generated the same signals from my PC with the sound card, the signals in the oscilloscope have a range from -2V to 2V (at maximum volume) this is 4V peak to peak amplitude. So my question is: If the speaker is able to support 4V of amplitude, will it be ok to use a 0 to 3.3V signal? Is there any way that this positive going signal can damage my speaker? <Q> The 0-3.3V signal has a 1.65V DC component, and this will effectively be shorted by the speaker. <S> To ensure that DC voltages do not reach the speaker, put the speaker in series with a capacitor. <S> With a capacitor, you need not worry about the microcontroller driving 3.3V, the capacitor will block DC voltages. <S> You should also ensure that your microcontroller's output pin has sufficient current handling capabilities to drive the speaker directly; buffering the output pin with a MOSFET would be a good idea, or you could get more sophisticated and add an amplification stage with high-pass filter to remove the DC component. <S> If you have a higher impedance speaker (more of a buzzer), then driving it directly may be perfectly fine. <S> You simply need to find the DC impedance of the speaker, and verify that the current draw is within the limits of the microcontroller GPIO driver. <A> Driving a speaker with DC-biased signals will cause it to convert more power into heat than driving it with bipolar signals. <S> In addition, speakers are usually designed for balanced mechanical stresses, and a DC bias will cause unbalanced stresses. <S> Speakers will often be more tolerant of this at higher frequencies than lower frequencies (as frequencies approach DC, the fraction of power converted into heat approaches 100%), and at 10kHz the effects might not be too bad. <S> On the other hand, adding a coupling cap will avoid the issue and at 10KHz the required coupling cap may be fairly small and cheap (the impedance of the speaker will likely increase with frequency, so even a 10uF cap might be more adequate). <S> As an alternative to using a coupling cap, or in combination with it, you could drive both ends of the speaker with separate port pins. <S> This would let you drive twice the voltage and could eliminate the need for a coupling cap (though a coupling cap could provide protection in case of software malfunction). <A> It's not ideal, but actually this is frequently done when you just want a noise from the electromagnetic transducer, and the 'speaker' is relatively high resistance. <S> I've designed several products with such small electromagnetic transducers controlled by an MCU and a single transistor. <S> It is not a good idea to put significant DC voltage across something like an 4-8 ohm speaker- <S> it will displace the cone, cause unnecessary power dissipation and could overheat the voice coil. <S> Use a blocking capacitor in that case.
| You should not connect the speaker directly to the microcontroller as the speaker presents a very low impedance at DC (assuming a regular moving magnet speaker with a voice coil), and it will essentially short the microcontroller.
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How to change -10;10V sine wave to 0;3.3V dor ADC I want to build very easy circuit to change -10;10V sine wave to 0;3.3 for ADC in my MCU. I know that I must reduce the amplitude and change offset. This circuit would be perfect (Output: 0-3,33V), but I have only 3,3V source (from MCU) and I made another circuit (in which I have output: 0,04-2,83V): What should I change in second circuit to have output (0-3,3V) like in first circuit? Maybe voltage stabilizer diode or op-amp?Thanks For those who would searching the same info to their project: I chose first circuit from this post. I changed R11 to 110K and R12 to 22K, In place of 2V source I put 2V voltage stabilizer (Sanyo LA5002) (between the 3.3V MCU output and R11). And it work great :) <Q> simulate this circuit – Schematic created using CircuitLab <S> From the Kirchhoff's Voltage Law: $$V_o = <S> \dfrac{\dfrac{V_i}{R_i} + \dfrac{V_1}{R_1} <S> + \dfrac{V_2}{R_2}}{\dfrac{1}{R_i} + \dfrac{1}{R_1} + \dfrac{1}{R_2}}$$ <S> In your case, we have three known input-output relationship: <S> $$V_i = <S> +10V \implies <S> V_o = <S> +3.3V <S> \\V_i = <S> 0V \implies <S> V_o = <S> +1.65V <S> \\V_i = <S> -10V <S> \implies <S> V_o = <S> 0V$$ <S> This gives you three equations with five unknowns. <S> If you fix some variables to reasonable values, for example \$V_1=10V\$, \$V_2=-5V\$, \$R_i=10k\Omega\$, you can easily calculate the remaining variables (i.e.; \$R_1\$ and \$R_2\$). <A> Maybe this: For flexibility, you can exchange R2+R3 with a potentiometer as well. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Try the following, its signal division is 0.15 (20 V to 3V Vss) -- but you can change values with R1==R3 and R2==R4 to any other ratio with R2/R1 and R4/R3. <S> I only choose nearest values from E12 family of resistors. <S> simulate this circuit $$ U_{out} = <S> U_{in}\cdot\frac{(R_1+R_2)\cdot <S> R_4}{(R_3+R_4)\cdot R_1} + U_{ref} <S> - U_{minus}\cdot\frac{R_2}{R_1}$$ Here <S> , we have \$ U_{minus} = 0\text{V}.\$ <S> So the last term can be omitted, we get:$$ U_{out} = <S> U_{in}\cdot\frac{(R_1+R_2)\cdot <S> R_4}{(R_3+R_4)\cdot R_1} <S> + U_{ref}$$$$ U_{out} = <S> U_{in}\cdot\frac{(R_1+R_2)\cdot <S> R_4}{(R_3+R_4)\cdot R_1} <S> + V2\cdot\frac{R_6}{R_5+R_6}$$When selecting \$ <S> R_1 = <S> R_3 \$ and \$ <S> R_2 = <S> R_4 \$ also \$ <S> R_5 = <S> R_6\$ <S> it simplifies to:$$U_{out} = <S> U_{in}\cdot\frac{R_2}{R_1} + \frac{1}{2}\cdot V2$$ <A> You can do something like this: simulate this circuit – Schematic created using CircuitLab <S> The output impedance is about 4K. <S> If you don't like the values I chose you can just scale by an appropriate factor. <A> Actually best is to use instrumentation amplifier. <S> It allows you to select gain and common mode voltage, so you just scale and offset your signal. <S> I am almost sure <S> that +/-10V is in fact a differential signal, so you will need a descent CMRR.
| But I think, the best way is to use an operational amplifier to get enough drive for the MCU ADC and to be able to DC couple the signal.
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Dissipating Heat from Tiny Components I need a 5v 2.5A output in a handheld device and have settled on the TPS61235P . Its 2.5mm QFN package was very hard to solder, but I finally got a board fabbed and tested it out. After roughly a few minutes of constant on-time @ 2.5A, the input traces started to burn up. Now for the next PCB I'm planning on using the wider traces with solder mask left off so I can tin the entire trace up to the chip. But once I solve this, I'm left wondering, how do I keep the chip cool? It seems way too small for a heatsink, and doesn't have a solder pad similar to other chips I've used. I assume a schematic won't be necessary, but I am attaching a pic of the current prototype pcb. <Q> Although I'm a little surprised as you at the lack of a pad, a quick back-of-the-napkin calculation makes me think you may be OK as-is. <S> The datasheet shows you stay above 90% efficiency through the entire operating range. <S> And with 5V@2.5A, you're dealing with 12.5W... <S> so you may dissipate, at worst, 1.25W. <S> Multiply that by your 28C/W Junction to Ambient, and you come up with 35 degrees C. Subtract from your 125C maximum junction temperature, and you get 90C ambient. <S> I can't tell at a glance what the copper thickness of that board is, but if you specify "2 ounce" copper for outer layers, it will result in thicker traces and a larger cross section to carry current. <S> Default is usually 1/2 oz or 1 oz. <A> You want to have the the pads connected to large areas of copper in this case to dissipate the heat. <S> I suggest you review section 11 on layout and thermal considerations when laying it out, and follow the recommendations as much as possible. <S> Notice how large the copper areas are, and how many vias are used. <S> These both help dissipate heat from the IC. <S> @1N4007 is right about the thermal calculations, but keep in mind <S> the 28ºC/W figure is usually based on a certain amount of copper area. <S> I wasn't able to find that specified in the datasheet, but often it's 1 sq.in of copper, <S> so you may not get 28ºC/W from the chip alone, i.e. in free air. <A> Vias with 1:1 ratio of circumference to height thus contain a SQUARE of Copper; the thickness may be 20 microns or 35 microns or otherwise, depending on how long the board house ran the via-plating step. <S> One square of 35 micron copper is 70 degrees Cent per watt. <S> Thus each via is 70 degrees Cent per watt. <S> And heat attempting to spread out lateral, on the surface, is also limited by that 70 degree Cent per watt per square. <S> Suppose the heat is generated in a corner of the PCB. <S> What is the thermal resistance to rest of PCB? <S> What if heat is generated in MIDDLE of a PCB? <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Consider heat, entering a single square of foil. <S> Perhaps from the 2mm tab of SOT-23, or 1cm^2 tab of TO-220. <S> What is the thermal resistance to the surrounding PCB. <S> Surround that single square, of whatever size, with a 3*3 grid, the heat injected into the middle square. <S> There are 8 squares around the center; the thermal resistance out of that center square is $$(70degree C/watt) / 8$$ or 9 degree C/watt. <S> Now consider a 5*5 grid. <S> What additional thermal resistance does the outer ring 4+4+4+4 squares contribute? <S> Just divide $$(70degree C/watt)/12$$ or 6 degree C/watt. <S> If the original IC (heat source) has Rthermal of 25 degree C junction_to_case, then we just add 25 + 9 + 6 = 40 degree C/watt. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> What if these 5*5 squares need to dump heat into the underlying GND plane? <S> The Rthermal of epoxy-fiber glass is about 200X that of Copper foil. <S> You can use the Rthermal of copper cubes, 1/(340 watts/degree C * meter) or 1/(3.4 watts/degree C * cm) and scale down further for 20 mils or 60 mils thickness. <S> Then scale up by 200X, as approximation for FR-4. <S> Get a quadrille pad, and start sketching heat flows through grids, laterally, and vertically through vias and FR-4. <S> Or generate 2_D grid of resistors in SPICE, with foil on the surface using low values, and FR_4 between layers using 200X higher Rvalues (in both x & y, to make a cube of FR-4). <A> To beter dump heat into the underlying plane, have the stackup use the minimum allowed thickness between layer1 and the GND plane (large GND region) <S> on layer2.And remember the large heat-dump pads that TI suggests: To dump heat even better, enlarge that SW region under the IC. <S> The TI appnote ignores the 70degree Cent per watt per square of foil. <S> There are 3 or 4 squares from middle of the "SW" pad to the "SW" region where the inductor is soldered. <S> Yet the SW pad is the DRAIN of the internal FET switch, the main heat generator. <S> simulate this circuit – <S> Schematic created using CircuitLab
| By the way, if you can afford the cost, you might be able to save yourself the trouble of manually tinning the traces by increasing the thickness of the copper on your outer layers. So as long as the air around the chip is less than 90C, you should be OK.
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How to plate pads in Altium I'm designing a PCB with allium and after route all the nets, I put allium in 3D view. (My first Double layer PCB) In 3D, it appears to me that the pads are not plated (connecting both sides of the PCB, through the hole), despite the option to plate the pads being selected in pads properties. Is this a 3D view bug? The right ones appears to be plated but the left ones looks like they are not. Left Pads properties: Right Pad Properties <Q> Look Closely From the picture, it looks like the holes are actually plated (as your settings indicate). <S> Rather, it looks like they are missing the pad on the top or bottom, because when you see that darkish almost-transparent core of the PCB, it's like you have no copper or soldermask there. <S> I've never seen Altium be as buggy with the 3D view as others here claim, even before I had a good graphics card. <S> In fact, I trust 3D view the most to see what's going on sometimes! <S> For example, it will even properly display tented vias. <S> My guess is either you have some sort of net tracing on <S> and it's dimming layers, so make sure in the bottom right <S> you click "Clear" near "Mask Level" That, or also <S> possibly you generated the footprint incorrectly at first, using the "top-middle-bottom" or "full stack" options, without really knowing what you're doing. <S> My guess is you made the top or bottom layer 0 sized. <S> Now you're showing us the footprint is fixed in the library, but my guess is you forgot to actually update the instantiated footprint on the PCB from libraries. <S> To do that, from the PCB view, click Tools - <S> > <S> Update from PCB Libraries . <S> Click OK and go through the wizard. <S> ======================================== <S> Sidenote: <S> An easier way to confirm NPTH <S> (Non Plated Through Holes), which sometimes are desired like for mounting holes or light pipe mounts that get glued, you should read through the board report. <S> To do so, click again from the PCB view: Reports --> Board Information . <S> Click Report, turn All On (really you only need Non-Plated Hole Size, but they can all be enlightening). <S> Then click Report at the bottom. <S> You'll get something like this, which you can count what you expect and find the error. <S> By the way, don't get thrown by the 32 0mil pads. <S> It's how Altium counts a top or bottom layer surface mount pad with no hole. <S> Kinda stupid, probably the way the tool came originally from through-hole days and that legacy code just continued on counting a SMT pad as 0-mil hole. <A> Can you show the settings dialog for both of them? <S> The left ones look like something is wrong with the stackup or you set the pad size inside the pad smaller than the drill hole. <S> "Plating" in Altium (!) <S> usually means if the barrel/hole is supposed to be plated with copper, not the pad itself. <S> There is a checkbox in the Pad properties dialog for that. <A> If this is twice the same component (left and right) and no differences have been made, it's quite possible <S> it's a bug. <S> However, it's smart to still check. <S> In 2D view you can double click a pad, and then it asks you if you want to see the options for the component or for the pad, choose pad. <S> Then there are options for layer stack (you can define the pad for all layers independently with a special setting) or for "simple", <S> if you chose simple all layers should be the same pad shape, if you have a full-layer-stack option selected, verify that all layers have a pad laid out. <S> It also has an option for "plated" which means the inside of the hole, not the pads, but if that's what you mean, there should be in the lower to middle left side a check at "Plated". <S> Of course it's best to go back to your library when fixing problems and update the parts on the PCB from there, since then you will never have to fix it again. <S> Finally, before you have your PCB made, especially your first projects, be sure to inspect the Gerber plots you will send to the manufacturer. <S> Anything that is shown on the Gerber data will be there when they send it back. <S> Unless they make a mistake, but then you have checked the Gerber data and you can prove it's their fault. <S> I prefer to use a separate program for checking Gerber data, so that I am sure there's no glitches in Altium when generating and viewing that cancel itself out. <S> ViewMate is my go-to application, you need to register to download it (or to use it), but after registration it is free and it has never let me down. <S> Good luck, and don't be afraid to google some of the above, because a lot of it is a bit fiddly and differs between versions of Altium. <A> Create a drill file for your board, and look at it with a text editor. <S> Hole plating information is not included in the Gerber photoplot files. <S> With only one drill file, all holes will normally be plated, regardless of what may be shown in Altium's 3D viewer.
| If the drill file has sections for plated and non-plated holes, or there are separate drill files for plated and unplated, your board may be made with some holes not plated. Altium's 3D live view is very buggy if you don't have a top-range graphics card.
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TIP127 + TL074 power supply doesn't work until I touch it I have such a schematic based on the TL074 opamp and the TIP127 PNP darlington transistor to control the fan voltage (shown as motor near the D3 diode). VCC = 12V, VDIG = 5V generated by a LM7805. The problem is that this schematic works perfectly in the simulation, yet it resists to work on the breadboard until I touch the opamp's positive input. After that, the schematic works perfectly until I try to turn it off and on again - it turns off fluently (using the pot), yet it does not turn on until I touch that positive input. I don't understand what is going on. My guess is that I should put a capacitor somewhere, but I don't understand where and - more imporantly - why to make it work. I am using a 12V 0.3A fan as a load. <Q> Here are my thoughts: - <S> The TL074 has a common mode input range that does not include the negative rail (0 volts on your circuit) <S> so, when you power it up the non-inverting input is breaking the rules and the op-amp cannot be expected to function correctly. <S> Typically, anything between 0 V and 3 volt on the input will likely cause problems. <S> This is probably what is mentioned in the question "until I try to turn it off". <S> Adding an extra level of voltage gain to an op-amp is usually a recipe for disaster (it turns into an oscillator) if not done carefully. <S> Q1 is a common emitter hence it has plenty of voltage gain from base to collector. <S> In this circuit, that gain could be in the realm of about 50 to the junction of R2 and R3. <S> If you look at figure 9 in the TL074 data sheet you will see that increasing gain by ten is probably enough to make the phase margin zero <S> (it becomes an oscillator). <S> Problems all round really. <A> Your supply is likely oscillating. <S> Collector output has its disadvantages. <S> Also, for really lot output, you will need a r2r opamp. <S> Or you cannot turn the PNP off, or not able to control at the low end of the output. <A> I've actually came up with an another, temporary solution. <S> The right solution is to use an another opamp with near R2R operation (like LM324) and use a schematic with a negative feedback. <S> The TL074 opamp, as some people have already mentioned in other anwsers, cannot handle the low-input cases. <S> When the schematic is off (potentiometer is at the lowest setting), the voltage on the positive input of the TL074 opamp is 0V exactly. <S> I tried after that turning on the motor by increasing the potentiometer value and therefore the TL074 opamp wasn't working in right conditions and failed to turn on the fan. <S> The solution I came up with is just to increase the voltage on the positive input of TL074 by adding a 100k resistor from the positive input to the VCC. <S> Therefore, when the motor is turned off, a small current will still flow across 100k resistor R4 and R2, making a "shift" - a small voltage of around 1V at the positive input of TL074. <S> In this schematic, however, the motor voltage cannot reach values larger than 10.5V approx. <S> because of the same "shift". <S> The solution for this problem I came up to is to use a diode like it is shown on the second picture. <S> I also added two diodes in series to higher up the voltage at the base of Q1 (which is also a very, very dirty solution). <S> Again, this is a temporary solution which should never be used. <S> Thanks everybody for their anwsers and pointing out the actual problem which I solved with the described hack.
| The problem wasn't in the oscillation (that's very strange - the schematic should oscillate because of the positive feedback), instead it was in the opamp. The TL074 has an output swing that does not reach the power rails so it's very unlikely that Q1 will switch off once it is on.
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Why does this common-source MOSFET amplifier work? I have built the MOSFET amplitude modulator shown in the bottom right onto another class C oscillator tube circuit, and it does work. I am having trouble understanding why based on what Ive read. MOSFET amplifier http://danyk.cz/avttc001.png It seems to be a common-source amplifier, but doesn't seem to have any Rd, or Rs, or even any power supply rail for that matter (the centre point of the tube filament should be at 0V, no?)!! Despite the fact that it seems to use a potential divider to provide bias for the gate. I also rebuilt it on a tube circuit that has a separate heater and cathode, and rectified the heater supply to provide the MOSFET a power rail, and took the output access an Rd that i added. This makes the MOSFET run much cooler (I assume this is by providing sufficient voltage for the divider to get the gate to a reasonable q-point. If my interpretation so far is correct, what is the consequence of not having an Rs between the the source pin and ground? Should there be one, ideally? Can I still add a bypass capacitor (I guess it's not a bypass capacitor since there's no Rs) between the source pin and ground to increase high frequency gain? (PS - An afterthought, does it make more sense to think of the whole circuit as a cascode?) <Q> The anode-cathode current for the GU-81M is modulated by the FET. <S> All the current from the valve flows through the FET <S> and it's DC configuration (using the potentiometer) <S> will set the operating point cathode voltage. <S> The center tapped heater <S> has no influence and the heater/cathode is not at zero volts. <S> Yes, this would be considered cascode operation as in this audio amplifier with two triodes: <S> In your case since the FET modulates the operating current (and therefore the C-G) of the pentode <S> it changes the amplitude of the RF output. <S> If you look at the datasheet for your FET: Excuse my mouse scribbling on top of the graph, but here you can see as the Vgs varies, the current through the valve will vary (and the cathode voltage). <S> I've no idea <S> where you operating point for the GU-81m is so cannot work out the valve current, but you should be able to. <S> Note that your grids are at very low voltages, so <S> as the cathode rises they are effectively negative biased. <S> It's doubtful you'd see more than 120-150 V on the cathode so well with the ratings of the FET. <A> There is a load for the mosfet - the whole anode circuit of the tube. <S> The connection is made via the electron stream through the vacuum. <S> The mosfet is an amplitude modulator. <S> The mosfet is biased for plenty of current Id when the audio input is 0V. <S> A guess: <S> The high voltage output generates a continuous plasma arc in a proper spark gap. <S> The audio signal makes that arc to speak or sing. <S> ADDENDUM: <S> In this circuit mosfet's load is an oscillator. <S> The cascode amplifier interpretations are bad just in this case, because the vacuum tube has his own signal and runs highly non-linearly. <S> The cascode mixer instead is a good simile. <S> The nonlinearity makes the amplitude modulation possible. <S> The oscillator takes all available current and the availability is modulated. <A> An afterthought, does it make more sense to think of the whole circuit as a cascode? <S> It's an example/spin-off from a cascode mixer. <S> You alter the cathode voltage (with the MOSFET or JFET or BJT or a.n. <S> other tube) and that modulates the amplitude of the signal on the anode. <S> Here's a more conventional example using JFETs: - The only difference with the pentode you are using is that it is self-oscillating with what looks like the lampbulb acting as a stabilizer. <S> But basically it's a cascode mixer. <S> what is the consequence of not having an Rs between the the source <S> pin and ground? <S> Should there be one, ideally? <S> The MOSFET is fairly stable due to the feedback of the 22 kohm and the 1 kohm so you don't really need a source resistor but adding one wouldn't make things worse <S> but, it should only be about a kohm or so maximum.
| The Tube is an oscillator that obviously generates a high voltage.
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How can I Further Reduce Power Leak from BSS84 MOSFET? I am working on a device which is running from a 3V coin cell. A system timer enables a MOSFET which starts the actual circuit (MCU etc.) for a very short time. The system timer (U1) only uses around 50nA of power while idle according to the specs. I checked this and it is true. When I measured current of the prototype I got something around 900nA. So it seems the used MOSFET (M1) leaks around 850nA current. The circuit diagram looks like this: simulate this circuit – Schematic created using CircuitLab Is current leakage of 850nA a normal value for a BSS84 MOSFET? Is there a simple and cheap way to reduce the leakage below 100nA? MOSFET data sheet: http://eu.mouser.com/search/ProductDetail.aspx?r=512-BSS84 IDDS=-15µA, IDDS@125C=-60µA <Q> I found the the "leakage". <S> The used multimeter: <S> Testo 760-3 should provide a resolution of 600μA/0.1μA, but now displays 0.9µA <S> with no load attached. <S> This 900nA are a simple error from the measurement. <A> I think you should be looking at IDSS parameter in the datasheet. <S> Which exact part are you using? <S> I mean, which manufacturer? <S> The specifications vary. <S> For example, the Nexpeiria version of the BSS84 specifies -100nA max under these conditions: Vgs=0V, Vds=-40V and Tj = <S> 25C. Fairchild version of BSS84 says -15uA max at 25C, with VDS = <S> -50V. <S> If you need guaranteed 100nA over temperature, you might need to find another way. <S> Maybe NMOS on the low side would have lower leakage (I'm not sure). <S> Just out of curiosity, have you calculated actual impact to battery life? <S> I would think that when you are under 1uA, you might be approaching the self-discharge rate of the battery. <A> 850 nA doesn't seem normal when you look at the Idss specified in several BSS84 manufacturer's datasheets. <S> Your low voltage coin cell makes it very unlikely that the leak comes from Vds driving a high Idss. <S> Instead, I'll turn to the gate voltage as the potential suspect. <S> If you're not pulling-up the gate with a resistor, your Vgs may not be actually zero and the MOSFET could leak some current. <S> Especially if your BSS84 sample happens to have a high gate threshold voltage (according to the datasheet, it can be as high as -0.8V). <S> Or maybe the device is damaged, who knows.
| Maybe using the Nexperia part will solve your problem.
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MOSFET reliability for critical applications I am working on a critical circuit that drives a DC motor in an aircraft landing gear system. The electric motor is a simple DC motor where one polarity extends the landing gear and reversing the polarity retracts it. My question is are there any reasons why using MOSFETs instead of mechanical relays would present a reliability issue? Obviously it is important that the gear comes down when it is commanded to so making sure this is reliable is high priority. <Q> I have some background in high reliability electronics for space applications. <S> The way you define "reliability" here is the key. <S> If you're thinking just in terms of the the random failure rate, a properly derated and protected flight-qualified MOSFET-based assembly can easily beat any equivalent relay. <S> But that's not the only thing to consider when chosing between those two technologies, of course. <S> That would be too easy. <S> :) <S> Are there any specific derating or reliable use guidelines for MOSFETs in aeronautical applications that make them unconvenient for your company? <S> Has this technology been used before for this very purpose? <S> Are there any intrinsic advantages of one technology over the other? <S> Such as higher immunity to single events (like lightnings, atmosferic radiation, etc.) <S> . <S> Is there the possibility of single-event latch-ups (i.e. destructive response to an environmentally possible event) in any of those technologies? <S> Are there any environmental parameters that impacts more to one technology than the other (vibration, shocks, temperature, thermally induced mechanical stress, etc.) <S> and that can precipitate accelerated aging and an early wear-out failure? <S> Which are the failure modes and their criticalities? <S> Does any of the two alternatives have a significant advantage over it? <S> Sorry for raising more questions that answers, but your question can't be easily answered without having a broader view of the design problem. <A> Here's an excerpt from the link below showing that the life expectancy of solid state relays (using MOSFETs) is much greater than electromechanical relays: "Using the daily number of operations from our previous example, this means using an SSR instead of an EMR could extend the life of the switching component in the oven [a hypothetical example in the app note] from 2 months (with the EMR) to 833 years (or 83 years at the lower end of the calculation, just to be prudent) <S> " Crydom application note <S> Of course the design has to be appropriate for the environment, including ESD, weather and temperature conditions, transient voltages, EMI, etc. <S> It's also possible to use additional FETs (or relays) <S> such that any single switching component failure does not cause loss of functionality, though that's much more complex and costly. <A> Well, relays can be easily replaced and can generally take much more abuse than mosfets can. <S> I also imagine that, at least depending on the mosfet, some condensation due to temperature changes during flight could cause the mosfet to act in weird ways. <S> And also, if you use a relay and it breaks, you can borrow another from another device on the plane temporarily to operate the landing gear in emergencies, whereas a mosfet isn't commonly found in other places. <A> Mechanical relays dont care of your smartphone's or other medium intensity rf fields. <S> Requires a microwave owen to make some effect. <S> Semiconductor parts are much more sensitive. <S> Maybe there's no radio transmitters built to be in a near contact with your relay, but some non-thinking idiot can carry in one just when an aeroplane is landing. <S> Mechanical relay tears and wears in the use. <S> But so does your motor. <S> This kind of stuff has a regular pre-emptive maintenance program. <S> The wear-out prone parts will be changed and refurbished before anything happens. <S> The relay and the motor could be in one swappable assembly. <A> As far as average lifetime goes, the MOSFETs will probably win if properly derated and protected with surge absorbing devices. <S> Reliability- <S> I would say it would go direct switch, relay, MOSFET, in order of decreasing reliability. <S> It takes very little energy to cause a MOSFET to fail 'on'. <S> There are plenty of sources of spikes in most aircraft. <S> Whatever you do I suspect you'll want a reasonable and well-documented plan 'B' available to the pilot. <S> What happens if the main bus power goes down? <A> MOSFETs would be more reliable than a relay due to total cycles, but you probably have a larger issue. <S> I would suggest using a BJT for the following reasons: The MOSFET is sensitive to ESD. <S> When you are flying around, you are actually creating charge. <S> Planes create lightening. <S> I would not be surprised to find that you blow out the MOSFET gate due to a static discharge. <S> Relays have a fixed amount of cycle times, cold metal shrinks, vibrations can destroy the aperture. <S> The advantage of the BJT is that is can self-heat down to about -80C, and it is largely immune to ESD. <S> The scariest part of any implementation is always the mechanical components. <S> You could also use the BJT current as an indication of up/down instead of limit switches by having a resistive shunt that would make it completely solid state.
| For example if lightning or whatever hits the plane, it is much more likely for the mosfet to break compared to the relay. A relay will seldom fail in the middle of its useful lifespan.
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What am I doing wrong with this LM393? I'm trying to learn how to use opamps as a comparator. uA741 was easy, but it's way too slow for an uncoming lab I'm going to do, so I switched it to LM393 instead which is rail-to-rail and very fast. Problem: LM393 is open-collector, and I have been pulling my hair for two hours now and can't get it to work. I follow datasheet guidelines (3k pull-up), but everything connected to the output (NPN-transistor, etc) does NOT behave as there is a "1" coming on that signal. So for now I give up open-collector, I will re-try this another day when I got more time. What I'm trying to achieve is to drive a NPN-transistor from the LM393-output. Using 3k pull-up, I measure around 1V on the output when opamp is "high". It's supposed to be 5V, but isn't. That's why the next step won't work either: as soon as the opamp has gone high, I'm trying to latch the output high by sending the output to another NPN which pulls in- to GND. in+ can only be 5V or 1V (never GND) so the output will always be high. But I can't get it to work because of the open collector confusing me. I get it to work out-of-the-box with a non-open-collector (like uA741). EDIT : See schematic. I added R10 as per the comments suggesting this. Q2 does now open, but the opamp-output is around 2V (why not 5?). If I raise R10 to 10k, opamp-output is ~5V which is expected, but I think 10k is a bit too high for the NPN-base, isn't it? If I connect the feedback-resistor R7 to the output, the output slams down to GND no matter what the output is supposed to be. The purpose of R7 + Q3 is to latch in- to GND, so the opamp-output is permanently high until powered off. I am probably making multiple errors here. I'm trying to learn, so please bear with me. ( Side-question : Which common opamp / comparator, without open-collector output, is a good replacement for LM393? "Common", as in "Go into RadioShack to buy") EDIT 2 : New screenshot (with irrelevant circuitry / components removed) This shows how the circuit works without the latch. The red line is the voltage supposed to be strangled permanently after the first hi-to-low-transition, i.e. the first time opamp goes low. The green line is the opamp output: High at first (correct), then low when triggered (correct), but then it goes high again, and this is incorrect. It's supposed to stay low here on. I've added a temporary Q3 to the circuit. In theory, this should bring the in- to ground, making in+ higher at all times. But, because the opamp starts high, it means Q3 will bring in- to ground immediately, which I do not want. I don't know if I can use Q3 like this, but I'm just trying to show a point here, how I'm thinking. Instead of Q3, I have also tried to add another LM393 (there's two in the IC), trying to make the feedback go through it instead of Q3, but the results are roughly the same. How can I solve this? That is, latch the opamp in low / "off" state as soon as one first transition has been done. <Q> The part of your circuit to drive Q2 from the output of U1A looks OK, although there is no need for R10. <S> The current thru the base will already be limited by R8. <S> R10 will only slow down the transitions. <S> Otherwise, Q2 won't be reliably turned off, or at all. <S> If the output of U1A really isn't going high, first check that the supply pins are at the expected voltages. <S> You don't show what voltage VM is. <S> Check the datasheet and make sure this is high enough for the LM393 to operate properly. <S> Then check the input pins. <S> Maybe the bug is in not driving them to a state that should make the output high. <A> If the output of the 393 is "on" (not pulled to ground through the internal transistor) and you're measuring the voltage directly on the output, then you're just measuring the output of a voltage divider made up of R8 and R10. <S> The output voltage would be (VM-0.7)*(1k/(3k+1k)) = <S> 1.075 volts assuming VM=5V. The 0.7 volt drop comes from the base-emitter junction of the NPN, which is effectively just a diode. <A> You can only expect about 1V on the output. <S> Your base resistor is 1k and the transistor appears to the op amp as a diode to the negative rail. <S> Increase your base resistors to about 47k on Q2 and Q3. <S> You may want to take a look at the other resistor values in your circuit. <S> 1k to pull off a mosfet gate is at least 10 times to small. <A> The LM393 is functionally a comparator by design. <S> It is configured as an open-collector device since its typical use is as a switch. <S> You may find it easier to analyse the operation of your circuit if you consider it in that sense rather than as a (general-purpose) op-amp. <S> So rather than thinking of the 393 in combination with the pull-up resistor as forming an op-amp, think of the 393 as a (open-collector NPN) switch which can pull down the base of Q2 to turn it off by removing the drive current being supplied to it by R8. <S> (As an aside, while it doesn't achieve your specified goal to "drive a NPN-transistor from the LM393-output", you may also like to consider how you might use the LM393 without Q2, which you may find to be an interesting exercise, too.) <A> around 1V on the output when opamp is "high". <S> It's supposed to be 5V, but isn't. <S> 1v is about right on the opamps output, because it got pulled down by the 1k and B-E junction of the NPN. <A> Not sure what your supply voltage is, but... <S> The common-mode input range for the 393 is 0 to supply minus 2 volts. <S> If both inputs are above this range, the output goes low. <S> The circuit might be latching as power is coming up. <S> You can try adding circuitry to ensure that it initializes to the correct state after power-on.
| If connecting R7 to the "output" - being pin 1 of the LM393 - causes the output to stay low then you probably have a wiring fault, or other circuit conditions are not as you describe/expect. Try adjusting R9 and/or R6 so that the voltage on the inverting input (with Q3 off) is at least 2V below the supply voltage. I didn't look up the LM393, but make sure that the open collector output is really with respect to the negative supply, which is ground in your case. It's a good idea to always have one input within the common-mode range.
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Discharging Circuit with LED indicators? I'd like to create discharging circuit with four LEDs indicating how much charge is left from the discharging battery. It needs to be a low cost solution (no Arduinos; only simple circuit components) I'm not entirely sure how to go about this theoretically; I can only imagine that there would be parallel branches of an LED in series with a resistor. I want each LED to be lit initially and indicate how much of the original voltage/current is left as each LED goes dim: 4 lit LEDs = 100-75% charge 3 lit LEDs = 75-50% charge 2 lit LEDs = 50-25% 1 lit LED = 25-0% 0 lit LEDs = 0% charge left EDIT: Let's says it's for a 9v battery Edit #2: Let's say it's a lithium ion rechargeable battery. <Q> Well, This is an interesting problem that can be tackled a variety of ways. <S> Typically to accurately measure the remaining capacity of a battery you will use a coulomb counter, which basically integrates charge flowing across a shunt resistor in both directions. <S> Essentially this gives you current/time bidirectionally <S> and you can solve for remaining capacity based on your initial specifications for the battery in your system. <S> This is from google images, so you will have to change the resistors to produce voltages at the junctions that correspond to points along the discharge curve that you will use, this is also a liquid level indicator, but the principle is the same <S> This however will cause some problems if the battery is a rechargeable type since aging effects will reduce accuracy of the gauge as the battery is repeatedly cycled. <S> The voltage/capacity relationship also changes as the battery is put under different loads, so either you can do the math to take that in to account or just measure the battery voltage when you are under a very light or no load, take a look at the picture for a typical AA battery discharge curve. <S> If you don't always need to check the battery, you can put a tactile switch in series with the circuit to kill power, <S> that way you will only ever have the LEDS enabled when you are actually checking the capacity <A> I'd probably use comparators. <S> Use simple resistor dividers as one input for each of the comparators. <S> Then use the battery voltage as the other input. <S> The output of each of the comparators will drive the LEDs. <S> Now, you may find using this technique could not be as cheap as an Arduino by the time you buy the IC(s), wire everything together, package it, etc. <S> Plus, like jonk commented, charge and voltage do not have a 1:1 relationship. <S> There are other factors like temperature. <S> It all depends how sophisticated you want your meter to be. <S> If the voltage only changes a very little for large changes in charge, you may need high precision resistors, some kind of amplifier, trim pots, or other sophistication. <A> Rechargeable makes it trickier. <S> Discharge rate, discharge depth, and discharge cycles are also factors. <S> If the discharge amperage is low, and not discharged too often or deeply then the voltage divider comparator circuit will do the job. <S> Otherwise it cannot be done well without a micro-controller. <S> Even at that it would need to be on 24/7 to count the number of discharge cycles and measure the depth of the discharge. <S> Li-ion Power Cell at a 2A, 10A, 15A and 20A discharge Comparator Circuit <S> Simple $2 device with current sourced LED drivers. <S> Use only the outputs needed.
| For less accurate gauging (which is what I am assuming you are doing) using the battery's discharge curve to relate the voltage of the battery with capacity remaining can be done with a resistor ladder and comparators at each branch, like was previously suggested.
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Just how does high VSWR damage RF amplifiers? Just how is it that a high VSWR can damage the final transistors in an RF power amplifier? Is the transmission line significant beyond the effect it has on transforming the impedance of the load at the other end? Or would an equivalent lumped impedance directly at the amplifier's output be just as damaging? Of all the possible impedances that result in a given VSWR, are they all equally bad? Is the reflected power "absorbed" by the amplifier? For example, if I'm getting 100W reflected power, is that more or less the same as putting a 100W heater on the amplifier? I've also read that excessive voltage can be the mechanism leading to damage. How is it that a voltage higher than the supply voltage can appear? Is there a limit to how high this voltage can be in the presence of an arbitrary mismatch? <Q> Just how is it that a high VSWR can damage the final transistors in an RF power amplifier? <S> Is it simply the wrong impedance (after transformation by the feedline) appearing at the terminals or is the transmission line in particular important? <S> It depends on the design of the amplifier you're using. <S> If the reflection coefficient seen by the amplifier is -1 (thus \$\rm{VSWR}\approx\infty\$), that's equivalent to driving a short circuit, and you can see why that would be an overload condition for just about any type of amplifier. <S> If the reflection coefficient is +1 (again \$\rm{VSWR}\approx\infty\$), that's equivalent to driving an open circuit. <S> If you're amplifier's output stage looks like a common emitter amplifier with resistive pull-up (for example a CML buffer), that's not going to be a problem at all. <S> In some other amplifier configuration with reactive elements, the increased output voltage could cause breakdown of the output devices, for example. <S> Is it reflected power being absorbed and dissipated in the transistors or something else? <S> If the output of your amplifier has a real part to its output impedance, then that would imply that it is absorbing the reflected wave. <S> However the reflected wave will likely be coherent with the outgoing wave the amplifier is producing. <S> Thus it's possible that interference effects between the two waves enhance or reduce the possibility of damage to the amplifier, depending on the phase relationship between them. <S> If you're driving a long line, then small changes in the signal frequency, or even the temperature of the line, could change the reflected wave phase significantly, so it would probably not be a good idea to try to design on the assumption that you can control the phase of the reflection. <S> If you're driving a short line, then controlling the phase of a reflection by controlling the line length is a common practice, done every time we use a stub or shunt as a matching filter, for example. <A> It's a reflection problem. <S> If the antenna in particular is not matched with the feed line power is reflected back down the feed line. <S> This leads to a standing wave on the feed line of nodes of high voltage where the incoming wave reinforces the reflected wave. <S> A VSWR meter reads the proportion of the transmitted wave that is reflected back giving you some idea of the size of the problem. <S> The higher the VSWR <S> the higher the voltage at high voltage nodes and it is this that does the damage to the driver electronics. <S> Most Higher power radios these days detect the VSWR and shut down or reduce power to avoid damage. <A> There are really only a few things that kill RF power devices: Over Current (You can burn out the bond wires) Over Voltage (A typical device running at 100V <S> (~50V rail) will fail if Vds exceeds ~130V even momentarily). <S> Over Drive (Especially MOSFET and LDMOS style parts, but also tetrodes), gate puncture or overheating the control grid. <S> Voltages and currents can clearly be increased by a reflection having the appropriate sign, as can power (Safe operating area) if the reflection produces high voltage across the device at the same time there is a lot of current flowing. <S> Over drive can have you via the reverse transfer capacitance or the feedback network if device stability is compromised by the fault. <S> Most rf amps lack the headroom to cope with a highly reactive load, because that costs money. <A> Usually, a radio-frequency power amplifier is followed by some kind of impedance-matching network (likely including inductors and capacitors) to transform the load resistance to something that the power transistor can cope with, considering its voltage and current handling abilities. <S> A transmission line may also be associated with this network. <S> But after all, at the operating frequency, the power transistor sees a desirable load resistance. <S> An amplifier designer also ensures that at all other frequencies the matching network presents an impedance to the power transistor that ensures no spurious oscillations. <S> An example 7 MHz amplifier . <S> A MOSfet drives a 50 ohm load through a matching network consisting of a low pass filter of L's, and C's. <S> It can deal with peak current of 3 A, and peak voltage of 90 V. With a 50 ohm load (blue), it operates within these limits. <S> But a 1 ohm load (green) causes peak current to be excessive and peak voltage to exceed MOSfet breakdown. <S> A 1000 ohm load (red) is acceptable in this case. <S> Note that this SPICE run neither creates smoke, nor shows what happens when drain voltage or current exceeds limits. <S> No transmission line is included here. <S> For a different matching network, or a transmission line whose length could vary, these results could drastically change, possibly exceeding limits for the 1000 ohm load. <S> A conservative designer might employ a MOSfet having larger limits, yielding a stable amplifier that remains within limits for any load impedance.
| Over heating, should be obvious, but high power devices often run the junction within a few tens of degrees of failure at full power.
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What should the minimum voltage rating of replacement power resistors be? I'm looking for four power resistors of the same type to replace R87, R88, R89, and R90 in the POWER AMP section of the amplifier of Klipsch SW-12 subwoofer shown below: What should the minimum voltage rating of the following new resistors be? Choice 1: 5W 1.5K ohm Choice 2: 5W 1K ohm Choice 3: 7W 1k ohm Choice 4: 7W 750 ohm Choices 1-3 all have their voltage rating listed (350V - 500V), but I can't find the voltage rating for Choice 4: http://www.newark.com/multicomp/mcprm07wjp751b00/metal-film-resistor-750-ohm-7w/dp/16R1987?exaMfpn=true&categoryId=&searchRef=SearchLookAhead&searchView=table&iscrfnonsku=false Datasheet: http://www.farnell.com/datasheets/1679919.pdf?_ga=1.219978905.1878132867.1486504283 If I can't find the voltage rating for Choice 4, should I drop it from my list of choices for the replacement resistors? [A picture showing R88 and R90 with four other white resistors: Added 2/13/17] There's a similar-looking row of six white resistors on the other side of the amp, too (bottom in the picture). [Added 2/13/17] I received a reply from a US distributor of the Choice 4 resistor: Received: Mon, Feb 13, 2017 9:33 amSubject: MCPRM07WJP751B00 Hi The max voltage rating is 500v. I don’t have a data sheet to send but the product manager was able to contact the mfg to get the voltage. Have a nice day. . . . . . . . . . . . [Added 2/14/17] This is the resistor's listing that prompted me to ask a question about voltage rating for resistors: http://www.newark.com/cgs-te-connectivity/sbche61k0j/wirewound-resistor-1kohm-7w-5/dp/16R6555 I hadn't been able to find it until this morning. <Q> I wouldn't worry about this too much. <S> The implicit minimum voltage rating of a resistor is V = <S> sqrt(W <S> Ω) where V is the voltage, W <S> the power in Watts, and Ω the resistance in Ohms. <S> If it were anything less, then the resistor couldn't dissipate the rated power. <S> The lowest of your choices is the 5 W 1 kΩ resistor, which must be able to handle at least 71 V to dissipate its rated 5 W. <S> This circuit has ±81 V supplies, with four of these resistors in series from one supply to the other. <S> Assume <S> worst case that the center point can be driven to either rail, <S> so 162 V across two resistors in series. <S> It looks like the resistors will divide that fairly evenly, so a bit over 80 V across any one of them worst case. <S> You're not going to be able to find a resistor that meets the power and resistance requirements that can't handle whatever voltage this circuit can possibly throw at it. <A> The question now says, "minimum voltage rating". <S> The answer is obvious: since the power supply is + <S> -80V <S> , the resistors could not be exposed to signals of more than 160V. <S> This is the formal answer - 160V. <S> Any resistor of this size would probably exceed this rating. <S> I am not sure though what the actual problem is that needs to be addressed. <S> What has failed? <S> Why to replace the resistors? <S> Too high temperature of the resistors, so they de-solder and fall off? <A> Each 750ohm part dissipates <2.5W or 50% of 5W suggested rating so with convection air flow and raised above the board should be ok with ventilation. <S> Expect it to be 100'C or 80'C above ambient. <S> Any smaller rating will need more air flow.
| Any resistor you find that can dissipate 5 W or more is going to be physically big enough to handle well over 100 V.
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Do analog circuits exist that are essentially mathematical functions in DC? I am curious if a class of analog circuits exist that would take say a 0-5 V input signal and output a mathematical function like a sine function mapping 0-5 V to 0-2pi, or a log, exponent, polynomial? I could see the associated coefficients also being DC analog inputs, say V_out = V1 * V_in ^ 2 + V2 * V_in + V3 I know op-amps can be configured as differentiators, integrators, adders, multipliers, so one way might be to have a high frequency oscillator working with these components to do a DC-AC-DC type conversion, but this seems overcomplicated. <Q> Analog Devices have papers on the subject, e.g. this one on log amps . <S> That paper links to <S> the AD538 : <S> The AD538 is a monolithic real-time computational circuit which provides precision analog multiplication, division, and exponentiation. <S> The combination of low input and output offset voltages and excellent linearity results in accurate computation over an unusually wide input dynamic range. <S> Laser wafer trimming makes multiplication and division with errors as low as 0.25% of reading possible, while typical output offsets of 100 microV or less add to the overall off-the-shelf performance level. <S> Real-time analog signal processing is further enhanced by the device's 400 kHz bandwidth. <S> It's important to note those limitations. <S> 400kHz bandwidth is not very high compared to doing the same arithmetic in the digital domain. <S> That's one of the reasons why almost all control systems computation is done in digital; the others are better linearity, noise tolerance, thermal stability and power consumption. <A> Do analog circuits exist that are essentially mathematical functions in DC? <S> google analog computers. <S> that's why "opamp" was invented and how they got their names. <A> Yes, quite a few elementary functions are available in analog circuits. <S> Heating of a resistor producesa thermal signal proportional to current-squared, and voltage bias ofa semiconductor diode (or transistor) produces an exponential current, and both theseeffects can have high precision (better than 1%) application. <S> square-law current meter <S> Logarithm converter <S> It is also possible to use voltage-clipping techniques, with Zener orrectifier diodes, or comparators, to modify a linear response witharbitrarily many 'breakpoints'. <S> The resulting networks are awkwardto design, but have, historically, been employed to make sine andother functions. <S> Many compensation schemes use (or used) suchtricks to linearize a thermocouple temperature scale.
| There are accurate amplifiers and summing circuits, which covers linear functionsof any available input.
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How to minimize the effect of threshold voltage mismatch of a current mirror? Consider the current mirror below. As you can see, the circuit is affected by a threshold voltage mismatch, modelled by \$ V_{mm} \$. I want to minimize the effect of the mismatch and dimension the circuit accordingly. So from my point of view, what needs to be done is adjusting the length of the MOSFETs, because of the short-channel effect, which causes a decrease of threshold voltage by lowering the length. Since \$ V_{mm} \$ is unknown, it is impossible to design the length in a way that the mismatch would be compensated. Therefore, the best thing would be to increase the length of both transistors to increase their threshold voltage, until the short-channel effect is not noticeable anymore. Then, a mismatch would have the smallest effect on the circuit. Is my above reasoning correct? Is there anything else I can do to diminish the effect of mismatch? simulate this circuit – Schematic created using CircuitLab EDIT: Concerning the influence of the effective channel length on the threshold voltage: An interesting phenomenon observed in scaled transistors is the dependence of the threshold voltage on the channel length. As shown in Fig. 17.5, transistors fabricated on the same wafer but with different lengths yield lower V TH as L decreases. This is because the depletion regions associated with the source and drain junctions protrude into the channel area considerably, thereby reducing the immobile charge that must be imaged by the charge on the gate (Fig. 17.6). In other words, part of the immobile charge in the substrate is now imaged by the charge inside the source and drain areas rather than by the charge on the gate. As a result, the gate voltage required to create an inversion layer decreases. Since the channel length cannot be controlled accurately during fabrication, this effect introduces additional variations in V . The implication of this phenomenon in analog design is that if the length of a device is increased so as to achieve a higher output impedance, then the threshold voltage also increases by as much as 100 to 200 mV. [Razavi] <Q> How to minimize threshold voltage mismatch of a current mirror? <S> I want to minimize the effect of the mismatch ... <S> minimizing the mismatch and minimizing the effect of the mismatch are two different things. <S> for the former, no choice other than a different topology, pre-selection / classification of transistors before putting them in a circuit, or in the case of an IC, process uniformity. <S> for the latter, more options are available, like degeneration. <S> You probably want to pick one of the two for a more in-depth discussion. <A> Your reasoning is correct. <S> However you have to take into account that by increasing L (or decreasing W), overdrive voltage will increase for a fixed Id current value. <S> In circuit where the current consumption is determined by a current mirror (classical differential pair, for example), a big current mirror Vod will reduce your output voltage swing. <S> There is normally a trade-off between the "quality" of the mirror and the Vod voltage. <S> In terms of design, the inversion coefficient of a current mirror should be set to approximately 10, which is translated in a Vod of 200 mV. <A> Firstly, let's look at the current equation (EKV unified): $$I_{f,r} <S> = <S> \frac{W}{L}2 <S> U_{T}^2\frac{\mu C_{ox}}{ \kappa}\ln^2 \left[1 + e^{\left({\kappa\left(V_g-V_{T0}\right) - V_{s,d}}\right)/\left({2 U_{T}}\right)} \right]$$ <S> If you want to change the threshold mismatch, you can: <S> make \$W/ <S> L\$ very large, move \$V_{T0}\$ in some manner, or move \$V_S\$. <S> Assuming that you do not have FAB access, I would just put two FETs at the sources of the currently existing nFETs, and then set their currents with separate bias voltages. <S> This will move \$V_S\$ on the devices so that you can match them. <S> In your image, \$V_{ss}\$ is just tweaking the offset between the two thresholds, but in practice, that would be hard due to noise.
| You can shift the threshold voltage, or effective gate voltage.
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PID : What does « ultimate gain Ku » in Ziegler–Nichols method mean? According to the PID controller Wikipedia page, in its subsection Ziegler-Nichols method , as in its Ziegler-Nichols page, it is said that using "ultimate gain Ku" could help to tune a PID. Unfortunately, I'm not sure that I understand well what is the "ultimate gain Ku". Is the "ultimate gain Ku", the highest value that could reach the system? Or, is the "ultimate gain Ku", the largest error value between the targeted value and the value delivered by the system? For example, if we want to target a value of 17 volts, and if we want to accept only a maximum of 24 volts, which means the highest error will be a positive +7 volts, is the "ultimate gain Ku": 24 volts ? or 7 volts? <Q> Let we take for example a P-controller, which is simpler to understand. <S> The output of the controller is the amplified error. \$Y_{\text{out}}=K(X_{\text{SP}}-X_{\text{PV}})\$ <S> Where \$K\$ is the gain, \$X_{\text{SP}}\$ is the setpoint value and \$X_{\text{PV}}\$ is the process value. <S> The error will tend to be lower as much gain you will apply, but because the system has a lag it will start to oscillate. <S> It is the gain where the system STARTS to oscillate. <S> With big letters it starts, because that is the real ultimate gain, giving more gain than ultimate <S> is it obviously that will oscillate, too. <S> But if you lower the gain from ultimate gain, then the system will be brought to a stable condition, back. <S> Well, in short: It's the gain when the system starts to oscillate. <S> An analogue depiction would be like black hole. <S> When you come close to it, it will attract and you will fall into it. <S> The speed that will allow you to spin around without falling into it, is the critical speed. <S> You can't go back, but you will not fall into, something like satellites that are orbiting around the Earth. <S> The depiction is the type of response: <S> a- <S> stable, b-non stable <S> , c- on the margin (controller response with adjusted ultimate gain) <A> OK. <S> Then, here are the answers to my two questions: <S> Question 1 was: Is the "ultimate gain Ku", the highest value that could reach the system ? <S> Answer is : no, the "ultimate gain Ku", IS NOT the highest value that could reach the system. <S> Question 2 was: Or, is the "ultimate gain Ku", the largest error value between the targeted value and the value delivered by the system ? <S> Answer is : no, the "ultimate gain Ku", IS NOT the largest error value between the targeted value and the value delivered by the system The "ultimate gain Ku", of the Ziegler–Nichols PID tuning method, is an amplifier of the error, but not the error, nor the highest acceptable value. <S> To explain this, according to classic Ziegler–Nichols PID tuning, when Ki and Kd are equal to zero, it shows what Ku is:commande = 0.6*Ku*(error) <S> , where error = (target - value). <S> According to this, "Ku" (as "Tu", the other parameter of Ziegler–Nichols PID tuning method) has to be determined regarding experience of the system, or using a "calculation model". <S> Here is an example of a calculation model using a spreadsheet in Libre Office. <S> The highest limit value is 24. <S> The target is at 17. <S> Equation of classic Ziegler–Nichols PID tuning method is the basis of the calculation. <S> Choosing different values of "Ku" and "Tu", gives a variation input in the value for the next round. <S> Depending to values of "Ku" and Tu", the curve will be different, and corrected values made by the PID, will be in an acceptable range or not. <S> Varying those "Ku" and "Tu", goes to a model wished. <A> I would like to add extra small thing on marko's answer. <S> Imagine we have this system <S> Its a type zero 2nd order system. <S> You are given only a proportional controller and you can't decide what value of \$K\$ should be in the system. <S> You probably know that in this prop. <S> controller changing the gain will change how the system responds to this step input and will also change the system steady state error. <S> This is the relation between the system steady state error and the prop. <S> gain constant \$e_{ss}=\frac{1}{1+K_p}\$ where \$K_p\$ is the static position error constant <S> \$K_p = <S> \frac{K}{15}\$; <S> Its quite obvous that increasing the gain is a good thing since it will lower the system steady state error but looking at the system root locus will tell you the gain limit that increasing your system gain more than this value will result instability. <S> You can tell from the root locus that a gain equals or more than 195 will lead to instability, but below 195 your systems is stable. <S> The value that the root locus intersects the imaginary axis at is the ultimate gain. <S> You can also get this value using Routh–Hurwitz
| The ultimate gain \$K_u\$ is the marginal gain, a point of non return.
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How to light a bulb from an audio signal? So this is my first post, hello everyone. :D Well, I am trying to make my computer/synthesizer light up a bulb - Reacting to the voltage. Right now this is working with an LED: (I using a 2n3904 and some random LED) But when I am using an incandescent light bulb (12v, 30mA) it does not work. I have removed the 1k resistor from the supply, and tried different lower value resistors for the transistor base. I even added another transistor to make more gain, but with no luck. Maybe I need more energy to light it up? A capacitor? More gain? Or am I totally forgetting obvious (Diode vs. bulb)? Please enligthen me wise people. :D <Q> The basic problem is that the light bulb requires a lot more current to light up than the LED does. <S> You said yourself that the bulb requires 30 mA. From the 12 V and 1 kΩ resistor in series with the LED, it's clear that the LED is drawing less than 10 mA. <S> You can probably see a visible indication with just 1 mA or so. <S> That's a lot less than a incandescent bulb requires to light visibly. <S> There is at least a order of magnitude difference between lighting the LED and lighting the "bulb". <S> To get higher output current with the same input signal, you need more current gain. <S> One easy way to do this is to use a second transistor: <S> Q1 works as your transistor does now. <S> But instead of turning on the LED or bulb directly, it turns on Q2. <S> With Q1 fully on, Q2 gets about 1 mA of base current. <S> It needs a minimum gain of 30 to support 30 mA of output current. <S> Many small signal transistors can do this, including the one shown. <S> This circuit will works to drive the LED, although you can now use a lower resistor in series with the LED to get higher current and therefore higher brightness. <S> Let's say you've got a common green LED that drops 2.1 V and can take 20 mA. <S> A 510 Ω resistor in series with the LED will limit to the current to just under the 20 mA the LED can handle. <S> This will produce more light than the "light bulb" at 30 mA, assuming you mean a incandescent bulb. <A> Current will only start to flow to the base of the bipolar junction transistor once the base - emitter voltage gets above + 0.7 V or so. <S> Thus only signals above 1.4 V peak to peak will get amplified and switch on the LED. <S> While LEDs will light up practically instantly, an incandescent bulb takes time to warm up before emitting light, so the transistor has to conduct for a much greater portion of time before light is produced. <S> Another problem with your circuit is that there is no path for charge which has flowed to the base of the transistor to replenish, as the transistor won't allow reverse current flow through the base at low voltages. <S> The audio input is most likely capacitively coupled at the source, so this DC bias current can't come from the audio source either. <S> The end result is that the circuit will build up a negative DC voltage at the audio line, which makes it progressively harder for positive peaks in the audio signal to overcome the combination of the base-emitter voltage drop and the generated input DC offset voltage. <S> Eventually just leakage currents trough the capacitor and transistor will be replenishing the charge at the base, which is apparently just enough to illuminate a LED. <S> The simplest way to fix this would be to add a 1uF or greater capacitor between the audio input and the base resistor, and a schottky diode between the ground (anode) and the base resistor - capacitor node (cathode). <S> This way the capacitor will charge trough the diode during negative peaks in the audio, and discharge trough the transistor base during positive peaks. <S> A normal diode will work too, but schottky diodes have a lower voltage drop (e.g. 0.2 V instead of 0.7 V), allowing for the circuit to be more sensitive. <S> If this isn't sensitive enough, search for "audio peak detector" on the web. <A> Since the circuit is working with LED, the 2N3904 might not have the sufficient DC gain to drive the incandescent lamp current, replace it with TIP127, which has a gain of 1000 compared to 2N3904 20 - 70 <A> it does not work. <S> That has very little information to help others help you. <S> A description of how it didn't work would have been a lot more helpful. <S> With that said, a few possibilities. <S> The bulb is broken. <S> The bulb is miss labeled. <S> The power supply cannot supply enough current. <S> The transistor didn't get enough base current. <S> The music isnt loud enough to get the transistor to conduct. <S> Thee music is too loud so the transistor is always conducting. <S> The music is devoid of low frequency content so the bulb isn't blinking - this type off circuits only works with low frequency signal. <S> It is always on to naked eyes if t is from. <S> High frequency signal. <S> ... <S> If I were you, I would drive the base high and. <S> Then measure the voltage on the transistor. <A> Use an LED with an Diodes Incorporated AL8805 <S> In the schematics below the CTRL is for a PWM signal. <S> This circuit drives an LED or string of up to 10 LEDs with 330mA to 660mA. <S> $7.95 Or RGB $14.95
| The bulb isn't placed correctly or securely in the circuit. Just about any cheap green LED you can find can do this.
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Wire color coding: not sure about the correct wire Today, I had a new PC screen at home, but without power adapter. I got the one from the old screen and I had to replace the original plug for another one, that matches with the new screen. The one problem I have is the color code. The power adapter has two output wires: one black and one white, and the other cable (the correct plug) has three wires: browm, yellow/green and blue. The problem is: I don't know wich wire is the correct one (+/-), not sure if black is + and blue is + too or these are the other wires. I would like to know what respresents each color code in this case (black, white, green/yellow, blue). I can't attach any photo of it because the phone's camera is broken, so I'm adding a little "description": From power adapter: - Black- White From the new plug for the screen: - Green/yellow (both)- Blue- Brown Thanks. <Q> The yellow/green is earth. <S> If there are no metal parts exposed (and some other requirements) <S> this wire might not be required. <S> The orientation of the other two wires does not really matter for getting things to work. <A> Probably it's as follow: <S> Black: phase White: neutral <S> On the other case: Green/yellow: earth <S> Blue: <S> neutral (usually very close to earth in monophase systems) <S> Brown: phase <S> Don't you have a polarity tester so you can check it? <A> but isn't required. <S> The brown is the live wire - you can remember this by it being the colour your trousers turn if you touch it. <S> The blue is the neutral wire. <S> Here is a page on the wire colours and their meanings for different parts of the world: Wiring Colour Codes
| The live wires are brown = black and the neutrals are blue = white. In the UK: The green/yellow wire is the earth cable - it is for safety and should generally be used if its there - It is required for safety but not for powering up the display.
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How can I reduce B+ from 460VDC to 420VDC in the power supply of a tube guitar amplifier? 1977 Deluxe Reverb. Schematic calls for a power transformer input of 120VAC, and a secondary HT winding producing 330VAC (x2, center-tapped). This gets rectified by a 5U4GB (which consumes 50VDC in the process), and is filtered to produce approximately 420VDC B+, and 415V on the power tube plates. My original power transformer appears to have been wound 'hot'. At approximately 120VAC wall voltage, my HT winding is at 377VAC, not 330. The downstream effect is that my B+ is closer to 465VDC. At the same time, the heater winding is right where it should be, 6.3-6.4vac, at 120-121vac input. So it seems isolated to the HT winding. The amp works fine, and has for years, but in the interest of learning and experimentation, I'd like to bring the B+ down to spec levels. I'd like to drop 35-40VDC from the B+. I've googled and come up with shunt regulator, bucking transformer, etc. I don't know what these mean in practical terms, in the context of a vintage tube amp. I'm hoping there's some component, or small analog circuit, I can place after the rectifier and filtering, and before the standby, to reduce the voltage. I'm aware there are other ways to reduce the B+. I don't want to apply a variac to reduce the input voltage, because it affects the heaters, too. And for this discussion, I don't want to swap the 5U4GB rectifier for a 5R4GB. While feasible, it would only drop a further 10-15V. Also, it's my understanding that this solution would introduce more 'sag' in the power supply. Finally, the capacitor input voltage of the 5R4 is only 20uf, and my (spec) primary filter caps are (total) 32uf. Probably not an issue, but still. In short, I'm trying to get the amp closer to spec, not farther. <Q> There are many ways of reducing the plate voltage on valve equipment .I <S> have done this on radios but not on a guitar amp .The <S> most common way I have done this is to place experimentaly 2 100 ohm <S> 5 watt wirewound resistors in the HT transformer <S> secondary .One <S> in series with each leg before the rectifier .You will notice the HT drop .Now you can ballpark calculate your final resistor value which in general wont be 100 ohms .The <S> resistors get pulsating DC currents with a high peak to average ratio which is why the correct value is lower than most people think. <S> The total power that the resistors waste is the same as a resistor in the DC side or a series Zener or a linear reg .If <S> your total filter cap is generous which is the expectation of high Quality Audio and you are doing a lot of class A <S> then your concerns about droop need not be concerns .This <S> AC HT resistor approach does have the advantage of providing additional protection .Although <S> I have not seen this I have heard about rectifier tubes flashing over and ruining HT transformers .The <S> AC resistors limit fault current and hence Arcing and they will fail open circuit .The <S> AC resistors make the rectifier current pulses slightly broader making the transformer copper slightly cooler . <A> Obviously these diodes will disipate a lot of heat and need a heat-sink. <A> there probably isn't a whole log of current going through those things. <S> If so, put some bulbs, or even LEDs in there. <A> Honestly, I would leave things just as they are. <S> One benefit of higher supply voltage is more output power before clipping occurs. <S> FWIW - I used to run my old tube-type amplifiers <S> Really hot, with excellent results. <S> So long as the filter caps are holding up, just leave it alone. <A> Put a couple of zener diodes in series with the red/yellow power transformer center tap to ground. <S> These need to be reversed polarity in order to pull the voltage down. <A> The power dissipation of a 100Ohm resistor with 20V dropped across it is: <S> P= Vsquared/R = <S> 20Vx20V/100R = 4W. <S> If you require a smaller resistor to drop the 20V the power dissipation will increase, so calculate the power dissipation and select a resistor accordingly.
| Putting resistors in series with each leg of the secondary windings of the PT will drop voltage but don't forget the power rating of the resistors - when dropping 20Vac on each each leg with a resistor, 5w may be insufficient. You can lower the voltage by adding a huge 40V Z-Diode into your B+ supply.
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Does Registers and RAM are same kind of memory? Recently, I found out a YT channel of a man who build a 8-bit computer and explain how. In his video on the registers, he uses a D flip-flop, and in his other video on the ram, I understood that he says he'll use the same system as the registers, and that's called 'Static RAM', and the 'Dynamic RAM' is faster. But I think that the registers are faster than RAM, so if dynamic ram is faster than static ram, which uses the same kind of system than the registers to store, they are slower than dynamic ram... Did I not understand the videos or something else? <Q> There is DRAM <S> which uses a capacitor to store a charge to represent the stored bit. <S> A row of transistors is used to connect a row of capacitors to the read/write circuitry to put the data into a set of latches to read it and also feed it back into the row to refresh the capacitors' charge which was depleted into the latch's input. <S> The capacitors will self discharge over time <S> so each row needs to be refreshed periodically. <S> In most DRAM you can only read or write a single row at a time. <S> This type is used for the main RAM because it is cheaper in silicon per bit. <S> Then there is SRAM <S> which uses flipflops to store the data which takes a half dozen transistors per bit. <S> It has a constant output available so can be read at any time using a MUX, instead of having to wait until the proper row can be selected. <S> This type is used internally for nearly everything on the cpu (including caching the current row in DRAM) because it is much faster. <A> I think that the registers are faster than RAM depending on the context under which the term "register" is used. <S> typically they are part of the RAM that the cpu has special access to. <S> cpus typically (=not always) have faster access to those memory addresses. <S> other times, "register" is nothing but ram that is specifically wired to certain peripherals. <A> the registers are faster than RAM Today? <S> Yes, very much so. <S> In the 80s era of 8-bit computers? <S> Off-chip RAM is nearly always DRAM. <S> On-chip RAM is usually SRAM, the same technology as the registers. <S> Confusingly, PIC architecture designers make no distinction between on-chip RAM and registers and just call everything a "register".
| No, it was often possible to access the RAM in one processor cycle because the processor was comparatively slow.
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Why MOSFET makes Motor ON directly? In my project I have to operate DC motor. For this I am using Arduino as controller and MOSFET for switching. I have connected MOSFET GATE pin to Arduino's pin 12. I have used this schematic.... Main problem is as I give +12v supply to one end of Motor and other end of Motor to Drain of MOSFET and other connection same as Image.But without connecting Arduino output pin to Gate of MOSFET motor staying on continuously. In my case I am controlling GATE pin using pin 12 of Arduino,which will send pulse to MOSFET to control motor(ON/OFF). What wrong am I doing? <Q> The circuit you show should work. <S> You probably don't have something connected correctly. <S> For example, if the drain and source of the FET were flipped, you'd get exactly the symptom you see due to the body diode of the FET conducting. <S> Check the FET datasheet and your connections carefully. <S> Measure the gate voltage with a voltmeter, and verify that it really is 0 V. <A> Two possibilities: <S> Careful measurements with a multimeter should show the Source-Drain connection as a diode - if it's wrong way round, the diode conducts, running the motor. <S> Electrostatic damage can destroy the MOSFET. <S> Be more careful with the next MOSFET and follow proper electrostatic handling procedures. <A> Additionally, but highly unlikely you might be dealing with a depletion mode MOSFET which is a "normally on" device as oppose to the common enhancement mode MOSFET which are "normally off" devices. <S> Depletion mode MOSFET are not as widely available as enhancement mode devices but there as still a few available and since you do not posted a actual part number for the device you are utilizing <S> I believe this could be highly unlikely but plausible scenario <A> What kind of current does the motor draw, & which MOSFET are you using? <S> If you draw too much current, the MOSFET will be ruined, and, they often go to low resistance on the Source to Drain, I've seen enough blown up power supplies to know this. <S> The Gate pin should read low resistance to the Source pin, but, high resistance to the Drain pin. <S> Also, many MOSFETs in the TO-220 package have the 3 leads Gate-Drain-Source, viewed left to right, <S> BUT, not all! <A> MOSFET has to be open if You want to drive a motor through transistor. <S> You are using MOSFET with N channel, that means gate voltage has to be positive (like +5V from Arduino), in other cases it won't run.
| There are MOSFETs with zener diodes built in, to reduce ESD damage potential. If it appears to be a short circuit both ways round - OR if excessive leakage to the gate means there's voltage across R2 - that's what has happened. Incorrect connection of the MOSFET - especially swapping Source and Drain will do this.
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How do I mix two audio tones from two 555s to make a polyphonic chord? I have set up two 555 chips in astable mode to produce a tone each through a speaker. However, I am not sure how to mix them together to produce a musical chord. If I directly combine the two outputs from the 555s into a single output, it creates a third tone, but it is just a single tone being some combination of the two frequencies. Instead I would like to hear each one independently, but on top of each other, just like a keyboard (or a hammond organ if you prefer). I have tried using an op-amp but with the same result. Is an op-amp the solution here? Or is there some better strategy? I just need a push in the general direction. UPDATE Both methods below seem to work, but it turns out the main problem was that the two 555s need to be "decoupled" by putting capacitors across pins 1 and 8. That is, when one 555 was sucking power from the battery, it was affecting the frequency of the other one. This (along with the large resistors from the 555s to the opamp) has more or less solved the problem, although the sound quality is not that responsive or stable. Thanks for the help! <Q> The reason you are getting something other than the sum of the two independent signals is because the two circuits are interfering with each other when you connect their outputs. <S> One solution is to put enough resistance in series with each output so that the signal generators can't interfere with each others' operation. <S> 100 kΩ resistors in series with each signal generator output should be high enough that it doesn't matter what is going on at the other ends of the resistors. <S> The problem now is that the combined signal is high impedance due to all those series resistors. <S> The solution is a unity-gain buffer. <S> Any rail to rail opamp that is unity-gain stable can take the high impedance signal from where all the resistors are connected and make a low impedance signal that can drive a power amp or some other audio equipment. <S> Here is what I described above: <S> The resistors are large enough so that the circuits on their left ends are not effected by whatever is happening at their right ends. <S> Together, these resistors create the average of all the input signals. <S> If any of the inputs are unused, leave them floating and they won't be included in the average. <S> The standard knee-jerk answer for something like this is a inverting summing amplifier. <S> I figured sooner or later <S> someone would post that. <S> I'm actually surprised it took as long as it did. <S> The reason I didn't go that route was for simplicity, especially in handling the DC levels. <S> Just connecting each signal generator to the inverting input via its own resistor does NOT get you what you want unless you are prepared to deal with the resulting negative voltage, and have a negative supply handy to provide to the opamp. <S> The inverting summing circuit can be modified to not require negative power voltage and to create a 0-5 V output, but that adds a little complexity. <S> Another minor issue is that you really want the average, not a sum. <S> The difference is only one of scale factor, so can be dealt with. <S> The circuit above inherently averages. <S> Any input left floating isn't included in the average. <S> The inverting summing amp has a fixed gain from each input to the output. <S> If the number of active inputs change, the overall gain needs to be manually adjusted so that the result is the average of the inputs. <A> AN op amp is what you're going to want, but you need to hook it up properly. <A> What is interesting to the ear (the richness) of a musical chord is not the sum of the two fundamental frequencies added together <S> it is all of the side band frequencies that are included as well. <S> You have two problems. <S> First the output of a 555 is a square wave. <S> This includes a number of frequencies from the fundamental to infinity (perfect square wave) which are going to interfere with the output. <S> The second is how do you effectively add these together. <S> For the first problem you need to shape the waveforms coming from the 555 what you do here is application dependent. <S> The chord from a violin is different from the chord from an acoustic guitar. <S> This is because of the difference in wave shape from a single string from both instruments. <S> Do some research here and decide on the sort of sound that you need. <S> The second problem has been addressed. <S> You need to add these voltages together in a manner that one does not interfere with the other. <S> I would suggest a capacitor to feed each signal to a high input impedance amplifier. <A> For two pitches to sound like a chord, the waveform that gets output to the speaker has to be the sum or difference of the individual raw waveforms. <S> Passing the outputs of the 555s through a voltage divider (which would yield the average voltage from the two outputs) and then through an amplifier would be one way of achieving that. <S> A simpler approach when using two 555s, though it wouldn't scale up to more than two 555s, would be to drive the two ends of a speaker with outputs from the 555s (likely adding a resistor and/or capacitor to protect the 555s, or maybe feeding each output through a high-powered inverter and then a resistor/capacitor). <A> I'm not sure if you are solely looking for a hardware fix (as my input is more software related) <S> but I'll throw it in anyway! <S> If your driver (sound source) is driven by software then as has been mentioned, you are probably looking at square wave tone synthesis so your hardware can play A# but cant make it sound like a violin as opposed to a cello. <S> LEDs use square waves but can trick the brain into seeing the light at different brightness's through on/off speed control <S> , have you tried this with your 555s <S> e.g. play tone A for 5Ms and then tone B for 5Ms, <S> that way you are truly playing top & bottom of the wave range (doubt if 5Ms is a long enough alternating time - too fast <S> and you'll just hear something similar to your existing problem, too slow and <S> it will just sound like 2 tones alternating)
| What you are looking for is a summing amplifier like this: If you put your 555's on Va and Vb, the output will do what you want.
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Change in transformer efficiency (leading power factor) with increasing load current Does efficiency increase or decrease when the load current increases (assuming copper losses equal iron losses along with a leading power factor load)? <Q> Does efficiency increase or decrease when load current increases (assuming copper losses equal iron losses, leading power factor load)? <S> The simple scenario of zero load current has to imply that output power is zero therefore power efficiency MUST be zero. <S> So, taking a little bit of load current MUST cause efficiency to increase because you have an actual output power (assuming the load is resistive). <S> If the load is leading by 90 degrees then under no circumstances will power be consumed therefore efficiency has to always equal zero. <S> However, if the load is somewhere between resistive and capacitive (i.e. complex) then any current taken will produce an output power and therefore efficiency can be seen to increase. <S> At some point there will always be turn-down of power efficiency but without a real scenario with a fully defined transformer equivalent circuit and load <S> it's hard to say where this turn-around will occur. <A> This core-loss term means low efficiency at low current, which is quickly rising with the current. <S> Copper losses increase proportional with the square of the current, which means that this copper-loss term reduces efficiency with increasing currents, especially for currents larger than the nominal operating point (but the effect is typically not very strong). <S> Therefore, transformer efficiency rises with the current to a maximum (typically nominal operation point), and then (slowly) falls again with further rising currents. <A> You need to be crystal clear on two facts before you reach to the conclusion. <S> i) <S> if you increase load current then definately output voltage across the load will increase. <S> Thus the output power should increase. <S> but wait a second <S> this isn't the whole thing... ii) <S> when output current increases then <S> the flux linking the secondary winding also increases. <S> This flux has a tendency to oppose the main flux linking throughout the the core. <S> Thus main flux decreases momentarily. <S> In order to maintain constant main flux the primary drives an extra current that is just sufficient to counter the flux created by seconday winding. <S> Conclusion : <S> if you increase load <S> current <S> then input current also increases thus copper loss also increases and hence efficiency falls.
| Core losses are not load-dependent, only voltage-dependent (so one cannot assume that core losses and copper losses are equal for different currents).
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What is your go-to prototyping connector? I have completed a few projects - sensors and an ADC circuit - and I find myself often connecting discrete boards together by tediously soldering wires to break-away headers like those below. I worry about poor connections and the time it takes is excessive. What are some good, flexible go-to prototyping connectors? I am considering getting some RJ45 connectors and using old cat 5 cable, but they aren't 0.1" spacing friendly, which makes prototyping with them harder. (Do 0.1" compatible RJ45 connectors exist? Could I rig these up anyway?) I realize this is highly dependent on the application but I was hoping to get a solution for the general case. Or maybe soldering ribbon cable to headers is the way to go. For a little more information, my current projects have involved carrying a ~2v signal ~1m to an ADC, carrying 1MHz SPI ~0.5m between an ADC and a PI, as well as carrying around 3-5V power and ground - so nothing terribly fancy. Update: I went with IDC connectors (2x5 pin and 2x20 pin) and I am happy with this decision. <Q> Why solder the wires to the header pins when you can use flat cable with IDC connectors? <S> No soldering at all, compatible with 0.1 inch spacing, and the cable and connectors are easy to assemble (and cheap as well). <A> When I have room for 0.1" connectors, I almost always use the KK-style connecters from Molex: <S> They are relatively cheap, have different pin configurations, and (importantly) have been considered "bin stock" in every lab I've worked in. <S> Also, if you choose the connectors with "locking ramps" (as shown), these ensure that you won't plug it in backwards, and also keep things from disconnecting accidentally. <S> You'll have to buy a crimper, but you'll use it for years. <S> A Molex-branded crimper will cost you a few hundred USD, and high-quality generic ones can be had for $20-30 on Amazon. <S> I'm currently using this crimper. <S> I bought it because it's cheap (currently $23), and I was pleasantly surprised at it's quality. <S> It hasn't failed me yet. <S> Another benefit is that you can choose connectors for different wire sizes. <S> Here is an answer I posted about using this type of crimper. <S> To answer your other question, I just ran into this at futurlec.com: <S> I've never done business with them, so I can't speak to their quality. <S> But these adapters are currently selling for $2.90, so it's hard to go wrong :) <A> They're available in a lot of different positions (e.g. 1, 2, 5, 8 pins and even double rows ), which is convenient when having to connect to pins that are not adjacent. <S> You have to buy the pin contacts separately , as well as a pin crimping tool . <S> But this approach has saved me a ton of time and effort trying to solder wires directly to these headers (and the cables can be re-used for other projects once you're done with the current one). <A> In addition to ribbon cables with IDC connectors (as mentioned by Enric Blanco), you can buy individual 'jumper wires' with female connectors (ex Here ) on each end that are compatible with the kind of pins you posted. <S> I've used them to great effect when there are many pins to connect <S> but they're not in a similar layout (eg connecting the gpio+power on an FPGA board to an LCD board). <S> MagJack Breakout or RJ45 Adapter
| For boards that already contain pin headers (like in the picture you provided), I tend to use connectors that mate into the existing headers - similar to this Harwin M20 connector . If you use a breadboard, you can also get similar jumper wires with female connectors on one end and male on the other to be able to wire from pins to the breadboard, and male on both ends for use on the breadboard. If you want something like RJ45 that are prototype-friendly, you can get breakout boards from many places - ex RJ45
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How to display 2 digit number in binary adder circuit? I've made 4 bit adder circuit using 4008 IC. And the sum output of the two 4 bit numbers from that IC was feed to one BCD to 7 segment decoder (74LS47) so that I can get decimal output. But since I can't diplay numbers over 9, I need two 74LS47 ICs. But what additional circuitry shall be made so that it can stop that overflow over decimal no. 9? And suppose my two binary numbers to be added are 101(5) +101(5) I will get binary output of 1010(10). How can I display this using two BCD to 7 segment decoder ICs? <Q> I had once done this as part of my digital circuits lab. <S> Your problem is that you have a 4-bit <S> no. <S> input (0-15) which needs to be given to 2 74ls47 ics. <S> Say the 4 bit input <S> no is a b c d , <S> inputs of 74ls47 ic for tens place are p1 q1 r1 s1 and that at for units place p2 q2 r2 s2 <S> (Here <S> a and p are MSBs). <S> You can now make a truth table with abcd <S> inputs and corresponding values you wish to get at both the pqrs . <S> Next, based on this table you need to solve the Kmap for each of the 8 variables p1-s2 . <S> Each of these variables depends only on the 4 inputs: abcd , So you have to solve 8, 4x4 Kmaps. <S> Looks tedious but all the Kmaps and expressions for pqrs will be quite simple. <S> Quoting from my lab report: p2= <S> a(c')(b')q2= <S> b(a'+c)r2=ab(c')+(a').cs2= <S> ds1=a.(b+c) <S> You don't need <S> p1 q1 r1 to display 0-15. <S> Implement this using basic logic gates. <A> I think this might help. <S> You need to make some changes in this and connect your IC but the use of two 7 seven segment displays can be seen here, get the idea. <A> Use hexadecimal . <S> A single hex digit (0-9,A-F) represents one of 16 states, so 4 bits. <S> This is exactly why hex (and in older times octal, not much anymore) were used. <S> Each digit directly indicates 4 bits (3 for octal). <S> For your own hobby project, and even displaying values in a computing context, you can often use hex directly. <S> If you have to display decimal values to end users, then use a microcontroller. <S> All this dedicated hardware just to generate 7-segment display patterns for decimal numbers is silly in today's world. <S> It can even multiplex multiple digits to reduce the number of connections, and therefore the chip size and routing complexity. <A> The DM9368 solves this problem directly - it displays correct hexadecimal codes, and has constant current drive for the display. <S> The drawback is that it is very power-hungry.
| A microcontroller can do the binary to decimal conversion, then convert each digit to the corresponding 7-segment display pattern.
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Does a 555 astable dissipate more power at higher frequencies? I'm planning on running a circuit on a 9V battery and it uses a 555 in the astable mode to generate pulses. I want this circuit to be as efficient as possible, and not have to replace the battery very often if it can be avoided. Looking at the 555 datasheet , the max power dissipation is 1180mW (page 4) -- but is this lower at smaller frequencies? Or is the chip designed to dissipate the same power at any frequency? I can't find any graphs on the datasheet (p6-7) regarding power dissipation with frequency, which is why I'm asking :/ <Q> Yes it does. <S> Reducing the oscillation frequency <S> \$f_{osc}\$ will have an impact on your circuit power consumption, because you'll be charging and discharging the capacitor less frequently, thus reducing consumption. <S> If you also want to reduce the current drawn by the 555 itself, then the first thing you should look after is lowering the supply voltage \$V_{cc}\$. <S> The efficient way to do this is with a switching regulator to bring down 9V to 5V. <S> A lower \$V_{cc}\$ has the extra benefit of reducing the amount of charge required for the capacitor to achieve \$\frac{2}{3}V_{cc}\$, thus reducing the power consumption. <S> You can have the best of both worlds and combine this with the reduction of \$f_{osc}\$. <S> EDIT: <S> As Jim Dearden has correctly pointed out in his comments: Using the CMOS version of 555 will help in reducing the consumption of the IC. <S> On top of that, the CMOS version can work with higher values of the timing resistor than the bipolar version, thus making possible to reduce the value of the timing capacitor while keeping the same time constant. <S> A lower valued capacitor stores less charge, thus consuming less current while charging up. <S> Double win for the CMOS version, which adds these benefits on top of those already achieved by reducing \$V_{cc}\$ and \$f_{osc}\$. <S> Note: <S> selecting the highest possible value for the timing resistor in order to lower the value of the timing capacitor will always yield a reduction in current consumption, regardless of the version of the 555. <S> So it should be considered for any design in which consumption is a concern. <A> Most of the time, you don't have to care about this value, because it's always depending on trermal resistance on your pcb. <S> Table 6.5 are the electrical characteristics, there's mentioned, that the chip itself can draw 6mA/5V, 15mA/15V in the application. <S> So you are far away from your 1180mW. <S> But please take care of the current flowing through the external components. <S> (Resistors). <S> Maybe consider taking a microcontroller for your application. <S> You can do lot of more things with your limited power. <A> Your bipolar 555 draws several milliamps when idle. <S> CMOS circuits of the same complexity usually draw very low leakage currents (<1µA) when idle, and only draw significant currents when switching. <S> Switching in CMOS is similar to a capacitor charge/discharge. <S> The amount of charge needed is proportional to supply voltage. <S> Therefore, average current draw is proportional to Vsupply multiplied by switching frequency. <S> It is thus very desirable to lower the supply voltage. <S> If you can use a chip that will run on an unregulated 1.8 ... 3.3V, then it will run on 2 AA batteries in series, which will take up about the same space as 9V battery. <S> However, the two AAs have about 4x the capacity in mAh, and lowering the voltage from 9V to 3V divides current consumption by 3x. <S> This increases your battery life at least by 10x, easily. <S> If you can do the rest of the circuit with 2 AAs, then do it. <S> Also, 2AA are much cheaper than one 9V! <S> You can use CMOS like the CMOS 555, 4000 series, CMOS 74 series, or a modern microcontroller. <S> Simply check that they are compatible with your supply voltage. <S> For example, 74LVC and ALVC operate From 1.65 V to 3.6 V. <S> Many modern microcontrollers will also run on 2 AAs until they're thoroughly exhausted... <S> even good old Atmega328P runs from 1.8 to 5.5, altough <S> it's not an ultra low power device. <S> Using a DC-DC converter (switching or linear) is not always a good idea, as its idle dissipation can be much higher than what the circuit actually uses, unless it is designed for this specific purpose. <S> You can also run your 1.8V device from 2 AA using a micropower 1.8V LDO, which will lower voltage, and thus current draw, even more. <S> However... we don't know if your circuit will have some power-hungry loads, so more details would be needed. <A> Examining the 555 schematic, all the front end circuits operate at constant currents using resistors or shared-drain current sources. <S> I don't think these will show much increase in current as Fosc rises. <S> However the output stage, uses a single node to control both pullup and pulldown circuits; classically that "single node" is the source for crowbar or shoot-thru charges for brief times where both circuits are transitioning between on/off and off/on. <S> That node is at the bottom of R11. <A> Does a 555 astable dissipate more power at higher frequencies? <S> generally <S> yes, for two reasons <S> : 1) the chip itself will dissipate more power. <S> the bjt version will need to annihilate charges / holes and the cmos version will have to charge up / discharge those little capacitance more frequently; 2) the output devices will dissipate more power: maybe you are driving a mosfet, or a bjt, or a resistor / capacitor.... <S> It is very rare that everything else being equal, power dissipation goes down with frequency. <S> so if you care about power consumption, look into newer / specialty chips; run it at as low of a frequency as possible; try to find the cmos version. <S> each comes with its own disadvantages, however.
| The 1180mW is the maximum allowed power for not turning too hot under certain conditions.
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How to figure out if a 25V capacitor can replace 50V in this circuit I'm following this video Astable 555 timer - 8-bit computer clock - part 1 where he builds this circuit: simulate this circuit – Schematic created using CircuitLab I just realized that the 1000uF capacitor I have is 25V and not 50V as the one he uses in the video. I suspect that 25V will do but I don't know for sure. I've tested the voltage where the capacitor would have been in the circuit on my multimeter and that reads 55.3mV which is much less that 25V but maybe the 555 timer can change that? Is this a valid way to figure out the maximum voltage the circuit will apply to the capacitor? <Q> The best way to know for certain would be to measure as you did (although you said 'where it would have been' which might suggest you had it removed). <S> In this case since the highest voltage source is 5V, and there is no boost converter topology present to increase the voltage (you would need an inductor to do that), the cap would never see more than 5V. <S> A <S> 25V rated cap would be perfectly fine. <S> In general, if you rely on measurement though, be sure that you check it under all possible operating conditions. <S> Ideally you should understand the circuit well enough to know what potential voltages (and current) may be present at any given component. <S> EDIT: If you plan to take measurements (of a non-steady-state circuit such as this) to confirm your expectations of peak voltage on a cap, it's probably best to use a scope if possible, since meters won't reliably capture brief pulses or transients. <A> Even 10 V would be fine. <S> The voltage on the cap can't exceed 5 V as you show the circuit. <S> Just because "someone on the internet" uses a particular value part <S> doesn't mean that the value is optimized. <S> In this case, it was probably just the value he had handy. <S> On a separate topic, using a 666 555 timer for a "computer clock" sounds like a bad idea. <S> These things aren't anywhere near accurate enough for keeping real time. <S> Unless you pay lots of money, the analog parts are ±10% at best. <S> For a clock, that comes out to nearly 15 minutes error per day. <S> To get 1 minute error per month, for example, you need 23 PPM accuracy. <S> That basically means using a 20 PPM crystal. <S> Note that 10% is 100,000 PPM, rather a lot more. <A> In general just measuring the voltage across the capacitor at one point in time probably isn't the most valid way to figure out the maximum voltage it would expect, though it certainly is a start. <S> If you want to go the measurement route, then you must ensure the voltage at all points in time (all possible operating conditions) doesn't exceed your value. <S> Practically this is somewhat cumbersome I think, unless you have access to an oscilloscope that can record the max voltage over some period of time and probe it with that? <S> Even then, if you do this in the dark (not understanding the circuit) you might not be sure you put the circuit into a condition where the capacitor would expect peak voltage.
| So, likely the "most valid" approach would be to gain a very firm understanding of the operation of your circuit to determine analytically if you should ever expect the capacitor voltage to exceed the supply voltage, if there is any voltage increasing circuitry present, and using empirical measurement to support your expectations. Yes, a 25 V capacitor is fine here.
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Filter-Amplifier circuit for Low Voltage I/P circuitry I'm designing my first ever circuit for a hobby project. I've a sensor that outputs in mV (1 to 2) and a controller with ADC. The ADC is 16 bits, VDDA = 3.0 V, Temp = 25 °C, fADCK = 1.0 MHz. Since the sensor voltage that is fed is to the controller is low I'm in a predicament! Should I use a low pass filter or a band pass filter in this application? Do you have a circuit that I can relate to? I've attached an example plot. Any advice is appreciated. Thanks :) <Q> You also have a secondary problem in that the sensor signal of 1 to 2 mV is way too small for the 0 to 3 V A/D range. <S> That can be solved with amplification. <S> It looks like your signal period is a bit over 20 s. <S> It's not a sine, so some harmonic content looks to be important. <S> You haven't given response time or frequency range requirements, but let's say anything past 2 Hz is noise. <S> For example, let's say you can sample at 10 kHz. <S> That's 100 µs per sample, which is a long time even for cheap and small micros. <S> Two poles of R-C low pass filtering at 500 Hz won't matter to the signal at all, but will squash signals above the aliasing limit of 5 kHz. <S> For example: The two poles are actually nominally at 480 Hz with these part values. <S> This filter would go between the sensor and the amplifier that drives the A/D input. <S> It's purpose is not make the final filtered signal, but just to ensure that there is no aliasing when the signal is sampled at 10 kHz. <S> In the micro, you then run a low pass filter that squashes frequencies above 2 Hz. <S> This digital filter can be more complicated if needed, and won't suffer from analog part tolerances. <S> I'd start with two poles of low pass filtering at 2 Hz. <S> It is unclear from your graph what signal to noise ratio <S> you will get this way, but try that and see. <S> If you substantially reduce frequencies past the valid signal and still don't have enough signal to noise ratio, then you need a better sensor, or you need to reduce the bandwidth requirements. <A> Your signal has too much noise to be useful. <S> I eyeballed a mean signal that might be a high order LPF and although there is correlation in some cycles , not in every cycle. <S> The results are almost meaningless for position sensing. <S> I suggest there are a few problems that can be improved greatly with much more disclosure on details. <S> Suggestions <S> Provide details on design,sensor,cabling, layout and sources of EMI for improvement on EMI reduction. <S> This includes radiated noise and conducted noise. <S> Improvements needed may range from higher gain at sensor, shielding, balancing, and filtering both the noise source and the sensor signal and averaging results. <S> Note that motor noise is likely and <S> ADC noise at Fadc=3MHz reduces accuracy on ARMs from 12 bits to <9 bits without quiet mode etc <S> so 1mV is too low a signal <S> There are lots of free filter design tools if you know what you need. <S> This must be defined 1st. <S> - gain- <S> Passband (PB) freq. <S> ( eg 1Hz or 10Hz)- PB ripple ( <1dB)- Stopband (SB) freq. <S> ( eg 10 Hz or 100 Hz)- SB reject @ <S> above f (-40dB ) <S> 2nd order is 40dB /decade <S> eg free from TI Large caps can be scaled down by 100 while R increased x100 so if you use Rail <S> to rail OA's rated for 10K load, you do not need lower values. <S> Also use 1M ~10M max depending on bias offset. <S> This is not a final design just a starting point. <S> If you use single supply 0~3V then use Vcc/2 as your ground ref with Resistors and cap. <A> Using Signal Chain Explorer, with numerous Efield and Hfield aggressors active under the Gargoyles mode, with 159Hz LPF before the single-stage gain, and 1.59Hz after that gain stage, we see 8.8 bits ENOB/55dB SNR. <S> Notice the FOI frequency of interest is 10Hz; this should be 2Hz or 1Hz. <S> (also, click on the topright I/ <S> C InterconnectsModel, so EFI/HFI have targets for their fields.) <S> Here are the active Efield and Hfield aggressors. <S> Reducing FOI (top right) <S> to 1Hz produces 69.6dB SNR <S> How can we trust predictions of Signal Chain Explorer, regarding interference?Examine this diagram of the PCB trace: 14mm long, 1.5mm above the plane, 1mm wide. <S> The Hfield aggressors couple into the Loop: <S> 14mm*1.5mm; the Efield aggressors couple into the trace area: <S> 14mm*1mm. <S> However, the Efield math needs to account for the HighPassFilter behavior of capacitive charge injection, using the vulnerable node(s) impedances. <A> You need an amplifier. <S> Your signal is so low that the ADC is having trouble reading. <S> That might be why it just looks like noise. <S> Or it's because you're unshielded. <S> You need to use a shielded coax and maybe balanced cable to a +40dB amp. <S> A simple op amp circuit would be fine but you might get significantly better results if you consider the source impedance of the sensor and select a part that matches that impedance and possibly at the right frequencies. <S> If you post a link to a datasheet for the sensor, you'll get better answers.
| It looks like the sensor may have insufficient signal to noise ratio over the frequency range of valid signals. Since 2 Hz is very slow for any microcontroller, you can use a rough analog low pass filter, sample much faster than you need the data, then low pass filter in the firmware.
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How can I drive a NEMA 17 stepper with a raspberry pi I purchased a NEMA 17 stepper motor for a small home project, and I want to drive it off a Raspberry Pi. I purchased a DRV8825 Stepper Motor Driver Carrier thinking that it would drive the stepper, but I think I got the wrong thing. The motor driver looks like it needs a higher voltage than I want to provide to the stepper, so I'm worried it would fry my stepper if I hooked it up to a higher voltage power supply. A tutorial I found recommends using a L293D IC to drive a stepper, and I can follow these instructions, but I'm not confident that the directions will apply to my stepper. I'm a total newb at electronics, so anyone that can point me at what the best way to drive this stepper would be my hero. <Q> Current matters far more on the ratings of a motor. <S> The voltage rating they list on the website is a DC voltage applied direct. <S> from wikipedia: https://en.wikipedia.org/wiki/Stepper_motor#Stepper_motor_ratings_and_specifications <S> "Stepper motors' nameplates typically give only the winding current and occasionally the voltage and winding resistance. <S> The rated voltage will produce the rated winding current at DC: but this is mostly a meaningless rating, as all modern drivers are current limiting and the drive voltages greatly exceed the motor rated voltage." <S> Go through the current limiting instructions on pololu's website and stay below the 1.2A rating. <S> Note the voltage ratings for the controller (as you need a minimum voltage to drive the circuitry). <S> There's some research to be done for supply voltage w.r.t. performance of the motor, and heat generation, but if you're not working the motor near it's rated wattage, you might not have to worry about it. <A> It is a PWM stepper driver. <S> It will ensure that the current required by your stepper is not exceeded. <S> Read the datasheet <S> You can watch a video or lookup the instructions for setup of the current limits for the driver board. <S> Typically a 12 V or 24 V DC power supply would be used. <S> Connection to the Raspberry Pi requires two digital I/ <S> O pins, one for Step, one for direction. <A> The stepper motor spec is 3.3V/1.2A, the driver <S> DRV8825 you have chosen has a minimum voltage range of 8.2V <S> so you cant use the motor with this driver <S> you have two choice <S> Change the driver to <S> DRV8834 which match your stepper motor specification <S> Or you can get Stepper Motor Nema 17 Stepping Motor <S> 26Ncm(36.8oz.in) <S> 12V 0.4A to match the DRV8825 driver
| The DRV8825 is ideal for your stepper motor.
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Running a 6V DC LED Bulb from a 8V AC current source? I hope that this question has not been asked earlier and is wasting anyones time. I don't particularly remember my ECE101 classes that well - and seek your help! I own my grandfather's hifi amplifier that has power light indicator that recently burnt out. The bulb is proving too hard to find right now. It was a tiny filament bulb and was powered by a 8V AC current (Schematic says bulb rated 8V 40mA). I just wanted to check if my math is still alright here. AC 8V RMS = 8V x Square Root of 2 = 11.3V for the DC LED. So by using a 6V 30mA led I would take 11.3v - 6V = 5.3V extra current across the circuit. Then I would then just use V = I x R; so 5.3V/30mA = 176 ohms resistance. By adding the 6V 30mA LED with a 180 ohm resistor would be a workable solution? I was also told I may need a rectifier diode and a capacitor to keep a smooth glow? Are either mandatory or just optional? Finally can I use a full wave rectifier to avoid using the cap? Any idea on how to wire this. Sorry for my basic question but every bit of help would be useful! Thanks in advance :) <Q> Your easiest wiring would be to do this: simulate this circuit – <S> Schematic created using CircuitLab <S> which gives full-wave lighting, but it'll be rather bright. <S> It can be adjusted by raising the R1 value. <S> White LED forward voltages are about 3V, red around 1.2V, so the LEDcolor matters. <S> Except for some very bright lamps intended forbicycles and lanterns, a "6V" LED is rather unusual (and probably expensive).Flickering at 120 Hz is unlikely to be a problem, so a capacitor is notnecessary. <A> That seems about right. <S> The capacitor is for smoothing the output. <S> Even with a full wave rectifier bridge, the DC output will vary widely without a capacitor. <S> The capacitor will remove the 30 or 60 hz flicker that is inherent in most AC signals. <S> The circuit above is how you would wire it. <S> The load is the led plus the appropriate resistor, which is the next closest standard value to 176 ohm, as you properly calculated. <A> There are three options: Use a bridge rectifier, like Passerby suggested. <S> Without a smoothing capacitor, the LED will flicker at 100 or 120 Hz, which isn't noticeable, but it will also glow dimmer without a cap. <S> If the drop on the bridge is 1.4 V, you can reduce the resistor to just 150 ohms. <S> This solution requires the most components and space, so it might not be the best. <S> Connect the LED with that 180 ohm resistor in series directly to AC. <S> It will flicker at 50 or 60 Hz, but I'm quite sure that's not noticeable for most people. <S> Make sure the LED will survive more than 12V reverse voltage. <S> Connect the LED to a DC rail instead. <S> Don't forget to recalculate the resistor. <S> If the voltage is really high, you may be wasting relatively much power at 30 mA through the resistor. <S> A 9V incandescent bulb that fits will works just fine, but be a bit dimmer. <S> The efficiency of a LED is about 10 times as high as for an incandescent bulb. <S> This means that at 30 mA, the LED will be (6V*30mA*10)/(8V*40mA) <S> = <S> ~6 times as bright as the original bulb. <S> So you may want to multiply the series resistor by 5. <S> 1 <S> Without cap: relative_current(t <S> ) = 0 <S> if sin(t)*8*sqrt(2) < 7.4 else (sin(t)*8*sqrt(2)-7.4)/(8*sqrt(2)-7.4). <S> And the integral of the relative current from 0 to pi is 1.129... <S> divided by pi <S> gives 0.114 which is already quite close to a sixth. <S> The resistance could actually be decreased in this case, but not to anything below 131 ohms. <S> 1 <S> With a large enough cap: This is always at peak voltage, so the resistor could be made 5 to 6 times as large to compensate for the efficiency of a LED. <S> 820 <S> Ohms. <S> 2: relative_current_when_forward_biased(t) <S> = 0 <S> if sin(t)*8*sqrt(2) < 6 else (sin(t)*8*sqrt(2)-6)/(8*sqrt(2)-6). <S> The integral of that from 0 to pi divided by two pi equals to 2.1. <S> 390 <S> Ohms <S> 3: 5-6 mA through the LED.
| A 6V bulb should also work, but lifetime will probably be reduced. The amplifier is most likely running on DC, so you could use the DC rails instead.
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Matching JFETs for differential amplifier application How to choose a properly matched pair of JFETs for differential amplifier? Is it enough to measure their Idss? <Q> To be on the safest side both \$I_\text{DSS}\$ and \$V_\text{P}\$ should be checked and matched, it is well known that those two JFET's main parameters usually spread widely. <S> What is less known is that they are strongly correlated, at least for same manufacuter and lot. <S> A look to this \$I_\text{DSS}=f(V_\text{P})\$ scatter plot done on over 200 different JFET may clarify ( source viva-ananlog.com) matching for \$I_\text{DSS}\$ will most likely accomplish good \$V_\text{P}\$ consistance too, and viceversa. <S> Then you have to known your circuit requirements, how much unbalance you may tolerate and eventually include some trimming if not confident. <A> however, it is much better to design your circuit so that it isn't critically dependent on matching jfets. <A> Worked for a guy who designed telemetry systems for aircraft flight tests. <S> One mode for customer's telemetry was programmable gain amplifiers, with need to reject the aircraft's 400Hz power electric/magnetic fields by at least 120dB.With differential inputs through differential FET switches, from hundreds of external sensors. <S> To achieve high-Zin, and thus very low voltage drop across the channel muxing FETS, thus minimal imbalance to sustain high CMRR, he used JFETs. <S> The PCB used Harris opamps and ADI matched JFETs. <S> He had the chief technician evaluate several of the matched pairs, over temperature and Idd, and record how the Drain voltage drops became imbalanced ----- searching for that magic Idd where the microVolts/degree Centigrade offset-voltage-drift became ZERO.
| Idss would be one of the most important parameters to match.
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Should a capacitor be considered functional if it's capacitance reading is correct? There are many different ways to test capacitors. Using a capacitance meter, using a DMM and an analog meter. In general, is it safe to assume that a capacitor is considered functional if it's capacitance measurement is +- 20% of it's declared value without doing the ohm/voltage test? Are these test overkill after the capacitance value has been verified? Just so we are on the same page, the ohm test I am referring to us to check to see if the resistance value is not constant and that it increases until OL is displayed. The voltage test is just to charge the capacitor and check that it is properly charging/discharging. Could these test be mutually exclusive? <Q> Short answer: No. <S> A capacitance measurement will only give you part of the picture. <S> You also need to measure the ESR, especially for electrolytic capacitors. <S> You could have an electrolytic capacitor that measures exactly what its rated capacitance suggests, but the cap will not work at all in the circuit because its ESR is too high. <A> Functional? <S> Maybe, it depends on the specified tolerances of the part. <S> But there are many other things to factor in (ESR, ESL, ripple current, etc.) <S> If you just want it as a quick functional test before soldering the part to a PCB, it may be good enough for you. <S> HOWEVER... <S> Reliable? <S> Who knows! <S> If maximum ratings are exceeded during part testing, it may well be the case that the part is mostly functional but not reliable any more. <S> Even benign operating conditions that are within the design range may precipitate a failure much sooner than expected from a random failure calculation. <S> Be very careful when testing parts on your own. <A> So, the question, as I understand it is, can a capacitor suffer a failure in a parameter such as ESR or leakage and yet still have a correct main value as measured by a hand-held tester (for example)? <S> The people who service things like central air etc who see a lot of failed motor-run and start capacitors tend to only go by the main value. <S> The hand-helds they have can only measure that. <S> However, these are mostly polypropylene film capacitors. <S> Other equipment has aluminum electrolytics that often have high ESR failures. <S> I think that measuring the capacitor uF may not catch these because there are hand-held ESR meters for these capacitors.
| The answer then may YES, but it depends on the capacitor type.
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Quick Question about ohms law and resistors This may have been asked before, but I'm currently I'm an argument about whether or not passing 12v through a .1 ohm resistor would result in 120 amps due to Ohm's Law. (I=V/R) Providing that the power source has a negligible internal resistance <Q> In theory, yes, exactly. <S> Provided the 12V source has negligible internal resistance and provided <S> the 0.1 ohm resistor does not change resistance (or melt or explode) from dissipating 1.44kW. <A> If you assume internal resistance is negligible <S> then yes that is exactly what will happen! <S> Of course you may start a fire or explode your resistor at that level of power... <A> If you can maintain 12v across a 0.1ohm resistor, then yes, it will conduct 120 Amps. <S> At these power levels and currents in the real world, we'd normally flip Ohm's Law around, and say that if a current of 120A was passing through a resistance of 0.1ohms, then it would drop 12v across it. <S> So for instance if you put your arc welder on the 120A setting, and used cables with a total resistance of 0.1ohms to connect to the torch and workpiece, 12v would get dropped on the cables, reducing the voltage available at the arc. <S> The cables would get quite warm as well. <A> Rather than passing a voltage through a resistor (or circuit) it is more usual to describe it as applying a voltage. <S> The term passing is more usually used in relation to current . <S> i.e. apply a voltage and pass a current. <S> The short answer is that according to Ohm's Law, V = I * R, so applying V = 12 V across a resistance of R = 0.1 Ohm, the resulting I must be 120 A. <S> Likewise <S> As others have indicated, there are various practical considerations related to a real-world application of this particular scenario. <S> The theory, however, is incontrovertible. <A> If the power of 0.1 ohm resistor > 1.44kW then you are right. <S> But based on my understanding, there is no such a big power resistor.
| passing 120 A through a resistance of 0.1 Ohm must result in a voltage of 12 V across it.
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Can I do anything to make this electric piano work? I have an electric piano but I don't have the power supply. It has a barrel connector with a negative tip, and wants 12 VDC, 1.5 A. I have a power supply that is 12 VDC, 2.67 A, but it has a positive tip. Is this supply sufficient? Can I modify my power supply in any way allowing it to meet the specifications? Sorry, I don't know much about circuitry and I'm hoping someone would be able to help. <Q> Yes. <S> All you need to do is cut, swap and splice the conductors of the power supply. <S> That's all that is needed to change a center positive to a center negative dc connector. <S> The alternative is to buy a modular power supply, like those offered by RadioShack with replaceable heads. <S> You can simply choose the right size head and plug it in so that it is center negative. <S> As far as the current goes, the amperage listed on a power supply is the max it can supply. <S> A device will only pull what it needs, so having a supply that can offer more is not a problem. <A> I have a power supply that is 12 VDC, 2.67 A, but it has a positive tip. <S> Just buy one of these: - They come in all shapes and sizes. <A> If you go to a distributor with a wide range such as Digikey and do a parametric search you can find a few supplies that will work. <S> Eg. <S> So long as the max output current is equal or better than the requirement, and the other requirements match (polarity, plug style, mains plug style, mains voltage, output voltage) you should be okay. <S> Or just cut and splice the cord you have, but it will tend not to be as strong as the original. <A> If the barrel connector on the power supply fits the piano, it would be fine to cut off the tip and reverse the wires, but be warned: Sometimes these power supplies, instead of using two wires, use a single wire with a shield. <S> If this is the case, you'll have to get creative when crossing the conductors. <S> Heat shrink tubing is your friend. <S> It should go without saying that you should test the fit of the connectors while the power supply is unplugged :)
| As long as the voltage matches, and the amperage is the same or greater, or should work.
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Internal resistance of a battery? The task is to find \$I_L\$ and \$R_i\$ given \$U_L\$ and \$R_L\$. \$I_L\$ is simple enough but I can't get my mind around how to find \$R_i\$ surely you must have more information? <Q> You are correct that you cannot completely answer the question with what is given. <A> Sure, it is very simple. <S> All you need is a big and powerful variable resistor (few kW) and voltmeter and amp meter. <S> Then you measure volts, amps with different load, but for very short time, because the current will be very high depending on type of the battery. <S> The you do a V <S> /I chart and you will find out that battery output voltage is decreasing with the load almost like straight line. <S> Few basic calculations will give you the internal resistance. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You actually need a Ri=ΔV/ΔI ratio or \$\dfrac{U1-U2}{I2-I1} = <S> Ri\$, not just a load voltage <S> Of course \$R_L\$ helps compute I1,I2 from U1 <S> /R1,U2/ R2. <S> Actually @Marko <S> I think the slope you show is the reduction of charge in the equivalent capacitance. <S> dV/dt= <S> Ic/C and <S> not Ri <S> Note in the graph below they this and show the negative V spike grows as the slope C reduces implying the ESR*C =T product tends to remain constant as the battery discharges and the ESR or Ri rises at the same time the voltage drops with a new slope. <S> Note that ESR above (or Ri) uses a steady current I1=10mA and a pulse current I2=500mA. <S> Thus Ri=(V1-V2)/(500-10)mA <S> The dual slopes of batteries show a "double-layer" effect found in Supercaps. <S> Yet batteries are 1k bigger C values than Supercaps (even Ultracaps which are same but may have lower ESR*C = <S> T values. ) <S> simulate this circuit – <S> Schematic created using CircuitLab
| You could fully answer the question if you disconnected the load resistor and measured the open circuit voltage.
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Why does my multimeter show a wrong voltage over a large resistor? I am having a hard time answering one particular question about our experiment.. In our experiment, R1 and R2 were set to 1Meg each and later to 10k... I understand the need for R1 and R2 a bit. Without R1 and R2, the voltage sharing wouldn't exactly be 50-50 for both D1 and D2 because no two diodes are completely identical. D1 and D2 will both have the same leakage currents (without R1 and R2) since they are just in series. However, they probably will have non-identical IV curves, so this particular leakage current will result to V@D1 /= V@D2. The question I am having a hard time is that, why is V@R1 + V@R2 /= 10v when R1 = R2 = 1Meg?... One the other hand, those two voltages add up (to 10v) when R1 = R2 = 10k... I included the 60 ohm source resistance in my diagram for completeness. However, as I can see, both D1 and D2 are reversed biased and thus, they offer a very large (reverse resistance) which should be much greater than the 60 ohms. Even with the parallel combination of 1Meg and D1 reverse resistance, it should still be much greater than the 60 ohms. I tried thinking of an answer in terms of the RD1reverse//R1 = Req1 and RD2reverse//R2 = Req2. Req1 + Req2 (series) should still be much more than 60ohms and I thought that the 10v should still show up at the node of D1 cathode. Yet in our experiment, V@R1 + V@R1 < 10v. Can anyone point me out if I am thinking this in a wrong way? Some tips/first step hint would really be appreciated Edit: question answered thanks to @CL. Assuming D1 and D2 are open during reverse bias for simplicity and noting that Rmultimeter = 10Meg, V@R2 (shown on multimeter) = 10v * (1Meg//10Meg)/((1Meg//10Meg)+1Meg+60) = 4.76v measured. <Q> The input impedance of your multimeter changes the circuit: <S> With 10k resistors, the difference would not matter, but the 1M resistors pass so little current that the additional current through the multimeter has a noticeable effect. <S> If you knew your multimeter's input impedance, you would be able to calculate the voltage that you would get without it. <A> In addition to the multimeter issue, your 1 \$M\Omega\$ resistor likely has 5% tolerance. <S> To figure out the range of voltages you may see, assume the values of the resistors may vary by as much as 10% (2*5%) <S> To see whether this or the multimeter is the problem, measure the voltage drop across the bottom resistor, switch the resistors and repeat the measurement. <S> If different, the issue is resistor tolerance. <S> Another possibility is if you're touching the probes while making the measurement, in which case you become a parallel resistor. <A> To get rid of the effect of the multimeter's input resistance, try making a measuring bridge. <S> You put something like a 1k precision pot across the voltage source and measure the voltage between its wiper and your measuring point. <S> Then you adjust the pot until the measured voltage is 0V. <S> At a voltage of 0V, there will be no current through the multimeter influencing the measurement. <S> Afterwards, you measure the voltage at the wiper as compared to 0V. <S> As the resistance of your pot is much lower than that of your multimeter, the result will be reasonably exact. <A> why is V@R1 + V@R2 /= 10v <S> when R1 = R2 = 1Meg? <S> ... ... <S> Yet in our experiment, <S> V@R1 + V@R1 < 10v. <S> Depending on how theoretical you want it to be. <S> In theory, V@R3 + V@R1 + V@R2 = 10v. <S> So in theory, V@R1 + V@R2 < 10. <S> However, since the current in the circuit is so small (approximately 10v/(R1 + R3 + R2) = 5ua), the voltage drop over R1 = <S> 5ua <S> * 60R = 300uv << 10v. <S> So V@R1 + V@R2 = 10v, for practical purposes. <S> that doesn't hold when R1+R2 is sufficiently close to R3, or your meter is sufficiently precise, or your experiment is sufficiently picky. <A> For practical reasons, assume that a modern multimeter (of the powered, digital type) will in itself behave like a 10 or 20 Megaohm resistor; this will change the pictured voltage divider by 5 or 10%. <S> Analog ones that work without their own power supply for voltage measurements will usually have a lower input resistance that is also dependent on what measuring range is set. <S> Voltmeters that have a far higher input resistance exist, but these are more typically lab grade than portable field equipment since they are easily confused (showing nonzero values with the probes connected to nothing) or even damaged by static electricity.
| If the measurement is the same, the problem is the multimeter impedance.
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What does it mean that a power supply is 'referenced to mains' and why is that dangerous? I was watching a few different videos by BigClive on youtube. He was mentioning the danger of a particular light chain power supply because it was 'referenced to mains'. The voltage across the output terminals was 5v but an electrician's safety wand was still set off by being near the '5v' output. Based on searching this site, it seems the neutral terminal of mains passed directly through to the power supply to be the neutral for the low voltage supply. Is this accurate? And how does this cause the detection wand to buzz? <Q> Most power supplies use a double-insulated transformer between the incoming mains and the outgoing extra low AC or DC voltage. <S> This means that you will not get an electric shock if you touch one of the output terminals. <S> Some very cheap power supplies do not provide that isolation. <S> One or other of the output terminal may be at the full mains voltage. <S> In some cases, it might be safe (well, safeish) with the plug one way round, but dangerous with the plug reversed. <S> Such power supply designs should never be used for "wall warts". <S> They should only be totally enclosed within the body of an appliance, and even then only if there are no exposed electrical parts (such as headphone sockets) that the user could touch. <A> "Mains isolation" is the approach used to make most power supplies safer. <S> So the real question is "Why do we use mains isolation?", which has been asked several times on this site with some very high quality answers already -- see here , here , and here . <S> There are some safety concerns when not using mains isolation, but there are many products that use non-isolated supplies and are still quite safe. <S> The specific example in the video is dangerous because it's not isolated and the power supply output is exposed to the users through the power cord. <A> A power supply has a pair of connections for mains input (it may also have a connection for mains earth). <S> While one of these connections is nominally neutral portable appliance standards generally assume that both of them are potentially dangerous. <S> Reasons for this include the potential unreliability of flexible wiring and frequently plugged connectors and the fact that some countries use unpolaised sockets. <S> * on the input may be passed to the output. <S> So the output terminals are a potential shock hazard. <S> Such power supplies are cheaper to build as no transformer is needed. <S> It is possible to design equipment with transformerless power supplies safely but it means treating anything connected to the output of the power supply as-if it was connected to live mains. <S> I doubt that the cables and connectors on those lights would pass muster as mains connectors at least in Europe. <S> The lights themselves <S> it's hard to tell as it depends very much on how thick the plastic is. <S> User expectation is also an issue. <S> Safety standards often talk about "reasonablly forseeable misuse". <S> If something looks like a typical wall wart then people are going to assume it is a typical wall wart. <S> Even if the equipment supplied with the non-isolated supply meets insulation and touchproofing standards for mains it would be all too easy for the wall wart to get repurposed to supply something else. <S> * to shock someone requires both sufficient open circuit voltage and sufficient short circuit current. <A> The 5V DC it outputs is relative. <S> Between DC <S> + and DC -, it certainly is safe. <S> But between DC output and earth of your house, there is either full mains voltage or nothing depending on how the lamp is connected. <S> If the lamp is referenced to neutral, it's safe. <S> But if it's referenced to live and neutral is connected to ground, it's not safe. <S> Just to make it more interesting, many countries have non-polarized sockets so you can't be sure which terminal is live and which is neutral. <S> Here's an illustration of one of the cases:
| "refrenced to mains" means that there is a low impedance connection between the input connections and the output such that hazardous voltages/currents
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Replace ALU With Lookup Table? Disclaimer: So this is obviously a silly question and I want to start by saying I don't want to discuss the financial costs of this, as I'm aware CPU cache is expensive. As this hasn't been made obvious enough, this is PURELY ACADEMIC - NOT FOR IMPLEMENTATION. Think "thought experiment" I've been wondering if it would be possible to precompute all the work an ALU would do and store the results in a lookup table. For this specific example, I've been looking at a subset of instructions the ALU is responsible for in a MIPS Architecture which is "AND, OR, add, sub, slt, NOR". In this architecture, these operations would take 4 bits to encode which we'll call the control, as there are only 6 operations. In addition, we'd have to take two 32 bit values as input and return a 32 bit value as output along with 3 1 bit flags. (Details listed here ) At a really high level, we'd use the 4 bit control, along with the two 32 bit inputs to return the 32 bit result and the 3 1 bit flags. So couldn't each control act as an offset to our cache, and use the input values to index into our lookup? We could even squeeze out some more memory for operations that have the commutative property (1+2 = 2+1) I'm aware this is goofy question but I was curious if anyone had any insight. Perhaps it could be faster? If not, maybe use less electricity or generate less heat? At the very least it's interesting. <Q> A single 32-bit x 32-bit to 32-bit lookup table would require an untenable amount of space: $$ 2^{32} \times <S> 2^{32} \times 32 <S> = 2^{69} \approx \mathrm{5.9\ quintillion\ bits} <S> $$ <S> A single microchip can store perhaps 2 40 bits (~128 GB); you'd need over 500 million of these to store the full lookup table. <S> (Halving the size by exploiting symmetries like <S> a+b= <S> b+a <S> still leaves the size in the implausible range.) <S> If you want to consider the implications on heat/power and speed, though: If we generously assume that each of the 500+ million microchips draws 1 mA at 3.3V, you're looking at a total power consumption of roughly 1.8 MW. <S> (Yes, that's megawatts .) <S> If we assume that each of those chips is 2x2 cm and 0.5 cm thick (including the circuit board), and that they require no other support circuitry, the resulting device will be roughly a 10 meter cube. <S> It takes light about 34 nanoseconds to cross 10 meters; even if we assume that it takes no time for one of these chips to look up a result, this would limit the speed of such a device to roughly 292 MHz. <A> The bottom line is that a lookup table is made up of logic gates, and it always takes fewer gates to implement the kinds of ALU operations you're talking about directly rather than use a lookup table. <S> So, no matter how much technology advances, it never makes sense to use lookup tables over direct logic using the same technology . <S> FPGAs are a special case because of how they are used. <S> In the first place, reconfigurability is their most important feature, and secondly, the tiny lookup tables that they use ( <S> typically 16×1 to 64×1) are very fast — faster than the interconnect logic and other details that contribute to their configurability. <A> At 32-bit level, its not practical. <S> But considerable excitement occurred in 1985, when 5-bit math was proposed for in-memory image processing. <S> By taking log(pixel_magnitude) <S> and storing that, dynamic-range was good and actual edge-detection should occur. <S> We never built it. <S> After all, who would want to perform 5-bit math in a memory?
| This is flatly impossible to build.
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Switching power to a sensor I want to control the power supply of a sensor with a microcontroller, so I can turn it off when it isn't being used. My goal is to minimize power consumption on a battery-powered device. Both the microcontroller and sensor are running on 3.3 V. Can I use a single MOSFET directly driven by the controller? Or is there a better method? atmega328p micro controller and GY 61 accelerometer sensor are used. <Q> Probably- one method would be to use a P-channel MOSFET as a high-side switch to control the 'sensor' power. <S> You might have to add a pull-down or otherwise ensure that the output of the 'sensor' was close to 0V with no power- if it is allowed to float around the MCU power consumption <S> might increase. <S> There are many inexpensive logic-level MOSFETs that will work fine from -3.3V gate voltage with guaranteed very low Rds(on) and reasonable guaranteed leakage. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You may have to ensure that the 'sensor' supply has some series impedance or that the MOSFET does not switch too rapidly if the 'sensor' has large bypass capacitance- <S> otherwise it could glitch the MCU power when it turns on and cause problems. <A> You may have a simpler option. <S> If the sensor's maximum current requirement is low enough, you can use a digital output (from your microcontroller) to power the sensor directly. <S> Check the following things: <S> The microcontroller's digital output, when set high, may not reach all the way to 3.3 V. Look for \$V_{OH}\$ in the datasheet. <S> Make sure that the this voltage is acceptable for your sensor. <S> The microcontroller has a maximum output current for digital outputs. <S> It this greater than the sensor's current draw? <S> If you update your question with the part numbers of your microcontroller and sensor, we can give you specific examples. <A> You did not specify sensor type or its power consumption, but here are some options: You can use a dedicated power switch chip, for example: TI load switches . <S> They can also come with overcurrent protection, so if your sensor is dangling on a cable outside and you short power to ground it will not disable your device. <S> If the sensor is consuming very little power (below <10mA - check your MCU pin specification) <S> you can just power it directly from a GPIO when needed ( CHEAPEST ). <A> I'll add a cheap ghetto solution: If your sensor draws a small current (like a few mA) <S> you can use a good old-fashioned logic gate as a switch. <S> Pick one with high drive strength, like 74AC or 74LVC. <S> Hell, you can even use an IO from your micro. <S> Or a pin from an I2C IO extender if you have one in your design. <S> This is obviously not for every situation, but when applicable, it does allow you to switch several loads with a single chip, and it is cheap and compact.
| You can use a separate LDO with an ENABLE pin to disable power delivered to the sensor (for example MIC5205).
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Why do single op-amp DIP packages have 8 pins, rather than 6? Single op-amps often come in 8-pin DIP packages, with 5 of those pins being standardized (or maybe it's just de-facto) like in the ones below. Older op-amps such as the LM741 (first image) have a very bad sense of what is 0 volts, so they require offset null pins to manually adjust them. Some newer op-amps such as the TL071 are much more accurate, but still make use of extra pins by having this feature, while their dual and quad siblings do not have this feature. Some modern single op-amps have three unused pins, and forego the offset null completely. So why don't many modern single op-amps come in 6-pin DIP packages? They would still conform to the standardized layout, and fit in 8-pin DIP sockets. But they would take up less space in circuits designed for them. edit: For examples such as TL071, I can that they would put the offset null on it despite it's absence in it's dual and quad siblings so that there's an option for those that do want to adjust it. But even then, many of these are available in different packages, so why not 6-DIP? That still doesn't explain it for the single op-amps that don't have offset null. <Q> The JEDEC MS-015 standard never had defined packages smaller than eight pins: <S> (Note that 10-, 12-, and 26-pin packages are also missing.) <S> With nobody actually manufacturing six-pin packages, the first one to do so would have had additional costs that would have been larger than just wasting two pins of an eight-pin package. <S> There actually are many (opto)isolators that come in six- <S> or even four-pin DIP packages (e.g., 4N25, PC817), but the manufacturing of isolators is somewhat different (typically, LED and phototransistor overlap vertically ) and therefore does not allow easy reuse for chips with 'normal' dies. <S> Packages other than DIP are commonly manufactured with fewer pins (e.g., SOT-23-5, SC70-5), so op-amps are also available in those. <A> Single OpAmps come in 5 and 6 pin packages these days. <S> You just won't find them in DIP anymore because most of the industry has moved to SMD packages. <S> It's easier and cheaper to work with them if you're into mass production. <S> For example 5 pin single OpAmps often come in the SOT23-5 package which looks like this: <S> There are 6 pin variants as well. <S> These often have extra functionality such as an enable pin, a null-offset trim pin etc. <A> CL is correct - it is a historical legacy - but you need to go back to the initial design and production of op amps in the 1960's to understand why 8 pins. <S> The first op amps required 8 pins to allow for external compensation components, even the 741 required at least 7. <S> Once production lines were set up it was a lot cheaper just to leave the packaging size and pin connections as standard (pin for pin replacement) <S> so that later (superior) <S> op amp designs would be easily substituted in existing (pcb) designs. <S> With the arrival of SMD this gives the opportunity to change production to other 'standard' packages with the 'through hole' packages phasing out <S> There's a good article here if anyone wants to read up further details: http://www.ee.bgu.ac.il/~angcirc/History/Solutions_2003_2004_B/SomeStuff/History18opamp.pdf <A> As Jim Dearden showed, those pins were used for offset adjustment and for frequency compensation (and sometimes to prevent latchup under overdrive). <S> Here is the UA709, with 4 different external C+C+R networks. <S> Notice the bandwidth even at 60dB gain: out to 500KHz. <S> Now for some typical applications. <S> With much respect to Bob Widlar, who started with discrete transistor designs at Ball Aerospace for satellites, he then moving to Fairchild and coercing the fab people to tweak the RTL resistor-transistor-logic process so he could develop the first silicon opamps. <S> Tho AMELCO already sold diffpairs with internal current sources + bias chains, all matching.
| In other words, creating six-pin DIP packages is just not worth it. Historically, it was more efficient to put multiple instances of the same function into the same chip, so DIP packages usually had at least 14 pins.
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Understanding Circuitry from a Datasheet I'm having a little difficulty understanding this section of a circuit from the datasheet for an MC33035 BLDC Motor Driver. Some info: Pin 7 is enable, send pin high to enable motor Pin 8 is a 6.25v source Pin 14 is a fault pin, active low during a fault (at VCC (pin 17) not during a fault) I'm assuming during a fault "Latch on Fault" is closed, pushing the reset button would therefore send pin 7 high, re-enabling the motor. However, following this logic then "Latch on Fault" would be open when the motor is running - but how is pin 7 sent high? Or is it default high, closing the switch next to it pulls it down, or if "latch on fault" closes it pulls it down? What is the 47uF cap for? Any help greatly appreciated. <Q> It looks like "Latch On Fault" is an optional switch via which you can set the behavior to display in case of a fault. <S> It does not change with the fault state. <S> The logic is quite simple: Pin 14 is low on fault, pin 7 is high for enabled. <S> Now if you connect these pins, e.g. via a closed switch, once a fault is detected it also disables the motor by pulling enable low. <S> If the switch is open, a fault will not disable the motor. <S> The reset button pulls the enable pin high again, overriding the fault signal of pin 14. <A> "Or is it default high, closing the switch next to it pulls it down, or if "latch on fault" closes it pulls it down?" <S> 25uA through a 4k7 resistor <S> gives you 117.5mV across it, which is comfortably below the logic 0 threshold. <A> Pin 7 has internal pull-up current source. <S> Don't know why it is a source and not a resistor. <S> Thus, you need to pull it down to disable motor control. <S> Haven't read the whole datasheet but 2.2k resistor and 47uF capacitor <S> look like RC network filtering small "fault" spikes. <S> If it is an automatic latch, latching will happen when sufficient "fault" spikes have been integrated over time. <S> Thus, 47uF capacitor acts as memory and retains the charge unless discharged via "reset" button. <S> I think a current source has been used to discharge the capacitor in a known period of time for auto fault escape.
| "Latch on fault" is either a manual switch or a latch that latches when the voltage across it crosses a threshold. (For this logic to actually work, during fault pin 14 must stay low even when enable goes/is low.) Correct, pin 7 is pulled high by that little symbol shown to the right of it on the diagram - it's a 25uA current source internal to the chip.
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If the integrated circuit die is very small what is the role of the extra circuit packaging? I noticed that the actual integrated circuits of processors, GPUs, ROMs, specific integrated circuits and other ICs are very small but they usually come in a package that is much bigger. What is the purpose of the packaging? And what are the materials that IC packages are made of? <Q> What is the purpose of the packaging? <S> Together with the leadframe take the connections of the IC further apart. <S> These can be as closely spaced as 100 um which is too close for standard cheap PCB manufacturing. <S> The leadframe + package expands this to something more usable like 0.4 mm up to 2.54 mm (DIP/DIL packages) <S> Make the IC easier to handle by humans. <S> A DIP package can easily be used and exchanged in a breadboard or in a socket. <S> And what are the materials that IC packages are made of <S> The leadframe: tinned copper or metal so that it can be soldered easily The black part: usually molded plastic, sometimes a ceramic material. <S> Some ICs can be bought in a CSP (Chip Scaled Package) which actually means no real package at all, on top of the chip a redistribution layer is made (which spaces the connections to what the PCB can use) <S> and the IC is then mounted directly on the PCB. <S> This technique is also called "flip chip". <A> In most cases, the extra packaging is needed only to attach pins and bond the pins to the die. <S> A lot of more modern packages are much smaller because they don't use older DIP pin size/pitch standards. <S> For example QFN, LGA, BGA, etc. <S> have small packages because the pins/pads/balls are close to the die. <S> Indeed some packages are practically balls bonded directly to the bare die. <S> This site describes the process in much more detail (and you can find a lot of other information online). <S> The encapsulant as it describes, are generally epoxies. <S> http://electroiq.com/blog/2005/08/materials-and-methods-for-ic-package-assemblies/ <A> If you have ever tried to set up an ultrasonic microwirebonder, you'll get it. <S> After days of fiddling to get the right temperature, humidity or something, stiffness of the custom made chip holder jig, it works and the thing machineguns away like a glorified sewing machine for gold wires. <S> Best get a million similar chips through before the weather changes, or a dark lord of Sith raises an eyebrow, or any of the other things which go wrong with these. <S> Since the exact successful process condition will change with different die sizes, one needs to get a lid and legs over it which won't fall off and won't need such inconvenient setting up to make connection to. <S> Lid and legs which will wet and contact with standard tin solder wave process are already plenty enough reason to want to work with packaged chips.
| Protecting the IC against light (light will induce current flow in a PN junction) Protecting the IC against moisture
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maximun number of inputs on a opamp adder circuit With 4 timers of a mcu I am generating 4 square signals , each of them have an amplitud of 3.3v and are generated at diferent frequencies (8KHz 9KHz 10KHz and 11KHz), to generate a multitone signal I am adding them with an op amp adder circuit, everything is OK, but I would like to try with more of 4 signals if it is possible, What would be the maximum number of input signals for an adder circuit?, the op amp is working with 3.3 volts (ad822). <Q> Consider the non-inverting op-amp amplifier: - <S> Now, regard Vin as the op-amp input noise with respect to 0 volts <S> but it gets amplified as shown in the formula. <S> Regard R2 is one limb of your mixer. <S> Every time you add a limb, R2 gets smaller and the noise on the output gets higher. <S> For instance, if you have one input, the output noise might be X but, with 10 inputs the output noise will be 10X. <S> It's not normally a show-stopper <S> but it's something to take into account <S> and anyway you did ask! <S> Clearly, the sum total of all the input voltages x gain has to produce an output voltage that avoids clipping or slew limiting as well. <A> There is no maximum number of inputs you can use, so long as the output does not exceed the op amp capabilities. <A> You could compress the signal, using diodes. <S> Keep the resistances low, so the node-capacitance (the summing node) does not cause excessive phaseshift and turn your summer into an oscillator. <S> simulate this circuit – <S> Schematic created using CircuitLab
| The more inputs you add, the more noise the op-amp produces - this is because the noise (regard it as in series with the grounded non-inverting input), gets amplified each time a new input is added.
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How to prevent MOSFET from operating in high resistance part of linear mode? I'm using several MOSFETs to control various loads, e.g. pumps, lighting, ventilation etc. In the implementation, I have several driver circuits on one board (simple emitter follower, which gives good switching speed with a gate voltage of ~5V; using pull-ups were too slow/inefficient), with jumper cables to a board with several MOSFET circuits. Each circuit is like the schematic. While testing, I touched the signal cable and it caused the MOSFET to turn on, but it got extremely hot in just half a second of on-time. Now I'm worried that a small failure in the driver circuit, e.g. if MCU is disconnected and the signal wire somehow picks up a small voltage, it will be amplified into the MOSFET gate, causing a meltdown in just a short period of time. Could this start a fire? Or will the device just fail closed? Ideally the circuit would not even allow the MOSFET to operate in high resistance part of the linear mode. Can I improve my circuit to avoid gate voltages less than, for example, 4V?Is there a way that I can detect/prevent overheating (besides a complicated setup of sensors). simulate this circuit – Schematic created using CircuitLab UPDATE: Pull-down on signal Adding a 10k pull-down on the base of the BJTs caused the noise to no longer turn on the MOSFET. QUESTION: However, it didn't really answer the core of my question, which is more that I'm worried about situations where a MOSFET may turn on ever so slightly, yet able to conduct, causing extreme heat and consequently potentially starting a fire. I can't use a fuse, since the problem is not overcurrent. Perhaps a thermal fuse mounted to the MOSFET, but that seems rather unorthodox.Is this a non-problem that I'm overthinking? <Q> A Schmitt trigger is (almost) guaranteed to produce a logic high or logic low even if the input is at an invalid intermediate level, weak, or floating. <S> Note that the integrated MOSFET driver that anrieff's answer mentioned has an internal Schmitt trigger in its input logic (denoted by the ⎎ symbol). <S> If your existing circuit is adequate at the actual job of of driving the MOSFET when the input is 0 V or 5 V as intended, then you can add a Schmitt trigger in the form of an inverter, buffer, or other logic gate that has a Schmitt trigger input, preceding your driving circuit. <S> As a further example of their use, Schmitt triggers are also commonly used in microcontroller inputs to protect the internal logic — ordinary logic gates can misbehave, and even draw excessive current and overheat, when the input is invalid, whereas a Schmitt trigger turns an invalid signal into some valid signal. <S> Another hazard specifically for driving MOSFETs (or any power transistor) <S> that is not solved just by using a Schmitt trigger <S> is if the input is is commanding the driver to switch on and off rapidly (whether this is deliberate, as in PWM drive, or accidental noise). <S> Then the MOSFET will spend more of its time in the middle of switching, and so dissipate more power. <S> To reduce this you must either avoid such frequent transitions reaching the driver (probably the better option, if needed at all, in your application since your loads might also be unhappy) or ensure that the driver switches very fast (low rise/fall time). <A> You can use dedicated MOSFET drivers like the TC4424. <S> You can use those instead of your Q1/Q2 pair. <A> I believe something as simple a pull-down resistor at the BJT will resolve this issue by providing a well define voltage level even when the MCU pin will go to high impedance or the cable goes open or disconnected. <S> I suggest trying like a 10K pull-down resistor and trying it out before attempting any more complex solutions. <S> As an engineer to must design and test your MOSFET driving circuitry to operate the MOSFET out of the linear region as much as possible while juggling all other trade offs. <S> In your case you found a fault with driving circuitry because under some circumstances its input was not in a well defined state. <S> If you were looking into making your device safer then you can start looking at thermal limit switches, temp sensor, more sofisticaed devices that have inherit thermal protection built-in etc. <S> Maybe even something as simple as proper heating sinking using the PCB copper is good enough to keep the device cool even in the linear region. <S> All depends on your application and its requirements <A> I am not sure your FET went in linear mode, it could also have been oscillating. <S> This can easily happen when the gate drive has inductance (ie, a 10cm wire). <S> I would advise a 50 ohm gate resistor to avoid this. <S> For your next board, it is better to put the drivers right next to the MOSFETs. <S> For low frequency switching, high gate drive current is not necessary as switching losses will be very low anyway. <S> Slow switching (1µs, not 10ns) is better, as it will radiate less EMI. <S> If your FETs are logic level, you can use 74HC logic gates as drivers also.
| A simple, modular, and robust way to get better behavior from a circuit when the input is in an invalid state is to place a Schmitt trigger between the exposed input and the driver. To answer your question is that keeping the MOSFET from operating in the linear region is the job of your driving circuitry. There are MOSFETs with integrated temperature protection as well.
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unknown component in mini cooler I am modifying a mini cooler/heater and there is one component whose role I don't understand. The Peltier cell is attached to a heat sink on one side and is in contact with a metallic block on the other side. On top of this metallic block there is a black plastic cylinder with two incoming wires. The two wires are set at 0 V both when cooling and when heating. What does this unknown component do? Thank you. <Q> My guess is that the device is a thermal limit switch. <S> The device opens the circuit up once a temperature level has been reached. <S> The main purpose of it is to ensure safety and prevent a possible fire. <S> The photo below is a thermal limit switch commonly found in your furnace <A> What you are looking at is a temperature switch like this . <A> That's why you are seeing the voltage change. <S> So if the voltage is decreasing when heating, the resistance is RISING with temperature, so it is a PTC (Positive Temperature Coefficient of resistance) thermistor. <S> You said it was feeding an LED, so likely the other side of that LED circuit was powered when heating, so the resistance dropped between the LED and ground as it reaches it's design temperature and the LED gets less voltage and thus goes dimmer as it heats.
| More likely it's a thermistor, a resistor that changes resistance with temperature.
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If electrons move slowly in an electrical circuit then what signal or energy is it that travels at the speed of light? I am reading "Radio Theory Handbook" and am confused with the statement that says that electrons are moving through the wire at snails pace.But it goes on to say the "electrical effect" is instantaneous. I assume he means the speed of light. Then what is this mysterious electrical signal in the wire that moves at the speed of light? Is it the EMF? By definition electricity is the flow of electrons. But yet the electrons are said to travel slow? There cannot be a contradiction. I am asking for help in clarifying this. Thank you. There is another question " Is voltage the speed of the electrons? This is not the same question. I am not asking about voltage. I am asking about the apparent contradiction given the fact that electricity which is the flow of electrons has an almost instantaneous effect while the electrons themselves more very slowly. Although it is not the same question there was enough useful information there and along with all the excellent responses that my question has been well explained. <Q> It is changes in the fields inside and around the wire that travel at the speed of light. <S> Imagine the wire as a hollow tube full of electrons. <S> When there is no current the electrons are all sitting there, repelling each other, but since there is nowhere to go, they just sit still. <S> When an electron at the back of the wire starts getting pushed (by a battery for example), it gets closer to its neighbors in front, which then pushes them forward. <S> These electrons start to move, which then causes them to start pushing on their neighbors. <S> Eventually all the electrons are moving down the wire. <S> The speed with which the electrons hear about their neighbors moving determines how quickly the signal travels down the wire. <S> Nothing about this process actually requires the electrons to actually go anywhere quickly, just that their neighbors feel changes quickly. <S> This is all a huge simplification, in reality there is a continuous electric field, and there is a magnetic field generated around the wire, and of course the wire isn't hollow, but this picture can help for getting the concept. <A> Could we analogise it to something like this? <S> Imagine a really long train with a huge number of coaches. <S> The engine begins to move, very, very slowly. <S> Instantly, the last coach also begins to move. <S> The speed of the engine has no relation whatsoever to the speed at which the information of the train starting to move is propagated. <S> The electrons are like the coaches. <A> Electrons themselves move at the glorious speed of fractions of a millimeter per second. <S> However, they are so close to each other that they are constantly bumping in to each other. <S> This makes an electrical signal travel down a wire at somewhere usually around 2/3rds the speed of light. <S> That speed can be effectively slowed by various circuit elements as well. <S> A signal inside a coaxial cable will travel more slowly than a signal in a free-hanging wire. <S> See wikipedia on the Velocity Factor for more details about this, specifically part about the velocity factor in a lossless transmission line. <S> As for "electrical effect", he's probably talking about the electromagnetic radiation that can be produced by certain oscillations in circuits (changes in the electrical field resulting in radio waves, electrons moving between energy states in LEDs, etc). <S> These travel at the speed of light in a vacuum because electromagnetic radiation is light and vice versa. <A> The electrons zip about at random due to thermal energy (more properly at the Fermi energy, once you start thinking in quantum terms). <S> However that random motion has no net effect on the large scale, other than to generate Johnson noise. <S> Superimposed on this random motion is the drift velocity, which is a snails pace, due to the macroscopic current. <S> Each 64g or so of copper has a faraday of charge in its free electrons (96500 coulombs), so they don't have to move fast to create a large current. <S> the current in the wire responds to the electric field, starting at the surface and working down into the bulk of the metal according to the skin-effect. <S> At radio frequencies all the current is carried in the outer most few microns of the conductor, pretty much all the action is in the space or insulator around the wire (or inside the waveguide) <A> Yes, 1) electrons travel extremely slowly, and in AC systems they constantly halt and reverse direction, 2) metal wires are always filled with enormous amounts of electrons, 3) batteries and generators are electron-pumps, and they don't create or 'generate' the electrons being pumped. <S> The 'mysterious signal' is exactly that. <S> It has various names: <S> electric signals, electromagnetic energy, electrical waves, EM waves. <S> It travels at the speed of light because it IS light/radio/EM. <S> If we have a ring of movable electrons (a metal circuit,) and if we suddenly apply a pumping-force at one spot, electromagnetic waves will spread throughout the circuit at the speed of light, until all of the metal's charges 'get the message' and begin moving. <S> Similar question: <S> what do electric companies sell? <S> Not current, since the path for current is a closed loop. <S> They sell 60Hz radio waves, but waves being sent over transmission lines and absorbed by distant motors and lights. <S> Inside your home wiring and inside the long transmission lines, electrons vibrate back and forth, but the EM waves proceed continuously forward. <S> Don't mistake the "medium" for the "waves. <S> " <S> In empty space, light and radio requires no medium, but if electromagnetic waves are traveling along wires, the movable electrons act as the "medium" for EM wave propagation. <S> Here's one confusing aspect: the flow of EM waves has no frequency limit, and operates all the way down to zero Hz. <S> The EM-fields description of 2-wire transmission lines will also apply to flashlights. <S> When a battery is lighting a bulb, EM waves are traveling from battery to bulb at the speed of light. <S> This effect becomes obvious for 100MHz signal generators, but it remains the same for 60Hz and for DC. <S> In DC systems we can only detect the high-speed energy-flow if we suddenly start or stop the flow, to produce a sharp edge which can be tracked. <S> And so, another name for your mysterious signal could be: "DC wave-energy!" <S> :) <S> Kraus: <S> EM energy flow in circuits DC transmission line fields <S> Book: ' <S> The Fields of Electronics' Electrostatics of simple circuits <S> More links...
| The electric and magnetic fields move at the speed of light in the insulating medium around the wire and this is what carries the signal and controls everything -
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Elusive SR Latch: 74118/19 – Hex SR Latch with common reset I am working on a circuit where I need to hold a few signals until my MCU reads them.Basically the MCU would read these lines at regular intervals (minutes? hours?) and if a line changed state at any stage during this time, it has to be recorded.I am opting for an SR latch, to be cleared by the MCU once the read has been completed. In this scenario a common reset channel on the IC would help maximizing the numbers of available latches in the same footprint (and make the circuit more elegant and simple).I have found a very elusive 74118/19 (possibly NOR vs NAND). However is practically impossible to find good supply of it and even a datasheet. Question:Do anybody have an idea of an IC that offers this capability (SR with common Reset)? Backup question (maybe deserving its own question):Any suggestion on how to implement this otherwise? Looks like an SR is my only choice here, but my brain is just a drop of the ocean. Thank you all for your help! EDIT – to clarify a few points in the design:CHEAP AND SIMPLE DESIGNThis is meant to be a quick, cheap and low complexity design.The most complex part (by design) is planned to be the MCU.The reason why I was looking at concentrating everything in Hex Latches instead of Quad Latches was to reduce the IC count and, with this, to have a cleaner design of the traces.As far as possible I want to keep it digital and without any high frequency line anywhere (or, better said, well confined in their own "realm": MCU, comms module and voltage regulation sections). MCU DEEP SLEEP VS INTERRUPTSI'd rather not give too much confidence at these MCU interrupts management. On top of that, when I will get into power-optimization for the MCU I may end up having to choose between keeping the interrupts alive or saving power.I want to keep it flexible, both capability and power-usage wise and this requires balance. <Q> You may be looking for this: CD4043/CD4044 3-state SR latch with common enable . <S> Connect all 4 R inputs to Vcc (HIGH), then use E as a common reset. <S> For this to work you need a pull-down resistor on every output. <S> Most MCUs inputs can't be configured with internal pull-downs, only with pull-ups. <S> So you may then want to consider another alternative. <S> Use CD4043 instead of CD4044, then connect all 4 R inputs to GND (LOW) and use E as an active low common reset. <S> You will then need pull-ups on every output instead of pull-downs, so just use the pull-ups of the MCU inputs by configuring it accordingly. <S> Why does this work? <S> If you look at the truth table of CD4044: <S> When E is HIGH <S> it's equivalent to set all 4 R inputs to HIGH <S> (= <S> you get the same output from either E or R). <S> When E is LOW all the outputs will be in high-impedance (open circuit) and the pull-downs of the MCU will force a LOW on them regardless of the S inputs (= it effectively acts as a reset). <S> You can derive a similar deduction for CD4043. <A> Any suggestion on how to implement this otherwise? <S> if you already have a mcu here why would you need a flip flop? <S> the state changes on those lines can trigger an external interrupt on the mcu <S> and you don't need to worry about missing a beat from polling. <A> I have never heard of a 74118/19 <S> but I found an obsolete part datasheet for a dual JK TTL 74118. <S> How about a quad RS latch 74LS279 or 74HC279? <S> Historical anecdotes on my other uses for RS latches. <S> Once I had a SCADA telemetry design with several low bandwidth (500Hz) <S> DC motor current signals and <S> I didn't need to digitize it and consume <S> whole synchronous channels BW. <S> To conserve bandwidth, I only needed 1 bit in a synchronous "sub-frame" channel to send the analog signal as a digital FM signal of 0 to 1kHz. <S> ( like going to sleep for your system) <S> Otherwise without an RS latch it might miss the 1 shot pulse or worse cause ALIASING (read inter-modulation or beat freq. <S> effect) using the RS latch. <S> ( so I had to invent a linear VCO 0~1kHz ( <S> 40 yrs ago) which turned to be a simple analog sol'n with a 1 shot IC <S> but then I could send 0 to 1kHz asynchronous pulses on a synch. <S> telemetry data channel as 1 bit in a word using a simple RS latch to get rid of aliasing and sampling the logic signal could be anywhere from zero ( like sleep) to equal the sampling rate of 1kHz with a 1us pulse. <S> Never say you are nobody! <S> You matter to me! <S> +1 to anyone who likes the anecdote.
| You might way to use the common enable in the CD4044 to implement the solution you're looking for. Some MCUs inputs can be configured with internal pull-ups/downs, so you might be able to do this with no additional parts.
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How to recognize USB to RS232 null modem cable I have been trying to establish connection between my PC and uC development kit over serial line. There is a standard serial port on the development kit. In oposite there is only USB ports on my laptop. So I have used USB to serial cable for the connection. The problem is that I am not sure whether it is a so called null-modem cable. Can anybody more experienced tell me how to double check it? Thanks in advance. <Q> It would be very unusual for it to be a null-modem cable. <S> A USB to serial converter normally allows you to add a serial port that is connected in, and works in the same way as the original COM ports fitted to desktop PCs. <S> You would use a null-modem cable to connect, for instance, two PCs together. <A> I manufactured a serial device for 20 years. <S> I have never seen a USB-SERIAL cable with null modem. <S> Look in your System => <S> Device Manager for Ports COM and LPT. <S> It should be there if the driver is installed. <S> The title or the driver may give more information. <S> Most of these cables use the Prolific PL2303 chip. <S> If you are having problems, its most very likely a driver issue. <S> The serial port and Windows was 97% of my tech support issues. <S> Try installing the Prolific Driver , the odds are good it will work. <S> IF that doesn't work Win 10 <S> does a good job of finding the drivers automatically. <S> It is most likely NOT a null modem cable. <S> It will be a DTE. <S> Not a DCE. <S> IBM designed the PC to be a terminal to their big iron <S> so made it a data terminal rather than data circuit-terminating equipment. <S> The USB cable's serial port is the same as if the PC had a serial port. <S> A development board made to work with a PC should be a DCE device using a straight through cable and not requiring a null modem unless the designer is an idiot. <S> The pinout will look like this: (To PC means to USB) <A> I usually don't bother to figure which way it is, I just connect it and if it doesn't work reverse term 2 and 3 on serial end.
| If you would need a null-modem cable to connect your development kit to a desktop, you would also need one to connect to your USB converter.
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Why would I need a low-dropout regulator? I am creating a device utilizing multiple components, all of which have an acceptable operaating voltage of about 5 V. One component has an operating voltage of about 6 V. I am looking at relevant schematics online, and noticed that this device utilizes a component* I didn't think I needed ( TPS763XX , a low-dropout linear regulator). This is the first time I've ever heard of this component; I've done brief research on it and don't understand it much. Is it used to regulate input voltages for certain components (i.e. were my device to have a 6 V supply, this particular component would create an output of 5 V to be the voltage supply for those devices requiring it?)? What is this component used for in general, and why does this particular application (a blood pressure monitor) require it? *Reference component U1 in Table 2 at the top of p.17. EDIT: fixed link <Q> Why would I need a low-dropout regulator? <S> To answer this you start by considering "Why would I need a voltage regulator?. <S> You need a voltage regulator to provide a constant voltage to a circuit. <S> More often than not, a constant voltage supply improves circuit performance and allows you to use components that cannot work on lower or higher voltages. <S> Not all circuits require voltage regulation however. <S> A voltage regulator isn't perfect - if you put 5.1 volts at the input, most will not produce a reliable 5 volts out under all (or any) load conditions <S> so, manufacturers specify the "drop-out" voltage such as: - With a 1 V (minimum) drop-out voltage the device can regulate the output from loads requiring 10 mA up to 1 amp. <S> This is just a made up example. <S> Low drop out regulators are generally classified as working with an input-output voltage lower than 1 volt. <S> Having said that, some manufacturers will call their devices "LDO" if the input-output voltage has to be 2 volts. <S> Notice also that in the above made-up specification I implied there was a minimum load of 10 mA - watch out for this as it can bite you <S> i.e. you buy an LDO regulator and hook it up but instead of producing 5 volts <S> , it's producing 5.75 volts - the smallprint usually informs that the minimum load is x mA and with no load connected, the output doesn't regulate very well. <S> why does this particular application (a blood pressure monitor) require it? <S> Some designs do, some designs don't. <S> Without going into great detail, as I said above, a voltage regulator can improve circuit performance and this can mean: - Better stability <S> Lower noise <S> More predictable signals i.e. more accuracy <S> Less drift over time <S> Using an LDO regulator offers exactly the same but allows the output voltage to remain regulated when the input voltage is quite close to the output voltage value. <A> "Low dropout" means you just need less headroom above your regulated voltage than a "normal" regulator. <S> For example, the minimum input voltage for a 7805 to produce 5V is about 7V -- or 2V of dropout. <S> Let's say that a "conventional" 3.3V regulator also needs 2V. <S> Well, if you have a 5V supply, and need 3.3V for some of your circuit, you're out of luck -- <S> unless you can find an LDO regulator with a dropout of less than 1.7V (which is no problem, these days) <A> LM317 drops 1.5 to 2V between 20mA and 1A for example. <S> "LDO" or "low drop out" regulators are designed for much lower dropout voltage, like 0.2V or even a couple tens of mV. <S> You don't need a LDO to make 3V3 from a 6V battery. <S> A LM317 would work just fine. <S> However, if you are looking for a voltage regulator with... <S> Low standby current Features (like an enable pin) <S> Low noise and such Stable with small output capacitors like ceramic Cheap <S> Then, it'll probably also be a LDO, because pretty much all chip makers only make LDOs these days. <S> And... in your case, if you want 5V from a 6V supply, you will need a LDO too. <A> I get the sense you are asking more about what a voltage regulator is as opposed to specifically a low-dropout regulator. <S> These are used very commonly in electronic circuits to maintain a fixed voltage, usually stepping down from some higher voltage source. <S> So in your case if you have a device that can't operate above 5V and you have a 6V power source, you would use a regulator to step the voltage down from 6V to 5V. Regulators <S> come in various fixed voltages, and some can provide adjustable voltage. <S> As others have noted, a low-dropout regulator is just one where the minimum input to output voltage difference is low. <S> In the 6V to 5V case, many standard regulators would not be able to do this because the dropout voltage is more than 1V. <S> A low dropout regulator would probably be needed. <A> A linear voltage regulator, when properly used, will provide a regulated (within limits) voltage from a varying input voltage (within limits). <S> They are used because: ICs/microcontrollers etc. <S> tend to have quite narrow acceptable voltage ranges, and power supplies can have quite a wide possible output range, it is often necessary to use a regulator to ensure that the supply to those components is within limits. <S> A typical design may have components that require different voltages. <S> (5V, 3V3, 3V etc.). <S> Different regulators can supply these voltage from a single input supply. <S> The "low dropout" part simply means that these parts will work down to a "low" difference between input and output voltages. <S> In the case of the heartbeat monitor you have linked to, they have just used 4xAA batteries to power it (roughly 6V), so the regulator was necessary to provide the correct voltage for the rest of the system. <S> If this were an actual low power product, the power supply would probably be very different.
| Well, a voltage regulator has an important characteristic called dropout, which is the minimum voltage between input and output for proper regulation.
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Selecting the correct input/output capacitors for a 7805 So I've got a 7805 Regulator and I've been looking it up and each website says different info. I am using an input of 12V DC 1A Wall Adapter as an input supply. www.adafruit.com Says This regulator does not require capacitors for stability. We recommend at least 10 uF electrolytic capacitors on both input and output . Datasheet Says 0.33 uF into the input pin and a 0.1 uF into the output pin. so I want to know which is better and if it matters. I am looking to make it stable to charge electronics and supply projects and such. <Q> Obey the datasheet, but there is no harm in adding extra capacitors. <S> The 330nF and 100nF <S> (non-electrolytic) capacitors are probably required to guarantee that the regulator is stable. <S> They should be as close to the regulator as possible. <S> The 10µF electrolytics suggested on the website may be beneficial to the rest of the circuit. <S> Eg. <S> ripple smoothing on input and a "circuit-wide" decoupling on output. <S> So, I'd suggest combining the two. <S> (Don't use just the electrolytics as they have strong parasitic properties.) <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The datasheet does not recommend any specific capacitors, it simply mentions that the measurements were taken with 330nF and 100nF. <S> IMO <S> this is a shortcoming of the datasheet. <S> If the regulator is stable without capacitor, then it should be written. <S> If a specific value/ESR is mandatory, then it should also be written. <S> A cap at the input usually helps stability, as regulators tend to dislike inductive supplies. <S> Now, the output. <S> Considering the history of the 7805 regulator, I would pair it with a decoupling scheme matching its age, like a 10-100µF aluminium capacitor with ESR between 0.5 and 1 ohms, and a 100nF decoupling cap close to the load. <S> I would avoid low-ESR caps. <S> If you guys are interested, I might be motivated enough to test one with the network analyzer. <A> simulate this circuit – Schematic created using CircuitLab <S> If you can analyze a pulse load noise response, you can choose your own Cap. <S> The LDO is internally compensated to be unity gain stable with a capacitive load. <S> But step load response may cause under-overshoot depending on rate of steps. <S> So choose Cout based on Ic=Cdv/dt for dv/dt= ripple and output ESR of emitter follower of about 1 Ohm without feedback (depending on current rating) and with feedback Zout is reduced by OA gain at DC and is implied by load regulation error in datasheet as the R ratio. <S> Assume OA BW is about 10kHZ. <S> Does this help you understand?
| If the main supply caps are more than a few cm away, adding the 330nF cap mentioned in the datasheet would be a good idea, or any small value modern aluminium electrolytic. If space is limited, I'd go with what the datasheet says only.
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Need alternative way to deal with LDOs minimum current requirement I am designing a circuit that uses an LDO to produce 3.3v from a 10 to 18v input.I sometimes will need 1.2 amps from this regulator and at other times I only need microamps. When I need even less I can shut down the regulator but there will be times when I need regulation and my circuit will only draw possibly .5 or 1 milliamp. My circuit contains a microcontroller. My datasheet says my LDO requires 5ma minimum load. It's a Micrel/Microchip MIC29150. So my question is, other than putting a resistor on one of the GPIO pins of my micrcontroller and burning off electricity at around 5ma when I want to, are there any other techniques I can use that are better? One of my thoughts were to put a zener in series with a resistor so that if the voltage goes over 3.5 or so that it burned off some milliamps but that seems dangerous to me. <Q> Most LDOs with minimum current requirements do so for stability. <S> Perhaps try to find a regulator which does not have such a requirement? <S> Check MIC2940A and MIC2941A <S> You could also use a switching regulator... <A> MCP1703 is specified up to 16V (tolerates 18V), has a \$<10\ \mu{A}\$ quiescent current and ground current \$<0.15\ mA\$ under maximum load conditions. <A> Per the datasheet for the MIC29150: The MIC2915x–2975x regulators are specified between finite loads. <S> If the output current is too small, leakage currents dominate and the output voltage rises. <S> The following minimum load current swamps any expected leakage current across the operating temperature range... <S> So it seems the minimum load requirement is determined by the worst case leakage from IN to OUT through the LDO. <S> 5mA seems like a lot - the leakage is probably far less than this in most cases. <S> Also, leakage is a very strong function of temperature, so if your application will not see high temperatures you could get away with less. <S> A zener at the LDO output might also be an option, as you stated. <S> This will prevent the output of the LDO from exceeding the zener breakdown voltage in the case that the leakage current is greater than the load. <S> In this case you would want to check that your load can survive the zener voltage, and also make sure that there is no overlap between the regulation voltage of the LDO and the breakdown of the zener - that would cause the LDO to dump current into the zener. <S> There are many with no minimum load requirement at all.
| Personally, I would recommend looking for a different LDO that doesn't have such an onerous minimum load requirement. If you need it, you can drop some volts using series diodes at the input. With too little current, the control loop may not have enough gain, or it may lose phase margin and get unstable.
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Capacitor of choice for low noise applications I saw in a schematic a while back that a polypropylene capacitor was used to generate a stable sine wave in an oscillator circuit. I suppose this gave a better 'frequency reference' for the oscillator. Are some capacitor types better for lower noise applications? <Q> HighK ceramics like X7R, Z5U etc have huge variation of capacitance versus voltage. <S> Using them in filters or any kind of coupling application guarantees humongous distortion. <S> They are piezoelectric: they are both good loudspeakers and microphones. <S> Decoupling a high impedance node with them results in a nice vibration detector. <S> Tolerance on values are not huge, rather they are hyuuuge: expect +20/-50% depending on DC bias. <S> Also, they drift a lot with temperature. <S> They are truly excellent for power supply decoupling, though, because they are small, have lots of capacitance per volume, low inductance, and are cheap. <S> For decoupling, who cares if it's 1µF +/- <S> 50%? <S> Now, for filtering applications, or when you run a signal through a cap as in your oscillator application, you want... <S> A known precise value Low temperature drift <S> No capacitance variation with voltage Low sensitivity to vibration <S> Dielectric absorption and leakage will not matter for your oscillator, but they will for other applications. <S> Film caps and NP0 ceramic caps are excellent on this, although: large thru-hole film caps tend to be microphonic <S> polyester has worse <S> dC/dV than the other films and NP0 <S> Your first choice should be NP0 ceramics if they are available in the value you need. <S> They are small and cheap, and almost perfect. <S> NP0 ceramics and High-K ceramics like X7R/Z5U are completely different materials. <S> High-K capacitance varies with DC bias, NP0 does not. <A> In audio, they're used for the signal path, whereas electrolytic or ceramic are used for bypassing. <S> Film capacitor lack the parasitic piezoelectric effect present in ceramics, and they also are very stable with respect to its bias voltage. <S> In ceramics it's the other way round: capacitance can change up to -90% with bias voltage, introducing enormous nonlinearities if used in the signal path. <A> I saw this in the LT6657 datasheet: <S> For very low noise applications, film capacitors should be considered for their lack of piezoelectric effects. <S> Film capacitors such as polyester, polycarbonate and polypropylene have good temperature stability. <S> Additional care must be taken as polypropylene have an upper limit of 85°C to 105°C. <S> Above these temperatures the working voltage often needs to be derated per manufacturer specifications. <S> Another type of film capacitor is polyphenylene sulfide (PPS). <S> These capacitors work over a wide temperature range, are stable and have large capacitance values beyond 1μF <S> And from a manufacturer Polypropylene is generally selected for its excellent dielectric characteristics (losses, absorption, dielectric strength, <S> insulation resistance: very low Tgd and dielectric loss, low dielectric absorption, excellent dielectric strength, high insulation resistance, temperature- and frequency-stable characteristics, excellent self-healing properties for metallized Polypropylene, etc. <A> We use a lot of C0G (NP0) dielectric ceramic capacitors. <S> Our typically application where use is for very narrow band and low noise filtering and single stage CE amplifier. <S> Input signals are typically below 1mv at 1MHz <S> Do keep what Enric said in mind. <S> Especially with DC bias. <S> That will quickly kill your effective capacitance of which ever part you choose. <S> This can be offset by choose a larger value capacitance and voltage rating part which potentially means more cost and larger footprint.
| Film capacitor are widely used in applications that require high stability of the capacitance value.
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Is it necessary to use pickit for pic MCU? Recently I bought PIC18f4550 dvelopment board by OUMEX. I started go through online tutorials and mess around MPLAB IDE to use it. While setting up a new project, I was asked to select a tool as shown in the picture below. I was not aware about these tools. I looked up online and its costs $70 plus (WOW) and takes about 5 days to deliver (In meanwhile I wouldn't have much to do). Back to my question is it necessary for me to use these tools to program my MCU? How does it benefits or affects my coding overall Thanks in advance! EDIT:Thank you everyone for you explanation. I think my research wasn't proper enough that I didn't even know that a I needed a PIC Programmer. Till now I worked on the development boards like arduino uno, Silicon labs 8051, etc which had it inside them so it didn't strike me that the development board I bought didn't have it. I just assumed that it would be embedded inside.Now I kinda regret not asking for suggestions on which kind of MCU I should have chosen. Since PIC18F4550 has been in the market for a good time, so troubleshooting it wouldn't take up so much time. Anyways lets hope for the best! Again thank you so much. Your insight has helped me. <Q> Almost all micro controllers need a dedicated tool to write the firmware in to the chip. <S> Atmel, ST, Microchip all have there own tool. <S> You can buy micro controllers with a boot loader that will enable you to write the firmware with a UART. <S> (example: Picaxe and Arduino) <S> There are 3th party clones available. <S> A quick search on Ebay for pickit 3: They sell for $11 to $20. <S> DM164140 - MPLABXpress Evaluation Board is a nice alternative. <S> It has a programmer build in. <S> It's easy to get started. <A> If you don't mind using another PIC microcontroller then you can get the Microchip Curiosity board which comes with a PIC16f1619 and a programmer built in for about $30. <S> If you specifically want to use that chip then you can get cheap Chinese knock offs of the pickit 3 for less than $20 in some cases. <S> I have one <S> and I can't tell the difference between the fake one and the real one that I use at work. <A> Back to my question is it necessary for me to use these tools to program my MCU? <S> To be sure, a pic can be programmed without thee use off any of those tools. <S> How does it benefits or affects my coding overall <S> They make programming easier, and. <S> They make coding more efficient, as you rarely need to wonder if your programmer is part of thee problem. <A> You need something that gets the bits into the target PIC. <S> This has to be able to communicate with a PC on one side, and wiggle the programming lines of the PIC appropriately on the other side. <S> Such things are usually called PIC programmers . <S> As far as I'm aware, my LProg PIC programmer is the lowest cost of all the ones out that that actually follow the rules for both the host and PIC interfaces. <S> However, one reason it can be made so cheaply is because it only supports those PICs that can be programmed with everything at 3.3 V and don't need a high voltage to enter programming mode. <S> This is true of pretty much all newer PICs, but not the old 18F4550 you are using. <S> Most of the old parts have newer replacements that are largely compatible, such as the 18F45J50, that do work with the LProg. <S> However, as much as I'd like to sell you a LProg, you're better off with one of the PicKits from Microchip. <S> This is because these also function as low end debuggers. <A> @PeterGreen Link: olimex.com/Products/PIC/Proto/PIC-USB-4550/resources/… – EEE <S> 6 mins ago Afaict <S> (the documenation is not especially clear) your specific board has what is known as a "bootloader" on it. <S> This allows you to reprogram most of the PICs program memory without needing an external programmer. <S> The downside of bootloaders are you lose a small ammount of program memory, you need to use special linker scripts that take account of the bootloader and you often have to use tools specific too the bootloader to do the actual programming. <S> They say the board comes with the "microchip USB bootloader". <S> I am not 100% sure what they mean by this, but I expect they mean the same bootloader that shipped on microchip's official "PICdem FS USB". <S> If so then you should be able to use the tools intended for that board to program it. <S> You can find the documentation/software related to the microchip "PICdem FS USB" at http://www.microchip.com/Developmenttools/ProductDetails.aspx?PartNO=DM163025 <S> I belive <S> you want the "USB Framework" download. <S> Unfortunately said tools are getting a bit long in the tooth now, <S> so you may well have to find a 32-bit windows system to use them.
| If you are making a living out of this, get the best tool you can afford and skim on anything else.
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Safety of Aluminium for enclosure earthing To earth an Aluminium enclosure, is it safe to use a welded Aluminium stud and then use a nut and washer to attach a hook-ended earthing wire to the stud? Aluminium oxidises to develop a thin insulating layer of Aluminium Oxide. Could this increase the resistance of the earth path and render this solution unsafe? Also, galvanic corrosion between steel and Aluminium seems to be an additional potential problem at the interface between Aluminium and Steel. Does this limits the types of materials usable for the nut, washer and hook? <Q> After some more research, I have made some progress towards an answer. <S> Here it is the key info of what we have, in case it can help others. <S> The surface of the Aluminium enclosure, or any other uncoated Aluminium part, is oxidised by atmospheric air, always ending up with a thin (a few nanometers) layer of Aluminium Oxide. <S> This layer is an electrical insulator, but since it is extremely thin, it will be scratched off easily when trying to measure its resistance with a probe or when applying several volts. <S> Yet, it can be a problem for long-term good earthing. <S> This problem is obviously very important in aviation, where Aluminium is often used and electrical earthing is improtant. <S> The suppliers of earthing studs (for aviation and other industries) have come up with a solution. <S> This is the key bit of info. <S> Aluminium earthing studs (with a Titanium coating) are welded to Aluminium enclosures. <S> Then nuts/washers/and hook-ended wires can be tightened to the stud making the connection to earth. <S> BTW, we have also learnt that only Aluminium welds to Aluminium. <S> This is why I am mentioning only Ti-coated Aluminium studs when discussing ways to connect to earth an Aluminium enclosure. <S> PS: There is of course another possibility, to simply use a through hole, nut and bolt (and washer). <S> No welding. <S> But in this case the oxidation of the Aluminium enclosure (in particular the area below the nut) could lead to high resistance. <S> It seems that welding would be the best solution. <A> I don't know about all the safety issues involved with grounding. <S> But we do use a lot of Aluminum panels with various grounds, both screws to connect to the Earth (Ground terminal on the AC plug) and also numerous BNC (or other style) panel mount connectors. <S> All of these have serrated washers with <S> I assume cut through the thin oxide layer. <S> (We do sometimes have issues with the coating on the Al panel, and will have to sand off either the plastic or anodized layer to get good contact.) <A> According to The Circuit Designer's Companion (Tim Williams, 1998), the best solution is a force-fit or welded stud, or failing that a shakeproof serrated washer in contact with the aluminium. <S> If the enclosure is in multiple parts, then you need to ensure continity between them. <S> I still need to find out what a force-fit stud is, and how to ensure continuity if I'm using an enclosure of slot together aluminium pieces.
| Aluminium studs are coated with a passivation layer (e.g. Titanium) that protects against oxidation but is highly conductive.
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Antistatic mat instead of antistatic wrist strap? Just to make sure, before doing something silly: can i place a 2nd antistatic (dissipative) mat on the floor (along the 1st mat on the table) and stand on it with bare feet? Instead of wearing an antistatic wrist strap? Of course, the mat is connected to mains earth thru a resistor. For safety. <Q> Any method you find that works to ground your body (for a wide definition of the word 'works') is OK. <S> 'Works' means it's safe, and you don't forget to use it. <S> I'd be a bit cautious about doing electronics in bare feet <S> , I drop all sorts of stuff on the floor. <S> That's why the ankle strap that goes under the heel of your shoe as well could be a good idea. <A> Mats will get dirty, and require periodic cleaning (more often than one might think). <S> They should be tested regularly. <S> They will wear out, and need to be replaced (even if it's just one person). <S> With that being said, we use mats, straps, and esd shoes. <S> Please, no bare paws while tossing solder, wires, pins, and electrons around. <A> That would probably work, you can also get heel grounders if you like your shoes. <S> They do this in ESD facilities where users need to walk around. <S> ANSI/ESD S20.20 Paragraph 6.2.2.2 ESD protective flooring used with approved footwear, may be used as an alternative to the wrist strap system for standing operations. <A> The problem with heel straps and what you are suggesting is you have to stay connected at all times. <S> If there is a chair the heels are not going to stay on the ground, not grounded, pointless. <S> Just the way humans are when they sit, heels bounce, or come off the ground. <S> Barefoot <S> you have a lot better chance as it is not just the heels. <S> But wrist straps, esp the beeping kind, no competition. <S> Should never rely on heel straps or anything while walking, heel straps are a fail there too. <S> If you use them at all you have to use two heel straps, otherwise pointless. <S> Just use the wrist strap. <S> Put the unit in an ESD bag or box, close it. <S> Then you can release the wrist strap walk/move/etc and reverse the process when you get to the next bench. <S> I would say that barefoot is significantly better than heel straps.
| In the long run, mats are more expensive than a wrist-strap setup, as wrist/ankle straps last longer, and are far less expensive to replace.
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Are electret microphones frequency mixers? I wonder if common electret microphones are frequency mixers. That is, say, when two signals pass through the diaphragm, will two new signals f1 + f2 and f1 - f2 be produced if the sound pressure is around 70-80dB-spl? If they are, how significant are the two signals - will the power levels be roughly the same as the input? <Q> No, at that SPL, the microphone will be operating linearly, and not producing any significant harmonics or intermodulation products. <A> Why not? <S> The more quadratic-type distortion a device has, the more efficient the mixing will be. <S> However, don't expect the mixing products to nowhere near to the power level of the inputs. <S> An efficient diode-based mixer can have a conversion loss of -6dB. <S> But a device not intended to be used as a mixer will be far far more lossy. <A> will two new signals f1 + f2 and f1 - f2 be produced only if the mic is non-linear. <S> it can become an issue when the mic is pushed outside of its envelope.
| In reality, those devices are designed to be as linear as possible so the 'mixing' is quite limited. Almost every device can operate as a frequency mixer (albeit an inefficient one) when driven into its non-linear region.
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How much power is wasted with a light dimmer? I have my rotational light dimmer and am wondering the logic behind it.If it is a potentiometer, when fully off, it should be wasting lets say 50V of power if not using a transistor. By logical design: 1) Should it have an extremely low voltage with a resistor and control a transistor to limit large power loss? 2) Is the dimmer a rheostat or a potentiometer? If anybody has a answer, that would be great for my understanding. Thanks! <Q> Dimmers traditionally involve a TRIAC that chops the ac waveform with a potentiometer controlling the duty cycle. <S> These circuits cause highly distorted currents leading to a low power factor. <S> Knowing this, you will easily find examples of such circuits online. <S> Varying current through an incandescent lamp with a series resistor (rheostat, potentiometer) wastes an unacceptable, even unmanageable, amount of power. <S> Note that this applies to incandescent lamps. <S> There is no single standard solution to dimming LED lamps because of the varying drive methods used in the lamps. <A> The are a triac and the rotating dial you see is the "user interface", emulating the actions of the older rheostat technology. <S> All the dial does now is advance or retard the firing angle of the triac done by the little firing board inside. <S> That's why you see dimmers now with sliders or capacitive touch plates, they are just different versions of the user interface as a way of applying an analog value to the firing board. <S> There is some waste heat in the switching losses of the triac, about 1.5W of heat rejected per amp passed through. <A> The potentiometer's function now is to change the "trigger" point of a triac. <S> The trigger points on the AC sine wave <S> The AC Sine wave with the missing part up to the trigger point. <S> Many LED light bulbs have triac dimming. <S> LED DRIVER WITH TRIAC DIMMABLE CIRCUIT. <S> Which looks like this located in the base of the light bulb .
| MOST modern light dimmers are no longer a potentiometer or rheostat.
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Is the 555 timer accurate and uniform enough for a metronome? I would like to create a simple metronome using the ICM7555 CMOS RC timer ( datasheet ), but I am not sure if the pulses will be "uniform enough". The datasheet states "astable timing accuracy" of 2% / 1.7 - 2.3 ms, but I am not sure how to interpret this. Does this mean the signal will have a variance of 2%, or that the resulting frequency will have a constant error or up to 2%? I can live with a slightly inaccurate metronome (i.e. beeps at 128 bpm instead of 130 bpm), as long as the pulses are uniform , but if the timing between pulses varies a lot, then there is no point in doing this. If 555 is not uniform, are there perhaps some other ICs which can provide uniform timing (apart from using a microcontroller of course, I am aware that's an alternative, but would like to avoid having to program it). <Q> Once going, the pulses from the evil <S> 666 555 timer will be fairly uniform. <S> However, the initial accuracy depends mostly on the accuracy of the analog parts, plus a little due to the comparator thresholds in the timer itself. <S> Unless you spend a lot of money on parts, the error budget will be 15% or more. <S> The capacitor contributes to 10% alone, and the various other resistors, internal threshold voltages, and comparator offsets will cause the remaining 5%. <S> A much simpler answer is to use a microcontroller. <S> Many are available with internal oscillators good to 2% or so. <S> Most are also able to drive a crystal or resonator directly. <S> That gets you to 50 PPM error or better for just a few 10s of cents. <S> In the micro, you can set up a periodic interrupt, like every 1 ms for example. <S> By using the error accumulator method, you can use that as the basis for creating any frequency with long term error as good as the oscillator. <S> Each tick would be within ½ ms of the ideal time. <S> That error is imperceptible to a human. <S> The speaker or whatever sound transducer you use can then be driven for the next 1 ms period. <S> That will make a nice "tick" sound. <S> This will take less parts, be more stable, be far more accurate, and cost about the same as the 555 timer solution. <S> It also has the advantage of allowing exact beats/minute to be entered by some user interface if desired. <S> You can also use a pot as input with markings on the dial calibrated to beats/minute, but the values won't drift with time and temperature as much as they would with the 555 timer circuit. <A> This will be driven by the capacitor. <S> For your application which I assume is adjustable the accuracy will fine. <S> Short term variation will be driven by temperature but will not be enough to worry you (of the order of hundreds of parts per million/degree C). <S> The 7555 includes an internal resistor divider chain which sets the trigger and threshold voltage levels. <S> Tolerances on this will give rise to some intrinsic error in the 7555. <S> The timing error caused by this will be stable in the short to medium term. <S> In my experience problems with timing consistency with 555 timers are usually caused by lack of decoupling capacitors. <S> The 7555 is not so bad and the data sheets says it is not strictly necessary. <S> However I would still use a .1uF across the power leads and from ground to the control input. <A> it is a relaxation oscillator by nature <S> so it would be terribly precise cycle to cycle - <S> the phase errors of a relaxation oscillator is a great random number generator. <S> for more precise timing, consider crystal-based oscillator + divider / counter, or a mcu.
| The accuracy and short term variations of a 555 timer depends largely on the accuracy and stability of the timing components that you connect to it.
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Circuit for connecting externally power instrument to MCU I have externally powered instruments that I need connected to the ADC inputs of an MCU. They output 0-5V analog, and the MCU accepts 0-3.33V. There is no guarantee of the voltage difference in grounding, although they should be connected to the same building ground (who knows where). I'm stumped on the most efficient circuit for this: Differential voltage measurement (to eliminate many sources of noise) Voltage divider (5.5 -> 3.3) Voltage clamp (to prevent overload to the MCU) Are not connected to the same ground locally With some tries, this is the best that I've come up with. However, I'm not sure it's even theoretically a good design to accomplish this. Updated circuit using AD623 <Q> This schematic is rather useless : <S> It has low input impedance: if the source has, say, 50-100R output impedance, which is common, then your input voltage divider will be off by quite a wide margin. <S> It has unbalanced input impedances, and thus nearly non-existent CMRR <S> If you don't need balanced, forget about it, use a voltage divider with much higher resistor values, and a follower. <S> Or no follower, if the bandwidth is now. <S> Its CMRR is humongous, and as a bonus, it will tolerate input overvoltage up to 40V above the opposite rail, so your inputs will not blow even if your board is unpowered and connected to a powered source. <S> However, it will not work on your 3V3 supply. <S> If you have other supplies on your board, then good. <S> If all you have is 3V3 then you could look for a similar chip which will be compatible with your supply. <A> The updated circuit looks functional additional this article explain the Mitigating ground loop offsets and also explain and suggest a differential configuration, which may be helpful to you <A> There is no guarantee of the voltage difference in grounding a few options: <S> try to convert the analog signal to digital locally and then transmit that to the mcu. <S> linear opto-couplers exist for cases like this. <S> convert it to an ac signal and transmit that via transformers -> like xdsl.
| If you need a balanced/differential receiver, a much better solution would be a real instrumentation amplifier like AD8421.
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Meaning of small circles and sharps (#) in pinout diagram? I'm trying to re-plumb a nice proprietary case for continued use with arbitrary motherboards. But I've no clear idea what the meaning of the small circles (shown in red, here) at the juncture of the traces is, or of the # chars on the descriptors. I thought the circle might simply mean negation, as it does in gate symbols, but I'm not really confident that that's true, and it still leaves me in the dark about the # that terminates some of the descriptors. Searching the web hasn't helped. [EDIT] Okay, someone else asked about the circles (229425), and now I know that they do mean negation. But I still don't understand the #s. <Q> The default in logic circuits is that a high level (logic one) activates a function. <S> The small circle denotes inversion, so we can have an external signal that is active low, which (when low) activates a function (that is still activated by the more-or-less theoretical signal after the circle, which is active high). <S> The fact that a signal is active low can be represented in various ways in the signal name, the # is one, and _ is another. <S> Don't be confused by the location of the symbol name, it denotes the external signal, not the internal signal after the inversion. <A> It's as you stated, negation gates. <S> The sharp symbol is exactly the same you can see that sharp symbols and circles coincide on the same pins. <A> Both the circles and the # imply negation, i.e. that the signal is active low.
| They use this to tell the user that the pin is active when applied 0V,so to keep it deactivated you must set a +Vcc voltage in the pin.
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Confused about polarity of Transformer I am confused about the Transformer dot polarity. I got design from ST's edesign Suite. Here is schematic for that :- Now I need to order the design for transformer and the transformer design centre asking for dot polarity/Start-End point of pins. According to circuit, there are only 4 pins used on AC side and 2 pins on DC Side. The ST recommends EE-10 size Bobbin. Here is layout for that :- What does 2 Pin in above figure donates? Is it start point or end point? Where to connect this pin in circuit? I only need 6 pins for my circuit, but the layout for Transformer shows 10 Pins. I am also confused about the polarity. The Transformer specs :- Design Specs :- <Q> 2 is Wurth's number for centre tap. <S> It's needed in push-pull configurations. <S> Actually there are 2 primaries in push-pull systems and the center tap is only a drawing habit due their serial connection. <S> The halves are used in turns. <S> You need only single primary, no need to make the centre tap. <S> "The parallelled 3" should NOT be ignored. <S> That can be 3 identical windings in parallel to make the losses smaller. <S> 3 wires is better than a thicker single wire due the skin effect. <S> Refer the documentation. <S> Wurth has thought of 2 secondaries. <S> Low loss full wave rectifier needs them. <S> You need single secondary, but again check, if it must be 3 identical windings in parallel. <S> Wind all windings to the same direction. <S> The dot is the starting point of the winding. <S> Fail in this = <S> > wrong pulse polarities = <S> > smoke. <S> Unused pins are harmful only because they need space. <A> It's just a termination halfway through the primary winding to "pause" it, wind the secondary and then resume the rest of the primary. <S> This is to reduce the leakage inductance, hence lower losses and better regulation. <S> Google interleaved transformer. <S> You can see it as the end of each winding too but this is an uncommon denotion. <A> If you PUSH current INTO a transformer lead with a DOT, then current will flow OUT OF, any other winding with a DOT.So, for example, <S> if you PUSH current into lead 1, current will flow OUT OF, leads: 4 and 7 and 8.
| As for the polarity, each dot denotes the start of each winding, assuming you wind all of them in the same direction.
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A Voltage Divider with Load Regulation and Efficiency Question? For this question, the voltage divider splits the 12 V battery into +/- 6 V and it supplies up to 10 W total to the load, split between the two output voltage outputs in an arbitrary ration. Load regulation is intended to be greater than 1% from 0 W to 10 W. Now, in this case, the resistors, R, aren't the load resistors, the load resistors would be connected in parallel to the resistors, R Now, what I need help on is approaching this problem. I think I managed to get the load resistors correct at 3.6 Ohms from 10 W / ((6 V)^2) but I'm not too sure <Q> The regulation of 1% should be taken into account when loads (not shown in your schema) are connected to this awful 'regulator'. <S> Since you need 1% regulation, the current through the resistors for your divider should be 100 times greater than the current through the load. <S> Each LOAD would be 3.6 Ohm in the worst case (10Wx2 transferred to the load). <S> The resistors on the divider should be 100 times smaller, namely, 0.036 Ohm each. <S> So your VERY hypothetical divider would be consuming 2kW!!!. <S> Your battery would be discharged in no time, the resistors of your divider would be bulky things heating the environment like a home radiator... <S> so it should be obvious that the voltage divider solution is not applicable for any application that requires a significant amount of power to be transferred. <A> Some op-amps like the <S> TCA0372 can output a high enough current to use them as a power source, hook it up like this <S> and it will source or sink current to maintain its output at the voltage that is on its non-inverting input (the middle of the voltage divider). <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Just make sure that whatever op-amp you choose can output about 1A. <A> simulate this circuit – Schematic created using CircuitLab <S> There are some ambiguities in the question. <S> One way to interpret and approach the question is to consider one side only and convert the source plus divider to its Thevenin equivalent. <S> Now with the required 1% difference of + <S> V for Rload <S> being 0W or 10W. <S> You can solve for R. <S> After you get R, you should do at least one other calculation. <S> Calculate the power dissipation of R even when there is no load. <S> The question is perhaps trying to show how ridiculously inefficient it is to use a voltage divider as a regulated power source. <S> Your calculation for Rload of 10W being 3.6 ohms is correct if you ignore that there may be an 1% output drop. <S> Since there would not be an 1% output drop if the +V and <S> -V loads are balanced, so the 3.6 ohms should be fine here.
| As a rule of thumb, if the resistors of your divider are smaller than 1K each you should probably consider another solution.
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Why does VGA have so many ground pins (compared to DVI-I for example)? If you look at the pinout for VGA, there are several ground pins: I was curious as to why, and I found this answer . To sum it up, the extra ground pins are so that each pin has its own ground in order to prevent interference in the analog signal. But here's a DVI-I connector that supports analog signals: The analog pins are on the right side. The big cross is ground, and the four smaller pins surrounding it are for the red, green, blue, and horizontal sync. What is interesting here is that the ground is shared by all three color channels, unlike VGA where each has its own. Why are the additional ground pins necessary to prevent signal interference when using VGA but not DVI-I? They're the same pins that send the same data, just with a different physical connector, so it doesn't really make much sense as to why the number of ground connectors are different. <Q> First: What's critical isn't so much that there's a ground pin for each signal as much as that there's a ground <S> pin near <S> each color signal. <S> The cross-shaped ground pin largely satisfies that requirement. <S> Second: DVI doesn't prioritize high-quality analog video -- it's a Digital Video Interface, after all. <A> The HD-15 (aka DE-15) <S> connector for VGA connection was compatible with hand-crimpingof pins to conductors of coaxial elements of the multiwire connectingcable. <S> That practically requires a two-pin set for each of R, G, Bvideo signals, to accommodate a signal pin and ground (coaxial shield) pin. <S> Those signals were not logic level, and lackedthe noise immunity conferred by the logic margin. <S> The DVI-I analog compatible signals may use the same wiring, but hand-assembling ofcrimp connections is no longer how the cables are constructed. <S> As forthe digital signals, in DVI-I and DVI-D cables those are twisted pair with shield, so require three wires each for up to seven fast digital signals. <S> To my knowledge, there aren't any hand-assembly crimp pin options, thecable connectors are intended for machine wiring . <S> In any case,twisted pair digital signals are noise-insensitive (because the digital signals have a significant logic margin).The shielding of those digital pairs prevents crosstalk, but the shields carryvery small currents. <S> Four pins (shaded yellow in the above question) are shared for up to 7 twisted-pair shields. <A> You are confusing the DVI connector with the cable. <S> That "single" analog ground is a big honking ground, but what goes into a cable? <S> The simplest is a DVI-A cable, which usually has a VGA connector on the other end. <S> Internally, it will carry 3 color channels (red, green and blue) and 2 sync: vertical and horizontal. <S> And guess what? <S> A DVI-A cable will typically have 10 conductors. <S> That is, 5 twisted pairs, 3 for color and 2 for sync, each with a ground line. <S> The 3 analog color grounds will be connected to the cruciform analog ground on the DVI connector and the sync grounds will be connected to another pin. <S> So, just because a DVI analog section only appears to have a single ground, that does not mean that the cable does as well. <S> In fact, the cable and the VGA connector will have the same number of ground pins (5) as a standard VGA.
| The small loss of quality incurred by using a single analog ground pin was probably considered acceptable by the designers.
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What is the impedance of a Microchip PIC I/O Port configured as input when Vpin > Vdd? I'm using a PIC18F1320, and it is powered with VDD=4V.There is an external power signal, with a higher value than VDD. The external signal is connected to the RB5 I/O port through a voltage divider: All internal pull-up resistors are disabled. Port is set as input via TRISBbits.RB5 = 1; Under those circumstances, I expected the node P to be around 6.95V . But instead I found it to be at 5.6V . This means that node P is leaking almost 35mA but: The node P is also connected to a MOSFET gate, whose leakage current is, according to specifications, 2uA. According to PIC18F1320 specifications, the leakage current of I/O ports configured as inputs is 1uA. But it also specifies that the test condition is "VDD < VPIN < VSS", which I'm not meeting. My question is: What is the impedance of the input port when the input voltage is slightly over VDD? simulate this circuit – Schematic created using CircuitLab <Q> You're outside the maximum absolute ratings, so nobody knows anything. <S> As far as the manufacturer is concerned aliens are messing with your warp field coils. <A> MOSFET gates are very sensitive to high voltages (if the gate oxide has a thickness of 20 nm, a voltage of 7 V results in a field strength of 350 kV/cm, which is likely to damage it; at 10 V, breakdown is likely).So most chips have ESD protection diodes from ground and to V DD that are intended to clamp voltage spikes. <S> (Sometimes, input pins are designed to be 5V tolerant; in this case, the diode to V DD is omitted.) <S> As long as the diode to V DD does not turn on, you still get the high gate impedance. <S> (This is why Microchip specifies "V DD + 0.3 V" in the absolute maximum <S> ratings.)But when the voltage goes higher, the diode conducts, and the voltage at the pin is clamped to V DD plus the forward voltage of the diode. <S> How large this is depends on the current, which is limited only by the resistor you have put in front of the pin. <S> Microchip's absolute maximum ratings also specify a clamp current. <S> You should stay far away from that limit, so 35 mA is way too large; consider making the resistor high enough so that you get a few tens of µA at most. <S> If that makes the signal too slow, you have to use a 'real' level shifter. <S> The input impedance of the pin itself is essentially that of the diode to V DD , but it's smaller than the series resistor which you need anyway, so that is your effective impedance. <A> What is the impedance of the input port when the input voltage is slightly over VDD? <S> practically zero <S> : the clamping diodes is all there is. <S> that's why it is always a good practice to put a serial resistor (or a buffer if allowed) on pins that interact with the outside world.
| Seriously though, most chips have a diode from their pins to Vdd, which is probably sinking your current.
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How to wire a 4PDT Switch to switch Rs and Caps for Hi/Lowpass filter Sallen Key Equal Component I'm making an active Low/High Pass filter based on the equal component sallen key design I found in the Active Filter Cookbook. In the book I read that by exchanging the position of the Frequency determining Resistors and Capacitors it is possible to switch between a high pass and a low pass response, using a 4PDT switch. I'm very confused about how to connect the switch to the components/board, I've been drawing out the connections for a good few hours! It's simple enough I'm sure, but I can't seem to get something which makes sense, or would work! My 4PDT is connected like this: Could someone please help me with a sketch of the connections or any tips on how to hook up the switch? :) Thank you very much, <Q> seriously, OPamps and even precision caps are far, far cheaper than 4PDT switches these days (it's been a couple of years since the active filter cookbook was written). <S> Just have the two filters in parallel and switch the output with a simple switch. <A> Look at the non inverting input and imagine an SPDT switch connected to it (using the switch's common terminal). <S> Now use that switch to choose which network of components feeds the non inverting terminal. <S> That should be all you need to do. <A> You should keep your expensive 4PDT switch for an application that needs it, and follow the advice already offered. <S> It is possible to use fewer than 8 poles with various messy kludges, but I doubt it can be done with 4 poles. <S> Or (without knowing what you're trying to achieve) you could use a state variable filter instead, which is more controllable and has simultaneous LPF, BPF and HPF outputs.
| If you MUST use the SAME components for the HPF and LPF functions, then in the simplest case you would need to switch BOTH terminals of all 4 passives, using an 8 pole switch.
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Low noise transimpedance amplifier (TIA) - why does the addition of a feedback capacitor cause voltage noise peaking? I am working on a low noise transimpedance amplifier (TIA) for the detection of weak optical signals. The aim is to achieve a 10MHz bandwidth with a white voltage noise floor of 10-20nV/rtHz. I am using the FGA21 photodiode and the OPA847 Op-Amp with a 10kohm feedback resistor operating in photoconductive mode. Key specifications include: Gain banwidth product: GBW = 3.9GHz Input voltage noise: e_n = 0.85nV/rtHz Input current noise: i_n = 2.5pA/rtHz photodiode capacitance: C_d = 100pF @ 3V bias The PCB design followed many of the suggested layout techniques (minimising track length, passing feedback components under the op-amp, isolating sensitive tracks from the ground plane etc.). Additionally, the voltage supply was heavily filtered using decoupling capacitors and the OPA820 Op-Amp was used to buffer the output. Two noise spectra were taken, one where the feedback capacitance was left open and one where it was set to 1.5pF: The dashed lines represent the corresponding theoretical noise curves. Clearly the capacitor causes the noise peaking to broaden and shift in frequency, this contradicts theory which suggests that a feedback capacitor dampens the transimpedance gain and reduces high frequency noise. To test this further a circuit was constructed without the photodiode, instead a 100pF capacitor was added to mimic the diode junction capacitance and the noise measurements were retaken: In this circuit the addition of a feedback capacitor causes the noise to dampen similar to how the theory predicts it to, suggesting to me that the simple photodiode model of a junction capacitance and current source may not be completely accurate. However, searching through literature I have not yet found discussions of the limitations of this model, nor have I seen any examples of this behaviour. So I am wondering if anyone else has come across this issue before or can understand how the addition of a single capacitor causes a large disparity between theory and experiment? (Please excuse the lack of circuit diagrams, I am a new user and as of yet can only attach two links per question) Edit: Here is the PCB layout for the TIA with photodiode: and here is the circuit schematic (it is worth noting the low pass filter between the op-amps wasn't used, the capacitor was left open): Edit 2: Note in the above circuit diagrams the photodiode is not reverse biased, in all noise spectra shown it is soldered in the correct bias <Q> I'm not sure specifically about the noise on your circuit, but here is a pretty extensive help guide for laying out TIA circuits: http://www.linear.com/solutions/5633 <S> I can't tell if you have voided the ground and power plane on your input trace from the photodiode. <S> However, you might try the following. <S> Stand the input cap and resistor on end (tombstone). <S> Solder <S> a very fine wire (40AWG perhaps) from one end that is airborne to the photodiode output pin. <S> This will minimize the input capacitance, and thus give you the best high frequency response. <S> Another less drastic thing to try is to trim the pads of Rf and Cf to the smallest possible, then solder them on the board sideways. <S> Parasitic input capacitance is your enemy at high frequencies, and both these ideas are aimed at minimizing it. <S> Although expensive in mass production, it may give you some ideas to make it better performance. <S> Some other ideas - use 0402's instead of 0805's or 0603's. <S> This will decrease input capacitance as well. <S> Another idea that was also in the LT literature was to run a ground trace between the pads of your input resistor. <S> This brings the field strength to 0. <S> I honestly don't have a good feel for how this helps, but they wrap some words around it in the link I gave above. <S> Good luck! <S> You should post some screenshots of your frequency response and let us know what you did - what worked and what didn't. <A> Your feedback capacitor should be more than 10pF as the photodiode capacitance is more. <A> The capacitor by itself will have a much lower impedance than the photodiode, and so I think it could be expected that the circuit will be less stable. <S> It looks like it is even resonating a little at the 10MHz peak so you may need a larger feedback capacitor. <S> If 1.5pF is around the right value then using an actual trimmer capacitor might be convenient for tuning if it does not increase path lengths and such too much. <S> I myself am not properly familiar with the theory so I can only offer basic advice. <A> In my experience it seems that you are experiencing capacitor flyback mode. <S> Which is causing your voltage to create peaks suck as yours.
| To fix this issue I would recommend using a larger Capacitor or adding additional resistance to the system before the capacitor.
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Can a high voltage rated push-button switch be used on lower voltages? I have a question regarding switch usage in a high voltage, very low current circuit (micro amps). Basically, we're taking the 120VAC out of the wall, converting it to DC (output can be either 5, 12, 15 or 24VDC) where it's going to a switch that is going to be the input to an SEL relay. Currently, it uses 120VDC as the input. What I want to know is can a push button with this high of a rating be used for a low voltage application? There are microamps going through so the power dissipated is <1W. Everything I've been taught tells me a push button will work regardless of the voltage running through it, as long as it's pushed (and the button switch isn't melted). The button I'm referring to <Q> You have to observe the minimum current and voltage for the switch. <S> Typically mechanical contacts aren't gold-plated so a thin oxide layer will build up with time and it needs some amount of energy to "burn" that thin layer. <S> See https://en.wikipedia.org/wiki/Wetting_current <S> A capacitator to generate some inrush current may help but this depends on your setup. <A> It is like an insulating film between 2 electrodes. <S> International standards such as IEC 60947-5-4 (2002) define the reference voltages for tests as +5 VDC or +24 VDC, where the current is chosen among the following values: 1 mA, 5 mA, 10 mA, 100 mA; <S> 10 mA is the preferred value. <S> Note that when tests are performed, there is a delay of 10 ms before actually checking if the contact is closed. <S> For additional information on the subject, you can also check the MAX13362, manufactured by Maxim , a 24-Channel Automotive Switch Monitor with selectable wetting current <S> (0 mA, 5mA, 10 mA, 15 mA). <S> If minimum voltage & current are not both present, even in normal conditions (T and humidity) <S> problems arise after a few years. <A> Yes, it should be fine. <S> Keep in mind that the switch is rated for AC, and not DC, so you need to derate it anyway. <S> But as long as you arn't trying to push 6 Amps through it, it will work fine for your case.
| You have to observe the required minimum voltage & current as indicated by the manufacturer. If this is a home project and reliability isn't essential just use the switch... Generally, people think: "if it can do more, it can do less", but this may not be the case with a switch in a humid environment, because of the oxydation phenomena.
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Is it allowed / possible to OR two opamp-outputs? See schematic. As from what I have been able to read, coupling two outputs together is a Very Bad Idea. I yet lack the knowledge to understand why, though. But; if one adds a diode in serial with each opamp-output, would it be possible for them to act like an OR-gate (so to speak), controlling the NPN-base? That is, if either U1 or U2 (or both) has a high output, and R1 is chosen properly, will the NPN-base be turned on here? <Q> Yes, what you show should work. <S> However, consider the stability of the opamps. <S> If they are being run open loop (used as comparators) or the feedback path is before the diodes, then it should be OK. <S> If the diodes are in the feedback path, then things could get unstable. <A> I'm playing around to learn, trying to grasp how opamps works <S> As an aside from the specific circuit in the question..... <S> You are using the op-amps like comparators and if you looked into a lot of comparators they have open collector outputs that can be wired together to form things like window detectors: - If either output transistor is activated it will pull RL down to ground (0 volts) and therefore you have implemented a logic function without the need for diodes. <A> The reason you are not (usually) supposed to connect two outputs together is that outputs are generally low impedance, and they are generally intended to drive relatively high impedances. <S> Typically op-amps have input impedance in the 100s of megohm range, and output impedance in the tens of ohms. <S> Inputs therefore allow only tiny currents to flow from (or to) outputs. <S> If you connect two outputs together you are connecting each to the output of the other. <S> Each output will therefore be working into a much lower impedance than it's designed to drive. <S> This may result in excess current, particularly if one output is going high while the other is going low, resulting in voltage sag <S> (i.e. "clipping" of the analog waveform) and possibly even damage to one or the other. <S> The diodes you have proposed will solve this problem, assuming that the resulting outputs are what you want. <S> For continuous (i.e. analog) signals, resistor-summing can be used to limit the current to safe values, at some cost in signal level. <S> A good example here may be drawn from common practice in audio work. <S> Line-level output impedance of just about everything is generally from 100 to 600 ohms, while input impedances are usually 10K or more, and use of "wye" adapters to mix outputs together is a bad idea. <S> (Using a "wye" to send one output to two or so inputs, though, is just fine.) <S> Similarly: power amplifier output impedance is usually a fraction of an ohm, while speakers are usually 4 or 8 ohms, and you do not ever connect two power amp outputs in parallel to the same speaker (unless someone else is paying for the repairs, and you are getting kickbacks from the amp repair shop). <S> For further discussion of this principle, including the use of resistors to "sum" analog signals, see the excellent article " Why not Wye? " at Rane Audio . <S> n.b. <S> : unlike in video and higher-frequency stuff, in audio we don't "match impedance" from source through cable to load. <S> The wavelengths are much too long and the cables much too short to have to worry about signal reflections. <A> So therfore coupling 2 stages could result in high current flowing between the 2 Opamps which result in lose of power. <S> But with diode it prevents this possibility as D2/D1 will be reverse biased when U1/U2's Output is high respectively. <S> But it would be important to choose proper R1 values to have the right bias current. <S> Also the same when implemented with MOSFET will result in a more efficient operation (because the gate of a MOSFET is high impedance (could be considered as open) whereas that of BJT is not).
| I think without diodes it would not be good practice because Opamps are meant to be having very low output impedance. Looking at the circuit that you have shown , yes it would be possible to have an OR function with this configuration.
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Why are BLDC motors so expensive? My general question is why brushless DC motors seem to be much more expensive than similarly sized brushed permanent magnet motors, especially considering that the construction of the latter is more complicated? In my concrete application, I'm looking for a brushless DC motor with a 10-15W max power rating. I was first looking for an alternator of that range, but this kind of product practically doesn't exist, I believe because a BLDC fills this role pretty well. Using a BLDC as a generator is a very attractive idea - since you're not driving it, you don't need to worry about ESCs, you just throw in a 3-phase rectifier and you're done (also, it's much easier to detect the running speed by detecting the zero crossings on one of the phases. And some BLDCs have Hall sensors as well, which makes it even easier). Comparing an in-runner BLDC to a brushed DC of similar size, you'd expect them to be practically the same in terms of construction, with the latter adding the brushes and the commutator - you'd expect it to be slightly more expensive. My research indicates that the reverse is true: a decent BLDC fetches upwards of $100 in most reputable electronics shops (e.g. Electrocraft in the UK), even in Alibaba they are $30+ (the price break for 100 units), and even RC-craft units optimized for onesies from eBay are $15-$20. Then you can definitely buy s very well machined japanese brushed motor (e.g., Mitsumi) for a few bucks. Why is that? Are BLDCs still considered a novelty and economies of scale do not yet apply to them? Or am I missing some technical detail? <Q> Brushless DC motors typically use strong permanent magnets on the rotor. <S> This does some nice things, like alleviate the need to get any electrical power across the rotating interface. <S> It also allows for Hall effect sensors to determine the position of the rotor. <S> However, these strong permanent magnets currently require rare-earth metals, such as neodymium. <S> Being rare but in demand makes them expensive. <A> Depends on which bldc motors you are talking about. <S> Those bldc motors used in computers are quite inexpensive. <S> Those ones used for drones are expensive due to their ability to handle large current and complexity / numerous windings but most importantly the lack of economy of scale. <A> My company once had a project to quote a high volume system based on stepper motors or bldc. <S> So on high enough quantities the only remaining difference is encoder, which in some cases isn't even required. <S> But the best part is that performance and closed loop itself sometimes allow even cost reduction compared to steppers, as the motor may be precisely selected for application, while stepper often tequired wide margins. <S> Bottom line, since the appearance of modern electronics bldc are not more expensive to drive and not more expensive per same performance. <A> I suspect that the brushless DC motors use neodymium magnets while the brushed ones use less expensive magnets. <S> Even though the Mitsumi motors are good quality, the brushless motors sold for quadcopter etc. <S> may be better quality. <S> There also may be less vigorous competition in the consumer market for brushless motors. <A> Comparing an in-runner BLDC to a brushed DC of similar size, you'd expect them to be practically the same in terms of construction, with the latter adding the brushes and the commutator - you'd expect it to be slightly more expensive <S> The construction of an 'in-runner' BLDC is quite different to a conventional brushed motor. <S> Instead of an armature it has coils arranged around <S> the inside of the stator. <S> Due to the restricted space, the windings on most small in-runners have to be installed by hand. <S> This is a difficult and time-consuming and job, which makes them more expensive than machine wound brushed motors. <S> you can definitely buy a very well machined Japanese brushed motor (e.g., Mitsumi) for a few bucks. <S> Well machined perhaps, but not in the same league as a good brushless in-runner. <S> Even cheap rc <S> hobby inrunners have neodymium magnets, ball bearings and cnc machined aluminum cases. <S> Peak efficiency is typically over 80% (compared to 60-65% for small brushed motors) which translates to lower loss and much higher power for the same size. <S> And they can spin at up to 60,000rpm, whereas brushed motors are limited by brush bounce and excessive wear at high rpm. <S> The biggest cost advantage of a brushed motor is often the controller, which can be made much simpler if bi-directional operation is not required.
| Also the brushed motors are manufactured in very large quantities for all kinds of inexpensive products.
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Will a buck converter drain 2 X 9v batteries with no output load I'm currently building a circuit used in a personal portable cooling system. The fan is a 12v turbine fan (0.1A) My plan is to use 2 9v batteries in series to give me 18v then use a buck-converter to step the voltage down to 12v. My question is, if I have a switch between the turbine fan and the output of the buck converter, and turn the switch off, will the buck converter its self still slowly drain the batteries? Or would it be best to have the power switch between the batteries and the buck converter? Ideally due to the way the product will be implemented and used, it would be more user friendly to have the switch between the buck-converter and the turbine fan however if this would drain the batteries even when turned off, then I will need to have a little re-think. <Q> you had mentioned placing the switch at the output is convenient, you can use the switch to shut down the buck converter also. <S> Most of DC-DC converter IC will have shutdown pins, which will put the device to low power mode in which it consume very low power, so you can combine this shutdown pin and the switch to reduce the power drain of the battery significantly <A> As said in the comments, yes it will drain the battery eventually. <S> However, there is a larger problem as switching regulators are not (usually) meant to be driven with no load. <S> Sometimes they can actually break, as I have done to my own projects by accident. <S> The safer option would be to put your switch between the batteries and the switching regulator, so you not only save some (albeit very small amounts of power as these things are rather efficient in terms of power draw) power, but protect your switching regulator from accidental damage. <A> Although to save the most power, you would want to use a switch before the DC DC converter, there are lots of them that are very very low quiescent(standby) current! <S> Since you are using two 9V batteries, which both have approximately 500mAh capacity, lets say you use a converter that draws 5 microamps of quiescent current. <S> 1000mAh/0.005mA ~ 200,000 hours before it drains both batteries completely! <S> Now this is only a very rough estimate, but you can clearly see that it's most likely your load would drain the batteries much faster. <S> (here is one app note for a very low quiescent current converter <S> http://cds.linear.com/docs/en/design-note/dn142f.pdf ) <S> Another thing you might want to consider is a buck-boost converter. <S> If you are just using a buck converter, then when your input voltage drops to 12Volts, you will not be able to use the rest of the energy in the batteries. <S> With a buck-boost, it can continue to maintain 12Volts even after and fully drain the batteries. <A> Very simple, if anything connected had an impedance / resistance it'll draw power. <S> If your switch is off ( provided not an electronic switch ), there will be no power flow. <S> Electronic circuits will draw leakage current and hence leakage power.
| Since the Buck converter will be running, the battery will get drained, if the switch is placed after the DC-DC output.
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Voltage rating of series connected fuses If a number of fuses are series connected, with the intention to achieve a higher voltage rating, what is the overall voltage rating? For example if we place two identical fuses that are rated for 120V dc in series. I suspect the overall rating will not be 240V dc as one fuse will blow before the other. Is the actual rating closer to 120V or to 240V? Is there some mathematical rule? The example given is two fuses in series but what about n fuses connected in series? <Q> The voltage issue with fuses is limiting the energy available to the arc as the fuse opens, it is not uncommon to see inappropriately chosen glass fuses explode when overloaded from a sufficiently butch source. <S> Fuses have limits on safe breaking capacity, and for small glass examples these can be as low as a hundred amps or so, which is also due to arc energy limits. <S> If seen in these terms the voltage rating of a series chain is clearly that of the lowest voltage fuse. <S> An important thing to note is that fuse voltage ratings for DC service are often very much lower then <S> the rated voltage for AC service as the DC arc will not self quench. <S> Where series connected fuses can be useful is when you have a very stiff supply, like say a telecomms battery bank, <S> if you want to fuse something close to the batteries at say 0.1A <S> , you have a problem, because the prospective short circuit current can far exceed the rating of the biggest 0.1A fuse you can find. <S> Lets say you have a PSC of 50kA... <S> What you do is place a much higher current fuse before the little one, coordinated such that the big fuse will catch nearly direct short circuits and limit the energy delivered to the arc in the much smaller fuse. <S> Fuses are NOT simple devices, read the datasheets carefully. <A> The rating of fuses in series will be that of the lowest . <A> As said above: The voltage rating of a fuse is based on the voltage <S> it can withstand when open <S> If the rating is exceeded, the fuse may arc then explode. <S> If you wire several low-voltage fuses in series, and if one blows and starts to arc, it will explode before the others have enough time to blow. <S> The only acceptable scenario to series connect <S> fuses or circuit breakers is as in house wiring, for example: 50A general breaker feeds a row of 10A-16A-20A breakers, each protecting a separate circuit. <S> In this case, you do have 2 breakers/fuses in series, but all breakers/fuses are rated for the full voltage, and the one with the larger current is only there to protect the wires between it and the next, lower current breakers.
| The voltage rating of a fuse is based on the voltage it can withstand when open (since the voltage across the fuse before it blows is insignificantly small).
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Appropriate charging current for parallel 18650 lithium cells I have 9 18650 cells salvaged from a laptop battery. They have all been tested to work and since they were always used together, I put them in parallel. I'd like to make a power pack, so I got some lithium charging circuits with all the protection bells and whistles, but with just 1A max output current. Now, I've been charging them with the circuit board from another old USB power pack, and it works, it's just super slow. Someone mentioned that the 1A max current simply would not work for 9 parallel cells, but I don't fully understand how it couldn't. As long as I'm charging the batteries with more current than I'm drawing from them, they should charge, right? I'm not super worried about the speed, right now it takes pretty much a full day to charge the pack. My question is, will the pack charge with a 1A total charging current (albeit slowly), or am I missing something? <Q> There could be some legitimate reasons why this won't work well. <S> These may not apply to you, and if you review the documentation of your charger, you should be able to tell whether these things apply. <S> Probably you are aware that it is not safe to float lithium ion batteries at 4.2V indefinitely. <S> Because of this, all Lithium ion chargers use some method of recognizing end of charge. <S> Typically, end of charge is determined by how much current flows into the battery. <S> For a single 18650 cell, end of charge may be defined as the point where charge current drops to 25mA during the CV stage. <S> But with 9 cells in parallel, it may never drop that low. <S> Which brings me to the second point. <S> Timeout. <S> This is a backup safety in case a defective battery is present, or something else prevents the primary charge termination condition from occurring. <S> Typically this timeout would be set to some number far larger than you would expect to ever happen unless something is wrong with the cells. <S> On a C/3 charger, you would set it to 5 hours or 8 hours. <S> If your charger has a feature like this, it could possibly prevent the pack from charging fully. <S> You would have to remove the pack, and put it back in to restart the charge timer. <S> Again, without seeing your charger details, I can't say whether these two issues apply to you. <S> But you may want to check it out or at least be aware and keep your eyes open. <A> Someone mentioned that the 1A max current simply would not work for 9 parallel cells Nonsense !!! <S> it will work but it will take a very long time to charge. <S> One cell of 2000 mAh (=2 <S> Ah) <S> (a typical 18650 is 2 - 2.5 Ah) takes 2 hours to charge <S> so 9 cells take 9 x 2 = 18 hours to charge. <S> Close to a full day indeed, not all the energy ends up in the cells and is lost so in practice 1 day sounds right. <S> As long as I'm charging the batteries with more current than I'm drawing from them, they should charge, right? <S> Yes, that is correct. <S> Next time someone gives an opinion about things always ask why and ignore their advice if they cannot give a good explanation. <A> For your information, your charger is based on TP4056 chip, programmed to 1A charge current. <S> There's no safety timer, meaning it will continue to charge until the charge current drops to 100mA: <S> Other than being slow, there are no apparent problems with your charger.
| Many charger IC's will not allow an indefinite amount of time for charge to complete.
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Can I use one voltage divider to bias multiple points in circuit? I have 4 nodes in my circuit that require virtual ground of half the power supply voltage. 2 of the nodes are just offset for input signal and other 2 are for op-amp biasing to virtual ground. Can I use one voltage divider to provide offset for all these nodes? This would greatly reduce BOM and save PCB area(eliminating 6 resistors and some capacitors), but I am not sure if this can't cause some kind of issues? There was similar question , but it was talking about one IC, in my circuit I have two separate input connections and two op-amps(though they are in same package). Also maybe a better question would be in more generic form - can one use single voltage divider to bias several points(no matter how many) to virtual ground? I understand that in some circuits the resistance on the output of voltage divider will basically make no sense, but if I for example have 16 op amps that need to be biased - will it work? <Q> It all depends on how you make that virtual ground and how you load it. <S> A virtual ground is just a DC voltage used as a reference voltage for (usually) opamp circuits. <S> This can be used to prevent having to use a symmetrical supply like +5 V and -5 <S> V. <S> Instead we use 10 V and make an internal 5 V as a "virtual ground". <S> You can bias 16 (or more) opamp circuits from your virtual ground as long as you take care that your virtual ground is not influenced too much by the circuits connected. <S> If you use inverting opamp circuits and directly connect the virtual ground to the + input of the opamp, no current is drawn from it <S> so you can bias many opamp circuits. <S> An inverting amplifier, here the + input is grounded but that can be a virtual ground as well: If you do load the virtual ground with each circuit then you have to make sure the virtual ground can supply all those currents. <S> but that increases your BOM. <S> Also it is not so much the DC biasing current as it is AC currents resulting from the signal being amplified which can cause problems with a shared virtual ground. <A> Yes, you can. <S> However, I would recommend that you look up the total of the input (leakage) currents for your 4 inputs then make the potential divider current at least 50..100 times that value while exercising some realism. <S> For example, four inputs with input current of 10 uA each would lead me to a potential divider conducting between 2 and 5 mA. <S> You can also consider putting a small capacitor on your potential divider output to steady it against rail noise, such as 10 nF. <S> The exact value depends on your divider resistor values. <A> 2 or 3? <S> You're probably ok. <S> "No matter how many" or 16? <S> Probably not. <S> Of course it depends on your biasing resistor values, and the bias current of your OpAmp. <S> But each one will introduce a small error into your bias point which will change where your "virtual ground" is.
| To prevent the circuits influencing each other you might want to filter (Resistor in series, capacitor to ground) each virtual ground
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Using arduino as a voltmeter, current drain with a voltage splitter? I'm trying to read the voltage of a 3S 11.1v LiPo battery with an arduino, so it can monitor the battery level and report when the voltage is getting low. The easiest way I've seen to do this is with a voltage divider. I've seen tons of "don't use voltage dividers" posts over the years, all for various reasons. My concern is the wasted power lost to heat that the voltage divider might draw. If I have a low current drawing circuit hooked up, with a voltage divider to cut the voltage down to the range the arduino can read via the voltage reference pin, how much current will be used/wasted by that divider circuit? Is there a way to estimate that, without actually building the circuit and testing it? This is going in a device that is to run long term, and I'd hate to shave a large percentage of time off the runtime just to monitor the voltage. Is there a better/more efficient way to read the voltage of a 11.1v lipo on a 5v arduino without using a voltage divider? <Q> Use a P channel mosfet to only enable your divider when you need it, (have a weak pullup to battery+ to switch it back off, and an N channel to pull its gate down to turn it on) <S> This is how its commonly done on other battery monitoring circuits, as its not like the battery voltage changes that much in say 10 seconds. <S> There are other approaches when you have control of the silicon in a chip, but for general user land, this is usually a good compromise. <A> A voltage divider is fine, provided you keep the current very small (as you indicated). <S> A reasonable way to do this is with a voltage divider of large value resistors (e.g. 2MEG and 1MEG will drop the 11.1V to 3.7V). <S> Does this help or do you need more detail? <A> Just use one, but decide when you have too much quiescent current. <S> simulate this circuit – <S> Schematic created using CircuitLab 5mA is probably fine with forklift traction batteries, but not in your iPhone. <S> While 5uA is fine in your iPhone, it may not be in your Airpods. <S> Of course, with higher and higher impedances it will become more challenging to maintain a high samplerate. <S> The first one won't need any buffering, but the last one will need a high impedance voltage follower to not change the voltage by measuring it. <S> If you find a better mosfet than BSS84 you may even get lower leakage.
| At the juncture of the divider, you will want a buffer of some kind, such as an op-amp follower so that the Arduino won't sink current from the supply.
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Reverse engineering the manufacturer's programming sequence of an FPGA As much as IP tends to make things easier, I would like to find was to learn more about protocols and interfaces by doing everything myself (I understand the difficulty of the task, and I have resources to aid me as in professors and books). From what I understand though, it is near impossible (without reverse engineering) to use any hardware features on an FPGA without IP. I have poked around online and found next to nothing on the subject as people seem complacent on using IP. Essentially I'm trying to find more open source HDL resources as there is a lack of it at the moment. Edit: to be more concise, is it possible to work around manufacturers' IP to program fpgas? <Q> It is certainly possible to program (using and HDL) and do useful tasks on FPGAs without using any third party IP blocks - if that is what you meant. <S> However you do still have to use the vendors synthesis tools. <S> Synchronous state machines etc written in appropriately structured VHDL easily synthesize onto the LUTs and flip flops and multiplexers in an FPGA cell and you can have these designs connect with the IO pins using a constraints file (which will be in a proprietary format). <S> The synthesis tools will recognise certain idioms in VHDL and map onto other resources on the chip automatically e.g. using global clock signals where appropriate and even mapping a pipelined arithmetical construct onto a DSP where appropriate. <S> On the other hand there are some resources on the chip for which you have to use vendor supplied macros for the synthesis tools to be able to map onto hardware. <S> One example is if using a PLL to generate clocks of different phases. <A> There may be a misunderstanding about what 'IP' means. <S> It's possible to program the fabric of an FPGA, <S> that is the programmable LUTs and things, from the ground up. <S> However, the bits and gates and tables in a Xilinx and an Altera are different, and are different from family to family, so it's like assembly coding a micro, they all have different low level machine code. <S> I don't think anybody does that, at least, not commercially. <S> It's generally easier to use VHDL and write <S> AND and OR and vector ADD assignments, and let the mapper turn that into LUT entries. <S> I wouldn't call that 'IP', I'd call that compilation. <S> When your VDHL that implies NAND gates and latches hits an Altera compiler, or a Xilinx compiler, as the fabric is different, the LUT tables will be programmed differently, but the low level function that you've described in the VHDL gets implemented the same. <S> If you want to use the specific hardware that Xilinx and Altera provide, like the dual port RAMs, or the 48x24 multiplier accumulator, then you need to use vendor-specific primitives. <S> But, you have full control over how you connect to them. <S> I still wouldn't call this IP. <S> If you want to use a packaged FFT, down-sampling FIR filter, or Viterbi decoder, or the embedded ARM, provided for or licensed by the vendor for their architecture, whether it's a free (with the tools) or paid-for license, that's IP. <A> You certainly can implement any interface protocol by yourself. <S> How do you think these IPs were designed in first place? <S> All you need is to carefully study particular specifications, and implement all state machines in accord to documentation. <S> Then you need to verify all your HDL constructions with test benches covering all corner cases. <S> Then you need to place and route your logic in such a way that it meets necessary timing for required clocks. <S> The provider of FPGA will provide automatic tools to do this job, but very frequently the placement of logic blocks can be tricky to achieve satisfactory timing, and great sophistication might be required to formulate proper timing constraints for the tool. <S> You will also need to design external environment to operate your interface under realistic conditions, either virtually (again, system/bus models,) or have physically operating environment, to validate functionality of your interface. <S> Of course, you can do it all alone, and it will take you just a couple of years to accomplish all these tasks for any modern packet-serialized interface. <S> Or you can use fruits of labor of several years of seasoned engineers who already did this design, and wrapped their work in a form of configurable IP. <S> You will only need a couple of weeks to understand external workings of it, and embed the IP into your design. <S> But the IP will cost you. <S> It is now your choice. <S> For efforts along the self-designing path you can start with OpenCores.org . <A> Depends on what kind of IP you mean. <S> Some IP is soft IP and is really no different from a design in HDL as it lives on the LUTs on the FPGA. <S> This stuff you can implement on your own, but it can be very complex. <S> Hard IP, in the other hand, exists as dedicated silicon outside of the FPGA LUTs. <S> Usually this will involve mixed signal components, such as in a PLL or a high speed serializer or deserializer. <S> Sometimes hard IP is purely digital, as in the case of a hard CPU core, PCIe interface, or Ethernet MAC. <S> If the hard IP has mixed signal components, then it is impossible to replace with HDL alone. <S> If the core is purely digital, then it may be more complex than would be feasible to replace. <S> Now, there are common FPGA primitives such as multipliers, LUTs, and block RAM that can actually be inferred with a bit of pure hdl instead of requiring an explicit instantiation. <S> For things like block RAM and DSP slices, this can be a very reasonable approach. <A> Now if you want to use specific hardware primitives like BRAM, Multipliers, PLLs etc, then you will need to use the tools, libraries, IP and interfaces provided by the manufacturer.
| If you want to make something like a FIFO, then yes, you can roll your own.
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Using battery and USB power at the same time I'm trying to make a Li-ion battery control circuit. I would like to be able to charge the battery via USB and power the load (at 5V) at the same time. I know how to make a circuit that disconnects the battery from the load using a MOSFET and powers it only from USB (something like http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf ). However, with this approach (if I understand correctly), plugging a USB cable connected to something that can't supply enough current (eg. 500 mA when 1 A is needed) would still switch off battery power completely. How can this be avoided? The ideal scenario for me is that in this situation the load is powered from both USB and the battery. If such a circuit is very complicated, simply ignoring the USB power and supplying the requiered current solely from the battery would still be OK. I will be powering a Raspberry Pi, so I would like to avoid a sudden current (or voltage) drop. I found some load sharing ICs on the Internet, but they are designed to drain equal current from two loads, which might not be the case here. I hope my question isn't stupid, I'm an electronics newbie :) <Q> In the case of the application note, you could replace Q1 with another diode, then the load would draw power from which ever source has the higher voltage, or from both if they are at approximately the same voltage. <S> But note that this circuit doesn't provide 5V output or any voltage regulation for that matter. <S> If nothing is plugged in you would get the battery voltage (~4V), if USB is plugged in you would get about 5V, and if the AC adapter is plugged in you would get whatever voltage that provides (minus the diode voltage drops). <S> You mentioned charging, and 5V output, so you might need a separate charger IC, as well as a boost regulator (assuming you use one LiPo cell) to get the 5V output. <S> There are some devices that have such capabilities combined, such as the LTC4090 device . <S> It will take a battery, USB input, and even another high voltage input, charge the battery, and give a 5V output (in the case of the LTC4090-5 device). <A> Linear LTC4067 is your solution. <S> It will "add" both the battery load and the power supply together. <S> the example below is taken from the LTC4067 datasheet. <S> If you install the "optional" mosfet, you will get on OUT both the battery load and the USB load. <A> Doing this is very hard and probably not worth the effort. <S> First up, it is very difficult to measure the amount of supply current. <S> What you can do is measure the voltage level of the supply rail, this voltage will dip when the supply current is exceeded. <S> However calibrating this will be difficult, some supplies will just provide low or noisy voltage levels. <S> When you do find a weak supply you will need to switch to being purely battery sourced, otherwise you will be back feeding the battery and explosively bad things will happen. <S> Most devices solve this problem by just not working when underpowered, a common Raspberry Pi issue. <S> Users fairly quickly learn that the weak power supply and device won't work together.
| You will also need to clearly communicate this state to the user somehow, being battery powered while plugged in is a non-intuitive state.
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Is it possible to jam wireless home alarm systems? Sorry if this is the wrong community for this question but in my mind, it's the best fit. Please close or move to a more appropriate community if it's off topic. The other day, several home alarm companies came to look at my home to quote an alarm system. They all gave me a wireless option since the wired option they said would be super expensive. I asked if it was possible for someone to jam the wireless sensors and break in. One representative said no but didn't elaborate why. The other said no because the communication between the sensor and the main panel is encrypted so people can't jam it. I don't believe this is true but I don't have an EE degree. I think it's not true because I've heard on the news people have built cell phone jammers so it probably isn't hard to jam these sensors too. I think these sensors operated in the 200Mhz range (if I remember correctly), if that matters, although he said there's some encryption going on between the panel and sensor. That confuses me because the encryption is digital but the communication is analog? <Q> A "denial-of-service" wireless attack is very easy. <S> It will disrupt radio communication between sensor and panel. <S> Hopefully, the panel is smart enough to detect that one (or more) of its sensors has failed to report-in. <S> A non-reporting sensor should be assumed under attack. <S> A much more difficult attack is a "spoof" attack, where the communication between sensor and panel is overpowered by an attacker with a valid message. <S> An "all-OK" signal is very difficult for an attacker to generate because of encryption. <S> Because these signals are regularly sent, it is vulnerable to a determined attacker who is willing to capture signals over a long period. <A> First of all, to clear some things up: All digital signals are built up by analog signals. <S> As already mentioned in the comments, all wireless communications can be jammed, encrypted or no. <S> And last but not least, jamming is not the same as hacking into. <S> Jamming is just "stopping" the signal. <S> Now, if the alarm central is good, I would expect it to expect a signal from each and every sensor on a regular basis. <S> If a burglar simply jams one or more of the sensors, the central should realize that it has "lost" a sensor, and sound the bell. <S> On the other hand, this could lead to false alarms if the reception is bad. <S> This would be a major problem, because over time the user will become annoyed, lose faith in the system, and switch it off. <S> That's not only an expensive paperweight <S> , it's also one that could be stolen. <S> The first rule of wireless is: Use a wired connection. <S> Thus, my advice would be to either get the wired version, or save up for the wired version. <S> But I don't know your house, your other wireless appliances (or your neighbors's), the thickness and materials of your walls, the exact alarm system in question, the warranty, insurance, or clauses in the contract, etc. <S> In the end it's gonna be your decision, so good on you for trying to make it an informed one. <A> Yes, it is possible to jam wireless alarm systems, probably even with low-cost, low-tech DIY devices (google "broadband jamming DIY"). <S> There are a couple of articles online which report successful attempts, for example this one by Cnet and this one by Forbes. <S> There is also a blog post by a producer of such systems. <S> Both Frontpoint and SimpliSafe, two producers of wireless alarm systems, claim to have proprietary algorithms in place which distinguish between random signal loss due to unrelated interference and an actual attack. <S> (This was a concern I had in a comment to one answer here.) <S> Of course it is impossible to verify these except by performing a comprehensive test. <S> What I take away from the articles and some other discussions on the net is: <S> Yes, it is easy to jam the signal. <S> Most burglaries are untargeted crimes of opportunity. <S> The system may or may not respond properly to jamming attacks. <S> Both systems raise an alarm via cell phone which is easy to jam as well, and this time the alarm system can do nothing about it. <S> Of course, cutting the landline is usually easy as well. <S> tl;dr: <S> Don't worry unless you are a high-profile target. <A> One of these has to be true: if the alarm controller receives no meaningful communication, it has to decide what to do, which can be either "nothing" or "alarm of some sort". <S> Depending on what the alarm does you'll then have to deal with either your neighbors annoyed by the noise, or police/security coming to your house. <S> If I were to buy such a system, the first thing I'd ask about would be how difficult is to trigger a false alarm and how much it would cost me. <A> Most off-the-shelf alarm systems <S> out there are outdated or insecure-by-design crap sold at a huge markup by salespeople who rely on your fears. <S> I've got a storage room full of access control systems and components at my workplace <S> and I haven't seen any single decent one (we ended up developing our own for our clients). <S> I would not expect any of those systems to either use encryption or protect against jamming, so unless you can have a demo of those features in action, don't buy it. <S> A good system should use good crypto for its communications as well as be able to detect jamming (the wireless devices send a heartbeat every X seconds, and if too many are missing then sound the alarm).
| If it is not possible to jam the system and break in , then it should be possible to jam the system and start the alarm . Such attacks are extremely rare, to the point that Frontpoint claims in the blogpost that no successful jamming attack on their systems has ever been reported. Ask your supplier what protocol is followed if your panel reports that a sensor has failed to report-in. Don't use wireless unless there is absolutely no other option.
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Lithium Battery Pack - Do I need BMS within parallel connections? Let's assume I am going to build a Li-ion battery pack with 12 18650s, where I connect four cells together in parallel and then the three sets of four in series. My understanding is that a BMS keeps an eye on the voltage and keeps it from going too high or too low. Thus, would I then use a BMS module that connects three batteries in a series, or would I need to have a BMS with 12 connections, including the cells that are connected in parallel. TBH, I am not at that level yet, but just trying to understand the nuances of li-ion batteries before I get my hands dirty. <Q> For the most part, putting cells in parallel just makes them behave like a bigger single cell. <S> So, if you take four cells and hook all of them together in parallel, it appears to a circuit to just be a single cell with four times the capacity. <S> BMS's are built to manage cells in series . <S> Along with current and voltage protections, it monitors each "cell" in the pack to make sure its voltage is within limits, and if any one cell dies prematurely it will cut off the whole pack to prevent the other cells from reverse-charging the dead cell. <S> Your configuration is "3s4p" - three groups of four parallel cells wired in series. <S> Thus, you need a BMS that can manage three cells in series - a "3S" BMS. <S> Generally speaking, it's irrelevant how many cells you put in parallel in each cell group, as long as all the groups have the same number of cells at similar capacities <S> (i.e. you do not want to put one parallel group of 3 cells in series with a parallel group of 4 cells), since the BMS will see your parallel groups as single larger cells and will manage accordingly. <A> The cells you put in parallel are no longer considered 4 cells in parallel but are now considered one cell with more capacity and able to source more current safely (if your bus is up for it.) <S> There is no need to put multiple leads to this grouping. <S> The leads monitor each series groupings only. <S> So if you have a battery that is 10 cells in series and 7 in parallel, You monitor the 10 and not the 7. <A> Someone please educate me if I'm wrong, but to give the correct solution he's looking for, he should make a 3s4p pack with only one bms connected to only the 1St set of 3 in series. <A> I'd err towards caution - everyone knows deep down that parallel batteries are not guaranteed to share charging and load currents evenly <S> so, I'd use parallel arrangements of series batteries each protected by it's own BMS. <S> So if you have a 3s battery then that has its own BMS. <S> Of course this is an expensive solution but it has to be considered as viable if the cost and risk warrant it. <S> If the cost and risk don't warrant it then just parallel 4 batteries and hope for the best with a single BMS. <A> My understanding is you have a number of options here with the most common being 1 BMS connected across the series cells as just described. <S> The issue I see with this is that in a large pack 1 or more cells could fail (lose capacity/ <S> voltage)and you would not know with this arrangement. <S> All that would happen is you’d lose capacity in that group of parallel cells. <S> If your BMS does balancing then this will probably lead to excessive balancing <S> You’d need some kind of capacity tracking or balance tracking in the BMS logic to trigger alarm/action if this occurred.
| If you have another 3s battery then that should have its own BMS: - With 4 parallel sets of 3s you'd have 4 BMSs and only make parallel connections at the ends of each series chain.
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What could be causing a very small voltage between mains grounding and a water pipe If I measure the voltage across the ground prong of a receptacle and a water pipe, I get around 42 mV of voltage. The ground prong is connected to a grounding conductor, which is connected to a ground bus bar at the main panel, which is connected to a #8 AWG conductor, and finally connected to an 8ft. ground rod. The power system is: 240 V, 60 Hz split-phase (residential) These are my questions: What could be causing this voltage? Is this an induced voltage? Or is this just a measurement error of my multimeter? Is this voltage a safety issue? Should I be concerned? <Q> The house ground is bonded to the supply ground at some point. <S> Although the ground circuit is not suppose to carry any current, it does due to capacitive or varistor leakage. <S> This current will cause a voltage drop and is what you measure. <S> Ask an electrician to ensure that the piping is bonded to the earth spike to be safe, if the regulations allow it. <A> Proper electrical work should be done by trained people. <S> That said, all conductors have resistance. <S> Using the equation: <S> V = I <S> X R ... <S> we see that only a little resistance (R) is necessary to develop a small amount of voltage (V) given a load is plugged into the outlet and is drawing current (I). <S> If there is no load plugged into the outlet, then you are likely measuring noise. <S> Most multimeters have a very high impedance (R). <S> So, again using the above equation, we see that only a very small amount of current (I) that we can likely attribute to noise will case a small amount of measurable voltage (V). <S> It should be noted that this observation may be attributed to ground loops . <S> A ground loop situation can range from annoying (for example the buzz heard from an audio amplifier) to deadly (for example performers electrocuted on stage). <A> That small voltage, 42 milliVolts, is why audio people suggest the entire set of equipment: amplifier, CDplayer, vinyl-record player+RIAA preamp, etc etc, all be powered from the same outlet. <S> Imagine this problem, but <S> 10X or 100X bigger, in a rolling mill with 1,000 horsepower motors moving the rollers. <S> How to instrument such a beast of noise and voltage drops and electric fields and magnetic fields? <S> Optical isolators and/or differential sensing.
| There might also be a ground spike and a connection to copper piping somewhere.
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Voltage regulator from first principles - why is power dumped in the transistor? I'm trying to further my understanding of electronics, so I decided to try to design a fixed voltage regulator capable of supplying an amp or so. I put this together from first principles without referring to any kind of reference on how voltage regulators are usually designed. My thoughts were: Zener and resistor to provide a fixed voltage reference. Comparator to detect when the output voltage was above the target threshold. Transistor to switch the supply on and off. Capacitor to act as a reservoir. With that in mind, I designed this fixed 5V regulator, which appears to work: What I did notice, however, is that it has certain limitations which I can't quite derive the cause of: The current from V1 (input) roughly equals the current at R2 (output), despite differing voltages. This seems to match the behaviour of linear voltage regulators (is that what I just created?) but I'm not sure why it happens. Why is so much power dissipated from Q2 considering it's just switching on and off? When V1 is less than about 7.5V, the output voltage never hits the 5V threshold, but instead hovers around 4V. I have tried this with varying loads but it simply does not function below that input voltage. What is the cause of this? <Q> I put this together from first principles without referring to any kind of reference on how voltage regulators are usually designed. <S> Not a good start, but you've actually ended up with almost the exact design of most linear regulators. <S> But the "first principle" you've forgotten about is the MOSFET linear region . <S> Have you tried this thing in a simulator? <S> The system will settle at a point where the transistor is half-on, dissipating power as a resistor. <S> When V1 is less than about 7.5V, the output voltage never hits the 5V threshold, but instead hovers around 4V. <S> I have tried this with varying loads but it simply does not function below that input voltage. <S> What is the cause of this? <S> This is called the "dropout voltage". <S> It's due to limitations in how close to the input rails the opamp is capable of driving; you lose approximately 0.7V in the output transistor of the opamp and another 0.7V because of the threshold voltage of the MOSFET. <S> You might be able to do better with a better op-amp than the ancient, obsolete 741. <S> Otherwise, you're trying to design what's called an LDO: low-dropout regulator. <A> Why is so much power dissipated from Q2 considering <S> it's just switching on and off? <S> Because it isn't a switching regulator circuit - it's a linear regulator that you have designed. <S> The current from V1 (input) roughly equals the current at R2 (output), despite differing voltages. <S> This seems to match the behaviour of linear voltage regulators (is that what I just created?) <S> Yes, you have. <S> When V1 is less than about 7.5V, the output voltage never hits the 5V threshold <S> You need about a couple of volts on the gate (with respect to source) to begin turning on the MOSFET. <S> This has to come from the op-amp and it probably "loses" about a volt on its output compared to the incoming power rail. <S> So, if you want an output voltage of 5 volts then you need an input supply of about 8 volts and that will be on light loads. <S> On heavy loads, the gate-source voltage might need to be 3 or 4 volts. <S> Now you will probably need an incoming supply that is about 10 volts to keep the regulator output at 5 volts. <S> Have respect for the simple regulator, especially those that are low drop-out types!! <A> Design is OK except the dropout of the FET LDO can be lower than the BJT LDO, but FET compensation may demand a limited range ESR for stability and allow some ripple for feedback. <S> You can make it up to 98% efficient by the good choice of inductor with a low RDSOn switch and low DCR choke. <S> Now you have a buck regulator. <S> Simulation here <A>
| The power is dumped in the transistor because it is the series element, so all the current for the load has to go through it, whilst at the same time it has to drop the difference between the input voltage and the output voltage.
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TRS female jack symbol in Kicad There is a device which is powered and also outputs signal by its TRS connector as shown below: This TRS connector needs a female correspondent for my circuit. I made a small board and need to add my schematics the female jack of this TRS. In KiCad there is the following symbol: But Im not sure if it is a TRS female jack, since 2 and 1 are seems connected. Is this how it supposed to be? <Q> But Im not sure if it is a TRS female jack, since 2 and 1 are seems connected <S> It's not the jack socket that you want. <S> The jack socket you show is suitable for a standard mono jack plug. <S> Notice the difference between the two plugs: - The socket symbol you have is one that has a contact operated when you insert the plug. <A> What you have is "jack_2P", which at first glance seems to be for TRS (Tip-ring-sleeve), except it only shows 2 pins from the cable and 1 extra. <S> The extra pin in the jack is like a switch that disconnects from pin one when the plug is inserted (it bends). <S> Below is an example of what you want; a jack for TRS (a 3-pin cable): Again you have an extra pin, of which is just there as a switch. <S> But now you have the ring (aka shield/ground), and the other two conductors. <S> The thing is, my diagram looks pretty much your like your diagram, except the ground has it's own pin. <S> It's possible that your "jack_2p" is the same thing, and that the ground is just not shown as a pin. <S> You may still be able to connect to it. <A> Pin 2 is an extra switch to disconnect something, for example the speaker when an earphone is plugged in. <S> Your symbol is for mono plug, there's no wire for GND. <S> Find a female stereo TRS jack. <S> Beware: <S> Hot +24V is extremely easily shorted, if your male connector is the output.
| Thus it almost looks like your jack is for a 2-pin cable. What you probably need is a three pin symbol like this: - In Hi-Z audio inputs the switch shorts the input to reduce the hum and buzz that a Hi-Z input easily collects when nothing is plugged in.
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How does inductor resist the increasing dc current? I understand the point that inductor will have high resistance for ac current.this is because the magnetic field produced during positive half cycle will oppose the negative half cycle since the energy during positive half cycle will be stored in magnetic field. But how would this happen there is an increase in dc current.please explain how would inductor react. <Q> You said it in your question: "an increase in DC current". <S> Sometimes people get all bundled up in the term "AC". <S> AC describes a special kind of changing current which is sinusoidal. <S> We know pretty much everything there is to know about sinusoids, so describing the nature of AC is pretty easy. <S> An increase in DC current is still a changing current just like the current changes in an AC signal. <S> The same magnetic effects apply (per your question) when there is any type of change in current . <S> It just so happens that AC is a special type of changing current. <S> Having said that, describing the true conditions of the inductor and its related fields under arbitrarily changing currents requires more general use of electrodynamics laws. <S> If you want some more elaboration, please speak up. <S> We'll talk about frequency and the Fourier representation of signals to go in a little deeper. <A> The inductor will react the same for AC and DC. <S> The energy stored in an inductor is $$ E_L = {1\over 2} L \cdot i(t)^2 $$ and in a capacitor $$ E_C = {1\over 2} C \cdot v(t)^2.$$ <S> The energy depends on the instantaneous current or voltage, AC or DC. <S> To change the energy or current, the voltage $$v(t) = <S> L {d i(t) \over dt}$$ is required on the inductor. <S> For a capacitor the current required is $$i(t) = <S> C {d v(t) \over dt}.$$ <S> The magnetic or electric field takes time to change and the size of \$ L \$ or \$ C \$ is what resists the change. <S> Tell me if I need to simplify this a lot <S> and I will try. <A> But how would this happen there is an increase in dc <S> current.please explain how would inductor react. <S> In order for the current through the inductor to increase, the magnetic field in the core has to increase. <S> This means the stored magnetic energy has to increase. <S> The magnetic energy can only increase as fast as the applied electric power allows. <S> For example, if you want 1 more joule of magnetic energy, you must supply 1 watt for 1 second, or 2 W for 0.5 s... <S> some combination that adds up to 1 J. <S> The net result of this requirement is the usual relationship that defines an [ideal] inductor $$\frac{\rm{d}i}{\rm{d}t}=\frac{v}{L}$$ <A> the magnetic field produced during positive half cycle will oppose the negative half cycle since the energy during positive half cycle will be stored in magnetic field. <S> You are misunderstanding how this works. <S> The inductor doesn't "wait" for the negative cycle before opposing a change. <S> It's doing it the whole time. <S> At the beginning of the positive cycle, it is opposing the rise in current as it builds to the peak. <S> As the positive cycle begins to return to zero, it opposes that, too (it wants to push it, so it doesn't decrease). <S> The same thing is happening on the negative part of the cycle. <S> As you pass through zero, the inductor is still trying to push current forward (using the residual field), but as the applied voltage is now going negative, the forward push is still in opposition. <S> And so on. <S> In a "dc" circuit, the moment the voltage changes (and the current tries to change), the inductor opposes it. <S> The inductor doesn't know if this change is the start of an AC cycle, or a change in the DC voltage. <S> And if I were to clip out this very tiny part of the waveform and show it to you, you couldn't tell, either.
| I the inductor is designed for AC, and then used in DC, it might saturate, causing the inductance to become almost the same as for a free-air winding.
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How to limit voltage charge of supercapacitor For a DIY project, I plan to charge an 100F 2.7V Supercapacitor using a 3.7V Lithium-ion 18650 battery. However, I need it to charge only until 2.25V. Is there a simple circuit that I can build to limit the voltage\charge? I basically need the stored energy to be ~250J. <Q> Here is a very basic circuit that would limit the cap voltage to 2.7V. <S> The zener diode is used so that the 2.7V reference is fairly constant regardless of the battery voltage. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If you use an op-amp as shown, the charging rate will decrease as the cap approaches the reference voltage. <S> If you were to replace the op-amp with a comparator, this shouldn't be an issue. <S> To use a comparator, I believe a pullup resistor would need to be connected from the comparator output to the positive end of the battery. <S> Here is a simulation that uses an op-amp . <A> The best would be a synchronous buck switching constant current source, with a voltage-limited output. <A> If you have a power PNP Darlington or can make one with a power transistor and driver then get 2 Ultrabright RED LEDs and any old NPN signal transistors. <S> These make good low voltage zener references. <S> Current sensing is on 330m should be 2 to 3 W. <S> The Upper LED turns on DIM when PNP Darlington is ON <S> and then when output LED turns on DIM it switches off the series regulator and <S> with 2A it takes <120 s. Java Sim design
| If you don't find a pre-made one, you can use a switching LED driver and a comparator to stop it when target voltage is reached.
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How to designate a specific portion of a design in altium I copy and reuse portions of a design, I'll copy and the designation numbers follow the copy (for example: I'll have U1 and R2, then copying you end up having to U1's and two R2's, I don't want that.). There are sections that I'd like to re-designate (or clear designations to U1->U? for example) or have Altium renumber the designators. 1) Is there a way to make a selection and clear the designators? 2) Is there a way to clear designators in a sheet or redesgintate only a sheet? I think I remember that you can use a 'special paste' and paste a design without designators labeled. I don't want to know about that way, I'm not a fan. I'm running 15 right now but could update. <Q> <A> If you go to Tools <S> >Annotate Schematics you can select a specific sheet or sheets (for just one, click All Off, then select the specific sheet you want) and then the Reset All or Reset Duplicates will apply only to that particular sheet. <A> You can use the rest designators on paste option as DerStrom8 answered, or if you already have the parts pasted (or don't want to do it that way), you can try Tools->Reset Duplicate Designators. <S> I find this is not always reliable (it sometimes resets the original one that you don't want to reset). <S> To reset designators on a single sheet, go to Tools->Annotate Schematics, then select the schematic sheet you want to reset in the bottom-left section where the sheets are listed. <S> On the right side the designator for that sheet will be listed. <S> Press Reset All, an accept the changes. <A> I'd suggest copying and reusing design portions before annotating them at all. <S> Saves you the hassle of doing the clean up later. <S> Only when you are done completely with your schematics, then you may annotate them altogether. <S> Also, like others suggested, you can "Reset Duplicate Schematic Designators" (sic). <S> But this method resets all duplicate designators (including the original ones, within that sheet) which you might not want. <S> So, in best practice, finish your schematic. <S> Once done, annotate the designators
| You may be looking for the "Reset Parts Designators On Paste" setting available in the DXP Preferences --> Graphical Editing pane:
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What is the advantage of a Sallen-Key filter over a normal second order filter? Wikipedia links to a Sallen-Key filter as a active low pass, so I tried it out with LTSpice. The frequency response and phase response are not linear, instead frequency response even gets higher after 10kHz. Why is that, and why would I use a Sallen-Key filter instead of a "normal" low pass filter? The Sallen-Key is on the blue line. <Q> What you call "normal" is a simple two-stage RC filter with very bad selectivity (two real poles only). <S> In contrast. <S> However, there is one big disadvantage of the Sallen-Key structure - if compared with other active filter topologies (multi-feedback, GIC-filters, state-variable,...) <S> : There is a direct path (in your example: C4) from the input network to the opamp output. <S> That means: For frequencies much larger than the cut-off frequency <S> the output voltage from the opamp is - as desired - very low. <S> However, there is a signal coming directly through the C4 path which creates an output signal at the finite output resistance of the opamp. <S> And this resistance is increasing with frequency! <S> As a consequence, the damping charactersitics of this filter are not as good as it should/could be. <S> And that`s what you have observed: The magnitude shows a rising characteristic for larger frequencies. <S> (This unwanted damping degradation is not caused by limitations of the gain-bandwidth product). <S> Improvement: The situation can be improved by scaling the parts values: Smaller capacitors and larger resistor values. <S> Comment 1 : <S> This undesired property of any opamp circuit with a feedback capacitor (between output and input circuitry) can be observed also for the classical MILLER integrator. <S> Comment 2: <S> So - are there any advantages the Sallen-Key filters have in comparison to other active filter structures? <S> Yes - there are. <S> Let`s compare the two most frequently used topologies: (1) <S> Sallen-Key has very low "active sensitivity" figures (sensitivity against opamp non-idealities) and rather high "passive sensitivity" figures (sensitivity against passive tolerances). <S> (2) Multi-feedback filters (MF): High "active sensitivity" and low "passive sensitivity" figures. <S> Both sensitivities are rather important properties of all filters because they determine the deviations between desired and actual filter response (under IDEAL conditions all filter types would have identical performance properties). <A> At really high frequencies, such as higher than UnityGainBandWidth, the opamp has lost control of its Vout. <S> Notice how this inverting single-pole lowpass has NON-INVERTING response to the fast input pulses. <S> The Cfeedback allows the input charge to appear directly on the output. <S> Here is the circuit, and the OpAmp params: The only reason the BODE (2nd screen-shot) has attenuation at higher frequencies is 'CL' 15pF forming LowPass with the 2 resistors into VirtualGround. <S> [ If you want better High Freq attenuation, install 470pF cap to ground at middle of the 2 input resistors.] <S> You'll have fun, by editing the Amplifiers ROUT. <S> And by enabling that input filter capacitor. <S> And editing out that 15pF Cload. <S> This example is one of those BUILTIN (no SPICE knowledge needed) to Signal Wave Explorer, free to download from robustcircuitdesign.com for 19 unique days of use. <S> And Walt Jung, of Analog Devices, discussed this frailty of LPF decades ago. <S> Here is <S> example of an opamp's MEASURED Zout (near 500MHz, looks like 10pF. 31 Ohms), for Active and for ShutDown modes: <A> You may choose from many configurations depending on your specs for group delay, Q, bandpass ripple, bandstop attenuation , skirt steepness. <S> Both Sallen-Key and Multiple Feedback can achieve the same results. <S> see below. <S> Both can achieve high gain limited by the GBW of the OP you choose. <S> This TI software can design any active filter and allows you choose from either configuration and choose resistor tolerances which selects the appropriate value. <S> It doesn't allow you to specify input impedance so you can scale all RC values to suit this. <S> I chose Bessel response, so group delay is flat. <S> Added From the other answer which exposes the limitation of Op Amp BW where the open loop output resistance or current limit of any Op Amp ( Rail -to Rail types much worse) <S> , I propose that the Sallen-Keys filter is worse for attenuation above the BW of the Op Amp and that the open loop high frequency ( > GBW) attenuation depends on input/Output impedance ratio above the GBW threshold where negative feedback reduction on Zout has no impact due to lack of gain. <S> simulate this circuit – <S> Schematic created using CircuitLab
| the Sallen-Key topology is capable of producing a second-order lowpass response with much better selectivity (higher pole Qp) and various possible approximations (Butterworth, Chebyshev, Thomson-Bessel,...).
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Equivalent resistance of a set of n resistors? Is the equivalent resistance across a set of n resistors connected in series greater than the equivalent resistance across them when they are connected in parallel? I want to find the relation between \$R_s\$ and \$R_p\$. My attempt: Let the resistance of the resistors be denoted by \$R_1, R_2, R_3,..., R_n\$ and the equivalent resistance across the series and parallel combinations of the resistors by \$R_s\$ and \$R_p\$. \$R_s=R_1+R_2+R_3+...+R_n\$ and \$R_p=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}}\$ From first look it may seem that \$R_p\$ is less than \$R_s\$ but then on careful inspection it is not possible to ascertain which one is greater. <Q> Let's see if this works. <S> Start with \$n=2\$: $$R_s = <S> R_1+R_2\\R_p = <S> \frac{R_1R_2}{R_1+R_2}\\\text{Then: }\\\frac{R_1R_2}{R_1+R_2} \stackrel{?}{<}R_1+R_2\\R_1R_2 \stackrel{?}{<} <S> (R_1+R_2)^2$$The last row can be derived from the one before if \$R_1+R_2>0\$ and it is true if both resistance are positive separately too, which is something ok to ask to a resistance. <S> \$n=2\4 is our base case. <S> Now... induction . <S> Hypothesis : \$R_p[n] <S> < R_s[n]\$ <S> Thesis : <S> \$R_p[n+1] <S> < <S> R_s[n+1]\$ <S> We try to prove that the sequence \$R_p[n]\$ is decreasing, while \$R_s[n]\$ is increasing. <S> Series resistance:$$R_s[n+1] = R_s[n] + R_{n+1} <S> > R_s[n] \text{ if } R_{n+1}>0\\R_s[n+1] <S> > R_s[n] <S> \textit{ q.e.d.}$$ Parallel resistance:$$R_p[n+1] = <S> \frac{R_p[n]R_{n+1}}{R_p[n]+R_{n+1}}\stackrel{?}{<}R_p[n]\\R_p[n]R_{n+1}\stackrel{?}{<}R_p[n]^2+R_p[n]R_{n+1}\\R_p[n]^2>0\text{ which is true, so that}R_p[n+1] < R_p[n] \textit{ q.e.d.}$$ <S> We finally chain things up: $$R_p[n+1] <S> < R_p[n] < R_s[n] <S> < R_s[n+1]$$ <S> It did work after all. <A> Let's try a very simple approach. <S> The collection of resistors will have a maximum value resistor \$R_{max}\$, and a minimum value resistor \$R_{min}\$. <S> By definition $$R_{min} <= <S> R_{max}$$ the equality is true if all resistors are equal, otherwise the inequality is true <S> The series connection will contain \$R_{max}\$, plus at least one other series resistor, which will increase its resistance, so $$R_s <S> > R_{max}$$ <S> The parallel connection will contain \$R_{min}\$, plus at least one other parallel resistor which will decrease its resistance, so $$R_p < R_{min}$$ as $$R_p < R_{min} <= R_{max} <S> < R_s$$ therefore $$R_p < R_s$$ <S> Notes <S> (1) <S> We assume all the resistances are non-negative. <S> This is true if they are all passive. <S> There are active networks that can exhibit a negative resistance, but they are not relevant to this question. <S> (2) Resistance is voltage/current. <S> The total resistance of any two resistors in series will have a resistance greater than either component. <S> The total resistor will have a voltage equal to the sum of that over each component, with all currents being equal. <S> (3) <S> The total resistor will have a current equal to the sum of that through each component, with all voltages being equal. <A> It's a purely mathematical result. <S> If we look at the aritmetic and harmonic means of the set of resistor values:$$AM (R_1,\cdots <S> , R_n) = \frac{1}{n <S> } <S> (R_1 <S> + \cdots <S> + R_n)\\HM (R_1,\cdots, R_n) = <S> \frac{n}{\frac{1}{R_1 <S> } <S> + \cdots <S> + \frac{1}{R_n}}$$ <S> We can observe that \$R_s\$ is \$n\$ times the arithmetic mean \$AM (R_1,\cdots, R_n)\$ of the resistors, while \$R_p\$ is \$1/n\$ times the harmonic mean \$HM (R_1,\cdots, R_n)\$ of the resistors. <S> The Pytagoreans worked out that \$AM \geq HM\$ if the values in the set are positive (as resistors use to be). <S> So, if \$n \geq 2\$ (at least two resistors), then: $$R_p = <S> \frac{1}{n}HM (R_1,\cdots, R_n) < HM (R_1,\cdots, R_n) < AM (R_1,\cdots, R_n)< n AM (R_1,\cdots, R_n) = R_s$$ Thus, \$R_p < R_s\$ always for any \$n \geq 2\$ and positive valued \$R_1,\cdots, R_n\$. <A> I think there's a simpler way to show this than the existing answers have given. <S> When thinking about resistors in parallel, it's often simpler to think about conductance rather than resistance. <S> In terms of conductance, we can simply write: $$G_p = <S> G_1 + <S> G_2 + \dots$$ <S> Since all of these G's are positive numbers, we know that \$G_p\$ is greater than any of the individual \$G_n\$'s. <S> Therefore we know (still taking into consideration that all the values are positive) that \$R_p=1/G_p\$ is less than any of the individual \$R_n\$'s. <S> Since \$R_p\$ is less than any individual resistor value, it must also be less than the series combination (sum) of all the individual resistor values. <A> This has been answered pretty thoroughly <S> but I like playing around with this stuff, the min/max answer is my favourite though. <S> So here's an unnecessarily awkward proof just using sums.$$R_s = <S> \sum_{1}^{n <S> } R_n\\R_p = <S> \frac{1}{\sum_{1}^{n} \frac{1}{R_n}}$$Multiplying top and bottom of fraction in Rp by Rs:$$R_p <S> = \frac{R_s}{R_s\sum_{1}^{n} <S> \frac{1}{R_n}} = <S> \frac{R_s}{\sum_{1}^{n} <S> \frac{\sum_{1}^{n} R_n}{R_n}}$$Now that inner sum always contains one term that is equal to its divisor, so that can be split out and simplified to 1:$$R_p = <S> \frac{R_s}{\sum_{m=1}^{n} \frac{R_m}{R_m}+ <S> \frac{\sum_{n=1}^{m-1} <S> R_n + \sum_{n= <S> m+1}^{n <S> } R_n}{R_m}} = <S> \frac{R_s}{n <S> + \sum_{m=1}^{n} \frac{\sum_{n=1}^{m-1} R_n + \sum_{n= <S> m+1}^{n <S> } R_n}{R_m}}\\R_p = <S> \frac{R_s}{>1}$$Now the divisor contains n, which is greater than one, and since no other part can be negative the denominator as a whole is greater than one, thus Rp will be smaller than Rs. <A> If we assume that all \$R_1, R_2, \dots, R_n \$ have the equal value \$ R \$, you get the following connectivity: \$ R_s = \sum\limits_{i=1}^n R_i = <S> n <S> \cdot <S> R\$ <S> (1) \$ R_p = \frac{1}{\sum\limits_{i=1}^n <S> \frac{1}{R_i}} = \frac{1}{n <S> \cdot \frac{1}{R}} <S> = \frac{1}{n <S> } \cdot R \$ (2) Resolve (1) to R = <S> > <S> \$ <S> R = \frac{R_s}{n} \$ plug in (2) = <S> > <S> \$ R_p = \frac{1}{n <S> } \cdot \frac{R_s}{n} = <S> \frac{1}{n^2} \cdot R_s \$
| The total resistance of any two resistors in parallel will have a resistance lower than either component.
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Why is not possible to light a bulb without closing a circuit by using two batteries I'm new to learning about electricity. I have learned that Electrons moves from - to + when there is a voltage differential. My conclusion is, that I can connect the plus side of one battery to a bulb, and connect it to the minus side of the second battery without closing a circuit. This way, the electrons will move, and the bulb will light? Is it will light for at least a few nanoseconds? Bonus: There is an online website that can help me create the diagram and test it? or just share here the image? Is it the truth that the bulb will be lighted for a few nanoseconds? (see answers below) <Q> You may have a look at this site: http://hyperphysics.phy-astr.gsu.edu/hbase/Chemical/electrochem.html <S> " <S> In order for the voltaic cell to continue to produce an external electric current, there must be a movement of the sulfate ions in solution from the right to the left to balance the electron flow in the external circuit" Because that ions can't move between the two batteries, your bulb will not produce light. <A> Battery terminals are marked (+) and (-) because the RELATIVE voltages are in that relation. <S> Until you connect the two batteries together at some terminal, you don't know the absolute voltage of battery #1's (-) terminal is more negative than battery #2's (+) terminal. <S> While one terminal of a battery is floating (unconnected) there's no way toapply the Kirchoff rule that all voltages in a closed circuit loop add to zero,so there's no way to identify a resistor (or lamp filament) as having anapplied voltage. <S> In a related note, if you attach the two (open) battery terminals witha very-high-value resistor (a million ohms?) <S> , the lamp will get some current,but not enough to heat it and make a glow. <S> An interesting question is,what resistor value WOULD make the light glow. <A> The misunderstanding is thinking that there is an imbalance of electrons. <S> There isn't on batteries. <S> Both the positive and negative terminals are entirely normal metal with equal numbers of positive and negative electrons present. <S> The difference is in the electrochemical potential between the electrolyte and the electrodes. <S> This creates an electric field with a potential difference between the two electrodes. <S> This is not something with an absolute reference, it's entirely relative. <S> (The only situation where you do get a significant mismatch in electron count is electrostatically charged items, like Ben Franklin's rods rubbed with wool. <S> In this case you can get a current to flow for a very short time between two charged items without having a complete circuit.) <A> that means they are connected either in parallel or series which in terms make the current flow and bulb will glow. <A> Hopefully this is something that helps you <S> The current needs a closed conductive path. <S> In the vacuum the electrons can propagate without a conductor, but not in the air. <S> Let's imagine that the middle scenario makes light (=current) <S> Then the total amount of electrons would grow at the other battery. <S> Finally they will push so strongly the new ones back that the current stops. <S> ADDENDUM: <S> But how long the current exists and how much electrons flow through the bulb if we put together the system in the middle? <S> For that exists the concept named"capacitance". <S> Imagine to put the system together instantneously (=in no time) <S> Then the electric fields of the batteries push and pull the electrons until new balance is found. <S> There really is a current until the capacitance between X and Y is fully charged. <S> Assuming the capacitance to be a few picofarads and the total resistance in the parts max. <S> a couple of Ohms, the current stops well before one nanosecond. <S> The total moved charge is well below one nanocoulomb. <S> If the battery voltage is 12 volts, then the total dissipation in the bulb is so low that no observable filament warmup exists in the bulb - no light, but a short current pulse yes, but in practice also not observable. <A> You have made a good start in recognizing that electrons <S> that are moving cause the lamp to light. <S> Your circuit is in static equilibrium (redrawn below with stackexchange's menu-bar circuit-drawing facility): simulate this circuit – <S> Schematic created using CircuitLab <S> No electrons move in the above circuit. <S> The negative end of BAT1 is at the same voltage as the positive end of BAT2. <S> So no electrons move through the lamp, and there's no light. <S> However, if you measure the voltage from the positive-end of BAT1, to the negative end of BAT2, you'll measure 6+9=15 volts. <S> Both batteries need to do no work. <S> Perhaps your error is in assuming that these loose ends are at the same potential voltage . <S> If you force them to be at the same voltage by connecting them with a low-resistance wire, electrons now flow (revised circuit below). <S> LAMP1 has 15 volts across its terminals. <S> This is a series circuit, with BAT1, LAMP1, and BAT2 experiencing the same electron flow (current). <S> simulate this circuit
| In your case as there is no closed path there will be no current (flow of electrical charge) flowing, thus the bulb will not glow Also in case of two different batteries the reference is different, unless if they both have common reference (e.g same ground) In order to draw the current the circuit must be closed meaning there should be easier path of current to flow in a circuit.
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How does an outlet know what wattage to put out? In the UK, if that makes any difference, I was wiring a motor and then I wondered (and kinda needed to know), how does a plug socket know how much power to put out? because all the input ampages on all the plugs sad different things but can all be plugged into the same socket. How does this madness work?!? XD <Q> The rating of a plug is the maximum current it can provide. <S> The rating of equipment is the current it will actually demand. <S> If the load tries to draw more current than the plug can handle, then in theory a breaker should trip. <S> In a properly designed system, the trip point of the breakers are set to a bit less than the current capacity of other parts of the system. <A> As Olin mentioned the socket does not technically put out power, it supplies amps (current) at a fixed voltage. <S> The load determines how much current, which translates to power, is needed and used. <S> If your load is too high for the fuse in the plug, or ultimately the breaker or fuse in the fuse-box, you will blow a fuse. <S> As for wiring your motor YOU need to figure out a few of things. <S> What is the load of the motor, or more specifically, what is the maximum current you need to start and run it. <S> The start current will be a lot higher than the run current. <S> Can the circuit supplying power to the outlet safely carry that much current. <S> When you calculate that you need to bare in mind, and add in, anything else that could be attached to this circuit. <S> e.g. If it's on the same circuit as say the air conditioning unit you are probably out of luck. <S> Is the main fuse-box fuse or breaker for that circuit big enough to handle that much current. <S> DO NOT INCREASE IT IF IT IS NOT. <S> The current breaker will probably be rated for the wire gauge of the circuit and should not be messed with. <S> Finally, is the outlet and plug fuse rated for that current. <S> If the answers to any of questions 2,3, and the outlet fuse in 4 is NO.. <S> Call an electrician to install a new circuit for your motor. <A> It is really very simple: When you close the switch the electric field between the wires travels almost at the speed of light towards the load. <S> When it reaches the load a current starts to flow. <S> This current travels back as a magnetic field between the wires. <S> When the current reaches the load the voltage drops. <S> This voltage drop travels towards the the load. <S> When the voltage drop reaches the load the current drops. <S> This current drop travels back to the source. <S> This process repeat, almost at the speed of light, until it becomes stable. <S> In this process some of electromagnetic field 'leaks' into the wires, heating up the wires, and some of the electromagnetic field 'leaks' into the isolation, heating up the isolation. ... <S> and this is the simple explanation. <S> James Clerk Maxwell described this in exact mathematical details in 1865.
| The plug provides a fixed voltage, and the equipment (the load) decides how much current to draw at that voltage.
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Setting up a new electronics lab I am setting up a new electronics lab for embedded systems related work. We need four or five workbenches/workstations, a soldering station, enough storage space for all the essential electronics supplies, and maybe some other useful equipments. We already have quite some equipments available such as oscilloscopes, logic analyzers, large cabinets, etc. But we need a storage solution for all those electronics components. I see there's already a SE question ( Best way to get components for a new electronics lab ). But that one is about getting components while I am thinking about the higher level things including lab layout, workbench design, flooring & lighting etc. I would like it to be a bright, modern, and safe lab environment. I guess I need some professional help here. I could quickly google some lab outfitting companies and contact them directly. But I would like to hear some recommendations from you, my fellow engineers. Thanks. Update 1: this is for a commercial company. There will be multiple projects going on in this lab. Update 2: To make this post less "opinion based", I would appreciate some suggestions on which firms offer this kind of consultant service. I don't have a lot of time to figure out all things myself. I just want some help to set up a good looking and functional lab quickly. <Q> Wood, or conductive flooring, and wood or conductive work tops are crucial to prevent static. <S> A PC and an oscilloscope (or PC-based Pico) is crucial for understanding. <S> Development kits are crucial. <S> These can be shared. <S> Heavy duty draughtsman's chairs are required. <S> Breadboards and jumper wires should be supplied by/to students/developers so that they can have some continuity if they cannot complete a project in one session. <S> They can use food/cellphone lockers or shoe boxes to store these. <S> (Shoe boxes are static safe :-) <S> Brainstorm and design this with stakeholders. <S> Sign off the conceptial design before proceeding. <S> (How many students in the long term, etc.) <S> For serious development an editor/gate keeper is required to ensure that the documentation is in step with the devolopment. <A> Storing electronic components depends on type and size. <S> Some need to be stored in electrostatic safe containers. <S> Suitable drawer systems are available for this. <S> You can use the cheaper variety for things like resistors and capacitors and other non ESD sensitive devices. <S> Larger items should be stored in bins. <S> All drawers and bins should be organised appropriately and labelled. <S> You should also keep all your parts in lockable cabinets. <S> Component inventory is not cheap and parts will walk if not supervised and locked up. <S> And metal cabinets make a lot of noise when someone tries to sneak into them. <S> Another thing you need to seriously think about is that you need a curator for all that. <S> Someone has to be in charge of organizing and keeping those shelves properly arranged and re-stocked. <S> Just filling the cabinets with parts and hoping for the best will not do. <A> Consider how workbenches are used. <S> Perhaps a workstation requires long periods of sitting, requiring a comfortable chair with back support. <S> But more often a workbench requires constant mobility with short sitting duration. <S> A deep workbench is needed if test equipment is large and deep...you still need a half-meter of space from front-of-equipment to front edge-of-bench. <S> And you need more power outlets than you imagine. <S> Embedded systems require static protection at workstations, and everywhere else they may go (powered or unpowered). <S> Floors should be plain colour, glossy smooth so that you can find small parts that fall. <S> Consider a good and fair way to store cables, cable adapters so that they are equally available to all workers. <S> This is a big problem in shared work areas. <S> Consider electrical noise sources (drills, LED lighting, switching power supplies) and keep them away from critical low-signal-level work areas. <A> I have discovered by chance that the format of a DigiKey/Mouser component bag is almost the same as a 4x6" (10x15cm) photograph. <S> I number the pages and keep inventory in a spreadsheet. <S> Finding a part is as easy as Ctrl-F, which gives the album and page number, then grabbing the album and flipping to the proper page. <S> This is so much more practical than a pile of bags inside a shoe box! <S> For passives, I use plastic boxes, shoe box size. <S> I put the bags in, sorted by value. <S> Between each value, I put a piece of card, cut to size, with the value written on it. <S> So, if I need a 100µF aluminium capacitor, I flip through the values, which are in order, and pull the bag out. <S> It's quick and much more practical than drawers, because values can be added at any time, and the mess remains sorted in value order, therefore easier to search.
| Plastic sleeve photo albums/binders are AMAZING to slip these component bags in! A bar-type (high) workbench with bar stools is more appropriate.
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Line of switches on a breadboard I'm connecting 8 switches on a breadboard, with a common ground (the already-connected line of the breadboard): Problem: these switches are always on, like if the 2 pins on the common line make them pressed . I thought that a solution could be to rotate 90° the switches, but then they don't fit on the breadboard anymore (one cannot use the common horizontal line for ground anymore). I can't believe there isn't a natural solution to connect 8 such switches on a breadboard. Is there a solution for this? <Q> Simple breadboard solution : <S> Get your pliers out and bend up one of the grounded pins. <A> 1) You failed to check which pins are common and bridged the switch to the supply rail 2) <S> You were not aware that these tactile switches are metric pitch and the short lead gap is the contact pair is probably 4mm or 0.157" and the breadboard has 0.1" centers. <S> These leads are plated round steel wire which is stiffer than copper wire and easily fits the sockets using 1/4W R's. <S> Note that the leads do not reach the holes. <S> stretching the pins sideways must be done with care not to damage the structure. <S> Thus the tactile switch body will not be firmly seated against the board and may move every time it is used. <S> However putting resistor wires on the outside can position the wires so they conveniently fit over the holes. <S> Using a solid spacer under the switch can support it so the body can sit tight on the board. <S> Or carefully mangle the leads <A> Just cut the diagonally opposite legs off, then it will work how you want.
| Solution Solder resistor leads to the pins then trim to proper length to prevent protrusion thru the bottom.
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