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Measuring Current in 12V DC Fan I am trying to measure the current of a DC computer cooling fan (Intel E18764-001) at various volts. Its rated specs are: 12V 0.20A. Check it out here . So I am suspecting that this is telling me: "give it 12V and you should get 200mA of current" - Am I right? Anyway, I supplied the fan with various volts using my power supply starting from 5V all the way to 12V. I measured the current with a multimeter connected in series. Here are the results: @5V --> 50mA (0.05A) @6V --> 30mA @7V --> 30mA @8V --> 30mA @9V --> 40mA @10V --> 40mA @11V --> 50mA @12V --> 53mA Interesting results. Here are my questions: (1) Why isn't there are proportionate increase in current with every increase in volts? In fact, after 10V, the current goes up as voltage goes up! Does it have anything to do with either the fan going all funny because it is not getting its designed voltage or the wires can't handle any current more than 50mA? (last one unlikely as that still doesn't explain the drop to 40mA) (2) why are the current readings nowhere near the power rating of 0.2A? <Q> Are you using a true RMS multimeter? <S> Remember that the fan is PWM controlled, so the current is not a DC current. <S> Maybe you are actually measuring the DC level of the fan? <S> Also, as Keelan mentioned, it vary very much depending on external measures. <S> I remember when I was designing a helicopter; If I held the copter down while applying constant throttle (voltage), the current was much higher than when I released it with the same throttle due to the drop in resistive force applied. <S> If the fan has an easy path of air flow, it may not draw as much current as expected. <S> Also, I think the label represents average current consumption. <A> Since DC motors are also generators, the reverse voltage raises proportional with the RPM. <S> So the higher the RPM, the higher the reverse voltage -> <S> less current. <A> With motors you actually have different current draws one being the starting current of the motor (LRA- locked rotor amerage) and the other being the running current (FLA- full load amperage). <S> Typically, and it depends on the type and design of the motor (there are a few) for a 3 phase squirrel cage motor <S> the LRA is 5-7 times the FLA. <S> So, if I were to guess, and I am because in the commercial electrical industry we don't use pc fans and the motors we use have very descriptive nameplates as per NEMA (national electrical manufacturers association) <S> I would say that maybe the current rating stated is the starting current.
| Also the ambient temperature is most probably not stable, which leads to fluctuating current measurement results.
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How to measure amperate going through a 220V cable? Meta objective: I want to pull up a TV with a winch, when the TV is turned off. I can already move up and down the TV invoked by my controller (Arduino Nano). The physical push-sensor stops the winch automatically. Good so far. Objective of this questions: I don't want the guest to push an extra button to make the TV go up or down. There is an indicator when the TV is used and that is the Watts that are going through the 220/230V cable to the TV. If amperage is high, I instruct the winch to let the TV down (viewing position). If it is low, I pull the TV up. See the sketch for a general idea. Now, I'm not an electronics engineer. My question is:1) How can I measure the high amperage (or watt) of the 220/230V cable and give that value to my 5V-controller (Arduino Nano). With my little experience I cannot think of an electrical component that could do that. However, there certainly must be a way to detect the big difference between 300 mili watts and 75 Watts. Nice if someone could point to a real component so that I could read up on that component. 2) Can one achive my meta objective without a controller? (This is nice to know. I'm totally happy if questions 1 is answered thoroughly.) <Q> I may earn downvotes for this, but I would NOT recommend dealing with 230V when having no experience! <S> Anyway, of course it's possible what you want to do. <S> http://www.lem.com/docs/products/lts_6-np.pdf <S> With that, you can pull the insulated phase wire through the hole, and get a voltage proportional to the current (the exact formula is in the datasheet) at the output to be measured by your Arduino. <S> It's most likely an RC5 or RC6 code. <S> Connect an IR receiver to your arduino and implement a RC5 decoder to know whether one turns the TV on or off. <A> You can buy a master-slave power bar/surge arrestor that will provide mains power on a slave outlet when the master is detected drawing power. <S> (image is of a US/Canada standard 120VAC/60Hz bar, but such devices should exist in your market). <S> With this approach, nothing touches the mains that is not properly isolated and safety-agency approved, yet cost is reasonable. <A> For those who are looking or hall effect sensor that can be hooked immeadiately to a controller like Arduino, search ebay for "5A Range AC Current transformer module Current sensor module". <S> Of course this is for hobby projects only!
| Then you can plug in a standard safety-agency approved "wall wart" supply into a slave outlet and use that to either directly control your winch circuit or derive a logic signal to feed to your processor. Another, less risky approach without 230V: A further indicator when the TV is off is when you push the "ON/OFF" button on your remote control. Not the cheapest, but maybe the safest way would be using a current transducer like this
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Replacing capacitor from unworking TV with some other capatitors which has not the same value I am repairing (trying to repair) my TV. I found 1 capacitor 25V \$1500\mu\$F which i think is bad (it has some weird shape at top)I don't have any capacitors with that value but I have some similar to that 25V \$470\mu\$F x3 35v \$1000\mu\$F Can I replace the bad one with this 35V \$1000\mu\$F, or can I combine them (serial or parallel)? <Q> Your original capacitor is very likely bad. <S> You can achieve a \$1500\mu\$F capacitance with one \$1000\mu\$F capacitor in parallel with a pair of \$1000\mu\$F capacitors in series -- the series capacitors have an equivalent \$500\mu\$F capacitance, and the parallel combination of that \$500\mu\$F and \$1000\mu\$F capacitance adds to \$1500\mu\$F. simulate this circuit – Schematic created using CircuitLab <S> It would probably be easiest to just buy a \$1500\mu\$F capacitor rated for \$\geq 25\$V, though, since the PCB will only have space for the one original capacitor. <S> They are pretty cheap on Amazon . <A> If you combine capacitors in parallel, their value will add. <S> (C1 + <S> C2 = Ctotal) <S> 2x <S> 1000uF capacitors in parallel will be equal to a single 2000uF capacitor. <S> If you combine capacitors in series, their value will divide: (1/C1 + 1/C2 = 1/Ctotal) <S> 2x <S> 1000uF capacitors in series will be equal to a single 500uF capacitor. <S> You should be able to replace the faulty capacitors with any that have: <S> The same (or higher) voltage rating <S> The same capacitance (with more knowledge of the circuit, it is possible that other values could be used) <S> The same polarity Replacing these can be pretty dangerous (read: physical harm could definitely result) unless you know what you are doing, though. <S> Make sure to unplug the TV, and also discharge the capacitors before removal. <A> If you put your three 25v 470uF caps in parallel, you'll get a 25v 1410uF cap with 3 times the current rating and 1/3 of the ESR (equivalent series resistance) of the individual caps. <S> Depending on the application, this will be close enough capacitance (original tolerance is probably +/- <S> 20% or so) and the lower ESR could be nice too...unless your jerry-rigged setup dwarfs the caps' ESR with its own actual resistance.
| Your new capacitor(s) must have the same or higher voltage rating (i.e. 25V or higher) but should have the same capacitance (\$1500\mu\$F). I had the exact same problem on one of my TVs and I repaired it by replacing the capacitor.
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How can the infinite reactance of either primary or secondary in an ideal transformer draw current? Assumption 1: An "ideal" transformer is said to have very large primary, secondary, and mutual reactance.. (self-inductance/mutual-inductance tending toward infinity), has a unity coupling coefficient (zero leakage flux), High or infinite magnetic permeability, absorbs zero real power (is lossless, 100% efficient). Assumption 2: From a pure circuit analytical and mathematical standpoint, and without the "real" model elements, the infinite primary and secondary inductances in the "ideal" transformer will draw current when secondary load is not open, and zero current when the secondary load is open or tends to infinity. Problem 1: How can the infinite reactance of either primary or secondary draw the current in assumption 2? Problem 2: The secondary load gets transformed and appears in parallel to the primary inductance, so if the primary reactance is virtually open, why even put it in the circuit? what good does this do?...there are an infinite amount of parallel opens in any given circuit. Thanks in advance! <Q> I think your confusion lies in your first assumption. <S> Thus, it doesn't make sense to consider inductance, or leakage, or less than perfect coupling. <S> All of these issues don't exist. <S> An ideal transformer simply multiplies impedances by some constant. <S> Power in will equal power out exactly, but the voltage:current ratio will be altered according to the turns ratio of the transformer. <S> For example, it is impossible to measure any difference between a 50Ω resistor, and a 12.5Ω resistor seen through an ideal transformer with a 2:1 turns ratio. <S> This holds true for any load, including complex impedances. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Since an ideal transformer can't be realized, considering how it might work is a logical dead-end. <S> It doesn't have to work because it is a purely theoretical concept used to simplify calculations. <S> The language you used in your first assumption is a description of the limiting case that defines an ideal transformer. <S> Consider a simple transformer equivalent circuit: <S> simulate this circuit <S> Of course, we can make a more complicated equivalent circuit according to how accurately we wish to model the non-ideal effects of a real transformer, but this one will do to illustrate the point. <S> Remember also that XFMR1 represents an ideal transformer. <S> As the real transformer's winding resistance approaches zero, then R2 approaches 0Ω. <S> In the limiting case of an ideal transformer where there is no winding resistance, then we can replace R2 with a short. <S> Likewise, as the leakage inductance approaches zero, L2 approaches 0H, and can be replaced with a short in the limiting case. <S> As the primary inductance approaches infinity, we can replace L1 with an open in the limiting case. <S> And so it goes for all the non-ideal effects we might model in a transformer. <S> The ideal transformer has an infinitely large core that never saturates. <S> As such, the ideal transformer even works at DC. <S> The ideal transformer's windings have no distributed capacitance. <S> And so on. <S> After you've hit these limits (or in practice, approached them sufficiently close for your application for their effects to become negligible), you are left with just the ideal transformer, XFMR1. <A> How can the infinite reactance of either primary or secondary in an ideal transformer draw current? <S> For two coupled inductors, we have two coupled equations: $$v_1 = <S> L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}$$ <S> $$v_2 = <S> M \frac{di_1}{dt} + L_2 \frac{di_2}{dt}$$ where <S> \$M = k\sqrt{L_1L_2}\$ is the mutual inductance and \$k\$ is the coupling coefficient. <S> Assume <S> perfect coupling, \$k = 1\$, from this point forward. <S> Using phasor notation, the above equations are $$V_1 = <S> j\omega <S> (L_1 I_1 + M I_2)$$ <S> $$V_2 <S> = j\omega <S> (M I_1 + L_2 I_2)$$ <S> Now, by (phasor) <S> Ohm's Law, it must be the case that $$ <S> V_2 = I_2Z_2 <S> $$ where \$Z_2\$ is the impedance connected to the secondary. <S> It follows <S> that $$\frac{I_2}{I_1} = \frac{j\omega <S> M}{Z_2 - j\omega L_2}$$ <S> So, for finite <S> \$L_1, L_2\$, the ratio of the secondary current to primary current is a function of frequency even when there is perfect coupling . <S> As the frequency tends to zero, the ratio tends to zero. <S> As the frequency becomes arbitrarily large, the ratio tends to $$\frac{I_2}{I_1} \rightarrow -\sqrt{\frac{L_1}{L_2}} = <S> -\frac{N_1}{N_2}$$ <S> Now, keeping the ratio \$\sqrt{\frac{L_1}{L_2}}\$ constant while allowing both \$L_1\$ and \$L_2\$ to become arbitrarily large, we have <S> $$\frac{I_2}{I_1} = \frac{j\omega M}{Z_2 - j\omega L_2} \rightarrow <S> \frac{j\omega M}{-j\omega L_2} = -\sqrt{\frac{L_1}{L_2}} = <S> -\frac{N_1}{N_2}$$ <S> The point is this: Even though the individual reactances go to infinity as the individual inductances go to infinity, the reactances 'cancel out' leaving the well known result true at any frequency . <S> In other words, the answer to your question is found by taking the limit as the inductances go to infinity and observing that the frequency dependent reactances in the numerator and denominator become a frequency independent, non-zero ratio. <A> Your question isn't completely clear, because assumption 2 says two things. <S> I'll address it both ways. <S> In the case where the secondary load is open there is no current. <S> So problem 1 isn't a problem in that case. <S> In the case where the secondary load is not open, there will be current, but now the reactance is no longer infinite. <S> For problem 2, I would say that the secondary winding is a real component, and as such we would want to have a model for it. <S> It's a way of saying, "I see that other coil there, and I've determined that it effect on the model <S> is vanishingly small, and here's how. <S> " <S> It's true there are other "parallel opens" (like the air leakage across a length of wire), but those don't appear as a component. <S> We'd get into an infinite effort accounting for those, and worse, an infinitely long philosophical discussion might erupt :) <A> It doesn't. <S> Current flows through the windings , but not through the reactance. <S> that is why it is shown in parallel, not in series (leakage reactance is shown in series, but an ideal transformer doesn't have any leakage). <S> If you don't put any reactance in the circuit then what do you have? <S> Zero reactance, a short circuit! <S> Any reactance less than infinity will cause some current to bypass the load. <A> Equivalent circuit of a transformer: - The area in the above circuit that appears to show a transformer with Es on the input and Ep on the output is, in fact, an ideal power converter such that: - \$E_P\times I_P = <S> E_S\times <S> I_S\$ and further... <S> \$E_S = <S> \dfrac{N_S}{N_P}\times <S> E_P\$ <S> Ignoring the small series secondary components (\$R_S\$ and \$X_S\$) <S> Any load impedance on the secondary will appear on the input side to the ideal power converter as: - Referred impedance of secondary onto primary \$= <S> (\dfrac{N_P}{N_S})^2\times Z_{secondary}\$ <S> So, if the magnetization inductive reactance (\$X_M\$) is infinite and the core loss (\$R_C\$) is also infinite then, apart from the small primary series components (\$R_P\$ and \$X_P\$), <S> the impedance seen at the real primary input is secondary impedance multiplied by turns ratio squared.
| An ideal transformer doesn't even have windings, because it can't exist.
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Do logic families use different type of transistors? Do the logic families such as TTL and CMOS represent only different logical structure? Or do they also use different type of transistors? Latest microprocessor chips with millions of components use CMOS. Is that only because CMOS "operating logic" is superior or also because CMOS family use entirely different transistors? So far I understood is there are transistors which are used in analog circuits and there are tiny ones which are used in IC chips. I guess analog use TTL level transistors? The transistors in ICs use many different logic families such as TTL and CMOS. So there are different transistor manufacturers for all these? <Q> TTL logic uses bipolar transistors and CMOS logic uses, well, CMOS transistors. <S> The structure of gates is also quite different between the two since the two types of transistors are so different and therefore need to be used differently. <A> Each logic family has different circuits for the same function (e.g. an AND gate), since each was an advance on earlier technology. <S> The earliest logic families used bipolar junction transistors. <S> Some examples are:RTL (1963) - resistor transistor logic (used in the Apollo Guidance Computer)DTL (1962) - diode transistor logic (used in the Minuteman II Guidance Computer)ECL (1962) - emitter coupled logic (faster, used in the IBM 7030 Stretch Computer)TTL - transistor-transistor logic (the most popular logic family prior to CMOS) 74xx (1964) - original TTL line (also 54xx for military) 74Sxx (1969) <S> - used Schottky transistors for speed 74Lxx (1964) <S> - low power 74LSxx (1976) <S> - low power Schottky <S> 74ALSxx (1976) <S> - advanced low power <S> Schottky <S> 74Fxx (1979) - fast than normal SchottkyThe following families used CMOS transistors:CMOS - complementary metal–oxide–semiconductor logic <S> CMOS (1970) <S> - CD4000 series CMOS HC (1982) - high speed CMOS, used same pinouts as 74LS family CMOS HCT (1982) - CMOS logic but as TTL logic levels (made combining them possible) <S> CMOS has the advantage over bipolar in that it uses no power except when it is switching from 0 to 1 and back. <S> The use of CMOS transistors has made it possible to cram millions or even billions of transistors on a single chip without requiring cooling, The most popular integrated circuits are made by at least two manufacturers (and in many case several); this is called second-sourcing and is required by large companies and the military to assure parts will be available. <S> However sometimes chips are made by only one manufacturer, and it is somewhat risky to incorporate them into a design. <S> They are by no means limited to the voltage supplies used by various logic families, be it TTL or otherwise. <S> In addition, it is not uncommon for them to have a negative supply, e.g. ±15v. <S> The main difference between analog and digital, is the latter operates on two levels only, 1 and 0. <S> For this reason, the transistors are turned completely off or on, the later is called saturated. <S> Whereas the transistors in analog circuits generally operate in what is called the "active" region between on and off. <A> Here is an awfully incomplete answer : By doping, oxyding, metallising a slice of sillicon, you can create on the surface several types of components : wires, bipolar transistors (aka BJT), metal-oxyde transistors (aka MOS), resistors, inductors, capacitors... <S> (but it is often easier to create a transistor than these passive components). <S> Components are created at once by applying patterns and exposing the die. <S> Transistors are not 'pick and placed', even for analog integrated circuits. <S> Bipolar transistors used in TTL gates are 'current controlled' and work very differently than MOS transistors which are 'voltage controlled'. <S> TTL (transistor to transistor logic) is traditionally defined as logic levels and <S> I/O characteristics, which were (in the '70s-'80s) optimal for bipolar transistors working with 5V power supply. <S> It is now possible to build with MOS transistors chips compatible with the TTL logic levels. <S> Logic gates can be created with bipolar and MOS transistors, but, nowadays, MOS is used almost exclusively for creating logic circuits like microprocessors, memories... <S> Bipolar transistors (and j-fets) are currently mainly used in analog components (for example operational amplifiers). <S> It is possible to mix bipolar and MOS on a single die, but the additional number of steps and constraints makes that technology more expensive and reserved to specialty component (for history, the first Pentium used BiCMOS which mixes bipolar and CMOS, this technology is nowadays used for analog or mixed signal components like ADC/DAC) <A> There's more to transistors than just the basic type (MOSFET vs. JFET vs. BJT). <S> CMOS, by definition, involves two types of MOSFETs (P-channel and N-channel). <S> And not every transistor of the same type will be identical. <S> For instance, the microcontrollers I work on use two voltages -- 3.3V for the IOs, and 1.2V for the core logic. <S> The core transistors are smaller and have thinner gate oxides because they don't need to handle as much voltage. <S> In analog circuits, it's also common to change the width to length ratio. <S> MOSFETs are not ideal for analog circuits because they have a high gate capacitance and low transconductance. <S> But they're better in digital circuits because CMOS logic doesn't consume any DC power[1]. <S> Plus, it's turned out to be very easy to shrink MOSFETs, and in logic more transistors always wins. <S> In mixed-signal ICs like microcontrollers, the analog circuits might use CMOS transistors just because that's what available. <S> (Adding BJT would mean a bunch of extra processing steps, which costs money.) <S> Aside from logic, MOSFETs are also used for switching large currents because of their low on-resistance. <S> BJTs have higher transconductance, but require base current to operate. <S> In integrated circuits you can do weird things like have a BJT with multiple emitters or collectors, but I don't think you can get discrete devices like that. <S> JFETs are a sort of middle ground, with medium transconductance but minimal base current. <S> They're commonly used to make the input stages of op amps. <S> In all cases, the transistors can be made differently to achieve the required voltage and current handling, switching speed, etc. <S> [1] <S> This is a historical advantage. <S> Modern CMOS transistors with gate lengths <S> shorter than 100nm have significant DC leakage currents. <S> On the newest processes this can be something like half of the chip's total power consumption.
| Analog ICs can use either bipolar or CMOS transistors, whichever is more suitable to the task.
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Physical significance of group delay What is the physical significance of the group delay of a system/filter?. Why is the slope of the phase response critical in a communication system. <Q> To understand the physical relevance of the term "group delay" one should know that the definition (slope of the phase response of a system) applies to a small group of frequencies only - centered at a frequency which is much larger than the mean value of this group. <S> Example: <S> Amplitude modulation with a carrirer frequency of 500 kHz and a signal bandwidth of - let`s say - 10 kHz. <S> Then, this group of frequencies centered <S> around 500 kHz is delayed according to the negative slope of the phase function of the corresponding system. <A> Note that the term group delay is only meaningful for a few special signals (see also this answer ). <S> In general, it's more useful to directly consider a system's phase response. <S> The relevance of phase in communication systems is that a non-linear phase on the channel causes intersymbol interference (ISI) , i.e. the symbols get mixed up in time and influence each other, and cannot be detected properly anymore by the receiver. <S> ISI is usually mitigated by using an adaptive equalizer prior to detection. <A> A pure delay is not that relevant in most systems. <S> It will affect timeouts, because the receiving system needs to be able to process and verify the incoming data before producing an acknowledgement, which in turn has to be transmitted back, so expected response times need to keep these things into account. <S> On the other hand, a non-linear phase response means that individual symbols affect each other, <S> so e.g. a 1 <S> following another 1 <S> has a different voltage level than if it were preceded by a 0 .
| As already pointed out in other answers, the group delay (at a certain frequency) is the delay of the envelope of a narrowband signal centered at that frequency.
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2.5mm Jack to USB data Cable I was recently diagnosed with Type 1 diabetes. With the need to test my own blood upwards of 8 times a day I have a ton of data to keep track of. My testing unit has the ability to upload its data to a mobile (OTG) or PC (USB). Sadly, the company who makes the unit does not ship either cable. They are an expensive accessory. I have enough parts, and should be able to make this, however it's proving difficult. The USB cable of theirs has windows drivers built into the cable. I was able to get a helpful person at the manufacturer to send me those. That person also told me that OTG wouldn't require drivers. The photo of both cables on their website show the unit end of the cable as being a 2.5mm (3 ringed) audio jack. So far I cut open a male USB, I also did the same with the end of an old 3.5mm headphone jack. I then purchased a 3.5mm to 2.5mm adapter. But I'm struggling with the wiring, and trial and error is not producing any results for OTG or USB. I'm also a little concerned that I'll damage my phone, laptop or blood glucose unit. Although all seems to be well after my tinkering. My USB has Green (Data), White (Data), Red (5v), Black (Earth) and some sheilding by the looks of it. My 3.5mm headphone cable has Red (Normally right audio), Yellow (Normally left audio) and White (Earth) I also have a USB to OTG adapter Note that my adapter has 4 rings, like an ipod shuffle cable or headphones with a mic. I'm assuming that's not introducing any crossed wires as using it for audio still delivers full stereo. Here are some pics Am I missing something in my approach? If not, how can I figure out which wires connect to which? I would be happy with either a working OTG or USB cable. My tests have shown no response in the mobile app, and only generated USB malfunction errors in Windows (as expected). Edit: Thanks for all the really good info. I have contacted CareSens for a NFC unit. I will update here when I find out more. This was going to be a fun project during my time off, but it looks like it's a bit more complicated than I first thought. <Q> Based on the manufacturer's youtube video for the app compatible with the cable: This is a video guide for SmartLog(Blood Glucose Management Software) <S> App represensted by i-SENS. <S> SmartLog App is a smart phone application which helps patients with Diabetes to monitor their health conveniently anywhere anytime. <S> This app works with CareSens N NFC meter. <S> CareSens N and CareSens N POP meters can also be used when using FTDI cable. <S> A FTDI cable typically refers to a USB to RS232 ( <S> TTL level Serial) <S> IC FT232 (or other generations of the FT232 chip) created by FTDI. <S> They also make other USB bridge ICs with similar functions. <S> as you have already seen. <S> If it's a simple straight through connector, it will have Ground, TX and RX. <S> FTDI's official 3.5MM cable uses Tx {To device from PC}, Rx {From Device to PC}, Gnd (Tip, Ring, Sleeve, respectfully). <S> With your multimeter, you can confirm the 2.5mm pinout by doing a continuity test between each section of the adaptor and your 3.5mm cable wires, then confirm the signal by checking for voltage between the three wires. <S> The voltage it runs at is a concern because using a 5v signal on a 3.3V port might be bad. <S> Of course it could be more complex. <S> TI calculators used a 2.5mm port for their Graphlink cables. <S> It was able to connect to a serial port, but required 6 pins, resistors and diodes between. <S> If you had a cable to hack up, or even a meter to hack up, it would be easier. <S> They occasionally pretty much aways give the device away for free, check with your doctor or the manufacturer's local sales rep. <S> Update: <S> Like a defacto standard amongst Diabetes Meter manufacturers. <S> You need the USB to serial IC for the OTG cable, but you could use a serial port for the PC instead (I am not liable if you fry your meter). <S> http://pinoutsguide.com/Electronics/bayer_contour_pinout.shtml <S> http://www.diabetesforums.com/forum/topic/65566-abbott-freestyle-freedom-lite-data-cable-how-to-some-other-info/ <S> Reading data from a glucose meter <A> There is no standard conversion between a 2.5mm audio jack and USB. <S> It's not like USB A and USB B, where these are simply different connectors doing the same thing. <S> Audio jacks usually carry audio. <S> USB is USB. <S> These are different things. <S> It makes about as much sense as looking for a 2.5mm audio jack to 110V wall outlet adapter. <S> While it's entirely possible that the meter's manufacturer simply used a 2.5mm audio jack for a USB compatible interface (maybe they hadn't heard of micro USB connectors?) <S> this is only one of many possibilities. <S> It could also be that the electrical protocol at the audio jack is a simple serial protocol, like RS-232. <S> Or, it could actually be audio, somehow modulated to convey the relevant data. <S> Unfortunately, without specifications from the manufacturer, or at least a working cable to reverse engineer, we can only guess at how it might work. <A> It becomes pretty clear after doing some research that there is some electronics involved. <S> FTDI, for example, makes SSOP devices(less than 1/4 in. <S> square) <S> that contain a microcontroller and programmable memory that can make the conversion between usb protocol to a serial interface. <S> In other words, it's not a matter of how to wire one end of the cable to the other; you need a cable that has the chip built into it that's been programmed to provide the interface between the usb port and the glucose meter.
| Connecting a USB cable straight through to a 2.5mm plug will most likely cause a problem. Point being, the cable accessory likely has the necessary electronics to make this conversion. Based on the two links below, the pinout is more likely to be Tx from Device to PC, Rx From Pc to Device, Ground (Tip, Ring, Sleeve). I have run into the same issue as I have also been diagnosed with diabetes.
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Can I get constant current at the output of a buck convertor (to drive a LED)? I want to drive a LED with a buck converter circuit as shown in the schematic below. In order to keep the circuit simple, the system will run open-loop (i.e.; there won't be any voltage or current feed-back). Is it possible to set the LED current to a fixed value by means of calculating the appropriate \$D\$ (duty cycle) value? Or, in this circuit model, will the LED current theoretically approach to infinity because of lack of a resistive element in the path? <Q> Yes, it is possible to calculate the duty cycle to acheive a specific LED current, given a specific set of conditions . <S> I actually did something like this in a commercial product, except that the power voltage was from a battery and could vary. <S> The formula from battery voltage to PWM duty cycle is non-linear and not easy to solve in real time in a small micro. <S> However, this was done with a table. <S> I set up the table to take the raw 8 bit A <S> /D reading directly and produce the value to write into the PWM duty cycle register. <S> All the conversions and calculations (which included a divide and a square root) were performed by the preprocessor using floating point math, and the result loaded into the table at build time. <S> It worked quite nicely. <S> The LED current stayed within 10% of the target over the full battery voltage range. <A> The design has too many problems to be worth trying to fix. <S> An N Channel MOSFET (as shown) must have its gate driven above its source by Vgs_operating. <S> Typically this is 3 to 6 volts and is seldom less than 1V with even very low Vgsth MOSFETS. <S> As the source is at V+ = <S> 12V potential when the MOSFET is on (as brhans notes) the gate must be driven ABOVE V+ by Vgs_operating. <S> ie to from 13 to 18 V depending on the MOSFET used. <S> You thus need either a higher voltage than +12V for the 555 to drive the gate (from a bootstrap driver or other source, or to use a P Cannel MOSFET. <S> Whereas a voltage source can be implemented open loop by using Vout ~= <S> Vin x Dt/T, attempting to use this to set an LED current rather than voltage is approximately impossible (for values of approximately close to "certainly"). <S> Chances are the LED Vf will be slightly more than 2V at 200 mA <S> but if we assume it is 2V as shown then if a current sense resistor dropping 0.1V is used in the LED Cathode to ground connection then an efficiency loss of ~= <S> Vsense <S> /V_LED = <S> 0.1v/2v <S> = 5% is incurred. <S> This is liable to be acceptable enough in reality. <S> The reasonably low cost and easy to use NCP3065 / <S> NVC3065 has a 0.235V sense voltage. <S> This can be reduced with a few extra resistors. <S> Overall efficiency is not marvellous (see fig 17 & fig 20 in <S> NCP3065 datasheet . <S> It will be higher than shown there as graphs are for 700 mA or higher. <S> Modern synchronous output ICs for a few dollars will allow lower Vsense. <S> A hall effect sensor could be used to provide approximately zero loss current sensing. <S> Overall efficiencies of 90%+ should be achievable. <A> Assuming the MOSFET was switching properly- <S> it would not work. <S> The circuit will approximate a fixed voltage source of voltage 12V <S> * <S> (ton/(ton+toff)) and a low output impedance. <S> That will not result in a controlled current through the LED- it will tend to lead to too much or too little current. <S> If you need it to supply a constant current, you'll need a sense resistor somewhere and some kind of feedback (not necessarily closed loop around the LED, but at least the switch current needs to be controlled).
| If you know the 12 V supply will stay fixed, know the LED forward voltage, and use a reasonable forward drop accross the diode, then this can be done. Better is to use a circuit that is designed to work properly and better still is to accept some sense losses in order to increase (not decrease) simplicity and designability.
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What would make me choose Verilog or VHDL over schematic design on CPLDs or FPGAs? I have absolutely no background in programmable logic, I use mostly microcontrollers in my projects but recently I needed to work with video and the microcontroller is just too slow for what I needed so I started playing with CPLDs. I was able to get good results with the CPLD only using schematic design but when searching for info on CPLDs I came across many examples using VHDL and Verilog. I am curious about what could make me want to define my device in one of these languages. What can they do that schematic design cannot? Are they used mostly for functions? Until now I have only used CPLDs, do FPGAs designs benefit more than CPLDs from using these languages? <Q> Schematic design is only useful when you're only tying together a few off-the-shelf modules (counters, adders, memory, etc). <S> But implementing an actual algorithm (say, a cryptography hashing algorithm) is nearly impossible to do without an HDL (like VHDL or Verilog), since there's no way to describe a system at a behavioral level with schematic symbols. <S> Most projects are done in behavioral-style HDL because they're too complex to be synthesized by hand and drawn using logic primitives schematically. <S> CPLDs are generally used for glue logic and less used for processing, and generally logic is easy to implement schematically, so I think you're right when you suggest that FPGA-based designs benefit more from using an HDL. <A> A couple of practical aspects in addition to Jay's excellent answer: Bugs. <S> The schematic tools tend to be buggier* than the rest of the toolset. <S> This is possibly due to the preference of Verilog/VHDL over schematics in the industry, and thus schematic entry is given less attention by the software developers. <S> Speed. <S> The schematic needs to be first converted to a HDL before passing it to the synthesis tool. <S> This can have a negative impact on build times. <S> The generated HDL also might not be very readable if in case you need to inspect it for some reason. <S> Portability. <S> Depending on the amount of vendor specific primitives used Verilog and VHDL are more or less portable between devices. <S> Porting schematics you either have to redraw everything or rely on the provided import/export capabilities (if any). <S> * <S> My favourite bug in Xilinx ISE was the inability to select vertical wires. <A>  <S> There are many advantages of a HDL (Hardware Description Languages) as a Design Entry standard. <S> The hardware description languages VHDL and Verilog were designed for modelling hardware with the intention of modelling at a higher abstraction level which includes features like, concurrency, timing, hierarchy, reuse of components, state behaviour, synchronous behaviour, asynchronous behaviour, synchronization and inherent parallelism. <S> Issues arise during synthesis, mapping the design description to a specific process and gate implementation. <S> This requires that you cannot use the high-level features of HDL - you must produce "synthesizable Verilog/VHDL" <S> So you have HDL for synthesis and HDL for Simulation and the subset that is synthesizable <S> is tool specific. <S> You cannot go from a Behavioural design description to a net-list/ layout. <S> But you can structure your design to have behavioural components that also have a synthesizable aspect that can be compared against each other. <S> You start with the behavioural and <S> then once that is working you rewrite for synthesis (which is a subset). <S> You go from the general to the specific and build test-benches along the way. <A> One more advantage is that HDLs get all the same advantages as regular programming languages in that they can be used in standard version control systems, diff-ed to examine changes, etc. <A> In addition to what already have been said: text representation are simply much more more manageable, especially in large projects. <S> You can (albeit with huge difficulties) convert any syntheizable HDL into into schematic, but hundreds of lines of plain text is easier to work with than hundreds of schematic elements.
| The description of the functionality can be at a higher level, HDL based designs can be synthesised into a gate-level description of a chosen technology, A HDL design is more easily understood than a gate- level net-list or a schematic description and HDLs reduce errors because of strong type checking.
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Does current flow stop when a capacitor is fully charged? If I take a 1V battery and a 1 farad capacitor and a bulb and connect them in a circuit. Then will after the capacitor gets charged (not fully charged as that is impossible) will the bulb gradually stop glowing? Then what would happen if you change the battery and make it a 2V battery? The question I want to ask is, that if there is a capacitor in any circuit, does current flow in the circuit stop once the capacitor is charged? Why or why not? <Q> For a capacitor charge Q = capacitance C multiplied by voltage V. Rate of change of charge is current and this leads to: - \$\dfrac{dq}{dt} = <S> C\dfrac{dv}{dt}\$ = <S> current <S> This quite simply means that a rate of change of voltage gives rise to a current. <S> If the voltage is rising linearly with time, the capacitor will take a constant current and once the voltage stops changing the current is zero. <S> In a real capacitor there will always be a small leakage current and this can be modelled by a resistor in parallel with the "perfect" capacitor. <A> However, DC voltage sources are seldom perfectly flat, and capacitors are far from ideal. <S> Ripple on the DC supply will pass through the capacitor <S> The ripple is basically a small AC component superimposed over the DC voltage. <S> This AC component can pass through the capacitor quite happily. <S> That's basically how smoothing capacitors work - they fill up with the DC, which can't then get through to ground, but the ripple can pass through the capacitor to ground, effectively getting rid of it. <S> Capacitors have a certain amount of self-discharge Charge a capacitor up <S> and it won't stay charged up forever. <S> It will gradually discharge. <S> So a very small amount of current will flow to keep the capacitor filled up. <A> yes, because it is the steady state(t= infinity) condition and after this capacitor act as a voltage source. <S> also, as the (impedance)Z=-jXc(reactance), where Xc= 1/WC and W=2(pi)f and f=1/t. <S> so at t=infinity, Z=infinity. <S> hence (current)I=0.
| If you have a perfectly flat DC voltage source, and an ideal capacitor, then yes, when the capacitor is fully charged then no current will flow.
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How long does ultrasonic rangefinder last I've built a parking sensor in my garage that uses the HC-SR04 ultrasonic rangefinder and an arduino to measure how far the car is from the back wall. The one thing that I'm worried about is if the sensor will continue to work for a long time. When it senses an object, it senses distance 10 times per second and once there's a period of inactivity, I scale it down so that it senses distance once per second. So it will be sensing once per second pretty much all the time, which is like 86000 per day, or 31 million times per year. Anyone know if these sensors go bad over time or if they're rated for a certain number of senses? Thanks for any advice! <Q> First off, sensing the distance once per second may be too slow even for a parking application. <S> Consider that 2 mph is equivalent to over 2.9 feet per second. <S> Almost 3 feet is a large distance in a garage. <S> In any event, I don't think you have to worry about the reliability of the sensor. <S> Quartz watches run continuously for years without any problems. <S> If the watch is analog, that means the parts associated with moving the second hand are operating once per second 24 hours per day or 31 million times per year as you have pointed out. <A> Just checked on the Sparkfun products (Maxbot)not Sainsmart like your HRC. <S> Their test show their line has a Mean Time Before Failure (MTBF) of 200,000 hours or 22 years. <S> Maxbot makes a big deal of their reliability so the Sainsmart might not be as good <S> but seems like a pretty sturdy kind of component. <S> Maybe Sainsmart has some figures on their products. <A> Some will fail much sooner than the MTBF and some will fail later. <S> The consequence of failure in your case is that you crash your car into the wall of your garage. <S> So you must have a way to deal with failure when it occurs. <S> One way to deal with failure is to use a redundant system. <S> 1) Build two sensor systems that operate independently. <S> 2) <S> Let the systems share data in some sort of fault tolerant way (isolated serial perhaps). <S> 3) Have each of the two systems indicate a failure if the sensor reading from the other system doesn't match within a certain margin. <S> If either system fails in any way you will now get a warning from the other one. <S> You are still not covered in the case where both systems fail at the exact same time, but that case less likely.
| Garage door openers have object sensors to prevent accidents that are also on continuously and last for years. Regardless of the manufacturer specified MTBF, all devices will eventually fail.
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Can a current be induced in an HDMI cable that runs parallel to a 120v AC cable for 15 feet? I just went to go unplug an HDMI cable and got shocked from the end of it! The only possible causes of this that I can think of are that the receiving or sending end of the long (50ft) HDMI cable is introducing a live current into the line, or that an electrical current is being induced in the HDMI cable because it runs side by side (parallel) to a 120v AC extension cable for 15 feet. Should I be concerned? What are the first steps I should take to diagnose this? For reference, there was a Sony PS3 connected to one side, and an HDMI Splitter to the other end ( http://smile.amazon.com/dp/B00B46XUQU ) If you need any clarification, please don't hesitate to ask! <Q> The coupling could result from a Y-capacitor in the HDMI switch power supply. <S> A Y-capacitor is placed between L-GND and N-GND and often used to filter out common-mode noise and to establish a common ground reference (with respect to "earth"). <S> They never fail as "short" and there are typically only a few uA flowing, but it is enough to feel a tickle. <A> It's not likely that a 120v power cord induced a significant voltage in the HDMI cable. <S> Voltage induced by wire A in a wire B is proportional to the amount of current flowing in wire A and is transferred by the magnetic field. <S> However your power cord has both hot and neutral lines, and they carry (nearly) the same amount of current in opposite directions. <S> This means the magnetic field is cancelled out, except for minute variances between the current in the lines. <S> What's more likely is that device on the other end is badly designed so that it's not grounded, not properly isolated from the HDMI plug or broken/damaged. <S> Since you are unplugging the cable, is it safe to assume that it's still connected at the other end? <S> Was it the PS3 or another device? <A> It didn't harm you so that is good <S> but, does it suggest a darker and more serious problem that could be classed as potentially life threatening? <S> My belief is that the power source that caused the "tingle" is a switched mode power supply and a little bit of AC current is leaching thru the EMI reduction capacitors. <S> If it is this it is quite common and not dangerous <S> but it ought to be a small wake up call for you to double check this by seeing if touching either of the wall wart's dc wires can produce a similar effect. <S> If it does then I'm fairly convinced that all is well. <A> A current carrying wire can not be at ground potential and inside buildings the neutral wire can vary in voltage from place to place depending on loads. <S> I have seen as much as 40 volt difference in large buildings. <S> It can be quite dangerous to run long extension cords and cables that connect to power sources on different circuits. <S> This is why there have been standards for ages that say doorbells are run from low voltage isolation transformers. <S> Check with a voltmeter to see the difference between the neutral side of the outlet (I'm talking U.S.A. here) <S> which is the wide slot and earth ground which is the round hole - and should not be carrying any current. <S> It is there in case of a short so the circuit breaker will blow, instead of you or a fire. <S> If it is old and there is no ground plug, the screw that holds the plate might be grounded. <S> Of more interest is exactly how did you get the shock? <S> What else were you touching? <S> What is the floor material and did you have shows on?
| If the outlet / plug does not provide proper grounding (only N and L connected) and you touch GND on the HDMI cable, there might be a small current between flowing from L-GND-YOU-EARTH. If it is 50 feet long and you are using devices fed from different circuit breakers or from a circuit supply a lot of power to other devices, it is more likely that "ground" is not the same at the two ends of the cable. Y-capacitors are tested to applicable standards to qualify them as safe.
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Can I cut up a proprietary barrel connector cord and make a USB power cord with it? I have a Yaesu VX-8DR radio , and its power input is a barrel connector. In the age of smartphones, where basically every single device I have uses a USB cable to charge at up to 2.4 amps, I'd like to make a USB cable to power this device as well. I have a small case I pack around with a charger and assortment of USB cables, and I'd like to just add one more USB-to-power-barrel cable to it, rather than two additional discreet proprietary car charger and wall charger units. (Or buy one ; I don't know the barrel connector size, though, and I don't have calipers, and I don't see it listed anywhere in the manual. Although this answer to another question I found here seems to indicate exactly what connector I may have, if Yaesu followed this standard. The plug does have the yellow insulating ring as mentioned in that answer.) The Yaesu car charger output current is 2A, and the included wall charger output current is 1A, both at 5V like all my USB devices. This means I should be able to use my existing USB car charger that outputs 5V/2A and my existing USB charging hub at home, right? To clarify, I want a full-size USB connector on the end where there is currently an integrated wall-wart or an integrated car power plug. The barrel connector will stay on the radio end of the cable, I just need to feed it from USB instead of feeding it from an integrated plug. Can I literally just chop off the integrated cord that runs out of the car charger and solder a full-size USB connector to the end and call it good? Secondarily, the radio treats any connection on this power port as "attached external DC power", and increases the output wattage on the 1.25m band from 1.5W to 5W until the external power is removed. If I connect this new USB connector to a battery (such as one used to recharge a dying cell phone) is there any possible problem with this increased output wattage connected to a battery instead of to an actual continuous DC power source? <Q> This means I should be able to use my existing USB car charger that outputs 5V/2A and my existing USB charging hub at home, right? <S> No. <S> Take note of the PA-48B charger: <S> Can I literally just chop off the integrated cord that runs out of the car charger and solder a full-size USB connector to the end and call it good? <S> No. <S> If I connect this new USB connector to a battery (such as one used to recharge a dying cell phone) is there any possible problem with this increased output wattage connected to a battery instead of to an actual continuous DC power source? <S> Since your device <S> ** <S> cannot* be charged by 5V from USB, a cheap lithium battery pack for USB phone charging <S> is not appropriate. <S> However , for continued operation at high power, you could use an external 12V lead-acid battery (car, motorcycle, marine) <S> connected with the correct polarity to a barrel jack to supply external power such that your radio will operate in high-power mode. <S> When using such a battery, be sure to use fuses to reduce the possible damage from the high-current capability of a lead-acid battery. <A> You can cut and splice the cords. <S> I do that kind of thing all the time. <S> You do have to take care to connect the correct wires so that you do not end up with a power polarity reversal. <S> Such reversal could spell disaster to the end device. <S> When you splice low voltage DC power cords it is wise to use a good scheme for insulating all the connections. <S> I always open up the individual wires to be joined so that they are long enough to let me slip heat shrink tubing over the wire on one side before lap splicing the two wires by soldering. <S> Also before soldering anything slip a somewhat larger diameter heat shrink tubing over the overall cable on one side that will be used to cover the wires and the small insulated joints. <S> Once you have soldered the individual wires slide the small heat shrink tubing over the solder joints and shrink using a heat gun. <S> After that you slide the larger tubing over the whole works and shrink that into place. <S> Some notes. <S> Heat shrink covered wire splices are not very flexible and if made too long or tool close to a plug end can make the cable very awkward to use or store <S> so plan the splice point accordingly. <A> Yes. <S> If a usb supply meets your device's needs (5V 1+ Amps) then a physical connector doesn't matter. <S> As for the usb battery pack, it depends on how your device deals with voltage drop on the input. <S> Some usb battery packs simply cut off when the battery + regulator drops below a predefined point, other cheaper ones simply keep providing a lower voltage. <S> Keep in mind <S> 5W is 5V 1A. <S> Your battery pack should be able to supply 1A for that to work well.
| A very handy tool for holding the wires in place while soldering the stripped ends together is pictured below. The VX-8R series radios require 11-14 V DC (Negative Ground, EXT DC jack) input to charge ("Operating with Charging" pg 166 of the Operating Manual ). Not only does your device not support 5V charging, but it is also not a qualified USB device and can neither negotiate power delivery nor respond to different chargers.
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What is the need for a pull-down resistor in this schematic? here at the second page: http://www.skyeinstruments.info/index_htm_files/W200P%20Windvane.pdf there is an external 470k resistor in parallel with the measuring device. it seems like the manufacturer wants users to connect this before measuring. but doesnt data acquisition boards and other measuring devices have already huge input impedance? what is this resistor for? <Q> [This does not contradict what EMFields wrote above . <S> This is just another point of view.] <S> Have a look at the figures from p.2 in the datasheet . <S> Assume that the wiper is at NW. <S> The output would be around 90%. <S> Now assume that the wiper is turning clockwise from NW. <S> The output will be growing. <S> Then the output will reach 100% when it reaches the point T1 on the circuit, which is at 358.25°. <S> The output will stay at 100% as the wiper keeps going clockwise between 358.25° and 358.85° . <S> At 358.85°, the wiper becomes an open circuit. <S> If there is no pull-down (or pull-up), the output would be floating. <S> It would act like an antenna for interference an leakages. <S> The reading is unstable. <S> The measuring equipment may even see 2.5V because of leakages and think that the vane is pointing South while it's actually pointing North. <S> So, the pull-down gives the output a determined state when potentiometer is open circuit. <S> Why 470kΩ ? <S> I think, it's a ballpark number, and it's driven by a compromise. <S> On one hand, you'd like a strong pull-down to reduce the effects of interference. <S> On the other hand, the pull-down is in parallel with the lower leg of the potentiometer, and it's introducing non-linearity into the measurement. <S> So, a weak pull-down is desired. <S> Weak pull-down means larger resistance value. <A> I'd guess it's to keep the impedance across the HI-Z stuff reasonable when the pot wiper goes open-circuit around North. <A> I'm currently working on this very device and the 470K resistor is not just plucked out of nowhere. <S> If you connect the device up to a micro controller you will notice that the readings close to or in the north gap are slightly erratic, even with a pull down resistor. <S> The smaller the resistor, the more benefit will be gained but the more non-linearity introduced. <S> If you can work out the non linearity then you can use a lower value resistor and get better readings. <S> Although the calculation is based on easy parallel resistance maths the formula for % non linearity is not completely obvious:
| Strong pull-down means smaller resistance value. The purpose of the pull-down resistor is to give the output a known state when the wiper is in the gap located at the North.
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What is the drawback of a high-value resistor for positive comparator feedback / hysteresis? I'm building a simple circuit which has an LM211 comparator at its heart. The positive input is connected to a 10k-10k resistor divider across the supplies, stabilized by a 1uF capacitor. Hysteresis is provided by 1.8MOhm from the output to the positive input. The signal on the SENS line is a slowly changing resistance hooked up to +5V on the other side. Everything works perfectly on a breadboard, but I wanted to check a few things before finalizing the board layout, and came across a TI application note which says in paragraph 6: It is a standard procedure to use hysteresis (positive feedback) around a comparator, to prevent oscillation, and to avoid excessive noise on the output because the comparator is a good amplifier for its own noise. In the circuit of Figure 2, the feedback from the output to the positive input will cause about 3 mV of hysteresis. However, if the value of Rs is larger than 100Ω, such as 50 kΩ, it would not be reasonable to simply increase the value of the positive feedback resistor above 510 kΩ. However, it doesn't say why a high value feedback resistor is "not reasonable". I built up my circuit on a breadboard, and it seems to work fine, I can definitely see good hysteresis behaviour with the 1M8 resistor as compared to without it. So, what is the drawback? <Q> 1 MΩ for the positive feedback should be fine in your case from a impedance point of view. <S> Whether it always provides enough positive feedback under all conditions is something you have to decide. <S> What bothers me about your circuit is that the positive feedback is fighting against C23. <S> That is going to slow it down, which probably in part defeats its purpose. <S> To fix this, you can add some resistance between C23 and the + input. <S> The immediate effect of the positive feedback will be based on the feedback resistance and this deliberate resistance. <S> In the long term, the 5 kΩ impedance of the power supply divider will be added to the deliberate series resistance. <A> Olin is right about the problem with C23: it lowers the bandwidth of the positive feedback rendering it useless. <S> Just decoupling the supply close to the divider will do the job. <S> In fact, that is probably also the reason why TI wants to keep the feedback resistance low in value. <S> Normally you would like the hysteresis to be faster than the response of the op-amp; this will be probably be the 510K TI talks about. <S> If you make sure the highest (noise) frequency that can reach the negative input will be lower than the feedback response, you should be in the clear; however, making the feedback faster then the op-amp can react will be the better option. <A> You can parallel the 1M8 resistor with 2.7nF and avoid the objections while still maintaining a 1uF filter cap. <S> That's maintaining the 1800K/5K ratio of feedback resistance.
| If the value is too high, the input capacitance of the op-amp alone will already lower the bandwidth of the positive feedback, so on a noisy signal the hysteresis will not work.
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Can I put 2 voltage regulators (5V) in parallel to dissipate the heat better? I connected around 10m of christmas light LEDs to a 5V voltage regulator (lm7805). It gets quite hot, put I can still touch it without burning my finger. So in is within its temperature range (120 C max). It's a self regulating voltage regulator so I am not affraid of it burning itself. But to ensure that my christmas tree wont burn, can I put 2 regulators in parallel to dissipate the heat better? <Q> No, but depending on what your power source is you may be able to do 2-stage regulation to spread the heat - for instance, if your source is 12V then lower it to say 9V first with a 7809 then drop it from 9V to 5V with the 7805. <S> If you're even higher than that, then you could use a 7812 first to drop to 12, <S> then the 9V, then the 5V. <S> Either that or just use a switching regulator, such as a UBEC used by RC vehicle enthusiasts (available for a couple of $ from eBay or most RC model shops). <A> Hot but touchable is entirely within the normal operating conditions of an 7805. <S> If you are concerned about the effect on your Christmas tree simply add a heat sink (which can be as simple as a 10 x 10 cm metal plate) to spread the heat over a larger area and thus reduce the temperature. <S> (This assumes you are using a TO220 version. <S> If you are using a TO92, swap it for a TO220.) <S> You could also spread the heat by dissipating part of the heat in a pre-regulator, series resistor, or extra power transistor, but that makes little sense in this case as it creates two hot-spots instead of one, which is a little bit better but not as good as a heat sink. <A> Yes, you can. <S> However if both regulators are not putting out identical voltages then one will draw more current and get hotter than the other. <S> The traditional way to fix this is to add a small resistor (eg. 0.1Ω) in series with each output. <S> If you can touch the regulator without it burning you then <S> it's probably not getting too hot. <S> If your Christmas tree is really sensitive to heat then you could put a heat sink on the regulator to reduce its surface temperature.
| Another option is to use a switching regulator, which will dissipate much less heat for the same current.
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keyboard very strange behaviour I have a USB keyboard which has a somewhat faulty wire. On shaking the wire, the USB keyboard sometimes gets disconnected or sometimes gets connected. This is normal. But I have another very strange phenomenon with my keyboard: On switching on/off the tubelight in my room, I hear the USB disconnecting and reconnecting sound clip in my computer. This only happens when my keyboard is connected to my PC. I do not have any light sensors in any device. Or is this due to some fluctuations in the voltage that the USB kb gets disconnected for a few milliseconds? <Q> Dry joints and dodgy connections can act as diodes. <S> Diodes can demodulate radio signals. <S> That's basically how old "cats whisker" radios (crystal radios) used to work. <S> It sounds like the bad connection is picking up the radiated noise (EMI) from switching your light on and off and interfering with the USB data stream. <S> Get yourself a new keyboard, dude... <A> The connection to the shield (the part connected to the rectangular metal part of the USB connector- the "shell", rather than the small pins) is probably open at one end or the other. <S> The keyboard will probably continue to function, but will be more sensitive to electrical noise (what you are experiencing). <S> There are at least one, sometimes two, shields in a typical USB wire- <S> the outer one surrounds the set of four wires, and an inner shield may surround the two differential data wires. <S> Both are connected shell-to-shell from one end of the cable to the other. <A> It's all about you keyboard, from what I understand. <S> If you have a keyboard that you really want to keep, you can still replace the USB head of your keyboards USB wire with a soldering gun and a spare wire <S> (See my post: How to "frankenstine" wires? ) <S> Also, the USB port of your computer might be silghtly damaged, try plugging other things in it and see what happened (I've never heard of sounds unintentionally coming from connected USB ports!). <S> Finally, you should just get a new keyboard... <S> XD.
| It's probably resetting the microcontroller in the keyboard or affecting the USB PHY on your motherboard.
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Circuit to tune multicolor LED colors with a variable knob (rheostat?) I am looking at getting some multicolor LEDs, which have Red Green and Blue inside of them. I'm doing this for space saving, rather than using 3 different LEDs--I am using just one. What I want to do is to use a multicolor bulb to produce all sorts of colors, mixing the channels (if you call it that) so I can the full color rainbow. I need to be able to use a single rheostat to do that, rather than using 3 independently so I can twist just one to change the colors up and down. If I need to use a small transistor, that's ok (I'd prefer to keep it simpler if possible) but I need the ability to use a variable controller knob to tune the colors. I can't seem to find any circuit design that can do this 1) for a multicolor LED, and especially 2) for a variable control knob in it. Sure, I can do this with an Arduino but I'm trying to use this in way more simply. I'm not set on using PWM or not using it, I just want the ability to set it up and be able to 'tune' it with a knob. If I can get away without using a transistor or having to program it, that would be ideal. Thanks! G <Q> In general you'd need three knobs to set the three LED light levels. <S> You could make a one turn knob potentiometer, read the voltage, and turn the voltage into some combo of LED's with a look up table. <S> But it will most certainly involve programming and PWM. <A> simulate this circuit – Schematic created using CircuitLab <S> this might just work if you sized your resistors and zeners properly. <S> it would cause one led to turn on at a time, then stay on without changing too much brightness while the next one turned on if you got the transistors to be saturated just at the point when the pot reaches the zener voltage of the next colour. <S> it might not work at all - just an idea <A>
| In order to create a full palette of colors, the intensity of each color must be manipulated independently from the others - from zero through some maximum for each color - so I think your proposition is impossible.
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What to do with unused pin? I am designing the Vishay VCNL4010 proximity sensor into one of my PCBs. This sensor has an interrupt pin that I don't need to use. Is it OK to just leave this pin floating ("unconnected"), or will this cause problems somehow? <Q> Normally I'd say it won't be a problem: it's an output pin, so it won't be floating as the device itself will drive it. <S> However, if we look at the datasheet , page 5, first note: <S> The interrupt pin is an open drain output. <S> The needed pull-up resistor may be connected to the same supply voltage as the application controller and the pull-up resistors at SDA/SCL. <S> Proposed value R2 [a pull-up for the INT pin] should be > <S> 1 kΩ , e.g. <S> 10 kΩ to 100 kΩ. <S> Proposed value for R3 and R4, e.g. 2.2 kΩ to 4.7 kΩ, depend also on the I2C bus speed. <S> For detailed description about set-up and use of the interrupt as well as more application related information see AN: “Designing VCNL3020 into an Application”. <S> If we look at this application note, Designing VCNL3020 into an Application , page 2: <S> The SCL and SDA as well as the interrupt lines need pull-up resistors. <S> I suppose that this is only needed when this line is actually used. <S> So I'd recommend you to use a pull-up resistor between 10kOhm and 100kOhm (as suggested in the datasheet excerpt) to the input voltage. <A> Well depends... <S> Input pins should NEVER be left floating else your circuit will do unpredictable things. <S> Output pin are OK to left unconnected. <S> They will not be "floating" because they are driven to some voltage by them self. <S> This pin in an open-collector, which is a output of special kind (that can be connected to other open-collector output) but can be left unconnected. <A> Unused pins are usually ignored. <S> They may be floating or unpredictable, but your circuit isn't using the pin anyway. <S> The floating value will never be read. <S> An example is the unused pins on the GPIO of a Raspberry Pi computer.
| In your case the "interrupt" pin can be left floating as it is an output pin. However, it's always good to follow the datasheet, and that extra resistor will still fit in your circuit, hopefully.
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What does manufacturer call PMSM motor? It is easy to find BLDC motor manufacturers on google, but it is difficult to find PMSM. Maybe there is a terminology issue here.There are many "Brushless Servo Motors" or "Synchronous Servo Motors" on google, do manufacturers use these terminologies to refer to PMSM motor? In theory:BLDC has concentrated stator windings and trapezoidal back EMF, the controling method is called "6-step commutation".PMSM has distributed stator windings and sinusoidal back EMF, the controling method is called "FOC". In reality:Most so called BLDC motors on the market have sinusoidal back EMF, and can be controlled by the same FOC method as PMSM motor. But I think they are still BLDC motor, not PMSM. My question is:Is "Brushless/Synchronous Servo Motors" more like the theoretical PMSM motor?What does manufacturer call PMSM motor? Thank you. <Q> The terminology for brushless, permanent magnet motors is confusing. <S> Be aware that brushless motors with a trapezoidal back-emf can be driven by sinusoidal waveforms and vice versa. <S> And also be aware that trapezoidal and sinusoidal back-emf's are ideals and you can never really get either one. <S> Of course, I've also seen IEEE papers that refer to BLAC motors and use other terminology, so this isn't strict across the board. <S> Industry hasn't really adopted this terminology completely. <S> You often will see companies refer to BLDC motors, as you've already pointed out. <S> And generally by BLDC <S> they mean exactly what the academics mean - a brushless motor with a trapezoidal back-emf. <S> However, I've also seen these referred to DC brushless (DCB) motors, brushless PM (BPM) motors, or even PMSM's. <S> With what academic literature refers to as PMSM's, I've seen them called PMSM's, brushless AC (BLAC) motors, AC servo motors, brushless servomotor (BLSM) and others. <S> Some manufacturers may not make a distinction between the 2 because in reality it isn't an either/or thing. <S> You can't make a brushless motor with a perfect trapezoidal back-emf <S> and you can't make one with a perfect sinusoidal back-emf. <S> Your best bet is to talk directly to manufacturers and tell them what you want to do and they will guide you in the right direction. <S> In reality: Most so called BLDC motors on the market have sinusoidal back EMF, and can be controlled by the same FOC method as PMSM motor. <S> But I think they are still BLDC motor, not PMSM. <S> This may or may not be true. <S> In my experience, BLDC motors do not have sinusoidal back-emf; they are much closer to trapezoidal. <S> Keep in mind that we are talking about the phase back-emf, not the line-to-line back-emf. <S> Sometimes the line-to-line back-emf looks close to sinusoidal while the phase back-emf doesn't. <A> For starters: PMSM: Permanent Magnet Synchronous Machine BLDC: <S> Brushless Direct Current Machine <S> BLAC: <S> Brushless Alternating Current Machine <S> Here is the thing. <S> These three types of machines are essentially the same. <S> PMSM is an umbrella term covering BLDC and BLAC as these two machine types are synchronous machines and differ only in their magnetic circuits (producing a trapzoidal shaped BackEMF or sinus) <S> You can control a BLDC machine with a SVM sinus controller and equally a BLAC machine with a quasi-squarewave controller. <S> They may not achieve the same performance results (as the stator flux will not match the airgap flux) but they will work <A> This is NOT my area of expertise. <S> But I don't think there is a terminology issue. <S> I think there are a lot more applications for so-called brushless DC motors than so-called permanent magnet synchronous motors. <S> Consequently, there are more offerings of them. <S> One small tip, if you are looking for PMSM's they are probably a subset of "AC motors," whereas the BLDC's may be under "DC motors," or their own category. <S> It doesn't seem like companies that make BLDC's make PMSM's or vice verse. <S> So finding both in one place may not be common. <S> But a lot of the places that sell induction motors also sell PMSM's.
| If you look in academic/technical literature like IEEE papers, then generally BLDC refers to brushless PM motors that have a trapezoidal back-emf and is driven by a six-step, trapezoidal drive, while PMSM refers to brushless PM motors that have a sinusoidal back-emf and are driven by sinusoidal waveforms.
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Is it possible to use a lead-acid battery as a power bank? I have an Oxygen 6v 4.5Ah lead-acid battery. Can i use it somehow as a power bank to recharge my cellphone? Your response and help will be greatly appreciated. <Q> There are many to choose from; LT1529 is a good choice since it comes in a TO-220 package, is stable with a single 22 uF output capacitor , and has a version with a fixed 5V output rather than needing two resistors to set the output voltage. <S> You can buy one, and the necessary output capacitor, on Digikey . <S> You may need to do something with the USB data pins ( connect a 200 ohm resistor between them , short them together, tie them to ground or +5V, or make an intermediate voltage with a potential divider) to make the cellphone charge at full rate. <S> Google should tell you what it needs to be for your particular cellphone; the first method I mentioned is the official USB standard but <S> not all cellphones support it yet. <A> Yes. <S> All batteries store energy, just like flywheels, compressed springs, inflated balloons, and food. <S> Power, being the time integral of energy, is what you get when you convert the stored energy in any of these things to some other form of energy. <S> Your battery has a nominal voltage of 6V. <S> All you need is a device that will convert that to a voltage compatible with your phone, and is compatible with the physical specification of your phone's charging port. <S> USB is popular and requires 5V, plus some additional signaling to indicate that the USB port is capable of more current (for charging) than typical (for running a keyboard). <S> 6V is close enough to 5V you might not need to convert the voltage, or it might force the smoke out of your phone. <S> It's probably out of specification, so it's hard to say without trying it, which I wouldn't recommend. <S> Or if you want to take a roundabout, inefficient path, you can connect an inverter to the battery, converting the battery's terminals to a physical specification compatible with that normally available from a wall outlet. <S> Then connect your phone's factory charger to that. <A> 6v is so close to 5V that a linear regulator will usually be about as good or better than a fancy switcher. <S> A simple lm323 regulator is rated for 3A max the 323 does in fact need more than 6v and won't work, but there's an LDO variant that I forgot the part code of that does. <S> You might have to short the data pins of connect a resistor to them or something. <S> Chargers do something odd with the data lines to tell the phone what they are. <S> Or, get another identical battery, put them in series for 12v, and use a normal car charger type thing.
| You should use a low dropout voltage (LDO) regulator, which is a type of linear voltage regulator, to reduce your battery voltage (which could vary from about 5.6V when empty to about 7.2V when charging) to the 5V required by your phone.
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Can I just connect two CR2032 batteries in parallel? I'm designing a device with a small current draw that will mostly sleep in a very low power mode and I need to use button cell batteries due to a limited thickness. I would like to double time between batteries needed to be replaced by using two or more CR2032 batteries connected in parallel, can I just connect them in parallel directly or do I need additional circuitry to prevent charging or discharging one battery from others while the device is sleeping? <Q> If this is for mass production, then no, you cannot do this. <S> The reason is that if a charged one is connected to a discharged one, the current will flow into the discharged one, and charging primary cells is a big safety no-no. <S> (For example, maybe you can accept the voltage drop of a Schottky diode). <S> But otherwise, use a higher capacity button cell. <S> There are many varieties out there. <S> If you are just goofing around at home, then by all means go ahead. <S> But you should make a point of removing both batteries before you replace them. <S> You might consider adding a resistor in series with each battery to limit the equalization current, just in case you accidentally put a fresh one in parallel with an old one. <A> It's a very common practice to do so. <S> No special circuitry is necessary. <S> If they are truly in parallel, both batteries will share the load evenly and there should be very limited (nearly zero) cross-feeding between the batteries. <S> Obviously, the two batteries should be at the same state of charge when you install them. <A> Cr batteries should not have more that 1uA charge current ... <S> and when you put these batteries in parallel they will charge and damage them selves internally making their internal resistance go up.... and possibly explode <S> It should not pass any compliance testing <A> Don't fear it! <S> Even for Automotive products (remote keys) there are series products (remote keys) that have two CR coins batteries in parallel. <S> Let me explain: In most parts of the world, these cells have to comply with certain standards, requiring some safety, e.g. they have to be short circuit safe. <S> It's not a problem for standard CR coin cells, as they will heat up until the electrolyte starts "boiling" (around 80°C); then they'll not be able to deliver any more heating current. <S> (Be careful with cheap "China" products - watch out for UL-certification or similar.) <S> So, what can happen when connecting two batteries in parallel: <S> The good case: connecting matching batteries (same brand and lot, both new - or similar discharge state): 1. <S> connecting correctly with same polarity: there might be a little balancing current at the beginning only, and the discharge will always distribute evenly among the parallel batteries. <S> 2. <S> if you connect one battery with reverse polarity: that's like a short circuit of two batteries in series, i.e. both will be discharged fast, with self heating up to ~80°C. <S> The bad case: the two batteries are significantly different in charge (e.g. one fresh and one empty) <S> 3. <S> both same polarity: the stronger battery will charge the weaker battery in this case, which may lead to internal short-circuit (from dendrite growth) of the weaker battery, leading to heating again and depleting also the strong/fresh battery. <S> 4. <S> polarities different <S> : that's the biggest problem, as in this short-circuit-discharge of the two batteries in series, the weaker battery will be empty first, while the stronger battery will still drive current, leading to an over-discharge (reverse-charging) of the weaker battery, that can make it leak due to dissolved housing. <S> (This is what you sometimes see in devices with multiple batteries in series when you do not switch off...) <S> I.e. <S> all scenarios, except #4, should be safe; for that case usually there is a mechanic reverse polarity protection (in worst case leading to a short-circuit of both batteries). <S> So, you have to decide how scatterbrained you might be handling your batteries. <S> Of course, you are free to add some protection circuitry to your device; I've seen this in the field also. <S> Alternatively you could use 1x CR2450 instead of 2x CR2032...
| Otherwise, the one with the higher voltage will initially attempt to charge up the other, which isn't desirable. If you can devise a circuit method to prevent charging of the cells under all circumstances, then you can do it. Yes, you can definitely put them in parallel. It is never recommended to parallel primary (non-rechargeable) batteries.
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What does a resistor do? OK, very basic question here. I read lots of books, searched quite a bit, and every description I read seemed to talk about the flow of electrons and right away go too deep in theory for me to grasp the basic principle of their use. I understand a resistor limits the "flow", so that an LED doesn't blow up for example. But I fail to understand exactly what a resistor does to current and voltage... Do resistors affect both current and voltage? In what manner? <Q> Electric flow is the motion of electrical charges through a material. <S> Resistance is the physical obstruction of these moving charges. <S> A certain amount of energy is required to keep these charges in motion, and since the energy drop is proportional to the amount of charge kept in motion, this results in a voltage drop across the material since electromotive force (in volts) is energy (in joules) per charge (in coulombs). <S> Since it is a physical obstruction, it also restricts the rate at which charges can move across a given point per unit time. <S> This results in a maximum current, since current (in amperes) is charges (in coulombs) per unit time (in seconds). <S> And as it turns out, if you apply more or less electromotive force across the same resistance, the current increases or decreases exactly linearly. <S> This gives rise to Ohm's Law, which states that electromotive force is proportional to the product of current and resistance, that is, \$E = IR\$. <A> It can be helpful to think of voltage as the pressure or force that is propelling the electrons through the pipe that is the wire. <S> Current is the number or amount of electrons passing a given point at any one time. <S> Think of electrons as ping pong balls passing through a tube, push one in and the ones already inside push one out the other end. <S> Doubling the length of the tube (series wiring a resistor) increases the force needed to push it through, so it limits voltage. <S> However, if you put the tubes side-by-side, then the same number of balls have to go through twice as many paths, limiting how many can go at once, and thus limiting current. <S> I know this is grossly oversimplified and does not account for all situations, but it can give your mind's eye a visual representation of the theory of electron flow and how resistors can affect such. <A> Hopefully this is simple enough: <S> Voltage arises from the potential energy in separation of charges (one node is positive with less electrons, one node is negative with more electrons). <S> Think about it like having a bowling ball (charge) on the ground, versus at the top of a ladder. <S> The ball at the top of the ladder has more potential energy, more voltage. <S> Current arises from the "flow" of charge. <S> Resistors let you choose how much current flows for a given voltage since you can think of wires as having no resistance (simplified). <S> In short:Resistors limit the flow of electrons, reducing current. <S> Voltage comes about by the potential energy difference across the resistor. <A> The mathematical answer is that a resistor is a two-terminal electric device which obeys, or you could say enforces, Ohm's law: <S> V=IR. <S> V is the voltage between the two terminals, I is the current flowing from one terminal to the other (through the resistor) and R is the value known as resistance. <S> For an ideal resistor, R is a constant and does not depend on V, I, or anything else. <S> Another way to describe Ohm's law is to say that the voltage across a resistor and the current through it are proportional. <S> The constant of proportionality is R, the resistance. <S> A fundamental consequence of physics is that resistors convert electric potential energy into heat. <S> So they tend to get warm when current flows through them. <S> Real resistors have maximum allowable power dissipation, and also, they may have R which depends on temperature slightly, and other shortcomings from the ideal. <S> As far as how resistors are made, well, real resistors are constructed from materials which have a conductivity somewhere in between insulators (dielectric materials) and conductors (such as copper wire). <S> If you can determine the path current takes through the resistor, making that path longer increases the resistance. <S> Making the cross-section wider decreases the resistance. <S> As far as what makes materials good conductors... <S> Well, generally good conductors have mobile electrons at the molecular level. <S> Good insulators do not. <S> Good resistors are somewhere in between.
| You can use them to limit either current or voltage, depending upon whether they are wired in series (one after the other), or parallel (sharing the same connection points, side-by-side. Resistors do just what their name says; they resist.
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May I power atmega32a-pu microcontroller from AA batteries? I need to provide +-4.5V to the microcontroller. That would be fine by having 4x rechargable batteries, but how is it with current? On each AA battery is written min.2550mAh. I am not sure but I think that max what I can put to the microcontroller is around 500mA. Could you please advice me how to do it? Should I use resistors? Am I right about the 500mA? <Q> Depending on your clock speed, you can run as low as ~2V. <S> The microcontroller will also use far less current at low clock speeds. <S> See the datasheet for more exact numbers, there will be a chart for the clock speed and minimum required voltage to operate. <S> The current consumed can be calculated per Megahertz, but may also not be linear. <S> Higher clock speeds (16-20Mhz) will consume far more than 8Mhz, and the difference is not linear. <S> Your microcontroller will not consume more than ~50mA <S> (this is what an Atmega328P at 16Mhz uses). <S> If it's consuming 500mA you are doing something horribly wrong, and probably a short circuit somewhere. <S> If you put 2 AA batteries in series, run at 8Mhz internal clock, you should get about 300-500 hours of run time. <S> This is about 2-3 weeks of constant operation, not considering any other circuit losses or silliness. <S> Also the capacity of a battery written in Amp-hours (or Milliamp-hours) does not really relate to it's ability to source current. <S> Internal resistances and different battery chemistry will determine it's current output and how much the voltage sags under load too. <S> For example I have a little 260mAh lithium ion 3.7V battery that can pump out 5 Amps if it wanted to - but this is not good for it <S> ;) <S> Your AA batteries, and batteries like the standard 9V "transistor" battery have terrible current output capability which newcomers to Electrical Engineering often don't know. <S> More in parallel will give more current output, and more capacity, but obviously only as much voltage as a single one. <A> The 2250 mAh rating of the batteries is a measure of the amount of energy they can store, not of the currrent that they can deliver. <S> The microcontroller will only draw as much current as it needs from the batteries - there is no need to add resistors, or otherwise limit the current supplied to the microcontroller (as long as the voltage is correct). <A> The "min.2550mAh" on the battery refers to its capacity, not its current. <S> The "mAh" suffix stands for "milliamp hours". <S> The label is claiming that the battery should be able to deliver 2.55 amps for one hour or 0.255 amps for ten hours or 0.0255 amps for 100 hours before it is exhausted.
| You would be lucky to get a few hundred milliamp out of a standard (NiMH) AA battery. The microcontroller will consume only as much current as it needs.
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Any simple power supply designs for a 24VAC transforrmer? I'm a beginner in Electronics Engineering. Any suggestions on simple power supplies with PCB designs which are run by a 24VAC transformer? Thanks! <Q> 24VAC is an inconveniently high voltage for many applications, with high line voltage and lightly loaded (the transformer regulation comes into okay) <S> its quite possible to get 45V or more at the filter output <S> (input to the regulator). <S> But if you do a parametric search for such regulators, they can be found, the details really depend on what output voltage and current <S> you want. <S> You would generally use a bridge rectifier and an electrolytic filter capacitor between the transformer and regulator. <S> It's easy to calculate the capacitor size and voltage rating from requirements. <S> For a switching regulator, something like an LT1767 would work, but there may be more appropriate parts for your requirements. <S> BTW if your 24VAC transformer happens to be center-tapped, the rectifier can use just two diodes and you get a nicer voltage to work with. <S> Very nice for analog projects. <A> How much current to you need for your low-voltage power supplies? <S> 24 Vac is the standard voltage used for HVAC applications in North America and parts of Europe. <S> That's what you have available, so that's what you use. <S> I have several power supply designs that I use for our various HVAC designs. <S> All use half-wave rectifiers so that the 24 Vac common line can also be the common line for the DC-powered logic stuff. <S> This makes interfacing to thermostats and solenoid valves easier. <S> The simplest designs use a two-stage power supply: the initial supply produces about 24 Vdc for driving relay coils, the subsequent supply is a simple resistor - Zener supply for the +5 Vdc rail. <S> The 24 Vdc rail is good for several tens of milliamps, the 5V Zener supply is good for 20 mA or so. <S> There is a separate power-up reset circuit that I haven't shown - <S> this also quickly discharges the main bulk storage capacitor when power is removed. <S> Neither of these DC supplies uses a SMPS because it simply isn't needed. <S> In addition, simple power supplies like this are far more reliable - this matters when you are building equipment that has an expected lifetime of decades. <S> But these techniques are NOT suitable if you need something like 5V @ <S> 1 or more Amps. <S> We get around that by designing our controller circuits to use only 5 or 10 mA from the +5V rail. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I'll check the component values when I get into work <S> but I'm pretty sure these are correct. <S> The 24 Vdc rail reliably supplies power for up to 8- JS1 relays or 3- <S> AZ2150 / AZ2150A relays (or a mixture of both). <S> Obviously, the relays have 24 Vdc coils. <A> You can use a bridge rectifier and a big capacitor to void most of the ripple, there you have about 32vdc. <S> If you need 5v or 3.3v you can use a switching regulator like the R-78 family from Recom .As <S> the unregulated voltage is over 28v, you cannot use the R78E family, I reccomend the R78C. <S> These are better than linear regulators because power dissipation is much much lower (no need to heatsink)
| If you use a bridge rectifier you can get both positive and negative voltages from a single transformer. Not many inexpensive switching of linear series regulators (definitely not LM78xx) are rated to safely operate with that high an inout voltage.
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simple way to measure wheel spinning speed with Adroid phone + magnet (and maybe Reed Switch) I'm trying to create the simplest/easyest/cheapest possible way to measure the speed at which a wheel is turning (like a bicycle speedometer) with a smartphone. On the phone the information will be further processed and used as the display of an open-rower-erg project . This is an open-source project. Although I can program a bit, I need opinions/answers from electrical engineers on the following options I'm considering for this measurement: 1) The simplest I could think of was attaching a small magnet to the wheel and placing the phone close enough to the wheel that the phone's magnetometer could record the magnet passing. Then the phone would process the signal to extract angular velocity. Is this idea feasible (the fact that no other tread mentions this makes me a little skeptical)?Could that magnet, systematically moving on the same direction, damage the phone or drain the battery?How precise can I be with the magnetometer? how fast could I sample it? 2) Another alternative would be to use Reed Switch, probably scraped from a cheap bicycle speedometer, and try to send the signal though the phone directly, either through the micro-usb or the headphone jack. Other treads( tread2 , tread3, tread4) suggest using the headphone jack would be better then the micro-usb slot. Is this really an alternative? How could I do that? 3) Others at the open-erg-tread1 suggested an Arduino, with the USB output. What would be the cost of that? This other tread5 suggests a similar idea to my point 1, above, sending information from Arduino to Android via magnetic field and the phones magnetometer. Any suggestions are welcome, I'm having a difficult time finding good references on google. Thanks in advance I don't have enough rep to put more than two links. The others are: tread3 https://stackoverflow.com/questions/20148791/android-app-monitor-reed-switch-usb-connected tread4 https://stackoverflow.com/questions/7806596/android-usb-bit-counter tread5 http://jdesbonnet.blogspot.com.br/2011/05/arduino-to-android-io-on-cheap-aka-poor.html EDIT: adding estimates for the max rotation speed (very rough, coming from imagining an erg, not actually measuring, please correct my numbers if you know better): x) dive lenght: 1.90m (~6"3') y) cog circunference: 91mm (= 29mm cog diameter * PI) z) turns per drive: x/y = 20.8 t) drive time: 0.6s - 0.9s (when rowing fast) finally: w) max RPS during drive: from 33.3rps (20.8/0.6) to 25rps (20.8/0.8) <Q> OK, I am brainstorming a little bit here. <S> One or two of the axes will have a nice sinusoidal wave as the wheel rotates. <S> This is an unmistakable signal. <S> You can count time between max or min, or count zero crossings to easily calculate RPM's. <S> You will have to know the wheel rolling diameter to calculate speed over ground. <S> The smart sensor on the wheel will transmit data to the smart phone via blue-tooth. <S> Could either transmit raw sensor data, or it could be processed into RPM's and only the end result could be passed to the smart phone. <S> You can also test it out with two smart phones. <S> Have one act like the sensor, and the other like the display. <A> Microcontroller measuring a hall effect sensor, a magnet, and a bluetooth module. <S> Unlike mechanical relays, they have very high trigger rates. <S> And frankly, this is how professional speed sensing is done. <S> From http://movableparts.org/category/arduino/ , which has multiple posts on a similar project. <A> The magnet and magnetometer idea is very simple and cheap. <S> It might not be fast enough though. <S> Look at this wireless anemometer for an example of a project that uses it already. <S> My phone samples its magnetometer at about 10Hz, I'm not sure how much faster you can read it. <S> Worth an experiment. <S> Don't worry about damaging the phone with a magnet nearby. <S> I wouldn't stick a strong magnet directly to the sensor, but at a few cm distance there will be no effect on anything apart from the sensor.
| You can figure speed based on Wheel Size and how many times the sensor is triggered in a predefined period of time. I believe the easiest way to get rotational rate is to put something battery powered and blue tooth enabled on the wheel (preferably near the hub) with a 3-axis accelerometer. Hall effect sensors measure magnetic forces, and are solid state devices.
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Lower 5V output to 2V max I am using a 5V 40 mA (max current) output on an Arduino to turn on (5V) or off (0V) as laser driver. The laser driver requires a max voltage of 2V to turn it on and flips at 0.8V. How can I lower the 5V to 2V for the driver? Thanks in advance! <Q> I think that you are reading the specification of the laser driver circuit incorrectly. <S> It would help if you can either post the exact specifications of the driver or <S> post a link to a web page that describes the driver. <S> Please don't link to a PDF file <S> - most people won't download a file to view it. <S> I strongly suspect that the laser driver circuit wants to see a TTL-compatible signal. <S> I'm inferring that from the 0.8V 'flip' voltage that you mention as well as the '2V max to turn it on'. <S> What that means is that the driver needs a voltage HIGHER than 2.0 Vdc to turn it ON and a voltage LESS than 0.8 Vdc to guarantee that it turns OFF. <S> Be sure to also connect your Arduino ground to the laser power supply ground. <A> I think the simplest solution is to use an adjustable voltage regulator like THIS one. <S> Use THIS site to find the value of the voltage adjustment resistor required to set the output of the LM317 to a specified level. <A> Based on what kind of laser you are talking about. <S> If it is a powerful 40A laser you will need some very beefy buck SMPS preferably deriving this 2V 40A directly from mains. <S> If you are just using small laser diodes this should work: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The TL431 is a voltage reference chip emitting a 2.495V nominal voltage, and diode D1 creating a diode drop of 0.7V, leaving about 1.8V to the laser. <S> The MOSFET is driving the laser.
| If I am correct, you don't need to anything except connect the Arduino output pin directly to the control input pin on the laser driver circuit.
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Eagle: Remove some text of a component label I routed a small board that will get connected to a BeagleBone Black using two 8x2 pin header connectors. As you can see on the picture there's two labels that are drawn out of the board layout so when I upload the board on OshPark the board is bigger that what it should be. How can I remove those two circled labels without editing the eagle component library? Do I have to generate my own gerber files and don't use the OshPark's website to generate them? I'm new to eagle. <Q> You could try to SMASH the part, select the label, and try to delete it. <S> Not sure if smash has that functionality, but I think it does. <A> Your best option is to edit the part. <S> Don't be intimidated! <S> Eagle parts are fairly straightforward, and you have the benefit of an already-built part to modify. <S> I've gotten to the point that I don't even use the libraries that come with Eagle. <S> It takes me longer to verify the library part than to create my own :) <S> If you do edit the part to remove the silkscreen labels, I suggest making Pin 1 a square shape. <S> This way you'll have some way of figuring out the orientation on the board itself. <S> Once you're done, you'll need to click Library -> <S> Update in your design. <S> Good luck! <A> Don't worry about OSH park or any other fab houses, whatever is outside of the board will get ignored. <S> You can use SMASH tool in Eagle, but it only works on some labels like >NAME and >VALUE, it doesn't work general labels. <S> When I design a connector, instead of using 1 for pin <S> I use >VALUE and then use 1 as a value when I place the component on the board, then I can SMASH it and move the 1 where I needed. <A>
| You can solve the error by going to: "Settings> Misc" and deselect "display via lengths", "display pad names" and "display signal names". Another option is to make the Gerbers, but even then you would need a gerber editor to remove the unwanted silkscreen...
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Manually Drill Vias into PCB As shown in the photo below, the PCB design was sent for fabrication without the drill data. Is it still possible to drill the via holes so we can save the PCBs? This is a 2 layer board, most of the traces are on the top, a few are below. Is it practical to send the drill file to a machining shop and have them drill the holes? Will the drill file already specify the holes, or is there a standard hole size which I can let the machining shop know? <Q> It will be possible to drill these if you have a very rigid drill press and carbide bits. <S> Back the PCB up with a sacrificial piece of laminate, get bright light and safety glasses and get very close to get the holes near the center. <S> Use the highest spindle RPM your drill press is capable of (30,000 RPM is not too much) and feed slowly, especially when the bits break through. <S> If you use steel bits (use the best cobalt steel if you try this) they may work okay, but they'll tend to skate around on the pad and not be in the center. <S> Replace them every 50 holes or so as they dull in the glass laminate. <S> You'll still have to find a way to solder them on the top or add additional jumper wires from top to bottom. <S> If that thing is a terminal block, for example, you'll have to add ugly jumper wires on the bottom to a number of the pads. <S> It won't be pretty but it might help you debug. <S> you're willing to pay a lot of money to compress schedule. <S> In the future, look at your files with a gerber viewer that includes NC drill files, make a checklist for what files have to be with a given number of layers PCB, and (IMHO) never do business with this supplier again. <A> The problem with drilling the holes is that they won't be plated. <S> Normally the top and bottom rings are connected through the board. <S> In addition to electrical routing, this allows strong solder connections. <S> In your case, these look like connector locations. <S> When you solder the connectors in place, you'll only be making electrical contact with the bottom side. <S> A few of your traces are connected to the rings on the top, so it's not going to work for you. <S> Another reason not to is if you are using a 4+ layer board. <S> In this case, the inner layers would have exposed copper at the edge of the drill holes, which could cause problems. <S> Sorry for the bad news. <A> Drilling will be very difficult, you are very likely to strip the whole copper pad. <S> Even if you drill it successfully you will have to solder a wire through each via on both sides. <S> Have a new board manufactured, it's not worth the hassle. <S> Plus you can improve your board, like adding labels to the connectors. <S> Speaking about connectors, you won't be able to solder them to the top layer because you will cover the pad with connector, so you really need to make a new one. <A> I say go for it, (what do you have to lose?) <S> As long as there are no internal layers that the vias need to connect to. <S> A nice little drill and high speed. <S> I've got some carbide tipped drill bits that won't dull when they go through the fiber glass. <S> But for a one-of just throw away the drill bit when you are done. <S> The copper layer may be interesting, (gummy) <S> If you had a little starter drill, maybe touch the tops off the copper with that. <A> You can definitely do this; I've done it before. <S> You just need patience and one of these: Sit down at the bench and very carefully drill a little bit, then reverse and back out. <S> Blow off the drill shavings and then repeat. <S> Try to keep it straight up and down :) <S> I've tried and succeeded using a standing drill press, but it's easy to screw up that way. <S> Sometimes the bit will catch your board or you put too much pressure on it and ruin it/break the bit. <S> By hand it takes a bit (no pun intended), but it lets you correct yourself. <S> Good luck! <A> The best facility to attempt to salvage these boards would be a PCB manufacturer. <S> Maybe you could send them back to the place that originally screwed up the order and ask them to drill + plate them in lieu of re-running the whole batch. <S> Or if they're off-shore, find a local place.
| If your drill press is not rigid enough or the bits are the slightest bit dull, they will break off and can go anywhere, so safety glasses are not optional. I doubt it's economical to get a machine shop to do this unless the boards are very, very expensive or
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MSP430 Launchpad connection of photoresistor and concerns regarding Launchpads protection I am using the MSP430 Launchpad and I have a question regarding photoresistors. The thing is that when you connect a LED with a pin in Launchpad, which you have configured, as output I read that you must use a resistor in series as not to damage the LED. When trying to find how to connect the photoresistor every source said that one must connect a 10kOhm resistor in series as not to damage the Launchpad. Regarding both the LED and the photoresistor as a simple resistor why in the first case we connect the resistor for the protection of the LED and in the second for the protection of the Launchpad? Also why can the Launchpad be damaged by its own voltage? I mean if a pin is configured to input and one tries to use an extremely large voltage then sure it will be damaged, but when you try to use 3.3V generated from its own Vcc whay may be damaged? <Q> I think you may be getting confused between the two devices. <S> With an LED a resistor is required to both protect the LED and the MCU's IO pin. <S> For a photo-resistor <S> (I assume you mean something like an old Cadmium Sulphide Cell?) <S> it is just a resistor which changes its resistance according to the incident light. <S> However, an MCU has no way of directly measuring that resistance, so a second resistor is required to form a Voltage Divider: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As the light levels change the ratio of resistance between R1 and LDR1 changes, and thus the voltage at the point between them (GPIO in this schematic) will change. <S> The MCU can easily measure that voltage using an analogue input. <S> The choice of resistor for R1 sets the light sensitivity of the LDR, and also the range of voltages the MCU sees for different light levels. <S> Different LDRs have different resistance ranges, and R1 should be sized to match. <S> You may need to experiment with different values to get the light range you desire. <A> MSP430G2 outputs can sink about 40mA from an LED at 3.3V, so a series resistor would be required if the LED was rated for less than this. <S> Also the maximum recommended current for all <S> I/O pins combined is 45mA, so resistors would be required if driving multiple LEDs. <S> The only restriction on inputs is that the voltage must not go above Vcc or below Vss by more than 0.3V. <S> A photoresistor is typically connected in series with a resistor between Vcc and Vss, so there is no chance of the input voltage exceeding these limits. <S> However a small photoresistor might be damaged if it had a bright light shining on it and the port pin was set to output. <S> Most CDS sensors are rated for 90mW or higher and so should survive such abuse (an MSP430 output can barely manage 2V * 40mA = 80mW). <A> An LED is not a resistor; it does not follow Ohm's law. <S> Basically, once you exceed the drop voltage, it will draw as much current as it can, to the point that it destroys itself. <S> This is what the resistor does: limit the current flowing through to a safe level. <S> Try connecting an LED directly to a 9V battery - it may last long enough to give a visible flash of light before it dies. <S> It may also become hot enough to be uncomfortable to hold. <S> As for the microcontroller on your Launchpad, again it's not the voltage but the current that will damage it. <S> If you were to connect an output pin to ground and drive it high (or connect it to Vcc and drive it low), you are basically making a short circuit. <S> Excess current will flow through the pin and can burn out the silicon of the chip or the internal wire connecting the silicon to the pin. <S> For the phototransistor, assuming you connect it only to an input, then the internal resistance of the input pin will greatly limit the current through the phototransistor and input pin.
| However, since GPIO pins can be programmed to either input or output, it's a good idea to put a current limiting resistor, just in case you accidentally set the pin as an output instead.
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What is the purpose of a PCB with no copper tracks, but only unconnected copper rings? I have a PCB which has the perfect size for a project of mine, so I would like to use it if possible. However, the copper plating on the back side of PCB only surrounds the individual holes (that is, no holes are interconnected). See picture right here: I find this strange. How can this be useful? I would definitely need some copper tracks with interconnected holes in there because some components need to be connected to each other. Am I supposed to make my own tracks somehow? I saw some stuff online about people who would insert multiple wires in the same hole to make interconnections, but this seems undesirable. I'd rather avoid that if there is some way to make tracks. <Q> What you've got is called a prototype board . <S> It is available at electronics suppliers everywhere and is obviously not meant for production. <S> Join things together any way that is convenient for you. <S> Many methods have been pictured. <S> Another common way is inserting a component lead beside its next connection and just bending it over to fit. <S> The results are typically quite messy, but it can take a lot more handling than a breadboard prototype. <S> Thus it is a common step before getting printed and etched boards made. <S> You can also find prototype boards in the same circuit pattern as the push-in breadboards, so you can simply transfer your circuit from one to the other, solder, and install. <A> The holes on a board like this are close enough and large enough that one can make solder bridges in order to wire components together. <S> It takes a bit of practice but can be decently reliable. <A> One technique is to use wires (or extra leads from resistors/LEDs etc.) <S> to create the "interconnected holes" you speak of. <A> In addition to using component leads and tinned copper wire to form connections between components sometimes I find insulated wire-wrap wire useful. <S> In the following example I've used a proper wrap tool and wire-wrap posts but it can also just be stripped and soldered. <S> Another good style of wire to use is the insulated solid core wire like you find in some styles of telephone cable. <A> Boards like this are typically used with jumper wires (the "multiple wires in the same hole" method you mentioned). <S> An alternative way to make tracks would be to use a circuit conductive pen as demonstrated in this video . <S> I personally like to use wire-wrap wire since it is thin and easy to insert several per hole. <S> Once you get used to using it, it can be very quick and efficient. <A> The presence of the rings will make things mechanically quite stable, so even uninsulated wires may be used in many cases without shorting. <S> In the absence of the solder rings, one might bend component legs in an effort to hold components in place, but such fastening would seldom be rigid. <S> When components are soldered together, the bent pins would keep them from moving very far out of position even if there were no interconnecting wires, but it would be the interconnecting wires themselves that actually held the components in position. <S> Such connections would thus be prone to much more mechanical stress than would be the case if components were soldered to the board, and would thus as a consequence be much more prone to failure.
| It's possible to solder a component to the rings, and then use wires or solder bridges to interconnect such components (in many cases I find wires easier).
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Convert a device that is powered by a 12V 2 amp ac-to-dc SMPS to a battery powered device I have looked up my requirements in all possible sources (Google, StackExchange, DiY boards, etc.) but I have not been able to find a clear answer to my problem. What I have is a device that needs 12V 2Amp DC Current and is currently powered by a AC SMPS which provides that. What i need to do is convert it to a battery powered device as where I am going is off grid but I have easy access to both Sealed Lead Acid Batteries and Li-On batteries that provide DC current well over 12V (13V-14V). I have been able to run the device by replacing the battery of an old UPS and running it off that but the size of the Unit becomes too large especially with the transformer (which is not such a big issue) and I need the device to be as effecient as possible. What I have found out is I need a DC-DC buck-boost SMPS (nobody has written this directly, but it is what I have gathered so far) to regulate, be effecient and to be able to handle the variation in voltages etc without harming the device (Please correct me if I am wrong here.) So basically the question(s) is: Is it possible ? (my current knowledge says, it should be) How would I go about it? What kind of a circuit (with components) am I looking at to do the job? PS: I humbly request all of you knowlwgable guys to be as forgiving as possible, as this is only my 2nd project involving power related stuff and 1st if u consider battries. <Q> Assuming you will still have automobiles in your off grid location; a very simple and cheap alternative is to buy one or two extra auto batteries. <S> The car already has the voltage regulated charging system in place. <S> And at a 2 amp draw, a typical car battery will run your device for quite awhile. <S> Nevertheless, aside of offering an alternative, back to the question itself: 1 <S> Yes it is possible. <S> Even with building your own battery fed regulator, you have some options. <S> You could make a battery pack that places several batteries in series til you have 3-5 volts above your rgulated output and use a cheap and simple linear regulator. <S> Or you could build a switch mode power supply as you mention. <S> 2 & 3 Pick a switch mode regulator chip and read its data sheet and app note. <S> Build the app note as a first step to gain experience building a power supply. <S> Here is a youtube video that will walk you through using the MC34063.The 34063 in boost mode will be easily able to handle your requirements and it has been used for some time in USB phone chargers and hubs, so there are a lot of schematics on the web easily found. <S> https://www.youtube.com/watch?v=qGp82xhybs4 <A> ... be effecient ... <S> Not at such a low drop. <S> Switching and linear will be neck-and-neck for efficiency here, and linear will be a lot simpler to use for the price. <S> Just make sure it is a LDO regulator, since you only have about a volt to squeeze by. <A> Then follow second or third stage. <S> 2) <S> If input is 12VDC, You can use DC-DC switching regulator with 2A or more current ratting(according to your requirement). <S> Use operating freq. <S> as high as possible to reduce space. <S> 3) <S> If you want to design battery charging/ <S> discharging(we say Battery management) circuit <S> then Simply you can use <S> MCP73833/4 <S> Stand-Alone Linear Li-Ion / Li-Polymer Charge Management <S> Controller(see <S> datasheet http://www.microchip.com/TechDoc.aspx?type=datasheet&product=mcp73833 ). <S> It will work very easily. <S> Keep in mind <S> Vbat track should be enough broad with PTH. <S> Be free to more help.
| According to your requirement, simple solution - 1) If input is 12VAC, You can convert it to DC by simple bridge and filter.
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Dual-supply op-amp LDO does not work, but single-supply does I'm designing an LDO with dual outputs (V out and -V out ), and only the positive output is shown in the pictures below. I'm having large trouble using dual supply op-amps in the design, and my single-supply one works fine and is shown in the first image. But my LDO using a dual-supply op-amp does not work at all in the same configuration, and I've tried a few different dual-rail op-amps. Am I missing something in the dual-supply design? The design with a dual-supply op-amp just saturates to the positive rail. The problem was that the dual supply could not get close enough to the rail. <Q> First, please check LT1007's datasheet , section "Unity-Gain Buffer Application (LT1007 Only)", it has BJT input with input voltage clamp diodes. <S> So if the output voltage goes beyond 1V + 0.7V, the input clamp diode will conduct, try to use another OP without input clamp, such as LF412, it has JFET input. <S> Another problem is just as others say, if your input voltage is two high, and your OP's positive output voltage swing is not high enough, then the OP can't modulate the LDO's output lower, the negative feedback loop is break then. <S> And the third question, if you want to make a "LDO", then the output voltage should very close to the input voltage <S> , right? <S> So \$V_{in}\$ should be close to \$ V_{out} = <S> V_{ref} <S> * \frac{R_{3} + R_{2}}{R_{3}} \$. <S> If \$V_{in} = <S> 10V\$, and the voltage drop on FET is low enough, the voltage feedback to the positive input of OP will be 5V or so, the OP will go to saturation, otherwise your voltage drop on FET will bigger, then it's not a 'LDO'. <S> I suggest to lower the input voltage or change the ratio of \$R_{2}\$ and \$R_{3}\$. <A> Dual supply opamps are not known for their ability to operate near the rails. <S> So, most likely is that the opamp can't get close enough to the \$V_{\text{dd}}\$ to control the FET. <S> FDS4465 has a \$V_{\text{th}}\$ of -0.2V to -1.8V and the LT1007 for example can only get to about -1.5V of \$V_{\text{dd}}\$. <S> If the FET you have has a nominal or less \$V_{\text{th}}\$, then try as it might, the opamp will not be able to reach a high enough output voltage to modulate FET conductance or turn FET off. <S> A high \$V_{\text{th}}\$ FET, and/or a drive circuit that can reach \$V_{\text{dd}}\$ should be used. <A> Neither one of those op-amps can reliably turn that particular MOSFET 'off' (by swinging within a few hundred mV of the positive rail), though one is better than the other. <S> I would look for a guaranteed swing within 200mV of the positive rail, since the -400mV minimum Vgs(th) has a temperature coefficient and you probably don't want it to go overvoltage on a cold day in Alberta. <S> You need to use an op-amp with "rail-to-rail output", and look carefully at the actual guaranteed specs.
| It's possible to add some parts (like diodes and a resistor) and work with what you have, but easier to just use a part that's made for the job.
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For a voltage divider circuit, why does \$V_{out} = I \times Z_{2}\$? For a voltage divider circuit , why does Vout = I*Z2 ? I understand that Ohm's Law says V = I * R, but I am new to electronics. Can you explain this better? Why is Vout not I*Z1? <Q> Because in \$V= <S> R\cdot I\$ the \$R\$ is the resistor across which \$V\$ is measured. <S> Since \$V_{out}\$ is usually measured against GND when the reference isn't marked, \$V_{out}\$ is the voltage across \$Z_2\$ in your example, and thus, the formula must be \$V_{out}=Z_2\cdot I\$. <A> Ohm's law <S> I <S> * Z1 gives the voltage difference (informally called the voltage drop or sometimes called burden voltage ) across resistor Z1. <S> So $$(Vin - Vout) = <S> I <S> * Z1$$ and $$(Vout - 0) = <S> I <S> * Z2$$ <S> By definition, the voltage of ground is zero volts. <S> (Vout-0) is the same thing as Vout. <S> For circuits where nothing is changing ( DC circuits ), we use R to stand for resistance . <S> But for circuits where the signals are changing ( AC circuits ), we use Z to stand for impedance -- <S> impedance is essentially a resistance that depends on the signal frequency and may have a phase difference between current and voltage. <S> , Ohm's law is usually first introduced as V = <S> I <S> * R instead of <S> the more general V = <S> I <S> * Z, to try to avoid confusion. <S> Electricity and electronics theory is hard to grasp because so much of it is abstract and invisible, without access to an electronics lab it can be difficult to develop a good intuition for how it works. <S> By the way, one of the features of electronics.stackexchange.com is very good integration with a third-party schematic editor / simulator website. <S> Look for the link below the schematic picture "simulate this circuit". <S> simulate this circuit – <S> Schematic created using CircuitLab <A> 'I' is the current flows through \$Z_{2}\$, because it doesn't give the reference of \$V_{out}\$ explicitly, \$V_{out}\$ should be relative to the ground, so it's just the voltage drop on \$Z_{2}\$, and you've known the Ohm's law, <S> so \$ V_{out} <S> = V_{Z2} = I * Z_{2}\$. <A> You can get there both ways. <S> The voltage across Z1 is I * Z1. <S> The voltage across Z2 is I * Z2. <S> Okay, that's nice. <S> Now what? <S> Z1's fixed reference is Vin. <S> So, the voltage at Vout = <S> Vin - I <S> * Z1. <S> A bit easier, Z2's fixed reference is ground. <S> So, that one is Vout = 0 <S> + I * Z2, or simply I * Z2. <A> It's all about how you define the voltage. <S> Voltages are not absolute, they are relative - without a reference, voltage is meaningless. <S> It's why voltmeters have two leads, oscilloscopes have a signal and a ground pin, etc. <S> You can think of the voltage as the force that is pushing the electrons in a circuit, but a force has to have something to push against. <S> Generally all of the pin voltages in a circuit are defined with respect to ground. <S> It is also possible to look at the voltages across specific components. <S> So Vout is really the voltage between the Vout pin and GND. <S> And Vin is the voltage between the Vin pin and GND. <S> Ohm's law relates the voltage ACROSS a resistor to the current THROUGH the resistor. <S> In this simple voltage divider, there will be a voltage across Z1 and a voltage across Z2. <S> Let's call these Vz1 and Vz2. <S> Since Vz2 is the difference between the Vout pin and GND, Vz2 <S> = Vout. <S> And Vin will be the sum of the voltages across the resistors, since when you put things in series the voltages get added together. <S> So Vin = Vz1 + Vz2. <S> Now, if the current out Vout is zero, then the current through both resistors will be the same. <S> Z1 and Z2 can then be combined into an equivalent resistance Zeq = Z1 + Z2. <S> The current through Zeq can be determined with ohm's law: <S> Vin = <S> I <S> * Zeq = <S> I <S> * (Z1 + Z2). <S> Then you can find Vz1 and Vz2 from this, Vz1 = <S> I * Z1 and Vz2 = <S> I <S> * Z2. <S> Since Vz2 = <S> Vout, <S> then Vout = <S> I <S> * Z2. <S> Substituting I yields Vout = <S> Vin <S> * Z2 / (Z1 + Z2). <S> Now, what about I <S> * Z1? <S> Well, there are several ways to look at this. <S> It is the voltage across Z1, Vz1. <S> It is not Vout. <S> However, it is possible to have the same numeric value as Vout if Z1 = Z2 and no current flows through Vout, since Vz1 = Vz2 in this case.
| It is also the difference between the input and output voltages, Vin - Vout.
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Recharging a Lead Acid battery I've an old lead acid battery with the following specs: 12V, 7.2Ah Valve regulated Sealed lead-acid type I've no idea about batteries prior. This battery has been in the attic so far and I would like to use it now. But, it is almost dead. I tried to recharge by applying 12V to it for one hour. But that doesn't seem to work. So, how can I recharge the battery or how to find out whether the battery is dead? <Q> Ouch. <S> My best guess is that your battery is toast. <S> Lead-Acid batteries suffer from several damaging chemical processes if they are allowed to remain discharged for long periods of time. <S> One of the real problems is something called "Sulfation". <S> You can use Google to search out that term and also look at the many attempts to build de-sulfation devices. <S> Wikipedia has a good description of the problem: Wikipedia <S> In general, I have good results with Lead-Acid batteries so long as they are kept charged. <S> We build our own chargers: <S> Equalize voltage is 15.0Vdc and float voltage is 13.5Vdc. <S> There are better 3-stage chargers available now but we have had great success with those chargers - some of that gear is more than 20 years old and the users routinely get 5 to 7 years of battery life. <A> If there is any crack or any physical damage to air inlet in VRLA battery, it will not work. <S> Please ensure. <S> There is still hope for its revival. <S> You have 2 charging options-1. <S> constant voltage at 14.4V for 8-10 hrs and then 0.3 A constant current for 4 hours.2. <S> constant current charging 0.3A <S> * 4Hrs+ <S> (-0.5A discharge * <S> 3hrs)+ 0.3A <S> * 2hrs. <S> The second option will improve the conversion of sulphate crystals to usable Lead dioxide. <S> Even after this if you don't get the voltage above 13V, scrap the battery. <S> Hope you will try. <A> Although sealed Lead-Acid batteries do suffer from sulfating another common problem <S> is that since they are operated under starved electrolyte conditions the electrolyte can end up as pure water that does not conduct. <S> (all the sulphate ions are in chemical combination with the lead in the electrodes). <S> To rectify this situation you may need to charge it with a high-voltage (20-30V) current limited supply (put 1k in series) and leave it for a few weeks. <S> If you are lucky there will be enough current to reconvert the water to sulphuric acid and restore the battery to operating condition. <S> After that you can charge normally. <S> Although everyone talks about "sulfation" as a problem, the normal operation of the battery converts the lead/lead dioxide in the plates to lead sulphate as it reacts with the electrolyte to create electricity, it is not in itself a problem condition. <S> What does happen if the lead sulphate is allowed to sit for a long period it can convert into a form that is not easy to transform into lead/lead dioxide during the charging process. <S> kevin
| The battery is charged with constant current until the current drops below the current limit value, then the charger drops into float mode.
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How to halve a frequency with analog circuitry while preserving signal integrity? The question is simple, the answer I'm not sure... How could I pick up an analog audio signal, divide it's frequency by 2 (one octave lower) and send it back on it's way? I could do this by sampling the input signal and generating an output signal that is half of it, but the generated signal would be a digital signal, not an analog, and when we speak of audio, tone means a lot, and the less differences there is between the input and output signal (besides the fact that the output is an octave lower, obviously...), the better... Is this possible in anyway? There can be digital circuitry, e.g. for sampling, but the least, the better. <Q> You could use a technique similar to digital processing, but without converting the signal to and from digital codes. <S> For example you could use bucket-brigades clocked in and out at different frequencies, or a tape recorder with two heads and transport mechanisms (real time recording, half speed playback). <S> However if you want to halve the frequency of an arbitrarily long and complex audio waveform then you have a problem - whether using analog or digital processing. <S> The signal is coming in twice as fast as it is being sent out, so in order to exactly preserve the original waveform you have to continuously store the input. <S> Eventually you must run out of storage space, then you will have to 'catch up' to real time and lose a chunk of the signal. <S> The resulting waveform might not be identical to the original, but it should sound virtually the same. <S> If you just want to change the frequency of a repetitive waveform (eg. <S> single note from a musical instrument) then you only need enough space to store a single cycle, or perhaps the duration of one note. <S> You then have to accurately detect the end of the waveform so that it can be repeated seamlessly, and decide what to do about its envelope (eg. <S> do you let the note play out at half real time, or force a faster decay?). <A> We are in a digital world and clearly this can be done digitally <S> so all that remains is for you to decide how many bits of resolution the analogue and digital conversion processes need. <S> 24 bits is plenty for me. <S> However, if you have the belief that any digital processing is unacceptable there is no useful analogue answer I could give. <A> DSP. <S> ADC, FFT, resample, IFFT, DAC. <S> You can translate frequencies in analog (add or subtract) but you cannot rescale them. <S> The only way to do that is in the frequency domain. <S> The only easy way to get to frequency domain with a reasonable amount of frequency resolution is with a digital signal processor of some sort. <S> You may be able to do all of this in one chip if you can get a DSP with built-in ADC and DAC.
| A sufficiently powerful digital system could simply include massive amounts of storage, or it could apply Fourier transforms to break the signal up into its component frequencies, halve each one and then recombine them.
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why does the parallel resistors with voltage source can be ignored as well as resistors in series with current source in term of equivalent circuit? I have a problem dealing with equivalent circuit. I don't know why when the resistors that is parallel with the voltage source can be ignore as well as one in series with current source. Of course the voltage in the load is no difference but what about the power. Having a resistor parallel with voltage source brings a difference in total resistance compared with having none resistor parallel with voltage source For example. If I place a resistor \$R_1\$ to 2 terminal a-b the thing will turn out for the voltage source having no parallel resistor $$I_1 = \frac{V_s}{R_1+R}$$ but for one having parallel resistor, it will be $$I_1 = \frac{R_p}{R_1+R+R_p}(\frac{1}{R_1}+\frac{1}{R_p}+\frac{1}{R})V_s$$ So although the voltage between a-b terminal is the same but the current is not. So how can it be considered equivalent as well as in the current source case. Please give me some proper explaination <Q> In your equivalent resistance equations, you are missing the fact that a voltage source has impedance of 0, and a current source has infinite impedance. <S> Anything in parallel with 0 impedance still results in 0 impedance. <S> Likewise, anything in series with infinite impedance still results in infinite impedance. <S> To help convince yourself, try putting some real number on your top example. <S> Pick some values for the voltage, Rp, and R. <S> Now put different loads on the result and compute what the voltage on the load is with and without Rp. <S> You will see that Rp doesn't matter. <S> The voltage across the voltage source is the specified voltage, by definition of what a voltage source is. <S> Put another way, a ideal 10 V source always gives you 10 V, whether someone else is drawing 10 mA, 10 A, or nothing from the same voltage source. <A> You run into a similar issue with Thevenin equivalents- <S> they yield the correct answers as to the voltage at the terminals, but the internal power dissipated will not be the same. <S> You just have to decide not to look too closely inside the 'box'. <A> I think( for voltage source) parallel resistor can be ignored because that resistor has same voltage as voltage source. <S> So if there is any possibility to flow current through this then total current of the circuit needs to flow through it as it needs to maintain voltage difference same to voltage source. <S> Then current doesn't have any other directions to flow so it's acted as series connection of resistor which is not acceptable. <S> That's why current doesn't chose resistor's path so it will act as open circuit which can be ignored.(For current source) <S> series resistor can be ignored because current flows through that resistor is just equal to the current source's supply and if other elements in that source's loop gain same current then series resistor can't drop any voltage,if it's drop any voltage <S> then current wouldn't be same as source for other other elements. <S> That's why series resistor can't drop any voltage as current source needs to maintain contact current flow. <S> So series resistor will work as short circuit which can be ignored in series connection.
| The current and the power delivered by the voltage source will not be the same with the parallel resistor, however the voltages at the nodes (and the currents beyond the resistor/source pair) will be calculated correctly.
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Does AC Power have Polarity? When I run 24VAC into a full wave bridge rectifier followed by a 220uF electrolytic capacitor to turn it into ~32VDC, the source has two wires. Does it matter in what order I connect the AC wires to the input to the bridge rectifier? If so how do I determine which wire goes where? I suspect that it's totally symmetric on the input side, but I'm am full of doubt when it comes to AC. Sorry if this is just a really dumb question. <Q> When combining two or more AC sources in series or parallel, the relative phasing is very important. <A> As long as you are dealing with a closed system (like a transformer secondary winding + bridge rectifier) then no, <S> AC is not polarized. <S> However, when dealing with outside power (like what comes out of the wall) we do consider AC-carrying cables to be "polarized". <S> There's the hot wire carrying the juice and the neutral wire carrying the return. <S> The hot wire should go directly to your device's switch / fuse, and any semi-exposed contacts must have the hot wire as protected as possible. <S> If you don't do this on production items you will fail UL certification and be wide open for lawsuits. <S> The common light bulb would likely fail many of today's standards, but it's been around too long to recall. <S> Devices that don't have a polarized wall plug will use a double-pole power switch to prevent the circuit being live up to the switch. <A> AC voltage has no polarity. <S> Therefore it does not matter how you connect the wires to the bridge rectifier. <S> Also, there are no dumb questions. <S> Feel free to ask about whatever is bothering you. <S> You will learn, as will others, and you will be safer for it. <A> At the atomic level, with the AC coming out of your transformer there is a herd of electrons which all run one way, then stop, turn round and run the other way half a cycle later. <S> Since the electrons are running in both directions the transformer output has no polarity. <S> The job of the rectifier is to round them up and make them all run in the same direction. <S> Now you know which way they're going they now have polarity. <S> That's why the input of the rectifier doesn't specify which way round to connect the wires, but the output does.
| When you're looking at an AC source in isolation such as in your question, indeed there's no polarity and you can connect the wires either way round.
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Why do all transistor latch circuits (that ive seen) use two transistors? I am looking at building a transistor latch to control a relay. I don't like soldering, so a few components as possible, especially those pesky transistors. I am looking around at various transistor latches online, and seeing things very similar to this: http://homemadecircuitsandschematics.blogspot.com.au/2011/12/simple-and-useful-transistor-latch.html I am having difficulty understanding the design. Why not just just have a single transistor, and feedback part of the output of the transistor back into it's base? EG: Please ignore the values, they are defaults. To be specific, I would like to use a BC547/8, and I am ommiting a lot of the components of this diagram. Why wont this work? Is there an elegant solution that only uses one transistor and passives? Thanks,Tom <Q> The base to emitter voltage is pretty stable at 0.6V, you are not adding any feedback with the resistor. <S> You actually built a emitter follower or common collector. <S> The main property of this circuit is that the emitter is pretty much at the same voltage as the base voltage. <S> If you are hinting at the non ideal behavior of the transistor, even the slight variation in BE-voltage is useless as <S> an emitter follower has a gain less than 1, it actually attenuates slightly. <S> So instead of adding feedback, you are actually just adding an extra load to the input. <S> By using two transistors for a latch, you actually add gain to the circuit and you add positive feedback. <S> The circuit in the image doesn't use the emitter follower, it uses two common emitters, which actually does have substantial voltage and curent gain. <S> Assume the input is low and the relay is not activated. <S> Because there is no current flowing through the relay, R3 is pretty much grounded. <S> When you start pulling the input trigger up slowly, at about 0.7V input voltage the leftmost transistor will start to conduct, causing T2 base to be pulled down. <S> T2 will start to conduct and causes an ever larger current through the relay. <S> Because of the relatively high current through the relay, R3 will be pulled up to a higher voltage and cause the input to be pulled up even further as a result. <S> Because of the resulting higher input voltage, T1 will conduct even more, pulling T2-base even harder down, causing more current through the relay, eventually swinging the relay voltage all the way up to the power supply rail. <S> Image found on Homemade Circuit Disigns Just for You <A> The analog-focused answers are correct, but there's also a digital way to look at it. <S> The circuit you linked to is basically two inverters connected in a loop, not unlike an SRAM cell. <S> Here's a simplified version to make this more clear: simulate this circuit – <S> Schematic created using CircuitLab <S> Homemade Circuit Designs added some resistors for current limiting and a capacitor for noise filtering, but the above is basically what's going on. <S> So another way to answer your question is to say that transistors invert when they amplify, and you need two inverters to store a stable state. <S> Adding feedback to a single transistor causes the output to either oscillate or attenuate, and neither of those does what you want. <A> In order for a circuit to behave as a DC-stable latch, it is necessary that there be a feedback loop with a positive gain at DC which is greater than one. <S> While bipolar junction transistors and field effect transistors may be combined with resistors in a variety of arrangements, all of them have a gain which is either negative or is less than one; consequently, no arrangement consisting of a single BJT or FET, plus any number of ideal resistors will be usable as a latch, at least in the absence of interesting parasitic effects in the transistor (e.g. a circuit which conducts when the transistor is warm but not when it's cold might behave as a "latch" if forcing the transistor "on" would cause it to heat up enough to become conductive, and if such conduction was sufficient to maintain it at that temperature). <S> Such a circuit could be used a single-transistor "latch" of sorts, using the presence or absence of oscillation as a means of storing data. <S> Indeed, one could in theory design a circuit which could store an arbitrary amount of data using one transistor plus a combination of resistors, capacitors, and inductors which could support a number of distinct oscillating modes. <S> The number of extra components that would be necessary to go beyond two states, however, would probably outweigh any real benefit from doing so.
| Adding certain kinds of non-linear components may make it possible to achieve greater-than-unity DC loop gain using only a single transistor, and adding capacitors and/or inductors may make it possible to design a circuit whose gain at some particular frequency would be greater than one when it is oscillating at that frequency, and less than one when it isn't.
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Calculating current draw from a battery using voltage and time information Is it possible to work out the current or power a device is drawing/using, based on the following information: Maximum capacity of a battery (48 Ah) A table of voltage readings over time (starting at ~13v ending at ~11v over a period of 40 days, sampled every day at 2pm, ignoring the end part of the test where the battery voltage drops off non-linearly) I am doing this to try and estimate how long the device would work on batteries of different capacities to the one tested, and I have a feeling it's not as simple as just dividing the battery's rated ampre-hours by the number of days it lasted and extrapolating...! Thanks in advance for your help. <Q> Measure the current directly. <S> Put a low ohms value resistor in series with the load and then measure the voltage drop across this resistor. <S> Trying to do this by the schemes you are trying are at best an experiment. <S> You are in selection mode not high volume production mode. <S> A few extra components and a little extra circuitry to do it the direct and correct way is the best. <S> The cost is inconsequential. <A> In principle you can, but you need to know the discharge rate of the battery depending on the load. <S> Also, its series resistance will cause a drop that depends on the load current, so you will need to compensate for some factors (ISR, discharge rate) in order to get a reasonably accurate estimation. <S> Note that the battery capacity will change depending on the load (not only the duration, the actual capacity). <S> You can see several rate-of-discharge graphs, and that should give you an idea. <S> That's what I suppose is done in mobile devices to estimate battery lifetime. <S> If you can measure the voltage and current together, your estimation will be a bit more accurate. <S> Also, a higher sampling frequency will help. <S> Do you have any reason to limit it to once a day? <A> First I have some comments / caveats. <S> Batteries have self-discharge, and discharge due to external load. <S> Also, the capacity depends on the load. <S> Larger loads will give you less Ah than smaller loads. <S> But as the load gets smaller, at some point the capacity becomes nearly constant. <S> And your load qualifies as very small, since the battery was able to support it for 40 days. <S> Lithium and NiMH battery capacities are slightly less dependent on load than lead acid batteries, but all of them are in the "small load" category when discharged over 40 days. <S> However, in a 40 day period, depending on battery type, there may have been some significant self-discharge. <S> I assume it is a lead acid battery, but what is the subtype? <S> flooded, or gel cell, or absorbed glass mat (AGM)? <S> The latter two have relatively lower self-discharge. <S> I told you that to tell you this: Very often, if self-discharge is negligible (not sure in this case), <S> and the load is small, you can absolutely just estimate the life of the second cell based on the capacity of the first cell (assuming same type or subtype). <S> In other words, if you get 40 days from 40 <S> Ah, you will very likely get 60 days from 60 Ah, and 20 days from 20 Ah as long as the battery type is the same. <S> One last thing: <S> self-discharge has a strong dependency on temperature. <S> A battery which will hold 80 percent of its charge for a year at 25C, may discharge completely in 3 months at 60C. <S> Those may not be the real numbers. <S> I am kind of making that up for illustration purposes. <S> But it is very dramatic, and applies to many battery types. <S> -McKenzie <A> The Amp-Hour rating of any battery is provided by the manufacturer and reported as a result following a very specific test documented within the datasheet, typically this tests consists of a constant load over time. <S> If your load is not constant (a periodic, pulsatile load), the Amp-Hour rating provided by the manufacturer should be viewed as guidance to a ball park estimate of its performance. <S> Since you said the test has already been done, and you are trying to extract meaningful data from old logs, you will need to identify the load which contributed to the readings overtime. <S> If you still have control of taking new data I would suggest to apply a small series resistor and measure the voltage drop across of it. <S> This will provide you with Current and Power information. <A> Maybe this isn't what you're asking, but if you have a battery with a given ampere-hour capacity and - with a constant load resistance - you know how long it takes it to go from fully charged to fully discharged, to a first approximation a battery using the same chemistry but with twice the capacity serving the same load will last twice as long, one with half the capacity will last half as long, and so on... <A> The right way to do this, is to use a Coulomb (charge) counter. <S> This circuit is available as IC and typically is very tiny, leading to a very small foot print. <S> The rationale is that most likely there will be spikes in current consumption (especially if the load is some uC/uP which alternates between states of high activity and low power idleness). <S> So you will most likely miss key events and underestimate the battery current, if relying solely on sampled values. <S> The Coulomb counter performs integration over time of the charge transferred, hence allowing for low frequency sampling. <S> All you need is a stable time base. <S> Then the reads can also happen a-periodically, as long as you keep track of the time passed inbetween successive reads. <A> TRY a hall effect current sensor to drive the coloumb counter <S> this will give good accuracy AND good dynamic range AND low volt <S> drops <S> If your currents are lower than the halls rating by a factor of 2 or more then run 2 turns etc <A> A lot of answers here are saying that you should include some current measurement or coulomb counting, which doesn't seem to answer the question, in my opinion. <S> If you only have periodic voltage measurements and the load current is small, you can approximate the state of charge of the battery with a SOC-OCV (state of charge - open circuit voltage) graph. <S> You can probably find this graph for whatever chemistry battery you have and find the SOC that corresponds to a given voltage.
| If you have a fairly constant load or an accurate enough model, you can indeed predict the performance of your battery depending on voltage measurements.
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Calibrating the oscillator of a microcontroller I would like to do precise timing of events which trigger hardware interrupts on a microcontroller. I am aware of the stability issues associated with microcontroller oscillators and I think a 30ppm (or smaller) stability in 50 MHz (or greater) oscillator will be good enough for my project. But an oscillator can be very stable and yet be inaccurate as a time keeper because the frequency of oscillator may not be what one thinks it is. So my question is: How does one go about calibrating the external oscillator of a microcontroller board? Thank you! <Q> If you are willing to do the measurement over a period of time (one midnight to the next), the frequency of the US power grid is maintained to exactly 60 Hz over a 24-hour period as discussed here . <S> The frequency will vary during the day, but it is trimmed at the end of the day so that exactly 5,184,000 (60*60*60*24) cycles will occur from one midnight to the next. <S> To count this, you will need a zero-crossing detector. <S> There are many such circuits on the web, here's a fairly simple one: <S> Count up how many cycles your clock is putting out in 24 hours (perhaps after going through a divider), and then compare those counts with the counts off the power lines. <S> Run this the first time <S> and it will give you the frequency. <S> Run it another day <S> and it will give you the drift. <A> Let me expand a bit on tcrosley's excellent answer from yesterday. <S> Let's define first <S> how accurate you wish to be. <S> You say 30 ppm, so that means more accurate than two seconds (exactly 2.592 seconds) drift per 24 hours. <S> You could follow tcrosley's idea of locking on to the power grid. <S> And with it, you could do continual adjustment of the device if it's continuously connected to the grid. <S> There's a couple of other alternatives. <S> You could use GPS time. <S> There's an excellent article "The Science of Timekeeping", which is Hewlett-Packard's Application Note 1289 and by quick googling, it's available for example here . <S> Did you know that Hewlett-Packard started by doing oscillators and frequency counters? <S> Using the atomic clock provided by GPS might be easier than you think. <S> Your computer might already use NTP protocol to keeping its time in sync with UTC. <S> If not, you can install a NTP client to your computer. <S> NTP servers generally derive their time from GPS atomic clocks and in any case the accuracy is way better than 2 seconds per day. <S> So calibrating (checking) the speed of the microcontroller's oscillator might be as simple as sending the microcontroller's "time" periodically to a PC and writing a small PC application to display the time and compare that to the PC's NTP corrected time. <A> Example of affordable tool is Stanford Research SR625 Frequency Counter . <S> The tool is accurate to 0.1ppb <S> (part per Billion), if I calculated this correctly. <S> The price tag is just $7k, but you can rent it for a lot less.
| The most standard way to characterize and calibrate self-made oscillators is to use proper calibration instruments, like frequency counters with a stable time base (say, Rubidium clock).
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What's the correct term to describe a conductor which doesn't always obey Ohm's law? Non-ohmic comes to mind, but that gives an impression that the conductor never satisfies Ohm's law. Is there a better term which describes a conductor which might satisfy Ohm's law at certain voltages and temperatures, but not at others. <Q> I've heard such things as thermistors called "non-linear resistors": <S> Thermistor <S> The idea is that a linear resistor has a linear V-I curve, but a non-linear resistor does not. <S> In the case of a thermistor, it's due to heating, but in the case of, say, a varistor, it could be due to voltage: Varistor <A> Since an "ohmic" conductor is defined as one with constant resistivity, and since all purely elemental conductors exhibit a non-zero temperature coefficient of resistance, all elemental conductors' resistivities change when the current through them changes their temperature. <S> It follows then, that - since resistivity of elemental conductors isn't invariant and that for a conductor to be ohmic, by definition, its resistivity must be constant - <S> there are no elemental conductors which are ohmic. <S> There are, however, some alloys which come close to being ohmic over wide temperature/current spans with Manganin , I believe, leading the pack. <A> I think there is no specyfic term that describes materials or circuits which "satisfy Ohm's law at certain voltages and temperatures, but not at others". <S> "ohmic" and "non-ohmic" are terms describing voltage-current relation. <S> An ohmic contact is a non-rectifying junction: an electrical junction between two conductors that has a linear current–voltage <S> (I-V) curve as with Ohm's law. <S> Low resistance ohmic contacts are used to allow charge to flow easily in both directions between the two conductors, without blocking due to rectification or excess power dissipation due to voltage thresholds. <S> By contrast, a junction or contact that does not demonstrate a linear <S> I-V curve is called non-ohmic . <S> Non-ohmic contacts come in a number of forms (p–n junction, Schottky barrier, rectifying heterojunction, breakdown junction, etc.). <S> Source: <S> http://en.wikipedia.org/wiki/Ohmic_contact <S> I think when you describe material or circuit - you should keep in mind circumstances/conditions and purpose of your description. <S> All depends on context. <S> Are you considering DC or AC? <S> Do you want your description to describe steady state only or should include transient response ? <S> For example - lightbulb in some circumnstances can be described as "ohmic". <S> If current is very low and filament temperature changes can be negligible. <S> However - lightbulb that works in normal (for bulb) <S> conditions is "non-ohmic", because filament resistance changes a lot. <S> So, if you want to describe some material clearly - you may have to use more words, like "thermistor behaves like ohmic for constant temperatures". <S> Sorry about my weird examples <S> (nobody considers lightbulb behavior at constant filament temperature or thermistor at constant temperature). <S> I had no better ideas. <A> First and formost, to avoid confusion, it would not be called a 'conductor'. <S> The term conductor is generally interpreted as a 'device' that either a (for the design) negligable resistance, or at least a small, predictable and ohmic resistance. <S> So, if the range in which it is (low) ohmic fals within the specification of the design, it will just be a conductor. <S> As such, in any other circumstance, such a device would basically be called something else... <S> Like a thermistor, semiconductor, etc. <S> There isn't a general term for it, apart form 'not a conductor' or 'a device not following ohm's law'. <A> I think, the best method for desribing devices with voltage-current relations that do not follow Ohm`s law is to use the mathematical description (formula) for this voltage-to-current relationship. <S> Well-known example <S> : pn diode with an exponential expression. <S> However, it is to be noted that, of course, Ohm`s law can and must be applied (in a specialized form): (1) Static resistance for quiescent values: R, <S> st=V <S> /I and (2) Dynamic/differential resistance for small-signal values: r,dyn=v/i <A> By dynamizing the initial ohmic resistance or adding a dynamic voltage <S> we can modify its initially linear IV-curve to obtain various (even negative) non-linear resistors. <S> Here is the philosophy of this approach.
| A better term which describes a conductor which might satisfy Ohm's law over certain temperature ranges, but not over others, might be: "partially ohmic" or, in some cases, "superconducting."
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Mounting hole on a PCB Here is a picture of 3 mounting holes on three different PCBs. The red one is what I prepared on Eagle CAD and used "hole" button on the left side menu , something like this: Do I need to use "via" in order to have the same style mounting hole on the other PCBs? As far as I can see, they are vias on the blue PCBs. Isn't it better to isolate the mounting holes on a PCB? What I am thinking to have a risk to get an unwanted signal or voltage to your ground plane as it is unisolated? or Does it improve grounding or tying grounds of unconnected PCBs? Is it a gold finish on the mounting holes? Why would I want to have a gold finish on a screw hole that would increase the cost? The mounting holes on blue PCBs have several holes on it, what is the purpose of having holes on the screw holes? Fabrication-wise, it seems it would slow down the production a little bit as each individual mounting(or screw) hole requires drill holes on it. The upper blue PCB is from a projector, the bottom one is from a harddisk of a PC. <Q> The mounting holes that have the large copper pads around the hole are fabricated that way because it is desired to have that connected circuit be conductive to the mounting screw. <S> In most cases it is the GND of the circuit that connects to the mounting hole pad and there is a desire to have such GND connect into the metal chassis to which the board mounts. <S> In the case of a multi layer board with an inner plane layer the vias may also be used to connect the mounting hole pad to that inner layer. <S> In the past it was much less common to see mounting hole pads with these via holes. <S> Instead the mounting hole was built as as a large plated through hole to connect to the opposite side and/or inner layers of the board. <S> However, as in the examples you show, note that the more modern type mounting holes do not have plating in the large diameter screw hole and thus the need for the vias to provide the electrical connection. <S> Plating removal in screw mounting holes is primarily done for one reason. <S> The sharp threads of a screw in the hole can cause small particles of metal to come off the plated hole. <S> This is particularly true when boards might be removed and re-installed during testing and/or repair. <S> These small particles can come out and appear on the circuit board or float around in the electronics enclosure. <S> In the advent of surface mount components with vary narrow lead spacings these small metal particles from the screw holes can lead to shorts on the circuit leading to either intermittent circuit operation or outright failure. <A> To get the same holes in your eagle project go to your schematic and click on add (or type add in the text box and hit enter). <S> Now to go the holes library, then you see there is a MOUNT-PAD-ROUND and a MOUNT-PAD-SQUARE library. <S> These are the library's you are looking for. <S> In the schematic you can connect the pad to a signal, which is mostly used to connect to ground (not sure about this). <S> And yes on the blue PCB's they are also functioning as via's probably because the screws are connected to ground. <S> The holes in your mounting hole is probably there so more current can flow through. <S> To get these holes also in your eagle project you probably need to make your own library for that. <S> The golden finish on the mounting hole is there because it is considered as a via or regular hole. <A> Another reason for the vias around the mounting hole is wave soldering. <S> When wave soldering solder covers all unmasked copper and therefore the whole mounting hole would be covered with solder. <S> Manufacturers usually mask those holes with some tape manually which costs money. <S> This way you have the electrical connection thanks to those vias that can be soldered through and the mounting hole is preserved and manufacturer didn't have to do extra work.
| The small via holes in the mounting hole pads are designed to electrically connect the mounting hole pad to the pad on the opposite side of the board.
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Turn pull-up into pull-down I have come across the following problem a few times in the past: You have a break-out board where some pins have already been pulled up or down for convenience, e.g. RESET pulled high to keep the board running even when you don't connect the input pin to your MCU. Now you want the opposite standard behaviour when the pin is not set (e.g. in sleep mode or when no power is applied to the MCU). How can you achieve this? The only answer I can think of is to (physically) remove the pull-up and hook a pull-down between the MCU and the break-out board. I would rather not mess with the SMDs on the break-out board, so is there another solution? A concrete example is an audio amplifier that is by default on and needs to be pulled down by a pin to switch off. Now when I remove power from the MCU, it automatically turns on again and pops audibly. Yours gratefullyJerome <Q> But, you could try just fitting a pull-down resistor of much lower value than the pull-up. <S> The two resistors will form a voltage divider, but with one resistor much lower in value than the other, the lower value will "win" and pull the pin down. <S> The disadvantage of doing this is higher power consumption then the MCU pulls the pin HIGH. <S> Make sure the pull-down resistor does not draw more current from the MCU pin than it can supply (check the datasheet) when pulling HIGH. <A> The best is removing the original pull-up or pull-down. <S> If not and if you have enough space, you can add another resistor. <S> There is a rule that the new pulls should be at least 15 time the original pulls. <S> That say, if your pull-up is 15K, then if you want to add another resistor to change it to pull-down, your resistor value should be 1K or so. <A> If I read you right, this should work: WAS <S> is on top, <S> IS on the bottom.
| The best way is (as you suggest) to remove the pull-up from the PCB and fit a pull-down elsewhere.
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How to increase range of zigbee? I have a Zigbee Pro with Arduino whose range is 2 miles as per specifications. But practically its maximum range is only 200 meters. I am using XCTU software. Is there any setting/configuration in XCTU to increase range?If not possible to increase range, is there any module which is compatible with Arduino? <Q> 1- Depend on the manufacturer of your ZigBee <S> maybe it provides a setting/configuration to increase the range. <S> You should provide information about manufacturer and model. <S> 2- <S> Technically it is possible to increase the range of your signal improving the: Antenna (gain, sensitivity to body effects <S> etc.)- Choose an external antenna with a good gain and sensitivity factors. <S> Usually external antenna connected through UFL or similar has better gain and sensitivity values than a on board antenna. <S> Sensitivity: <S> Lowest input power with acceptable link quality. <S> Channel Selectivity: <S> How well a chip works in an environment with interference. <S> It must be checked by your self or there are zigbee modules with scan function to determinate the best chanel. <S> Output power - Usually fixed by the model. <S> Choose a Zigbee model with enought ouput power. <S> see Environment (Line of sight, obstructions, reflections, multi-path <S> fading)- Place your devices in the best possible location. <S> Data rate: to reduce the data rate can also increase the rage of your zigbee network. <S> Wavelenght: <S> Zigbee module with longer wavelegth (low frequency) have more range. <S> There are also expensive commercial solutions to extend the zigbee network even more than 10 milles ( old article ). <S> 3- <S> Simple formula to calculate theorically the range <S> , so you can get an estimation about the effect of your antennas gain and Zigbee model (Tx & Rx Power & wavelength) with the range: <S> Pr is the power available from receiving antennaPt is the transmitted power. <S> Gr is the gain of your reciever antennaGt is the gain of your transmitter <S> antenna.lambda is the wavelength -> <S> c/freq of your Zigbee <S> 4- <S> You should not have problem with different models of ZigBee, if you have Arduino compatible with IEEE 802.15.4 <S> ( List_of_Arduino_boards_and_compatible_systems ) <A> Probably ZigBee devices' range is 2 km only in deep space. <S> Depending on the power of your module (high-power ones can transmit up to 20 dBm (100 mW, although it's not legal in all countries). <S> Note that this is the same power used by WiFi, so it's reasonable to expect a similar order of magnitude for range. <S> Also note that your range will decrease due to: obstacles; reflections (may increase or decrease actually); background noise; temperature; <S> humidity. <S> For longer ranges you might be better off with lower frequencies (868 MHz?) and possibly higher power. <S> Finally, even if you can achieve the specified range, don't expect to have a reliable communication. <A> I've been fooling around with the HopeRF RFM69HW module, which supports lower frequencies (433MHz penetrates obstacles better than 900MHz or 2.4GHz), lower bitrates (down to 1200 bit/s for robust reception), and reasonably high power (18dBm, or 20dBm at a reduced duty cycle). <S> The best result I've gotten so far has been 320m range with simple quarter-wave wire antennas, but other folks have reported ranges of 540m at 38400 bps, and 2400m at 1200 bps, so my setup is probably not very good.
| There are circuits like the SKY65336 which increase significant the range ( see ).
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Why do we need transforms (Fourier, Laplace, Z and wavelet etc.) for a signal to analyse? Why do we need transforms (Fourier, Laplace, Z and wavelet etc.) for a signal to analyse? Is it necessary for practical calculations and analysis? <Q> No, transforms aren't "necessary", but they do make some types of calculations much simpler and more convenient. <S> It is possible to do all computation and analisys of a signal in either the time domain or the frequency domain. <S> However, some operations are much simpler and more intuitive in one than the other. <S> This can be illustrated with something as simple as a R-C low pass filter. <S> If it is on the input of a signal that you are measuring, then you may want to know how long of a delay it adds for the result to settle within some error of the input signal. <S> That is best done in the time domain by writing the equation of the output signal in response to a unit step input. <S> This could be done in frequency space, but would be quite convoluted, and you'd have to express some of the things you inherently know in the time domain in the frequency space. <S> On the other hand, if this is a audio application, you may want to know how the filter effects the amplitude of different frequencies. <S> That is most easily and intuitively done by expressing the filter in frequency space. <S> The response to any particular sine input could be computed in the time domain, but it would take much more computation and not be as intuitive. <S> In summary, both time domain and frequency space are whole and consistant ways of looking at a signal, but each gives different intuitive insights, and each makes different types of problems harder or easier. <A> All transforms are tools to make analysis easier. <S> They are tools that engineers, scientists, and mathematicians have developed over the years to help make their jobs easier or to help them gain a greater understanding of the phenomenon they are looking at. <S> The Laplace transform, for example, makes solving differential equations easier. <S> The wavelet transform helps you analyze both frequency and time domains at the same time. <S> I think the word you used - "practical" - is key. <S> These transforms are used to take cumbersome problems and make them more practical. <A> There are some cases where frequency is directly important, such as radio communication and audio reproduction. <S> But in general, the Laplace and Fourier transforms are nice because they convert certain difficult mathematical operations into easier ones: Differentiation -> <S> Multiply by s Integration - <S> > Divide by s Convolution of two response functions - <S> > Multiplication of two transfer functions <S> The last one in particular is very important for dealing with feedback control, so much so that Laplace transforms are used even when talking about time-domain phenomena like step responses. <S> This is applicable to many areas of engineering, not just signal analysis. <A> I'm sure you can do it in any domain, but it will be much more complex. <S> Think of a filter. <S> What does a filter do ? <S> Think about how difficult it would be to explain to someone or analyse the circuit in the time domain. <S> Sometimes, its just easier to work with one domain than the other. <S> You can solve an RLC in the time domain but it will be a 2nd order differential equation equation. <S> You can absolutely solve it using calculus and take the derivative of this and that. <S> Or you can transform it into the S domain (Laplace transform), and solve the circuit with simple algebra and then convert your results from the S domain back into the time domain (inverse Laplace transform). <S> Some domains are just the digital equivalent, like the Z domain is to the S domain. <A> Transforms (Fourier, Laplace) are used in frequency automatic control domain to prove thhings like stability and commandability of the systems. <A> These transformations are mainly adopted to solve differentaial equations under different boundary conditions or you may call limits. <S> For Laplace u can go for positive limits up to infinite but in case of Fourier limits may be from minus to plus infinity. <S> Moreover the kernal also vary for both the functions as it contains iota in exponential of Fourier kernals whereas not in Laplace.
| Transformations are useful because it makes understanding the problem easier in one domain than in another.
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Electrical switch/relay that will stay on once triggered (regardless of future inputs) I am looking for an electrical device that will allow for me to take an input (that will change from high to low to high again) and have it trigger an output (turn high when my input is low) but stay at this high state regardless of any future inputs. Does something like this exist? Sorry if this is a duplicate, I just don't know what terminology to search with. Thank you in advance, Julia <Q> What you describe is simple latching behavior. <S> There are lots of ways to get that. <S> You can wire up a relay so that it keeps itself energized once it is energized. <S> A SCR is a electonic device that inherently latches once triggered, and stays on as long as some minimum current is supplied. <S> There are various ways of wiring up transistors to do this. <S> Basically anything with a gain greater than 1 can be arranged to latch. <S> There are digital gates specifically for this purpose, and of course a microcontroller can be trivially programmed for this function. <A> It's possible to buy magnetically latching relays, if you don't want to do this electronically. <S> Or maybe a set-reset latch (SR flip-flop) would do it, if you don't need to drive a relay. <A> There are a huge number of hits here , which should get you started, and if you want to go the mechanical route, here are a bunch of latching relay circuits. <A> The easiest way to create a latching relay is to use one of the contacts of the relay to hold it energized, like this: Pressing the NO (normally open) pushbutton S1 <S> operates the DPST (double-pole, single-throw) relay K1, and contact P1/S1 <S> keeps it operated, even after the pushbutton is released. <S> Pressing the NC (normally closed) <S> pushbutton S2 release the relay. <S> The second contact P2/S2 is used to control whatever circuit is needed. <S> One could of course also use a DPDT (double-pole, double-throw relay as well.
| If you're OK with an electronic latch, an SCR (Silicon-Controlled Rectifier) circuit would do the job. Things like that do exist, and they're called, generically, "latches".
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What things to think about when selecting a relay for controlling small valves? We are creating a novelty drink dispenser that controls a set of 8 pumps with a Siemens Simatic Step7 1200 PLC. The decision to use this PLC was based on our need to have a very stable prototype. Now we are nearing a stage where we are going into mass production, and the PLC feels like overkill. It has many features that we don't use and it costs a lot. We would like to replace this with a simple relay-card. We have limited experience with making electronics and we will most likely use a professional service for the development and production of this card. However given the importance of this card for our business, I would like to ask some general advice here first. For example, what are the most important things to think about when selecting a relay for controlling small valves, and what features should we know about in a valve that can make our product better? Of course, we want them to be able to open/close as many times as possible without failing, and when failing, never failing in an open state. Also if there is any way to detect wear/failure that would be awesome. For reference, our valves are 24VDC 5W solenoids. <Q> Short answer: don't use relays. <S> One of our past projects was a liquor dispensing system and we drive the solenoids with TO-220 MOSFETs. <S> I'm not aware of any failures in many years of operation. <S> Be sure to include the back-EMF diodes across the solenoid terminals. <A> The most important things to consider are the specifications which will determine life (which must be adequate) and to make sure EMI does not affect the rest of your circuitry. <S> This is particularly important with a bespoke design with limited production because respins will be very costly. <S> When I did the controls for a highly configurable commercial espresso etc. <S> machine some years back <S> , the relays were selected to have a life of more than 10 years (there were PID boiler temperature controls, grinder controls, steaming controls etc.). <S> A LOT of different parameters to be set. <S> Sometimes a mechanical relay backup and a solid-state relay combination is called for. <S> If you do use mechanical relays, be sure to use adequate suppression on the solenoid valve coils (a diode or TVS at a minimum) and preferably use relays designed to meet European safety regulations as they tend to give better coil-contact isolation. <A> For those low power valves, I would consider switching them with transistors instead of adding intermediate relays (the relays will still need transistor switches) if possible. <S> As Spehro mentioned, you can also use solid state relays, but those tend to be much more costly than a switching transistor. <S> A common ULN2803A Darlington array device will switch 8 solenoids at up to (IIRC) <S> 300mA each and it costs around $1.50 in single quantities. <S> To verify proper operation, you may consider a flow switch <S> so if there is a pump or valve failure, it can be detected. <S> One other thing to consider is duty cycle. <S> Some solenoid valves will be rated by how long they can be powered before they must be shut off to allow cooling.
| You can also consider solid-state relays at that current level, with careful design they should be quite reliable, but note that when they do fail (and everything eventually fails) they will fail 'on' generally and some system-level considerations may be necessary to override a failed relay if property damage could be an outcome.
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Of what use can a logic gate be to a circuit that contains a micro-controller I am wondering if a logic gate would be of use to a circuit that contains a micro-controller.Are there cases when an analog logic gate is preferred over a micro-controller and what are some of the gadgets that use analog logic gates in the age of micro-processors and controllers. <Q> I think you mean a 'discrete' logic gate. <S> Yes, there are reasons you may wish to use an external gate. <S> I'll take an actual example: I have a microcontroller that controls a clock signal to some external boxes. <S> It must turn the clock (several MHz) on at a time determined by the microcontroller and must not feed out any 'runt' pulses. <S> By synchronizing the enable with the clock generator (a flip-flop) and gating it (an 'and' gate), the specifications can be met. <S> There's no way to do it with just the micro. <S> In another case, an external signal from a comparator should be enabled by a timer in the microcontroller. <S> The microcontroller peripheral has the ability to precisely (in time) toggle its port pin, but no ability to 'and' signals with that pin state. <S> In some cases, microcontrollers have been fitted with some configurable logic to take care of this sort of requirement (the 'CLC' or 'configurable logic cell') on Microchip parts, for example), but there will always be applications where the micro maker did not anticipate the application. <S> In fact, sometimes we tie an entire FPGA with hundreds of thousands of gates to the micro to get enough external logic to meet the requirements. <A> Digital logic gates ("analog logic gate" makes no sense) are still sometimes used with microcontrollers. <S> Usually the reason is that something needs to be done at the speed of a logic gate that firmware is too slow for. <S> Another possibility is expanding the I/O capabilities of the micro. <S> If you have several devices on a bus, for example, you will probably have external logic that latches data to and from the bus. <S> There may also be external logic to handle arbitration, since this often has to happen at the speed of bus cycles or fractions of a bus cycle. <A> On one hand, "discrete" logic can perform faster and more reliably operations that could otherwise be done by a microcontroller. <S> And, which sometimes is even more important, logic gates can operate concurrently, while a uC is inherently sequential. <S> Also, if you have a crowded board you can save I/ <S> O pins on the microcontroller if you can perform such operations externally. <A> Another reason: ultra low current consumption when not switching. <S> You could use this to respond to simple input when the micro is off, or decide whether to wake it up to perform processing. <A> Microchip has determined that enough designers need to add some peripheral "glue" logic to their designs that they have come out with two microcontroller families -- PIC16(L)F150 and PIC10(L)F32X -- which include up to four Configurable Logic Cells (CLC), much like a miniature CPLD. <S> There are eight different logic functions available: • AND-OR• OR <S> -XOR• AND• S-R Latch• D Flip-Flop with Set and Reset• D Flip-Flop with Reset• J-K Flip-Flop with Reset• Transparent Latch with Set and Reset <S> For example here is a J-K flip-flop: <S> In the case of the PIC10 (which has one CLC), that's pretty amazing for a six-pin device that costs under 40ȼ in quantity. <S> At that price, the cost and space saving over having to include several separate logic chips adds up. <A> For instance, on a board I'm working on at the moment, a phase detector - simply an XOR gate - with a resistor and capacitor allows the MCU to read out the relative phase of two signals using its ADC, instead of having to sample the whole high speed signal. <A> Everybody has ignored op-amps here; analogue logic has many functions even in modern uP / micro controller circuits. <S> Proprietary signals via long wires between micros would be just one application. <S> Personally I work on security and other electronics all the time where I have to sort out bad designs by highly qualified Engineers because of their lack of understanding of op-amps and discreet logic. <A> They are also used for safety logic, to keep the whole complexity of software out of the critical path. <S> In this laser cutter board for example, to turn off the laser.
| To add to the list of applications, they're also useful when dealing with input signals faster than you could process directly in the microcontroller.
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Understanding purpose of components in a half bridge driver circuit I'm currently trying to learn and understand about creating an H-Bridge using two half bridge drivers. The half bridge drivers I have are IR2184. I found a circuit diagram someone else has designed using the same driver ICs but I have some questions about it. I've seen several similar circuits where resistors are placed in line with the gates of the mosfets - What are those resistors for? Also on the circuit above the designer has placed some diodes parallel to those resistors. Is there a reason for doing that? <Q> There's another important reason to the presence of the diode antiparalled to the gate resistor. <S> In a bridge configuration, when a transitor goes on the switching, the voltage acroos the other transitor has a increased dV/dt. <S> In a MOSFET, that's the case of your H-brigde, there's three intrinsic capacitances composing its bory, like in the image below (that shows other stray components of the Mosfet bory): <S> When the Vds voltage increases, all the three capacitances need to be charged, like in the schematic below: simulate this circuit – <S> Schematic created using CircuitLab <S> Where the red arrows are current flowing through the circuit (when Vds are increasing). <S> So, if this resistance of the discharge isn't low, the Cgs capacitange could had a peak of charge, creating a short-time short-circuit, named punch through . <S> More information at this app. <S> note http://www.irf.com/technical-info/appnotes/an-936.pdf <A> The resistors in series with the MOSFET gates are damping resistors. <S> They reduce the tendency of the circuit to oscillate or ring when switching. <S> So, they help the MOSFETs switch off faster. <A> You should always provide a gate resistor. <S> At the very least it actually helps define the current that will flow in and out of the GATE during switching transients (to charge and discharge the gate capacitance). <S> Without any gate resistor you could have very large (relatively speaking) peak currents flowing. <S> Best-case... <S> its an EMC concern, worst-case <S> ... you potentially burn out the gate of the FET. <S> You also run the risk of creating a Pierce Oscillator. <S> In this instance you can see a series gate resistance with a diode in parallel. <S> This limits the current and thus the switching time for TURN-ON. <S> The diode then "shorts" the resistor out permitting a higher charge transfer for a TURN-OFF. <S> This is a cheap/simple means to mitigate shoot-through's of a H-bridge leg during PWM transitions as you now have fast TURN-OFF and slow TURN-ON <A> Because FET input has gate capacitance, and the driver usually has strong current source capacity. <S> So, when with no resistor, the current spike may cause EMI problem, and other problems. <S> So adding series resistor to slow-down the switching. <S> But usually, we needs to shut down the FET faster, adding of the damping resistor will slow down the discharging of the gate capacitance too. <S> With reverse paralleled diode, when the gate capacitor discharge, the current will flow in the diode, but not the damping resistor, this makes the shut down faster than turning on.
| The diodes in parallel look like they are intended to speed up the discharge of the MOSFET's gate capacitance when the signal goes from HIGH to LOW.
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Is probability theory useful in electrical engineering I have a few days off after finishing this semester and I am hoping to study something. Ideally, I would like to be able to benefit from it in any upcoming courses I might take. I'm thinking of either Fourier Series or Probability Theory. Currently I'm leaning towards Fourier because it's essential to Electrical Engineering, however Probability Theory seems intriguing (also opens up interesting areas such as Machine Learning). So how relevant is Probability Theory to an undergraduate Electrical Engineering student? <Q> Probability theory is very relevant to Electrical Engineering -- in fact several universities have courses just on this topic, such as University of Utah's ECE 3530/CS 3130 - Engineering Probability and Statistics and USC's EE503 Probability for Electrical and Computer Engineers . <S> Take a look at the syllabi for these courses and you will get a feel for the many different topics covered. <S> Probability is particularly relevant in the area of quality engineering, and being able to design reliable circuits where you have to take into account the tolerance of various components, and the predicted lifetimes. <S> What is the mean-time-to-failure (MTTF) of your system? <S> Where can you get by with a 20% tolerance capacitor instead of a more expensive 5% one? <S> In a recent design of mine, I had to use a 0.1% resistor in a current sensor instead of the usual 1% resistors I always use. <S> I have used a Poisson distribution to predict the maximum amount of memory needed <S> so I would have adequate buffers to handle peak incoming traffic in an embedded system based on the expected average traffic entering the system. <S> Communications theory and error correction are all about probability and statistics, where the maximum amount of information is jammed into the smallest amount of bandwidth, while dealing with possibly very noisy channels. <A> Both Fourier and Probabilities are extremely useful and relevant to Electrical Engineering, so there is no doubt that you will need both any way. <S> As you are asking specifically about the relevance of Probability Theory I am going to talk only about it, but as I have already mentioned there can't be a comparison between them as you are certainly going to need both of them, one more that the other depending on what area of Electrical Engineering you decide to focus on. <S> One can write countless lines about where you can use probabilities but I am writing the first things that come to mind. <S> In most Universities you are taught an Introduction to Probability Theory and Statistics and then in other classes you are learning about more specific areas depending on the needs of each class. <S> For instance if you are going to choose Control Systems Signals Information Theory <S> Communications <S> then you are certainly going to need probabilities in most of your classes. <S> But that is not the case if you are going to choose Energy or Motors. <S> (The areas I refer to are just examples and there are a lot more, both in the list that needs probabilities and also in the one that doesn't.) <A> You are already considering the most relevant topic. <S> If you are considering looking into probability theory <S> it is useful in Engineering in general <S> The most relevant part is FMEA - Failure Mode and Effects Analysis where the failure of a functional part is determined & then the end effect at system level. <S> Another aspect is on complete circuit analysis. <S> Its all well and good to design (for argument sake) <S> a voltage divider, but <S> in practice it is never just a 10k:10k... <S> Those 10k themselves have tolerance, drift with life, temperature variance. <S> Then there are the other aspects of such a circuit. <S> The variability of the PSU, the range of the ADC it might be hooked up to. <S> Probability theory is directly related to FMEA while a healthy knowledge of probability will help with circuit analysis <A> Given a few days off between EE undergraduate terms, should you study Fourier series or probability theory? <S> I would prioritize probability theory ahead of Fourier analysis, especially for reasons well stated by Adam. <S> Both subjects are quite valuable for a well rounded EE. <S> My MSEE focus was communications theory, which relies on both. <S> Only you can provide the best answer for yourself. <S> My general advice to engineering undergraduates is to get a head start before each new class meets, for example by reading every textbook preface and chapter 1 ahead of class. <S> Most students are unable to absorb all the material presented; preparation gives an edge. <S> Best wishes for you.
| Probability Theory is one one the most important courses in Electrical Engineering.
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How to Check Rechargeable Battery NiMH Status with Microcontroller? I created a device, based on an Arduino Uno, which runs on 6 rechargeable NiMH batteries. Now I would like to add a check, if the batteries have enough power left, to warn with a signal when the batteries needs recharging. As I understand, the voltage of the batteries will slowly go down, until they drop under the minimum required level. How is a battery load level check implemented? What kind of elements do I need? <Q> The most accurate way to know how much energy is left in a battery is to monitor the voltage, current, and temperature over time, then use knowledge of that particular chemistry to estimate remaining energy. <S> There are ICs which do parts of this, sometimes called battery fuel guage ICs. <S> Of course you can do the same thing with a microcontroller, but it takes constant A/D readings and the algorithm can be complicated, depending on how accurate you want to be. <S> A much simpler but less accurate way is to just monitor voltage. <S> NiMH cells start at about 1.4 to 1.5 V right after being charged, quickly drop to 1.2 or so, go down only slowly over most of the discharge cycle, then drop quickly at the end of charge. <S> Usually you stop discharging at 900 mV or so. <S> Letting the voltage of any cell get less than that can risk permanent damage. <S> The best levels depend on your load. <S> The manufacturer will give you discharge plots at various currents, tell you how low you can go without damage, etc. <S> As always read the datasheet . <A> I realize that this is a little late, but I have had this same problem and found a solution that looks like it will work great. <S> Basically, the reference voltage on the Arduino is based off of Vcc (unless you provide an alternative Vref), which makes measuring Vcc very difficult. <S> To fix this, you can base your measurements off of the internal Vref of 1.1v. <S> Even though the internal Vref is smaller than Vcc, you can still use it to calibrate the Vcc measurement so that it is accurate. <S> This has a lot of advantages for things other than battery measurement as it can keep the voltage readings steady even if you don't have a clean Vcc source. <S> Please note that I did not come up with this solution <S> , I merely found it here: <S> [ https://provideyourown.com/2012/secret-arduino-voltmeter-measure-battery-voltage/][1] Best of luck! <A> You should use a voltage divider using 2 resistors to decrease the voltage to the range that mcu can meager, and connect the output to the ABC pin of mcu.
| Of course you need to consult the datasheet for whatever particular batteries you are using. You could simply pick a voltage around 1.0 to 1.1 V and decide to warn the user when the battery gets that low, then go dead at 900 mV.
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When do we prefer isolator over optocoupler? Today I have learned that a device called "isolator" exists (eg: SI8712CC-B-IS ). Its function is very similar to the optocoupler's. On the good side, it is much much faster. While it is very hard to find an optocoupler with propagation delay less than 1\$\mu\$s, almost all isolators have less than 100ns delay. On the other hand, they are slightly expensive though. If such a device exists, why do we keep using optocouplers? Is the price the main factor of choice? Can we always use an isolator instead of an optocoupler? How do they differ in general? <Q> What you show and call a "isolator" is a optocoupler with digital output. <S> The term isolator refers to various means to isolate a signal such that there is no common connection. <S> A opto-coupler or opto-isolator (different words for the same thing) is only one method. <S> Opto-isolators can be broken into a few broad catagories. <S> The most common is a LED shining on a phototransistor. <S> As you found, these usually have a few µs to to 10s of µs propagation delay, particularly when turning off. <S> Another type uses a photodiode as the detector, with active circuitry to amplify the relatively small signal from the photodiode into a digital signal. <S> These can have less than 100 ns propagation delay. <S> Note however that they require separate power on the receiving side, and generally cost more. <S> As should be no surprise when you find two different things that address the same problem, each has its advantages and disadvantages. <S> Everything is a tradeoff. <A> An optocoupler is an isolator, but not all isolators are optocouplers. <S> As you've already indicated, when the specs of an optocoupler indicate that it won't do what you need, move on. <S> Another special case is when you're trying to isolate a bidirectional bus. <S> In that case, some manufacturers have specialized IC's just perfect for those jobs. <A> They’re also much easier to implement in a design since the inputs and outputs use standard digital logic. <S> Standard digital isolators have different numbers of input and output channel combinations and can operate up to 150Mbps. <S> Specialized products also exist for specific serial interfaces, such as USB and I2C. <S> You can see the entire digital isolator product line available from Analog Devices at our web site: http://www.analog.com/en/interface/digital-isolators/products/index.html <A> What you provided is actually different. <S> Yes, it is functionally equivalent to an optocoupler (and it is designed to actually be a pin to pin replacement), but the part you provided does not use light as the means of making the output go low or high. <S> Instead, an "isolator" uses electrical means. <S> In this case, it is probably electrically coupled via a coil (think relay or transformer). <S> The other way is to do it with capacitive coupling. <S> You see this a lot in chips that are used for isolating rs232 or usb (Analog Devices and Linear Technology, which is now Analog Devices, make these chips). <S> Why do we keep using optocouplers? <S> Because the technology is trusted and has been used for a while. <S> It is the same reason why we still use those old rs232 to USB converter chips. <S> Yes, they have newer and improved versions, but you still use the older ones because the designs are documented and well understood. <S> In reality, if you do not need the features of an isolator, then why spend the extra dollar? <S> It may not seem like a big deal if you make small qtys, but in huge qty, a dollar more is a lot. <S> Also, even if they do advertise that they can be a direct replacement to optocouplers, that is not 100% correct. <S> With isolators (remember, they are connected via an electrical mean, not optical), you sometimes need to make sure the input is not left floating. <S> This means you need to use a pull down (or up) or even a schmitt trigger to ensure the input is always at a low or high.
| Digital Isolators are a more modern alternative to optocouplers, offering improved performance with lower power dissipation.
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Total resistance with resistors bridging two branches I know how to calculate resistance in parallel and series connected resistors, and how to reduce most circuits to these, but I missed my lecture on cases where you can't just reduce the circuit to series and parallel and now I'd like to catch up... simulate this circuit – Schematic created using CircuitLab How to calculate total resistance between A and B in such case? R5 rains over my parade because I don't know a'priori which direction current will travel through it, and I don't know how to transform this into something I'd know - I could try to calculate potentials at points p1 and p2, to get the current that flows through R5, except that current modifies these potentials... How do I solve a circuit like this? <Q> I think you can try Y-Δ transform , transform half of the upside or downside 3 resistors to "Y", then it'll be easier to analyze. <A> From the voltage and the current you can figure the equivalent resistance. <S> With that mental model, you can replace R1 and R2 with a Thevenin source, and R3 and R4 with another Thevenin source. <S> Now you can compute the current thru R5 because it is simply connected between two Thevenin sources, which themselves can be reduced to a single Thevenin source. <S> Once you have the current thru R5, and therefore the voltage accross it, the current thru the remaining resistors should be easy to find. <S> There are lots of ways of attacking this problem. <S> The above is actually somewhat roundabout, but it provides a good conceptual understanding of what is going on each step. <S> If you're clever, you can use that method to derive a more direct method. <A> I'm probably in the minority here <S> but I don't agree that you should do the Y-\$\Delta\$ transform. <S> That is sort of a formal way of solving these problems. <S> Since you are asking this question I think you'll learn more by working through this in gory detail. <S> Write down the loop laws and enforce that the sum of the currents into each node is zero. <S> Set the potential between \$A\$ and \$B\$ to \$V_{AB}\$. <S> Once you calculate the effective resistances you will find that \$V_{AB}\$ drops out of that calculation. <S> The currents should all be proportional to \$V_{AB}\$. <S> When you apply the current law you have to assume a direction for the currents. <S> It doesn't matter what directions you assume. <S> You just have to apply the laws consistently. <S> For example if you assume that the currents at \$p1\$ all flow into \$p1\$ you will find $$i_1+i_2+i_5=0$$ <S> When you apply the current law at \$p2\$ assuming \$i_3\$ and \$i_4\$ flow into \$p2\$ you will have $$i_3 + <S> i_4 - i_5 = 0$$ <S> The drop through \$R_4\$ then is $$V_4 = <S> i_4 <S> R_4$$The drop across \$R_3\$ from \$p2\$ to the top of the diagram is $$V_3 = <S> -i_3 R_3$$ ( <S> because you wrote the current laws with \$i_3\$ pointing into \$p2\$), etc.
| One way to look at this conceptually is to imagine a voltage applied accross A-B, then calculate the current drawn.
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Is it Electrically Correct to Direct Driving Seven Segment in Multiplexed Manner? I want to design a voltmeter with minimum no of components so the pcb becomes as small as possible.For this purpose i wish to direct drive the seven segment display. Considering the fact Typically for a standard red coloured 7-segment display, each LED segment can draw about 15 mA to illuminated correctly, so on a 5 volt digital logic circuit, the value of the current limiting resistor would be about 200Ω (5v – 2v)/15mA, or 220Ω to the nearest higher preferred value. if I am going to use PIC16f676 ,The data sheet says PORTA & PORTC Can Source & Sink 200mA combined ,with maximum source /sink per pin 25mA How will i calculate source/sinking current for a seven segment display working in a multiplexed environment( say 4 seven segments ) ?? <Q> Assuming a common-anode display, if you want 15mA per segment you'll need a segment current of 4 * 15 = 60mA per segment driver (7 or 8 needed with decimal point), so the anode current will be as much as 7 or 8 <S> * 60mA = 420mA or 480mA total. <S> You can forget about driving a multiplexed display that's that crummy (or is required to be that bright, if it's a high brightness type) directly with any micro. <S> If you use a good (high brightness) display, live with more subdued brightness, and use transistors to drive the digits <S> (anodes in my example) you can drive the segments directly from the micro at perhaps 100mA-150mA total safely, which will give you an average of about 3-5mA per segment. <S> In such a case, I suggest a common-anode display with the PICs because the outputs are asymmetric and can sink current better than they can source it. <A> If you are using Multiplexed Seven segment displays in Sink mode the current won't exceed 15 mA per pin for the segment port. <S> But for the selection port if they are in source mode then the current sourcing will be at max 8X15 mA i.e 120mA per pin which will damage the pin. <S> So its always good to use transistor drivers for driving 7 segment LEDs. <A> In common Anod the power comes from external source. <S> So the MCU port will work as a drain for leds current not as source. <S> That will be safe for MCU .
| You can use BJTs with internal bias resistors or dual MOSFETs to drive the digits, so only a couple parts, plus you should have a resistor in series with each segment.
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Using a P-channel MOSFET as an on/off switch The project I want to build is shown below. I want to operate the P-channel mosfet on and off with variable frequency from 20-100Hz and I choose P-channel to do this so that the +350V will not be present on the load when the control circuit is off. I want also the high power supply to the mosfet and the load to be separated from the control circuit. My questions are: Will this circuit works? please give me suggestions or revisions. What if the load is disconnected or fails and becomes "open circuited" - will be high voltage present on the drain or will the mosfet still switch on and off in that case? I am new to electronics. <Q> As Ignacio is trying to say, you need to pull the gate of the FET up to the 350V supply rail to turn it off. <S> This makes the VGS ~= 0V. <S> Which means the FET is off. <S> To turn it on, you need to reduce the voltage on this gate, perhaps the full 350V is not healthy for it. <S> Perhaps drop the voltage at the gate to +330V to turn it on. <S> This will make the VGS = -20V which is plenty. <S> But always check datasheets. <S> This is a very high voltage switching circuit and your BJT will need to sustain the 350V as well, like I said. <S> I suggest an opto-coupler to isolate the signal from the 350V supply. <A> No, this will not work, it will first destroy the mostfet, and then <S> (depending on whether the mosfet gate-source failed open or short) destroy the 2n3906 and possible the rest of the circuit. <S> The reason is that the gate of the mosfet MUST be within +/- <S> 20 V of its source. <S> You switch it between -350 and -338, which is WAY too much. <S> Note that this makes touching the circuit dangerous. <S> Otherwise use a standard level shifter with a high-voltage transistor. <S> Or one that uses an opto-coupler: isolation from 350V is A GOOD THING. <S> The fact that you are unaware of such basic principles and yet you are building a 350V switch makes me a bit nervous about your future health. <S> But Darwinistic selection has its place. <S> Leave us an address where we can send the award. <A> I would like to add to Kyran's comment. <S> You can consider using SiHfrc20 offered by Vishay. <S> It has a Vds = 600 V and you could drive your load at around 1.3 A at Tc <S> = 100 Deg. <S> C. <S> When the load is open, it will be able to handle the voltage. <S> However, I propose that you should be take some feedback to avoid the stress on the MOSFET.
| If it is possible to connect the ground of your 555 circuit to a different voltage you could drive the mosfet (I would prefer an N type) directly.
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Unexpected resistor for oscillator Please see these two part of this board : In your opinion, What's the usage of the R32 resistor? also in the AN2867 aplication note has been mentioned: Then we have to put a resistor to limit the inverter output current (I mean Rext). then What's really the usage of the R32 resistor? An unrelated question, Should I always connect the RTC crystal(32.768Hz) to the ground?(I mean the shield of it) <Q> The answer depends on the kind of inverting amplifier you are using. <S> case 1: If the active unit has a finite output resistance rout (medium-to-high valued) <S> the additional resistor is indeed not required (however, it may be included - and it influences the time constant for CL2). <S> case 2: If, however, the inverter is realized with an opamp (low output resistance rout with an unknown value) such a resistor is necessary to determine the associated time constant. <S> Think, for example, of an ideal opamp inverter with zero output resistance - the capacitor CL2 would be useless because it would act simply as a load without any influence on the feedback path. <S> Additional explanation: The working principle of the oscillator circuit is based on a feedback path consisting of a third-order lowpass. <S> This lowpass contains a first order lowpass section (the first two elements) in series with a second-order L-C lowpass. <S> (The inductance L is realized with the crystal). <S> This 3rd-order lowpass is able to produce the necessary phase shift of -180deg at fixed frequency (as required by the Barkhausen criterion). <A> R32 (the resistor in parallel with the crystal and forming feedback from output to input of the inverting amplifier) is vital to provide negative feedback at DC - this sets the bias point of the inverter into its linear area hence it can amplify the AC signal through the crystal and produce oscillation based on the crystal's frequency. <S> Without R32 the input is not biased correctly and oscillations are at best sporadic and at worst non-existent. <S> The phase shift caused by this resistor and the capacitor CL2 is a part of the picture with the other phase shift coming from the crystal itself and CL1. <S> Regarding the grounding of the crystal's case, I always do this as it adds a little more EMI protection. <A> The equivalent circuit of a crystal is basically a combination of series and parallel LC circuit with some resitance affecting it Q-factor (very small series resitance or very high parallel resistance). <S> The very high Q-factor ( <S> e.g. 10^5; <S> i.e. series R very low/parallel R very high) is good for a small frequency bandwidth (desired) but bad for starting the oszillation. <S> An additional external (high) parallel resistance assists to start the oszillation more readily and reliable.
| The resistor in series with the output is needed (whether it is internal or not) to produce an overall 180 degrees phase shift from output to input.
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Can I use a switch rated as 250VAC 25A to switch 50VDC 25A? I am looking for a switch that can handle 25A continous current, the voltage will be maximum 50VDC, most often about 46VDC, and it will be used to switch on two dc-dc converters that have no load connected. The load will be switched on after the main switch is on. Is this safe? Switch: http://www.ebay.com/itm/291078954518?_trksid=p2059210.m2749.l2649&ssPageName=STRK%3AMEBIDX%3AIT <Q> Mechanical devices (switches, relays, etc.) rated for AC use need to be derated strongly when used with DC since lack of zero crossing means that arcs are harder to extinguish. <S> In some cases the derating required may effectively make it unusable with DC under any conditions. <S> Consider finding a part properly rated for DC instead. <A> You most often see dual ratings on relays rather than switches (though you do see them on some switches). <S> In general the DC rating of a relay is about 10%-15% <S> the AC rating for the same current. <S> For instance, taking a random selection of relays from my bits box: 16A, 250VAC <S> / 30VDC <S> 10A <S> , 250VAC / 30VDC <S> 12A <S> , 250VAC / 24VDC <S> So for 50VDC and 25A the equivalent AC switch would be around 500VAC and 25A. <S> So no, you can't really use that switch. <A> You might be able to get away with this switch if you can arrange it so that it never actually switches (breaks or makes) <S> the 25A. <S> It is okay for it to <S> carry 25A DC. <A> Here is how I figure my own rating conversions:Your 250v switch with 25amp rating:The dc equivalent of 250vac is 250 <S> x .707 = 176.75 vdc. <S> So to find an equivalent rating for 50vdc:50/176.75 = .2829.2829 times the 250vac 25amp rating = 7.0725amps @ 50vdc. <S> Arcing is more prevalent at higher voltages and currents, so you are safer using a higher voltage switch for a lower voltage application. <S> However switch life is directly associated with how much current is being switched, and your dc loads which typically causes more currents will cause shortened life. <S> I have used these calculations with few problems. <S> Just try to switch when currents are low when possible, otherwise just be prepared to replace a switch from time to time...
| This might be possible if you could use enable inputs on the DC-DC converters to only enable the converters after the 50V is present and to disable them before (or within a few milliseconds after) the switch is flipped 'off'.
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How to run electric match and microcontroller from a single power supply We have a fairly large system that's been built up over the years for running a fireworks show from a computer (see here for the first generation software and hardware) using a pretty simple XBee-based wireless system. I'm in the process of working on a replacement system and I'm trying to tackle one of the annoying things in the current system...which is current. Ideally I'd like to be able to run both the local microcontroller (I'm using a version of the XBee that has an on-board Freescale) and fire the electric matches with a single power supply. The problem, of course, is that firing the matches takes all current away from the uc, causing it to reset. I've tried thinking about how to isolate the matches, which are essentially shorts in the circuit from the uc, but I'm afraid I'm woefully out of my depth in this area. I'm assuming some capacitors would be involved but I don't know exactly what it would look like. The firing current of an electric match is around 1A and extremely rapid. Ideally we'd have up to 16 matches per system. Bonus points if the solution could also be used to power an on-demand flash :). <Q> You not only need to store enough energy to run the MCU, you also need to prevent it from being used up by the match as well. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This circuit will allow the supply to charge up the capacitor, but not allow it to feed back into the supply when the match is triggered. <S> Note that the voltage to the MCU will be reduced by the forward voltage of the diode. <A> Treat the matches as high-current loads and drive them with power transistors or MOSFETs in low-side driver configurations. <S> Connect the low-side drivers to the microcontroller via optocouplers <A> A fairly large decoupling cap right at your microcontroller power pin. <S> Without knowing how big of a dip your power rail takes and for how long or how often, it's hard to suggest a decent one. <S> You could get away with just a 100 or 1000 uf or both in parallel. <S> An oscilloscope to measure and graph the actual drop is required.
| You need a separate power supply system for your electric matches, entirely isolated from the microcontroller.
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extern and the XC8 C compiler Perhaps I don't fully understand extern . With the XC8 C compiler, it seems like I can get away without using it. In the file keypad.c I have a function signed char keypadGetPressedKeyLabel(void) . In the header file keypad.h I have: signed char keypadGetPressedKeyLabel(void); I thought I would have to use extern , ie: extern signed char keypadGetPressedKeyLabel(void); The project builds without extern . And it works. Might this due to the fact that keypad.c and keypad.h are all part of the project? I'm using MPLABX. <Q> extern is not strictly necessary for function prototypes in .h files - whether or not the function is actually used in one or many different .c files in your project. <S> Whether you have extern int foo(void); or just int foo(void); in your .h <S> your compiler will read is as a function prototype either way. <S> Its necessary for global variables though since unlike functions they don't have prototypes. <S> So if you have an int thing; in one .c file <S> and you want to use it in another .c <S> file <S> then you'll need an extern int thing; to tell the compiler that it does exist 'somewhere' in your project. <A> Functions are extern by default in C ; that's why it works without the keyword. <A> extern is really a sign to the linker, that the symbol is defined in a different object file. <S> As you only seem to have one object file, it doesn't make any difference whether it's there or not. <A> extern is largely obsolete for functions in other CUs in many of the smaller simpler compilers. <S> If a symbol isn't found locally it will be searched for in other CUs and libraries. <S> You provide a function prototype, but not the actual function, in the CU where you want to use it. <S> However, extern is required for variables that are shared between CUs. <S> You can't define a prototype for a variable, so
| the extern keyword in this situation has to be used to create the variable equivalent of a function prototype.
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how to know how many bits a microcontroller has? How can I know what kind of architecture a microcontroller has?For example where can I read in the datasheet of the pic 16F873 how many bits architecture is? 8, 16, 32? Next question what does:What does mean: Up to 256 x 8 bytes of EEPROM Data Memory.Mean that there are 256 slots of 8 bytes? Or (256 x 8) bytes? So 256 slots of 8 bit each?I don’t think it’s explained really well :S.Thx for answers :) <Q> With PICs in particular, sometimes you find what you need in the Family Reference Manual as opposed to the data sheet. <S> In this case, the mid-range family ref manual has a chapter on architecture that you might find of use, and its in a different tone than the chip's datasheet. <S> For PICs datasheets tend to have a ton of chip specific info,but the Family Ref Manuals tend to show a more global picture. <A> Page 1 top of the page then further down in the architecture <S> there are the clues: <S> - 256 x 8 usually means 256 bytes <S> each byte is 8 bits wide. <A> The datasheet states categorically the core CPU size on the title page: 28/40-Pin <S> 8-Bit CMOS FLASH <S> The EEPROM is arranged in 8-bit bytes, and there are 256 of them. <A> It's a Harvard architecture 8-bit processor with a 14-bit instruction width. <S> That's why a 4096-word program memory is listed as 7.2K (7168) bytes. <A> The 16F873 is an 8-bit microcontroller. <S> Its internal registers are 8 bits wide. <S> The data sheet should say so, but maybe it assumes you already know that -- the person writing it may have thought it was too obvious to mention. <S> The internal RAM and EEPROM are also 8 bits wide. <S> As for the EEPROM, 256 x 8 in this context means 256 8-bit locations in the memory. <S> 256 slots of 8 bits each. <S> It's not well explained, by the sound of it. <A> There are many Microchip microcontroller families. <S> 8-bit: <S> PIC10, PIC12, PIC16, PIC18 16bit: PIC24, PIC30, PIC33 <S> 32bit: <S> PIC32 <S> If you go to Microchip website and enter microcontrollers ( link ) you will see these families on the left side. <S> Regarding 256 x 8 bytes - I agree that this may mislead. <S> Each byte has separate address. <S> address <S> data0x0000 xxxxxxxx <S> (byte 0)0x0001 <S> xxxxxxxx <S> (byte 1)0x0002 xxxxxxxx <S> (byte 2)0x0003 xxxxxxxx <S> (byte 3) <S> If that would be 256 x 16 - thay would be 512 bytes, but only 256 directly addressable <S> (you have to store 16 bits at once) <S> and you cannot directly access each byte. <S> address <S> data0x0000 xxxxxxxx xxxxxxxx (byte 0, byte 1) 0x0001 <S> xxxxxxxx xxxxxxxx (byte 2, byte 2) <S> In Intel x86 - most of registers are 32-bit wide. <S> You cannot read only one byte, you have to read whole register to get one byte from it.
| 256 x 8 means: 256 directly addressable 8-bit registers of memory (256*8 bits).
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will two plls wander with shared reference clock I'm trying to learn about pll wander or drift. My reading leads me to believe one of the reasons plls were developed was to fight wander so maybe it does not effect plls? Although I've seen some things about the wander of the source becoming the wander of the pll or perhaps noise higher in frequency than the pll can handle doing the same. If I have two PLLs in two separate chips, both of which are fed from the same oscillator. Then I set them both to the same frequency I feel like they will jitter, and that I won't know the phase difference, but that they should not wander away from one another over time because they both share the same reference clock. So the plls will constantly be trying to correct their output to the same reference, thus keeping them close (within 1 bit time). If they both used their own reference clock I could understand how they could wander because one might be slightly faster than the other. Am I right or wrong about that? Do plls wander from their reference? Thanks! <Q> Not wandering from their reference is a PLL's entire purpose in life. <S> There will be some phase noise / jitter that is uncorrelated between the two of them, but they won't drift apart over time as long as they both stay locked (as mkeith said). <A> You said you feel they will have a phase difference. <S> You're right <S> , they will have a phase difference but it will be stable. <S> If you measured it you could then compensate for it in your system. <S> The two PLLs will be out of phase because the two different chips will "see" different phases of the reference clock due to propagation time differences. <S> Therefore, the outputs of the two PLLs will not be in phase. <A> You're right. <S> A PLL will have some slight phase error (due to timing mismatches in the phase detector and current mismatches in the charge pump), which remains roughly constant. <S> Add to that some non-deterministic jitter induced by various noise sources. <S> However the frequency will be, on average, perfectly correct. <S> That said, when you place two PLLs in the same system they can interact with each other (through their power supply and shared ground). <S> This can cause extra phase jitter. <S> For example, if one PLL naturally tends to output its clock edge slightly before the second one, the first one can prematurely trigger the second one by jolting on its power supply. <S> The second PLL will sense this as a "too early" error, and it will slow down its clock further (until the edges are spaced at a sufficient interval to no longer interact). <S> However, it will sense this as a "too late" error and speed itself back up until they start interacting again. <S> In this way, the PLLs' interaction can be a cause of deterministic jitter. <S> On average, however, the frequency should still be correct. <A> A vital point which hasn't yet been mentioned is that even though the two PLL outputs won't wander far from the reference clock, or from each other, neither may safely be used to safely clock signals clocked by the other unless deliberate measures are taken to ensure that signals do not change anywhere near the time they are sampled. <S> If two devices clocked by the two PLL outputs try to feed data to each other, and each device samples at about the same time as the other outputs new data <S> , it may appear that the two clocks are jittering horribly relative to each other; the problem may be ameliorated either by having both devices sample on a rising clock edge but delay their outputs until a falling clock edge, or by having one device sample and change its outputs on rising clock edges, while the other samples and changes its outputs on falling edges
| Two of them that are locked to the same reference will not wander away from each other.
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High AC voltage affects the pcb I had a PCB board printed that has some relays that are connected to 220 VAC circuit.Now unfortunately some traces passes near the relays and are affected by this high voltage causing the circuit to fail. Is there a way to stop the effect of this high voltage on the near traces? Regards <Q> If you are just trying to make your current board work, can you just solder an RC Snubber circuit across the actual contacts switching the AC? <S> For 220VAC, I'd go about 150-250 ohms in series with .1uF. <S> That might help clean up some switching noise (and reduce contact wear). <A> Move the traces. <S> There are separation and creepage distances for a reason. <S> Mostly to do with safety, but EM interference sounds like the problem you're experiencing. <A> General answer: <S> If the high voltage is "tracking" across the surface of the circuit board then make physical slots in the PCB to make the surface path much longer. <S> If the problem is induced emf due to currents then try using some ferrite sheet but make sure it's insulated against the AC voltages (ferrite can conduct a little). <S> Small decouplers on the sensitive tracks to ground may also help. <S> The problem may also be alleviated by adding series resistance to the sensitive tracks. <S> However, it sounds to me like you may have to redesign the PCB. <A> As you have a low and a high voltage part you could perhaps separate them by putting the simpler one on a separate prototype PCB. <S> Then you can connect the two boards with twisted wires, eventually soldering some wires directly to the relais connectors. <S> As always - keep an eye on security, especially when other persons will use the finished product too.
| If high AC voltage is capacitively coupling to the sensitive tracks small decouplers on those tracks to ground may alleviate the problem.
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Hot wire cutter. Can I use copper wire I am not from an electrical background and this might sound silly.What I am trying to do is create a hot wire cutter to cut thermocol(Styrofoam). The sugestions from internet is to use Nichrome but I cannot find one. I have bunch of copper wire though. Can I heat the wire good enough to cut Styrofoam by shorting with 3 AA batteries ? <Q> Not really, it does not self heat very well and when hot it anneals and gets soft. <S> Nichrome is good, steel or stainless steel should also work ok and may be preferred if you need less heat and more strength. <S> Sources of resisting wire would include fishing leaders (or other fine wire rope), piano wire, guitar strings (music wire), salvaged heating element wires (brittle if previously heated), florists wire, and pretty much anything except copper and aluminium. <S> Using 3 AA cells to heat wire will be disappointing. <S> They made some gimmick bag sealers that heated a short 10mm length of nichrome wire with 2 AA cells <S> but I doubt it would have got hot enough to cut Styrofoam cleanly. <S> If you should find some very fine filament that will get hot enough to melt Styrofoam with the 3 cells you may find that it will cool down in air and on contact that your cutting speed will be measured in frustration. <S> Transformers supplying 6 to 24V are often suggested for the wire heating and this for lengths of about 300mm (12 inches). <S> The current may be a few amps so would drain small cells fast. <S> Note that you need an isolating transformer for your scenario - using an autotransformer may have dire consequences (including electrocution). <A> Copper wire will quickly deform even when not heated as you apply pressure. <S> Further, as a very low resistance, it will not heat well given most power supplies, and particularly using batteries. <S> The internal resistance of the power supply will dominate the circuit, and most of the heating energy will be lost in the batteries or power supply, rather than the wire. <S> You might be tempted to choose a thinner wire to increase its resistance, but you will only be frustrated by the lack of mechanical strength in the copper wire. <S> You can still make this work if you support the wire. <S> Adhering it to a thin support, such as glass, will allow you to heat the wire significantly without worrying too much about breaking it while using it. <S> Unfortunately you then run into another problem - oxidation and uneven heating leading to fusing. <S> Even if you find wire that is perfectly conductive throughout and has no change in resistance along its length, over time it will oxidize unevenly, causing hotspots and cold spots, and in order to get the whole thing up to temperature you will likely find one spot get hot enough to turn into a fuse, and cause a break in the wire. <S> This isn't going to happen immediately though, so if copper is your only choice and you've solved the power supply internal resistance problem, then you can simply replace the wire whenever it stops cutting well. <S> However, there are many other metals with superior mechanical and electrical attributes that would better fit your needs. <A> I don't think it's silly. <S> According to NIST data , copper has about 3 times the resistivity at 800K compared to 300K. If we pick a reasonably sturdy wire such as an AWG 20 , the resistivity of 6" (150mm) will be about 15m\$\Omega\$, so for a current 50% of fusing ( about 30A ), we'd need about 0.5V. <S> That's about the current that six fresh Alkaline AA cells in parallel can supply at 0.5V terminal voltage- <S> easy enough to try. <S> Of course you need to use thicker wire to connect the calls (maybe AWG 8). <S> Don't expect the batteries to last long, and the batteries will get very hot <S> so keep them separated from each other and shielded physically in case they burst. <S> Unfortunately Nichrome gets a bit brittle after use <S> so it's difficult to scavenge from, say, a hair dryer. <S> Edit: see KalleMP's suggestion in the comment to consider music wire (steel) which could be used with a PC supply. <A> Copper wire will stretch and deform quite quickly, jump onto ebay and purchase a small pack of nichrome wire, it's the most appropriate material for the task <A> Model R/Cer's having been cutting styrofoam wings for decades with ni-cr wire and a variable power supply. <S> (A model RR transformers can serve as a variable power supply and they're cheap on EBay.) <S> Steel piano wire also works but will need a little more voltage. <S> Search videos on "cutting foam wings". <S> Use the lowest power that cuts and CUT SLOWLY.
| Copper wire that is strong enough will require a fairly low voltage, but it may work out for you. Longer term maybe procure some Nichrome wire (eg. eBay) and use a repurposed PC supply.
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Industrial flash (LED?) for rolling shutter I am developing industrial monitoring system, at this moment with Raspberry Pi and its camera (this might change if it is not satisfactory). The camera will take still pictures of the scene once in a few seconds. Since the objects are moving, it has shutter time max 1/1000s. After some experiments, I determined I need around 10kLux light source which is either turned on all the time (though without producing excessive heat), which is turned on and off programatically, bracketing the camera action, or one which can be synced with the camera shutter using the camera's chip flash sync. The camera unfortunately has rolling shutter, so xenon flash tubes are out, AFAIK -- their light interval is probably too short, there would be partial exposure -- but that's just what I understand from readings sources, so I will be glad to be corrected here). What other flash options do I have? I will be glad to use LED-based flash, but need reliable circuitry around it as well. Ready-made industrial cameras systems costing over $4000 are something I would like to avoid, if possible. I would like to ask for pointer in which way to go, keywords to search for, or even marketed products which would be suitable. I am personally only moderately skilled in low-level electronics, but have colleagues who know that stuff. <Q> I'm not quite sure I understand the issue, but I'll take a stab at it anyway. <S> I'm assuming you can get a light source that you can turn on and off with digital logic, and your problem is that the flash sync signal isn't long enough to keep the light on for your image acquisition. <S> If this is the case, just use the flash sync to trigger a monostable multivibrator (i.e., a one shot) with a pulse long enough to allow the image capture, then use the output of the one shot to turn on your light source. <S> Not sure you can achieve the same functionality with the Pi alone. <S> It would be simple on a microcontroller with full interrupt functionality, but then you lose the imaging and file handling convenience of the Pi. <S> You might consider a Beaglebone Black, which has its own microcontroller (though I haven't read much about using it). <S> In any case, the one-shot is a one-chip circuit, and shouldn't be burdensome to implement. <A> The flashes will only illuminate the subject for the short 1/1000 second exposure time (make sure the flashes are that short). <S> If the ambient light is low enough, then very little light in comparison will be captured during the rest of the time the shutter is open but the flashes aren't illuminating the subject. <S> The extreme case of this is a totally dark room and the shutter always open. <S> The scene will only be captured during the brief time of the flash. <S> This is exactly how Doc Egerton made those famous high speed pictures, like a bullet thru a apple, so many years ago. <S> The mechanical shutter wasn't anywhere near as fast as the exposure time, which came from the speed of the flash, not the speed of the shutter. <A> We replaced an old (strobbing) Xenon flash lamp with a bright white LED. <S> Flashing it for a few milliseconds every second or so should not be a problem. <S> For the circuit, I'd do an opamp as voltage controlled current source with a FET as the pass element. <S> With a little sense resistor to measure the current. <S> I can scribble out a circuit if you don't know what I mean. <S> (something like this , but it needs a bit more around the feedback path.) <A> Rolling shutter with fast moving objects is just not a good mix. <S> You don't mention some key elements of your requirements (like imager format, frame rate etc.) <S> so it will be hard to recommend a solution. <S> It's likely you will need to go to a ILT (Inter Line Transfer) CCD for a global shutter. <S> You also don't say what control you have over the the rolling rest and <S> the rolling read <S> so again it will be hard to give any recommendations. <S> The one fall back that you do have is to do the following.- Set you exposure time to be longer than a frame time (or as long as possible)- Figure out how to determine when the Imager has finished the rolling reset- <S> Keep everything in the dark- Fire your flash once the imager has finished the reset but has not started to read out the imager.- <S> Read out the imager Note:- <S> this will adversely affect your frame rate.
| Depending on how much ambient light there is, one possible solution is to use the fastest X-sync speed the camera is capable of, then fire a xenon flash or several of them in sync.
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Why does there seem to be a missing component in many PCBs? I've seen some PCBs were some components have been sketched , but they don't seem to be mounted on the board. See, this image (of a PlayStation 2) where the transformer has been drawn: Why is this done on some circuits? <Q> There are a couple of possibilities. <S> One is that they're using a single PCB for multiple designs. <S> Another (that's still technically sort of the same idea, but mostly different in practice) is that they started with one design. <S> Then (for example) a new part became available that (for example) integrated more functionality onto an IC, so some of the passive components that used to be required aren't any more. <S> For example, a part that required external pull-up resistors might be replaced with one that's otherwise identical, but has internal pull-up resistors. <S> Your board has spots for pull-up resistors, but with the new part you get exactly the same functionality by simply omitting them. <S> For example, let's assume it costs $500,000 to design a PCB, and by designing a new one you could save $1 on each finished item. <S> Obviously you need to sell at least 500,000 of the modified design for the re-design to break even. <S> If you're selling fewer than that, you're better off sticking with the existing design. <S> In addition, most suppliers give (often substantial) price breaks for producing a larger number of identical items. <S> For example, let's assume you have two designs and expect to sell about 5,000 of each of those designs. <S> All else being equal, one of the PCBs should cost $1 less to produce than the other--but if you buy 10,000 of a single design you get (say) a 10% discount. <S> In this case, if each PCB costs at least $10 to produce, you end up saving money by using the same board for both designs. <S> Having fewer PCB designs also simplifies production. <S> You only have to track one part in inventory instead of two. <S> That part (the PCB) may easily be the only one that's truly unique to each design, so if your guess/prediction about how the two designs will sell is wrong, keeping your inventory balanced can be quite a bit simpler. <S> For example, ordering more of a custom PCB will typically have considerably longer lead time than ordering more of standard capacitors, resistors, off the shelf ICs, etc. <A> Largely, cost controls. <S> You have one board, or a small family, make them flexible, and populate those parts that you need for a given product. <S> Sounds cheesy, but in large production, every nickel can count. <A> You can leave some filter parts out for product destined for countries that are not so picky about EMI and save 20 or 25 cents a unit. <S> There are many reasons to have minor variations on a theme. <S> That particular board in the photo is likely a stamped board (not drilled) <S> so there is some capital cost in the manufacturing, and (more importantly) <S> the case will be sized to take the biggest board it has to so there are no savings in making multiple blank PCBs, only cost and another custom part to inventory (after stuffing the blank part numbers fork into multiple sub-assemblies, of course, so <S> the savings may not be that great). <S> For example, in Altium, these are referred to as "Assembly Variants". <S> The same schematic and PCB source documents can be used to produce BOMs for each variant. <S> Parts may be omitted, part numbers changed, or values altered for each variant. <S> In the case of the item in the photo, it looks like the radial disk part is used in one variant, and the magnetic part in another variant. <A> As other answers state, multiple products may be economically provided by the same PCB, or manufacturing flexibility in the face of varying supplier reliability ('second sourcing') requiring alternative parts to be accommodated from alternative suppliers. <S> The circuit in your photo is part of a switched-mode power supply (SMPS). <S> The part in that has been omitted looks like a a transformer (one side has two terminals, the other has three, so it's unlikely to be a common-mode choke). <S> The transformer has been replaced by either a capacitor or transient suppressor. <S> Note that the capacitor lies within the outline of the transformer - indicating the designer fully intended it would be an either/or design. <S> Now, the PCB is also single-sided. <S> This is the most popular choice for consumer products due to it being the lowest cost. <S> So the location of the transformer or the capacitor with respect to the copper tracks underneath forbids both parts being fitted. <S> Most products these days have 'universal' power supplies removing the need for voltage selection switches in different parts of the world [EDIT - for the same reason, using different EMI components for different regions is very unlikely for a global product. <S> The USA, Canada and EU are very demanding in this regard so once your design is capable of satisfying those needs, it's more costly to do EMI versioning on geographical grounds than it is to have a common design]. <S> So the 'versioning' implied by the transformer or capacitor is that the capacitor is being used to provide a DC barrier that would otherwise be provided by the transformer. <S> My guess is that this is part of the primary side of the supply to bias the SMPS primary controller, and the capacitor is simply a cheap alternative to the transformer. <S> In other words they found during production and testing that the capacitor would do as good a job as the transformer in this particular case <S> but they left the transformer on the layout in case <S> it turned out not to be the case, or because there is more than one product that uses this PCB and the capacitor-bias is inadequate for that product. <A> PCB's are pre-fabricated <S> , So in case of circuit changes in the long run it is often incorporated in the same board until and unless there are drastic changes to the design.
| It's mostly a question of balancing the cost of designing a PCB against the increased cost of producing the design using the existing PCB design.
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What is the best way to plate VIAs? My local hackerspace has a CNC machine. I can give them my Gerbers and they can print out the PCB. The machine does not plate VIAs. I need to do this after the CNC machine is finished. What is the best process and materials to use? <Q> The only way to literally plate vias is with a somewhat fussy multi-step electrochemical process (even commercial manufacturers had issues with it not that many years ago) <S> that I doubt you want to get into, and in any case it's too late when the boards come back- <S> you'd have to get them to drill the holes, plate the vias, then give it back to have them mill the spaces betwixt the traces. <S> That leaves you with at least these three options: <S> Use component leads as vias (obviously they have to be parts that can be accessed from the component side. <S> This is a real constraint on layout in general- if a component lead cannot be soldered from the top you'll need to either put a thin wire through before it is stuffed (risky) or add another (accessible) via. <S> If all your parts are SMT you need to make sure that the vias are not under parts on either side. <S> and users of machines such as LPKF's mill or laser cutters could use them). <S> You'll need to make the via holes and pads of appropriate size for the rivets you can buy. <S> Use a bit of wire (perhaps a strand from stranded wires) to solder top to bottom on vias. <S> You don't need thick wire for even relatively high currents as the length is so short. <S> Which, if any, is "best" will vary depending on the type of board (density, type of parts, etc.), your situation, what you are trying to accomplish (such a prototype is relatively far from what you'd get from a commercial board house, so it's not as useful for some purposes as a commercially made prototype <S> would be .. for example, to check things out before you order 100 boards). <A> One more option : where you want plated vias, email the Gerbers to a company in the UK, North America, or China, and wait ten days. <S> There are some surprisingly low-cost options available. <A> Assuming only a double-sided board, hand-solder appropriately sized buss wire to each trace, through the via, and then cut the ends off flush with the traces. <S> One caveat: it's easy to lift the traces when they're hot, so be careful.
| There's at least one on-line vendor who sells the chemicals and electroplating supplies required for the task, but I think if you were willing to do that, you'd be etching the boards yourself. Use rivets (there are small rivets available for this purpose-
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Is there any material from which lenses are made which are fine with IR signal? I mean really cheap ones I am making a prototype where I need to direct the flow of the IR signal at some specific point. I found out that glass material doesn't work properly in combination with IR. Is there any cheap lens which can do that with IR? The main purpose is to not have wide range of the IR but directed one, like in laser, but wider. It can be any LED which can work at 10m+- distance. But now I am having these:5mm IR infrared 940nm LEDAnd sensor which receives 38kHz - TSOP4838 IR Receiver Module <Q> No need for pricey special filters at that wavelength, ordinary acrylic plastic ( Plexiglas is one trademark) is quite transparent to 940nm IR, no problem (well, up to stupidly thick pieces anyway). <S> If you also want to filter out visible light, there is G 3142 or 1146 Plexiglas that absorbs visible light but transmits near-IR. <S> Should be available from plastic distributors in various thicknesses, and it's easily machined if you need a custom shape. <S> If you need an actual lens or prism rather than just a window, acrylic is a common material for making molded lenses- <S> it's optically quite transparent at visible wavelengths too, and fairly hard (but brittle). <S> Edit: <S> Also, polycarbonate (one trade name is Lexan) is also fully transparent to near-IR. <A> Way back in the 1970s, I saw a thermal imaging camera made, I think, by EEV in Chelmsford. <S> It had a lens made of germanium, which looked like a shiny metal lens. <S> Totally opaque to visible light, of course. <S> Sure enough our spectacles looked solid black in the thermal images produced, because it was working in IR wavelengths that glass cannot transmit. <S> I don't know if IR lenses are still made from germanium, nor do I know if such things could be made cheaply. <A> These aren't cheap: https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=7999 <S> They're designed for mid infra red. <S> I suppose the price is relative to the cost of the prototype and how important it is <A> I think AR coating on a BK7 glass lens will work for you. <S> Thorlabs have those: https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=3280
| So, pretty much any plastic optical component will be made out of acrylic (lenses, some sheet plastic) or polycarbonate (many light pipes, also available in sheet), and you can use it without problems at 940nm.
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What is the Q-point of the zener diode? Is the Q-point f the zener diode in figure equal to its breakdown voltage of 4 V ? <Q> "The operating point or Q point is defined by the intersection of the load line equation and the i-v curve of the diode" http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-071j-introduction-to-electronics-signals-and-measurement-spring-2006/lecture-notes/18_diodes2.pdf <A> The breakdown voltage Vz is the voltage that corresponds to a current Iz <S> that is is 50% of the maximum alloed DC current Iz,max. <A> According to http://en.wikipedia.org/wiki/Biasing : The operating point of a device, also known as bias point, quiescent point, or Q-point, is the steady-state voltage or current at a specified terminal of an active device (a transistor or vacuum tube) with no input signal applied. <S> Assume <S> the zener is off$$V_{zener} = V_{R3.6V} = \frac{3.6k\Omega}{11k\Omega+3.6k\Omega} \approx 4.9V$$So, <S> zener diode should be on, becuase \$R_{z}=0\$ (that means when the voltage on it greater than the breakdown voltage, it can flow through any current). <S> The Q-piont is just equal to the zener voltage, that is 4V.
| The Q-point (that is the DC operating point) is defined by the quiescent current Iz,o through the diode and the corresponding voltage Vz,o (which slightly can differ from the "breakdown voltage Vz).
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Calculated output is greater than the op amp can produce? I had to analyze a differentiator and integrator op amp. I am now asked the question of what would happen if the calculated output is greater than the op amp can produce? I am given a hint that it is usually a little less than the magnitude of the power supply voltage. I think the answer is that the op amp would damped on suppress the output, so that the output is within the range that the op amp can produce, but I can't find anything that can confirm this for me. Can someone tell me if I'm on the right track or not? Or if I'm going in the completely wrong direction. <Q> Take the integrator. <S> Suppose you apply a voltage to the resistor, and the capacitor starts off with 0V across it. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The voltage at the output will rise from 0V towards +10V at +1 volt/second, so after 10 seconds the output voltage will (ideally) reach the power supply voltage. <S> At that point the op-amp will saturate or "rail" at the (ideally) positive supply voltage. <S> The op-amp can no longer operate in its active region so the output looks like its shorted to the +10V supply. <S> At the moment of saturation, the voltage on the inverting input is 0V, but after that it looks like a classic RC circuit and you'll see the kind of RC response you'd expect on the inverting input (while the non-inverting input obviously will remain at 0V since it is grounded). <S> After a very long time the voltage at the inverting input will thus be V1. <S> Since it's driving the op-amp output towards the positive rail (negative input on the inverting input wrt the positive input) <S> the op-amp stays saturated at the positive rail. <S> You can do a similar analysis for a positive input to the integrator and similar inputs to the differentiator that cause the op-amp output to saturate. <A> Yes, you are absolutely right. <S> Op amp cannot produce voltage greater than the supply voltage. <S> It will give it's maximum possible voltage on the output that is a little less than the supply voltage. <S> But there is one important point. <S> And to enter its' normal working mode then it takes some time to leave over-saturation mode, and thus for this time it can not give correct output voltage corresponding to the input signal. <S> (If the op amp belongs to "rail-to-rail" class, its' maximum output voltage is more close to supply voltage or it is actually the supply voltage itself.) <A> There are mainly 4 blocks of op amp 1)input stage <S> which is a dual input balanced out put differential amplifier. <S> This differential amplifier provides the voltage gain of the op amp. <S> Then intermediate stage which is directly coupled to the differential amplifier. <S> This intermediate stage is a dual input unbalanced output differential amplifier. <S> Because of direct coupling the output of this stage is above ground potential. <S> A level shifter is used to bring the output to ground level. <S> The fourth stage is a push pull amplifier which increases the output voltage swing. <S> If the level shifter is not making the DC level exactly to zero and other non linearities associated with these stages causes the output voltage to be above our expected values I believe .
| - After the input differential voltage multiplied on the op amp gain coefficient (Uin_diff*k) exceeds the maximum output voltage, the op amp enters so called over-saturation mode.
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Why the voltage drop is so high in my Half-Bridge SMPS? I have been building an offline Half-Bridge SMPS of about 300W with a friend of mine. We need a variable range from about 10V to 30-35V. We finally got it working and we could achieve 10A at 30V. The problem is when we have no load but a 1K resistor at the output of the SMPS, the voltage goes up to 45-48V and we cannot go below 32-33V, but when we put a load of only 1A the voltage suddenly drops and the maximum we can get again is about 30V but with high current as needed (but we got around 8Vpp of 100KHz ripple). The SMPS is working at 100KHz, we drive the MOSFET through 1:1 driver toroids and we get around 12V at MOSFET gates (we are using IRF840). The duty goes from 5% to 50% with a 375ns dead-time. Our problem is that a drop of about 15-20V when we put a load is too much, we get around 50V with no load and when we put a load we can't get higher than 30V (even when we calculated the transformer for a 40V output). We used the book "Power Supply Cookbook from Marty Brown, 2nd edition". Here you can download a PDF with the complete schematic and the PCB. The transformer has 33 turns on the primary side and 12 turns on each side of the secondary. L2 is a toroid from a PC power supply with 70 turns. Also, the efficiency oscillates between 40% and 50%, I think the problem could be there but I don't know what is causing such a low efficiency. EDIT: Here are the measurements with a 20% duty cycle: T4 Primary side: T4 Secondary to ground - 100mA load (minimum load): T4 Secondary to ground - 3A load <Q> Is your transformer secondary 12 turns total or 12 turns on each side of the center tap? <S> If it's the former, that's why you can only get 30 V under load. <S> I calculate it this way: Input voltage is 220 V RMS <S> full-wave rectified to about 310 V DC . <S> This means that your half-bridge is driving the transformer with a voltage whose peak is half of this, or 155 V. <S> The 33:12 transformer is going to turn this into a peak voltage of about 56 V. <S> If the secondary is center-tapped, then you're only hitting the rectifiers with a peak of 28 V. <S> As for the excessive rise at low loads — well, that's why lots of SMPS specify a minimum load. <S> It's actually quite difficult to design an efficient one that also has a huge dynamic range. <S> One problem might be excessive leakage inductance (i.e., less than perfect coupling) in your transformer. <S> EDIT: Since I can't put this drawing in the comments, I'll add it here. <S> Your transformer drive waveform <S> always needs to be symmetric. <S> At 50% duty cycle, it should look like a square wave, with a small amount of "crossover distortion" created by the dead time: -------- <S> -------- | | | <S> | | <S> | <S> | | | <S> | <S> | | | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | <S> | | <S> -------- <S> -------- <S> But at lower duty cycles, it still needs to be symmetric, with longer "off" periods between the alternating pulses. <S> It should look like this: -- -- | | | | | | | | | <S> | | <S> | | | | <S> |- ------ <S> ------ <S> ------ <S> -- <S> | | <S> | <S> | | <S> | | <S> | | <S> | | <S> | <S> | <S> | | | -- -- <S> This is the sort of waveform that the drivers on the SG3525 are designed to produce. <A> Inspect the leading edge of the secondary waveform for short duration overshoot or spikes or ringing. <S> The rectifiers will see these transients, and in the absence of load, store their peak value on the output reservoir capacitors. <S> This is why, as Dave T says, many SMPS specify a minimum load. <S> If you don't want to do that, I suggest reducing those transients as the next move. <S> R-C snubbers across the secondary windings, to compensate for the leakage inductance, may be one way to do so. <S> I can't give an exact solution though someone may have a better answer. <S> Meanwhile but I hope this pointer is helpful. <S> As for the low efficiency I can't see the reason for that, but this looks like quite an unusual circuit <S> and I wonder if the fact that you are only putting half the DC voltage across T4 primary has anything to do with it? <S> The traditional push-pull circuit (ala valve/tube amplifier) with a centre tapped primary is simpler, possibly more efficient, drives both power FETs with low side drivers, and eliminates the need to split the DC supply (C6,C10). <S> It does have the disadvantage that the power FETs need to be rated for twice the DC supply (you would obviously need twice the turns on each leg of the primary, i.e. 66+66T) <S> Is there any reason why you did not pursue this traditional design? <A> I get a 404 on your link to the schematic. <S> But in general the reason for voltage rise on switched mode converters under light load is that the converter enters "discontinuous mode". <S> When operating under "heavy" load the converter operates in what is known as continuous mode. <S> The converter switches between two states, a state where the inductive component is being charged and a state where it is being discharged. <S> The output voltage depends on the time ratio of these two states. <S> Under light load however we get a third state. <S> A state where the inductive component is fully discharged. <S> This shortens the discharging state changing the ratio of charging to discharging and raising the output voltage. <S> Exactly what "light" and "heavy" mean depends on the component values in the converter. <S> So what can be done about this? <S> there are three soloutions all with downsides. <S> Add a dummy load. <S> This solves the problem but obviously wastes a lot of power. <S> Use a second active switch rather than the diode. <S> This allows current flow in the inductor to go "negative" and solves the problem of large voltage rise under light load <S> but it means under light load charge <S> is continuously flowing back and forth between the input and output side incurring losses. <S> Control for the second switch also increases complexity. <S> Actively change the duty cycle to maintain output voltage. <S> This works pretty well in general but there is a limit to how much pulse width can be reduced. <S> So to make this work at very light load requires increasing the delay between pulses. <S> That can cause problems with low frequency noise.
| This tends to be an issue with any switched mode converter that uses a diode as one of the switching elements.
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Critique for my short circuit indicator and guidance to improve it This circuit was designed to save electronic components in the assembled board from damage. It gives warning beep if there is a short circuit in the assembled board. Thus it helps to switch off the power supply immediately to save valuable components. If the circuit board is OK, Green LED lights indicating that power supply is normal. If there is a short in the PCB tracks or pins of components, Green LED turns off and Buzzer sounds indicating the short circuit. The circuit uses two NPN transistors BC 547 to sense the short circuit. If the assembled board is normal, current flows from the power supply through polarity protecting diode D2. The Assembled board gets power and Green LED lights. At the same time Q2 forward bias and its collector goes to ground potential. This makes Q1 off due the absence of base bias. Buzzer and Red LED connected to the collector of Q1 remains off. If there is a short circuit in the board under test, D4 reverse biases and Q2 turns off. Now the base of Q1 becomes high and it conducts. This turns on Buzzer and Red LED to indicate short circuit. Now, I'm using general purpose components here, what I want its to make this circuit "universal", or at least usable for currents up to 10 Amps, I have some questions, Would this circuit be able to replace slow blow fuses up to 10 Amperes? How can I adapt this circuit for negative voltages? Adding 1mH choke inductors right after D2 could help me out to stop transients from destroying my circuit in a worst case scenario? Any feedback about how can I improve this is welcome <Q> Very nice circuit! <S> To answer your questions: This circuit is not able to replace a fuse, because it does not protect in case of a short circuit, it only gives a warning. <S> Your diode D2 still needs to handle a forward voltage of 12V in case of a short circuit and will break as a result. <S> Or maybe even safer. <S> A circuit with memory, that turns off the power supply when the load is short circuit and until a button is pushed stays off. <S> To adapt this circuit for a negative power supply, you could connecto the 0V to the location of your current 12V and the -12V connection to your current GND connection. <S> You might however want to have the diode not in the GND connection, and move it in the path of the -12V supply. <S> In that case your diode D2 would forward from D5 and JP1 to your -12V supply, and you could leave out the inverter made with Q2 and R1. <S> What kind of transients do you mean here? <S> Short over voltage of your 12V power supply? <A> I see a couple of problems with your circuit: Without a current limiter somewhere, D2 must be able to handle the full short-circuit current of the power supply. <S> Even with a current limiter, D2 must be able to handle the normal current required by the load (and then the current limiter must be adjusted to pass the normal load current). <S> While some short circuits will cause excessive power supply current, which could be detected with this circuit, many faults ocurring during manufacture result in shorts between signal lines, or otherwise cause faulty operation without causing excessive power supply current. <A> If it's a short circuit, won't that pull the power supply down near zero... <S> nothing turns on!? <S> (Have you tested this circuit with different power supplies?) <S> When powering up a circuit for the first time (that may have shorts) <S> I like to use a current limited supply. <S> (Sometimes that's one with a changeable max current, but other times it's just a wimpy supply that can't source too much current. <A> (This doesn't really answer your question, but I think its relevant to other readers...) <S> You can get electronic fuses that are designed for this purpose. <S> For example, the TPS25921 . <S> Devices such as that have traditional current limiting circuits, but with integrated thermal protection. <S> On a short, the resulting behaviour is that they can 'test' the line repeatedly until the fault is cleared while keeping the current below the design current, making them suitable fuse replacements. <S> The TPS25921 is not rated for 10A, but there are many variants. <S> I'd guess they'd be roughly double or triple the cost of the components you'd replace in your diagram, but compared to the time to implement, debug, PCB space saved, etc, probably still be quite economical.
| As a solution to this, you could think of a circuit that turns of your power supply to the load, whenever the load is short circuit. In addition, the more advanced ones like the TPS25921 will feature over and under-voltage protection for transients (though you then need to protect the IC itself from them when it switches!), and fault indicators (for your LEDs).
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Using Smart Phone displays in Raspberry Pi or other computers Many smart phones aren't used after 2-3 years but the displays are good and i was wondering why no one is using those display within electrical components? Wouldn't it be great to have a module which makes it possible to use smart phone displays as general purpose displays for any computer? Is there something like thyt? <Q> Such LCD screens are often very difficult to use: <S> the connections are via a very small and flimsy flatcable the intelligence is not in the display but in the main processor <S> the interface is not documented there are lots of different versions on the electrical side, the interface often requires weird voltages and complex waveforms <S> The same arguments apply to the use of e-ink screens found in e-readers. <S> Note that the LCDs used in the older not-too-intelligent phones like the Nokia 5510 are sold and used a lot. <S> These LCDS are black-and-white and low resolution (48x84), but they are cheap, easy to interface, well-documented, and there are only a few versions. <A> The Nokia 5110 is one example. <S> I've used the Nokia 1202 display, which is 96x68 pixels, monochrome: http://dangerousprototypes.com/2012/01/18/cheap-i2c-1-3-96x68-monochrome-lcds/ <S> But the real problems are the tiny connectors and the lack of documentation. <S> There are a couple of documented modules available with 0.1-inch headers (break-out boards). <S> One is a 2.2-inch LCD with an ILI9341 controller chip, giving 320x240 pixels. <S> Really new smartphone displays will be higher resolution, but may have parallel interfaces that are even more difficult to connect to a microcontroller. <S> They use these (undocumented) parallel interfaces for higher update speeds, which are required if the phone can display video. <A> The MIPI DSI standard is used by quite a bunch of smartphone displays and there is a shield to use it on Raspberry Pi or anything else with HDMI out: <S> MIPI DSI Display Shield/HDMI <S> Adapter by twl <S> There is a video demo of a RPi driving an iPhone 4/4s screen with this shield. <S> The board also fits on Arduino GPIO pins and it is intended that one can eventually use drawing commands on Arduino to paint into the framebuffer. <S> For now, it's HDMI input only though. <S> The project is open source, PCB design and source code is available via Github (though as of 2015-01, it's the rev 2.0 software but the rev 1.0 board). <S> So you can build it yourself if you really want (not too easy though) but as of 2015-01, the boards are not sold anywhere but it is intended for the future.
| Some older monochrome phone displays are available as ready-made modules for Arduino and Raspberry Pi use.
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Level Translator for AVR and microSD I want to connect a microSD card in my AVR ATMEGA32, through SPI, which its operate with 5V.The microSD card i think that operates with 3.3V so, it is must to connected via bidirectinal level translator. I found this TI's translator http://www.ti.com/lit/ds/symlink/txs0101.pdf . which its accepts any voltage from 1.65V to 3.6V in input A, and 2.3V to 5.5V in input B. My matter is: Can i direct wire the chip's I/O between AVR and microSD?or it's only, as the datasheet say, "for open-drain and push-pull applications"? <Q> The SD specification stipulates 2.7V - 3.6V for all IO and power. <S> Using any other voltage will most likely cause death to your SD card. <S> Assuming you are running the SD card in SPI mode (most common for small microcontrollers), then you don't need to translate the signal from the SD card to the ATMega32, since 3.3V is a valid HIGH input even when running at 5V. <S> A common one is the 74HC08 AND gate, with both inputs of a gate tied together to form a buffer. <S> Run the chip from 3.3V, feed the 5V signals directly into the inputs via a small series resistor (say 1KΩ) and it should translate the 5V signals to 3.3V quite happily. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The 74HC08 (in common with the other 74HC series chips) has quite robust input clamping. <S> Any voltage that is more than Vcc + 0.5V, so anything above 3.8V in this case, gets routed through input clamping diodes to the Vcc rail, clamping it at Vcc + 0.5V, so 3.8V. <S> That current should be limited to below the "input clamping current limit", which is 20mA limit. <S> The 1KΩ resistors limit the current to a tiny (\$5 - 3.8 = 1.2, \frac{1.2}{1000} = 0.0012\$) <S> 1.2mA. <A> SD Card's supply voltage is a "standard", <S> in SD specifications, part 1, Physical Layer Specification version 4.10 , it states the operation voltage should be 2.7V~3.6V, so you can safely use it with 3.3V. <S> If your microSD's I/Os are 5V tolerant, then you certainly can directly connect them, but i've never seen such a card. <S> And for voltage translator, TXS0106 should be OK, and the old 74LVC4245A may be a choice too. <S> In NXP's doc: http://www.nxp.com/documents/application_note/AN10911.pdf , most I/Os are push-pull, so use a translator with push-pull output maybe better. <A> your atmega should be 3.3V high state tolerant so you can use incoming data without level translating and to the sd card you can use simply resistor voltage divider. <S> I used that setup with uart with pics and worked well.
| You will need to translate the 5V from the ATMega32 to the SD card though, and that will only be one way. A simple 74-series buffer or logic gate of almost any kind will do the job.
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Up to ~1000 A constant current supply 10-40 VDC So a little background to explain the need. Occasionally we need to test starter motors (for engines) and provide enough power when the load (engine) causes the starter to pull in excess of sometimes 1000 A. Granted, this is for quite large starters. The test would need to last a minute or so, but foregoing the need to alter designs, we'd like to just go ahead with a continuous demand rating. So I've got a fair understanding of linear power supplies and some common designs of higher currents, such as multiple transformers/taps, bridges, parallel transistors ( 2N3055 , etc.), but in my own experience I have only "engineered", mathematically, relatively simple sub-circuits, RLC, and other simple circuits ( 555 timers , opamp 747, etc.) using Kirchoff voltage/current, Thevenin, systems of ODE/PDEs, phasors, etc. But really I rely on the community for a suggestion on what to look for. (And ideally a schematic!) Some of the design requirements are 10-40 VDC (some are 24 VDC starters and a few are 36 VDC) Maximum of about 1000 A at 24 VDC ~ 25000 watts/VAC Be a current supply more than a voltage supply, similar to the idea of having the load in series with a much larger resistor. Lastly (I think), is to be able to use single-phase, two-phase, or 3-phase input. For larger loads, 110 VAC at the main is not going to work. For some idea of portability (we are a start-up business and have plans to grow and move) it would be nice to be able to run off "single" phase or three phase. This way each leg could supply a good amount of the load. (Can these be joined after the rectifier bridge or smoothing cap?) By the way, companies such as DynaPower, though awesome, are just way too far out of our budget for right now. Besides, it's much more fun to build it yourself! For what it's worth, we had thought of purchasing a used portable welder but were worried about the voltage adjusting itself to give the required current, and therefore power, to the load. I wouldn't know how to limit voltages at these currents... <Q> So, use exactly what the big trucks use - big batteries - these can be charged (I think "trickle charging" might be a little of an understatement of course). <S> Don't waste time and effort designing something that might take a year to properly engineer - use what the trucks use (or the buses, lorries, trains, planes or whatever). <S> You say that replacing the batteries is expensive - I bet it's cheaper than the effort to design a 40kW DC power supply. <A> I don't think that you will be happy with a regular power supply aproach, e.g. adjustable from 10V to 40V. For 40V and a regulation you need about 42V before the regulation. <S> When you put it to 24V and draw 1000A you will have (42V - 24V) <S> * 1000A = 18KW waste power. <S> You could mitigate this situation using a transformer with different output voltages for 12V, <S> 24V and 36V. Already taken this step i would also consider if your starter motors realy need a regulated power supply, this would reduce your circuit to a transformer, rectifier and some capacitors. <S> As a last step i would build a separate power supply for each voltage, then you could do some tests in parallel. <S> If you realy need a regulated power supply i would think about a switching one and considering the impression that i have from your experience in relation to the requirements i would perhaps buy one. <A> I would suggest using the the battery pack from a forklift as the basis for your project. <S> they should last for quite a few cycles.
| Batteries sound like a simple relatively inexpensive approach as well.
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When do I twist the wires going to transformer winding terminals, and why do I do it? I sometimes see the wires of winding are twisted outside of the transformer (or inductor), as seen in the images below. What is the purpose of this? When should I twist winding wires and when shouldn't I? As two conductors come closer and closer to each other, the capacitance between them increases. Wouldn't this parasitic capacitance between the wires cause problems at high frequency ferrite core inductors and transformers? Driving them will be the same as driving an LC tank circuit. There will be overshoots and ringing fluctuations. What trade-off does make us accept these side effects? (Note: This is not about using multiple twisted wires as a single wire inside a winding (i.e.; Litz Wire).) <Q> <A> Don't forget that the coil itself has a parasitic capacitance which may be much larger then the capacitance of the connection wires. <S> In high frequency applications you have to look for short connections anyway. <S> Additionaly the twisted wires make clear which ones belong together and it looks neat. <A> Say you have 4 wires on a transformer, or a balun with 3 windings (6wires). <S> Chances are, unlike your first picture, they are simply enamel coated wire... <S> Which one is which? <S> When you are installing the transformer, you can reach for a multimeter - but even then you will need to constantly check and re-check. <S> You COULD add insulated color wires, as per your first picture. <S> Then you just need to write down, or memorise, which colors are pairs. <S> Or, you can quickly twist pairs together, with the plan to UN-twist them (and cut them short) when you install the transformer.
| Twisting the wires exiting a high frequency wound component (that might be carrying several amps) is a way of reducing the electromagnetic interference generated to other parts of a circuit.
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what could be the visualization equivalent of the resistors in series? Why do resistors in series add-up? Is there any visualization equivalent of that? I have in mind electrons as cars driving on the highway, now a resistor could be a narrow spot (R1) where cars are moving slower, if I will "add" another congestion point R2 (resistor), after the first one, the cars that had already past by R1 will continue on R2, in case R2 is narrower they will slow down even more, in case R2 is wider it will be no problem since the cars will move freely,in case R2 is the same again the cars will move freely after the first congestion point R1 Can someone explain in that way why should the total R is equal to R1+R2? |Many thanks in advance! <Q> The cars analogy doesn't model what's going on. <S> Let's say you have a pump (like voltage source) that keeps a constant pressure from one end of a pipe to the other. <S> With just a section of open pipe, the flow rate (like electric current) will be quite high. <S> Now imagine installing a constriction (like electric resistor) <S> somewhere in the pipe. <S> The flow rate will be less. <S> Now install a second restriction further down the same pipe. <S> The flow rate will be even less. <A> Your car analogy is almost there, but not quite. <S> Instead of a single length of road, imagine instead a racetrack. <S> That racetrack is packed with cars, end to end, wall to wall. <S> No space between them. <S> Now there are some narrow points on the racetrack. <S> Each car is an electron. <S> The cars have to queue up to get through a narrow point. <S> Not just because it's a narrow point, but because there are cars already in there, and cars filling the next section of road queueing to get into the next narrow point. <S> That's the crucial difference with your analogy - you're assuming the "wire" after the resistor, and before the next resistor, is empty, but it isn't, it's full. <A> I don't know if some sophisticated math could explain the situation also with cars. <S> An often used analogy for electric current is water flow. <S> Think about a big pipeline, with low pressure you can push a lot of water through it. <S> If you replace part of this pipeline with a small tube the water flow will be strongly reduced. <S> Adding another such stage the amount of water flowing through will be about halved. <S> Putting the second tube in parallel to the first one you will get the water flow doubled. <A> Why do resistors in series add-up? <S> No fancy analogies, just ohms law. <S> The voltage across a resistor is the current through the resistor multiplied by resistance <S> i.e. V = IR. <S> Now, if two resistors were put in series (with the same current flowing) <S> the voltage has to become twice as big because you have <S> individual "V=IR"s adding up. <S> so now you have the equation 2V = <S> IR + IR = I*(2R). <S> 2V = I*(2R) <S> I think that should be clear but then again I've known and understood ohms law for a million years <S> and it's easy to forget what it's like for someone starting out on these things.
| A better analogy is water in a pipe. Each narrow point is a resistor. So the more "resistors" you have the more cars will be queuing, and the bigger the tailbacks will be.
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Purpose of 2 resistors in high pass filter? In the picture below, what is the purpose of R1? Compared to a high pass filter with 1 resistor, how does this affect Fc? Thank you. <Q> This is basically a voltage divider between \$R_2\$ and \$R_1, C_1\$ in series, so the transfer function is the following: $${V_{out} \over <S> V_{in}}={R_2 \over{R_2+R_1+{1\over <S> j\omega <S> C_1}}}={j\omega R_2C_1 <S> \over <S> 1+j\omega C_1(R_1+R_2)}$$ <S> The cutoff frequency is smaller than without \$R_1\$: instead of $$ <S> f_c={1\over <S> 2\pi <S> R_2C_1}$$ <S> it is $$f_c={1\over 2\pi(R_1+R_2)C_1}$$ <A> Your voltage source is an "ideal" source. <S> In practice, voltage sources vary, depending on how much current they need to source. <S> This is modelled with an external series resistor - \$R1\$ in your diagram. <S> When you draw a lot of current, the the output voltage will drop by \$V = <S> IR\$.If \$R_2\$ is more than \$10 <S> \times R_1\$, you can generally ignore \$R_1\$ for ballpark measurements. <S> However, since \$R_2\$ and \$C\$ form a high pass filter, changing your selection for \$R_2\$ affects your value for \$C\$. <A> Purpose of 2 resistors in high pass filter? <S> This is a (1st order) <S> high-pass attenuator . <S> That is to say, the high-frequency asymptotic gain is less than one (unity). <S> An ordinary RC high-pass filter has a high-frequency gain that approaches unity. <S> What if, instead, one desires that the high-frequency voltage gain be less than unity? <S> One approach would be to use the circuit in your question. <S> At high enough frequencies, where the impedance of the capacitor is insignificant, the voltage gain is approximately $$\frac{V_{out}}{V_{in}} \approx <S> \frac{R_2}{R_1 + R_2} <S> < <S> 1$$ <S> Of course, one must take into account the resistance of \$R_1\$ in the calculation of the corner frequency: $$\omega_c = <S> \frac{1}{(R_1 + R_2)C}$$ <S> Note that, for a consistency check, as \$R_1 \rightarrow 0\$, we recover the equations for the ordinary RC high-pass filter. <S> In summary, this high-pass filter circuit gives an additional degree of freedom: the high-frequency attenuation can be now be specified as a constraint. <A> I want to show another method without math: <S> In high frequency, the C is shorted, so the transfer function should have "high frequency <S> " gain $$\frac{R_{2}}{R_{1}+R_{2}}$$ <S> So, the high frequency gain now won't be 1, and will be smaller, dependents on the ratio of \$\frac{R_{1}}{R_{2}}\$. <S> The circuit's time constant basically is$$\tau = R_{total} \times C_{1}=(R_{1}+R{2}) \times C_{1}$$ And the well known first-order high-pass filter's transfer function is $$H(s)=\frac{1}{1 <S> +1/(s\tau)}$$ <S> Combine the high frequency gain and substitute the \$\tau\$, we get the whole $$H(s)=\frac{R_{2}}{R_{1}+R_{2}} \frac{1}{1+\frac{1}{s(R_{1}+R{2})C_{1}}}$$
| The first resistor limits the current into the capacitor, therefore it can charge slower, with a time constant of \$\tau=(R_1+R_2)C_1\$.
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Flashing leds from bicycle tail light I'm not sure if this is the right place but I'm a bit puzzled by this (cheap chinese) bicycle tile light. I opened it up, but I don't seem to get it working again. It has three components: the housing with the batteries the circuit with the LEDs a rubbery blue button but with a hard black centre piece By simply pushing the button you could change the flashing modes. From all on, over all flashing, flashing on at the time, to all off. There were 8 modes (including off). It's somewhat frustrating that with just these three components that I can't get it to work again (I haven't tampered with them). And now I'm also curious as to how it works. So does anybody have any ideas? the blue button black centre piece makes some connections but how does it work? what's the black dot on the circuit board? ... Thanks for any pointers, help, ...! Some pictures: <Q> It creates a connection between the PCB traces that are exposed (no green solder mask) on the right of the PCB. <S> This is a common cheap way of making buttons, and is found everywhere in small / cheap devices, such as your TV remote control. <S> This is the main control of the circuit - the "brains" if you will <S> - that tells the LEDs just what to do and when to do it. <S> This "chip on board" is incredibly cheap to manufacture in huge quantities, and you'll usually find it in such things as calculators, watches, kids hand-held games, etc. <S> It can be quite fragile though, and prone to breaking if the PCB is flexed, which might be what has happened in your case, especially as the PCB seems to be cheap SRBP (Synthetic Resin Bonded Paper), which is quite flexible as PCBs go. <A> Others have answered the questions relating to the parts and the functioning already sufficiently. <S> Before discarding the light you could try several things. <S> It looks like the whole thing is very susceptible to connection problems. <S> Perhaps you have an alignment problem. <S> It seems as well that the PCB is not clean in the corresponding region. <S> Normaly traces used for buttons are gold plated to prevent oxidation. <S> Chances are good that sooner or later the bare copper will oxidice and the light will stop working correctly. <S> One can not see in your pictures how the battery is connected to the PCB. <S> Potentially you may have a problem there as well. <S> Perhaps you could connect the battery with two soldered wires. <A> You missed the fourth component. <S> But that's not surprising, seeing as how it's barely there. <S> Under that blob of black epoxy on the bottom of the board is an integrated circuit, a bare die with bond wires connecting it directly to the PCB. <S> This construction is called "Chip-on-Board". <S> If that IC is dead then you won't be able to get the device working again. <S> The black center piece on the button is carbonized rubber. <S> It makes a connection between the "fingers" on the PCB below it. <A> That's right, below the black epoxy, there is a IC, I suppose that is a specific driver IC for LED, because for example a IC555 would need more passive elements like R and C, that is not in this PCB Try to bypass the battery to the led, and make a short in the wires of the buttons. <S> If it didn't work, i suppose that the IC is dead.
| Connecting the batteries the wrong way will most definitely not work, and in a worst case scenario can actually kill the chip, especially in cheap devices like this where there is no money spent on "hardening" the circuit against misuse. You could try to clean it with alcohol and perhaps (gently) with scouring powder. First the black part of the blue button should connect two traces of the PCB when pressed. The "hard black centre" of the button is carbon impregnated rubber. As has already been mentioned, the fourth component you didn't mention is the black blob of epoxy with a microchip underneath it. It's also worth noting that all these components, including the LEDs, are sensitive to the polarity of the power.
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LED bulb still emits light when turned off I have recently purchased an IKEA LED bulb with G9 fittings, to replace a halogen one. This one . The bulb is installed in a lamp with 3 other halogen G9s in a 3 + 1 configuration with a 3-state switch. "up" lights up all four bulbs, "down" lights up only the socket where the LED is. I have found a strange behaviour. When the lamp's switch is in the off position, the LED bulb still emits a very dim light, whereas the halogen ones do not. What is happenning here? Is this normal? From my limited knowledge I would guess the lamp's circuit has some residual current which is not enough to light up the halogen bulbs, but makes the LED emit the dim light observed. <Q> There are two possibilities the switch is inserted: Switch switching the voltage line. <S> Switch switching the GND line. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The capacitors shown in the circuit are the capacities <S> the (more or less) long lines form to GND. <S> If the 2nd circuit is what you have the lamp always is connected to alternating voltage. <S> In that case there is a possibility of some very low alternating current flowing via C10 to GND even if the switch is open. <S> You can find out if your configuration is the 2nd circuit by testing with a one-contact neon test light . <S> It is the case if the test light lights up when touching one of the two connections in your lamp socket even if the switch is turned off. <A> I would guess that your off-low-high switch has an indicator light so you can find it in the dark, so the LED lamp ends up in series with the indicator when the switch is 'off'. <S> Edit: For the lamp to emit visible light, current must be leaking across the switch. <S> Possible reasons are: Indicator light (ruled out by OP) Capacitor or RC snubber across switch to avoid EMI when switch is flicked (not seen often in North America, but might be a possibility in Europe). <S> Dimmer circuit that has an internal snubber and is not switched off entirely damage (arcing and tracking or moisture) to the switch. <S> A mechanical switch by itself should not leak enough to light an LED. <S> It will not draw more current than it would with the halogen lamp, and all <S> but the last cause are nothing to be concerned about. <A> I took a hint from a previous comment and flicked the plug 180 degrees on the wall outlet. <S> It seems that the lamp's circuit was interrupting the GND line, like illustrated in @Curd's answer. <A> Added for safety value - even though a very old question: Andre said: I took a hint from a previous comment and flicked the plug 180 degrees on the wall outlet. <S> The bulb doesn't emit light when in the "off" position anymore. <S> It seems that the lamp's circuit was interrupting the GND line, like illustrated in @Curd's answer. <S> WARNING - ELECTROCUTION HAZARD <S> If reversing the polarity of the circuit fixed the problem then you probably have Phase / Live connected to the circuit at all times and it is improperly wired and a "death trap". <S> If only lighting equipment without grounded metal accessible to users is used then you may not have anyone killed. <S> But if you use equipment on the circuit where neutral and ground are connected (as happens even though it shouldn't) then such equipment will be "safe enough" on a properly connected circuit BUT lethally dangerous on this one. <S> Getting it fixed while everyone is still alive is liable to be "a good idea".
| The bulb doesn't emit light when in the "off" position anymore.
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Relay alternatives that are silent and support 500mA+ What can I use as a silent "general-purpose" relay alternative? There are various scenarios where I will need a relay-like device that allows me to control one circuit with a separate isolated circuit. I have 3.3-5V microcontrollers and I want to use them to control, say... A 12V DC 1A circuit (large-ish current) A button on a radio (low-voltage and current) A 24V AC circuit A simple 5V magnetic relay+diode+transistor would work for nearly any situation (and no calculations necessary), but they are too loud. I see two alternatives: Optocouplers and solid state relays. Unfortunately, most of them seem to have a maximum output of around 50mA ( I generally need 500mA or more ). They are also harder to find and more expensive. Finally, they are not a simple coil and contact setup, and there is much less documentation on how to implement them with a microcontroller. Are optocouplers and solid state relays the best solution? I'm looking at these: http://www.mouser.com/ds/2/149/FOD3150-95595.pdf http://www.mouser.com/ds/2/427/vo14642a-279692.pdf http://www.toshiba-components.com/docs/opto/TLP152_en_datasheet.pdf <Q> The simplest is to use a solid-state relay. <S> You can find them from a couple of hundred mA up to 100 amps, in both DC and AC. <S> I would look for one with a logic-level input; their input is an LED hooked to an opto-isolator, so you drive it just the way you would an LED; with a resistor hooked up to output from the microcontroller. <S> They aren't cheap, however. <S> You can use an opto-coupler to drive a transistor to switch most DC current. <A> <A> A simple MOSFET can deal with most of your DC needs. <S> They can switch (if you get the right ones) <S> many amps (even hundreds of amps if you look around). <S> For the higher currents you may need to have a multi-stage driver which uses a small BJT to switch the larger MOSFET, and of course there may be heatsinking needed. <S> Unless you really need galvanic isolation I wouldn't bother with relays (solid state or otherwise) and optocouplers and such. <A> If you're only concerned about digital outputs, you can use the op amp as a comparator , where your input signal is tied to the non-inverting input and some other threshold voltage (eg 1/2 Vcc) is tied to the inverting input (with a transistor again at the output). <S> In theory this gives you complete isolation from anything attached to the output. <S> What I mean by that is that regardless of what's attached to the output of the op amp, your input signal will just behave as if it was connected to ground through a fixed, and very large, resistance.
| You could perhaps use an operational amplifier configured as a voltage follower and use its output to drive a transistor capable of switching 500 mA. Realistically, you can expect an op amp to provide about 2 Mohms of impedance across its inputs (a more accurate value should be stated in the data sheet for your particular op amp), and it shouldn't produce any acoustic noise. A reed relay should solve all of your problems.
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How to properly discharge a high voltage capacitor? I have recently blown a capacitor on my home made emp. I have got the perfect replacement which is an electrolytic capacitor, 330v, 140 micro farads. I have charged it but I want to learn how to safely discharge a high voltage capacitor. I was going to short the capacitor but my friend said that high voltage capacitors can blow when you short them. I don't have any resistors that I can discharge with. Any ideas? Or can you short the capacitor without it blowing? <Q> A resistor is called for. <S> If it's charged to 300V and you want to limit the current to 1-2A <S> you need a 150-300 ohm resistor. <S> DX has a suitable Dale-style 100W 200 ohm resistor for less than $4 including shipping, or you can get a genuine one for a bit more. <S> Peak power will be 450W <S> But a typical resistor of this type rated for 100W continuous can handle about 100J pulsed, and you have at most <S> CV^2/2 <S> = <S> 7.6J. <S> As always, check that it actually is discharged before handling (and after that, maybe short it just to be double sure). <S> Note that because of dielectric absorption, capacitors that are discharged briefly can appear to re-charge themselves somewhat. <S> Probably not an issue with such a low voltage, but it can be enough to cause a jolt on high voltage capacitors. <A> Connect it across a hair dryer or toaster or soldering iron etc. <S> I tend to use soldering iron. <A> So your friend is correct as discharging by directly shorting it could be bad. <S> The previous answer has given a good solution already, discharge it through a resistor could do it. <S> This is one of the safety demo my professor love to do. <S> He will touch the two ends of a cap (which is the size of a bottle) with a metal rod. <S> You will see a huge bright arc at the contact and a huge sound (worse then firing a gun). <S> It is very very very frightening . <S> Also, if you accidentally touch the two ends of the cap while trying to short it. <S> you are basically cooked.
| For the last question, it is possible to discharge the capacitor by directly shorting it without blow up the cap.
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Why 555 timer IC has three 5k resistances and not any other values? Why does 555 timer IC has three 5k resistors and not other values, like 10k-10k-10k or something else? <Q> The original 555 with 5K resistors: <S> And here is a CMOS version with 40K resistors: <S> The choice of resistors for R7, R8, R9 (bipolar version) would be influenced by two things- 1) <S> The desire to minimize power consumption (as high value as possible without using up too much chip area) <S> 2) <S> The desire to minimize temperature variations due to beta changes of Darlington pairs Q3/Q4 and Q12/Q13. <S> The second point does not apply to the CMOS version. <S> It's easy to see that the Thevenin equivalent source resistance for either node is 2/3 of the resistor value. <S> the circuit is symmetrical (horizontally) and the currents will be the same as the trigger and threshold currents. <S> The currents are quite different, probably because of the low beta of the lateral PNPs. <S> Hans Camenzind says the comparator offset can be as large as 30mV, which implies a large offset voltage on top of the 7mV maximum due to the input bias current, but the input bias current is quite variable with temperature (maybe 3:1 over operating range). <S> If we assume it changes from 0.7uA to 2uA, at 5V that would be a change in threshold of 0.25% or about 15ppm/K. Overall actual accuracy is about 24ppm/K, so the resistors are not overly dominant (the offset will change at something like proportional to absolute temperature). <S> Back in the 70's, 10mA at 15V or 3mA at 5V was considered reasonably low power, so HC probably chose the resistors as being "reasonable" -- not too big and not too small, and this was all pre-computers <S> so he would not have had the option of running an optimization routine to get odd value that minimized some arbitrary cost function. <S> Here's the actual die photo ( as taken by HC and published in IEEE Spectrum ), with the resistors highlighted. <A> The value is a tradeoff among various design constraints. <S> On one hand, you want the value to be large, to minimize the quiescent current requirements of the chip. <S> There is also the consideration that you want the input bias currents of the comparators to be a tiny fraction of the current in the resistors. <S> Taking all of this into account, the designer settled on a value of around 5K. <A> Let's look at the silicon! <S> The three 5k resistors are the horizontal bars at the top of the chip. <S> Making resistors in silicon is a pain; the available materials are all fairly conductive, so it's hard to make large-value accurate resistors. <S> At the time of design of the 555 <S> the minimum feature size was quite large, large enough to be seen with an optical microscope as in that photo. <S> There's the additional design constraint that those resistors affect the accuracy of the timer. <S> That probably determines the choice of the material, which will have a certain resistance in ohms per micrometer. <S> From there, we can see that the 5k resistors could not be made much larger in the available space. <S> Perhaps they could have been made 6k, but choosing 5k makes it simpler for the users of the chip to compute timer values by hand. <S> (I think the "5.0E" on the chip there is actually a registration mark indicating that that's layer 5, like the smaller ones at the top of the chip. <S> Not a component value.)
| On the other hand, large-value resistors take up a lot of physical space on the chip. We can easily divine what the production limits are on the currents drawn at those nodes from a 555 datasheet- It doesn't matter what the exact value is, as long as all three resistors have the same value.
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What is the name of this connector from a fan? I am EE hobbyist and I often used recycled parts in my projects. I came across this connector to a Heat Sink Fan and would like to know the name of it. Thanks for the help. <Q> Found it in section 4.1.5 on page 17 of this doc: <S> http://www.formfactors.org/developer%5Cspecs%5Crev1_2_public.pdf <S> Here is a Molex product page: http://www.molex.com/molex/products/datasheet.jsp?part=active/0470541000_CRIMP_HOUSINGS.xml <A> The standard 3- and 4-pin fan connectors are Molex KK series connectors. <S> According to Wikipedia the part numbers for the 4-pin version are: Female Housing: 47054-1000 <S> Female Contacts: <S> 08-50-0114 Male Header: 47053-1000 <S> Of course, many manufacturers produce compatible connectors, and most 0.1 inch pitch headers will work with these connectors provided they are long enough (approx 0.3 inch). <A> I don't know very much about electronics, but I recognize this as the 4-pin connector of a fan that uses PWM to regulate the speed. <S> It's a newer technique where previously varying the voltage was used. <S> It will work just fine if black is connected to ground and yellow to +12V. Green is used to measure the RPM and blue regulate the speed.
| The three-pin version (non-PWM) is referred to by some computer technicians as a "TX3 connector," but this is something of a misnomer (similar to the moniker "Molex connector," referring to the once-common 4-pin ATX power connectors originally produced by AMP). I think that it could be Molex 47054-1000.
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Embedded system for image processing project Is it possible to do image processing (to extract numbers and text) using a microcontroller? If not, what are the alternatives to do this task? <Q> Yes, image processing can be done using Microcontrollers and Microprocessors. <S> You can either do image Processing using Arduino with OpenCV or MatLab. <S> RPi has an inbuilt GPU which is better for your applications <S> but BBB can also be used for image Processing application considering the fact that it has a better and faster ARM processor. <S> Since BB and RPi both run on linux you can use the most common OpenCV or simpleCV to do the task. <S> You can use Image Processing using BBB and OpenCV or RPi and OpenCV. <S> OpenCV is an It has C++, C, Python and Java interfaces and supports Windows, Linux, Mac OS, iOS and Android.. <S> I would suggest you to use OpenCV that uses C++ for BBB (C++ is faster compared to RPi and BBB doesnt have GPU <S> so there is a chance to slow down processing) and use OpenCV that uses Python for RPi <S> (python is much easier to code and RPi <S> has a GPU) . <S> I have used both RPi and BBB in the past <S> but I would suggest you to buy a RPi for your application since its cheaper and has a huge Documentation online. <S> Update <S> : There are many OCR based algorithm's available online for OpenCV but are not that reliable. <S> I think best Open Source OCR Engine is Tesseract. <S> You can get more idea about this from the thread Tesseract or OpenCV for OCR . <A> It is really depending on which kind of microcontroller and peripherals we are talking about, as well of what kind of image processing. <S> You should ask yourself following questions (and answer them of course) 1) <S> First problem is to get the image data into the processing unit - how will it be done? <S> Where is the data is coming from? <S> Where is it stored? <S> 2) <S> How fast the processing should be? <S> 3) Does the image have to be fully loaded to processing unit memory in order to be processed or just part of it is enough? <S> It will lead to memory requirements 4) <S> What are you going to do with the processed data? <S> In general, image processing is done on a special kind of microcontrollers/processors, called DSP (digital signal processors) that are optimized to process large arrays of digital data. <S> So after you have answered the above questions, look into these. <A> Extracting numbers and text from a document is called Optical Character Recognition (OCR). <S> This is a fairly complex topic, and there are several approaches to performing OCR. <S> Although the term OCR implies using some sort of camera to image the text, more often than not the characters are "read" from an existing document, such as a Word or PDF file, a FAX, or a scanner output (which maybe an image file such as a jpeg). <S> So in all of these cases, the format of the file has to be taken into considerations. <S> In any case, there is usually a pre-processing stage, in which the characters are de-skewed <S> (aligned so they are vertical for example), and then edge detection is performed. <S> If the characters are known to be created from a particular font, such as Times Roman, or from a few known fonts, pattern recognition can be used. <S> This is fairly easy but requires quite a bit of fixed memory (which can be in flash). <S> If the characters are from a variety of fonts, this becomes more difficult. <S> In that case feature extraction may be used instead. <S> The hardest challenge is if the characters are free-form handwritten. <S> If it is known that only numbers are being input in a particular field, this also makes things much easier. <S> Likewise, a dictionary may be used to help recognize words (once again, this would require a lot of flash memory.) <S> Given enough memory, both flash and RAM, there is no reason a microcontroller cannot perform OCR, although it probably would not be able to do so in real-time -- it would need to be an off-line task. <S> In addition, a 32-bit processor is highly recommended. <S> OCR as mentioned above is primarily a pattern recognition task, and does not necessarily require DSP; however DSP could be useful in cleaning up the characters before OCR post-processing, such as removing noise. <A> Basically, no. <S> Do it with an embedded computer such as the Raspberry Pi or Beaglebone. <S> These are great for embedded computer vision, are very cheap, and the Beaglebone has reference schematics <S> so you design your own system after prototyping stages are over. <S> Text recognition algorithms probably have many open source libraries available now, but they will all be for proper linux OS, rather than a real-time microcontroller OS. <S> It's just much easier with a full CPU style architecture. <S> On the raspberry pi (using OpenCV 2/3 with Python bindings) I got 20-30 frames per second on object recognition and tracking with slightly down-sampled images from the Pi camera module.
| Or if you are more interested in Microprocessors you can use a embedded computer such as the Raspberry Pi(RPi) or Beaglebone(BB) which is more suitable for powerful image processing projects.
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Can 2 boost converters outputs be put in parallel at the same voltage? I have a slight doubt here. Can the outputs of 2 boost converters set to output the "same" voltage (isolated, they would never be exactly the same obviously) be safely tied together, as long as there is a capacitive load? If so, is there a minimum to that load? Context: simulate this circuit – Schematic created using CircuitLab I need at least 4Ah of battery capacity, and I was only able to find single cells charger+boost modules , so I have to put them in parallel. It's for long term logging, the current is very low: 30mA peak, and around 5mA 99% of the time. <Q> I can not tell if you can put two of these modules in parallel without seeing a circuit diagram. <S> However if one module can deliver enough current for your application, you can do it with only one module. <S> LiPo cells are a bit picky when it comes to voltages above 4.2V. <S> When you put two cells in serie the maximum charching voltage would be 8.4V. Now consider one of these cells is defective and ceases to charge at 3V, the other cell will now get 5.4V - too much. <S> Thats the reason why in a serial configuration each cell needs it's own charging or balancing circuit. <S> The key here is that you like to put the two cells in parallel, they will be charged up to 4.2V, no problem. <S> They will not catch fire or explode because of over voltage. <S> As usual there are drawbacks, in this case two. <S> The charging current will be split, the charging time doubles. <S> If one cell is defective it can discharge or even demage the second one, they are not isolated from each other. <S> As a sidemark, before putting them in parallel I would bring them to the same state, e.g. fully charge them separately. <A> Given the diagram you have drawn in your question I would be tempted to look for two buck converters that each produce 2.5 volts on their outputs and then wire the two converters in series to give you 5 volts. <S> You say they can be ran "isolated" and therefore this should work. <S> When it comes to charging them, two charging power sources will be needed and these also need to be isolated. <S> Alternatively, find a charge-control circuit that is suitable for 2 x LiPo in series and then use a buck regulator to down size the raw output to 5 V. <A> I know this is an older post, but I wanted to mention that parallel sources like this do not inherently split the load current very well. <S> Supplies are often paralleled in commercial power systems but they have circuitry built in to evenly split the current demand between all supplies. <S> There is often an extra connection between each supply that allows the supplies to match voltages very accurately and thus supply currently evenly. <S> It's true you can do OR-ing diodes, as they are called (they work as a logical OR function) <S> but as you already said, this causes power loss, and it still does not result in real sharing. <S> At minimum, OR-ing diodes are a good idea, but if you consider the topology of a boost converter, it has a diode inherently in the design. <S> I don't think you would get reverse currents when one goes dead. <S> If you aren't bothered by the fact that one battery would likely go dead before the other, then maybe this scheme could work. <S> There could be efficiency tradeoffs in some ways also. <S> I think you will get overall worse efficiency at low load conditions, because you will have two converters switching even if only one is needed. <S> There are losses in switching converters any time they are switching the inductor. <S> One last note for the future, 3.3V is really suited well to running on Li-Po and a lot of devices are coming with Li-Po charging circuits built right in :). <A> Made and test this yesterday with two of these very excellent boards: <S> https://nl.aliexpress.com/item/wholesale-2-pcs-lot-DC-DC-2A-Adjustable-Step-Up-Boost-Power-Supply-Converter-Module-2/32266587114.html <S> To power an Amp (2x50W, class D, very efficient) with 3.7V 8000mah (4x2000mah) battery to 24V (4A) <S> and it's working great! <S> Really. <S> On each board there is a 3A schottky diode present that protects the output so it is safe to connect these boards together <S> (diode is not shown on the aliexpress product picture, dunno why). <S> For best results you NEED to be SURE <S> both modules are connected to the same source (and switched on at the same time) and output EXACTLY the SAME voltage. <S> In this case it is 23.8V <S> and I use a large 25V capacitor 4700uF connected to both output 'lines'. <S> Before you connect these modules to eachother, you have to setup first the voltage! <S> I made a jumper on the back to be able to disconnect or connect the boards to eachother to be able to setup or to troubleshoot. <S> Tested the boards in this configuration and measure the voltage at high load and it seems to be pretty stable (almost no voltage drops until you reach INSANE because the Amp is able to output 100W and this config can't reach this). <S> Very reliable, very nice. <S> I will try to connect another board to get at least 5A so it can handle the full potentional of the Amp. <S> So yes, it is possible.
| Sometimes this is done for simplicity in redundant systems where either supply can handle the load independently, in the case that one fails.
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Why doesn't PCIe and similar signaling systems use full-duplex links? I'd like to get more PCIe bandwidth for GPU compute applications. It occurred to me that PCIe bidirectional links are really dual simplex (a pair of unidirectional links). That means if there's no data to transmit in 1 direction, 1/2 the links are idle! So, my question is obviously, why not allow the links to be bidirectional (half-duplex)? I can't think of any other networking physical layer that's bidirectional, except I2C and Wifi, which I doubt can reach GhZ speeds. Is high speed incompatible with bidirectionality? Update:I did think of a high speed bidirectional interface: a memory bus. XDR memory uses differential signaling and claims 20 Gbits/s per pin! Why does it work for memory but not for PCIe? <Q> I think you are assuming here that the number of physical lanes is the bottleneck, and that if you only had more bandwidth then the logic inside the GPU would no longer be held back. <S> I don't think that is really the case. <S> Then your whole pipeline has to be designed to support this new double dataspeed. <S> Not to mention a high speed serdes is a complex beast, so now you'd be making each piece even harder by trying to support both RX and TX. <S> When designers want to double bandwidth it's simple to just add lanes, just step and repeat. <S> When PCI-SIG wants to increase the max speed for the next generation <S> it's easier to just add features and increase frequency. <S> PCIe in particular is based on the concept of each lane pair being bi-directional. <S> From the time it starts up and negotiates speed and TX/RX parameters, to later on for flow control. <S> If you decided to take away one TX and change it to an RX, now another TX in a separate part of the chip would have to take over it's duties <S> and it all gets pretty messy pretty fast. <S> Technically though if someone wanted to design a HS link system that allowed you to change your TX to RX for extra receive bandwidth it could be done. <S> But it really would only make sense if pins were expensive because why add all the complexity to your system and design to support this when you could just add a few pins. <A> Is high speed incompatible with bidirectionality? <S> Yes, because of turnaround time. <S> It takes roughly 1ns for a signal to travel 30cm, plus delay through the TX and RX at each end. <S> So it will take at least several nanoseconds, possibly hundreds of nanoseconds, to reverse direction. <S> Then you have to work out how to control the switchover, which makes the signalling more complex and imposes a latency of its own. <S> Note that PCIe lanes may not have exactly identical propagation delay, so signals that set off at the same time may arrive at different times. <A> I feel bad about answering myself, but unlike most other people who think it can't be done efficiently, I think the answer is more qualified. <S> I agree that it can't be done for PCIe. <S> The main assumption is that the communication is peer to peer. <S> Then, it would be pretty difficult like SomeHardwareGuy claimed because if you're trying to send and there are no available TXs, you would have to buffer your packet. <S> But bidirectional communication at GHz speeds is possible on a memory bus. <S> The key difference has to be that only the memory controller initiates communication, so it's known precisely in which clock cycles who should transmit and who should receive (Wow, so those memory timings (e.g. tCAS) aren't just for bragging - they're actually for synchronizing the memory controller with the memory). <S> That means no need to buffer packets and no need to negotiate whether a links is to be used for RX or TX.
| To give you more bandwidth in one direction you need more area, power and resources dedicated to say receivers.
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Electrically control material to channel light from specific angle I am wondering if there is any material that can be electrically controlledso that light from a specific angle or diretion is received. Is it possible to manufacture such materials using semi-conductormanufacturing processes? <Q> DLP does something like that, using an array of micro-mirrors made with semiconductor (MEMS) processing techniques, but it only has two two stable directions. <S> Possibly a combination of static optics and a micro-mirror array could be used for whatever it is you have in mind. <S> Here is a TI datasheet . <A> Use of polarisers plus a Pockel Cell or Kerr Cell or liquid crystal materials may allow what you want. <S> A Pockel cell uses material which changes optical rotation angle linearly with increasing applied electric field while akerr cell responds to filed squared. <S> What sharpness of cutoff do you require. <S> Pockel pictures <S> Wikipedia - Pockels effect Pockels primer Tutorial - 1986 but looks OK Wikipedi - Kerr effect <S> Related: http://fastpulse.com/pdf/pcp.pdf <A> What about doing something similar to optical image stabilization where a lens is moved laterally to compensate for rotation? <S> With the correct elements, it should be possible to get a decent range. <S> Another option is some sort of optical phased array. <S> If you only need one dimension, then you could build an array of waveguides, each with a phase shifter, then combine all of the outputs at the detector. <S> Adjusting the phase shift offsets will change the receive direction. <S> Probably not very much fun to build, though. <S> These techniques are both rather experimental at this point. <S> They are also narrow band. <A> A pair or trio of servo motors and a light sensor with a tunnel/light blocker can work. <S> One controlling each axis. <S> Think motorized mirrors on a car. <S> With micro-servos or actuators (solenoids) it can be pretty compact. <S> Not solid state, but <S> your question is pretty sparse with details.
| Another option might be some sort of phase shift MEMS or phase shift liquid crystal (LCoS) to generate a phase gradient in two dimensions.
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Why is resistance increasing while I'm measuring using a multimeter? I have two copper strips with a one-megaohm resistor between them. That makes the resistance between the two strips 1 M ohm. I am measuring resistance between two copper strips using a multimeter. With this arrangement, the reading on the multimeter shows 985 K ohm. Now I put one water drop in between the strips, then as it is in parallel with the one-megaohm, and since water has some resistance, the equivalent resistance between those two copper strips will reduce. That is what is happening, and the multimeter reading shows 473 ohm; that is correct. But the problem is, it starts increasing slowly like 475, 478, 482.... to 736 K ohm after some 5 to 10 minutes before I turn the multimeter off. I thought it may be that the water drop might not be at the same position and there may be a little bit of spreading so that might be the cause of resistance change, but the question is why it is always increasing and not decreasing, and why so much of change like from 475 to 736 is a huge change of approx 200 K ohm. I believe there is some other reason. Can someone give a solution to this problem? <Q> After you run this test for 10 minutes, wipe off the water and take a close look at the copper where the water drop was. <S> You will see some discoleration. <S> As you probably notice, bare copper left in the elements is no longer that bright copper color after a while. <S> The same thing happened to your copper electrodes, except that the electric current speeded up the process. <S> The reason the resistance goes up is because the corrosion layer has significantly more resistivity than copper. <S> This is one reason the mating surfaces of electrical connectors aren't made from copper. <S> They are usually made from material that doesn't oxidize, like gold or nickel, or forms a conductive oxide, like tin. <A> My personal interpretation. <S> Clean water, e.g. destilated, is a poor conductor, so I assume that you have used tap or mineral water which is a much better conductor. <S> The salts solved in normal water provide the charge carriers in form of free ions that allow a current flow. <S> With that in mind i can think of two possibilities why the resistance rises: <S> Electrolysis takes place and elements of the salts where set free, for example chlorine. <S> This would reduce the amount of available charge carriers. <S> In the electric field between the two copper strips, the ions will become separated and can not longer float freely in the water. <S> That would also limit the current flow capabillity. <S> Eventually someone with an (electro)chemical background can say more. <A> A number of things may be happening here. <S> The first is that there may be evaporation of the the drop. <S> The second is related to Olin's assertion of "corrosion", but more specific. <S> Electrons aren't conducted through the drop of water -- <S> Ions are (which is why distilled water is a poor conductor). <S> The ionic reactions at the water/copper junction are non-reversable, so you have what's called a "polarizing" electrode. <S> Thus your water drop is slowly reaching a state where it will stop conducting electricity. <S> The "corrosion" is a byproduct of these electrochemical processes. <S> I don't believe you're seeing much electrolysis of the water -- which is another possibility. <S> Is the water bubbling? <A> I think Olin's answer is correct, but there's another factor that may account for at least part of the effect you're seeing. <S> I'm <S> pretty sure water has higher adhesion to clean copper than to epoxy (which forms the surface of most PC boards). <S> When you first drop the water on the board, it'll form a relatively round drop on the surface because of its cohesion. <S> Over a short time, however, the higher adhesion to the copper will cause it to "pull" into two drops over the copper. <S> As the water migrates toward the copper traces, you'll end up with less and less water between the traces, so you have a thinner connection, leading to increased resistance (and eventually, all the water will migrate to the traces, and you'll be left with the original resistor as the only connection between the traces). <A> The question which led me here was in regard to the increasing resistance of an agarose gel petri dish over time. <S> As suggested by Olin, I tried changing my electrodes from copper to another material, but saw no difference. <S> After some more research, and taking into consideration that agarose gel contains salt ions which carry the current, it is likely that the Ohmmeter is in a way depleting the ions available for conductance since they have been drawn to their respective poles.
| Essentially the copper was corroded a bit where the water drop was.
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What is the function of a diode connected to a GPIO? I'm reading a schematic (done by a former employee) with a microcontroller, and I came across this circuit: What is the function of D1? Also, I think that R2 isn't necessary, because the control input come from another GPIO that always has 1 or 0 as its output. Besides this one, there is another block of this same circuit but connected to the RESET pin of the microcontroller. This part of the circuit is used to flash the microcontroller via ISP . R1 = 47kR2,R3 = 10kD1 = 1N4148 <Q> It's there to protect the B-E junction of Q1 from reverse breakdown. <S> For positive input voltages, the B-E junction of Q1 will conduct, and the voltage at the base will be limited to about <S> +0.65V. <S> As long as R1 is sized appropriately to limit the current, fairly arbitrary positive voltages can be applied to the input. <S> D1 provides a similar path for negative input currents, guaranteeing that the base voltage never goes below -0.65V for the same range of negative voltages. <A> That is called an "input clamping" diode. <S> It's often done with two diodes, not just one (one to Vcc as well as one to GND), and is used to "clamp" the incoming voltage to ground minus the forward voltage of the diode (or Vcc plus the forward voltage for the upper diode). <S> In your circuit it is specifically to remove any negative voltages from the input. <A> Question poster mentioned the critical info in his comment "Originally this control pin was receiving signal from a DTR pin, of the old RS232 serial port" - <S> that's the real answer to "why"; RS232 uses negative voltages for signalling (as low as -15v) <S> , hence the (fairly large) current limiting resistor R1 and diode D1 to stop the transistor getting fried by being reverse-biased across the base-emitter junction. <S> You don't normally see (or need) <S> this kind of diode if it's being driven by a CPU; typically just a current limiting resistor will do. <A> As others have suggested, but not explicitly stated, the diode is basic input protection for an externally accessible input. <S> I.e. <S> the pin is likely on a connector which the user of this device drives. <S> On the other hand, if this control input is not external, the diode is likely not required unless it's being driven from a circuit that provides a negative voltage under normal operation (i.e. maybe its a negative going pulse like <S> +/-5V <S> square wave, or being driven by an AC coupled waveform). <S> Protecting a transistor like this if negative voltages are not expected is probably pointless since something far worse has happened if your circuit is driving a negative voltage unexpectedly, and in which case the base resistor R1 likely provides enough protection anyway. <S> Edit: <S> Just noticed you mentioned that a microcontroller drives the CONTROL input, so <S> Also just as you alluded, if the microcontroller is always driving the output then R2 is also pointless. <S> However, in reset state many microcontrollers will tristate their outputs, and possibly enable pull-ups. <S> The enabled pull-ups could cause a small current to flow in the base, multiply that by a couple hundred <S> and that's the current that would flow through the transistor. <S> This depends on the strength of the internal pull-ups (~50 - 100kOhm is typical).
| I'd say the diode is pointless, since you've got a much bigger problem if the microcontroller is driving a negative voltage.
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Is the SPI Interface just a protocol or actual hardware? I want to use the SPI interface between a PIC and an ENC28J60. However, there are no pins on the PIC called MISO or MOSI that I can use for SPI. So is SPI just a protocol, which I can use with any PIC pin (by just using any I/O pin, for example, in the same way to communicate with a DH22 sensor), or is SPI hardware (like UART)? <Q> SPI is a protocol [ de facto ] standard developed by Motorola. <S> It doesn't define any special hardware, only how signals are used and interpreted. <S> SPI can be implemented using software (known as "Bit Banging"), or using dedicated hardware, sometimes as part of a "USART" interface (Universal Synchronous/Asynchronous Receive/Transmit). <S> A hardware implementation is invariably more efficient, since the CPU is free to do other things while a byte of SPI data is being transferred. <S> Also, in chips with DMA the whole SPI subsystem can be completely decoupled from the CPU to run with absolute minimal overhead. <S> By the way, "SPI Interface" is an example of RAS syndrome . <S> But I digress. <A> Try looking for SCK, SDI and SDO - these are commonly used as names for the clock, input and output ports on some micros. <S> I'm not saying that all micros have an SPI interface but a lot do <S> and they are implemented (usually) in hardware <S> just like a UART is implemented in hardware. <S> Of course, just like a UART you can make a soft UART and you can make a soft SPI interface. <A> The term "SPI" is a non-trademarked term describing the a connection scheme where one master controller sends information to an arbitrary number of slave devices using a common clock wire and data-output wire, as well as one select wire per slave device, and receives data back using a common data-input wire. <S> A "real" SPI slave device will only pay attention to its clock and data inputs when selected, and will asynchronously float its data output whenever it's deselected (allowing the same pin to be used by other devices). <S> The kind of hardware that's useful for a processor to serve as an SPI master, however, is useful not only for talking to "real" SPI slave devices, but to many other things as well. <S> For example, if the master only needs to talk to one slave, it won't matter if the slave doesn't float its data output wire when deselected. <S> Further, in many situations it won't matter if a slave continuously captures data which is intended for other devices if it includes a signal which indicates when it should do something with the most recent data it has captured. <S> Consequently, it's very common to use SPI ports to talk to a wide variety of devices which use a clock wire and a data wire, but don't otherwise really "fit" the SPI communications model.
| SPI stands for "Serial Peripheral Interface", so "SPI Interface" would equate to "Serial Peripheral Interface Interface".
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What kind of "locking" barrel connector is this? I am looking for a replacement for a barrel connector; this is a locking connector (PN: PWRS-14000-260R for a Motorola FX9500 RFID Reader Country of origin: CHINA) however the locking 'fins' (for lack of a better description) are located on the sleeve instead of more commonly (?) nubs on the plastic tip Even knowing the plug specifics (2.1mm x 5.5mm inline plug (Female)) and providing measurements of these 'fins'(are about .5mm tall x 2.7mm wide and stick out of the barrel sleeve about 7.5mm from the end of the plastic tip) and contacting Digi-Key and Condor Electronics with pictures, I am unable to find anything on this plug type. I am beginning to think that this is a proprietary plug made specifically for Motorola by LEI and Motorola is NOT good about providing vendor info (I'm finding out). Any ideas on what this particular connector is referred to as and if they can be found anywhere commercially? Barrel front: Barrel side: <Q> It looks like the company in the next link sell these connectors: http://www.cliffuk.co.uk/products/dcconnectors/ the specific datasheet: <S> http://www.cliffuk.co.uk/products/dcconnectors/DCPlugSocketLockable.pdf <S> You can check if they have an distributor in your country and get it from there. <S> Hope this helps you! <S> Have a nice day!Dennis Tessels. <A> Condor Electronics came through for the win, their part "CA205LP-18IN-S/T-WWP-18AWG-105C" appears to fit the bill after viewing pictures. <S> They do not have this part on their website but the image they supplied (no tech sheets unfortunately) appears promising. <S> Maybe I shouldn't be answering this question without having the piece in hand <S> but I have good feels about it <A> You probably already found this, but if not, try Kycon part number KLDX-PA-0202-A-LT.
| We were also searching for this because off a Motorola FX7500 RFID reader.
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Via's in switch mode power supply I'm using a TPS54386 and currently designing a PCB layout. The data sheet says to avoid via's within a certain loop; I was wondering what would happen or if anything would if I went ahead and placed via's anyway to save space. <Q> The short answer is, as long as your components are spec'd properly, it will probably work just fine. <S> However, if you ever want to sell the circuit, you're going to need to do some compliance testing. <S> A 600kHz switching regulator with a poor layout will give you lots of radiated emissions problems when it comes EMC test time, especially in the AM band. <S> In a buck regulator, that is through the switch, inductor, and output cap. <S> You can make that loop smaller, and control it better if it all remains on a single layer. <A> You should look at the example design layout on page 31 and 32 of the datasheet, and be informed by it. <S> Look how tight the layout is. <S> Yes, they did put vias in the output power loop, but not really in any of the power switching signals. <S> In fact the vias all go to ground or return. <S> No switching voltages involved. <S> Via structure used is 8 parallel vias per via section. <S> Since the vias are in parallel they present very low inductance. <S> For example, vias typically have 2nH or 3nH of inductance, so with 8 in parallel any added circuit inductance will be negligible. <S> This means that the characteristics of the output capacitor C17 won't be changed and will have no impact on loop stability. <S> These vias only augment the return path in the output circuit by allowing return routing on the bottom layer. <S> That bottom layer copper provides a low inductance (small loop area) path for the power switching of SW1, D2, and L2. <S> If you want to add vias like that, you should have no problem. <S> If you want to add vias to the power switching routes that connect SW1 to D2 and L2 there could be trouble. <S> Added inductance, especially between SW1 and D2, will make snubbing more difficult and increase noise. <S> Also, in your schematic, L1 and L2 are not the same. <S> Is this intentional? <S> The combination of L1 (47uH) and C13 (100uF) with \$f_o\$ of 2300Hz isn't a good match to the built-in compensation of the TPS54386 which has zeros a ~3300Hz. <A> It says in the design notes: - Maintain a tight loop of wide traces from SW1 or SW2 through the switch node, inductor, output capacitor and rectifier diode. <S> Avoid using vias in this loop. <S> However, in the PCB layout they defy their own instructions: - Both C17 and D2 utilize vias in order to make connections in the PCB layout above and, they made an error in the PCB layout that I fixed (they had C17 disconnected from L2). <S> So how do you rationalize this? <S> Picture errors on the layout are easy to see and correct <S> but when it comes to vias they have used many parallel vias to overcome any hot-spot problems caused by single vias. <S> To see the error open the data sheet and look at the PCB guidelines: - That's the simple truth of the matter.
| The key to limiting the radiated emissions is minimize the switching current loop area.
|
Does the color of Jumper cables matter? Does the color of jumper cables in a circuit matter? I'm starting out with Arduino and can't get it to work at all with a breadboard! I'm thinking it's something simple like this! I know Black is Negative and Red is positive, but what about green, blue, orange and white etc? Is it just to make the circuit clearer or does each color have a meaning? <Q> No, the colours don't matter. <S> Using red for positive and black for negative is just a convention. <S> It helps everybody understand the way the circuit is wired, but the current will flow just the same whatever colour the insulation is. <S> Now, you may find it helpful to devise your own colour-coding scheme. <S> Maybe inputs to the Arduino are blue, and outputs are orange, something like that. <S> Non-working Arduino breadboard circuits are usually caused by a wiring error. <S> Be sure that you understand where the breadboard holes are connected, and where they're not. <S> If you can, get someone else to check your wiring, because it's very common to miss something if you check it yourself. <A> The wire colour is just an aid to help you keep track of what is connected to which, and will have no effect on the operation of the circuit. <S> You can even use red for ground and black for positive if you wish. <A> It is only colored for organization like you said. <S> Remember that the holes on the outside of the breadboard are connected vertically, and that the ones in the middle are connected horizontally. <S> Good luck! <A> It doesn't matter to the electrons. <S> It does matter to you. <S> Adopting a convention for the various colors can help you determine circuit functionality at a glance and aid diagnosis of problems if/when they come up. <A> For a simple circuit the color of the jumper cables may not be that crucial. <S> But when you deal with big complicated circuit you may have problem in tracing out your errors. <S> It is always a good practice to connect the cables or jumpers according to color code. <S> Thus if your project is handed over to another person, he/she may know how to trace out the connection.
| The color does not matter. Anything that helps you keep the wiring clear and organised will help. In North American house wiring black is hot, white is neutral and green is safety ground.
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What facilities/methods exist to debug and simulate programs for PIC microcontrollers In the university, even though I did do some PIC programming and project, the project was quite simple. We could always put in a blinking LED as a sign of life into our program or have the code write to the serial port every now and then to let the programmer know that the code is executing ok. However, I am sure that there are some real proper standard ways to (1) debug and (2) simulate the PIC code. This becomes important when we have really huge and complex programs. I think there is some sort of simulator for PIC but I can't find a link which says "The simulator does ABC and this is how you use it...". So what do the experts here say? <Q> The PICkit3 , ICD3 and REAL ICE <S> programmers/debuggers from Microchip all support various levels of debugging. <S> All of these allow you to pause execution, set breakpoint(s) and view variables / SFRs / CPU registers / etc. <S> REAL ICE also allows you to stream data out at high speed through DMCI. <S> There are also third-party tools like Proteus VSM . <A> PICsim is an open source program which provide a way to simulate PIC PIC16F628/16F877A/18F452 <S> MCUs. <S> It simulate a complete development kit with useful components such as leds and keypad. <A> I have used Proteus 8 from labcenter electronics and found it to be quiet useful for my graduation project. <S> It provides a veriety of circuit components and devices which can be used to great advantage. <S> It also has a virtual compiler for microcontroller programs. <S> http://www.labcenter.com/index.cfm
| Microchip also has their MPLAB SIM simulator, which does a pretty good job of simulating the parts (with exceptions of course).
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Communication between 3 Atmega controllers I am studying a bio chemistry analyzer, in which there are Main controller, printer board controller and detector board. To establish communication between these three, RxD pin of one controller is connected to same (RxD) of others and same with TxD (all ic's TxD is interconnected. Is this type of communication possible (Normally we connect TxD to RxD and vice versa)? How they are communicating then? Which protocol is used (I2C,SPI or USART). How will be master and slave configuration? <Q> Hard to say. <S> If these TXD and RXD pins are really dedicated to RS232 serial, then there would be a multi-driver conflict with the TXD pins. <S> Also, they would not be able to talk to each other. <S> There are a couple of possibilities, though. <S> I2C, CAN, and half duplex <S> RS-485 are all shared busses that are connected in this way, though the pins are not TX and RX. <S> In I2C you have clock and data. <S> In CAN and RS-485, you have a single differential pair for the data. <S> Possibly they are disabling the UART on the transmitter end, turning the RX pin around, and bit-banging it. <S> Or perhaps the microcontrollers can swap the TX and RX pins. <S> Or perhaps what you said is not correct and the controller TXD is connected to RXD on the others, and vise-versa, with the TXD pins disabled when not transmitting to avoid conflicts. <A> I've seen systems where RXD and TXD are all connected to a single node but only one TXD pin in the system is enabled at any time.it's not RS-232, <S> it operates at UART voltages, CCTalk is one example of this setup. <A> I'm not entirely sure why you'd do that, but definitely I2C is an option. <S> Most mirocontrollers nowadays have plenty of I2C libraries and peripheral implementation documentation in their resources. <S> Another thing you could try doing is multiplexing, so you choose only one of the 3 lines as a valid one for whatever you're communicating to. <S> In any case, I understand that you didn't design that <S> but that you're only studying it right? <S> Have in mind that communication can follow standard protocols (I2C, RS232, SPI, etc...) <S> or you can just make your own <S> so it all depends on how the controllers are programmed. <S> For instance, if the three TxD are all interconnected (properly pulled up or down, and connected through buffers or resistors) but thanks to the programming only ONE can actively pull the line's voltage Up or Down, then it would work. <S> Try studying the data-flow a bit more, like what should be happening inside those processors and how should they be "talking" to each other, <S> and then we can help out more
| It's also possible they are using a UART, but they would have to have some method of crossing over TX and RX.
|
MOSFET PWM switch jamming on - why? The circuit presented here is sized to switch 100W at a minimum pulse width matching 250Hz. But actually it's currently only being asked to switch around 10-20W. The load is an array of LEDs. Q1 and Q2 also have small heat sinks on them for extra measure. The input circuit actually includes one more BC557 (making the input active-low) and is driven by an Atmega328P PWM output @ 3.3V. There are two of these circuits (two channels) in a single enclosure, driving remote LED lights. The lights are part of a solar-powered lighting system, and the lights are often turned on during the day to regulate the charging voltage to the battery. When this circuit works (which is normally the case) it works very well. The problem is that occasionally the LEDs are jamming on, eventually draining the battery completely flat. I'm pretty certain the problem is in this circuit (rather than code), because: the condition appears physically very fragile. Just opening the (normally sealed) enclosure seems to be enough to have the LEDs turn off - as does picking it up and shaking it! Makes it hard to perform any circuit diagnostics! But I can tell (wirelessly) that the Atmega has not rebooted. And because both channels have the same fault, it makes me think it's a design flaw and not a component fault or bad connection. Also, sticking my fingers all over the circuit (my quick test for high impedance circuitry) doesn't cause the LEDs to react. And Q1 and Q2 don't appear to be very hot when in this condition, at least not after being in this condition for a fair amount of time - I've never caught it in the act... On reflection, R3 could also be smaller. But I find it hard to believe that's the problem either. One possible mistake is the lack of a snubbing diode on the load, since the wire lengths are quite long. But if that was the problem, would I expect to see it turning full-on? Another mistake could be that I've undersized R1+R2 for suppressing some kind of oscillation? Another mistake could be not allowing for some kind of thermal event's effect on Q1 and Q2? Possibly exacerbated by being out in the sun? I'm sure someone here can tell me what's going on :) <Q> When using paralleled MOSFETS in the output stage as you are, your should install small ballast resistors (0.2 Ohm) on each source terminal to ground. <S> This will ensure that the MOSFETs share the current equally. <S> ... <S> Also, its best to give each MOSFET its own independent gate resistor, this can eliminate any potential oscillations (at least thats recommended in amplifier circuits with paralleled output transistors). <A> The problem could be the code. <S> When there is over voltage, you are telling it to discharge the battery, but when it is under voltage, you are not telling it to stop discharging . <S> If you are telling it to stop, then most likely the input is floating (up or down), causing the LEDs to stay on. <A> I'd do several things. <S> 1) Forget parasitics. <S> 2) <S> For the love of all that's holy, put in some decoupling capacitance! <S> 10 - 100 <S> uF electrolytic and 0.1 uF ceramic. <S> 3) Put an LED in series with R1, and install it so it is visible outside the enclosure. <S> When your fault condition occurs, you can tell whether it's in the MOSFETs or the gate drive circuitry. <S> 4)When <S> you say that shaking it can fix the problem, is it the shaking? <S> Or is it the touching the case when you pick it up? <S> If it's the latter, it suggests that you don't have a proper ground connection back to your command source. <S> 5) Replace R6 with a 1k, and replace R3 with a 10k. <S> But these are minor. <S> 6) <S> Instead of R5 driving both gates, run a separate resistor to each gate. <S> This shouldn't be critical in this case, but it's good practice. <A> I would change the R1 position to be at right of R2 instead of left as showed in attached schematics, and change its values to 1k. <S> In this way, when PNP is turned ON, the MOSFETs voltage at gate will be 6 V (12/2) which is enough to fully turn it on. <S> In the original schematics, it's 12 V, and there is no extra benefit at this bigger voltage, on the contrary, there will be more charge stored in the MOSFETs input capacitance. <S> When PNP is OFF, MOSFET's capacitance can discharge through 1k, and therefore will have a faster discharge than through 1k8+180 in the original schematics. <S> Also, any leakage current from PNP "trying" to charge the MOSFET capacitance will behave better than in the original schematics, as I try to elaborate below. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Let Ileak be the leakage current from PNP. <S> If we do Thevenin Equivalent analysis from MOSFET gate, with the proposed schematics, the Thevenin voltage will be Ileak·1k and the Thevenin resistance 1000 Ohms. <S> With the original circuit the voltage will be Ileak·1k8 and the equivalent resistance 1980 Ohms. <S> If, for example, the leakage current is 500 uA, the new schematics would be charging the capacitor to 0.5 V (not enough to turn it ON). <S> In the original, while slower, would be charging it to 0.9 <S> (close to turn it ON). <A> A couple observations: 1) Is the ground common throughout the design including <S> ATMega? <S> or do you allow a lot of return current to flow through the ground? <S> 2) <S> You are way over driving the bases on all the transistors. <S> Just look at the lower limb of Q3. <S> R1 will have ~ 1 mA flowing through it, this just needs an Ib on Q3 of 50 uA to hit 10 <S> mA Ic. <S> But you are pulling the base of Q3 hard to ground through R4 (2K) <S> = <S> > <S> ~ <S> 6mA. <S> Look at the values from the Moto data-sheet, page 3 , figure 3. <S> This will certainly slow it down (but won't cause it to stick). <S> While you're at Recalculate the values for R4 and R6 and normalize R3 and R5 to the 11K value. <S> 3) <S> The fact that shaking it changes the state may indicate that you have an intermittent short. <A> As has been said before I'd double check the bias points of the BJTs as they are being driven quite hard, but that shouldn't be an issue. <S> There's a very, very, very remote chance of their parasitic junctions being turned on but the circuit wouldn't recover unless it was powered down and <S> the BJTs would likely be ex-BJTs by that point. <S> It would be worth taking the gates of the MOSFETs and giving them with separate gate resistors (i.e 180R each) and then connecting to R1 to pull them down. <S> Though it doesn't normally show itself with this kind of power level or bog standard MOSFETs <S> it's perfectly possible to get the gates 'ringing' (i.e. you could be seeing Q1 and Q2 turn on back and forth during the 'stuck on' situation) between each other with their C's and L's along with any inductance lying around the connections and circuit acting as the reactive elements. <S> As I said, it's a slim chance but easy to try and from a good practice point of view I just wouldn't share gate resistors among MOSFETs.
| The way this works is if one MOSFET starts taking more of the current, this will lead to a lowering of the Vgs of that MOSFET which turns the MOSFET off slightly and re-balances the current between MOSFETs. At the drive levels you're using, these are not going to be a problem. Voltage drops from high current flow will shift this circuit's voltage relative to the controller.
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How capacitors connected in series are affecting the total capacitance of the network, since there is no actual flow of electricity due to dielectric? Can someone explain how is it possible to calculate the value of caps connected in series? I am not asking about the maths, I just try to understand how can such a network be created since there is the gap(open) of the circuit, because of the dielectric(insulator). I can understand how the capacitance works in one capacitor because of the AC, in case of DC I know that the cap will charge and then if we short-circuit the legs of the capacitor, it will discharge. But what about when we have 2 or more capacitors in series? In my understanding it will be used only the one plate of the first cap(the plate closer to -) and the other plate of the other cap(closer to the +) of the power supply. How the in-between caps will affect the total capacitance since they are not even "connected" to anything due to the dielectrics of each one? I mean they are insulated if there is no current between them. Any insights much appreciated! Please make it more visual, use some kind of analogy if possible :) <Q> It's quite simple really: One plate of the first capacitor is negatively charged - <S> that is it has excess electrons. <S> Those electrons repel the electrons in it's opposite plate. <S> Those electrons have to go somewhere, so they go towards the first plate of the second capacitor - so that plate then has excess electrons, which then repels the electrons in the second plate of the second capacitor. <S> And of course those repelled electrons have to go somewhere ... <S> ad infinitum. <A> Whether this varying electric field is created near-instantaneously (a direct connection to a battery with very short leads) or via a resistor, there will be a rate of change of electric field applied to the bank. <S> This is not just a DC phenomenon but an AC phenomenon as well. <S> You say: - I can understand how the capacitance works in one capacitor because of the AC <S> If you do understand what you say you understand <S> then it's not a big step to realize that current isn't just the transfer of charge (conduction currents) <S> but it can also be a "displacement current" and this is due to the change of electric field. <S> Displacement currents are also responsible for EM wave propagation (no conductors in space to carry ordinary conduction current of course). <S> In a single capacitor, it is displacement current that passes thru the insulation and becomes conduction current on leaving the capacitor plate. <S> With several equal value caps in series, the final static electric field across the lot of them is applied voltage divided by total accumulated width of the dielectrics i.e. volts per metre. <S> With equal value caps this field is shared equally. <S> It is also shared equally when charging and hence the total capacitance of n <S> equal-value series capacitors is C/n where C is the capacitance of an individual capacitor. <A> I did some more realllllly basic reading and is all about electrostatic force that makes the existence of charge in between the plates of the capacitor. <S> So when electrons are building on one plate, they repel the electrons of the other plate, they do not "physically" push them but due to the elementary principal of the "repulsion of same charges and attraction of the opposite <S> " this is what happens, just like the charges between the huge capacitor of our natural space. <S> Sky and earth where the dielectric is the air in between. <S> This is a video that explains it better https://www.youtube.com/watch?v=3TmuYAz2_B8&list=PLJVQe4CzhfPaPLOZri-DVziUF6pmtxKKe <S> Really appreciate everyone who answered my question!
| If you have a bunch of caps in series then the act of charging them is to apply a varying electric field across the ends of the bank of capacitors.
|
Automation schematic software for linux Is anyone aware of a good linux program for drawing automation schematics?Not PCB design, circuit simulation and fine electronics. Industrial automation. Example: I looked at a couple of different EDA applications, but did not find any functionality for multiple sheets in a single schematic. I will also need cross-reference symbols to handle nets/symbols that go across multiple sheets. Perhaps someone knows about an EDA application with multiple sheets and good scripting capabilities to create the cross-reference system? Solution:I ended up using gschem with some custom scripts for cross-references and titleblock stuff. They can be found here . <Q> There is a free and open-source CAD package called LibreCAD. <S> It may do what you want, if AutoCAD is what you're expecting. <A> There is gEDA for Linux or KiCAD or eagle <S> All three can do schematic capture & multi-sheet. <S> They equally come with PCB layout (which you say you do not require). <S> There is also qucs which is a circuit simulation tool and <S> while it will draw circuits, what you draw for simulation is rarely exactly what you draw for schematic capture (no need for decoupling, adding additional components to appease the simulation engine etc...) <S> My recommendation is kicad as it has a lot of 3rd party symbols available for it <A> I haven't done this is a very long time (more than 10 years now) but what I used to do to draw cabinet wiring schematics is to use a PCB design tool and build my own custom component library. <A> Although it can do PCB layout, you don 't have to pay for that or the autorouter; the schematic editor is available separately . <S> The standrd version costs $315 and can handle 99 sheets of schematics, with cross-references.
| Eagle software is used by a lot of folks around here and runs on Linux.
|
Replacing Crystal Oscillator with DDS. In a circuit like this, can the crystal oscillator be replaced with a DDS such as a AD9850? If so can it just be dropped in or would other components have to be changed? Second question, how sensitive are simple transceivers like this to slight changes in capacitance, for example, a 110pf capacitor instead of the 120pf one below the crystal? Final question, if I want to change the DDS frequency(to 14Mhz for example) do all the capacitors, etc need to be changed? Thanks so much for your time. <Q> No. <S> In this case, it seems like the crystal is being used as a bandpass filter. <S> The idea is that the gain of the amplifier will be the largest in the passband of the crystal. <S> You might be able to get away with a DDS for the transmit function only, but it's not going to work for the receiver. <A> My approach to solve this would be: - Try simulating the above circuit (LTSpice or PSpice etc..) <S> and then when happy it seems to work, inject a waveform (as if from the DDS) onto the base of T1 (xtal removed of course). <S> Mess around with levels until you are satisfied it works across the range of frequencies you desire. <A> It would be tough to do the RX because it is actually a regenerative or self oscillating detector .This <S> is why the simple circuit works well <S> .The <S> TX is easy .Why not take a reasonably simple direct conversion reciever and try the DDS as the LO.
| You cannot just replace it with a DDS as a DDS does not act like a filter, it acts like an oscillator.
|
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