source stringlengths 620 29.3k | target stringlengths 12 1.24k |
|---|---|
What is a "solder mask opening" in a footprint? In the Luxeon LED datasheet under Solder Pad Design section there is a diagram with 3 crosses around the edges of the LED. They are identified as "4x Soldermask Opening (hatched area)" but I am not entirely sure what it is exactly or why it is used. What are they, what do they achieve and how should they be drawn in Eagle or other software? <Q> It is for easy alignment with respect to the outline of the package. <S> The Luxeon outline shows perfect alignment and easier for optical correction with a vision system on the pick N place machine. <A> The machine probably need the bare copper to get a good image of the alignment mark <A> Soldermask is used to cover the copper to prevent solder from adhering to it (hence its name, "soldermask"). <S> Soldermask is the green stuff you see on most PCBs, though it can be just about any color (red, purple, blue, white, black, etc). <S> By creating a soldermask cutout you will allow copper to show through the mask. <S> This is often used for fiducials, which are used as optical markers for the position of components and used by pick-and-place machines to properly align components. <S> I expect that's what the crosses are used for in the component layout you posted. <S> They are either fiducials or, more likely, indicators of polarity. <S> Having a non-symmetrical mark to indicate where the cathode and anode are can be useful for the assembler as well as anyone debugging the board, reworking it, testing it, etc. <S> I can't quite tell, but there is also the possibility that they are being used as actual solder pads. <S> I see the LED has matching marks, but I can't tell if those are metal tabs or just marks on the case. <S> I'm pretty sure the crosses in the PCB layout are just there to make sure the assembler aligns the LED correctly on the board.
| In the case of the cross mark you mention however, the opening is probably specified to provide an alignment mark for the pick-and-place machine that will use the cross to figure out when the part is aligned properly before placing it down. Soldermask openings are CAD layers usually meant to specify to the PCB fabrication house that they shouldn't put any other covering on the bare copper so that when the soldermask layer is applied, the solderpaste can be placed on the direct copper.
|
How to calculate the output amperage of transformer I am going to attempt to wind my own transformer, but I need to know the how to simply calculate the output amperage. I know how to calculate the voltage: Np/Ns = Vp/VsPower = Voltage x CurrentVp = 240Np = 100Ns = 2100/2 = 240/X100/2 = 50240/X = 50240 x 50 = XX = 12000240/12000 = 0.02voltageSecondary = 0.02V But I don't know the amperage calculations. Keep in mind I am very young.I cannot find it anywhere on the internet. Thanks! <Q> You start by specifying what power you want through the transformer. <S> A 1kW transformer needs much more iron than a 6 watt. <S> Given the power, you then choose a suitable core. <S> Broadly, power goes as weight of core, over quite a wide range. <S> Core manufacturers have tables that shows you core size required for any given power rating. <S> Only after you have the core size, then you can work out the primary turns. <S> Too few and the transformer won't work, it will blow fuses or burn. <S> Too many, and it will be inefficient. <S> Calculate the primary turns based on your input voltage, and swinging the core over a 'reasonable' range of flux, typically +/- <S> 1.5T for iron. <S> Size the primary wire to fill half of the winding window. <S> Now you can calculate your secondary turns, as (Vout/Vin <S> ) * primary_turns. <S> Size the secondary wire to fill the other half of the winding window. <S> If you have started with the right size iron core, then filling the winding window with copper (allowing for insulation, the bobbin, and a few gaps) will 'just work' to give you enough current handling capacity to meet your power specification. <A> Amperage depends on load and thickness of secondary wiring. <S> Power going to primary is the maximum power you can get from secondary (minus loses). <S> So in short Vp*Ip=Vs*Is, then Is= <S> Vp*Ip/ <S> Vs <A> For ratio = <S> a:1 <S> a= <S> Vp/ <S> Vs <S> a= <S> Ip/Is such that for ideal lossless <S> Vp Ip= <S> Is <S> Vs L inductance of primary must be high enough to only conduct 10% of rated current. <S> This is the current used to couple primary and secondary through a steel laminate core. <S> rated current is mainly from resistive loss, core saturation limits and temperature rise <S> Vs*Is= <S> VA rating of output which may be 10% less than input due to excitation current which occurs with no load. <S> side with the most turns uses the smaller gauge wire. <S> V*I or VA ratings depend on linear R load, but if using diode bridge and cap, it must be derated at least 30% depending on load and peak ripple current. <A> Ideally, if the turns ratio is 100:2 then the voltage step down ratio is also 100:2 hence 240 V RMS on primary becomes 4.8 volts RMS on secondary. <S> If there is a load on the secondary of 4.8 ohms then the current in that load (due to ohms law) is 1 A RMS and this will be seen as a much reduced current on the primary of 20 mA i.e. when you step down voltage you use the step down turns ratio and this ratio is also used for determining the primary current for a current in the secondary. <S> In your calculation you got the algebra wrong i.e. <S> Vs should be 240/50 <S> Your bigger problem will be the 100 turns on the primary. <S> Almost certainly this is too small for most modest home-appliance type transformers and will cause transformer core saturation. <S> Probably a figure of about 1000 turns is more realistic.
| You have to specify amperage yourself and then use wire suitable to handle that current.
|
How well do microcontrollers deal with unstable power supplies? Assuming the VCC voltage never goes below the minimum of the specific chip (1.8v for the ATmega48PV) or above the maximum, how well do microcontrollers deal with noise? Are they guaranteed not to malfunction? Does anything happen, apart from I guess ADC readings might get a bit funny and inputs and outputs fluctuate? <Q> In theory they should be fine since you aren't violating any specs. <S> However, there is probably a upper frequency limit on supply voltage variations that will cause trouble. <S> You can only guess what that might be since datasheets don't tell you. <S> Do what you can to limit the frequency of the fluctuations. <S> You always need a good high frequency bypass cap on every power lead to the nearest ground lead. <S> If your supply voltage might otherwise fluctuate at high frequency, add additional capacitance. <S> For example, a 1 µF ceramic for every power lead, and then maybe 10 µF or more near the chip where all the power leads are tied together. <S> Consider adding a ferrite chip inductor in series with the supply before all the capacitors mentioned above. <S> This should be followed by 10s of µF capacitance, plus the individual bypass caps. <S> The inductance and capacitance form a second order low pass filter that attenuate high frequencies significantly. <S> The drawback is that there will be some voltage drop across the ferrite chip inductor. <S> These are designed for this kind of use, so have only a few 100 mΩ at most, but that can still be a problem in some cases. <S> It helps to keep the heavy current draw off the microcontroller power feed. <S> If the micro needs to drive LEDs, for example, have the micro control transistors, which then control the LED current coming from some other supply or from before the chip inductor. <S> Ultimately, you'll have to talk to the manufacturer to get a real answer, if you're ever going to get one at all. <S> The best strategy is to avoid being in this situation in the first place. <A> I just went looking for it in the Atmega328 datasheet and couldn't see it. <S> For analogue parts this is specified as PSRR (power supply rejection ratio). <S> What that datasheet does specify is the affect of VCC on pin thresholds and internal oscillator behaviour. <S> That suggests that if your power supply is sufficiently noisy you will may glitches on input pins that may manifest as errors in e.g. fast SPI. <S> You may also disrupt the internal oscillator. <A> They probably will. <S> However, the bypass capacitors will place some kind of limit on the maximum dv/dt on the power supply (assuming the current drawn from the power supply is known) <S> so you can at least quantify the problem. <S> It's probably a reasonable guess that, at least, the digital parts won't care much if the rate of change is less than a certain amount- <S> say 10,000 clock cycles to slew the worst-case range from 1.8V to 5.5V peak-to-peak <S> (of course we're assuming the clock is at or below the maximum 4MHz for 1.8V operation ) so that would be about +/-1500V <S> /second or a ripple frequency of about 500Hz for 1.3V RMS (full range).
| They are not guaranteed not to malfunction.
|
What is the difference between smart antennas and phased arrays? Wikipedia defines smart antennas as follows Smart antennas (also known as adaptive array antennas, multiple antennas and, recently, MIMO) are antenna arrays with smart signal processing algorithms used to identify spatial signal signature such as the direction of arrival (DOA) of the signal, and use it to calculate beamforming vectors, to track and locate the antenna beam on the mobile/target. And phased arrays as follows In antenna theory, a phased array is an array of antennas in which the relative phases of the respective signals feeding the antennas are set in such a way that the effective radiation pattern of the array is reinforced in a desired direction and suppressed in undesired directions. From these definitions it seems that two terms define similar things. And yet the article on smart antennas do not reference (in the text) phased arrays and vice versa. To contribute to my confusion, there is a book Adaptive Antennas and Phased Arraysfor Radar and Communications by Alan J. Fenn, which has two major parts: one on adaptive antennas and one on phased arrays. These parts address different problems: smart antennas for nulling out interference sources and phased arrays to scan an antenna field of view. And again the core structure looks the same to me. So what exactly is the difference between smart antennas and phased arrays? <Q> Simply put: both are different ways of harnessing the properties of using multiple antennas. <S> In a Phase array the antennas are usually placed in a repetitive pattern like a matrix or circles or hexagons. <S> Often the distance between the antennas is always the same. <S> Phased arrays are used to make these many antennas behave as one antenna with certain properties like a more narrow beam pattern (where the antenna is sensitive and/or emits more power). <S> I find smart antennas a misnomer as the antennas are not smart at all, the "smart" bit is in the signals which are used with the antennas. <S> This can be used for improving reception by beam-forming or the reverse: enhanced sensitivity in a certain direction. <S> This "smart antenna" signal processing is also used with Phased Arrays. <S> You could say that some smart Antenna systems are a simple form of Phased Array. <S> Where a simple Phased array might use 9 antennas, a smart antenna system might use 3 antennas. <S> but I would not call any smart Antenna system a phase array. <S> for example some laptops have 2 or 3 Wifi antennas for beamforming and Mimo. <S> You would not call that a phase array <S> but I think it would be OK to call that a smart antenna system. <A> There is much convergence between the two types of antenna, and the main difference is one of history. <S> Phased arrays have been around for a long time, as an electrical method of making the equivalent of a rotating or steerable beam, both for transmission and reception. <S> The military have developed these for a long time, much of the development being secret, hidden inside radomes. <S> Typically a phased array will have 100s if not 1000s of elements. <S> Cost was no object. <S> Smart antennae have been around only a decade or so, as an evolution of mobile communications. <S> One antenna is subject to fading, this is much improved by having two, and two antennae were standard for a long time. <S> DSP was used to use the two receive chains intelligently. <S> Initially, a receiver would switch to the strongest signal, but as more DSP became available cheaply, using both receivers to estimate the data independently, then soft-combining them, gave a better error rate. <S> This was the first 'smart' antenna. <S> More antennae got added to the system, as it was realised that crude beamforming could increase the data rate. <S> Under ideal conditions, a 4 antenna MIMO system may give as much as three times the channel capacity of a single antenna by 'spatial multiplexing', employing different scattering paths to the other end of the link. <S> As the number of active antennae increases, the hardware certainly begins to look like the classic phased array hardware. <A> Smart antennas and phased arrays do their magic by taking the RF signal and linearly combining copies of it with different phases and amplitudes. <S> Depending on the combination, it points its beam to a different direction. <S> A phase array antenna usually does that choosing the phases and / or amplitudes <S> (usually it is only the phases) from a fixed, finite set of values. <S> This is implemented with a matrix, called a feeding matrix or feeding network, of passive elements like phase shifters arranged in a suitable manner. <S> A Butler matrix is an example of such a matrix. <S> The matrix has N inputs and N outputs, and the choice of the input determines the direction of the beam. <S> Smart antennas, on the other hand, is more versatile. <S> It chooses the phase and / or amplitude values according to a given algorithm and implements them with electronics, and hence is able to get a much larger number of combinations. <S> This translates to the smart antenna being able to point its irradiation beam to many more directions (potentially, infinite) than a phased array. <S> On the other hand, it costs more. <A> Not knowing anything about "Smart Antenna" before reading the Wiki article, But working the RF Communications Field for the better part of 2 decades, I would suggest that Electronically Steerable Phased Arrays are synonymous with the term "smart antenna. <S> " The difference is likely similar to saying you got a 3dB diversity gain by using 2 antennas, versus using a 2 omni element array. <S> They are for all intents, the same thing. <S> The fundamentals of array processing are that you use amplitude and phase to either change your antenna response, both rx and TX (i.e: <S> Beam-steer, null steer,sidelobe rejection), or to glean info about the incoming signal (i.e: direction, velocity.)
| You could call any Phased array with signal processing a Smart Antenna system
|
How to build a noise generator based entirely on logic gates? I would like to build a noise generator based on logic gates. It could be white noise generator or a random bits generator that sound similar to a noise source. I need a circuit recommendation or an idea how to make it. Updated: The purpose is to make an electronic music device. Any kind of noise can be interesting for me. <Q> How about one of these: - Taken from here . <S> It is described as a white noise source: <S> - 24 <S> Stage Linear Feedback Shift Register (LFSR) <S> This circuit can be used to generate a pseudo-random sequence of 0's and 1's -- that is "white noise" or "static. <S> " <S> The schematic shows a 24 stage shift register with XNOR taps at registers 7, 16, 22, and 24 using the Fibonacci configuration. <S> With these taps, the circuit generates a "maximal length sequence" of pseudo random values which repeats only after (2\$^{24}\$ - 1) clock pulses. <S> Or maybe just use an analogue one based on the emitter-base breakdown voltage of transistor Q1: - <A> The output of a PRBS generator, especially for the longer sequences (PRBS31), can have a very wide bandwidth. <S> It's also a pretty simple circuit to build; just shift registers and XOR gates. <S> See https://en.wikipedia.org/wiki/Pseudorandom_binary_sequence and https://en.wikipedia.org/wiki/Linear-feedback_shift_register . <A> You can make a digital white noise generator with a small 8-bit 8-pin micro such as a PIC12F series. <S> It's basically a code implementation of a PRNG based on one or more shift-registers, very easy to do. <S> Since you are interested in audio you might actually want pink noise. <S> There are samples available on-line if you want to compare- white noise sounds much harsher. <S> You can get that from white noise via an analog filter that approximates 3dB/octave or you can use a much more powerful micro and implement a digital filter. <S> Here is a somewhat more sophisticated circuit than the minimum that can produce Gaussian noise and pink noise.
| Probably the simplest method for generating "noise" with digital logic is to generate a pseudorandom binary sequence (PRBS) with a linear feedback shift register (LFSR).
|
Why is there a transistor between the MCU and LED? In the answer for this question ( Does my circuit need decoupling caps? ) there is a transistor between the attiny85 and the led.I understand that if the mcu isn't capable of providing enough current for a sensor, you have to do it the way the schematic shows, instead of controlling it directly trough the mcu. But I think that is not the case.So, why? Is there any advantage? <Q> There are several reasons for using a transistor like that. <S> One is if the load (in this case, the LED) requires more current than a microcontroller I/ <S> O pin is capable of providing. <S> Another is if the load is powered by a voltage significantly different from the microcontroller's power supply. <S> Perhaps the microcontroller is powered by a 3.3 volt supply while the load is powered by 12 volts. <S> Another reason could be to isolate the microcontroller from various secondary effects, such as back EMF in the case of an inductive load such as a motor or relay. <A> Taking the context into consideration, he was probably just using a transistor to have something to decouple . <S> Otherwise: Attiny datasheet states Absolute max current per I/ <S> O pin: 40mA, and <S> source/sink capabilities symmetrical. <S> So, if you're going anywhere over 20~30mA you wouldn't want to drive the load with the MCU (50mA in his example). <A> If the current you want to put through the LED is high, the output voltage of the MCU is not well defined. <S> If the difference between supply voltage (at the edges of tolerance) and the LED voltage is small you may have a problem. <S> A transistor, driven hard into saturation, will have a very well defined forward voltage at typical LED currents. <S> So if you care about reproducability the transistor could be better. <S> Otherwise you might get one unit that has an LED current that is double or half the next. <S> At 10mA the output drop in the ATTiny w/5V supply is less than 0.6 or 0.7V depending on sinking or sourcing (it's not quite symmetrical). <S> Using a MOSFET you can make that number almost zero- <S> eg. <S> AO3400 has <32m ohm Rds(on) with 4.5V drive, so at 10mA the drop would be 320uV. <S> One way to see if this is likely to be a problem (other than the proper way of running all the sums on a spreadsheet or whatever) is to note whether the voltage drop calculated to be across the LED series resistor (yes, there should be one!) <S> is getting very low, say much less than a volt or two. <S> This is more of a problem with 3.3V supplies when you want an LED that isn't red or IR. <S> Another possibility a transistor affords <S> you is to run the LED from a higher or lower voltage- perhaps an unregulated battery voltage before a boost converter or an DC unregulated input before a buck converter or linear regulator. <S> This can reduce the power dissipation in your regulator, at the expense, perhaps, of some brightness variation with the input. <S> Finally, if you want a high-reliability design- or one that must work over the military temperature range or worse- <S> it's best not to run too much current through IC outputs. <S> At the maximum rated current there can be enormous current densities on the chip (equivalent to passing thousands of amperes through an AWG20 wire) which can result in eventual failures (eg. metal migration) especially at high temperature. <A> In the case of a single visible-light LED and not much else there's no advantage if the MCU can handle the current required. <S> However, if you have many LEDs connected then it may be possible to overload the port or even chip itself, and transistors or even a dedicated driver may be warranted.
| It can also be useful to isolate relatively high current (but still within the I/O pin capability) but possibly high frequency signals that could cause ground bounce or similar issues, such as when the microcontroller is driving an LED with a high frequency PWM signal. Another possibility is when your load needs a Voltage rail higher than supported by the MCU (not the case in his example).
|
74HC595 Rows/Columns to drive LED matrix Let's say we have one 74HC595 and we want to light 16 leds (common anode) connected as a 4x4 matrix exactly as in the following picture : So, 4 first outputs to controls 4 rows.The remaining 4 output to controls 4 columns. From what I understand at this moment: In the picture, there is a NPN transistor on the 4 columns to allow more current than what the 595 can sink. Let's say only ROW1 is active, and all columns are active (LED1,LED2,LED3,LED4). Column 1, Column 2, Column 3 and Column 4 on the 595 will indicates a very low current, just the current set at the base of each transistor by the base current limiting resistor. However, at ROW1, would it indicates at the 595 the sum of each led current of this row? So 80mA, if we assume each led at 20mA ? If it's the case, in my case, there is much more 595 and much more leds and I don't want to operate near/over the 595 maximum current (75mA), would I simply add appropriate PNP transistors + base resistors to each ROW to reduce the current of each 595 rows pins ? I want to keep the 74HC595 IC and low-cost single transistors. I know there are shift registers with higher current capacity and transistors-arrays IC that saves wires and resistors. Also that I can reduce the current of each led with higher resistors to keep it under 75mA, but I would like to understand how to properly work with a matrix arrangement that needs more than what the 595 can work with, using these simple components only. In other words, is 4 PNP for rows and 4 NPN for columns, the best way to handle >= 75mA on a single row ? <Q> You can do it the way you suggest, however you should move the resistors to the NPN collectors, and you will need PNP (or emitter-follower NPN) transistors on each row output. <S> To get a rough idea of the current required- if 5mA is enough for the LED to be bright enough when supplied with DC you will have to supply 20mA for 1:4 multiplexing, and each row driver will have to source 80mA with a 25% duty cycle. <S> Each column driver will have to sink 20mA with 100% duty cycle. <A> You are proposing to do the multiplexing on a row basis, where 0 to 4 of the LEDs in a row are on at the same time. <S> I don't know which LEDs you are using, but a rough estimate is that each takes 15mA, so you have to source 60mA from the '595 (which is well beyond its specification). <S> With this arrangement you also have another problem. <S> Only one column will be on at any time, with any number of LEDs in that column turned on. <S> Each ROW pin only has to source enough current for 1 LED, which the '595 is just capable of doing. <S> The combined LED current for the column of up to 60mA is handled by the column transistor (T1, etc). <S> You also have 1 current limit resistor per LED (R1, etc), so the LEDs have constant brightness. <A> Simon, your calculations are incorrect. <S> The 220 ohm resistor on each 595 output limits current drawn to about 10 mA, no matter <S> how many column LEDs are turned on. <S> I'm assuming that a LED drops about 2.5 v when lit. <S> As Steve G has noted, turning on another column will simply share that 10 mA between two LEDs - and perhaps not evenly. <S> So your HC595 is safe from overcurrent, but multiple columns will cause dimming. <S> Likely not what you wanted. <S> Be aware that internal Rds of HC595 adds about 40 ohms to each 220 ohm. <A> You have probably worked this out already but this is what I propose you configure your LED matrix. <S> Each LED has a resistor therefore guaranteeing each LED get the same current regardless of how many are on. <S> Of course in this configuration the rows have reverse logic to the columns. <S> In other words, is 4 PNP for rows and 4 NPN for columns, the best way to handle >= 75mA on a single row ?
| Since you have a single current limit resistor per row (R1 for ROW1, etc) as you turn on more LEDs in the row the current through each LED will drop, as will the brightness of each LED. I would say yes in my opinion but others might have different suggestions that are just as valid. The correct way to do multiplexing of this type is to do it on a column basis. As you point out, the current required from the '595 ROW pins becomes excessive.
|
How to distinguish the type of transistor amplifier I tutor students studying for an exam where they are likely to be asked to recognise a circuit, with a single transistor, as one of: a) a common base amplifier; b) a common emitter amplifier: c) a common collector amplifier. Is there a mnemonic or other easily remembered way to distinguish them? <Q> Think about it. <S> What would be the point? <S> The purpose of naming some common transistor circuits is to quickly communicate them to others, and after some experience, to short-cut some analysis of the circuits. <S> Simply being able to name a circuit without understanding what that means is pointless. <S> Teaching students to do that is doing them a disservice. <S> Instead, they should learn some basics of how bipolar transistors work, and then how they can be used in some common circuits to achieve certain goals. <S> There is no substitute for actually understanding these things. <S> The fact that some of these circuits are common enough to have standards names should only be a minor point once the understanding has occurred. <S> After the students work with these common circuits enough, they'll eventually remember their names. <S> Learning by memorizing mnemonics and half-baked rules of thumb is no learning at all. <S> Do your job right and teach them what they really need to learn. <A> CE does not use E as the direct input CB does not use B as the direct input CC does not use C as the output <S> The word "common" means that the pin-name following cannot be either a direct input or an output. <S> For all three types you can always say B is never an output and C is never an input. <S> Thus CE (for instance) implies the input is at B and the output is at C. <A> First off, I must agree with Olin's assessment, and think a mnemonic is causing students to just remember an algorithm to solve an exam problem (it happened to me!) <S> instead of truly understanding the task, <S> but I don't find the mnemonic evil and utterly useless. <S> I'm doing some tutoring on that area as well. <S> As mentioned, while I was studying it myself, I wasn't really thinking about it. <S> But it's really logical when you think about it. <S> Think of an amplifier as a two-circuit... circuit. <S> You have an input circuit (where you have an input signal generator, for instance), and an output circuit (where you send the amplified signal to the output device - a loudspeaker, for instance). <S> A transistor has three terminals, or pins - I believe it is safe to think of it this way: one in the input circuit, one in the output circuit, and one that serves as an interface of a kind between the two. <S> In other words, it is shared between the two circuits, and it is common to both the input and the output circuit. <S> As per alephzero's suggestion, another way to look at it would be this: if you "pull out" a transistor from an amplifier schematic, the input and output circuits will have two open connections each. <S> That's four. <S> Since the transistor has only three terminals, and both circuits need to be closed, one of the terminals must be common to both circuits. <S> So! <S> CE <S> - The emitter is common - it closes both circuits. <S> The base has the signal generator, and the collector has an output device, for instance. <S> Emitter - input, collector - output. <S> CC - <S> The collector is common - it closes both circuits. <S> Base - input, emitter - output. <S> Hope that helps. <A> An amplifier has an input port, and an output port. <S> A port has two terminals. <S> They should practice to identify the terminal that is common to both the input and the output port.
| Looking for a mnemonic or some mental hack to distinguish different transistor circuit configurations is the wrong thing to do. CB - The base is common - it closes both circuits.
|
5V DC over Coax I need a sanity check. My house is wired with a Coax cable that is currently completely useless to me. I want to use the coax wiring to send 5V DC power for small appliances (like Raspberry Pi's) throughout the house. I want to use this PSU: https://www.amazon.ca/gp/product/B00N2RW72C/ref=ox_sc_act_title_2?smid=ASNOLMMI4SF6N&psc=1 that will feed into a DC step-down and a splitter like this: https://www.amazon.ca/gp/product/B00DIGACBU/ref=ox_sc_act_title_3?smid=A3DWYIK6Y9EEQB&psc=1 I am aware that more length = more resistance so I am wondering if I am good with slightly increasing the voltage on the source side and measuring on the output until I have an exact 5V. It just feels like I am missing something. EDIT: Looks like (and makes total sense) going with the 12V and dropping it at termination point is the way to go. Thank you everyone. <Q> A "better" way is to send 12 V (or more) over the coax and have local regulation to 5V at each RaPi outlet. <S> You can use cheap buck regulators available on Ebay (a few GBP or dollars) to take the 12 V DC and efficiently convert to 5 V locally. <A> The splitter in your link is for distributing RF cable TV signals - it will seriously attenuate your DC power, if it passes DC at all. <S> If you are only using the coax to distribute DC (or low voltage AC), you can join several coax cables as if they were simple two-conductor cables. <S> There is no need to worry about impedance matching and other RF complications that your suggested TV splitter deals with. <A> I am aware that more length = more resistance <S> so I am wondering if I am good with slightly increasing the voltage on the source side and measuring on the output until I have an exact 5V. NONONONONO! <S> (Did I say "no"?) <S> Current draw is very not constant with electronics. <S> In normal running, you might have enough current to get exactly 5V, dropping 1V on the cable. <S> At startup when it takes more current, you might be up to 2V drop on the cable so the supplied voltage will be 4V and <S> your kit browns out. <S> When you change the code so it goes into sleep mode, the 1V drop might go to virtually nothing, so your electronics will see a full 6V and be fried (i.e. permanently dead). <S> If you want to do this, I suggest using a higher voltage such as 12V for your DC supply. <S> Each Raspberry Pi or other device must then contain its own 5V regulator. <S> The 12V supply will wander up and down as each device draws more or less current and voltage is dropped in the cables, but you should still have more than 7V which will give you a solid supply for a 5V regulator to run on. <S> You also need smoothing capacitors to protect against brown-outs. <S> And reverse-polarity protection to stop the 12V supply trying to suck charge back through the 5V regulators. <S> A good tip is to use a regulator like an LM2940 which has this kind of protection already built in, instead of a more basic regulator like a 7805.
| With 12 V being sent down the wire and with local switching buck regulators, the overall current down the coax is less than 50% of the current had you put 5 V on the line and this immediately drops less voltage and makes the whole system more viable. You might even consider using DC-to-DC (isolating types) converters at each RaPi connection to avoid "earth" issues - they would also give a measure of protection against local (not direct) lightning strikes.
|
How to prevent sealed in product battery drain on store shelf? How can a microcontroller-based device prevent battery drain while the product is on the store shelf? We are creating a waterproof, sealed in product without any buttons. This means the battery is attached to the PCB in production and can't be accessed after production. What I'm looking for is a way to prevent battery drain while the product is still in the box and on the shelf. Here's what we've doing so far and why I don't like it: We're using a normally closed magnetic (reed) proximity sensor on the board and a permanent magnet in the box. The magnet breaks the circuit and no power is applied from the battery to the circuit. Take the product out of the box and it should automatically go live. What I don't like: Normally closed reed switches are scarce and expensive. They are also often too big for our small product. On top of that, the ones we've been using have very fragile SMD "legs" that break from their solder pads at the slightest impact. Edit : We have two GPIO pins accessible on the outside of the product and would be replacing the reed switch inside the product. <Q> If your product has a microcontroller which consumes tolerable amounts of current in sleep mode, you could simply wake up every minute (or whatever interval is appropriate) and check if the box is open using whatever sensors you like (Hall for magnetic field, photoresistor for light, etc). <S> The check can take very little time, and if it's negative you'd go back to sleep for another minute. <S> For example, drawing 6 mA for 1 ms every minute produces the average current of 100 nA. Assuming a current consumption of 500 nA in sleep mode, you'll have a total average discharge current of 600 nA. <S> You'll have to find a compromise between battery shelf life, wake-up delay and the appetite for current of the sensors which tell you to wake up. <A> I've seen several on-the-shelf products lately that have a pliable plastic strip that is between one pole of the battery and its contact. <S> The plastic strip is exposed to the outside, usually with "PULL ME" printed on it. <S> Pull the plastic piece out, the battery contacts close <S> and it's on. <S> However, this would break your waterproofing unless you had a clever design for it. <S> Maybe have a plastic tab on the outside that you push that pushes the insulator out of the way - that could be sealed. <A> How about a mechanically latched button under the waterproof seal (possibly a soft rubber over an opening on the product?) <S> that is pressed by the package and has to be depressed (or vice versa) to get the product out of the box? <S> Once the button is latched the micro comes out of deep sleep. <S> No need for magnets or relays. <S> Just need to design clever packaging. <A> I would try using a photo sensor to wake up your microcontroller. <S> You will need to make sure your packaging can be opaque and your device will need a window. <S> When the user opens the package and exposes it to light, the device wakes up and starts operating. <S> I see some parts on digikey with 20 nA or even 1 nA dark current. <S> Just wire up the emitter to an interrupt pin with the biggest pull-down resistor you can get away with. <S> If you go this route, make sure to check the temperature dependence of the leakage current -- it is likely that it goes up quite a bit at high temperature, this may cause a problem with higher current drain or even unintentional triggering. <S> Also, for this option you need to really make sure your packaging is light tight. <S> If for some reason the phototransistor doesn't work for you, you could even use a photodiode in photovoltaic mode. <S> It provides its own power, so you don't need to worry about battery drain. <S> However, silicon photodiodes can only generate a few hundred mV, which is probably not enough to trip an interrupt so you will instead need to wake periodically and measure the diode with an ADC. <S> Alternately, you may be able to use an LED, which will have higher output voltage depending on color, but quite low output current. <S> The nice thing here is that you will have a bit more leeway in terms of stray light since it won't be drawing leakage current from the battery. <A> Keep the magnet in your box, but use a small SMD hall sensor IC to activate your product. <S> There are hall sensor ICs on the market that consume just a few micro-amps of average current. <S> Put the output of the hall sensor into the gate of a small SMD MOSFET that sits between your battery and the rest of your circuits. <S> When the hall sensor deactivates your FET will turn on, and so will the rest of the board. <S> If you want the product to stay on forever once removed, then use a latch circuit so that once the product turns on it remains on. <S> There are a few obvious options here. <S> 1) Run the output of the hall sensor and the MOSFET into the inputs or an OR gate and then drive the FET gate from the OR gate. <S> The gate itself is powered directly from the battery. <S> 2) Run the output of the hall sensor and a spare I/O pin from your micro-controller into an OR gate and then drive the FET gate from the OR gate. <S> This is the same as above, except that you can add some software configurability. <S> The microcontroller can make those kinds of decisions in software as long as it can set a flag in EEPROM or flash. <S> The AUP series of logic gates from Texas Instruments only draw nano-amps of supply current, and could be a good choice here. <A> Use the method you are using now, except substitute a cheap and widely available normally-open reed that holds your circuit in reset. <S> You'd need a large value pullup resistor to keep the current much less than the battery self-discharge rate. <S> This would also provide a way to hard-reset your circuit in case <S> it 'hangs' in some unpleasant manner, as you have now. <S> There are existing products that use a similar method to read an external reed switch- bicycle accessories.
| The simplest option would be to get a low-leakage phototransistor. For example, you may want the system turn off the first time you put it in the box, but remain on thereafter if a magnet is present.
|
How can I add a signed vector with an unsigned one in VHDL? I would like to know how can I add an unsigned vector with a signed one. The reason is that I am creating a MIPS processor and I would like to add the program counter which is unsigned with the immediate field of I-Type instruction which is signed. Below you can see the datapath and the addition of those 2 vectors. I've already tried the following resulting in error: -- Signals usedPC_BRANCH_DEC_OUT : out std_logic_vector(31 downto 0);PC_PLUS4_DEC_IN : in std_logic_vector (31 downto 0);signal signed_imm_s : std_logic_vector (31 downto 0);-- Addition of the signals abovePC_BRANCH_DEC_OUT <= std_logic_vector(unsigned(PC_PLUS4_DEC_IN) + signed(signed_imm_s)); The aforementioned addition works only if both are unsigned or signed. Is there a workaround to make this addition possible and how? <Q> There are to_integer functions for both signed and unsigned. <S> You can also try simply typecasting the unsigned signal into a signed signal, however that might turn 15 ("unsigned 1111") into -1 ("signed 1111"), and might mess up your addition. <A> Why are you keeping the program counter in a <S> std_logic_vector ? <S> This data type is meant for a bundle of independent std_logic values, while the bits in your program counter are not independent as there is carry during addition. <S> I'd implement the entire program counter as an unsigned , which allows using addition operators directly. <A> All you need to do is extend the unsiged program counter to make it a signed value, perform the addition with the signed immediate value, then chop off the sign bit: signal pc_signed : signed (32 downto 0);signal <S> new_pc_signed : <S> signed (32 downto 0); ... <S> pc_signed <S> <= signed ('0' & PC_PLUS4_DEC_IN);new_pc_signed <= <S> pc_signed + signed(signed_imm_s);PC_BRANCH_DEC_OUT <S> <= <S> std_logic_vector(unsigned(new_pc_signed(31 downto 0))); There is obvious potential here for something bad to happen if an instruction tries to move the PC below zero, but I suppose it would be up to the programmer to make sure that this did not happen. <S> You could add checks for this in hardware if you wanted; just check for the most significant bit in <S> new_pc_signed being set. <S> As someone else pointed out, you would be better off with ports and signals that represent numbers having a numerical type, so your program counter would be unsigned throughout. <S> This would avoid having so many type conversions everywhere. <S> It seems to be a common complaint that awkward looking type conversions are 'required', when the real solution to this is often to just use more appropriate types in the first place.
| The easiest way might be to to convert both signals to an integers, add them, and then convert back to signed.
|
LTC3106 solar harvester example - open drain on power good pin The part of the circuit of interest is highlighted on the image. The block with the antenna is just some device being powered, not important for the question except for providing the generic Vdd , EN and GND pin. I'm trying to understand the concept of an open drain from the example. The description of PGOOD states: Power Good Indicator. Open-drain output that is pulled to ground if VOUT falls 8% below its programmed voltage. The PGOOD pin is not actively pulled to ground in shutdown. If pulled high the PGOOD pin will float high and will not be valid until 3.5ms after the part is enabled. Assumption The 10MOhm resistor connects V_out and PGOOD . My thoughts are that this serves as a pull up for the EN pin because PGOOD can only provide a fixed ground potential and is floating otherwise. Is this a correct assumption? What is open draining in this example? What current and voltage are used for calculating the pull-up resistor value? <Q> I don't think I understand what you mean in your assumption, so I can't tell you if it's correct. <S> The way I think of an "open drain" is: there is a switch inside the IC that either pulls that line to ground, or doesn't pull it to ground. <S> If the signal is to be low, the switch is on, shorting the line to ground, if the signal is to be high, the switch is open, letting it float up to the relevant voltage level. <S> The output is reliant on the voltage rail that the pull up resistor is connected to and the value of the resistor. <S> A very large resistor will mean that the line will take a long time to increase (due to a large resistor providing a low current to charge the parasitic capacitance of the signal line). <S> Hence, when you ask for "current and voltage" to calculate the pull up value, that isn't quite right. <S> What you're really after is working out the pull up value based on the rise time you want and the capacitance of the signal line (including the pins of connected ICs, the data sheets are usually good at providing this). <S> As a rule of thumb, the charge time is about 5 times the RC constant (5*R*C, where R is pull up resistor and C is the capacitance of all items on the signal line). <S> In this event, the open drain is on the PGOOD signal, which means that the signal is always going to be low until the rail you're pulling up to is high, and then it will only become high once the power rail you're creating here is within tolerance. <S> As these signals don't need to be particularly quick, the resistors are quite large (often 10K) to reduce power consumption when pulled low. <S> Calculating the resistor value is more critical when dealing with an open drain communication link, such as I2C (or version thereof). <S> Quick summary: <S> The value of the pull up resistor is based on the rise time you desire and the capacitance of the signal line and devices on the signal line. <A> Open drain means that the output is taken from the drain of a mosfet but the drain itself is not connected to Vdd. <S> Hence, if the gate voltage is high enough and the mosfet is saturated, the drain will be a tenths of a volt above ground, and so it is effectively grounded. <S> If Vg is low, though, the mosfet is in the cut-off region and the drain is disconnected from ground and since it is not connected to Vdd, it is floating. <A> MOSFET Nch outputs require pullup for logic level indication. <S> The Switch is active only when sufficient supply is available yet more than 8% below threshold. <S> When an internal comparator turns off logic threshold MOSFET inside, PGOOD out is open circuit and pulled up with high R value.
| Open drain is either pulled to ground when low, or left floating for high.
|
What is the difference between monolithic and non-monolithic dc-dc converters? What is the difference between monolithic and non-monolithic dc-dc converters?For example this is monolithic dc step down controller: LTC3609 <Q> There are cases when you have to purchase the switches as external components. <S> In my opinion, a truly monolithic converter would be one that also integrates the inductor and possibly even the capacitors. <S> I mean, look at all the stuff you have to add for the converter to work! <A> Monolithic means that everything is on the same rock. <A> To be a little more specific, monolithic in this sense usually means that the power switches (FETs usually) are on the same silicon die as the control circuitry. <S> There are also converters that have several die in the same package, e.g. a controller and 2 FETs for a sync buck. <S> This might be transparent to the user as from the outside <S> it seems like a single IC, but it is not referred to as a monolithic part.
| Monolithic in the sense of the part you have refers to the fact that the power switch(es) is(are) integrated into the controller IC as a single part.
|
How does pulse or burp charging work? I've been doing some research about how to fully charge a lead-acid and AGM battery. However I have some questions and need some guidelines for proper implementation. Basically I am trying to find out if: There is an optimal voltage or current to use during pulse charging. I know the voltage can be higher but does the .1C charging current rule apply for example? If I were to setup a solar PV system, would you recommend modifying the circuit in such a way that pulse charging technique is applied to the battery array for charging? Would really appreciate some sort of open source reference for this especially from someone who's tried it as it seems a good amount of documentation around this is proprietary or patented. <Q> There are two fundamental rules of battery charging: you shouldn't let the battery get too hot, and, you shouldn't let the voltage get too high. <S> The .1C charge rule is a rule-of-thumb to protect you from both of the real rules. <S> Particularly when you have cheap charging equipment that can't shape the charge curve, <S> and/or Particularly when your real skills are at manufacturing batteries at a competitive cost, not battery chemistry, and you don't know any better. <S> If the ambient temperature is low, you can use medium-higher rates for medium-length-periods when the battery voltage is low, because the battery voltage won't go too high until is is partly charged. <S> Most battery charging is done with pulses, from switch-mode power supplies. <S> Most solar-system charging is done with systems where the charge rate is limited by the solar panels, so you couldn't get too much pulse current if you tried. <S> Burping the batteries is probably a bad idea. <S> Experiments which have tried it report reduced battery life. <A> Lead-acid batteries suffer from two annoying problems. <S> The first problem is that if voltage exceeds about 2.4V per cell the water in the electrolyte will break down into hydrogen and oxygen ( electrolysis ). <S> Apart from the risk of explosion, the lost water needs to be replaced or the battery won't work properly. <S> A 'flooded' or 'wet cell' battery can be topped up with deionized water, but sealed batteries cannot. <S> The reason this is a problem is that to prevent gassing the voltage must be kept safely below the point of electrolysis, so the battery can only be charged at maximum rate for a few hours. <S> Current must then be progressively reduced to stop the voltage from rising, and the battery gets charged slower and slower (taking 20 hours or more to get a full charge). <S> AGM <S> (Absorbent Glass Mat) sealed batteries get around this slow charging issue by recombining the hydrogen and oxygen rather than venting gas. <S> However this increases pressure inside the battery, so it can only handle a limited amount of gassing. <S> A charging system that is designed for AGM batteries can charge them a bit faster than normal SLA (Sealed Lead-Acid). <S> Pulse charging <S> may be able to charge at even higher rates without gassing if it can get a burst of current into the cell before it starts to gas. <S> However it is possibly more useful for dealing with the second lead-acid battery problem - sulphation. <S> This is the formation of lead-sulphate crystals that occurs when the battery is left in a partially discharged state for a long time. <S> The crystals have low solubility and are not easily dissolved by normal charging. <S> A sulphated battery loses capacity because there is less acid available, and it may also have reduced current carrying capacity due to the crystal's poor conductivity. <S> Pulse charging aims to break the crystals down so they will dissolve faster and restore capacity. <S> The technique does work, and can even use the battery's own power to do it! <S> However there isn't much point applying it to a battery that is in good condition. <A> Summary Standard well tried well documented methods usually are "not too bad". <S> Super 'patented' chrome plated pulse methods MAY work better. <S> Or not. <S> Unless they are certainly superior and have no hidden problems, there may be no good reason to use them. <S> Pulse charging tends to lie somewhere in the realms of some mix of: Advanced real world science understood only by the illuminati (or technocrati) / <S> Common sense / <S> Accidental discovery that works, sometimes, sort of / Magic and-or Scam <S> (oxygen free copper-crystals-phase-of-moon-bank-holidays-dead-fish-nsfw-ymmv-dnttah-ianal , ...) <S> / Other. <S> Deciding which is which can be difficult to impossible. <S> IF a method is of clear capability, reasonably priced and does not damage the battery, then consider it. <S> Otherwise, tried and true methods are in widespread use and probably preferable. <S> Whatever <S> Trojan Battery USA say is a good starting point. <S> (I have no association with Trojan - just admire their products) <S> Technical support <S> Downloads Battery university are usually not terrible although some choose to slang some of their stuff. <S> Super patented chrome plated pulse methods MAY work better. <S> Or not.) <S> Reputable battery manufacturers usually have a good idea of what works with their batteries. <S> If the spec sheet is deficient in this area then, if you care, changing brands may be in order.
| You can use much higher rates for short periods because (1) If it's short, the battery won't over-heat, and (2), If it's very short, it's just charging the battery capacitance, so the voltage won't go too high.
|
Maximum SPI clock speed specified in data sheet For integrated circuits that use SPI interface, they specify maximum clock frequency in their datasheets such as attached below. The question is that if the datasheet speficies the maximum frequency as 6MHz, can I decide that it will be safe to set up clock speed exactly at 6MHz(accuracy determined by crystal used for MCU)? Or It is safer to have some margin? <Q> No, 6 MHz is guaranteed to work by the specification. <S> In practice you will very likely find devices which can work up to 8 or even 10 MHz. <S> But it is not guaranteed that that will always work for all power supply voltages, temperatures and all chips from any batch. <S> The 6 MHz however is guaranteed over all conditions. <A> Note that the clock high and low widths are specified as 83ns min, which works out to just over 6MHz, question is can you guarantee to meet those timings (Which requires your SPI clock to be pretty much an exact 50% duty cycle).... <S> When figuring out how to meet timing you generally need to look at the data sheets for both ends of the link, and then do some sums to see what the maximum frequency is with worst case timings on both ends, just looking at one end only tells you what may be possible, not what will be possible (as that depends on both sets of timings). <A> Actually Bimpelrekkie 's answer is not correct. <S> You should consider the propagation delay between master and slave either. <S> You can find the detail information in the link Using SPI protocol at 100 MHz .
| So no, you do not need a margin as long as your 6 MHz is accurate enough (and it probably will be as it is usually derived from a crystal based clock).
|
Why is power so often associated with current and not also voltage? I was reading one of my textbooks the other day about linear BJT emitter-follower amplifiers (non-relevant) and came across the following passage: Although the small-signal voltage gain of the emitter follower is slightly less than 1, the small signal current gain is normally greater than 1. Therefore the emitter follower circuit produces a small signal power gain. But I've learned that the power can be expressed: \$P = IR = \frac{V^2}{R} = I^2R\$ Which means that power is directly proportional to both current AND voltage. Wouldn't this mean that a large voltage gain also provides a power gain? This is not the only place I've seen this discrepancy either. It seems that whenever people talk about power, they are only really concerned with current and not voltage, even though the math seems to suggest that is not the case. Can anyone elaborate on this? EDIT: One explanation I can think of is that there can be voltage across an open circuit, so increasing that voltage would theoretically increase power, even though nothing is getting hot... whereas if there's current flow increasing power by increasing current would make the component dissipate more energy... <Q> It seems that whenever people talk about power, they are only really concerned with current and not voltage <S> Ignorant people perhaps, but every competent electrical engineer knows that both voltage and current must be taken into account. <S> Although the small-signal voltage gain of the emitter follower is slightly less than 1, the small signal current gain is normally greater than 1. <S> There is no discrepancy here. <S> The authors are pointing out that if there is current gain <S> then voltage gain is not required to increase power. <S> So the myth they are trying to dispel is that you can't have power gain without voltage <S> gain - exactly the opposite of what you think people are concerned about. <S> If the output is open circuit then it will draw no current so there cannot be any power gain. <S> However it may still be useful to consider the voltage gain if that voltage can be maintained with a load. <S> In some circuits (eg. video amplifier) <S> the source and load impedances are matched, resulting in half of the output voltage and power being lost in the source. <S> In this case you would normally only consider the voltage gain under load (so a video amp with open circuit voltage gain of 2 is actually a unity gain buffer). <S> In others (eg. audio amplifier) <S> the load impedance is normally much higher than the source impedance, so the voltage gain remains (almost) constant whether driving the load or open circuit. <S> If the load impedance is reduced then it will draw more current and power at the same voltage. <S> This is not important for low level signals so usually only the voltage is considered. <S> The output power of an audio power amplifier is very important, so speaker impedance is always considered - but output voltage and current are rarely mentioned. <A> You are correct that power is a function of both current and voltage. <S> There are circumstances where one is more significant than the other. <S> Using the quoted example of an emitter follower (aka common collector) <S> amplifier, the voltage gain is almost 1 (not very interesting) so all the power gain comes from the current gain. <S> But wire the same transistor as a common base amplifier - the input is supplied to the emitter. <S> Now most of the current appears at the collector (i.e. the current gain is almost 1; you lose a tiny fraction to the base) <S> so all the power gain comes from voltage gain. <S> And of course the common emitter has both voltage and current gain, hence relatively high power gain. <S> However there are cases where power is most usefully expressed as I^2R - often wasted power in a motor or a long cable. <S> Consider a transmission line supplying voltage V and current I to another city. <S> The power wasted in transmission is independent of V, and proportional to I^2. <S> Now, associating wasted power with current is useful; we can see that to minimise waste, we want to reduce current. <S> So to transmit the same power we want to increase voltage. <S> Which is why high voltage transmission lines are used, despite the expense of transformers at either end. <S> Naturally the power transmitted to the load is VL * <S> I (where VL is measured at the load end), while the power required is VG <S> * I (where VG is measured at the generator) - and the difference (VG - VL) is the voltage drop along the cable = <S> I <S> * R. <S> A similar case is seen in electric motors, where the main power loss is I^2R losses in the winding resistance R. <S> There is no similar loss due to voltage, so it follows that a motor is most efficient run at relatively high speed (speed is proportional to voltage) and relatively low torque load (torque is proportional to current). <S> In both these cases, and perhaps others, thinking about wasted power in terms mainly of current leads to useful insights. <A> I think you answered your question in the "edit" part. <S> It's right, <S> \$P = VI\$, but in the equation you wrote, \$P = <S> IR = \frac{V^2}{R} = I^2R\$, is more evident that there's also a resistance that play in the game. <S> Power is the work done in the unit of time, or better is the amount of energy consumed per unit time. <S> Voltage is potential energy , and I can have a big voltage source, but if we don't put a resistor across the source, the energy remain "potential", don't do any job. <S> When we connect a load a current flow and "do something". <S> But... you have already written this in more succinct way, in the "EDIT". :-) <A> Although voltage can/does vary, many things are designed for a fixed voltage, so to change power you change the current. <S> The AC power coming out of the wall, sure that is AC, but its amplitude and frequency are fixed, it is current that is tied to power assuming V fixed. <S> A battery based system <S> although the voltage can vary it is the current that does most of the changing. <S> Just about everything is designed around a fixed-ish voltage, and current varies, so if you hold V constant then P is related to I and you normally only need to talk about I. <A> I think this may be because power is directly proportional to both current AND voltage drop <S> (not just "voltage", in the sense of voltage to ground). <S> Power dissipated at any device is <S> \$P= <S> I \Delta V\$, where \$V\$ is the voltage drop across terminals. <S> When you look at different devices in series in a given circuit section, each voltage drop may vary a lot, but the current is common to all of them. <S> In this situation, current helps better to get a handle on power. <S> See also this other question .
| One explanation I can think of is that there can be voltage across an open circuit, so increasing that voltage would theoretically increase power
|
Location of diode bridge on transformer circuit? I'm designing a full bridge rectifier that takes 110vAC to 4-16vDC for input into a 5v switching regulator. I opened a few 5v power supplies and they place the bridge on the high voltage side. Why not use the low voltage size so the diodes are smaller/cheaper? Also, I notice they use an IC on the high voltage side, what is the purpose of that IC and is it required? Most circuits I see online are basic like this one which should give me full wave recification: This is the transformer I'm using: http://catalog.triadmagnetics.com/Asset/FS10-600-C2.pdf <Q> You're confusing the "oldfashioned" linear mains supply like this: Reasons to use this design are: <S> It is a simple design As long as you don't touch the primary side of the transformer, it's pretty safe <S> With the more modern switched mode power supply: As you can see the switched mode supply is much more complex ! <S> Reasons to use this design are: the transformer is used a a much higher frequency than the 50 or 60Hz mains frequency meaning that it can be much smaller and moreefficient the smoothing capacitors after the transformer can be smaller <S> being a switched supply means that it is much more efficient it can be made more compact (because of reason one) <S> it can be made cheaper (no expensive transformer) <S> lighter in weight, easier to carry. <S> For Bonus points: Here's a "circuit level" example of a simple switched mode mains supply, note that it has 2 outputs, 5 V and 12 V. <S> You could leave out the 12 V output to make it even more simple. <S> Now compare that to the first schematic ! <A> The 5 V power supplies you opened use a different topology to the one you are trying to build. <S> Since there are various ways to make a power supply, the real question is why you would expect a random mass-produced one to be the same as yours. <S> Commercial low-voltage DC power supplies usually rectify the incoming AC to make high voltage DC. <S> This is chopped at much higher frequency than the incoming AC, then run thru a transformer. <S> This allows the transformer to be smaller, lighter, and cheaper than your big, klunky, and expensive 50 or 60 Hz transformer. <S> The chopping circuit also modulates the output, usually with feedback via a opto-isolator. <S> Big fat power transformers that run at the power line frequency haven't been mainstream for a couple of decades or so. <A> Diodes are on the secondary. <S> IC on the primary side is most likely power factor correction. <S> That in itself is a switching power supply ( AC/AC convertor ) that keeps the voltage and current in phase as far as the primary is concerned. <S> I don't think you need that. <S> but you will need output capacitance to smooth the rectified AC.
| The circuit shown is correct.
|
Can PWM safely be considered as effectively lowering voltage? Specifically with LEDs which need constant current more than voltage. I'm using those 30V, 100 Watt LEDs (which strangly draw less than 1A) and due to using four 11V batteries in series to power them they need to be "dimmed' to 30 volts. My voltage regulators are only rated for 35V, so would a microcontroller's PWM with a MOSFET be a safe form of protecting the LEDs from would could otherwise be over voltage/current? I don't know if flickering could be an issue, but it doesn't matter too much to me and I could research how to include capacitors and/or inductors to reduce it. <Q> PWM does not lower the (peak) voltage. <S> PWM reduces the average current (and in consequence the average power). <S> In case of LEDs - they don't care much about the voltage, it is the current that can destroy them (both forward and reverse). <S> So you have to take care not to exceed their peak current rating. <S> To reduce voltage you would have to add an inductor, catch diode and capacitors to make a buck converter <S> which does lower the voltage. <A> PWM alone will not 'effectively lower voltage' for this purpose. <S> The problem is, PWM makes a voltage which is a function of time, it has MANY DIFFERENT VALUES, not a limited range (such as an LED can tolerate). <S> What endangers an LED may be an average, RMS average, peak value of V, or a peak value of current (which is a nonlinear function of voltage applied). <S> Assuming 90 lumens/watt, and a "100W equivalent" lamp <S> means 900 lumens <S> , you wantabout 10W into your circa 30V LED load. <S> 330 mA DC is the desired current. <S> A linear regulator (LM317) could work; it needs only to be rated for thedifference of input and output voltages <S> (44V - 30V = 14V) and not forthe 44V. <S> The most effective way to regulate current is to use a current sense resistor, as: simulate this circuit – Schematic created using CircuitLab <S> The resistor and regulator will get warm, so you'll want a heatsink. <S> For better power efficiency, PWM followed by a filter inductor and a catch diode is called a 'buck' voltage dropper. <S> It filters but won't regulate (lacking any kind of feedback), so <S> the battery condition and ambient temperature will affect thecurrent. <S> Buck regulators are available, that will regulate, dissipate nearly no heat, improve battery lifetime. <S> Many suitable modules are based on LM2596. <A> The peak voltage will be the PWM peak and it will determine the peak current. <S> Check that your design does not exceed specified LED peak voltage. <S> Check that if using peak current that the maximum duty cycle at that current is not exceeded. <S> e.g., 1 A for 5 ms at 20% duty cycle.
| PWM will reduce the average voltage and current to the LEDs. Check that your design does not exceed specified LED peak current.
|
How to identify this particular SMD component? It is used on a small board RFID antenna circuit based on the TMS3705: At the first check it seems to be a 0 ohm resistor, but the dimension 4x4mm its strange. Any ideas? L. <Q> You can find them listed in the resistors section of a vendor's parts catalog. <S> It is typically used to turn on or off configuration options on the circuit board. <S> The "0" marking is intended to make this stand out among a group of actual resistors. <S> They come in many sizes and even through hole versions. <S> It is possible that this one is large in order to make hand-soldering easier. <A> It is a 0 ohm resistor. <S> They probably planned to have an inductor there, but replaced it with the 0 ohm resistor during testing/production preparation. <S> It might be replaced to reduce cost. <S> Or they might have it as a way of correcting the impedance of the antenna if needed. <S> You can of course measure it using a multi meter. <A> As others have said, it's a 0\$\Omega\$ resistor. <S> The reason why the unusual package is used is because they wish to minimize the inductance since it's in an RF circuit. <S> Here is a datasheet for resistors (including 0\$\Omega\$) in wide (side terminated) packages.
| It is a "zero ohm link".
|
Drive relays with optocoupler and a Raspberry pi I'm currently designing a PCB in order to drive relays (5V) with an optocoupler and a Raspberry pi (a CNY74-4 here is the datasheet ). So I made this schematic: My questions are the followings: Is my scheme right? The 5V come from an DC/DC supply which deliver 2.1A and 5V, it's not a problem for the optocoupler and the transistor? Which CMS transistor can I use in this case? Is the optocoupler good for this use? THanks for your answer! <Q> You are missing catch diodes on the relay coils. <S> Back-EMF from relays will blow the transistors when they are turned off. <S> You can use a BC547 transistor or a driver chip like ULN2003 or ULN2803 (they have built-in diodes and base resistors). <S> Optocoupler is fine. <A> The 5V come from an DC/DC supply which deliver 2.1A and 5V <S> I suspect that there isn't a need for the opto couplers. <S> By the sounds of it, the 5V is rated also for the RaPi (2.1 A) and so there is no benefit from adding the opto because the relay will provide isolation from the load however, you do need to use flyback diodes across the relay coils to prevent damage to the NPN transistors. <S> Looking at the spec of the opto, it has a minimum CTR (current transfer ratio) of 50% and therefore you are roughly obliged to feed (waste) <S> more current driving the opto than you would driving the NPN directly. <A> In the "does it work" respect, it will work. <S> You just need to find out what resistors should be placed in the transistor bases. <S> ( This is really important in order to have safe and fully functional circuit ) <S> But some small problems that you may encounter and you <S> must check: <S> The optocouplers EASILY burn if the LED is lit with excessive current. <S> Check that optocoupler LEDs are series with right resistors. <S> Also if the resistor is very high in value, the optocoupler phototransistor might not be completely in ON-state. <S> Check datasheet of the optocoupler. <S> It might be able to sink enough current to drive the relay coil, so that you could remove those 4 transistors attached to relay coils. <S> The relay coil has noticeable mutual/self inductance. <S> As we know from power electronics, Switching inductors and motors using transistors without use of catch diodes (Sometimes called free-wheeling diodes) will put the poor transistors in risk of having excessive voltage applied to their Collector-Emitter pins and being blown. <S> You have to insert a freewheeling diode with Cathode connected to collector and Anode connected to emitter. <S> As @filo mentioned, you could use ULN series which already have the diode. <S> I agree with Andy. <S> If you want isolation of the Raspberry Pi from the outside work ( You must be wanting that either way), the relay is already giving you that isolation. <S> Optocoupler is just one more step of precaution. <S> You can remove the optocoupler and power the coils using those 4 transistors by directly connecting the bases to Raspberry Pi (via resistors surely!). <S> But make sure that you have a good 5V supply. <S> Some 5V Relays just fail to turn on with low voltages as low as 4.2V. <S> (Notice that when you turn on the transistor, a little amount of voltage known as VCE(SAT) ~ 0.2V drops on the junctions) <S> Also the 2.1A power supply is surely sufficient. <S> Relay coils have small resistance but they usually draw about 30mA~150mA (depending on type and quality and also design of the circuit). <S> Check the relay datasheet.
| Power supply from DC/DC converter is perfectly okay for relays.
|
Testing points in PCB I am designing a prototyping platform called evive . It is a learning cum prototyping platform for students and makers. Since many times we requires debugging of our circuits, so I thought to include some testing points in PCB. I came across this image: It is used in Dr.Duino (debugging shield). Can anybody explain what it is? Or any other idea about how to give test points in PCB for users, such that user can easily attach the alligator (crocodile) clips or Banana multimeter cable tips. <Q> They are called Test Point Terminals, believe it or not, and here is an example: <S> http://uk.farnell.com/vero/20-313137/terminal-pcb-red-pk100/dp/8731144 <S> Many people make them, in many different colours and shapes. <A> I have used these on several boards. <S> The ones I generally prefer are the Keystone 5000- and 5100-series. <S> Part number 5000 is a red test point, 5001 is a black test point, <S> 5002 is white, 5003 is orange, <S> 5004 is yellow, 5115 is brown, 5116 is green, etc. <S> http://www.keyelco.com/product-pdf.cfm?p=1309 <S> Digikey selections: <S> Keystone 5000 Series Miniature PC Test Points <A> The ones you have are through-hole parts similar to the Keystone 5000 series that Derstrom8 mentions. <S> You can also get SMT types that take up a lot less space (none on the opposite side) and can be machine assembled. <S> Despite the simplicity of these parts they are relatively expensive (over $150 US for a reel of 2,000 compared to more like $6 US for 5,000 resistors, reflecting the low demand (most high volume products will use other methods). <A> Probes that can stay in the holes are abundant, or you can solder a pin in and use an alligator clip for that. <S> Especially for learning/prototyping, the user should have pins laying around anyway, or you can throw some in with the kit. <A> As everyone says, those are test points, and there are many kinds available. <S> To choose the right type for your needs, and where you need to put them, try to anticipate what you're going to be doing with them. <S> Are you going to clip scope probes to them, or just touch the probe without the clip to them? <S> Do you need to hook the grounding clip of a scope probe to them? <S> Are the test points for important signals within easy reach of a place to hook a scope ground clip? <S> It seems like none of this makes a difference, but if you take some time to figure this stuff out, they'll be much easier to use. <S> Take it from someone who got it wrong once.
| While there are plenty of specialized products for small-spaces, just putting holes that are sized for your usual .1" pin header pins works for larger sized board layouts.
|
Is it possible to have two switches controlling the same motor? Basically I want one (3 prong)switch to allow the motor to spin in one direction, and the second (3 prong)switch will allow it to spin in the opposite direction. Is there any way to do this? <Q> The closest you could get is to use the switches as a H bridge. <S> Certain combinations of switch positions would then give different motor operations: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> With both switches in the same position the motor will be stopped. <S> With the switches in alternating positions the motor will spin, the direction dependent on which switch is "up" and which switch is "down". <A> You could have a DPDT relay acting as a motor reversing switch: - http://www.quadomated.com/wp-content/uploads/2012/11/Multi-Direction-DC-Motor-Control-with-Two-Relays.jpg <S> The above picture should give the general idea - just imagine that both SPDT relays are one combined relay. <S> - OK, the lamp is the relay coil in this example. <S> If the relay coil is DC then replace the AC supply with the appropriate DC supply for the relay coil. <S> Alternatively you could use a H bridge and activate the direction using the same switch arrangement shown in the 2nd diagram. <A> If you want to use just a switch, you can use a double-pole, double-throw (DPDT) switch. <S> DPDT switches are available with and without a "center off" position. <S> As you can see from other answers, it is not very convenient to use two switches. <S> Here is a way to do it with momentary-contact pushbuttons and relays.
| Next you can wire two switches together so that either switch will reverse the direction of the motor by activating/deactivating the relay coil:
|
Why is "hole" conductivity called so, isn't that a case of electron movement? Why is it that "hole" conductivity is called so, isn't that just a case of electron movement in the opposite direction, or as in 'Vacant Seats in Theater' analogy mentioned in the Electron Hole Wikipedia , it can be thought of as people moving in opposite direction? I am not looking for a long explanation, just curious why would textbooks/theory prefer mentioning something fictitious rather than something real which is electron movement? Is it because it helps distinguishing the conduction due to normal electron movement vs movement due to lack of it? Probably when you visualize it, it does indeed look like movement of holes as well. I was really not able to understand the rationale, hence this somewhat basic question. As a side note, did anyone have this question while learning? I hardly saw any discussion on however I thought it is a good question to ask. <Q> You are correct that the only thing moving in a metal or semicondcutor is electrons, but also that the behaviour of a hole and an electron is different. <S> A couple of ways in which this is important for electronic engineering (as opposed to solid state physics) are: At a PN junction. <S> In the semiconductor far from the junction, the behaviour is pretty much the same whether there are hole or electrons doing the work. <S> But at the junction they can meet, and when they do, the electron falls into the hole. <S> Or to use the technical term, the electron and hole recombine. <S> In a hall sensor. <S> Electrons and holes which are moving are pushed sideways by magnetic fields. <S> They have opposite charges and are flowing in opposite directions, so they are pushed the same way. <S> But because they have opposite charges they create opposite voltages. <S> The hall voltage in a P type semiconductor is in the opposite direction to the hall voltage in an n type or a metal. <S> Also, if you get stuck into the physics of conduction in metals and semiconductors, you quickly find that the quantum nature of electrons mean that you can't really think of individual electrons or holes moving about. <S> The real physical insights come from something called a Fermi surface. <S> The idea of electron surfaces and hole surfaces is key to those insights, so after learning about them, the "electron" and "hole" terms seem more natural. <A> I remember discussing this question during my studies for a long time. <S> The 'holes' are electrons 'missing' in the valence band, and not in the conduction band. <S> So an atom with all its electrons in the valence band 'gives' its electron to that one with a 'hole'. <S> The free electrons on N materials, on the other hand, are in the 'conduction' band. <S> Since those electrons in the valence band are closer to the nucleus than the electrons in the conduction band, their mobility is more difficult and that is why for the same size, an N-channel transistor is a much better conductor than a P-channel transistor. <A> The answers here are skipping one important feature: <S> A hole is not a missing electron. <S> A hole is a semiconductor atom which has extra protons in relation to its electrons. <S> A hole is a genuine positive charge. <S> When a hole moves, a positive charge has adopted a new location. <S> Take the simple case of neutral hydrogen atoms. <S> If we remove one electron, we're left with a bare proton: a positive charge. <S> A proton is not a "missing electron." <S> Positive charge is not a "lack of electrons," instead it's a genuine type of charge in it's own right. <S> The OP question would be valid if silicon was made of electrons only. <S> Then, any "hole" would just be a gap in the population of electrons, and if the hole moves, it's really just electrons moving backwards. <S> But silicon is made of cancelled-out charge: equal positive and negative charges. <S> A "hole" is a gap in the electron population which exposes a hidden proton. <S> A hole is a genuine positive charge. <S> With the movie-theater analogy, the complete picture would be to place red light bulbs on all the seats. <S> These are the protons. <S> Now let the audience of electrons sit on the seats and cover the lights. <S> If there are any gaps in the audience, then in the dark theater we will see brilliant red lights at the location of "positive charges. <S> " <S> These are the red lights which haven't been canceled out by an electron. <S> If one person changes seats, then the location of the red light will move the other way. <S> But from a distant view, there are no electron shifts. <S> They are indetectable, unimportant. <S> It's the changing position of the positive charges which dominate. <S> All we see in the dark movie theater are some extremely powerful red lights which suddenly leap from one seat to the next. <A> Basically, hole is vacancy of electron. <S> So when electron moves hole is created. <S> Electron is negatively charged(at negative or low voltage) and hence attracted(move) towards positive terminal of battery, which means holes moves towards negative terminal of battery which is same as the direction of current(+ to -). <S> That is why we say <S> holes(vacancy of electrons) are contributing to the current.
| So that's the answer: at the micro-level, yes, electrons are shifting in order to cover or reveal the positive charges.
|
LM358P output saturation around 6.05V (root cause turned out to be related to Arduino protection diodes) For a constant current drain using a PWM input I am using the following setup: simulate this circuit – Schematic created using CircuitLab For small currents it seems to work as expected - the op-amp producing a voltage to the MOSFET which causes the current flow over the battery be such that the voltage drop over the 3.3 Ohm sense resistor becomes exactly the configured voltage, as expected. But there must be something I have missed, because the current I can get maxes out much earlier than expected. This happens around 150 mA corresponding to an input voltage to the op-amps positive input (via PWM) of around 0.45V. At this point, the output from the opamp to the MOSFET seems to saturate around 6.05V. The problem is that further increases in the op-amps positive input has no effect. The opamp keeps producing a 6.05V output even though there is a large difference between the positive and negative input. I can see the opamp is getting, say, 0.8V at the positive input and is only seeing 0.45V at the negative input, yet it refuses to output more than 6.05-6.1V. The Vcc to the op-amp is 12V so there should be plenty of headroom for it to output higher voltages even though it isnt a rail-to-rail opamp. I have looked in the datasheet but I cant seem to find a limitation that would explain this behavior, so I hope someone can explain what I am missing. Update! I am a bit closer to resolving the mystery. I had a wire (not on the drawing) going from the output of the op-amp and to an analog input on the Arduino. I thought the analog inputs were high impedance so it wouldnt affect anything. But with that wire there I am seeing the 6V saturation (as measured by a multimeter, the Arduino only measures to 5V), but if I disconnect it everything works. Is this some type of oscillation involving the Arduino? Or is the Arduino draining the voltage? NOTE! The root cause of this problem turned out to be different. I had connected a measuring wire from the op-amp output to the Arduinos analog input. My thinking was that it was high impedance and hence wouldnt affect outcome. I had noticed the measurements topping out after 5V but thought that was simply due to it being unable to report more than value 1023 and that it wouldnt actually affect the behavior of the circuit. This assumption was wrong; due to a protection diode in the Arduino inputs, the extra voltage was drained (the LM358P has a short-circuit voltage of 40-50 mA) causing the drop in the measured voltage (even with a multimeter) at the op-amp output. But the real culprit was the protection diode in the Arduino. Removing this measurement wire resolved the problem. I am leaving the original problem description in place so other people can find this report in case they encounter similar issues. <Q> I believe your circuit has gone unstable and is oscillating. <S> You have a 1.5 kohm resistor feeding the gate and the gate capacitance will be "fractionally" seen relative to ground at this point <S> and this will add a fair amount of phase shift. <S> You close the loop and you get an oscillator. <S> Normally you can rely on the FET acting as a 1:1 voltage converter from gate to source <S> but, with such a low value resistor in the source, this might degrade to more like 1:0.7 and this will mean 0.3 X gate-source capacitance is seen to ground at the gate. <S> If it were a perfect 1:1 gate-source voltage converter then there would be no capacitance seen at the gate theoretically. <S> The specified gate capacitance is 1.9 nF for the IRFZ44 and so <S> the 1.5 kohm resistor will see about 30% of this i.e. ~600 pF. <S> This forms a 3dB low pass filter at about 178 kHz - at this frequency <S> the signal is shifted by 45 degrees and thus reduces the phase margin of the LM358 causing it to sing its little heart out! <S> It's probably oscillating somewhere <S> between 200 kHz and 1 MHz. <S> A more modern op-amp would have a graph of its open-loop phase response and this frequency would be easier to predict. <S> Try reducing the 1k5 to about 100 ohms or less and see if the situation improves. <S> In fact I would just short out the 1k5 entirely. <S> It's a lot easier to drop the 1k5 in value. <A> I had first imagined that your problem was that the opamp was oscillating and suggested that you try something like this (thanks to Spehro, too): simulate this circuit – <S> Schematic created using CircuitLab <S> Per further discussion, where you added that you were supplying the output of the opamp to an MCU input pin on your Arduino, it's also clear that even if you did (or would have had) <S> an oscillation problem (and you would <S> , I'm pretty sure) <S> it is more likely now that your measured limit of around \$6V\$ was due to a heavily loaded protection diode at your MCU's analog input pin. <S> The protection diode limitations for any given MCU can usually be found under the "Absolute Maximum Ratings" section of the datasheet, where that section will include a line described as something similar to: "diode current at any device terminal. <S> " <S> Usually, the value listed there will be around \$2mA\$ or less. <S> If you must measure the opamp output voltage, then you will need to consider a different circuit arrangement to condition the opamp output (without disturbing its MOSFET control behavior) and make it suitable for input to the Arduino analog input pin. <A> This is an example of over-driving your Arduino analog input. <S> A voltage divider should have been provided in the first place (and actual connection to Arduino showed to allow meaningful discussion
| You could add an integrating capacitor and input resistor as pointed out in another answer but this will significantly reduce the dynamic capability of the circuit to maintain a constant current.
|
Splitting DC Supply Rail with High Output Currents I'm currently working on a audio amplifier circuit that requires split supply rails. I have a wall converter that can supply 19V, and I ideally want to be able to create a ± 12V supply rail out of the 19V and still be able to output somewhere around 3amps. Does anyone know an easy solution? Most solutions I've found can only supply up to 500mA. If it's much easier, I can also just split the 19V to ±9.5V. But again, I have not been able to find a circuit or IC that can output up to 3amps. <Q> If you could do what you want, you would win the Nobel Prize. <S> Your supply puts out 19 volts at 500 ma, which is a power level of 9.5 watts. <S> If you could use that to produce +/- <S> 12 volts at 3 amps, a single side would produce 12 volts at 3 amps, or 36 watts. <S> So you could then use 9.5 of those watts to feed the input of your converter, and have 26.5 watts for free. <S> Of course, you could then cascade converters to run your house, run an electric car, or anything else a perpetual motion machine is good for. <S> As you might guess, this is not going to happen. <A> This becomes a make/buy decision. <S> I suggest you shop for the best bipolar <S> +/- <S> PSU to match the bass current you need with regulated outputs and line filters to prevent hum. <S> Otherwise hum or switching regulator noise will result from unregulated or regulated wall worts with dual sources for both polarities. <S> Consider surplus PC PSU's with adequate -12V current. <S> So the driver is inefficient and carries all the speaker current. <S> But if you used a single 24V supply with 2 bridge amps (differential) per speaker then the speakers will be at 12Vdc on either side. <S> or use 12 V supply with differential like many 22W car radios. <A> You have to define what you mean by an easy solution though... <S> It is possible, yes, but your power supply(the 19V one) has to be able to supply nearly 2 ampéres and above.... <S> Because there is no such thing as a 100% effecient system. <S> Search it up and you'll see what I mean. <S> You have to experiment around and blow some stuff up;; Murphy's Law always holds in all things :')
| To make a floating 24V supply into a bipolar supply requires a high current linear amp with V/2 reference and output gets grounded. What I did was to make a mini power supply that changes your 19V to 12V dual rail using a regulator + 555 combination to drive a MOSFET to drive a switch mode supply transformer.
|
This relay drives me crazy I'm currently learning electronics and right now, I'm trying to use a relays... without success. Here is a picture of my circuit: First of all, don't mind the 2 cables on the top right and the relay on the bottom right, they are not used. So, from what I understood about relays, here are my thoughts about this circuit. When the button is not pressed, the relay does not receive any tension, so the pins connected to the lightning LED are active, which is why the LED on the top is lightning... That I get... Now, when I press the button, a 12V DC tension is going inside the relay, so I was kind of hoping that it would make it switch and then the second LED would be lightning but... no... The relay is an AXICOM IM26, you can find the datasheet of this one in here: http://www.te.com/commerce/DocumentDelivery/DDEController?Action=showdoc&DocId=Specification+Or+Standard%7F108-98001%7FV%7Fpdf%7FEnglish%7FENG_SS_108-98001_V_IM_0614_v1.pdf%7F4-1462039-1 However, a friend made me notice that this datasheet "talks about" a switching current of 2/5A but I have trouble to imagine that this small relay really requires that amount of current to switch, especially because it is writing "12 V DC" on it, which, for me, tend to indicate that it switches at 12V. But it would explain why it is not switching when I press the button. Can someone explain me what is wrong with my circuit ? Thanks <Q> Your problem is very simple. <S> You are using your breadboard incorrectly. <S> The top pins (pins 1 and 8) are being shorted out by the contact strip. <S> The spacing between rows of contacts is 0.3 inches, which is a standard DIP IC pin spacing. <S> Your relay, however, has a lateral contact spacing of 0.2 inches, and so cannot straddle the row separation. <S> Since it didn't fit that way, you plugged the relay in to a set of holes that did fit, and that is the source of your trouble. <S> To confirm this, set up an LED and resistor to monitor your +12 volts. <S> Unplug the relay and you will see that the LED lights up. <S> Now plug the relay in and the LED will go out. <S> With this breadboard you really have no good way to mount the relay. <S> If you have access to a soldering iron you can make extenders for the relay pins with short wires and plug them into the breadboard, but other than that <S> I'm afraid you're out of luck. <A> Your relay has an 'ultra high sensitive coil' which suggests it is biased with a permanent magnet. <S> If this is the case then you must apply voltage to the coil with the correct polarity, otherwise the coil will cancel out the bias magnet's force instead of adding to it, and the relay won't operate. <S> this datasheet "talks about" a switching current of 2/5A <S> but I have trouble to imagine that this small relay really requires that amount of current to switch <S> That is the rating of the contacts, ie. <S> how much current it can switch . <S> The coil requires 10.2V DC and has a resistance of 2880Ω, so it needs at least 10.2/2880 = <S> 3.5mA to operate. <S> At 12V the coil should draw 12/2880 = 4.2mA. <A> AM26 is rated for 12V at 50mW <S> It comes in 3 ranges of sensitivity, this being most sensitive. <S> If coil Power = 50mW @ <S> 12V then current is 50mW/12 ~ <S> 4mA <S> This is pretty easy to drive and you might be able to make it click with the 9V battery across the coil with + on pin 1. <S> Although due to hystereis, the guaranteed or MUST SWITCH threshold is 10.20V and the MUST release threshold is 1.20V <S> but it MAY switch in between. <S> FYIThe lowest contact rating is 2A resistive. <S> So this relay has a net switch gain at max load of 2A/4mA = 500 or you could say a a coil sensitivity of 0.2% or rated current for a 12V coil.
| If the white stripe on your power supply cable is positive then it appears you have the polarity reversed (positive should go to pin 1 of the relay, and negative to pin 8). In this case, the breadboard is incompatible with the relay.
|
Does a capacitor connected directly to a battery consume any energy? In this example simulate this circuit – Schematic created using CircuitLab After the initial charging of the cap to 3V, current gets blocked, but over time does it consume any energy from the batteries?Is this safe to make? <Q> In steady state (after a long time) an ideal capacitor does not draw significant current from a battery. <S> A real capacitor will draw some small leakage current. <S> The amount of leakage current will depend on the type of the capacitor, electrolytics will have higher leakage than films and ceramics. <A> Leakage current will drain the battery, most likely not that significantly compared to the internal self-discharge of the battery. <S> An aluminum electrolytic might leak 100nA long term, which is not much compared to the self-discharge of even a button cell. <S> The guaranteed maximum of a typical e-cap of this size is 0.002CV or 400nA (whichever is greater) after 3 minutes. <S> Most parts will beat that significantly. <S> Some SMD parts are not nearly as good. <S> Your second question was whether this safe to make. <S> Generally, yes, however there are almost always exceptions in engineering. <S> If your 3V battery has a large current capacity (perhaps an unprotected 18650 Li cell) and your capacitor is something like a 6.3V tantalum capacitor <S> there is a significant risk of an 'ignition' event upon connecting the capacitor to the battery (picture flames shooting out, a bright light and some noxious fumes). <S> The risk can be considerably reduced by adding some series resistance of some tens of ohms. <A> An ideal capacitor would be open circuit to DC, so no current would flow, and no energy would be consumed after the capacitor is fully charged. <A> You should check something called "insulation resistance" I quote from Murata: The insulation resistance of a monolithic ceramic capacitor represents the ratio between the applied voltage and the leakage current after a set time (ex. 60 seconds) while applying DC voltage without ripple between the capacitor terminals. <S> While the theoretical value of a capacitor's insulation resistance is infinite, since there is less current flow between insulated electrodes of an actual capacitor, the actual resistance value is finite. <S> This resistance value is called "insulation resistance" and denoted with units such as Meg Ohms [MΩ] and Ohm Farads [ΩF]. <S> I checked a datasheet I had (part number: GRM32ER71H106KA12 ) for an example to approximate how much leakage gets to pass. <S> Check the image below: To fully understand the behavior of the capacitor at steady state (as in directly connecting a capacitor to a battery) <S> I highly recommend reading the following article: http://www.murata.com/support/faqs/products/capacitor/mlcc/char/0003 <A> If the polarity of the battery is reversed in this scienario, then even an ideal capacitor will consume current to change it's polarity in tune with the battery. <S> But in this case only a real capacitor will be able to consume energy due to springing effect i.e leakage of charge from the edges of the capacitor. <S> However it'll depend on the type of capacitor and the material used in making the capacitor.
| However, real capacitors do have some small leakage current, so, in Real Life, energy would be consumed from the battery very slowly after the initial charging.
|
Using a Micrel MIC2940A-3.3 LDO regulator without additional circuitry? I'm designing a basic table-top toy based around a dsPIC33 MCU. It'll be battery powered and I'm using a Micrel MIC2940A-3.3 regulator to produce the 3.3v for the chip. (I've fitted a PP3 clip, with the intention that my device can run from a 9v battery or 3x or 4x AA batteries in a cage with a PP3-clip.) I'm currently using it in the basic configuration with a couple of capacitors to make the MCU happy. (The typical applications from the datasheet just give it used like this.) That works, but I've read elsewhere that the 3-terminal regulator can be used as a voltage reference in a feedback loop to a power transistor, so that the power current comes from the supply rather than the regulator. Is it common practise to use these regulators bare in a simple application? And is the more complex configuration only necessary when a large amount of power is drawn? (And would using it as a reference be more efficient/stable than the regulator by itself?) Edit - More info: With the MCU running in its 40 MIPS mode and LEDs and other outputs all on this 3.3v output, it comes to 85mA max draw. <Q> Your circuit draws 85mA maximum, so the maximum power dissipation is 490mW. <S> That is acceptable for a TO-220 package with no heatsink. <S> However, you could save your users quite a bit of money by using something like 3-4 AA cells and an LDO. <S> There is also more choice of LDOs if the input voltage is lower, and some have low enough quiescent current that you could avoid a hard on/off switch. <S> Suggest <S> you calculate the cost per hour of operation each way. <S> By the way, the capacitors are not just to keep the MCU happy- <S> this LDO, like most, will oscillate if you don't provide the appropriate capacitance using the appropriate type of capacitors on the output. <S> Not paying attention to this stuff is a shortcut to a lot of headaches: External Capacitors <S> A 10μF (or greater) capacitor is required between the MIC2940A output and ground to prevent oscillations due to instability. <S> Most types of tantalum or aluminum electrolytics will be adequate; film types will work, but are costly and therefore not recommended. <S> Many aluminum electrolytics have electrolytes that freeze at <S> about –30°C, so solid tantalums are recommended for operation below –25°C. <S> The important parameters of the capacitor are an effective series resistance of about 5Ω or less and a resonant frequency above 500kHz. <S> The value of this capacitor may be increased without limit. <S> At lower values of output current, less output capacitance is required for output stability. <S> The capacitor can be reduced to 3.3μF for current below 100mA or 2.2μF for currents below 10mA. Adjusting <S> the MIC2941A <S> to voltages below 5V runs the error amplifier at lower gains so that more output capacitance is needed. <S> For the worst-case situation of a 1.25A load at 1.23V output <S> (Output shorted to a 22μF (or greater) <S> capacitor should be used. <S> They also recommend at least a 0.22uF capacitor at the input and you only have 100nF. <S> I suggest a 100uF/10V electrolytic in parallel with at least 100nF. Batteries greatly increase in internal resistance as they deplete. <A> Do the math: How much current does your device draw, and what is the voltage drop across this regulator? <S> This determines how much power the regulator must dissipate. <S> Does this fall within the ratings in its datasheet? <A> You say the regulator gets 9 V in and produces 3.3 V at 85 mA out. <S> The voltage it drops, 9.0 V - 3.3 V = 5.7 V, times <S> the 85 mA is what it will burn up in heat. <S> (5.7 V)(85 mA <S> ) = 485 mW, nearly half a Watt. <S> That's more than a SOT-23 or probably a SOT-89 package can handle. <S> A TO-3 would be OK in free air, but would get noticeably warm. <S> Since this is a battery powered device, presumably power drain is important because it directly effects battery life. <S> Using a 3.3 V linear regulator powered from a 9 V battery is quite wasteful, as shown above. <S> And, the typical 9 V batteries with the two clips on one end have low energy density to begin with. <S> (3.3 V)/(9 V) <S> = 37%, which is how efficiently the battery energy is getting delivered to the where it's used. <S> Even a simple buck regulator can do much better. <S> Such chips are available off the shelf. <S> At 85% efficiency, you get over twice the battery life for doing the same thing.
| Or you could use a switching regulator, but the cost per kWh of 9V batteries is higher than that from AA cells, so it would be even better to use a SMPS from AA cells. Depending on how much power you need, one that does PWM but can fall back to PFM mode at low output currents could be useful.
|
What is the "screen of the cable" for medium/high voltage cables? I heard that for medium and high voltage cable, there is something called "screen of cable". What does this "screen" do in the cable? Why is it part of the cable construction? <Q> In high voltage cables, especially underground, the cable is shielded, or screened, with an earth conductor. <S> If a digger, or something similar, should cut through the cable it will generally connect the conductive inner core directly with the protective screen around the outside causing the fuse to blow at the substation (or whatever feeds the power) rather than travelling up through the digger and killing the operator. <S> In those cables, from the inside out, you have the power conductor(s) <S> (aluminium), insulation (white), plastic sheath (black), copper shielding (orange), more plastic sheath (black), armouring (steel wires), and finally the outer plastic sheath (black). <A> To supplement Majenko's answer: <S> This is a labelled diagram of all the layers in a three-core, steel wire armoured cable. <S> The diagram is from Olex Australia's HV catalogue. <A> Ref Wiki : A shielded cable or screened cable is an electrical cable of one or more insulated conductors enclosed by a common conductive layer. <S> ... <S> The shield minimizes capacitively coupled noise from other electrical sources. <S> It also minimizes EMI from a fault current as well as minimizes induction from major class X solar flares that create <S> magnetosphere wobble and may create havoc on the grid ( and may damage satellites and other communication electronics.) <S> In addition to this , tri-phase grid transmission lines have phase wires rotated every km to also reduce the same CM noise from lightning and magentosphere wobble currents. <S> This is share equal distance to earth and is similar to the benefits of twisted pairs. <S> There are many variations and other benefits of HVAC shielding.
| It also provides mechanical strength and some moisture protection to the dielectric insulation from Partial Discharge and transient faults from spring thaw in buried cable due to tiny amounts of moisture absorption in plastic sheath.
|
What will cause if I put a larger inductor to a step down converter like MP1584 or MP2307? For example my calculation for MP1584 inductor is 10uH, but i could put only 47uH to it. It is a problem?I'm doing this wrong? <Q> Let me get this straight. <S> The datasheet gives you a formula for inductor value. <S> You crank thru this, and discover that the inductor should be 10 µH. <S> Then you decide <S> "Eh, what do they know, I'll ignore all that and use something off by 5x from what it says." . <S> Now you're seriously asking if you're doing something wrong!? <S> Using a component nearly 5x off from what the datasheet specifes is a bad idea. <S> This really should be obvious. <S> Inductors have 10% or 20% tolerance. <S> That kind of error is most likely built into the formula and the compensation network. <S> But 5x is way beyond reasonable error. <S> Some datasheets tell you the tradeoffs with different inductor values, but most don't. <S> You don't know what considerations went into designing the compensation network and guaranteeing stability while living up to the ripple and transient response specs. <A> For a given switching frequency, when voltage is applied to the inductor during the "on part" of the duty cycle, current will ramp up at so many amps per second. <S> This rate (di/dt) is determined by the voltage applied across the inductor and the value of the inductor, given by the formula V = <S> L di/dt. <S> So, if inductance is too large, the current rate of rise is too shallow and the final current at the end of the "on part" of the duty cycle won't be enough to impart sufficient energy into the inductor (energy = \$LI^2/2\$). <S> The energy needed is determined by the load and the switching frequency <S> so, if the load requires more energy per switching cycle than what the inductor has stored, it won't be enough to keep the output voltage in regulation. <A> The inductor and capacitor on the output side form an LC lowpass filter that smooths out the voltage and current through the load. <S> For these, \$F_{cutoff} = <S> \frac{1}{2\pi\sqrt{LC}}\$ and <S> \$Z = \sqrt{\frac{L}{C}}\$ <S> So increasing the inductor value will lower the cutoff frequency and increase the characteristic impedance. <S> The main effect you will see is decreased efficiency and worse behaviour when current demands change suddenly. <S> I recommend load step testing your converter.
| If the datasheet doesn't give any guidance about different inductor values, then you have to assume the values they specify are requirements within reasonable inductor tolerances.
|
Circuit to detect presence or absense of current I'm trying to come up with a bad-bulb detector for a circuit that will control the turn signals and brake lights in a car or truck. I'm defining "bad bulb" here as an open circuit, which is what typically happens when a light dies in a vehicle. The allowable current range per output is pretty large: 100mA to 10A. This will allow the installer to use a single LED bulb on each output or several old-fashioned incandescents in parallel. In order for the MCU to detect a bad bulb, it only needs to know that current flowed or didn't flow, not necessarily how much. Ideally, the circuit will have an extremely low part count and very inexpensive components (but still work in an automotive environment). This is mostly because the circuit will need to be duplicated several times on the PCB, so any excess will grow the size of the PCB and the budget. And the budget for this project is extremely small. Here's one idea I've come up with so far using only passive components: simulate this circuit – Schematic created using CircuitLab When M1 is turned on, C1 charges up and power is available to the bulb. When M1 is turned off, one of two things may happen: If the bulb was good, C1 will supply additional current to the bulb for a very short amount of time. NODE1 will very quickly drop to zero volts, which will cause NODE2 to also drop off quickly. If the GPIO pin samples NODE2 a few milliseconds after M1 was turned off, it should see a digital low signal. A low signal indicates the bulb is good. If the bulb was bad, the only load to drain C1 will be the resistor divider R1 and R2. But those resistors are large, so the voltage across C1 will stay higher longer. If GPIO samples NODE2 a few milliseconds after M1 is turned off, it should see a digital high. A high signal indicates the bulb is bad. Everything outside the dotted box is already committed in the schematic, so I'm just adding two resistors, two capacitors, and two diodes per output (D1 and D2 are just to protect the MCU from load dumps). These parts might add up to 25 cents per output, which is great. This is an adequate solution, albeit untested outside of simulation so far. I'm hoping to get some ideas for clever ways to do this using even less components or for making it more robust. I would also appreciate critiques on the circuit I've presented above. <Q> You might want to consider an 'old school' method - using a reed switch with several turns of wire wrapped around it. <S> When current flows through the wire the magnetic field closes the switch. <A> You will need current sense resistor you will put between bulb and M1 MOSFET. <S> Then you will monitor voltage across this sense resistor. <S> If M1 is on and this voltage is lower or higher than some threshold then that bulb is faulty. <A> M1 the P channel Drive mosfet does have on resistance so you already have a sense resistor <S> .It will be "Accurate " to about 30% which for this job is fine .This <S> is not a new method .Some <S> texts refer to this as channel resistance sensing .You need a highside comparetor which is not rare or expensive .I <S> have implemented such a comparetor with discrete components 2 decades ago for an indicator lamp application where accuracy was not important .This <S> scheme has the bonus of being able to add overcurrent protection for M1 which the attractively simple reed switch does not .
| The switches do come as spdt types or you can reverse the 'logic' with a simple transistor inverter or logic gate.
|
How can aluminium cables be lengthened for automotive environment? My company's business is stretching and armouring cars. In the former process we must also lengthen all the wires. We've just come across aluminium cables, which pose a problem — until now we used to cut and solder the cables together with the extension. The sections we have to deal with are (in mm²): 0.5 0.75 1 1.5 2.5 I'd like to know what [industrial] options are available to stretch aluminium cables. There is one possibility that is out of question: replacing entire cables with longer ones. We don't have all the connectors since not every single of them is available, some are proprietary and brand-specific. So we're left with actually "stretch" the cables. So what are the available options? <Q> You can solder aluminium, or at least certain grades. <S> I can solder to cooking grade aluminium foil for instance. <S> The trick is to buy a specialist multicore solder, with a particularly nasty aggressive flux in it. <S> It needs some practice to get the solder to take, especially getting the temperature right. <S> I don't know whether it is still available, I bought mine 3 decades ago, and still have some left! <S> One alternative is screw clamps, another is crimp. <S> I'm not sure either would be too robust against possible corrosion of the mating surfaces, unless scrupulously encapsulated after joining. <S> Both would be larger profile than soldering. <A> This is another solution: Tulsonic <A> The problem with aluminum wire is that the conductor "cold-flows" when crimped with standard crimp splices. <S> The most reliable splice that I use is a standard electrical Marrette splice cap. <S> This has a conical spring inside that tightly grips the conductors. <S> These are reliable for decades. <S> Although I regularly solder aluminum using specialty solder and Flux, it just doesn't seem to be as reliable as a Marrette splice cap. <S> The downside of Marrette splice caps is that they are physically large.
| I find it easiest to 'tin' all the surfaces, then use conventional solder to complete the joint.
|
A threshold version of a CMOS 4050? I want a buffer that outputs Vcc (of 5 V) when I give it voltage about a threshold, say 1.5 V. And it outputs Vss (ground) below that threshold. I have a 4050 CMOS thing but I think it does not have a threshold - it just outputs whatever it sees on the input. Bonus points if it's pin-compatible with a 4050. Many thanksRich --- edit --- I'm not after a specific threshold, it's TTL levels. If it looks like an 'ON', output an 'ON'. Else output 'OFF'. Sorry, could have been clearer.Rich <Q> You should use a comparator for this (or a single rail op-amp). <S> Most digital logic gates require more than 30% of VCC to trigger reliably and not burn more static power. <S> You could check some datasheets, if you wanted. <S> The key term you are looking for is \$ V_{IH} \$ (for input-high). <S> This represents the lowest possible input which will register as a clean '1', or high logic. <S> For most 4000 series IC's, this value is ~4V. <S> The \$ V_{IL} \$ is likewise the highest input voltage which will register as a clean '0'. <S> For CMOS 4000 series, this is ~1V. <S> The difference between these values is ~3V. <S> Normally, higher is better, because this means you would require more noise to cause an output error. <S> The situation you described, with \$V_{IL} = 1V\$ and \$V_{IH} = <S> 1.5V\$ only leaves 0.5V for noise. <S> So for general purpose gates, most will not have these characteristics. <S> 4000 series datasheet <S> But, you can use a comparator (or just plain old single rail op-amp) with the V- = 1.5V and the V+ as your input. <S> You can give the comparator IC supply and ground of +5V and 0V, so it will produce your 0-5V output signal. <S> (Like LM293) <S> Edit : As pointed out by Peter Bennett, the LM293 has an open collector output. <S> So connect the output to the supply voltage with ~3.3kOhm or so resistor. <S> Most op-amps do not require this pullup resistor; just check the datasheets. <A> Note that the edited version of the question clarifies that it is TTL-compatible levels that the OP is after, so \$V_{IL} <S> = 0.8V <S> \text{and} V_{IH} <S> = 2.0V\$, so that standard TTL output level limits of 0.4V/2.4V yield a minimum of 400mV noise immunity. <A> What you are looking for applies the the 74HCTxx family with T for TTL input thresholds, but the CD4050 was never migrated to HCT. <S> Closest beast is http://www.ti.com/lit/ds/symlink/cd74hct365.pdf . <S> 74HCT365
| The MC74VHCT50A is one possibility, but I don't think the pinout matches the olde CMOS parts, especially stupid ones like the 4050 with Vcc on pin 1.
|
Relay contact sticking while driving a capacitive load I am using a Panasonic relay ( ALQ105 ) rated for a resistive load of 10 amps 5A at 250V to drive LED bulbs. These LEDs have capacitive circuits in their ballasts and hence draw high inrush current when switched ON. Steady state current is quite low (less than an amp at 220 Vac). This inrush current (and possible spark during switching) is causing contact welding in the relay. I cut open one and found the contacts stuck together. A gentle prying with my finger nails separated them together. I couldn't find any black deposits (which I suspected would be there due to sparks). The value of capacitance in the circuit is unknown and different in different cases. As such I am looking for a general method to get rid of this issue. I could think of these three possible ways: 1) Use a series resistor and possibly an inductor to limit the high inrush current - Possible negative effects would be steady state power dissipation in resistor. Also, I am not sure about what values of R and L should I choose. 2) Use NTC thermistor like this - http://www.cantherm.com/media/productPDF/MF72_JUNE_2016_1.pdf This looks a better idea as compared to L-R circuit but I am not fully sure of the calculations that I need to do for selecting the correct part. 3) Switching to solid state devices like triacs. Since there won't be any mechanical contacts, there won't be any welding. I am currently using BTB-16-800-BW triac in other applications. ( ST BTB16 Triac datasheet ) This has a steady current carrying capacity of 16A which is much more than what my requirement is. It also has a surge current limit of 160 A. However I am not sure whether 160 A is good enough for the load I am dealing with. Please help me selecting the best feasible solution for this issue. [Updated relay contact rating to correct value of 5A at 250VAC] <Q> You might want to use an inductor after all. <S> Besides limiting the inrush current, it will improve your circuit's power factor which right now should be reduced by the capacitive nature of the load, at least judging by your description. <S> Of course, you'll need to find out how much of capacitance you have to counter to calculate the right inductor value. <S> You'll have to either obtain that information from the datasheet or measure it. <S> If you can't, just pick a value and try it out (I'd expect something in units or tens of mH to work). <A> You might want to also consider the triac and then turning it on at the zero crossing either using either a specialized trigger circuit, or monitoring the line with a microcontroller and sending the turn on signal near the zero crossing. <S> The reason this will help is because the turn on is at a very low voltage in the AC cycle and then the current is naturally limited as the AC cycle ramps up. <S> Anyway, just another option that may work for you. -Vince <A> You could try use a soft-start module to control the inruch current. <S> These are common in large (audio) power amplifiers with linear (transformer) power supplies. <S> A few examples here . <A> If the load is actually a fully discharged capacitor, its initial inrush current will certainly be very high and well above the capacity of the relay contacts. <S> As the other answer suggest, in-rush current limiter is needed. <S> The most easiest of all is the low ohmic high inductance in series with that contact. <S> High inrush current will thus be limited by the inductor because of its inherent property of the device of opposing large changes in current. <S> This is much better than the resistive current limiters with relayed bypass because the latter has power loss and additional switching components, which makes it little complicated than a simple passive inductor.
| Choose the inductor current capacity based on nominal circuit current and calculate the inductance so that the peak current is within the capacity of the relay.
|
How can I calculate current(I) when I don't know voltage drop So everyone knows Ohms law (\$I=\frac{V}{R}\$) . Let's say I have a 9V battery and a LED. Furthermore the LED only needs 1.9v so the voltage drop of my resistor would be: \$9-1.9 = 7.1\mathrm{V}\$ I have a 1k resistor (1000 ohms) so now to calculate current we have to plug it all in... $$I=\frac{7.1}{1000} = 0.0071\mathrm{A}\space\space (7.1\mathrm{mA})$$ But what if I change my mind and I don't want my 1k resistor to drop 7.1V but 7.2V instead, so now it would look like this: $$I=\frac{7.2}{1000} = 0.0072\mathrm{A}\space\space (7.2\mathrm{mA})$$ And now this is the part where I get confused. Because I can't control the voltage drop of my resistor so it can either be 7.2V or 7.1V. But I don't know which one is the right one. I could say I want my 1k resistor to drop only 1v so then again: $$I=\frac{1}{1000} = 0.001\mathrm{A}\space\space (1\mathrm{mA})$$ So and if I don't know what my voltage drop is then I don't know current either. So can someone explain this to me? <Q> Normally, you don't want a fixed voltage for your LED. <S> You want a current. <S> (\$20mA\$ not uncommonly.) <S> So you are focused on the wrong goal. <S> But let's say you really do want to set up some fancy voltage regulator for your \$9V\$ battery. <S> (Instead of doing the sensible thing about setting up the current, instead.) <S> You can try this. <S> Just use a \$100\Omega\$ resistor where the LED goes, at first, and adjust \$R_2\$ until you measure a voltage across the \$100\Omega\$ resistor that you like (as shown, where the (-) and (+) appear in the schematic and where you place your voltmeter for measuring.) <S> Then you can replace it with the LED and see how that goes. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You can play with the potentiometer, \$R_2\$, to then adjust the voltage at your LED. <S> Should only waste an excess of \$2mA\$, regardless of the set voltage. <S> The maximum output voltage will be \$6V\$ with a current compliance of up to \$100mA\$. (\$Q_4\$ may need to be a TO-220 packaged type, just in case.) <S> The problem will be that you will have a very hard time adjusting the potentiometer without having a serious impact on the LED brightness. <S> But it can be a learning experience. <S> Or you could just use a variable voltage regulator IC. <S> But you really should just shoot for setting a current using a simple resistor, using the formula \$R=\frac{9V - 1.9V}{20mA}= 355\Omega\$. <S> Then go get a \$330\Omega\$ or a \$390\Omega\$ resistor. <S> It will work just fine. <S> Then measure the voltage across your LED, too. <S> See where it is. <A> There is nothing you can do except maybe get a variable resistor to control your current Also if you were to drop 7v with the 1k resistor then you would be expecting 7.0 mA <A> But what if I change my mind <S> and I don't want my 1k resistor to drop 7.1V but <S> 7.2V <S> instead If you're using a 9V supply and a 1.9V LED, then any resistor will always have 7.1V across it. <S> Perhaps you're not familiar with the "voltage divider" concept? <S> The nine volts of the power supply will automatically distribute itself across the chain of components connected to it it. <S> If we have a 1.9V LED and a resistor, the resistor will always have 7.1V across it, where 7.1V plus 1.9V adds up to be 9.0V, the same as the power supply voltage. <S> In other words, you don't get to change your mind and have 7.2V instead. <S> The LED gets to decide, not you. <S> (Well, I guess you could instead raise the power supply up to 9.1V instead of 9.0V. <S> Or, instead use a 1.8V LED with your original 9V supply.) <S> Note that all of the above is describing a simplified case. <S> It's using an ideal LED: an ideal LED which always shows 1.9V across its leads, even at zero current. <S> A real LED won't have a fixed voltage, but instead will exhibit an exponential V-I curve like any real-world diode. <S> A real LED might have 1.9V at ten or twenty <S> mA ... <S> but this voltage will be much lower at a microamp, and much higher at 500mA. <S> If you want to really understand everything that's going on, you'll have to knuckle down and study the exponential diode equation of voltage versus current. <S> That, or go and play around with a circuit simulator which uses a full-blown mathematical model of an LED, and <S> not just a simplified approximation where the LED Vd is fixed at 1.9V. is it possible to drop all 9v with the 1k resistor so that the led would get 0v Just replace the 1K with an infinite-ohms resistor! <S> In other words, an open switch. <S> But this won't work for a simplified ideal LED, where the voltage across it remains 1.9V forever, even when the battery is removed. <S> :) <A> You can't "change your mind". <S> Both relationships can be expressed as an equation of the form I=f(V). <S> Thus, there are two equations for the system. <S> The simultaneous solution of both equations tells you where the system will live. <S> From my answer, https://electronics.stackexchange.com/a/151653/11684 , which points to <S> http://i.stack.imgur.com/1cUKU.png <S> , you can see that this can be solved, or at least conceptualized, graphically as the intersection of the two lines. <S> You change the resistor, that changes the slope of the line, and the new intersection point tells you your new equilibrium point. <S> You need a certain voltage drop? <S> Draw a vertical line at that voltage. <S> Connect the point where the vertical line intersects with the diode curve, connect that point to Vcc, and now you know what resistor you need to acheive that from the slope of the resulting line. <S> In other words, you pick your parameter, and the set of simultaneous equations fills in the rest. <A> Find out what current, I , your LED needs from its data sheet. <S> Do the calculationR = <S> (9 - 1.9) <S> / <S> I Use a resistor of value R in series with the LED to provide the correct current.
| There is a linear relationship between voltage and current for a resistor, and there is a non-linear relationship between the two for a diode.
|
What are primary and secondary sides of PCB? I am designing a 2 layer PCB with components on both side of it. SMT components are one one side of PCB, but TH (through hole) components are on both sides. I have to go for mass production of say 300 pieces and for assembly. So, I was reading about component placement and minimum spacing. I need to understand something mentioned in guidelines ( Dave Jones and Spirent ) called primary and secondary sides of PCB and secondary side components needs more spacing. Also some very good source of guidelines for volume production will be very useful. <Q> The primary side is the side which has most of the components, especially the ICs. <S> The most common soldering technique these days is reflow soldering, and it is simply easier to get precision temperature profiles on one side of the board only. <S> The secondary side can also support components, but a) it's harder to control the temperature precisely, and b) gravity works against you as the solder paste begins to melt and the components on the underside of the board want to fall off. <A> the secondary side could also be called the bottom side, or more exactly the solder side, with a through-hole board. <S> The thing is: when designing for manufacture, you usually want to design the board with all through-hole parts on the same side. <S> Because through-hole parts are soldered using a solder wave, and you can't use wave soldering on both sides without risking components from the other side falling (unlike reflow process used for SMT). <S> And that what is said in the document. <S> I'm quoting: "The secondary side has the possibility of going through the wave so has a much greater restriction than the primary side". <S> This means to me that the primary side does not have the possibility to go through the wave. <S> Which means all through-hole parts must be on the same side (the primary one). <S> SMT should be alowed on bith sides, though. <S> You may want to confirm this with the factory, but I'm afraid you may have to redesign your board according to this. <S> There could, however, be exceptions to these rules, for example if you have a single through-hole component on the secondary side (like a connector), in which case it would be manually soldered anyway, rather than going through the whole wave soldering process. <A> Chances are the through-hole components will have to be placed manually on at least one side of the board. <S> If the assembler uses a solder wave one one side, it would be necessary to plug the holes for the parts which are to be placed later- <S> this can be done by printing a mask which is later removed to open the holes out, but for only 300 pieces it maybe easier to place all the through-hole parts manually. <S> Of course if you have just a few parts on the bottom and there are massive connectors with hundreds of pins that can be wave soldered then it may be worth it. <S> Since your SMT parts are only on the 'top' (primary) side of the board, no worries, but if they were on the bottom they would have to be glued on to withstand the wave soldering and it would be fairly hard on them (immersion in molten solder). <S> It's even supposed to be possible to stencil print solder paste for through-hole components, possibly in several swipes to get enough volume of paste into the holes, totally avoiding the solder wave, but that may not be feasible in small quantity. <A> There are really no "guidelines" as a design engineer if you place tracks close together bad things can happen. <S> This is dependent on voltage and other problems that arise such as solder bridges, there are calculations for this. <S> There are also limits on manufactruability, typically 7mil but can be tighter depending on the PCB manufacturer. <S> The thing that will place requirements on PCB's is regulatory specifications. <S> Most countries require a regulatory body such as UL, TUV or CE. <S> There is testing and requirements that have to be met before you can sell your product as well as a 16k$ to 8k$ fee for testing. <S> One wishing to go through the regulatory process is best served by finding a consultant or minimum buying the specifications that apply to their product. <S> As far as High speed goes, it depends on what kind of transmission line you want, which will be determined by the termination in the IC. <S> If you want a 50Ω transmission line on your board, then you will need to find the PCB material and thickness, then you calculate the trace width above the ground plane. <S> I'd get a book and start studying.
| Given the references to "primary" and "secondary" in the document you linked, it appears that: the primary side could also be called the top side, and corresponds more exactly to the component side when doing a through-hole board with all components on the same side.
|
Drive LED with 220 VAC I want to light up an LED with 220 VAC using least components. These two circuits come to mind: R1, R2 will be around 200K - 300K depending upon the brightness that I will need. Not much brightness is required so I can even go higher if I am able to get some amount of light out of the LED. Which one (if any) will work? (I am not much concerned with efficiency as these will be used as indicators when a high power device is turned on. For ex - a geyser. This circuit will be wired in parallel with the geyser to accomplish that. If I use 200K resistor, I'll be using up around 0.25 watts which will be negligible as compared to 1000 watts being consumed by the main device.) <Q> Instead of using a resistor to drop most of the voltage, you can use a properly rated capacitor. <S> If you use this in a device with a mains plug, add a 1 Mohm resistor in parallel with the capacitor so that it is discharged after use ! <S> You could use a high efficiency LED <S> so it needs less current, then you can lower the value of the capacitor to 100 nF for example. <A> Both are bad efficiency-wise <S> but the second one is playing with fire. <S> If D12 has a leakage current which is comparable or higher than D11, D11 will drop half of the mains voltage or more and probably will get damaged. <S> If you really insist on using the second circuit (as it has somewhat better efficiency and lets you use a smaller resistor), put both diodes in it: simulate this circuit – <S> Schematic created using CircuitLab EDIT: <S> Now that you got me thinking, another circuit comes to mind if using a small capacitor is OK for you: <S> simulate this circuit <S> This one adds some capacitive load to your mains, but doesn't dissipate any significant power itself, so technically it is more energy-efficient. <S> Note that FakeMoustache has a better version of this circuit in his answer , mine is more of a concept. <S> That resistor he has is not needed in continuous operation, but it protects the circuit from inrush current at startup. <A> Expect something like your #2 to be inside. <S> Another is to buy a special bicolor LED with both LEDs the same color and just use a series resistor. <S> They're not that easy to find, so I would tend to avoid that option. <S> All the half-wave options look flickery, even with 60Hz but much worse with 50Hz power- like cheap Xmas LED sets. <S> The minimum component count that is very safe and is full wave is this: simulate this circuit – <S> Schematic created using CircuitLab <S> The bridge has very loose requirements- <S> the lowest voltage and lowest current you can buy will likely be okay - for example, the < 10 cent DA4X series- <S> it only sees a few volts reverse voltage and the current is limited by R1 to a few mA typically. <S> R1 should be a flameproof type, but it's less likely to ignite than the capacitor dropper circuit where the cap could go shorted, and the LED should have a lens or other additional insulation.
| One possibility is to simply buy a panel mount indicator LED designed for 230VAC input, which will come with a plastic lens wires, and a slew of safety approval markings which can help with approvals of your equipment (since individual LEDs will require additional insulation for safety).
|
How can I decrease the brightness of these BA-15 LED bulbs I'm looking at these LED replacement bulbs for the interior of my sweet conversion van. I'm looking for a more efficient alternative to the BA15 bulbs that are in there already. These bulbs appear to be really bright, and I can't seem to find a lower lumen LED light in warm white. I'm wondering if it is possible to unsolder or intentionally damage some of the LED panels to decrease the brightness of the overall bulb, or if that will disrupt the entire bulb. <Q> You could wire a resistor in series with the base. <S> The lamps are claimed to be 2W so they should draw around 170mA from the 12V bus. <S> Try about 10-50 ohms 1W to get the brightness you want. <S> Of course you will no longer be able to safely install an incandescent bulb, in fact doing that could cause the resistor to get dangerously hot. <S> As a bonus, the lamps should last much longer, especially in a warm climate. <S> They are typically driven to within a millimeter of their lives because that saves the manufacturer money. <A> These assemblies have 18 LEDs on them, which I guess are connected in series of 3. <S> All series are in turn connected in parallel. <S> Desoldering 1 LED will disable the whole 3-LED block. <S> Try desoldering one of the LEDs and test to confirm. <A> What you likely have here are 6 strings of 3 LEDs connected in parallel. <S> If you disconnect one set of 3 that should work. <S> If there were an LED driver then the current would be diverted to the other 5 sets. <S> > <S> At this price point I am doubting there is a driver. <S> Each panel is likely independent of the others <S> Remove one LED from one of the panels. <S> If it dies then dissemble it to see what's under the hood.
| You could add a small resettable fuse (polyfuse) to prevent that situation. I'm willing to bet only that one panel goes out and everything else will be fine.
|
How to have multiple devices getting measurement from a single thermocouple I have a system with redundancy on the control unit using thermocouples.My question is : is it possible to get from different devices the measurement from a single thermocouple ? I am thinking about sending the signal after the cold junction is applied into 2 different ADC, but as the thermocouples involves very little voltage, I'm not sure if this would screw up my measurement. <Q> Thermocouple voltages are so low it's unlikely you'd connect one directly to an ADC in practice. <S> Instead there's going to be some kind of buffer amplifier between the thermocouple and the ADC. <S> Since the buffer amplifier has a low output impedance, connecting multiple ADCs in parallel shouldn't present much of a problem. <A> Yes, there is no problem at all- except the devices must accept a T/C input that has some arbitrary common mode voltage on it. <S> You are really getting into diminishing returns looking for redundancy after the CJC. <S> I would suggest your #1 choice is two completely independent sensors and signal conditioning circuitry, followed by two completely galvanically isolated front ends. <S> There will be a small effect from the burnout current from each input, but you can design the front ends so that any likely failure (for example, put two resistors in series) will not unduly shunt the sensor and an open will be detected as an unsafe condition. <S> The most likely failure mode for a properly installed and functioning thermocouple system would be the wiring (and any connectors) to the sensor, possibly the sensor itself if abuse is possible, followed by damage due to lightning or EMP type events, which would likely affect the front end itself. <A> Thermocouples are cheap; install several. <S> There are a LOT of different devices that accept thermocouple input, and some of them ground the (-) lead. <S> Some ground the (+) lead. <S> Someapply a test current through the wiring periodically, to checkfor continuity. <S> Lacking complete knowledge of the controllers, one safe wayto mount a single thermocouple, but give all the controllers a separatethermocouple circuit: <S> you could make a heat-bath, with N thermocouplehot junctions and one cold junction. <S> Each of your N controllersconnects to one of the hot junctions, your 'input' thermocoupleconnects to the cold junction, and a heat-bath-control thermostatheats the bath until the 'input' thermocouple circuitreads zero temperature difference. <A> If you want redundancy, you need several controllers and several thermoelements. <S> To keep things simple, each controller should be connected to its own thermoelement. <S> If you connect one thermoelement to several controllers, you might get additional noise and errors.
| Thermocouples are very low impedance devices (less than 100 ohms) and one instrument will hardly affect another at all.
|
Refrigeration compressor circuit, help trying to reverse engineer it I have disassembled the motor controlling circuit for my refrigerator.I am trying to understand how it works, so I hopefully reduce the start surge-current and make it work with my 1000w inverter. Currently the fridge sometimes draws 1100w on start, othertimes only 600w. The fridge is brand new, high-efficiency, "BEKO" with a Donper AG100CY1 compressor. I've checked the circuit a hundred times, and it really is like this: simulate this circuit – Schematic created using CircuitLab But it seems to go contrary to what I've read about refrigerator controller circuits. The start winding is permanently connected (normally it would be disconnected after start, e.g. by a PTC) The run winding is current runs via PTC at start, then via capacitor once hot My main surprise is that the surprise winding is permanently connected. (And additionally, I am hoping to add a start capacitor...) Why isn't the start-winding disconnected? Does this make any sense? <Q> Why did you check the circuit "a hundred times"? <S> This is a technical site and "a hundred" means 100. <S> ;^) <S> The 43 Ω winding is the primary one as it is directly across the mains. <S> The 26 Ω winding is the start winding as it has the series capacitor. <S> I can't guess at the inductance <S> but we could do some guesses about the currents. <S> The impedance of the capacitor is given by \$ Z = \frac { <S> 1}{2 <S> \pi f C} \$. <S> At 50 Hz this will be 1061 Ω and at 60 Hz 884 Ω. Worst case on 230 V <S> 50 Hz is 230 mA and on 120 V is 135 mA. <S> These seem a little low to me <S> but maybe they're enough to give rotational torque to the motor to get it started but low enough not to waste significant power. <S> The start winding does, after all, do some useful work even when up to speed. <A> That is a permanent split capacitor (PSC) motor. <A> Motors in consumer products are typically carefully designed to do exactly what they need to do at the minimum manufacturing cost with no safety margins. <S> I believe that refrigerator compressor motors often require less starting current when they are started after not running for some period of time. <S> I have seen a refrigerator with a "minimum off time" circuit. <S> If this refrigerator does not have such a circuit, consider adding that. <S> That might result in the motor requiring only 600 watts every time rather than 600 some times and 1100 other times.
| Any alteration of the circuit is likely to result in the motor not starting reliably.
|
Calculating the dropper / series resistor for an LED I'm so sorry if this is a stupid question. I have read some of the questions previously asked but in afraid the maths and physics are way over my head. I have made a wooden template and would like it to have light up eyes. I am hoping to use 1.5 V AA battery to power them. The eyes will be initially yellow LEDs. I don't want to fry them so if I wanted to power two separate 'eyes' which I believe are 2.1 V each, what resistor would I need to add to my circuit please? In the future, I may want to use other colours which are different voltages so if anyone can explain how to work this out in dummy terminology I would be incredibly grateful. Also, does it make any difference to the equation if you change the amount of LEDs being powered? <Q> First of all, AAs are 1.5V, not 5V. <S> The equation is simple: V = I * R, where V is voltage, I is current, and R is resistance. <S> This is the amount of voltage that your resistor must use up. <S> Then determine how much current you want to flow. <S> Probably between 0.001 and 0.020 amps (i.e., 1-20 milliamps, but the equations are for amps). <S> Now, just plug the numbers into Ohm's Law and go! <S> So, if you have a 9V battery, an LED that eats 2.1V, and you wanted 1 milliamp (0.001 amp), you would do: V = 9 - 2.1 = 6.9. <S> Ohm's Law: <S> V = <S> I <S> * R <S> 6.9 = 0.001 <S> * R R = 6.9 / 0.001 R = 6900. <S> If you have two LEDs (let's say they are both 2.1, and are in series with each other), then you have to subtract them both. <S> So V = 9 - 2.1 - 2.1 = 4.8. <S> Now do Ohm's Law again: <S> V = <S> I <S> * R <S> 4.8 = <S> 0.001 <S> * R R = <S> 4.8 / 0.001 <S> R = 4800 <S> If you need more current, just change out the current in the equations. <A> First off, you'll have trouble finding a 5V AA battery, but you could put four AA's in series to get 6V. <S> When a diode (including an LED) turns on, there is a voltage drop across it. <S> Let's say that you start with your 5V supply, a resistor, and your LED. <S> When the LED is on, its voltage drop - AKA forward voltage (from the datasheet) could be something like 2.6V, with the positive side at the anode. <S> The voltage drop across your resistor, added with your forward voltage is equal to the supply: V_supply = <S> V_r + V_LED <S> therefore, V_r = 5V - 2.6V <S> = 2.4V <S> Using Ohm's Law: <S> V=IR, you can find out what current you want to pump through that LED. <S> That will allow you to select your resistance. <S> Say you want to pump no more than 12mA through that LED, because the datasheet says that over 20mA, and you'll fry it. <S> R = <S> V <S> /I = 2.4V/(0.012A <S> ) = 200 Ohms. <S> simulate this circuit – <S> Schematic created using <S> CircuitLab <S> I hope that helps! <A> For sure, you need more than 1.5V to do the job; easiest is 3 series AA for 4.5V. <S> As Johnnyb said, then you have to lose 2.4V in the resistor; so that's (4.5V-2.1V)/10mA = <S> 240 Ohm. <S> My suggestion would be to use separate parallel branches for the two LEDs; battery life will be much longer with AA batteries than with 9V. <S> The value of the resistor is really not that critical; anything in the several hundred ohm range should be good; could be an interesting experiment for you.
| Start with your battery voltage and subtract the voltage drop of the LED itself.
|
Can I reduce the start current of a fridge compressor? Not a duplicate question: this question is about reducing start current , not about identifying the motor type . I have a refrigerator that has a permanent split capacitor (PSC) motor, which I am powering from solar panels/battery/inverter. My inverter is rated at 1000w (2000w peak), but unfortunately, the motor occationally trips the overload protection of the inverter, with the risk of spoiling the contents of the fridge if it goes unnoticed. I am considering replacing the motor with a DC compressor (Danfoss BD35F), but it's quite expensive (about $350). Before I do that, I want to experiment first, to see if I can get it to work with a consistently lower start current. I realize it might break the motor, that it might not start consistently, but it's worth a try. I've read about the following options: a) use a start capacitor, 150uF, for the first 1000ms b) use an NTCthermistor c) power resistors, for the first 1000ms (similar effectas NTC) d) voltage transformation to a lower voltage, e.g. from 230Vto 200V but: a) I'm having a hard time finding a start capacitor to try with, and random experimentation could be expensive, since they seem to cost about $10-15 a piece and can break if powered for more than a few seconds. b) I've tried a 10ohm NTC, but that made no difference at all. c) I'm going to try with various resistors at 50, 70, 100 ohms (rated at 50w). d) I'm not sure which type, which voltage to try etc.. ALSO: I tested the motor with the following results: Disconnected for 24h, ~850w Connected for 20min, disconnected for 5min, ~600w Connected for 45min which included a self-restart, ~1100w Disconnected, then reconnected after 15min, ~800w I'm puzzled by these results, because it seems that the motor draws more if left connected. But the thermostat disconnects the circuit.The only way this makes sense to me, is if the thermostat somehow causes an arc that draws an additional 300w when reconnecting.But I tested it by disconnecting for 15min, turning the thermostat to 0, reconnecting, then turning the thermostat to 3. The max power draw was only 782w. Very strange... Any ideas? <Q> I have a similar situation, off grid PV / battery storage system with a 1Kva 12-volt inverter <S> and I run a small 230-volt standard fridge. <S> It used to knock out my TV & Satellite at start-up and hit the battery bank voltage quite hard. <S> I connected a VDR (voltage dependant resistor) in series with the live feed. <S> This just takes the sting out of the start-up inrush current and overcame the problem for me <A> I solved this with a 2 farad 16v super capacitor on the input of the inverter> <S> Although it doesnt reduce the start current it provides enough electrons to start the compressor without shutting down the inverter <A> Clearly, if your inverter cannot provide enough power for the fridge to start, the proper solution is to get a more powerful inverter. <A> Reducing startup power is not needed for on grid use. <S> The measures would only increase the cost of the fridge. <S> For big compressors this is done by lifting the suction valve(s). <S> Sometimes a softstarter is also included. <S> A better way in your situation would be to change to a socalled absorption type of fridge. <S> They work with a small heating element and have no inrush current at all. <S> It might be possible to work even without an inverter if you select the right type (12 or 24 VDC also needed for the Danfoss compressor). <A> First make sure that your inverter is performing as advertised. <S> Some brands, such as the newer AIMS, have the surge time so short (40ms) <S> that it is useless for pretty much everything. <S> The second is to make sure that the battery Voltage isn't going low enough to shut down the inverter during start up. <S> Now the answer: Modern energy efficient motors seem to have very low resistance motor coils, which makes the start up surge much larger. <S> These low resistance permanent split capacitor compressors seem to be able to restart immediately while the refrigerant is still pressurized, which is something that always used to lock up the compressor. <S> A 1 Amp running compressor may draw 10 to 15 Amps with a locked rotor. <S> I haven't tried it, but I would recommend the resistor in series with the run winding with a 1 second timer relay, which shorts out the resistors after 1 second. <S> You shouldn't use a capacitor as it would change the phase and defeat the effect of the permanent run capacitor. <S> If you hook up both windings (by hooking up to the whole compressor supply wire) you should be able to use either a resistor or a capacitor to limit current while the 1 second relay is open. <S> Also, if your inverter puts out a 2 step square wave or it puts out a 3 step square wave with very little dead time I don't recommend using it with a high efficiency motor. <S> It will make the motor run hot and waste a lot of electricity. <S> A 3 step square wave with 25% dead time is okay, but with most inverters, especially cheaper ones, as the battery Voltage drops and other loads are turned on that 25% gets reduced to increase the AC Voltage. <S> Such inverters have no other way to regulate the output Voltage. <A> I think your best bet would be one of those power factor correction capacitors. <S> A large motor run cap might be a suitable alternative. <S> The only method I have of calculating the required size is from someone else (I haven't tried it) and is really hit and miss. <S> It goes like this: connect the capacitor in parallel with a filament bulb and connect both of these in series to the appliance while it is running. <S> Keep increasing the capacitor size till the bulb goes out. <S> Or you know you could use the saying: the bigger the gob the better the job. <S> Alternatively you could use a larger inverter to start it and use a change over switch to go to the correct size. <S> Finally. <S> Use a DC inverter compressor refrigerator. <S> Obviously you'll have to buy a new one. <A> You need to limit inrush current, it seems. <S> Perhaps do a search on "mains inrush protection". <S> One such item from the search is this <S> This device is useful where the current drawn from the mains supply by a piece of equipment at switch-on (eg. <S> an audio power amplifier) is considerably higher than the steady state current consumption of the equipment. <S> This current surge can be troublesome in that it may blow fuses or trip circuit breakers. <S> The unit has IEC input and output connectors and limits the current at switch-on to an acceptable level for 200mS–400mS. <S> Two LED’s on the front panel indicate the status of the unit. <S> A voltage dependent resistor is included for transient suppression.
| That said, if you know your inverted can provide the extra current for a short time during startup, you may just be able to replace its protection circuit by a slow-blow circuit breaker, which is specially designed for loads like motors and such. To reduce the startup power some form of unloading is required. Soft-starter can be another solution, but only if your compressor supports it (many big compressors do).
|
Can you identify this component? This may be obvious to some, but not to me. I am an IC design engineer, as such I rarely do any practical work. This morning I had to fix some electronics which were custom made by a third party for my company. On removing the front panel I saw this gray box clipped to a bundle of wires (images below), I have no idea what it is or what it does. After 10 minutes of fruitless googling for "gray box clipped to wires" I find myself here asking this question. The inside looks like some type of metal or ceramic. I thought it might be magnetic but it is not. So can anyone identify it and tell me what it does? I fixed the problem, which was not related to the wiring. <Q> Here is a complete datasheet for a similar product. <S> The ferrite is lossy so it is better characterized as an impedance at a given frequency. <S> In the case of my example, every wire that goes through the core behaves like it has 241 ohms in series @ 100MHz but only for common mode voltage. <S> Unbalanced current (say to ground) will behave like it has the impedance in series. <S> It is very important to make sure that all the wire pairs that carry significant current are contained within one core, otherwise the core will saturate and the benefit will be lost. <S> A typical application of such a core would be for wiring to a noisy source such a VFD (motor drive). <S> Looping the wires \$n\$ times through the core will increase the impedance by a factor of \$n^2\$. Again, every wire must be looped the same number of times to get the full benefit. <A> It's a ferrite choke. <S> It is a passive electric component that suppresses high frequency noise in electronic circuits. <A> Rarely you see the snap on type. <S> They are very common in round ring type ferrite cores where the set of wires are looped once. <S> I see them mostly inside all type power supplies.
| It's a snap-on ferrite core which, when clamped over a number of conductors, becomes part of a common-mode choke. If current is going in and out through a pair of conductors the magnetic fields cancel and it acts pretty much like a pair of wires for that current.
|
Alternative method for tying bunch of wires I'm making control panels on which some heavy duty PCBs are mounted, with many wires (approx 1 meter long) coming out of it. Wires are currently tied together using zip ties at every regular distance. It is time consuming, so does anyone have any alternative? Note that I don't want to increase the cost, so I prefer a cheaper alternative if possible. <Q> However, look at DigiKey or other electrical/electronic supplier for "wire management" or "cable management". <S> Some possible products: expandable tubing, split tubing, nylon spiral wrap. <A> The old-school method of lacing with cord is still possible, and pretty fast once you get used to it. <S> They are available with vibratory feeders to feed cheap loose cable ties <S> so you don't have to buy special types of ties. <A> Can you use this type of wire harness tubing: <S> It can be obtained in many diameters and several colors are available. <S> This tubing can have electrical tape wrapped around it at points where bending stress would tend to pop the slit of the tubing open. <S> Note: In the above picture something has been placed in the slit on the far side to hold it open. <S> Normally the slit sits closed up. <S> This tubing is made as a pleated tubing and then slit afterwards. <A> $28 budget version or spend up to $1k for power assist and spool feeder <A> Velcro is cheap. <S> Even cheaper if you use the green ones used for planting gardens. <S> Roll of 50ft for 2 bucks vs the computer kind 7 dollars. <S> Holds then just as good as if t he computer cables were biological.
| More likely, you might want to invest in a fully automatic cable tie application machine. I think zipties are quick, easy and flexible - you can take wires out of the bundle at any location.
|
How to find stall current of servo motor? I am using 18 of this hobby servo s for a hexapod, that are going to run simultaneously. Now I need to decide the power source, and need to determine the current needed. I know that the maximum current that a motor draws is its stall current, and it should run with lower current than the stall current. But no datasheet mentions the stall current. All of them mentions the stall torque, 1.6kg-cm(at 4.8V) in this case. And I cannot find any equation that calculates the stall current from stall torque and applied voltage. Some people suggested to actually stall the servo and find the current with an ammeter, but I'd rather find it by calculation. So how can I find the stall current? <Q> V <S> = IR <S> To the first order <S> , stall current = <S> Applied drive voltage divided by the DC resistance of the coil. <S> The stall current occurs when the motor is stalled (not moving), so there is no reactance and no "back electromotive force (EMF)". <S> It's literally only the resistance of the wiring in the servo motor that is resisting the flow of current... <S> just like a resistor in a DC circuit. <A> I cannot find any equation that calculates the stall current from stall torque and applied voltage. <S> There is an equation that calculates motor current from torque and voltage, but you also need the motor's torque constant (Kt) or velocity constant (Kv). <S> Kv can be roughly derived from the servo's operating speed at a particular voltage, so you might think that the published servo specs are enough. <S> Problem is these specs were not intended for calculating current, so their relevance is suspect (eg. <S> torque rating may not be at stall, operating speed includes starting and stopping time) and being cheap Chinese servos their specs probably aren't accurate anyway. <S> Then divide voltage by resistance to get current. <S> However this will probably overestimate the stall current because:- <S> The H bridge transistors have some loss, so the motor won't get the full supply voltage. <S> The servo doesn't power the motor continuously, but with pulses whose width is proportional to the difference between commanded and actual positions. <S> At normal frame rate (50Hz) there is always a gap between motor drive pulses, so even at stall the motor may only be on 70~80% of the time. <S> So if you want to find out how much current the servo <S> really draws, you will just have to measure it. <S> Measuring supply current while stalling the servo at one end of its travel and commanding it to go to other end will give you the worst-case average current at stall, but not the peak current. <S> This test is also very hard on the servo. <S> A better way is to put a current shunt in one supply lead and measure instantaneous current while the servo is operating. <S> The motor in the HXT900 <S> I tested measured 4.1Ω. <S> When the servo was powered with 5V the motor pulses were 3V high (2V drop in the H bridge and wiring!). <S> This equates to 0.73A peak current. <S> Here's a scope trace of supply current measured with a 0.1Ω shunt resistor, while the servo was making rapid small reversing movements. <A> Power=torque x angular velocityPower in wattsTorque in NmA. <S> Velocity in rd/sec <S> That is the mechanical power that the motor is delivering to move an arm of the robot <S> So its not that easy to calculate <S> but if you know the angular velocity from the datasheet you can calculate the power <S> Bow <S> the electrical power consumed by the motor is given by:Power= <S> V x IV: <S> VoltsI: <S> amps Ofcource <S> the mechanical power isn't ecactly equal to the electrical power as some of the electrical power is lost due to friction (especially because of gears) and heat (elextrical resistance of coils inside motor) and the ratio of the used mechanical power over the total consumed power is the effeciency of the motor. <S> Actually there are special expensive equipment needed to find the torque and velocity of the motor.. <S> They use these equipments to model the motor. <S> So it's not as simple as you may have thought Now to find the stall current, the electrical energy supplies at stall is all converted into heat due to the internal resistance of the motor. <S> But you can't actually access the leads of the actual motor because there is a driver inside the servo and that driver is connected to the motor. <S> Theoratically you could have stalled the motor at a low voltage and simply use u=ir to find r. <S> But you can't do that for a serco as it's actually made up of a microcontroller, a driver, a feedback sensor(potentiometer) and a motor. <S> The stall current may not be the actual stall current but maybe a limit imposed by the microcontroller in the servo motor. <S> Summary: stall the motor and measure the current!!
| The easiest safe way to calculate stall current is to take the bottom off the servo and measure the motor's resistance (Rm) with an ohmmeter.
|
hfe classifications in data sheets This is a snippet of the data sheet of the BJT transistor 2SC828A. What do they mean by Q R and S classifications? I know that 2 in the code means transistor. What do the other numbers and letters mean (SC828A)? Link to the data sheet: 2SC828A data sheet <Q> Let's dissect the word 2SC828A ... <S> 2 <S> This is part of the Japanese Industrial Standard where they denote the markings of transistors with 2SC, 2SA and more. <S> SC <S> Again part of the JIS, the "SC" stands for NPN high frequency transistor. <S> 828 <S> So far I am unable to tell a pattern between the 3 numbers and the other characteristics. <S> All I know is that if the number is different from another 2SC transistor, some characteristics would vary. <S> A <S> The A in 2SC828A denotes the voltage rating of the transistor. <S> Generally, higher values like B and C have higher voltage ratings. <S> Like the TIP35C. <S> But wait... <S> There is an extra letter that will show up on your transistor. <S> So your transistor will actually say something like, 2SC828AQ, 2SC828AR or 2SC828AS, being 3 variants of 2SC828A. <S> This explains your question with the Q/R/S part of the hFE shown in your datasheet. <S> The suffix your transistor has determines which hFE gain it will have, as concluded from the datasheet. <S> Q in 2SC828AQ has a hFE of 130-260. <S> R in <S> 2SC828AR <S> has a hFE of 180-360. <S> S in <S> 2SC828AS <S> has a hFE of 260-520. <S> These suffix letters are also Hfe ranks. <S> See here . <S> Source: http://www.transparentsound.com/transistors/vintage-transistors/JIS/jis01.htm <A> [...] BJT transistor 2SC828A <S> What do they mean by Q R and S classifications? <S> These are the different ranges of \$h_{FE}\$ values, as shown in your original question, which each of those transistors could have. <S> Look on the transistor for one of the letters Q, R or S often marked just underneath an abbreviated part number like 828A. <S> If the transistor has no such letter Q, R or S marking (sometimes called a Current Gain Group) <S> then all you know (without measurement) is that its \$h_{FE}\$ value should be within the limits shown on the datasheet. <S> Your linked datasheet shows the \$h_{FE}\$ minimum = 130 and maximum = 520: <S> As expected, those values fit with the lower and upper limits of those classifications/groups. <S> I know that 2 in the code means transistor. <S> What do the other numbers and letters mean (SC828A)? <S> This is the JIS marking standard. <S> In the case of the 2SC828A, they mean: <S> SC = NPN HF bipolar transistor 828 = <S> the JEITA-EDEREC number assigned to this type of transistor originally; the individual digits have no additional meaning <S> A = a later version than the original (i.e. the one without any letter suffix); multiple versions are possible for some devices e.g. B, <S> C suffixes etc. <S> The general rule is that you can safely replace an earlier suffix letter device, with a later one (e.g. you can replace a 2SC828 with a 2SC828A) but not vice versa. <S> (Same thing applies with the 2N2222 and 2N2222A, for example, where the 2N2222A has some improved maximum voltage ratings. <S> However the meaning of a suffix can vary; if in doubt, find better datasheets for that device which explains them.) <S> For this particular transistor , the "A" suffix versions have some improved voltage ratings, as shown in this datasheet which covers both the 2SC828 and 2SC828A. <S> The differences are summarised as: <A> You can look at this link for an explanation of the hFE ranking. <S> http://engineering.rohmsemiconductor.com/index.php?/Knowledgebase/Article/View/1/1/what-is-the-transistor-hfe-ranking <S> In case the link does not work anymore! <S> Here some further information. <S> The ranking code is mentioned as a suffix after the basic partnumber.hFE Ranking CodeshFE RangeA 16 - 32B 25 - 50D 60 - 120E 100 - 200F 160 - 320M 39 - 82N 56 - 120P 82 - 180Q 120 - 270R 180 - 390S 270 - 560E 390 - 820U 560 - 1200V 820 - 1800W 1200 - 2700The complete number would be 2SC828A (suffix R,Q,S)The suffix A indicating the voltage range.
| The "2" like you said means it's a transistor (where "1" means diode).
|
How to know if my voltage regulator is still working or not I have a voltage regulator TPS788 from TI. The output voltage is 3.3v, and the output current is 150mA. I connected a LED to it, and I found that the LED is blinking really fast. Does that mean the Regulator is not working properly, or the LED it self is not standing this amount of current ? Note: I am using another regulator TPS789 (1.8V,100mA), and The LED is working fine. <Q> If you connect a normal LED directly to 3.3 volts, it will try to draw a near-infinite amount of current, due to the LEDs I-V curve. <S> This is from a completely random green LED I found: <S> Follow the horizontal axis to 3.3 volt, and you can see that the current, if possible , would be hundreds of amperes, or more. <S> This is a pointless exercise, because the LED would burn up in flames before that. <S> Instead, what likely happens here, is that your regulator has a few protections: <S> The TPS788xx features internal current limiting and thermal protection. <S> During normal operation, the TPS78833 limits output current to approximately 350 mA. <S> When current limiting engages, the output voltage scales back linearly until the overcurrent condition ends. <S> While current limiting is designed to prevent gross device failure, care should be taken not to exceed the power dissipation ratings of the package. <S> If the temperature of the device exceeds approximately 165 °C, thermal-protection circuitry shuts it down. <S> Once the device has cooled down to below approximately 140 °C, regulator operation resumes . <S> I don't know if it is the current or temperature protection that kicks in here, it depends on the frequency of your "blinking". <S> Make sure it can handle 0.5 watts. <S> Then measure the voltage. <A> Please be aware that led's idealy should be current driven. <S> For that reason a serial resistor is mostly used to limit the current within a save range. <S> Therefore if you have an led with a nominal voltage of 1,5V and a current of 20 mA <S> and you wish to use a 3V supply <S> then you need a series resistor of (3,3 - 1,8)/ 20 = 0,075 kOhm or 75 Ohm. <S> Testing you circuit following this principle will show you directly if the regulator is working properly <A> The TSP78918 is working because the voltage is less than the forward voltage of the LED and therefore not drawing excessive current. <S> The TPS78833 outputs enough voltage to turn on the LED with the regulator being the only current limiter. <S> The value of the current limiting resistor is dependent on how bright you want it to be and how much current is required to make it that bright. <S> Use the desired current according to the LED's datasheet and divide 3.3 by the desired current to get the ohms. <S> If the forward voltage of the LED is 3.0V, then the voltage across the resistor is 0.3. <S> The wattage will be in the milli-watt range, So a very small wattage resistor will suffice.
| If you want to make sure that your regulator is still working, the easiest way is to connect a 22 Ω resistor from the output to ground.
|
Is it mandatory to include a pulse detector in order to design an edge-triggered JK Flip Flop using logic gates? I am trying to design a positive-edge triggered JK-Flip-Flop (using 7400 and 7410 NAND gates) on a breadboard. The circuit diagram is in the picture (I would replace all AND and NOR gates with NAND gates). However, I see that there is a pulse-detector circuit from the clock-pulse input. My question is:1) Is it mandatory to include that pulse-detector circuit for proper working of JK? 2) If yes, then can I create the pulse-detector circuit using NAND gates as shown in the 2nd picture? (As usual, I would replace the NOT and AND gates by NAND) <Q> The four-gate AND+NOR circuit you see is actually a latch with respect to the control input. <S> That is, state can "flow" from J and K to Q if the control signal is always logic high. <S> A solution to this with D latches is to put two latches in series with \$180^{\circ}\$ difference in clock phase in the Master–slave edge-triggered D flip-flop configuration: Public Domain work from the Wikimedia Commons <S> However, since the J-K operation function isn't simple like the D case, some other method of making the flip-flop edge sensitive is needed. <S> A solution to this is to keep the control input high only on the edge of an input: <S> thus the edge/pulse detection circuit. <S> To your second question: yes, the NOT+AND circuit you show implements the desired pulse-creation circuit. <S> However, you need to make sure that the output pulse is wide enough that the J-K circuit stabilizes after a transition in state happens. <S> The timing required will depend on the speed of your circuit. <S> You don't want to partially update the state! <A> No. <S> You can easily make a master-slave jk flip flop. <S> Adding a pulse generator to a latch makes it a "pulsed latch" ( http://ieeexplore.ieee.org/document/5719582/ ) not a flip flop, even though it acts like one. <S> See the schematic below: <A> I'm just at the same problem and find this demonstration perfectly instructive of the issue. <S> Ben Eater: JK flip-flop racing
| Yes , the edge detector/pulse generator is needed to operate the circuit as a J-K flip-flop .
|
Why do we use current mirrors? Sorry if the question seems stupid, im not much into electornics just trying to get into it...I'm trying to figure out the purpose of current mirror, I read everywhere that it copies reference current to another node but why we're doing it? why can't we just use that current source without connecting them to N-channel transistors? also why 2 transistors are used? <Q> Current mirrors take a positive current (source) and generatea negative current (sink). <S> Current mirrors have low input impedance, so the 'reference' current can be from a poor current source (output impedance 1k ohms) and give rise to a mirroredcurrent that is a better current source (output impedance 100k ohms). <S> They isolate the input from variations in the output, so you can wiretogether a dozen current mirror outputs to make a summing junction, withoutcausing any signal flowing back into the input nodes. <S> The illustrated current mirror makes an excellent pull-down load foran NPN emitter follower, sinking equal current at a wide range of outputvoltages (unlike a resistor pulldown). <S> The use of two transistors is not required, of course; you could use dozens, andthe single reference input can drive two parallel output transistors tomake a gain-of-2 mirror, or four to make a gain-of-four mirror (thereare D/ <S> A converters built this way). <S> Matched transistors are very easy to integrate, andare more accurate than resistors in most IC processes. <S> Current mirrors with sense-resistor inputs are used as current monitors high side current sensor as well, with highly attenuated output current. <S> The low input impedance, like an ammeter, is a major benefit. <S> Current mirrors have a wide range of uses. <A> There are many answers to your questions. <S> Regarding why don't we use the current source directly, one of the reasons is that the 'original' current source may not be strong enough to drive the load. <S> Another reason to use current mirrors is to stabilize the current against changes on temperature. <A> Other possible reasons: You only have one source but your circuit needs many copies of the same amount of current <S> You need several copies of the same current to match. <S> MOSFETs from the same die built with proper layout techniques can match very well. <S> In many cases, better than your ability to find multiple sources that match each other. <S> You want the currents to "track" each other. <S> If the master current moves up or down <S> Your current source has a compliance limit that is not useful in your target circuit. <S> You may have to look this term up for more details
| Integrated circuit operational amplifiersoften use a single current reference and current mirror to generate many bias currents. , your circuit will be more robust if all the current "copies" follow the same trend
|
Operating a 60W incandescent bulb I want to dim a 60W incandescent bulb using PWM on a MCU. The bulb is rated at 220-240V 50-60Hz at 60W. I am planning on using a optoisolater connected to the PWM pin. I want to run it on DC, so since the bulb is 60W and P = VI then am I correct in saying that I can get full brightness at say a voltage of 30V and current of 2A ? <Q> A 220 V bulb isn't going to light with <S> 30 V applied. <S> If that's what you want, get a 30 V bulb. <S> Better yet, get a bunch of LEDs and run them from the 30 V at much less current. <A> To add some explanation to the other answers: I want to run it on DC, so since the bulb is <S> 60 W and P = VI then am I correct in saying that I can get full brightness at <S> say a voltage of 30 V and current of 2 A? <S> If you were designing a bulb <S> then you could choose any voltage and work out the current required as in your question. <S> Unfortunately someone has already made that decision for you <S> so two of your three variables were specified: <S> Power: 60 W. Operating voltage: 240 V. <S> That only leaves the current as a dependent and will be given by \$ <S> I = <S> \frac { <S> P}{V} = \frac {60}{240} <S> = <S> 0.25~A <S> \$. <S> 30 V is an odd choice for a lamp. <S> If you were prepared to go for 24 V then you would find a wide selection of truck lamps to suit. <S> A headlamp is typically 55 W. <S> You know enough to work out the current! <S> Watch the cold resistance - as mentioned by pericynthion - as the current will be much higher when the lamp is cold. <S> Measure the resistance <S> so you know what to expect. <A> No, the bulb will not draw 2A at 30V. <S> If you were to model it as an ideal resistor, it would have a resistance of (230V)^2 / (60W) = <S> 880 ohms, thus drawing 35 mA. <S> In reality lightbulbs have rather lower resistance when cold so it would draw several times that current, but it will still put out much less than 60W. <S> You would need 220-240V (AC or DC) to get 60W from it. <S> Use a lightbulb rater for lower voltage, or use 220VAC and a triac - look up how a household dimmer switch works.
| No, the bulb will need 220 to 240 V to fully light, whether AC or DC.
|
How to light a strip of 32 LCD backlight LEDs Ok, for starters I haven't been able to find a lot of solid details about the LEDs I have here. All I know is that there are 32 of them; 8 LEDs per strip, in series, and 4 strips, each also in series. Through my poking and prodding, I've managed to determine that the LEDs are probably around 3.6-3.7v and 150mA. The power circuit board in the TV was very well labeled and it specified 118V and 260mA at the LED connector, so I have to assume that means that each LED gets 3.6875v and 260mA. Unfortunately, the power board was broken (which is probably why the TV was in the dumpster...) So, my question is, how the heck can I get 118v and 260mA? My initial thought was to use a resistor and rectified AC voltage from the wall, which would give me around 170v. That wouldn't work though, because $$\frac{(170 - 118)}{260} = 200Ω$$ and $$0.260^2 * 200 = 13.52W$$ Good luck finding a resistor that can handle 13.5W, and that's SUPER inefficient. I have a basic understanding of electronics, but this kinda stuff confuses me for some reason. I'm learning more every day, I'm just stuck figuring this out. What would you do in this situation? Edit: If it helps at all, I found one of these in my parts bin: https://www.fairchildsemi.com/datasheets/FQ/FQA11N90_F109.pdf 900 volt N-channel MOSFET. Could this be useful? Edit 2: I bought a DC-DC buck converter, 1.5-30V adjustable. Hooked it up to 32V, adjusted it down to 28 volts, then hooked it up to one of the light strips. 3.5V each and it's using around 200mA. Oddly, when I put more of the strips in parallel the current doesn't increase linearly. All 3 strips only draws about 400mA. <Q> I expect that the supply has been a switched mode constant current supply. <S> Output 260 mA resulting in a 118 V when all the leds are in series. <S> The open voltage of such a supply is most probably higher. <S> An LM317HV in constant current mode with a supply of 170 V could be a simple solution but does not solve the inefficiency. <S> If a high voltage switching supply is difficult you could try to connect the strings in parallel with 4 LM317 in constant current mode ( maintaining 260 mA) for each string and use a 33 to 35V DC power source for all of the strings. <S> In that case the efficiency would be acceptable. <A> Easiest solution would be to cut the strips at their 8 led sections. <S> With 4 sections, that's 29.5 Volts, just right for any common led driver. <S> And even some common wall warts. <S> Alternatively a boost regulator or a boost led driver can be used with any suitable lower voltage supply. <A> It would seem that a series buck regulator to drop 60V from a bridge cap source with CC regulated is the most efficient. <S> Adjustable CC. <S> edit <S> $$\frac{(170 <S> - 118)}{260} = 200Ω$$ is a linear supplybut if you switched a $$1Ω . <S> MOSFET$$ rated for more than the difference of voltage of 52V with a duty cycle of 1/200 , it would now be $$200Ω$$. <S> Then with a series choke to smoothen the current, you now have a Buck regulator except the duty cycle is modulated to match sensed current with a PWM
| You could drive them at a lower voltage as well, so a 24V laptop power can be used.
|
Purchasing a battery for a 12VDC, 5A device I have a 12V 5A CPAP device that I'm looking to use while camping for 2 nights. Therefore, I'm trying to determine a deep cycle battery to meet my needs without burning a hole in my wallet. Based on calculations I've seen while browsing the net, I need a battery that can deliver 200AH. Calculation is based on C'' found here: http://www.powerstream.com/battery-capacity-calculations.htm However, searching various car batteries at different stores, they are not listed as deep cycle, but rather CCA (Cold-cranking amps). Searching for a 200AH Deep Cycle battery results in batteries that are over $300. Am I calculating this incorrectly? I was hoping to spend far less considering I won't use the battery very often. <Q> To calculate the capacity you need, you just multiply the number of amps by the number of hours. <S> So 5A for 40 hours is 200Ah, simple as that. <S> You should build in some safety margin, as the capacity will drop as the battery gets older. <S> The car batteries you've been looking at should really have a capacity marked on them somewhere (it's required by law over here, maybe not in the US), but they are probably in the 40-80Ah range. <S> 300USD seems not unreasonable for a 200Ah version. <S> Because the car batteries are mass produced, it may work out cheaper to get several car batteries and put them in parallel, a quick look at my usual suppliers suggests that could be done for £180 (about 240USD). <S> Whatever you do, check your equipment can run OK from a car battery. <S> The actual voltage of a car battery at 5A will drop a bit below 12V, and when charging from a car may go as high as 14.5V. <A> It is good practice to limit lead-acid type battery discharge to no more than 50%, otherwise you significantly reduce the battery life (charge / discharge cycles). <S> This means that you want a battery with an amp-hour rating of twice your expected usage. <S> Lithium batteries can generally be discharged 80%, and they are lighter, so these may be a better portable solution. <S> Unfortunately, Lithium batteries are more expensive than lead-acid. <S> For portable use, with lead-acid batteries you want gel-cell or AGM (Absorbed Glass Mat <S> ) batteries as these don't spill and can be operated in any orientation <S> There is a newer lead-acid battery design that allows for much deeper discharge -- this is called "carbon foam". <S> Firefly makes these batteries. <A> 5A load for 40 h requires 200 Ah according to your assumptions for 2 nights and daytime hours. <S> You are reasonable in your cost calculations. <S> Maybe consider a small gas generator. <S> I just read CPAP is a Continuous positive airway pressure ( <S> power assisted breathing machine )
| Batteries which are not advertised as "deep cycle" or "deep discharge" will age faster as you use them, but if you're not using it often, it won't really matter too much.
|
What are the benefits of a PCB over other methods? Besides the obviously poor choice of a breadboard for a product, what benefits are there to a PCB over things like protoboard with wires soldered? Is it just the size and amount of "wiring" that's already done on a PCB, or is there a benefit to the thin conductive lines on a PCB? To clarify, sorry if I was vague, this is primarily a technical question about the electrical properties of circuits on a PCB compared to other methods but I'd be pleased to read about other criteria that differentiate the PCB from other circuit implementations. In short, why should I go to the trouble of making a PCB? <Q> PCBs typically have lower parasitic inductance and capacitance than a breadboard. <S> Impedance can be controlled for high frequency signals. <S> Circuit density can usually be higher. <S> In circuit and functional testing is easier. <S> Manufacturing is WAY easier. <S> Conducted and radiated EMI will be less for a properly laid out board. <S> Reliability of a PCB should be better, especially if there's shock and/or vibration. <S> There are other benefits that I'm not thinking of at the moment, but I'm sure others will jump in. <S> Of course it's possible to make a complete mess of a PCB with poor layout and manufacturability. <S> PCB layout is a skill that requires a lot of experience for all but the simplest least critical circuits. <A> If you're working with pure DC, relatively low current levels, fairly low voltages, etc. <S> , then you can get by fine with protoboard and soldered wire, etc. <S> The biggest disadvantage there is that it's potentially more work to build your board (assuming you have the PCB's fabbed by a fab and aren't etching your own or whatever), and it's almost certainly harder to get a reproducible <S> build that behaves the same way. <S> Altogether, the protoboard and wire approach doesn't scale as well. <S> If you're building some one-off thing for yourself or just building a prototype, it might not matter. <S> But if you're building something that you want tens or hundreds (or more) of, you will want to use a PCB. <S> crosstalk / inductive coupling, noise, etc., and minute changes in impedance and capacitance become critical factors. <S> In those cases, it's probably easier to build a PCB that takes those things into account. <S> Another "non electrical" factor is that a lot of newer components aren't available in old fashioned through-hole DIP packages. <S> If you want to use the latest IC's you may almost be forced to use a PCB, or at least to use a small shim / PCB "adapter" that the SMT component solders to, that lets you treat it as through-hole. <A> Protoboards with wires take a much longer time to assemble than PCBs. <S> They also lend to assembly mistakes that take time to debug. <S> They get real nasty for anything above modest complexity. <S> There was a time when wire wrapping was the preferred method for complex prototypes, as PCBs were expensive and slow. <S> Now I wouldn't dream of wire wrapping. <S> Cap all this, and all other answers, with the fact that many ICs are only available in SMD packages, and the fact that PCBS are now dirt cheap and fairly fast. <S> When soldered using a tiny smidgen of attention, they come out error free. <S> Clear winner.
| The other case where a PCB is clearly better is when you're working with higher frequencies, especially up in the RF range, where you start having to worry about Electrically, there's not a huge difference in many common scenarios.
|
How can I build a simple USB signalling device? I would like to construct a simple device that plugs into a USB port and sends the computer a signal (i.e just a 1 ) when a user presses the button or switch etc. I imagine there must be some way to take a USB cable and modify the side that doesn't plug into the computer so that it sends a signal when activated, perhaps by tapping a battery or something. My goal is to build a simple button that I can use to perform a regular series of actions with a simple press. I know they sell things like this but it would be cooler and perhaps cheaper if there was a DIY way. <Q> ATTiny + V-USB acting as HID keyboard. <S> ATMega32U4 (e.g. Arduino Leonardo) acting as HID keyboard. <S> PIC18F14K50 acting as, yes a USB HID device. <S> There are a host of other USB enabled microcontrollers, those are just three I can think of off the top of my head. <S> You will need some microcontroller of some description capable of interfacing with USB. <S> You can't just cut the end of the cable and hope to simply connect the wires together and get it to print numbers without some form of processor. <S> There are certainly sample codes available for the Leonardo board which can be used to send keystrokes to the PC. <A> Aside from the practical solutions above, there is more diy hacky ways. <S> Take apart a usb keyboard and remove the key matrix. <S> Replace a specific matrix combination with a single push button (say an unused one like f8). <S> Then use a software hot key program to carry out the rest based on that specific keyboard and key press. <S> The other way would be taking a mouse apart and doing the same. <S> But a keyboard is simpler and with any recent cheapo one, the pcb would be about the size of a flash drive. <S> But a microcontroller solution can be done completely without any computer side software needed. <S> Pure USB HID keyboard goodness. <A> I would recommend you use a USB to Serial cable (or adapter, as they are often called). <S> The drivers are easy to come by, and are typically installed automatically. <S> You can then toggle the state of one of the control pins, such as CTS or RI using your switch. <S> This will allow you to easily read the lines with a simple program on your system. <S> You can toggle output lines to your circuit similarly, should you need to. <S> I did this myself to monitor a magnetic relay mounted on a door to monitor and timestamp entries to the lab. <A> I am assuming you want to work with existing USB device drivers across common operating systems, e.g. Windows (several flavours), OS X, and Linux. <S> This would enable a program running on the host PC to open a file, and read the 1 . <S> To send a 1 which is recognised by the host PC, you will need to exchange several message packets. <S> It is not a simple one-bit, or even fixed multi-bit, signal. <S> Instead the host OS will request enough information from your USB 'button' device to figure out what type of USB device it is, and start the corresponding device driver. <S> You need to understand how USB works. <S> This USB tutorial USB in a nutshell may help. <S> It describes the protocol in enough detail that it is understandable how the host PC' will interact with the USB device. <S> To be easy to use, you will probably implement a HID device, which looks like a keyboard to the host PC. <S> The button press could be made to look exactly like a keyboard 1 . <S> There are several projects which you can find by searching the web. <S> They all use a microcontroller to manage the quite complex interaction with the host. <S> One example, which implements USB totally in software using a low-cost microcontroller is V-USB . <S> There are several other projects, <S> some which do what you are describing, like PJRC's Teensy , and I digispark <S> Their are lots of MCUs which include a hardware USB interface. <S> Typically they cost more than 1GBP. <S> If you want to go down this path, most of the manufacturers of the USB-enabled MCU provide some sort of library, and their is also the Open Source 'LUFA' USB library . <S> Summary : the USB protocol is so complex that you will need a microcontroller to implement anything useful. <S> USB is not a simple binary bit pattern. <S> Edit: This V-USB stompbox looks like a close-fit solution, if my other assumptions are correct. <S> It has a single button, emulates a keyboard, and is built on a piece of stripboard (veroboard) with an ATtiny and a few components. <S> You could use your Arduino to make an AVR ISP programmer. <S> That would enable you to load the firmware onto the ATtiny, so it might be quite a quick, low-cost project. <A> You could use something like this FTDI breakout board that allows you to easily interface via serial protocol over USB. <A> USB has complex signaling indeed. <S> But what many overlook is that USB also employs some very simple signalling . <S> If you fully control the software on the computer, depending on your goals all you may actually need is to have your button connect a resistor between either of the data lines and VBUS. <S> Such a pull-up resistor is how USB peripherals signal their presence and which of the two base speeds they utilize. <S> The operating system USB stack would then normally attempt to talk to the device using normal complex-state differential USB signalling and read out basic information such as VID/PID to determine which driver is needed. <S> Of course in the simple-switch case this will fail but the attempt and failure is detectable from system log messages or via custom kernel code, and thus may satisfy your need. <S> Expect some latency and a low rate of repetition if using the stock USB stack, but custom kernel mode drivers can probably do this at rates that would feel prompt in human terms.
| USB is a fairly complex protocol, so the best bet is to find something that will either do what you want (like a keyboard), or a microcontroller with a prewritten USB HID stack which you can then customise the code for to send key press info when a push button is pressed.
|
Why not include a thermal fuse in electrical outlets? I have found myself reading reports of fires lately, including some that started due to high-resistance connections in electrical outlets and switches. A load on a high-resistance connection can generate enough local heat to ignite building materials, while not tripping any upstream safety devices. This is a problem that may not manifest until years after the original installation. (Having not thought of this before, obviously I am now scared out of my mind...) Why not integrate a thermal fuse into the outlets? It seems like a cheap way to prevent some fires. Is there some technical flaw in the concept? Or is this just an actuarial cost-benefit computation? <Q> The other answers seem to be misinterpreting the 'thermal fuse' part of your question. <S> A 'thermal fuse' is an electrical overload sensor that uses heat as an indicator of an electrical overload. <S> It sounds like you are asking about a thermal cutoff like the kind included in motors to which open a circuit when the locally detected heat (ie. <S> not generated by the electrical current itself) exceeds a set parameter. <S> The reason this is not included in electrical outlets has to do with the cost-benefit of including such a complex sensor (~$0.75) in such a simple and inexpensive device like an outlet (~$0.30). <S> Electrical codes require all wiring devices to be installed in a UL listed box or enclosure. <S> The same codes require these boxes to be flame resistant. <S> The idea is that the effects (heat, fire, etc.) of the high resistance connection will be limited to the box. <S> Fires certainly do occur as a result of this <S> but this is infrequent compared to the much more common ways electrical fires start. <S> Codes are updated every couple of years and are getting better and better at addressing less and less common occurrences. <S> For example the 2014 NEC requires AFCI (Arc-Fault Circuit Interrupters) in many locations that do a much better job of detecting events on the more dangerous, fire-starting end of the spectrum of 'high resistance events' you are describing. <A> There is a very simple reason why this is not used. <S> The main function of the fuse or circuit braker is the overcurrent protection of the enitire wire section. <S> Wires with certain cross section are protected at the node where a higher cross section is distributed in many sections with smaller cross section wires. <S> All sbsequent devices like wall sockets, connecting cables,... <S> etc need to have the same nominal current capability. <S> With your proposal, even if the wall socket had a fuse, this won't prevent the wiring taking on fire, in case that wire cross section is too small. <A> The use of thermal fuses in domestic electric systems does not help to improve safety. <S> In general the design of switches and outlets is done in such a way that any temperature rise due to the flowing current is absorbed by the mass or contacts and connections. <S> This is possible since the dissipated energy is very small (I2R).Considering that outlets and switches are properly designed <S> and if professionaly connected the risks here are very limited. <S> The connections in junction boxes also can form a risk if the junctions are not professionaly made in a diy situation. <S> If an outled or switch has a bad connection the device gets hot and gets burned long before there is any fire hazard. <S> If an outlet is burned it should not be used and replaced by a new one. <S> If such an outled remains in use the risk of fire increases <A> The easy answer is cost, and the main reason is standards and certification bodies. <S> Cost is one of the constraints for an engineer. <S> Standards and certification bodies rely on engineers and subject matter experts, that generally have to come from the same industry they are regulating. <S> They don’t want standards to become prohibitively onerous, and they have to justify and document their decisions. <S> Placing a 5km/hr speed limit would considerably increase safety, but would also be unnecessarily burdensome to the same society that it’s intended to protect. <S> A cold actuarial compromise has to be used. <S> No system is 100% safe. <S> The scenario you describe implies a degradation of the plug and/or socket that puts it way out of the limits of the required standard. <S> It is a very rare scenario, that could be caught in time by an informed consumer and is commonly included in device documentation. <S> Consumer education (particularly lack thereof) is the main part of the equation. <S> There are inexpensive electronic safety devices that can be added to table saws that prevent them from slicing through your fingers. <S> But these are not required. <S> And consumers pay more attention to the item price than to a safety feature that dares to imply they are clumsy, that enters the capitalistic equation. <S> But standards and regulations can be changed. <S> It just needs people championing them. <S> And that generally requires an informed consumer.
| In case that an electric device has internaly smaller cross section than connecting cable, then it needs another fuse inside. Cost of mass produced items has to be minimized, adding a safety feature that is not required by certification and increases production costs can be seen as engineering malpractice. And modifications are made all the time to include new causes for concerns.
|
Can part of Integrated Circuit be damaged from heat gun usage? I tried to re-use a costly Integrated circuit which includes dual port ram inside. I removed it from unused previous board by carefully using heat gun. The problem is that the other part of the ram works fine, but the dual-port ram part always reads as 0x00. The interface is SPI and other part of the ASIC works just as fine, and the same firmware on the other board that was build from new IC works fine. Is it possible that only that part of IC damaged from using heat? <Q> carefully using heat gun LOL! <S> Do you carefully use a lawn mower to trim your fingernails too? <S> Hot air can be used for soldering, but has to be properly controlled and applied. <S> Without a hot air soldering station with the right tip and the right temperature and airflow settings, you're just slicing deli meant with a chain saw. <S> Consider anything you removed from the board with the chain saw heat gun to be junk. <S> Toss it and move on. <S> Don't waste any more time and frustration with abused parts. <A> Yes. <S> Manually using a heat gun (even a temperature controlled one) is a risky operation. <S> The pins of an IC are designed to dissipate heat to the PCB <S> so it is quite possible to have one side of your chip below the thermal failure tolerance while the other side is above that threshold, leading to potential damage to the silicon substrate in an isolated area. <S> Sounds like you got lucky in that you can see the damage straight-away rather than having an intermittent failure that might have only become apparent in the field. <S> There are products out there that make manual desoldering easier by dropping the melting point of your solder by forming a new alloy during desoldering. <S> One such is ChipQuik <S> but I have never used it myself for large chips. <S> This makes it possible to melt all the solder simultaneously at a much lower temperature than normal (temps as low as 90 Celcius are possible, which is not likely to damage any IC I can think of). <S> (I don't work for the ChipQuik company by the way :)) <A> It's dead Jim. <S> "Careful" and heat gun don't usually belong in the same sentence. <S> You can buy nozzles for desoldering heat guns that match various chip types which are significantly less likely to cook your chip in the progress. <S> Hot air is directed to component pins. <S> Just regular lead solder liberally applied over all the component pins <S> makes it easier to desolder. <S> It's not going to instantly coagulate with more volume and a pb based solder has a lower melting point. <S> Another basic tool is a PCB heater that will raise the PCB temp up to desired temperature <S> so it's easier to (de) solder things. <A> If anything you need a heat gun with variable heat settings and many different attachments.
| I use a smaller has pen gas torch if its a big jobThere are a few heat guns reviewed here but be carefull a heat gun can burn a hole in a wall let alone a circuit board.
|
Sensing BLDC motor torque using a current sensor As it is commonly known, the torque supplied by a motor is directly proportional to the current it draws. So when a motor is under heavy load, the torque required will be more hence it would draw more current For a BLDC motor, the current is supplied via three phases which are switch by a motor driver. But this motor driver is supplied with a single DC supply. Can I reliably measure the torque provided by my motor by measuring the current drawn by the driver? My motor is currently mounted as part of a robotic arm and there is PID control loop for position control. I only need to get an idea of the current load and detect load anomalies (like an impact i.e. higher load or loss of payload i.e. lower load etc). Given that I can calculate the nominal load at any position and hence calculate the nominal current drawn, is it a good idea to use current sensing for this kind of load sensing? Is there any research on this which analyses this method of torque control/load sensing which I can refer to? <Q> It depends on the motor driver, but in the general case you can't expect the driver input current to equal the motor current. <S> This is because good motor drivers are essentially switching power supplies, using the inductance of the motor as a integral part of the supply. <S> The current thru a winding can be higher than the supply current, just like the output current of a buck switcher can be higher than the input current. <A> It doesn't give you motor current directly, but it does tell you how much power <S> the motor is drawing. <S> If you know what (average) voltage the controller is applying to the motor <S> then you can easily calculate the current (since current = power / voltage). <S> The problem is getting that voltage. <S> If the BLDC driver is controlled by 0-100% PWM and directly applies that same PWM ratio to the motor, then (assuming negligible voltage drops in the driver) <S> motor voltage is simply supply voltage / PWM ratio. <S> However if the driver is controlled by eg. <S> a servo signal <S> then there probably won't be a direct 1:1 relationship between the input control signal and motor voltage, so your calculated voltage will be less accurate. <S> At low speed (low PWM ratio) there may be significant current ripple and extra power loss in the driver and motor. <S> Also motor current doesn't tell you exactly how much torque the motor is producing, since it includes current used to overcome internal losses, and <S> when the motor is accelerating it needs extra torque to deal with its own inertia. <S> If you can account for these effects then you should be able to get a reasonable estimate of output torque. <A> minus losses. <S> You need to be able to accurately estimate the losses. <S> For a system that produces less that 100 watts, losses might be as much as 50 to 100 watts. <S> There will be some fixed losses, some that vary with speed and some that vary with load. <S> You will also need to calculate the average or RMS value of current over an appropriate time period. <S> Estimating torque in this manner is not a trivial task, but it can be done quite effectively. <S> However, it is not necessarily much less complex than using motor current measurements. <S> There is a lot of research on doing this sort of thing. <A> You could use a non-contact Hall effect current sensor (from e.g. LEM, Allegro or Amploc) around one of the motor phase wires. <S> Low-pass filter it with a pole somewhere between the motor electrical frequency and the controller PWM frequency <S> , then either sample it with an ADC or use one of those "true RMS" converter ICs. <S> That should give you a measurement of RMS motor current which is ideally proportional to torque. <A> You can use the power equivelence as above (torque X speed) = <S> (voltage X current) - Losses as stated, but beware of your DC link large capacitance. <S> This will heavily filter the DC signal and so can't really be used for any transient nature torque measurements, only for steady state.
| What you can absolutely depend on is that the mechanical power delivered by the motor (torque X speed) is equal to the input power (voltage X current)
|
Is there any solution to high-precision high-side sampling current measurement? I want to do some high-precision current measurement. (Voltage up to 20V and requires current to be shown in 0.1mA resolution, current can be up to 4A). The simplest is low-side sampling by MCP3421: simulate this circuit – Schematic created using CircuitLab This one is easy and accurate. But this breaks the ground connection. Using MCP3421 for high-side sampling is not possible, because it will break its input voltage limit(VCC + 0.3V). Our upstream voltage can be up to 20V. Another possible solution is by using OP-Amps: Link to pdf Page 18 describes how to use OP-Amps to do the conversions. But I wonder the resistors can be inaccurate and ageing is a killer of accuracy. There's one more solution: using high side current sense amplifier like TSC101 or LTC6101. But this sort of ICs often has a high Input Offset Voltage (Vos) up to approx. 1mV. 1mA on 15mOhm sampling resistor is 15uV, so the offset seems to be unacceptable. Are the above solutions possible? Or are there any other solutions? <Q> Look at the Allegro Microsystems current sensor ICs. <S> These use hall-effect sensors and can be used in high-side applications. <A> There are tons of high-side current-sense amplifiers which already include the precision resistors needed for good CMRR and offset. <S> https://datasheets.maximintegrated.com/en/ds/MAX4376-MAX4378.pdf <S> http://www.analog.com/media/en/technical-documentation/data-sheets/AD8221.pdf <S> etc... <S> I suggest you spend a moment using DigiKey's search engine, <S> look here : <S> Then select "Current sense" in the "amplifier type" box. <S> For ultra low offset, you can search for a chopper (or auto-zero) instrumentation amplifier like LTC1100, however you are unlikely to find one which can sense a voltage beyond its rails. <S> If the load can be switched on/off, then a simple way to null offset is to use double sampling: measure the current with load ON, then with load OFF, and substract both. <S> This works extremely well. <A> With high CMRR it will do the job. <S> Ad8222 is great. <S> A drawback: you may need to create a voltage rail higher than the input voltage. <S> For higher voltages you can use isolated ADC. <S> Meaning you put Isolated power and ADC <S> right near the measured resistor, so the common mode component is 0 related to them. <S> A good example is Ad7400. <A> The simplest is low-side sampling by MCP3421. <S> This one is easy and accurate. <S> But this breaks the ground connection. <S> If you are content with this method then use one more chip that can provide isolated power and isolated buffers for SCL and SDA. <S> I'm thinking here of the ADuM54xx series from ADI: - http://www.bdtic.com/adi/images/ADuM5401_fbs.gif <S> You can get versions with different IO configs to suit clock in and data out from the ADC. <S> Possibly the ADuM5402 will be of most interest to your application. <S> Basically use the isolation chip and float your ADC up to 20V to make the measurement. <A> Go ahead and use a high-side monitor, and if the offset voltage bothers you,preload the current, then apply an offset at the output to null the result with zero load. <S> The TSC101 , for example, with a 0.050 ohm sense resistor, can handle the 4A. Up to 2.5 mV of offset in the chip can be treatedby shunting 0.0025V/.050ohm <S> = 50 mA from the output <S> (so that the offset voltage will never saturate the sense amplifier). <S> A negative power supply will simplify the level-translation (and allow youto continue to monitor current below the 2.8V 'input common mode' limit).I'd draw the 50 mA with a grounded-base NPN transistor. <S> simulate this circuit – <S> Schematic created using CircuitLab
| For PRECISION measurement i see two options: Since your voltage is not very high, use a good instrumentation amplifier.
|
How do I describe this circuit? simulate this circuit – Schematic created using CircuitLab Yes, this circuit diagram is correct. No, I did not design it. How do I describe this circuit in words? I thought about "a switch and light bulb in parallel, instead of series" but the circuit is so bizarre that I want to make sure what I am saying is unambiguous. I cannot insert a schematic/picture/diagram/drawing, and I have a fairly tight word limit. If anyone knows the tags for this question, add them. EDIT (due to interest from the comments): The circuit was made by my 3rd grade teacher for an electricity test (when she thought the questions from the curriculum were too hard for us...) EDIT 2: No, this is not a fancy NOT gate. The question was "What will happen when the switch closes" and the teacher-accepted answer being "Lightbulb turns on " <Q> How do I describe this circuit? <S> Do you want subjective opinion? <S> Or technical description? <S> Subjective opinion would be: this circuit is badly designed, no matter what problem the designer tried to solve. <S> Technical description would be: it is a battery killer with stand-by indicator. <A> The best description for this circuit is "How to NOT control a light with a switch." <S> This was clearly not designed by anyone with even basic knowledge of electricity. <S> Or else it was incorrectly "interpreted" or reproduced by someone with absolutely no understanding at all. <S> Obviously, closing the switch will put a DEAD SHORT across the power supply. <S> That is NEVER a good idea, and can be quite dangerous in many cases. <S> Clearly, the light bulb (or LED, it makes no difference!) <S> turns OFF when the switch is closed. <S> Anyone who thinks otherwise has zero understanding of electricity and should not be attempting to teach others. <S> In any grade. <A> "Battery short circuit stress tester". <S> This circuit will allow you to observe how the supply source reacts when short circuited. <S> Also tests the switch under extreme fault conditions as well. <S> An expendable person should be deployed to operate the switch. <A> Current tends to select line with lower resistance. <S> Short circuit resistance is 0 and open circuit resistance is unlimited. <S> The current of lamp when switch is closed = <S> zero <S> then you have no light but when switch is open the current of lamp is (1.5v/100ohm <S> = 0.015A) <S> then you have light like this picture : <S> and something about the battery : <A> I'd describe it as: Switch and bulb in parallel connected to a voltage source. <S> I think the voltage source should be mentioned (without it you cannot reproduce the circuit, although most would think implicitly that a voltage source is there). <S> And I don't think a description should contain any hint that this is a quite useless contraption <S> , stuff in school is often not useful or practical but only a test to see if the concepts are understood. <S> An explanation of how it works can contain all the comments on how dangerous and useless this might be. <S> However the solution is wrong - like pointed out by everyone else, if the switch is closed the light will go out. <S> I've seen something similar while restoring my old Vespa. <S> This answer is more of a try to explain why the teacher might ended up with a wrong circuit. <S> The generator of a Vespa works like a Dynamo on a bike, so it acts like some sort of current source. <S> And the horn was controlled by a switch which was hooked up in a similar way. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If you press the button (break the short), the current goes through the horn and you get a nice sound. <S> In the old schematics, there was no nice current source and the switch was also not depicted like in this schematic but as a normal switch. <S> While analyzing the whole thing I was baffled at first until I drew a complete schematic and measured the switch to realize it's normally closed and pressing it does not close it but open it. <S> So maybe your teacher came across some circuit like this and in an attempt to make it easier, the circuit was changed in a way to make it wrong and <S> the switch being normally closed was not clear. <S> Most people think of pressing a button or flicking a switch will make a contact. <A> Or if you want to describe the circuit construction through words: "Bulb+switch in parallel" <S> Though, you could also describe it as: "Simplest bulb NOT gate." <S> If you short an alkaline AA 1.5v battery, it's probably not going to heat up as much as you think, especially that it's a small BATTERY which is current limited. <S> See here .
| I would describe it as a: "Inverted bulb short ckt w' switch."
|
Triac Operation: Gate connects to A1 not working? I have tested the above circuits and found that I can switch on the triac with the circuit at the top, but not with the bottom circuit. In my understanding, the Triac can be triggered on no matter the gate is positive or negative, the main difference is I may need a larger gate trigger current (Ig). I have lowered the gate resistor value but still cannot switch on the Triac, why? Note: I know it is not related to the optocoupler, because when I am testing it, I even shorted pin 4 and pin 6 of the optocoupler and it still does not work <Q> Your triac is upside down in the schematic. <S> Your "not work" schematic shows you shorting out the trigger pin to T1 so that does nothing. <S> In your "work" schematic turning on the opto-coupler connects the trigger to the voltage across the triac. <S> Since this is high enough the triac turns on. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> The WORK and NOT WORK schematics redrawn for clarity. <S> You are confusing operation of a trigger derived from the mains to one where a separate voltage is used. <S> Figure 2 shows how this would work but note again that the trigger voltage is with reference to terminal T1 (the same side of the triac as the trigger pin). <S> simulate this circuit Figure 2. <S> Switching a triac with a separate trigger supply. <S> Finally, for reference: Figure 3. <S> Triac triggering modes and quadrant numbers. <S> Source Wikipedia (with addition of numbers). <S> Note that your circuit works on quadrants 1 <S> (both mains and trigger positive) and quadrant 3 (both negative) only. <A> Due to the position of the gate in the PNPN and NPNP layers of the triac the device must be triggered between the gate and T1 and not between the gate and T2 <A> The triac is triggered by a current flowing between the gate and MT1. <S> Connecting the gate to MT1 does not make it turn on, rather it makes it a bit less likely to turn on. <S> It's like a relay coil- <S> shorting it out will not make the relay turn on. <S> MT1 and gate will always be within a couple volts of each other on a (non-damaged) triac, because of the internal structure.
| To trigger the triac you have to bring the trigger positive or negative with respect to terminal T1 (the terminal on the same side as the trigger on the schematic).
|
What does the term 'lifetime' mean for electrolytic capacitors I am making a circuit for a car and a regulator is used to get from 12V to 5V. This regulator is surrounded by two 10 uF electrolytic capacitors rated at at 25V. These capacitors have a 'lifetime' of 2000h. Does this mean that there is a 50% probability that the capacitors will fail after 2000h of usage? Since the circuit is for a car the longevity and integrity of the circuit is important. This is the datasheet . <Q> The "lifetime" is the total time the capacitor can be in service while subjected to the maximum stresses specified in the datasheet, after which the manufacturer no longer promises anything about the cap. <S> In some cases, the manufacturer even tells you which parameters it no longer guarantees to what level. <S> Generally the ESR (equivalent series resistance) goes up. <S> Manufacturers are therefore stuck with a tradeoff of specifying low voltage and temperature or low lifetime. <S> Most of the time, they push the voltage and temperature and quote very low lifetimes, like 2000 hours (not even 3 months). <S> Notice that caps specifically sold as having a long lifetime have lower voltage and temperature specs for the same size case. <S> Lifetime goes up quickly as the voltage and temperature are lowered, although most datasheets unfortunately don't give you a lot of guidance on this. <S> A good rule of thumb is to keep the steady voltage on a electrolytic cap at about 2/3 of the maximum. <S> It's OK to go near the maximum occasionally, but don't ever exceed it. <S> Similarly, you want the steady temperature to be 20 to 50 °C below the maximum. <S> The only way to get realistic lifetime values is to ask the manufacturer or find the rare cap where useful data on this is included in the datasheet. <A> The datasheet guarantees that capacitors will keep their characteristics within limits after spending 2000h in quite extreme conditions (maximum rated voltage and 105°C). <S> This says nothing about MTBF or expected lifetime, that information is simply missing in the datasheet. <S> Caps lifetime is often limited by chemical electrolyte degradation, which is simply not going to manifest itself in 2000 hours. <S> You'll have to ask the manufacturer to provide such information if you have MTBF requirements on your product. <S> Also note that the datasheet says: This product qualify for AEC-Q200, but it has some deviations <S> So if you plan to use it in automotive applications, you'll have to contact the manufacturer anyway to know what those deviations are. <A> Datasheet states: <S> Endurance: 2000 h to 5000 h at 105 ° <S> C <S> That means that parameters will be within specification defined in datasheet:
| Electrolytic caps wear out rather quickly at maximum voltage and temperature, but the higher the allowed voltage and temperature, the more possible uses, and therefore sales, there are for a capacitor.
|
Start Delta Motor Starter, at higher ampere in Delta Configuration We are using 3 Phase Six Wire out standard ABB foot mount motor of 25HP. We are using star-delta starter with delay of about 30 second from star to delta. We are seeing around 40 Ampere when machine starts in star, but that amperes jumps to 130 ampere when the starter switched to delta, which trips the whole circuit in six to ten seconds. How do I diagnose such behavior ? As I understand, the ampere should be reduce when it is used in delta configuration. Thanks. <Q> If the rated voltage for the delta connection is applied to a star-connected motor, the resulting torque and current are about one third of the characteristic torque and current for the delta connection. <S> As a result, at the transition from star to delta, the change in current and torque is about a three times increase. <S> Therefore, an increase from 40 to 130 amps is within the normal range. <S> Star-Delta Starting Characteristic Curves and Explanation <S> The torque and current vs. speed curves shown below are for a typical IEC Design N or NEMA Design B motor with star-delta starting. <S> The load torque requirement shown increases as the speed increases as it would for a fan or centrifugal pump load. <S> The substantial torque requirement at zero speed would be a requirement of a pump with a substantial static head. <S> Note that the torque available to accelerate the motor and load is the difference between the steady-state load torque requirement and the available motor torque shown by the motor cur torque vs. speed curves. <S> In this case, there is very little star-connection acceleration torque available between 40% and 70% of full speed. <S> If the load torque was a little more, or the motor torque a little less, the motor capability torque curve would cross the load torque requirement torque curve and the motor would not accelerate the load to a higher speed until the star-delta transition. <S> The load and motor curves are shown as crossing at a little above 90% of rated speed, a little below the star-delta transition speed. <S> At the star-delta transition, the torque and current increase to the points at which the vertical transition line crosses the delta torque and current curves. <A> If the delta formation is improper, the motor will first start and run ok. <S> But after it changes over the motor will stop if the contactors are in closed position in the starter. <S> Place the relay below the delta contactor while after the change is over. <A> Check the motor connection plate. <S> If the motor connection plate's U1, V1, W1 side is ok, may be U2, V2, W2 side have some mistake. <S> In that case, the would start normally but after motor goes into delta, it would trip the circuit breaker because of high current. <A> When you switch the motor on in star connection, it has already the nominal voltage. <S> P=0.4kV*1.73 <S> *40A = 27kW, which is already the rated power of described motor. <S> For example a delta/star motor of 400/690V, suitable for star/delta starter should have a power 1/3rd in star connection, compared to nominal power when delta connection is established, so something arround <S> 7kW. <S> But it hard to say without any nameplate data.
| If the current doesn't drop to a normal running current in six to ten seconds after the transition, the motor is not adequate for the load, there is a problem - possibly an unconnected phase in the delta connection, there is a mechanical problem with the load or some other problem. First look at the motor name plate, (in star delta starters check the six leads connected properly to their terminals). Perhaps your motor is designed for example 230/400V delta/star.
|
Testing a circuit systematically and efficiently I need to determine a circuit's response to variations in component values. What are common go-to methods for this? Background : I'm a growing novice in electronics design. So, my question is about common, de facto strategies accepted by experienced hobbyists and professionals alike. What for? To analyze how a circuit behaves for different combinations of its components' values (and other parameters). And to do so in a structured, systematic, efficient way. My first thought is a SPICE simulation. But I would like to confirm whether this is 'it' or whether there are other approaches that I'm unaware of. Context : I have a simple optocoupler circuit as shown below. The left side (1) operates on a different potential than the right (2). And a MCU on the right needs to read an analog signal on the left ( V_ana_1 ). Hence the optocoupler. I use it to convert V_ana_1 to V_ana_2 which then is fed to the MCU. Now the optocoupler's response is not linear. So I would like to choose R1 and R2 such that V_ana_2 's range falls within the sensitive (as opposed to the saturated) region of this response for a given V_ana_1 range of, say, 2-3 V. What's a good way to go about doing this without physically and manually trying out different resistor values? (Which is what I did, which is why I'm asking this question - that was a pain.) simulate this circuit – Schematic created using CircuitLab <Q> Generally it is preferable to use intuition and basic calculations rather than blindly swapping components in and out. <S> For your example, you have an optocoupler, and you want to take a voltage on one side and transfer it to another. <S> As I mentioned in the comments, this is not the greatest tool for this (a linear optocoupler would be much better), and the part to part variations make this likely not work. <S> Here is the FOD852 optocoupler. <S> It was the first datasheet the popped up when I searched for optocoupler. <S> On the 4th page is a number of figures that show the characteristics of a particular device. <S> Figure 8 shows a plot of collector to emitter voltage as a function of collector current, for a number of different forward currents of the diode. <S> We know that we want some range of currents through the diode to equal some range of currents (or voltage) through the transistor: <S> We can kind of squint and see that a current of ~3-10 mA will lead to a collector current of ~80-100 mA. <S> So, we want to generate a 3-10mA current through the diode, and we want a 80-100mA current through a resistor to give us a Vce of ~1V. <S> The 3-10mA is accomplished with a resistor of (Avg(V_ana1) - V_f)/6mA, <S> The 80-100mA is accomplished with a resistor with a value of (Vcc2 - 1V)/90mA. <S> By thinking a little, and using the data available to us, we can make a pretty good guess. <A> Often specs for worst case V are given for CMOS at Vcc-10% at 25'C and over T range. <S> if design cannot meet worst case then Monte Carlo methods for parameters like dielectric constant of PCB on controlled impedances, trace accuracy etc. or pay for TDR coupon tests done for Zo. <S> The best way to verify a design is test to functional failure (HASS) during DVT to determine margins and root cause. <S> e.g. vary Xtal f over spec range incl tolerance at 25'C and temp. <S> by injection and Vmin at cycle temps with high RH with vibration to board. <S> Every test should have a margin budget verified, e.g. comparator asymmetry, regulator error margin etc, surge or ripple current/rating, noise margin, timing margin, real-time stack margin, buffer overflow margin etc etc. <S> the best design has DFT and these parameters defined in the design spec. <S> then DVT is easier to perform. <S> DVT= Design Validation Test. <S> DFT=Design for Testability). <S> This enables good fault detection/isolation for self-Test, Functional(FT) and In-cct test(ICT) <S> then sys test. <S> Once the functional design is verified, solderability is the #1 cause of poor yields, which gets into bad design, process, or procedure <S> Each test should fit onto 1 page with method illustrated and equip. <S> used <A> The most common method is to use SPICE simulator. <S> Go here, http://www.linear.com/designtools/software/#LTspice Download and install. <S> It is free. <S> It has many examples for starters, including optocouplers. <S> Briefly, every element of schematics can be represented by a realistic parameter (SPICE) model, extracted by part manufacturer. <S> The simulator then calculates EVERYTHING depending on type of stimulus applied to circuit inputs. <S> For commercial-grade simulator it would calculate everything for all temperature and voltage corners of the parts. <S> LTspice will give you results for typical parameters. <S> Read tutorials and enjoy the journey.
| The best way is to apply worst case tolerances over T and V with some margin for vendor variation, aging must be done for reliable production designs.
|
Interpretation of Laplace transformed function and Laplace vs Fourier I'm used to Fourier transform, i.e. I have a good unterstanding of how to interpret a Fourier transformed function. But I'm struggeling with the Laplace transform. What's the meaning of its complex parameter \$s = o + iw\$ (where \$w\$ is the 'normal' frequency)? I also don't see the advantage of the Laplace transform when it comes to transfer functions. It seems to me that in the end one is just interested in the spectrum of the analyzed system which coincides with evaluating the transfer function just on the imaginary axis ('normal' frequencies). So why one wouldn't simlpy use the Fourier transform to define the transfer function like this: $$G(w) = \frac{F\left(y(t)\right)(w)}{F\left(x(t)\right)(w)}$$ where \$w\$ is real, instead of: $$G(s) = \frac{L\left(y(t)\right)(s)}{L\left(x(t)\right)(s)}$$ where \$s = o + iw\$. <Q> A very nice property is that the Laplace transform evaluated along the jw-axis is equivalent to the Fourier transform, which is less abstract and easier to understand. <S> This of course brings us back to question, why we didn't use the Fourier transform in the first place. <S> The answer is quite simple. <S> We often have to analyze and work with systems that are unstable or where we want to determine whether they are stable or not. <S> In such cases the Fourier transform fails (=does not exist). <S> This is the reason the Laplace transform was introduced. <S> It includes an additional term that helps the integral to converge and therefore the Laplace transform can be applied to a broader range of problems and applications. <A> What are the Fourier transforms for the step and ramp functions? <S> As well stated by Prof. C.P. Quevedo: " The idea of saying that such functions are periodic, with infinite period, no longer applies (the function never returns to zero and do not have the opportunity to repeat, neither in the infinity ". <S> This is where enters the Laplace Transform. <S> By introducing a real term \$\sigma\$ <S> in \$s = <S> \sigma <S> + j\omega\$, it is possible to make the "Fourier Transform" integral (now a Laplace Transform) to converge. <S> $$F(s) = <S> \int_{^{^0{-}}}^{\infty}f(t)e^{^{-st}}dt <S> $$ <S> In fact, the main goal of the Laplace transform is to convert a differential equation into an algebraic equation (like logarithms). <S> After, to operate in the complex domain, expand the result with partial fractions return to time domain using (normally) transformer tables; not only to a sinusoidal signal inputs. <S> Also, It should be remembered that the frequency response (sinusoidal) is just one of the transfer function applications. <S> Fourier series : <S> A periodic time signal is seen as an infinite sum of sinusoids (discrete frequencies, harmonics). <S> Fourier transform : A not necessarily "periodic" time signal is seen as an infinite sum of infinitesimal scaled sinusoids (continuous frequencies). <S> Sum -> Integral. <S> Laplace Transform : <S> A not necessarily "periodic" time signal is seen as an infinite sum of infinitesimal exponentially scaled sinusoids (continuous frequencies). <S> ADDITION:\$\sigma\$ comes into play when when the concept of frequency is generalized to "complex frequency". <S> When you write\$Ae^{kt}\$, the exponent should be dimensionless; <S> so that \$k\$ should be <S> \${second}^{-1}\$. <S> Note the similarity with "Hertz", ie it is a type of frequency. <S> For example, in \$Ae^{2t}\$, the \$2\$ is the frequency (events per second) with which \$A\$ will be multiplied by \$e\$. <S> The unit of this "frequency" is Neper/second. <S> Thus $$s [complex Neper/ second] = <S> \sigma <S> [Neper/second] <S> + j\omega [radians/second]$$Now an exponential function has a frequency, even though it differs from the traditional concept. <S> In the Laplace Transform, the \$\sigma\$ is an appropriate value (but not only) for the integral to converge. <A> So typically I use the Fourier domain to analyze a signal and it's frequency components. <S> In the Fourier domain you can do tricks to manipulate the signal (like multiply by a step function to do a LP or HP or BP filter). <S> Although last time I was faced with this I just did the convolution of the signal against a sinc function (easier to write in a program and faster). <S> The Laplace domain is more so used for system analysis and control theory. <S> Typically you will start out with the transformed version -> <S> like a capacitor is Z = 1/(sC). <S> And once you have figured out the transfer function of your whole system, you can see how it responds to inputs (often you will do a frequency sweep or a nyquist plot). <S> These concepts are not mutually exclusive. <S> I believe Fourier is a specific case of the more generalized Laplace case (please correct me if I am wrong or elaborate as necessary because I really don't remember this point).
| The Laplace transform has some nice properties that help to get more insight into the behavior of linear systems.
|
Spectrum analyzer as source of electromagnetic interference I was wondering if someone could enlighten me on this. I would like to know whether spectrum analyzers can be inherently the source of interference to other components or not? Following is the block diagram of spectrum analyzer which is copied from (Frenzel, L. (1995). Communication electronics. New York, N.Y.: Glencoe.) <Q> Certainly could be. <S> There is nothing special about spectrum analyzers. <S> Edit: let's work through your block diagram. <S> Input attenuator. <S> Passive component. <S> Shielded for dynamic range. <S> This will not produce any EMI. <S> Input amplifier. <S> Active component. <S> Shielded for dynamic range. <S> This should not produce any interference unless it is perhaps badly misbehaving. <S> Mixer. <S> Could be active or passive. <S> Shielded for dynamic range and LO leakage. <S> Could let out some of the LO frequency if not shielded correctly. <S> Sweep generator. <S> Active component. <S> May or may not be shielded. <S> Could possibly produce low frequency EMI, especially if it is driving the coil of a YTO. <S> Local oscillator. <S> Active component. <S> Shielded to prevent LO leakage. <S> Should not produce EMI if shielded correctly. <S> Filter. <S> Passive component. <S> Shielded for dynamic range. <S> Should not produce EMI. <S> Detector. <S> Passive nonlinear component. <S> Shielded for dynamic range. <S> Should not produce EMI. <S> Video amp. <S> Active component. <S> Shielded for dynamic range. <S> Could possibly produce low frequency EMI. <S> Your diagram is missing some very important components, though. <S> Namely, power supplies and control circuitry. <S> These components are less likely to be as shielded as the RF signal chain and therefore are likely to produce far more EMI. <S> Here's something interesting to note, though: user123753 mentions using a spectrum analyzer to measure it's own EMI. <S> If you try that on a spectrum analyzer that has a leaky LO, you will not be able to see it on the analyzer! <A> Spectrum Analyzers can definitely cause EMI. <S> In fact, if the EMI is high enough and within the SPA's frequency range, connecting an antenna to the SPA's input will allow you to see the EMI on the SPA's display. <S> A low noise preamp between the antenna and the SPA input can further aid in letting the SPA examine its own emissions. <S> I've personally done this numerous times. <A> Note that in a recent design, a spectrum analyzer is a lot more likely to look like: attenuator/amplifier, ADC, microprocessor doing an FFT, and an LCD for output. <S> Depending on the signal levels you care about, the attenuator/amplifier section may well be built into the ADC chip. <S> That doesn't eliminate EMI sources (by any stretch of the imagination) but it does mean that a lot of what you're dealing with is entirely different sources. <S> Digital signals are doing their best to be square waves, so they usually have quite a bit of energy well above the rated clock speed of the bus (or whatever) involved.
| Any piece of electronic equipment could potentially be a source of interference.
|
Do 110-250V power supplies waste power at higher voltages? I'll provide a standard Mains->USB adapter as an example. We know that the output of the USB will remain the same no matter what voltage between the allowed range is applied at the input. If we're using this adapter in the UK, then we'll be applying 230v across the input. Are we wasting any power at this voltage compared to using the device at a lower voltage, say 110V? <Q> You talk about transformers and power adapters as if they're the same but they're not. <S> An (oldfashioned) transformer based power adapter will be large and heavy and usually only suited for either 110 V or 240 V AC. <S> Not both unless there is a setting for it which selects a different tap on the transformer. <S> Modern power adapters are quite different, much smaller and lighter in weight (for the same power rating) and can handle a wide input voltage range like 80 V to 240 V AC. <S> These adapters are isolated switched mode power adapters and contain a very small transformer operating at a high frequency. <S> Your USB adapter will undoubtedly be of the second (switching) type. <S> Does it fit in your pocket easily ? <S> Then it's a switcher. <S> It depends on the design of the switched mode power adapter if it is more or less efficient at 110 V or 240 V AC. <S> The design could be optimized for 110 V and thus be less efficient at 240 V. Or the other way around. <S> There is no general truth here. <S> Maybe you think that the excess voltage when using 240 V instead of 110 V would be "burned off". <S> The way that such switching converters van handle such a wide input voltage range is a result of the way these converters work. <S> The electrical energy from the input is converted into magnetic energy in the transformer and then back to electrical energy again. <S> A transistor at the input side of the transformer switches <S> the input power (at 100 kHz or so) and thereby determines the amount of power going into the transformer. <S> So no more power than what is needed (at the output) is fed into the transformer ! <S> This is a very efficient solution to control the power and thus voltage at the output. <A> The switching currents are lower at higher voltages (as they must be for the same output power), so conduction losses will be less at higher input voltage. <S> Supplies designed for only 120VAC input often have an input doubler to produce a 300V rail, however 600V is a bit too high for comfort <S> so that's rarely, if ever, done for supplies that can handle 240VAC input (including wide range input supplies). <S> The lowest efficiency of a switching supply (for a given output power that is within normal operating range- <S> say more than 10% of rated output power) is typically at the minimum input voltage . <S> For a given input voltage, the efficiency usually peaks for a load current somewhere in the normal operating range (it is a convex curve with a maxima). <S> To make this clear, however, this is a second order effect, the efficiency might be (for a certain load) 80% with 240V in and 75% with 120V in. <S> There should not be an enormous difference, assuming the supply is well designed (all bets are off with counterfeit junk). <A> The Mains to USB adaptor almost certainly is a switch mode power supply and probably contains a circuit similar to this <S> If it were a simple transformer it would be heavy and either get really (dangerously) hot when used with 250Vac or would have a switch to wire two primaries in parallel when used with 110V mains and in series when used with 230V mains. <S> This type of circuit is called a Flyback power supply and works by the switching device turning on and current builds up in the primary. <S> When the switch turns off the transformer flies back and current can flow into the secondary. <S> The transformer and opto-isolator provide isolation for safety reasons and the transformer is switched at relatively high frequency (high tens or low hundreds of kHz) to minimise the size, and cost, of the transformer needed. <S> This type of circuit is usually more efficient with high mains than low, meaning it will waste less power at high mains. <S> That said its highly efficient at any mains. <S> The control IC need not be a TOPSwitch: many manufacturers make ICs for this.
| Well it isn't, switched converters handle this in a more efficient way so that only a small amount of power is lost with any input voltage and any output current.
|
2 adapter plugs melting, fire hazard, clueless to why Any forensic skills here? Trying to figure out if this fire hazard is due to crappy adapter plugs or difference in voltage/hertz in Japan. Got this plug in China (cheaply...) where they use 230V/50hz, and have been using it in Japan where they run 100V/60hz to connect a (sturdy Chinese) extention cord, but yesterday when using an electrical kettle (Japanese so made for 100V/60hz, but pretty high W), then this happened. I didn't see it so I tried with an identical plug and saw electrical arcs from inside the plug, seemingly between the two connection points of the prong and connecting plate. Does anyone know what happened? Update: The kettle is 1250W, and the adapter plug lists 10A ...so with the help of the forum the problem was seemingly P/V=I , 1250/100=12.5A, which was just too much for my crappy adapter plug. Thanks! <Q> There are a few factors here. <S> Kettles are high power loads, often close to or right at the limit of what the domestic socket standard in the country in question can safely supply <S> Items bought cheaply in china <S> are likely to be of low quality. <S> Your adapter looks like the "pin" part of the adapter is only connected to the socket part by pressure from the case. <S> Put it all together and you have a recipe for failure. <S> Such adapters are sometimes a necessary evil when travelling <S> but I would strongly advise avoiding their use for long term or high current applications. <A> It seems you exceeded the rated current with 12.5A on 10A rating which was not shown on your original question. <S> With additional contact force on the receptacle, more current can be transferred with less heat loss at the expense of increased plating wear. <S> Some Chinese outlet plugs and sockets do not provide the necessary friction fit with tension and possible <S> are too smooth. <S> The result is low contact area and high contact resistance. <S> Instead of <0.05 Ohms it might be > 0.1Ohms and <S> thus heat loss at 12.5A is 15W, which can heat up a small contact area pretty high with no air flow... <S> enough to melt plastic eventually like a soldering iron tip. <S> Some plugs have a circular hole in each blade in North America to scrape off carbon or oxide each insertion. <S> Ensure plug and receptacle is rated for appliance in future, because it is the receptacle contact force and contact area that enables higher current. <A> I have seen this before on mains power plugs running hot. <S> I opened one up and the wires were loose.
| The socket holes on that adapter look to be some kind of multi-standard compromise, accepting a bunch of plug types, but none of them well. Possible that one or more wires was not properly secured inside the housing, leading to a reduced cross section for current flow.
|
Reducing the Voltage of Car Battery I've been looking around for an easy way to convert car battery 12v to 5v. I have seen some people saying that a simple resistor is all that is needed. What I've tried so far is the DC to DC regulator. It works, but it's pretty expensive for such a cheap project. And not to mention it is twice the size of the project. Following the ohm's law, it should mean that a single resistor with appropriate wattage rating should be fine to convert the car battery into a 5v supply. But the project used a max of 350mA, will the resistor or the project burn? EDIT: The project will run even with ~100mA supply, but at some points it could go way up to ~350mA. Haven't tested it thoroughly but atleast 500mA would be sufficient. And yes, since it's a Microcontroller, a stable 5VDC would be preferred. And about destroying the OBD2 ports... Really I never had any thoughts about it. <Q> A single resistor is not appropriate. <S> The voltage a resistor drops is proportional to the current thru it. <S> Even then the resulting voltage will vary with the input voltage. <S> At 350 mA out, a resistor or linear regulator will dissipate a lot of heat. <S> When the car is running, figure the input voltage could be as high as 14 V <S> (13.6 V is a common value). <S> That means the linear pass element will drop 9 V. <S> At 350 mA thru it, it will dissipate 3.2 watts. <S> That's going to require some space and expense one way or another. <S> That's too much, for example, for a TO-220 in free air. <S> The forced air cooling or extra dissipation surface will be big and expensive. <S> The best answer is a buck regulator. <S> These are much more efficient, and are therefore smaller and cheaper since they don't have to deal with getting rid of all that heat. <S> There are many commonly available chips from a number of manufacturers (Microchip, ST, TI, Linear, etc) that come with the controller and switch integrated. <S> You add the inductor, input/output caps, and a few extra external parts. <S> A properly designed buck solution will be smaller than anything that can safely dissipate 3 watts. <S> Consider that car power can have a few 10s of volts spikes on it occasionally. <A> Convenience stores and discount retailers sell these things for as little as $1, they are called USB Car Charger Adapters. <S> They are made to plug into the cigarette lighter port and provide output to a USB jack, but you could adapt it for your purpose. <A> Firstly I want to make it clear that I agree with Olin that a DC-DC converter is the way to go and that DC-DC converter should be a Buck regulator. <S> But in the interests on completeness assuming you want to go with a linear solution you will still need to dissipate 3.2W maximum but this does not all need to be in your TO-220 regulator. <S> If you put a 12 ohm 2W resistor between the battery and the regulator you can reduce the dissipation in the regulator but this will still require a heatsink but you can get away with a smaller one. <S> Calculations for this: <A> try a switching regulator with a small footprint. <S> something like MURATA's OKI-78SR-5/1.5-W36H-C <S> It costs only 4$ and has a very wide input (7V to 37V) which may allow it work when cranking or when a load dump happens
| You need to get a buck switcher with a sufficiently high maximum input voltage, or put some kind of clamp in front of it.
|
Why does the die of a desktop CPU need to be so small? The amount of space on the PCB would allow for the die to be much larger and would make designing it much easier, what is the reason for this? <Q> Cost is proportional to die size, so for that reason the die is usually optimized to be as small as possible to meet the performance requirements. <S> There are other reasons as well, like routing very high speed signals, but cost is the main driver. <A> Yield! <S> Defects are unavoidable on a wafer. <S> Imagine that a number of darts are thrown at it during the production process. <S> Where they hit, it fails. <S> Larger square dies are more likely to be hit, so you get fewer usable chips per wafer. <S> Newer process suffer more of the slings and arrows of outrageous fortune, making more unusable areas. <S> If the chip has large regular structure on it, you can sometimes work round this by disabling parts of the chip. <S> More often used on GPUs, which have bigger dies anyway. <A> what is the reason for this? <S> I can't say for sure that what I've answered is the reason <S> but it's a pretty significant one... <S> So if you just made everything bigger by 2, parasitic or gate capacitances would double in value and speed performance would halve at high speed. <S> Power consumption at high speed would also double because of the number of times per second each gate capacitor is charged and then discharged - this energy is totally lost. <S> Capacitance is proportional to area so double the dimensions, quadruples the capacitance but, of course the thickness to the substrate would double and this halves the value of C.
| The number of die you can get per wafer is extremely important in a high cost process like those used for desktop computers.
|
Can I use a coaxial cable as a DC power cable (without using it as a signal cable)? On my balcony I have a satellite dish and two coaxial cables attached to it. I don't use that dish, and I want to remove it and reuse those coaxial cables to power some other device on the balcony (a kind of IoT stuff). Can I send DC power over that coaxial cable? If yes, then can I use a single cable and its centre wire as positive and outer wire mesh and foil as ground? EDIT: my dish has LNB sign on it if it matters: <Q> Yes, this should work fine. <S> Even in its original application, the coax was probably carrying a DC bias to power the LNB on the satellite dish. <S> Assuming it's RG-6 cable, it should be good for at least 3 amps. <S> It's considered conventional to use the shield as ground and the core as +V, though it doesn't particularly matter from a functional point of view. <A> RG-6 coax has a 18-gauge center conductor, and exhibits a resistance of 6.385mOhm per foot. <S> Its insulation is suitable for a max of around 10A, but it may get too hot before that. <S> Typical satellite TV operates at voltages between 13v and 18v. <S> At 18v, a 200ft long RG6 will allow up to ~14A. <S> However at 10A, the voltage drop is about 7.8v, so your mileage will vary. <S> That is after all what it does in a satellite TV setup. <S> The set-top box powers the LNB through the coax. <A> Yes, you can. <S> In fact, using coaxial cables to deliver power to small-scale electronics is a common engineering practice, with all corresponding approvals (EA, UL, etc). <S> You can combine DC on coax even with AC signal transmission if you need. <S> All TV cable amplifiers (usually located on attic) use this method of delivering DC over a single coax cable. <S> Using outer shield of coax cable as negative end of supply makes a lot of practical sense since it is frequently exposed at connectors and junction boxes, and it is advisable to maintain it as common ground.
| So, yes, it is entirely suitable to carry power.
|
How dangerous are disconnected bulging batteries? It is well understood that if you have a device that uses a Lithium-Ion battery, and if the battery starts to bulge, then something is wrong with the battery and you should properly dispose of it by taking it to a battery disposal/recycling center. Improper handling could result in a fire being started by the faulty battery, so it's best to dispose of the device and have it replaced. However, I am curious when the battery begins to deteriorate and starts to bulge, is it still dangerous when it's removed from the device and set aside? To be clear, none of the metal leads are coming in contact with any metallic, or conductive material, so the circuit is not complete on the battery. It's just sitting on a non-conductive, safe surface. Context I bought a piece of hardware yesterday at its only problem is that the battery is bulging. I've taken out the battery and I want to keep it until I can find an OEM replacement, which could take a few weeks to arrive from the distributor. I want to keep the old battery so that I can compare it against the replacement which, I'm afraid, might be a KIRFy piece of hardware. I'd want to return it and get a refund if that were the case. Keeping the old battery around for a while does have it's purpose. <Q> I don't know, and I suspect there are not any solid numbers that will answer your exact question. <S> If it's bulging there is clearly pressure inside, but the risk is not obvious. <S> However, there is a way to make the battery less dangerous. <S> The worst that it can do is burst into flames. <S> You can find videos illustrating pretty much the worst that is possible. <S> So keep the battery in a place where, if that ever occurred, serious consequences would not result. <S> For example, inside a metal box of some kind with no flammable materials nearby. <S> Don't store it in a cardboard box on a shelf full of papers, for example, rather treat it more like a thin vessel filled with gasoline or a container of gunpowder. <S> By the way, lithium cells are not allowed to be shipped by air in a good many instances, so this kind of thing tends to take longer than you might otherwise expect. <A> Exactly what are you going to compare to the replacement? <S> Can pictures or numbers or some kind of mock-up be used instead? <S> Unexpected fires are bad. <S> Bad things can happen <50,000 youtube videos could be linked here <S> > ; the thing might start burning pretty hot and long, and can skitter about while it throws a couple feet of flame. <S> Is your "safe surface" still safe? <S> Even so re-read your fire insurance and check that it's up to date. <S> Bad things often find a way of getting worse. <A> On a lithium ion cell that has been overheated or mechanically damaged, self-discharge through internal shortings can be in progress. <S> This can lead to thermal runaway at any time, even if the cell is not used/connected/moved/heated/whatever. <S> Such a thermal runaway will typically annunciate with a self-heating but you are not safe if the cell is dead cold now. <S> This may change within minutes! <S> Do yourself a favour and keep such a cell in a safe location. <S> A metal case on a non flammable surface will do. <S> There are few consumer goods that contain materials with the potential to self-ignite from cold temperatures. <S> In fact, chemical fire starter is a safer material than your phones battery. <A> If you have access to electrical tape, carefully tape the contacts to prevent any accidental short circuiting. <S> Place it in a non-sealed container outdoors, ensuring it cannot be punctured. <S> If it it stored in a sealed container, and the battery starts releasing gases, pressure could build up and cause an explosion. <S> Also perhaps photograph the battery incase it catches fire before the other one arrives (it shouldn't, but it's better to be cautious), so you can still make a comparison between the old and new batteries. <A> I feel that there is paranoia about Li batteries. <S> This is the result of the news reports of various fires/explosions attributed to these batteries. <S> However, when you take into consideration the millions of them that are in use, it seems to me that they are very safe. <S> I would recommend that you put the battery in a metal case, about 10 times the volume of the battery, and place the case in the shaded part of your backyard. <S> If there is no shade on your backyard, make some by using cinder blocks and a piece of plywood. <A> Its probably harmless if its stable (not heated, deformed, charged or short circuited). <S> If by some chance it does overheat, be extremely wary of the fumes. <S> Its possible that hydroflouric acid can be produced - this is very nasty even though the immediate effects are not noticable. <S> The risks are low, but its better to be aware.
| No need to worry unless it is damaged, but if it is ... be careful!
|
Why do receivers have a choke but transmitters don't? Here's a receiver I found on the web: And here's a transmitter: [ Why do receivers have an inductor specifically designated as a choke (RFC) in their circuits but transmitters don't? <Q> Why do receivers have an inductor specifically designated as a choke (RFC) in their circuits but transmitters don't? <S> Your assertion is incorrect. <S> Chokes / RFCs are used in transmitters and receivers when the designer thinks they are appropriate. <S> Providing a single example of each does not make your point well. <S> The purpose of an "RFC" is to block the passage of RF energy but to allow DC or low frequency AC to pass essentially unimpeded. <S> This will never be achieved perfectly but the (usually) large difference in frequency of audio and RF means that an unductor can have a large impedance at RF frequencies and minimal impedance for audio, and also have a low DC resistance so DC current causes minimal voltage drop. <S> ____________________ <S> _ <S> Example <S> Here is a classic "RF final stage with choke" diagram. <S> 12 VDC is connected to the transistor collector via the RFC. <S> The RFC prevent RF "escaping" into the power supply. <S> This circuit is intended for 88-108 Mhz operation. <S> At say 80 <S> Mhz the 1 uH RFC has an impedance of about 550 Ohms. <S> The following 0.1 uF capacitor on the 12V supply has an impedance of about 0.02 Ohms <S> so the RF reaching the 12V supply is "divided" by a factor of about 0.02/550 = <S> a factor of about 25000. <S> (It's more complex than that, but that gives a guide. ) <S> From here <A> The purpose of a Radio Frequency Choke is to 'choke' ie. <S> restrict radio frequencies from getting into parts of the circuit where they are not wanted. <S> It does this by making use of the property of inductance , which causes the choke's impedance to increase as the frequency increases. <S> RFC's are not restricted to receivers (transmitters often have them too) and not all receivers use them. <S> Whether an RFC is necessary depends on the particular circuit configuration and how the designer decided to implement it. <S> The RFC in your superregenerative receveiver actually has two jobs. <S> As well as stopping RF from getting into the audio amplifier, it also raises the impedance at the Emitter of Q1, so the signal from the antenna won't be shorted to Ground by the 0.001uF capacitor (that capacitor is required to remove the ultrasonic quench frequency which is used to maintain superregeneration). <S> Your example FM transmitter does not need an RFC because audio is fed into the Base of Q1, which does not have any RF on it (ensured by C8, which shorts any RF present to Ground). <S> L1 could be considered to be an RFC, except that in this case it forms a tuned circuit in combination with VC1, C9 and C7. <S> However, many transmitters instead use a Pi filter between the output transistor and the antenna, and therefore do require an RFC to stop the RF signal from being shorted out by the power supply. <S> Here is an example:- <A> A radio frequency choke is a way of providing a low DC impedance to a circuit node (possibly to allow DC current to flow) whilst simultaneously ensuring that an RF signal present on that node does not become significantly attenuated. <S> The RFC works across a wideband of frequencies but at a high enough frequency will become a resonant parallel LC circuit due to its parasitic capacitance. <S> At higher frequencies it will progressively behave as a capacitor: - Compare this with L1 in the transmitter circuit - it is tuned into parallel resonance because of VC1 and C9 but, other than that, it behaves exactly the same as an RFC - it allows DC current to pass to the BJT (Q1) and offers a high impedance to RF signals (on the collector) at the operating frequency. <S> It will (should or must) have a self resonant frequency (due to parasitic capacitance) significantly higher than the operating frequency of the circuit so that tuning with VC1 and C9 is not compromized <S> but, other than that, it is a RFC.
| Many transmitters use an RFC to feed DC to the "final" amplifier while blocking RF.
|
Choosing capacitors for a linear voltage regulator I'm trying to use an LM1117 linear voltage regulator to convert to 3.3v (input voltage will be 9 or 5 volts; not yet decided). The datasheet suggests using 10uF tantalum capacitors on the input and output. While I could just go with the suggestion, I find most of the tantalum capacitors that are available are considerably more expensive than other capacitors, and in a SMT form factor (I'd prefer through-hole). I'm thinking of using aluminum or ceramic caps instead. This excerpt from the datasheet specifies what ESR ranges are acceptable: 8.2.2.1.3 Output Capacitor The output capacitor is critical in maintaining regulator stability, and must meet the required conditions for both minimum amount of capacitance and equivalent series resistance (ESR). The minimum output capacitance required by the LM1117 is 10 µF, if a tantalum capacitor is used. Any increase of the output capacitance will merely improve the loop stability and transient response. The ESR of the output capacitor should range between 0.3 Ω to 22 Ω. In the case of the adjustable regulator, when the CADJ is used, a larger output capacitance (22-µF tantalum) is required. As I understand it, ceramic caps have a very low ESR. Could I just add a 0.3 Ω to 22 Ω resistor in series with a ceramic capacitor? I've read what ESR is and isn't, and I don't understand it one bit, so right now it's just a range of numbers I'm trying to match. However, ceramic caps tend to have smaller capacitance, so it might not be as easy to get 10uF ceramic caps. If I recall correctly, aluminum caps have a much higher ESR (though exactly what "higher" means I'm not sure). Additionally, it can be difficult to get a datasheet for a capacitor simply because they're such common components (I'm in China and purchasing online, but don't read Chinese). So I'm not sure how to make sure I'm within the ESR range specified. Finally I should note that I don't have any of these components yet, but I'll probably get an assortment of capacitors when I do order. So my question: What capacitors can be used in this situation, and with what value resistors (if any)? Note that I'm interested in the theory, not just this single situation, but not being experienced with capacitors some specific examples could be helpful. Related: ESR and CSR of Capacitor <Q> Yes, you can simply put a resistor in series with a ceramic capacitor. <S> The lower the better from the point of view of bypassing, so I would aim at 0.5 to 1 ohm. <S> If you have lots of space, the electrolytic is fine (in fact you can parallel the two), and they are cheap. <S> There are low ESR electrolytics and ones that are not-so-low, read the datasheet. <S> If no datasheet, no buy. <S> You should be able to read the numbers on the datasheets even if some of it is in a foreign language. <S> If you're going to the market and picking shiny parts off of vendors displays without looking at datasheets you will get bitten. <S> I've always been able to get answers to questions such as the load capacitance of a crystal without being the most amazing linguist around. <S> The x1117-3.3 is extremely cheap and very available in China <S> so I don't see any reason not to use it. <S> If you don't need the power dissipation there are better choices in SOT-23. <A> Yes, you can add deliberate resistance to a low ESR cap to replace a tantalum cap in this case. <S> However, a much better answer is to use a more modern regulator that is stable with a 0 ESR output cap. <S> A LM1117 makes no sense here, since you don't need really low dropout. <S> Even then there are plenty of modern LDOs available that are 0-ESR stable. <S> Use the right linear regulator, and this problem goes away. <S> If you are really going to use 9 V in, then even a 7805 would be fine here. <S> They are very robust and tolerant of significant capacitance on their output without any resistance in series. <S> After the next asteroid impact, the only thing left will be cockroaches and 7805 regulators. <A> Increasing capacitance value gives better transient response and allows use of a lower ESR capacitor while keeping the regulator stable. <S> For this kind of regulator, aluminium electrolytics work fine unless you use the low-ESR kind. <S> I usually use no-name 105°C 100µF 25V caps which cost 4 cents a piece. <S> They have an ESR around 0.5 ohms. <S> (should be 200-400 µF) <S> not specified for low ESR or low-Z brand name or 105°C if you care about service life. <S> It'll work.
| Basically, for this 3V3 regulator, grab an aluminium electrolytic through-hole cap with the following charatceristics: small and convenient package (6.3mm diameter) largest capacitance value you can find for 6.3V rating
|
What happens before main()? I am working on embedded systems as beginner and have come across files like start.s or cstart files that run before main() function begins. What is the purpose of these or similar files? What information we are telling the system? I've heard of initialization but don't know exactly what that is. <Q> It is completely dependent on the compiler and architecture, but generally that code initializes the most basic hardware required for the rest of the code to run. <S> The code for example: Defines the reset vectors Defines the layout of data in memory (many systems use a linker script instead) <S> Defines the addresses of interrupt service routines in a big table (the interrupt vector table) <S> Configures <S> the core clock <S> In addition, that section also serves the runtime needs of the programming language used. <S> It: <S> Initializes whatever function parameter passing system used Initializes global variables by e.g. copying flash contents to RAM and zero-initializing memory <S> If dynamic memory allocation is used, initializes the heap <S> If floating point math is enabled, initializes the FPU (if available) or initializes the floating point library <S> If exceptions are used, initializes exception handling. <A> Somewhat related question: <S> Who receives the value returned by main()? <S> main() is an ordinary C function, so it requires certain things to be initialized before it is called. <S> These are related to: Setting up a valid stack Creating a valid argument list (usually on the stack) Initializing the interrupt-handling hardware Initializing global and static variables (including library code) <S> The last item includes such things as setting up a memory pool that malloc() and free() can use, if your environment supports dynamic memory allocation. <S> Similarly, any form of "standard I/O" that your system might have access to will also be initialized. <S> Pretty much anything else is going to be application-dependent, and will have to be initialized from within main() , before you enter your "main loop". <A> On a typical embedded system, startup code will at minimum will have toload all initialized variables with their defined values and zero out alluninitialized variables. <S> Depending upon the hardware platform, it mayalso have to configure the CPU stack pointer [on some hardware platforms, areset will automatically set the stack pointer to the top of memory, but onother platforms it must be set manually] or configure various other featuresin the CPU or memory controller. <S> The startup code is usually pretty short and simple, and some platforms maydocument how it works and allow a user to substitute something else (e.g. ifan embedded system will need to have a user-supplied startup routine copysome code from a serial flash chip into RAM and then execute it, it may makesense to have initialized variables be part of the code image, rather thanhaving their initial values be part of the code image which is copied toanother area of RAM on startup but then ignored thereafter). <A> In a traditional GNU/Linux desktop toolchain, most of such code is implemented by glibc and is contained in objects such as crti.o , you can get the full list with gcc --verbose main.c . <S> Basically, glibc, is setting things up so that further glibc calls will work properly. <S> Therefore, you can learn what is happening by reading the source of glibc, e.g. sysdeps/ <S> x86_64/crti. <S> S . <S> TODO: <S> how to step debug that code with the source? <S> Modifying those objects itself is a major pain unless you know dark arts .
| Initializes CPU registers, e.g. the stack pointer
|
How will supply ripple affect OCXO performance? To feed 3.3V/1.5A to my IQOV-164 , I wonder if I should go for a low noise LDO with 6uV RMS noise or, to the other extreme, if I can simply feed supply voltage from a step-down converter with 20mV ripple. Design goal is good short-term stability, long-term is not a concern (I am aware that IQOV-164 may not be the best choice for this objective). <Q> Edit based on Spehro's answer- After re-reading the datasheet <S> I think it's correct to say that the 50% is 50% of 50ppb. <S> My original interpretation doesn't seem like it would be a very practical product. <S> However, the recommendation for an LDO still seems valid. <S> I would say based on the wording below in the datasheet that this part is very sensitive to supply variation: Supply Voltage Variation (measurement referenced tofrequency observed with TA=25°C, Vs varied from 3.13V to3.47V, VC=1.65V/NC and load=50Ω/15pF): <S> ±50% of frequencystability <S> There is likely a frequency dependence to the supply rejection, so we can't assume that the DC rejection ratio applies to switching supply ripple (it could be better or worse) but by extrapolation 20mV*147 %/V = about +/-3% frequency variation. <A> From the datasheet I see: Assuming a linear supply variation response, you could get 0.5/(3.47V-3.13V)*20mV <S> = 0.029 or about 3% variation. <S> This sounds like a lot of short term stability variation to me. <S> I would recommend an LDO. <S> I am curious however, where did you get the ripple spec for an LDO? <S> It's not a switching circuit. <A> I interpret the datasheet as saying the variation is worst-case 25ppb (50% of 50ppb) for variation between 3.13 and 3.47V. <S> That's not a whole lot, but I guess if you are using an ovenized oscillator you do care about stability. <S> It is, however, typical that the sensitivity to high frequency noise will be much higher than low frequency <S> ('DC') changes. <S> At a minimum I would expect more phase noise. <S> Here is a brief app note from another maker- they do suggest that LC filters may be required in the power supply and that good layout is important. <S> 20mV is pretty bad. <S> 6uV RMS (maybe 40uV p-p) is very low. <A> Considering the time to heat up OCXO to operating temperature, any ripple even if 10% >1000Hz would have no effect on temperature. <S> Consider power dissipation is a function of DC not AC ripple and thermistor feedback <S> regulates the OCXO temperature will be fast <1kHz <S> so any noise above this has no effect. <S> But if DC steady state is stated as Say the OCXO stability is rated at 1e-9 (1ppb) and 50% of this is 5e-10 <S> Then the DC nominal (3.47+ <S> 2.13)/2= <S> Vnom=2.8V <S> The difference is (3.47-2.13)=1.34V <S> The Sensitivity is then 5e-10/1.34 or 37 [ppb/Volt] for DC <S> The sensitivity for Fout above <S> x kHz Vs ripple is null. <S> ( except for phase noise perhaps ) - The assumption is thermal shift is zero using assumed large thermal time constant of at least 30 seconds measured over xx microseconds of 1 ripple cycle. <S> warmup time is 3 minutes max.
| I'd go with a good low-noise LDO.
|
Why diameter of telephone cable is exactly 0.44mm and 0.63mm I am wondering why telephone cables are having the diameter or exact 0.44mm and 0.63mm? Is that any special reason? I have seen some of the cable have up to 2400 twisted pairs, could I know is the number of pairs will affect performance of cable? Thank you. <Q> This is achieved by a side-tone cancellation circuit: - <S> When the line impedance matches Rw in series with Cw there is no microphone signal transferred to the earpiece connected to winding A. <S> The microphone feeds the centre tap between S and P windings. <S> So, for this to work effectively <S> , the impedance of the telephone line has to "look like" Rw in series with Cw across the range of useful speech frequencies. <S> Here's what a cable impedance might typically look like: - As you should be able to see, the nominal impedance at 1 kHz is about 600 ohms in magnitude but is in fact similar to the complex impedance mentioned above. <S> The characteristic impedance of any cable at low frequencies is \$\sqrt{\dfrac{R}{j\omega C}}\$ where R is the series resistance of the loop per metre and C <S> is the parallel capacitance per metre. <S> Any change in conductor spacing or diameter means the capacitance per metre will change and the approximation of Rw in series with Cw won't be as ideal. <S> Capacitance for twisted pair for example: - \$C_{twistedpair}\left ( \frac{pF}{inch} \right ) <S> = <S> \left <S> ( \frac{.7065}{\ln \left ( \frac{2s}{d} \right )} \right ) <S> \cdot \epsilon_r\$ <A> The diameters are almost certainly for historical reasons. <S> The sizes you've given correspond to 1/40 inch and 1/60 inch. <S> Probably these looked like good round numbers to someone who was making the decision a long time ago, and we've stuck with them ever since. <S> Any new cable would have to have the same characteristic impedance as the old, fit into the same connectors, have the same or better bend radius etc. <S> More pairs, more signals. <S> More pairs can slightly reduce the performance of a cable due to crosstalk, but it's generally worth it to carry the extra signals. <A> In order to maintain a specific impedeance all properties of a telephone cable are specific. <S> So the wire diameter,the wire distance to each other, the insulation thickness and type. <S> The way of twisting and so on. <S> This results from the past in the thickness indicated
| The characteristic impedance of a telephone cable is very important to get right when it comes to minimizing side-tone (that's the handset microphone producing an amplified sound in your ear). Each twisted pair in a cable carries a signal.
|
What exactly happens at short circuit? I'm new to electronics and I don't fully understand how short circuiting works. For example I have this battery LiMn 3000mAh, it states: Nominal Voltage: 3.6V Standard Charge: 2A Continuous discharge current: 20A First thing I don't understand is if I connect a wire between the positive and negative terminal, it would create a short circuit, correct? Can the battery blow up? My goal is to simply make a hot wire, but I don't understand how will the wire 'determinate' how much current it needs or will it draw full 2A or even 20A from the battery? If I use resistors, what's their purpose? to protect the battery from all the 'unused' electrons or to protect the wire? What about taser gun? is it 'short circuiting' and what effect does it have on the power source? <Q> Battery specs are often stated assuming a constant average current over the span of decreasing voltage range. <S> All batteries have an internal resistance that defines what the short circuit current is which is never specified since the internal heat can be dangerous ( explosion) but can be estimated from a load test from say 1A to 20A std discharge 3.6V/20A= <S> R = <S> 180 mOhms Using AWG calculator or table use AWG 20 to 30 and determine length of wire needed to make 380 mOhms and coil it up and you have a heater or Igniter wire which gets very hot. <S> Better wire is NiChrome or "heater wire" which is high resistance in a shorter length. <S> Exactly what do you intend to blow up with this wire? <S> ;) <S> everyone is watching you ;) <A> A short circuit is, generally, just an unwanted connection between two or more points in a circuits, permitting current to take a shorter path than the deisgner intended. <S> A short circuit between power supply leads will cause a large current to flow. <S> The current will be limited only by the power source's internal resistance, and the resistance of the wires carrying the short-circuit current. <S> If the wires, printed circuit tracks, or other components carry excessive current, they may overheat, melt insulation, burn the PC board, or otherwise cause the circuit to emit the magic smoke that electronics depend on. <A> 1) Yes, connecting a voltage source directly to its return with a wire creates a short circuit.1b) <S> Yes, discharging a battery at too high a current draw will overheat the battery and can result in a catastrophic failure (blowing up, boiling electrolyte, fire, other bad things). <S> 2) <S> To just heat a wire, ideally, you would connect it to a controlled (regulated) current source. <S> Then you can adjust the available current to control the heating. <S> Most mid-range and up bench power supplies have a current limit mode. <S> 3) Resistors server many purposes, you will need to be a little less broad. <S> Resistors can limit current, but at the cost of heating up. <S> They won't "Protect" the wire actually, just reduce the current available, and make the resistor a heater. <S> 4) <S> A Taser generates very high voltage pulses, at a very very low current. <S> There is a lot of circuitry between the taser contact points and the battery... <S> Again, really too broad to answer here.
| If a short circuit occurs between two signal lines, it probably won't cause a large current to flow, but it will prevent the circuit from operating correctly.
|
Raspberry pi with two USB 2.5 inch Hard drives and powering them in parallel via one 5V PSU I'm a hobbyist/tinkerer and want to make sure I don't fry my equipment by doing it wrong. I'm trying to connect two/three 2.5 in USB HDD's to a Raspberry Pi 3. The power supply I have is 5V 2.5A and I plan to upgrade that to a 5V 4A or 5A whichever is available/needed. (At startup each drive can take between 600 to 1100 mA and then falls back down to 400 mA during normal operation.) I imagine that just by providing higher PSU to Pi will not suffice and I should look into separating the Data and power of my HDD's so that the Data is still connecting to Pi, but the power is provided by the PSU directly. To oversimplify - I'm thinking of taking the 5V 4A PSU and using a Y connector splitting it into two. 1. goes to Power the Pi.2. goes to power the HDDs. My question is:- Is this the correct way to do it ?- What kind of circuitry do I need to build so I can power these reliably? Thanks in advance. <Q> Raspberry Pi (all variants) are not well-engineered for correct power delivery on USB ports. <S> Your best bet is to use a good quality self-powered (via a wall adapter) hub, preferably USB-IF certified, and <S> USB 3.0 type (for higher supply capability). <A> To oversimplify - I'm thinking of taking the 5V 4A PSU and using a Y connector splitting it into two. 1. goes to Power the Pi. <S> 2. goes to power the HDDs. <S> You could do that <S> and it would work. <S> So if a component fails in short-circuit mode, it can draw the whole 4A (through the 5V line of the HDD's USB cable) and potentially start a fire. <S> You can eliminate that risk by cutting the 5V line in the USB cables leading from the Pi to the HDDs. <S> However, do not cut the GND line, otherwise your HDDs won't have a clean ground reference (only indirectly through the PSU) and communication will likely fail. <S> So to summarize: Connect 5V and GND from PSU to Pi. <S> Connect 5V and GND from PSU to HDDs. <S> Connect D+, D- and GND from Pi to HDDs, but not 5V. <A> I'd be modifying the raspberry pi. <S> soldering new wires for +5 and 0V to the underside of the USB sockets and then connecting those wires to a 5V power supply. <S> you loose the protection given by the polyfuse <S> so a fault on the GPIO header could burn up tracks on the raspberry pi, <S> a fault on the USB is most likely to cause the power supply to shut down (so long as it's a reasonable size, and not a 20A monster). <S> this power supply would run the USB peripherals and the raspberry pi itself.
| However, you would bypass the fuse between the Pi's USB-B jack and its 5V power rail.
|
Charging battery simultaneously from 2 sources This is my first time posting here.I have an issue, I want to charge battery from 2 sources simultaneously, but I am kinda a worried about how that might work.If I add the voltage feedback directly to battery then one charger might trick other charger. I guess I could put diodes, but I have a feeling this might make me another trouble. Of course instead of diodes I can use Mosfets for lower power losses in this load sharing. Have any of you guys tried something like this? Any advises? <Q> Current shared Voltage regulators that do not sense remotely are inherently stable with Diode OR isolation. <S> ( high supply delivers the current.) <S> Other related info but same as your situation <S> Unlike above Current Shared DC supplies with shared remote sensing must be balanced within 10% typically and often require a preload of 10% to be stable due to change in loop gain high low load. <S> I recall 20 yrs ago in RB, Cali,USA (Unisys Server factory) <S> we had redundant Current shared dual supplies with major brand OEM source who had to come out and view stability oscillation for themselves to know their design failed without a 10% preload, so we added this to the Production initial HASS test process. <A> Well, if you are developing those chargers then what they need is not a voltage feedback, but rather an output current feedback. <S> This way you'd have a control loop that measures the output current and, assuming the charger is some kind of SMPS, the duty cycle would be adjusted to keep the output current constant. <S> This means that the chargers would pretty much balance out their currents without problems. <S> And by the way, some SMPSs are inherently protected against reverse currents due to their topology. <S> For instance, if those chargers are boost converters, then you don't need to be worried about one charger trying to "charge" the other one, because the boost topology already uses a diode directly at the output: <A> Why can't you diode OR the DC supplies and use one charger? <S> Advantages: <S> Most chargers can take a wide input voltage with no problems. <S> If your charger is switch-mode and DC input is higher, then the current is lower through your diodes by putting diodes on the input side. <S> Saves you the cost of another charger simulate this circuit – Schematic created using CircuitLab
| Of course, when the battery starts to get to 100% charge, you need to go into constant voltage (CV) mode, and when this happens the currents involved will typically be small in relation to what you had before going into CV mode, so it wouldn't matter which charger supplies the current, as long as the charger is protected against reverse current.
|
How to count the number of people in a room with a specific sensor? I want to count the number of people in a small room. According to my research, there are person-counters in the web but I want to use the number of people data for coding hence I must transfer the number so I found occurence sensors can be useful for this goal. However, I cannot trust these sensors much. So is there any practical way to count the number of people in a single room at any time ? What should be my steps to calculate this ? Thanks... <Q> Laser tripwire <S> You would have to install 2 wires in each doorway if you need to know if someone entered or exited. <S> One laser on the outer side of the doorway, and one on the inner side. <S> Double Pressure plates this solution will take a bit of extra engineering, but if you put a mat over a pressure plate, then you can also count how many people move in and out. <S> Something like a force sensitive resistor will be a sufficient pressure plate. <S> The problem with this one is that people may like to stand on these pressure plates for various reasons, and it may be hard to detect who is moving where. <S> Good luck! <A> The best solution would require to know more details about the setting. <S> This is the setting that I imagine. <S> It may not be the cheapest way to do, but with the following assumptions I think this would work rather well: small room, people are moving. <S> In this case you could have a ceiling camera and detect people by detecting motion on the picture. <S> Of course, if there was an big dog among the people, it could be recognized as a person. <S> Then, you could process those ROIs to see if the person is there. <A> Probably a little late, but here is my approach. <S> I placed a sound based distance sensor into my door frame and tilted it, so it measures 30° relative to the floor into one possible walk direction. <S> If someone walks through the door from the side the sensors is pointing to, you will see a sudden decrease of the distance measured until it suddenly falls back to the default distance (the floor or whatever).When someone walks through in the other direction, the whole thing reverses. <S> First a sudden decrease, followed by a slower increase, back to the default distance. <S> Not so easy to get the software to work reliable and the sensor to produce less noise, but still doable. <S> If you are interested in my software (Python), just leave a comment. <A> CO2 PPM Correlation <S> If it is a relatively closed room (e.g. no open windows or strong drafts) you could use an air quality/VOC sensor like the IDT ZMOD4410-EVK to get a CO2 ppm approximation and then correlate this to number of people in the room. <S> If you won't be developing an embedded solution maybe a Vernier CO2 sensor with a Go!Link might be a viable sensing solution. <S> The practicality of the solution lies in it being a non-mechanical method of measurement.
| While I won't go into too much detail, it is possible to count how many people are in a room using a laser tripwire in the doorway to count how many people enter/exit. For you I'd recommend a 5mW ultraviolet laser, as UV lasers are typically the safest lasers to handle, they're discrete (invisible), and photo diodes react well to them. In case it was a classroom, you could dedicate regions of interest in the image were the person is expected to be (such as a chair).
|
Hacking characters in custom LCDs Say I have a device/instrument that has custom LCD for display with mix of text and graphic/icon on the corners. I want to log the values and feed it to the computer or say automate a response when the characters display a set of numerical values. How do I hack the LCD that way? Is there even a generic approach here? Is it just a matter of tapping into the the contacts for each line or pixel (for low resolution characters)? <Q> It would be a lot simpler and require a lot less wires if you could tap in to that communication link rather than the signals going to the physical display. <A> In general you would have to figure out the backplane and segment arrangement (usually at least triplexed if there are many display elements) and drive it with the right waveforms. <S> It's possible to generate the correct waveforms (DC component should not exceed a few tens of mV and RMS waveforms need to turn the required segments on) with ordinary GPIO pins and resistor networks, at least for simpler displays. <S> All told, it's not as straightforward as figuring out an LED display. <A> Digital cameras are available these days!! <S> Just take pictures of the display as it shows the varied content. <S> Then place the camera storage card in a USB card reader and extract the photos to the computer. <S> Job done!!
| It's hard to say without details of the exact hardware being used but in general there will normally be a processor that is sending data to an LCD controller IC that then drives the pixels.
|
Undervolting PC Fan from 12V to 7V I have a 3 pin 12 V Fan, rated at 2000 RPM max. Its nominal voltage is 12 V which will make it spin at full throttle. That's what is happening when I plug it in the motherboard. However I don't like that, because the fan is making a lot of noise. I would like to lower its speed to about half. I have figured out I should supply it with 7 V instead. There are methods online here , here and even a famous video on YouTube here demonstrating the procedure. I've read a lot about this and I'm unsure about whether this is safe. I use a modern and pretty good and reliable 80 Plus PSU. So I'm skeptical now and I want to ask here to understand better. What we have to do is pretty much shown on the picture below. Instead of supplying the fan with 12 V, using the yellow 12 V pin and the black ground pin from the Molex cable, we instead supply it with 12 V for V CC and 5 V for "ground". So the fan sees voltage difference of 12-5=7 V, but shouldn't there be a path to 0 V ground? This is the part where I'm confused. Is this secure, is this possible? <Q> This is not a great solution. <S> The 5V rail is not intended to sink current in most multi-output power supplies, and it may not be able to do that effectively. <S> It may do weird things like cause the 5V line to float up to a higher voltage or even damage something. <S> You can get them cheap on Amazon . <A> But I hate to do that sort of thing. <S> As far as fan startup, I think it'll probably startup at 7v. <S> It'll depend on the fan... <S> I've put various voltages on fans before (from a variable supply) and if I remember correctly, they turned on there. <A> No problem here. <S> The fan voltage is differential 7V and exceeds startup voltage. <S> Just dont try it on an LDO pulling up the 5V output to 12V ( emitter follower) with no load. <S> as Vo will also rise on LDO. <S> Here its no problem. <S> I guess no 3 pin fans with 3 pin headers and BIOS controlled fan speed eh? <S> ;)
| Technically, yeah it should work (the voltage difference is 7v seen by the fan). A more solid solution would be to use a DC-DC converter to step the 12V down efficiently.
|
How to stop an electro magnet from sticking I am using an electro magnetic door lock which I am powering with 12 volts. It is switched on and off using a transistor array (uln2003.) I have also included a flyback diode. My problem is this: after I have switched off the magnet, there appears to be a lingering field that holds the door shut until it is pushed open. It is not a strong field, and it goes away after the door is pressed open. I need to build a circuit that can neutralize this field. How? My inexperienced mind is telling me that I need to send a short burst of current through the magnet in the opposite direction. <Q> I am using an electro magnetic door lock which I am powering with 12 volts. <S> It is switched on and off using a transistor array (uln2003.) <S> I have also included a flyback diode. <S> My problem is this: after I have switched off the magnet, there appears to be a lingering field that holds the door shut until it is pushed open. <S> It is not a strong field, and it goes away after the door is pressed open. <S> I need to build a circuit that can neutralize this field. <S> How? <S> (1) <S> A small airgap when at closest approach will (should) make a substantial difference. <S> This can easily be achieved with a small layer of non magnetic material if desired. <S> Try tape, paper etc initially. <S> A more permaent solution can follow. <S> This applies even with a non-magnetic material for the door as the plate the magnet attracts may suffice. <S> (2) <S> A low force spring should be able to be made to work. <S> Magnet-on attracts plate and compresses spring. <S> Magnet off - spring is designed to exceed remaining attraction. <S> Again, start informally - you can eg use bubble wrap air bubble with suitable mechanical placement and sellotape to hold it. <A> There is some level of reverse pulse <S> you could put through the lock coil to fully demagnetise it. <S> However, what that level is would be dependent on a number of factors, some you could not control, and would vary between samples of the same lock. <S> However, even if not perfectly demagnetised, some demagnetisation may be beneficial to you. <S> Go do an experiment on some. <S> This takes the sample round the hysteresis loop in ever decreasing amplitudes. <S> Unfortunately the means to provide it could be fairly complicated. <S> In CRT TVs, it used to be provided by the mains, fed through a PTC resistor, that steadily heated up and reduced the current. <A> Remnant magnetism is a concern to relay designers. <S> Figure 1. <S> The addition of a brass "pip" on the armature prevents magnetic latching. <S> Source: David McAudrey . <S> When a small electric current passes through the energising coil, the armature is magnetically attracted to the end of the rod against the bias force of the spring and, if the Ampere-turns are sufficient, it will rock to close the gap in the magnetic circuit. <S> A similar approach is what is required in your application but, as others have said, a thin layer of non-ferrous material may suffice.
| In practice a small brass pip on the underside of the armature prevents the gap from becoming fully closed; otherwise it would tend to stick in the energised position when the energising current is switched off. A more robust way to demagnetise things is to use an initially large, then decreasing steadily down to zero, AC current.
|
What kind of electronic device is this? I have been taking apart a old television circuit board and I found several of these things. They look like resistor but when I test them with the multimeter the resistance is very low, around 10 to 20 ohms. <Q> That would be an inductor with a resistor style color code. <S> Here's a picture from that Wikipedia page showing some similar 100 µH axial lead inductors: Vahid alpha at English Wikipedia CC BY 3.0 , via Wikimedia Commons <A> It seems very likely that it is a small-value inductor. <S> There is an open-source LCR meter available from dozens of vendors on Ebay for USD 15~20. <S> Quite ingenious, dirt cheap, and excellent for identifying not only what a mystery component is, but also measuring all the key parameters. <S> (Like inductance and series resistance for your inductor, for example.) <S> Recommended for anyone who recycles electronic components. <S> Search for "Transistor Tester Diode Triode inductor Capacitance ESR Meter LCR <S> " There are many to chose from. <A> This is what a 10mH looks like. <S> BRN-BLK-BLK <S> Since there are so many turns, it looks fat. <S> The resistance determines how many mA can be passed thru with DC for filtering purposes in a low noise LC LPF supply. http://www.digikey.com/product-search/en/inductors-coils-chokes/fixed-inductors/196627?k=&pkeyword=&pv7=3&pv19=90&FV=fff40003%2Cfff80013&mnonly=0&newproducts=0&ColumnSort=0&page=1&quantity=0&ptm=0&fid=0&pageSize=500 <A> That's a 10 μH inductor (brown-black-black = 10e0). <S> Low-value axial inductors of this sort are colored green to distinguish them from resistors . <S> With the increasingly common use of surface-mount technology, axial inductors (and resistors) aren't common in low-power circuits these days, except in some very cheaply made, hand-soldered devices where the factory doesn't have a pick-and-place machine (the stuff often associated with super-low-cost Chinese labor). <A> Given that the colour bands appear to be Brown Black Black = <S> 1 0 <S> x 10^ <S> 0 <S> = 10 <S> Ohms then concluding that it MAY be a resistor may be a good guess. <S> However, it may also be an inductor or "choke". <S> If you are happy to destroy one, carefully scrape the outer layers off.
| If it is an inductor it will probably have a ferrite bobbin core plus many turns of very fine wire.
|
Where do I perform circuit simulation and testing of the circuit schematic before actual implementation? I am new in circuit simulation.In the given circuit obtained from TI'S WEBENCH , DC to DC buck conversion (110V to 5V) using flyback opto-coupler feedback topology is used. I want to perform circuit simulation and testing of the circuit schematic before actual implementation. But which is best free software available on internet where I can do circuit simulation testing with Texas instruments UCC28C40 IC (PWM controller) included in its component list? <Q> but if you have some money to spend I highly recommend Everycircuit . <S> It is like circuitlab but with more features . <S> You can see the current flow in real time and you can also adjust the parameters in real time. <S> It is very intuitive ! <S> But you wont find the UCC28C40 on those 2 tools. <S> You can try to find or make a substitute of the UCC28C40. <S> LTSpice is a desktop based circuit simulator that is very popular. <S> It has not the UCC28C40 in his default component list <S> but there is a chance that you can find a tirth party model of UCC28C40 free to download. <S> If I'm correct (not sure): the file extension of third party components of LTSpice are ".SUBCKT". <S> Thank you @Arsenal for correcting me, here is his post: "LTSpice doesn't care much about the file extension of the circuit. <S> You have to use the spice directive .libor <S> .include <S> to get it into your simulation. <S> The .subckt defines a new subcircuit." <S> Screenshot of everycircuit: <A> You will want to use TI's model in spice simulation http://www.farnell.com/simulation/1746635.zip?_ga=1.248798085.341376100.1470941132 <A> Looks like you obtained the schematic directly from TI's WEBENCH. <S> If your aim is not putting a voltmeter/oscilloscope/ammeter etc. <S> then WEBENCH itself is sufficient for you. <S> At the top of WEBENCH window you can find "Op Vals", "Schematic", "Charts" etc. <S> They're all simulation-related tools. <S> EDIT: There's a spice-based simulation software called TINA-TI ( http://www.ti.com/tool/tina-ti ) <S> but I'm not sure if its library contents and Flyback simulation capabilities can meet your expectations.
| Circuitlab is a very nice online cloud based simulator
|
How should I layout voltage regulators (VRs) to separate Analog and Digital? I need to create two power and ground domain: Analog and Digital. I've a common power supply that I'll call VCCBattery that power 2 voltage regulators. Since each voltage regulator have decoupling Caps and GND pin, on which plane should I connect this GND pins?After that, how should this 3 planes (Battery GND, Digital GND, Analog GND) shoul be connected together? (like first Digital and Analog and next to Battery, or a star joint between them) Here's voltage regulators that I use and their Pinout with essential capacitors: <Q> Whether you separate the analog ground from the digital ground REALLY depends on how isolated the analog and digital functions are. <S> If however, there are numerous control signals between the digital and analog circuits then the long current return path through a separated grounding system <S> causes issues and a uniform grounding system is more appropriate. <S> If there are only a few interconnections, then careful layout and PCB design to have those circuits as close as possible to where the grounds separate may be sufficient to allow you to continue to separate them. <S> Most of the time though, one really good ground plane is sufficient with the occasional localized ground "island" for a particularly delicate analog circuit. <A> They all got to Vbat- but path is explicitly used only by Analog loads or digital loads by layout and not shared. <S> Layout is your choice, research how ADC boards do this. <S> Since Inductive current noise must be minimized, and Inductance is determined by length/width ratio of tracks, and V=L di/dt, you want wide tracks or copper pour to high switched currents. <S> Also consider stray coupling capacitance in layouts. <S> For high current switched loads, consider ESR of output caps and min/max range if specified in datasheet. <S> Usually one or two caps ensures good switched load regulated noise, if there is any on these 200mA devices. <S> Having two separate regulators gives low Zout and excellent isolation on Vout and these parts have excellent common mode Vin noise noise rejection, but two IC'/ may not be necessary in some cases.... <S> unless one becomes the Analog Vref for say an ADC. <S> Since n* CMOS gates switching at once is <S> n* C load with low ESR/ <S> n <S> it's <S> Voltage Source and ground return must be as ideal as possible. <S> Namely, for 5% ripple, 1/20th of the load ESR using a small C >yet 20x. <S> Bulk E-Caps and batteries do not have this low ESR in the 0.35/risetime equivalent frequency range. <S> Often high SRF high Q 100pF to 1nF caps are necessary close to the point of distribution. <S> THis translates into 50MHz ~1GHz and depends on Cap SMD size. <S> The reason the above is critical is that a small T= RC load , with a transient response is effectively the ratio of Z(f) for source/load up to the frequency of 0.35/Tr. <A> If you really know why you are using split-grounds, then alright. <S> If you don't know why <S> and it's more of a tradition, or someone told you so without knowing <S> , then you should wonder why you are actually splitting the grounds. <S> IMO, placement matters a lot more. <S> Now, a regulator produces a voltage which is referenced to its GND pin. <S> Say it's a 3V3 regulator, then it will output "Local GND voltage" + 3V3. <S> The output capacitors do the same at higher frequencies. <S> Noise on your supply is shunted to ground by the capacitors, but noise on GND is also injected into your power supply. <S> This is why, if you have analog and digital ground planes, and separate supplies, then the regulators for each and their output caps should be on the ground plane they power.
| If they are 100% separated, then running different grounds with them only connected together at the battery terminal connection is better.
|
Regenerative Power on DC bus between DC motor and voltage supplier Currently, I am looking for a back-EMF solution for brushless DC motor drive. My input voltage for motor driver is 48V (which starts at 50V level and starts to decline until 35V as a feature of fuel cell). However, as I try to accelerate DC motor back and forth during motion, I observe a regenerative power on the input of the motor driver. (after 56V the motor driver disables itself) Do you know if there is any strategy to compensate this back-EMF on DC bus?Is there any suitable product which I can use for this purpose? Here is what I saw on the screen of oscilloscope (I just supplied the motor driver with 28V) Supplied with 48V. The driver shut downed itself <Q> If you can put enough capacitance on the DC bus to absorb the energy without going overvoltage that would be the simplest solution without wasting energy. <S> If not, a common approach is to use a bank of resistors and a comparator on the DC bus. <S> When the voltage approaches the maximum allowable the comparator will turn on, which turns on a MOSFET and puts the resistor bank across the DC bus to dissipate the energy as heat. <S> The comparator should have some hysteresis to avoid high frequency chatter. <S> Lastly there are more complicated regenerative schemes to store the recovered energy in a battery or regenerate it onto the AC line. <A> This is only partially related to the BackEMF of an electrical machine. <S> Assuming you have a 4-quadrant controller & you are ACTIVELY decelerating the rotor, what is occurring is the transfer of energy from the rotor ( \$\frac{1}{2}J\omega^2\$ ) and the stator inductance ( \$\frac{1}{2}LI^2\$ ) onto the DCLink capacitance (\$\frac{1}{2}CV^2\$) <S> The voltage on the capacitance must rise as the energy is transferred. <S> Three ways to minimise or deal with the increased voltage <S> Increase the DClink capacitance close to the H-Bridge. <S> With an increase in capacitance the final voltage for the same energy transfer will have been reduced Decrease controller bandwidth <S> It is not stated whether you have a form of PI closed loop control, nor whether you have a speed loop, but reducing its bandwidth and the rate it can decelerate will reduce the rate of energy tranfer which will reduce the end voltage Incorporate a resistive brake circuit <S> By placing a Resistor + FET across the DClink (plus a freewheel diode across the resistor) & a DClink hysteretic monitor (comparator), the FET will "chop" the DClink between predefined thresholds (say .. 54V -53V). <S> As long as the resistor value and power rating have been appropriately selected, the DClink will be maintained below the level of concern <A> Could you put one of these inline with your DC supply to the motor? <S> That way when the current reverses on the line you would block it and allow the motor to play out the stored energy. <S> These are used a lot in automotive industry to deal with coil collapse induced back EMF. <S> Look for Flyback Schottky Diodes and see if this is what you are looking for.
| I believe what you are looking for is a flyback diode on your DC motor.
|
Using parallel capacitor-diode pairs to limit current in an AC circuit I want to limit current from an AC source using capacitors. As I am using dielectric (polar) capacitors, I thought I could use 2 capacitors with opposite polarity in parallel with a diode in front of each capacitor to protect it from reverse bias damage (see circuit diagram for clarification). My thinking was that every time the source's polarity switches, the current will go through a different capacitor-diode pair, always with the capacitor's reactance limiting the source's current. When attempting to do this in practice, no current moves at all, so I must be misunderstanding something very important. Would any kind sage be interested in pointing out my error? I'm a physicist who likes to tinker, so if this is something embarrassingly stupid, that I should have learned in AC 101, I apologize. I'm mostly interested in understanding the general idea of why this won't work, but if you'd like details, the AC source is mains (120VAC, 60Hz), the capacitors are 22uF, 400V, the diodes are 5A, 1000V rated. <Q> Here's what happens as time goes by... <S> For power applications, a capacitive dropper isn't always a good choice, but when it is it's far better to use a single film capacitor, as shown below, than to use a pair of electrolytics hooked in series opposition, as suggested earlier. <S> The reason for that is that the electrolytics will always dissipate more power in their electrolytes and dielectrics than the film caps will in their dielectrics, and the electrolytics will run hotter and vent their magic smoke sooner. <A> I'm mostly interested in understanding the general idea of why this won't work ... <S> Because there is no discharge path once the capacitors are charged. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Rearranging the diodes provides a discharge path for the capacitors and reverse polarity protection. <S> How it works: <S> When L goes positive D1 is reverse biased and C1 charges. <S> D2 prevents C2 being charged more than 0.7 to 1.0 V reverse. <S> When L goes negative the reverse happens. <S> The effective capacitance will be the value of one of the capacitors since only one is in use at any time. <S> I used this arrangement almost 40 years ago to replace a failed starting capacitor on a family friend's 230 V well pump. <S> (It died on a Friday evening on a bank-holiday weekend. <S> How do capacitors know what day it is?) <S> It worked but I was concerned about the long-term reliability due to the small reverse bias on each cycle <S> so I replaced it with an unpolarised capacitor as soon as possible. <S> Update after comments. <S> simulate this circuit Figure 2. <S> Original redrawn with GND applied between the two capacitors to assist in visualisation with the simulation. <S> Figure 3. <S> Results of simulation of Figure 2 when V1 is set to 100 V p-p . <A> Each C would conduct current during one half of the first cycle and then would be charged to the peak +/- <S> voltage then stop conducting current.
| I used a couple of valve amplifier capacitors and suitable diodes.
|
How to reliably secure stranded wire in terminal block? My 18AWG stranded wire continues to come out of the terminal block it's screwed into. What are some ways that I can more reliably secure this in an industrial environment? Strip extra wire and loop/fill more of the chamber before fastening down? Twist strays and then tin the ends of each wire tip before fastening down? How much of a solder blob are we talking about? Twist down harder? (I'm worried about cracking the solder and/or ripping off the terminal block) <Q> (This is info already mentioned in the comments on the question — but nobody wrote an actual answer, so I'm doing that. <S> It does also agree with my small experience.) <S> Do not use any solder at all. <S> Caveats: <S> Use the proper crimp tool, which will leave a textured surface that is readily gripped by the terminal and resists sliding out more. <S> Adding the ferrule will increase the size of the wire end, possibly too large to fit in the screw terminal. <S> (This should not be a problem if one of the options you have considered is doubling over the wire.) <S> This picture shows several sizes of ferrules, two ferrules crimped on 22-gauge stranded wire, and the crimp tool I used ($21 when I bought it). <S> The plastic part of the ferrule guides the strands in, covers any exposed metal, provides some strain relief, and identifies the size of the ferrule. <S> Caution: <S> ferrule makers do not all use the same color scheme! <A> Twist it together and double it back on itself. <S> That is, strip twice the terminal depth, twist and bend over. <S> Ensure that the wire then approximately fills the screw terminal. <S> If it doesn't, you need a smaller terminal. <S> Good termination practise should be second nature. <S> Bad terminations may well initially work, but are liable to fail, heat up and perhaps cause fires. <S> A good termination will last a lifetime. <S> Basic electrician termination training, this. <S> At least it was when I did it about 30 years ago :) <A> You MUST NOT solder tin the whole portion that goes ino the connector. <S> Tinning the VERY END is OK to stop it unravelling. <S> See my prior answer on this here <S> A method of "strain relief" that works well but which is "very naughty" and which people will criticise is: Strip wire several times longer than depth of terminal block hole. <S> Bend wire back over outside of insulation and wind it in a spiral <S> so it wraps completely around the outside of the insulation several times and extends back by about the depth of the terminal block hole. <S> Insert wire into terminal block and screw down clamp or screw onto combined wire end + insulated end. <S> The screw or clamp both makes contact with the conductor and also clamps the insulation rather than just the conductor. <S> I'll add a diagram if this does not get run out of town on a rail. <S> I was shown this method decades ago by a man who designed/built/installed taxi meters in taxi fleet cars. <S> I tried it. <S> It worked. <S> This method is contrary to a number of things you'll have been taught. <S> It does not seem to be an utterly terrible idea in theory and proves to be a good one in practice. <S> A more conventional method is to either add a sleeve which can be clamped to the cable end or run the insulated part of the cable just before the outer ends under a clamp which is screwed down onto the outer. <A> It depends on if vibration exists and causes the wire to move how rugged the attachment must be. <S> Normally a 5 to 10 lb. <S> pull test is the criterion for any connection, whether it is twist wire clamped or crimp lug attached with proper two-stage crimp on insulation and conductor. <S> Even molded connector junctions can and often do fail due to designers under-estimating how much strain users actually apply to their DC power cords on laptops or iPads <S> (I have had to replace mine often from grandkids and my frequent use). <S> If this describes your situation, then do not tin but neatly twist, <S> shape (with tool or needle-nose or plastic barrier wall) and clamp under screw head for an air-tight pressure contact with sufficient but not excessive force to pass pull test. <S> This will translate into a fixed screw torque level with calibrated production tools or training. <S> Better terminal strips have barriers and crimp washer under screw head. <S> However in my own practice, I prefer a thin solder dip or coating so that it wicks under the insulation and a star washer. <S> The solder wicking adds strain relief. <S> The star washer exerts more contact pressure and the solder mitigates loose strands. <S> Careless loose strands can only be avoided by proxy inspection or a better design with crimped terminals and routine pull tests. <S> If rugged strain relief is needed to prevent frayed wire breakage, then additional cable plastic clamp strain relief is needed such that the stiffness on strain relief is about <5x stress/strain of the wire for a graduated relief. <S> This can be a moulded jacket as well.
| Use a wire ferrule, which is a metal sleeve that slides over the (straight) strands and is crimped in place, making a solid end that can be clamped securely by a screw terminal of the type you are using. There should not be any bare wire (stripped insulation) beyond the terminal itself.
|
MOSFET delay time when transitioning between triode and saturation regions. Charging and discharging mosfet Summary For a mosfet, transition from saturation to triode region happens much quicker than from triode to saturation. Why? Details UPDATE: the load resistor was removed. Fig.1: Schematic Fig.2 Transient analysis Notes: 1) BSIM4 SPICE model of MOSFET is used. Model parameters used (file cmosedu_models.txt) can be downloaded here . Initially, the transistor is in saturation region (Vgs=350mV > Vth=280mV, Vds=V(out)=400mV > Vdssat=50 mV). Idssat=10uA. After Vgs has increased, the transistor moves into triode region and its Vds goes down from 400 mV to 8 mV (below Vdssat). Then, the opposite transition happens. Transistion from triode to saturation takes around 6 ns , while transition from saturation to triode happens almost instantaneously .Why such a difference? Appendix Fig.3: Transistor parameters (for the operating point corresponding to Vgs=350mV) . Source: CMOS Circuit Design, Layout, and Simulation, Third Edition. R.J. Baker. Page 300. <Q> The different transition times are a result of your particular test setup. <S> In your case, for a falling edge discharging is done by the transistor itself and a large current is possible. <S> In case of rising edge the transistor turns off and the current has to be supplied by the 10uA current source. <S> Because of the resistor at the output this results in an exponential characteristic. <A> I'm going to post an answer <S> but it's really adding more details to Mario's correct answer <S> and, if he wants to strip my answer of anything I'll just delete this. <S> A randomly googled MOSFET output (DS) capacitance versus drain voltage: <S> - It doesn't matter what MOSFET you use, the drain capacitance (\$C_{OSS}\$) at 1V (as per Sergei's table in his question) will increase to something like 4 times that value when the MOSFET is fully on in the triode region. <S> The drain voltage is 8 mV <S> and, as you can see \$C_{OSS}\$ rises to about 20,000 compared to about 5000 at 1V. <S> It's all relative and these could be farads, pico farads or fractions of femto farads. <S> So if the drain capacitance is 6 fF at 1V then it is likely to be in the realm of 24 fF at about 8 mV drain voltage. <S> At 450 mV (as per the waveform in the question), the capacitance could be about 12 fF. <S> If 24 fF is charged with 10 uA the dV/dt will be 10 uA / 24 fF which is 417 volts per us <S> OR 417 mV per nano second. <S> Here's what it will look like against Sergei's picture (orange is the line I've added spanning between 6 ns and 7 ns and rising from 0 mV to ~417 mV): <S> - Clearly it's about the same sort of rate as he is seeing and the exponential asymtotic shape is going to be due to non-infinite drain resistance. <S> It's going to be more complex than this because the \$C_{OSS}\$ falls rapidly as drain voltage rises and, if the equivalent drain saturation resistance were infinite (i.e. a tending towards a true flat line), I would expect more like an exponential rise rather than a linear or asymtotic rise. <S> Anyway, @Mario, strip out anything you want and let me know to delete my answer. <A> note that the effective rise time constant { dt=Coss/(Ic) <S> * dV} is large with 10uA current source and fast with ( RdsOn*Coss )= <S> tF, fall time <S> Since turn off slew rate is non-linear function of this circuit, an approximation of Coss is Coss= 1ns <S> * 10uA/300mV= 33x10e-12 or 33 pF <S> When low 8mV/10uA implies an RdsOn (rON?) of 800 Ohms for the switch. <S> 400mV from 600mV Vdd with <S> 1M load implies a leakage of 0.5 MOhm in the current source ?? <S> Coss is a common parameter in MOSFET output capacitance specs. <S> Some parameters must be different in your model! <A> Stealing ideas from Ali Chen, Andy Aka and Tony Stewart and possibly Mario. <S> If you like this answer, pls upvote their posts. <S> Summary <S> After the large increase of the overdrive voltage, the channel resistance drops substantially providing a fast route for discharge current. <S> After the overdrive voltage goes down again, the mosfet can only be charged with the current source what determines the MOSFET rise time. <S> Second order effects are: 1) <S> During the charge phase there is a leakage through the MOSFET , thus estimating the rise time simply as \${{C\Delta U} \over <S> I}\$ is not entirely correct. <S> Instead, the rise time should be longer because some portion of the current from the current source bypasses the capacitor (transistor). <S> 2) <S> The transistor output capacitance increases several times when moving from saturation to triode. <S> The both second-order effects explain an exponential growth of Vds instead of a linear one as predicted by the formula. <S> Details Fig.1 Transition from saturation to triode. <S> Fig.2 Transition from triode to saturation. <S> Note a different scale. <S> 1) <S> The assumption is that channel resistance adjusts very rapidly to Vgs, which makes sense, I think. <S> 2) <S> Since capacitance changes with Vds (V(out)), it was assumed the capacitance changes relatively "slowly", so in the simulation it was assumed to be constant and as at <S> the beginning of the transition. <S> A more precise simulation would be with capacitance value linked to the value of Vds. <S> Numerical delays (tens of picoseconds for discharge and several nanoseconds for charge) in this simplified model are close to the ones presented in the question, what should validate this answer. <S> Appendix <S> Fig.3 IV curves used for calculating channel resistances.
| The transition time depends on the parasitic capacitance at the output and the current to charge or discharge it.
|
Is this a triphase or monophase AC motor? I have got this motor from a local junk yard. It says it is a monophase motor, but its wiring looks like a triphase one. Please help me classify it. It reads Monophase dual-value capacitor asynchronous motor Model ... Power 2200W , RPM 2800 r/min Voltage 220V , Frequency 50Hz Current 16A , B grade insulation Operating 45uF/450V , Starting 300uF/250V IP44 Implementing standard: JB/T9542-1999 Serial ..., Date Aug 2008 (Left) CCW , (Right) CW (Wiring Diagram) <Q> This because a single phase motor can not turn by itself. <S> So a rotating field of some kind is required. <S> The connection diagram is clearly indicated on the id plate. <S> By placing the shunts on the left side in the vertical sense (U1-Z2 and V1-U2 the motor ccw. <S> Placing them in the horizontal sense (U1 - V1 and Z2-U2) makes the motor cw. <S> Since the motor has two capacitors in series ( 300 uF for starting) and 45 for running there should be also a centrifugal switch inside to short out C2 when running <A> Phase leads are U1 & U2. <S> You make connections as per left hand diagram to get clockwise rotation (according to diagram symbol and opposite of text you supplied.) <S> You make connections as per right hand diagram to get anticlockwise rotation (again, according to diagram symbol and opposite of text you supplied.) <S> The capacitor values required for C1 and C2 are stamped on the specification plate. <S> These are essential for normal operation. <A> Figure 1. <S> There are two power connections: U1 and U2. <S> As shown in Figure 1, there are only two power connections. <S> It's single phase. <S> Power 2200W, RPM 2800 <S> r/min <S> It's an induction motor running at less than 3000 RPM, the closest multiple of 50 Hz above 2800. <S> Operating 45uF/450V, Starting 300uF/250V <S> A three phase motor will not require starting capacitors.
| The motor is a single phase motor with an extra winding combined with a capactor to create a rotating field. Single phase, as it says.
|
How do I make a good electrical connection with wire I can't solder? A few days ago my DSL connection kept tripping. After abusing the service provider for a few days, I looked at the stats and noticed signal attenuation was sky high and attainable rate was very low(<1000 Kbps). I checked the wires and found that one of the wires going into the ADSL splitter was hanging on by a few strands and was about to break. The problem The main source wire has solid cores, and is joined(twisted together) to a thin stranded wire which has a RJ11 jack to the splitter. I wanted to improve upon this by soldering but later found out that the soldering iron's heat burns off the strands in the thin wire. How do I connect it reliably to the solid cored wire? I'd prefer if I get to connect the two wires together directly rather than by a jack or something similar(since introducing jacks could cause signal loss). Edit: My source wire(untwisted) looks similar to this: The RJ11 cable is something similar to headphone wire - it has 4 non-enameled ultra-thin wires in each colored insulation. It has a red and a green insulation in there(so 4 strands for red, 4 for green). The thin wires also make it a nightmare to strip the insulation off - the wires break too easily. Here's an actual image of the source cable: The top and bottom wires are just copper solid-core wires and the printed part is sheath. An electrician came yesterday and 'fixed' the thing. It stopped the tripping problem but I want something more elegant than this. The other end of the red 'fixer' goes to the splitter. I don't think it'd do much good by opening it up - unless I'd just solder the source wire under the ADSL jack on the splitter's PCB. <Q> Figure 1. <S> An insulation displacement connector (IDC) box. <S> Wires are pushed down into the terminals using an expensive professional push-down tool with built-in cutter or a cheap plastic pusher. <S> The blades in the IDC connector displace the insulation (by cutting through it) to contact the wires. <S> Figure 2 and 3. <S> The "pro" tool and the cheap plastic pusher. <A> I'm back with an update. <S> First of all, thanks a lot @pjc50 for your advice that I'd be better off crimping the wire rather than soldering it: otherwise this'd never have occurred to me. <S> And now that I think back, I feel kinda stupid <S> I didn't do this from the start. <S> Before starting, this is how the RJ11 cable I was talking about looks like: <S> That thing is hard to strip, and impossible to solder. <S> As is evident, I stripped out two of the four strands on the green cable - stripping is made harder by the white 'filler' that's also poking out of there. <S> It was way easier to just use a full male-male RJ11 cable and attach it to the female port on the above box. <S> Time to open up the fixer(I dread to see what's in there...) <S> Just as I expected. <S> And guess what happened when I pulled the insulating tape off one of the joints? <S> Yup, the source wire broke. <S> The main reason I hate fixers. <S> I marked one of the wires before starting, which would connect to the red wire: And crimped on the spade-connectors. <S> Test complete! <S> Time to finish the job. <S> Finally, something that looks cleaner: Here's the result: Not as good as I expected, but now I think it can't really get higher than that. <S> Maybe I should move closer to the exchange or something(:P) <S> Thanks everyone so much for your time and support, and I am extremely glad <S> I asked here, otherwise I'd still be stuck soldering that impossible joint. <A> Note that those are probably not stranded or solid wires. <S> Examine under a magnifier. <S> For flexibility, some phone cords are made from extremely tiny copper-foil ribbons spiraled around a core of ? <S> nylon? <S> strands. <S> A soldering iron melts the plastic fibers onto the copper, forming a crunchy carbonized glob. <S> I've soldered these in the past just fine. <S> Unspiral the ribbon, snip off the white fibers in the center, then twiddle it back together. <S> Tin it with low temperature (NON-leadfree) flux-core solder. <S> Then solder normally. <S> Note that the soldered ribbons are fragile, so it might be a better idea to use a spade crimp; the kind with insulation crimp for strain-relief.
| I decided to crimp the source wire to the two spade-connectors above.
|
Options for soldering a through hole PCB (e.g. Teensy) flush/flat on another PCB I am looking to incorporate a Teensy board on my PCB. I want to solder the Teensy PCB onto my PCB. But I also want to minimize the height above my PCB as much as possible. Hence, I DO NOT want to solder using regular header pins -- i.e. I DONT want this: I want the Teensy to sit as flush as possible on my PCB. Unfortunately, the Teensy does not come with castellated holes (like the Photon or most wireless boards): So I was thinking of the following options: Using low profile header pins like this one -- https://www.pololu.com/product/2663 (plastic height is 1.5mm instead of usual 2.5mm). But I want to avoid the 1.5mm as well Use straight pins like this - http://www.digikey.com/product-search/en?mpart=3560-2-00-15-00-00-03-0&vendor=54 -- and soldering the Teensy flat on my PCB. Is this a viable option?? A google search doesnt really show me examples of this being done. What other tricks/tips can I use?? (I understand that the best way is to not use the Teensy board directly, and instead to design the Teensy ckt on my PCB. But for some reasons I do not want to get into that right now.) <Q> Stick the headers through your main board from the bottom, with the plastic on the bottom. <S> Set your Teensy on top, with the headers going through its holes. <S> Solder the Teensy to the headers. <S> Using a small screwdriver, pry the plastic off the headers. <S> You'll be surprised how easy it comes off. <S> Solder the headers to the main board. <S> Trim headers as necessary. <A> Have you considered simply surface-mounting the module onto your PCB? <S> The header holes are big enough for solder to flow down and wet onto the main board. <S> Or you could use some solder paste and reflow on a hot-plate or an oven. <A> A Kapton tape layer between the two boards to prevent accidental short circuits. <S> (Must be Kapton to withstand soldering temperatures) <S> This isn't something you'll do in any quantity, or else it would be worth re-laying out the board (or cloning the Teensy with castellated pins to suit your purpose) <A> Have your board made with a cutout that the "teensy" board fits into, and through-holes around the edge of that. <S> Make a bunch of U shaped wires to make the electrical and physical connection. <S> Add epoxy or hot-glue after <S> soldering <S> if you want more physical connection.
| Just use resistor legs (or other sources of tinned wire) through each hole, soldered both sides, trimmed to length.
|
How can I identify transistors with a low Vce-sat (collector-emitter saturation voltage)? I want minimum voltage drop between collector and emitter as I have very small Vcc voltage to supply the load. My Vcc is 3.3 volts. I want to use this switch to turn on/off 3 LEDs connected in parallel with 1K ohms resistor in series at the collector. Is it possible to find transistor with low collector voltage? <Q> What you want to do is saturate the transistor, which means you want a base current (very approximately) <S> 0.1 <S> * the collector current - around 0.3mA, or higher currents won't hurt. <S> Assuming you're switching from a 3.3V MCU output, you'll have about 2.6V across the base resistor after subtracting Vbe, so Rb should be less than 2.6/0.3 = <S> 8.67kilohms (choose 8k2 or even reduce to 4k7). <S> What voltage does that give between collector and emitter (Vce)? <S> Check the datasheet for your chosen transistor; it'll probably guarantee "below 0.2V". <S> e.g. for the BC847, Table 7 shows " <S> VCEsat : IC=10mA; IB=0.5mA 90(typ) 200(mak) <S> mV"So Vce would be under 0.1V at currents below 10mA - and note the base current is only 0.05* the collector current. <S> Also see the graph in Figure 3 which shows measured performance of a sample transistor, where Vce (at 25C) is only around 50mv under these conditions, rising to 100mv at 30mA. <S> If that's not enough, the higher gain BC847 reduces Vce to about 30mv under the same conditions (see Fig.11) <S> Most small signal NPN transistors should have similar info in the datasheets, I've just used the BC847 (aka BC107 for old timers) as an example. <A> You are intending to use a 1 kohm resistor and this means that the maximum current (into a short) will be 3.3 mA (from a 3V3 supply). <S> With the LED in circuit this might drop to about 1 mA. <S> If you have three such LED circuits in parallel then, you would need the transistor to supply about 3 mA. <S> However, if you are intending to drive the transistor's base from a GPIO (or other similar source), it makes sense to throw the transistor away and directly drive the 3 mA load. <A> You say you want to switch 3 LEDs in parallel from a 3.3 V supply, and that you intend to have a single 1 kΩ resistor in series with the parallel LEDs. <S> There are several issues here: Putting the LEDs directly in parallel is not a good idea. <S> Instead of one resistor for the whole bunch, use a separate resistor in series with each LED. <S> You didn't say what kind of LEDs, so I'll use typical green as example. <S> These drop about 2.1 V. <S> That will be quite dim, but I'll have to take on face value that this is what you want for some reason. <S> This also means the parallel LEDs aren't for getting more light, but because you want them in physically different places. <S> Your ideal average per LED is then <S> 400 µA. <S> (1.2 V)/(400 µA <S> ) = 3 kΩ, which is the resistance you should use per LED. <S> This also makes sense just by seeing that 1/3 the current should be thru each resistor compared to the single resistor, so it should be 3x larger. <S> As Andy said, with only 1.2 mA total, you should be able to use the digital output directly. <S> Check the datasheet, but 1.2 mA is quite likely within its capabilities. <S> In this case, the saturation voltage of a low side NPN transistor isn't a big deal. <S> With only 1.2 mA collector current, you can easily run it well into saturation. <S> 200 mV is a typical value of saturation voltage in a case like that. <S> Even if it is as high as 500 mV, you can easily design for that just be lowering the resistor values. <S> For example, if the switch drops 500 mV, and the LEDs 2.1 V, then that leaves 700 mV across the resistors. <S> To get 400 µA per LED, you need (700 mV)/(400 µA) = <S> 1.8 <S> kΩ. <S> If you really needed very low on state voltage for some reason (none is apparent here), then use a FET instead. <S> The IRLML2502, for example, is guaranteed to be 80 mΩ or less with 2.5 V gate drive. <S> At 1.2 mA, this will only drop 96 <S> µV. <S> The slop in the forward voltages of the LEDs will be more than that.
| Even if the switch drops no voltage at all, that leaves 1.2 V across the resistor, which means you intend the bunch of LEDs to be driven with 1.2 mA total.
|
What is the problem with my LED panel switching circuit? I have soldered this circuit on PCB. When I switch Q1 via MCU pin-1, I see LED glowing on Q3 (not too bright though). What could be the reason for this? Even though, MCU pin 2 is pulled down to GND! It works fine on the solderless breadboard but not on the PCB. <Q> The problem was very simple. <S> My PCB circuit had some remnant soldering flux on it which created some conductivity. <S> I cleaned it properly with hair dryer and it worked completely fine. <S> Thank you all for your answers. <A> There must be an unwanted connection between the Q1 circuit and the Q2 circuit. <S> If you made the pcb yourself make sure it is clean. <S> Even an unexpected high resistance can make the led glow light. <A> The TIP122 is a Darlington. <S> I don't know what you are driving it with. <S> But if it is a \$5\:\textrm{V}\$ Arduino or anything with \$3.3\:\textrm{V}\$, then you are terrorizing the Darlington with milliamps at its base! <S> This isn't exactly good, either. <S> One of the pairs is likely to hog more current than the rest. <S> Finally, I don't know what kind of LEDs you have. <S> But even if they are high efficiency reds, you still are only going to provide (probably) less than a milliamp through them. <S> That \$2.2\:\textrm{k}\Omega\$ resistor is pretty limiting here. <S> Even if that poor Darlington is able to sink \$2\:\textrm{mA}\$ (and that's not even on the spec sheet, it is so low), that's still divided into four chains of LEDs you can only hope will get the current shared among them. <S> If equally lit, \$500\:\mu\textrm{A}\$ each. <S> If not, perhaps more in one pair and less in the others. <S> Darlingtons are very sensitive to base currents. <S> I'm not surprised that you are experiencing odd behaviors, given the lack of design. <S> The TIP122 collector is designed for \$\ge <S> 100\:\textrm{mA}\$. <S> Assuming you really are getting bright results when ON (and I'm still having a hard time with that), then try adding a \$330\:\Omega\$ resistor (or up to \$680\:\Omega\$ resistor, if that smaller value dims the LEDs too much) from the TIP122 base to ground. <S> Do that for both TIP122 Darlingtons. <S> Just for starters. <S> See if that helps any. <S> That should cut the current down to something marginally reasonable and it should provide a path for any reverse biased leakage currents, as well. <S> I'm kind of curious how that works out. <S> If it does work, then I think you really need to get a decent design worked out. <S> Write up the exact details of what you want to achieve and what you are using. <S> This means: (1) what is driving the transistor switch; and, (2) what voltage output represents ON (and OFF, too); and, (3) what kind of LEDs are you using, what voltage do they drop when ON, and what current do they require; and, <S> (4) what voltage rails do you have handy? <A> You need a resistor in each series string of LEDs in order to limit the current in each string to the required value, as shown below. <S> The LEDs you're using have a forward voltage drop of 2 volts with 20 milliamperes through them <S> You're starting with a 9 volt supply, and two LEDs in series drop 4 volts, so the ballast resistor has to drop the difference between the supply and the Led drop, which is 5 volts. <S> Then, from Ohm's law, \$ R = \frac{E}{I} = \frac{5V}{0.02A} <S> = 250 \text{ ohms}\$. 270 ohms is a standard E24 value and should easily give you good brightness from the LEDs.
| You also are placing your strings of LEDs directly in parallel.
|
Contact sparking even with flyback diode I'm building a circuit to measure both voltage and current given to a solenoid valve 24V DC 15W. In parallel to the valve is a voltage divider with 470k + 100k Ohm resistors. In series with the valve is a 1 Ohm 4W resistor. In parallel with both valve and 1 Ohm resistor is a 1n4007 Diode correctly positioned in order to actuate as a Flyback Diode. But yet, whenever i remove the +24V contact of the valve, simulating a opening of a Relay, I still see some sparks flying around. They are easily noticeable. I've measured the peak voltage and indeed it has lowered: from -10V (minimum readable value) to -250mV. But there are still sparks! Is it ok or will it damage a operating relay? simulate this circuit – Schematic created using CircuitLab thanks! <Q> If you are simulating the relay by simply pulling a wire out (of a breadboard for instance) there will be a period of connect/disconnect as the wire moves. <S> This will give a series of forward connections to the circuit, each time the inductance will cause an inrush current of some magnitude. <A> It is preferred to place the flyback diode directly over the valve coil connections. <S> Verify the health of the diode. <S> If it is ok then the remaining spark does not harm the valve coil. <A> Sparks are… maybe fine! <S> It depends. <S> What kind of relay is it? <S> Do you see sparks inside your relay? <S> Your relay for driving a solenoid valve may be strong enough to prevent this from happening. <S> You can limit the initial current with a soft-start circuit. <S> This page from Elliot Sound Products has more than you ever wanted to know about soft-start circuits. <S> https://sound-au.com/project39.htm <S> Side note : <S> The forward voltage of the 1N4007 is 1V. <S> This may be fine for your application. <S> I have seen other situations where 1V is too high. <S> Enough noise was created to cause a microcontroller to reset! <S> In this case, a Schottky diode (with a forward voltage of ~0.3V) fixed the resetting problem. <A> 24V sparks, what can you do? <S> It's fine. <S> I mean, if this is a small spark. <S> If you see something like 1mm bright purple spark, it's something else, like no diode. <S> Normal spark from 24V is very (length) short and visible only if you are staring at the contact point. <S> As per spark period- depends on the contact. <S> With good contact it happens once for short period, with bad one- <S> it's a burst of dozens of sparks for 20msec. <A> If the contact is arcing I would investigate a snubber circuit across the contacts. <S> If the arcing is just when the contacts open, I would start with a 10 or 20 ohm resistor in series with a 1uF cap. <A> A 1N4007 is too slow. <S> It takes time to turn on and the spark could occur in that time. <S> A faster diode like a schottky will help. <S> Even a faster diode takes time to turn on so if that is still not enough, add in an RC snubber which is even faster "always on", but will not clamp the voltage. <S> But an RC snubber does slow the transmition-times which a diode will not do.
| In my experience if you can see sparks inside, your relay contacts are probably going to melt together eventually. Remember the flyback diode only protects the valve coil from high induction voltages.
|
Is antenna length for a baseband signal infinity? I am studying RF communications and learned that antenna length is proportional to a wavelength of a signal. And many literature say that a high frequency is preferred in wireless communication because a length of antennas can be short, hence, a size of a device can be small. From here I can deduce a reason that a baseband signal can't be sent through the air. The reason is that, a length of antennas for baseband signals will be infinity, but in a real world, infinity length of antennas is impossible. Hence, sending a baseband signal through the air is impossible. However, as I am a learner, I can't tell the above statement is valid. <Q> Sort of. <S> Certainly antenna length is proportional to wavelength, so if you take the limit going to DC then you'll need an infinitely long antenna. <S> However, even if you make an antenna infinitely long, it won't work very well. <S> Antennas have limited bandwidth. <S> Say you design a 7.5 cm long antenna for a 1 GHz signal. <S> This antenna will work very well at 1 GHz, but at 2 GHz it's too long and the radiation pattern will change. <S> At 500 MHz, it will be too short and won't radiate efficiently. <S> One of the features of RF transmissions is they usually have a small bandwidth relative to the transmit frequency. <S> Say, a 10 MHz wide band sitting at 1 GHz. <S> The frequency only varies by 1% in this case, so you can pretty much ignore that variation when designing the antenna and design it for a single frequency. <S> So, if you wanted to design a 10 MHz baseband antenna that works from DC up to 10 MHz, you need to design an antenna that works properly not over a 1% bandwidth but over 7 orders of magnitude, which is pretty much impossible. <S> Actually, it's not even 7 orders of magnitude-that would be 1 Hz to 10 MHz <S> -it's infinite orders of magnitude. <S> The infinitely long antenna that you would need for DC will not work well at all at 10 MHz. <A> Hence, sending a baseband signal through the air is impossible. <S> This statement is nonsense ! <S> Ideally an antenna should have a length of 1/4 the wavelength you want to transmit/receive, or a multiple of that 1/4 the wavelength. <S> If the antenna is shorter or folded to be smaller, it will still work yet less efficiently. <S> So you can transmit/receive low frequencies (with large wavelengths) through the air (or space) <S> but it is not efficient. <S> Also, if you send a 1 kHz signal, no-one else in range can use that frequency. <S> Additionally propagation of such 1 kHz signal is such that that range could be very large. <S> The solution is to modulate that 1 kHz (baseband) signal onto a carrier . <S> For example 100 MHz. <S> That 100 MHz travels less far as it behaves more like light, it has more difficulty in going around corners etc. <S> But someone else in the same region as you could use 101 MHz and another 102 MHz each of you without disturbing the others (assuming the receivers can filter out the other signals). <S> So it is possible to transmit a baseband signal through the air <S> but it is very impractical and not limited by antenna size but by efficiency. <A> Any antenna, no matter how short, will radiate energy. <S> The key point is that as the antenna becomes small compared to the wavelength of the signal, the efficiency of the antenna decreases. <S> Thus, the short antenna is not useful. <S> AM radio (with frequencies of 550 KHz - 1.6 MHz or so) is effectively baseband for a system like Bluetooth. <S> (Bluetooth uses a baseband bandwidth of about 2 MHz). <S> AM requires large radio towers for an efficient antenna. <S> Bluetooth - sitting on a 2.4 GHz carrier - only needs a small piece of metal in your cell phone. <S> Of course, the other problem with transmitting at baseband is that all signals will be on the same "channel" and interfere with each other.
| The ressaon is that, a length of antenas for a baseband signal will be infinity, but in a real world, infinity length of antennas is impossible.
|
Thermistor and Voltage Divider to ADC I am looking for some advice or pointers as to whether I need to solve this problem and if so how difficult it is to do so. I have read a post which made an off the cuff remark about needing to using an OP-AMP to be able to use the full range of an ADC is certain circumstances. I think I am in this situation but I have never used an OP-AMP before. I have a temperature sensor (thermistor I guess) that I am reusing from a pool heating system. I have tested and measured the resistance of the thermistor at 2deg C at around 493k and at 72deg C at 398k. When I hooked it up to my little voltage divider circuit that will feed into a MCP3002 ADC I used a 1M ohm resistor and this gave me 2.21V at 2deg C and 2.36V at 72deg C. So the issue is that I will only be using a small amount of the ADC range. Using online calculators and reading other posts I see I should drop the fixed resistor to around 450k which I will do but this only increases the voltage range by a small amount. I can probably get by with this range as I don't need much more than 2 degrees of precision but I thought if it isn't too difficult I might see if I can add to my currently simple circuit (voltage divider and ADC) to increase the used range of the ADC. Clearly I am new at this so please be gentle with your explanations! Thanks! <Q> Thermisters are generally used for rough measurements around a narrow operating temperature. <S> If you are willing to build an OP-AMP circuit why not just purchase an MCP9700? <S> This is a nice linear temperature sensor in a TO-92 (or SMT) package which is pre-calibrated and outputs a temperature proportional to the voltage. <S> For example .82V = <S> 82degF. <A> Thermistors are problematic in calibration, linearisation and replacement, should one fail. <S> I recommend a semiconductor sensor instead. <S> Source: Arduino-info . <S> The LM35 is a very simple temperature sensor with 10.0 mV/°C output. <S> See the datasheet Figure 2 for details on extending the range to -55°C. <S> Should the sensor fail or get damaged it can simply be replaced and calibration will be guaranteed. <A> You can use an op amp to boost the voltage. <S> You could, for instance, have the op amp set to amplify the voltage by 2. <S> In the case of your example this would have the voltage go from 4.42V to 4.72V. <S> This may or may not help you all that much. <S> This website has a nice explanation of how to configure an op amp: <S> http://www.electronics-tutorials.ws/opamp/opamp_3.html . <S> Here is a recommendation that I have used and works well - https://learn.adafruit.com/thermistor/overview . <S> It is advertised as having "100 ohms or more of change per degree" <S> So with a 10k or so resistor you should be able to measure everything you need pretty easily.
| I would suggest that you probably need to get a different thermistor that is designed to have larger variability in the resistance over the temperature range in which you are interested.
|
Can I use ethernet instead of DMX for DIY led lighting? I've read that while most pro-lighting is uses dmx, some intelligent lighting rigs use ethernet. I'm interested in building some MCU controlled L.E.D. footlights for my stage show, and currently know nothing about dmx and damn little about lighting period. Would love to hear pros and cons of just using ethernet protocols instead. From my perspective as a pro-coder/hobby-EE, the path of least resistance would seem to be OSC over ethernet between my controlling machine (likely a Pi or something) and the lighting brains, (likely something like a Teensy). I am definitely on Mount Stupid in this area, input appreciated. I have built controllable l.e.d rigs on costumes already but nothing powerful enough for stage lighting. Edit for clarification based on comment:In our case, interoperability with standard controllers is not a concern at all, but reliability definitely is. We are jugglers, and we are almost never a priority with the lights and sound guys, so we are trying to build a system where we roll in our stuff, drive off our controllers that we start, and tell the regular light/sound guys to do nothing until we're done. In a corporate gig situation we don't trust them to hit our cues EVER. lol thanks! <Q> You can do what you like; you could use the old Rank Strand analogue multiplex if you like <S> (it's a joy to watch on an oscilloscope, for a start). <S> The issue is whether you want your light units, whatever they are, to also be able to interface to DMX. <S> If not, knock yourself out. <S> DMX may be more robust than ethernet in an electrically noisy environment, though it is a bit long in the tooth (it started coming in about 30 years ago, if the length of my teeth is a reliable guide). <S> It's also designed to be transmitted over robust connectors rather than weedy <S> RJ45s. <S> It probably depends on how pro you are, or want to be, or to look. <S> Using a non-standard system is fine at first but if you invest a lot in it, then need to standardise, you're throwing a lot away. <S> NB I used to be a theatre lighting engineer but haven't been for a long while. <S> Back in my day, real men used multicores with +10 or -10 control voltages. <S> edit Bear in mind that interoperability is not just about interfacing to other rigs. <S> You may want to purchase a unit yourself to add to your own rig, which would obviously need a standard interface. <A> Most theatres already have miles of suitable DMX cabling, plenty of ports and the distribution infrastructure to get the signals around. <S> Many will also have a DMX capable lighting desk, even if it's very basic. <S> DMX is so popular because it's basic and relies on very simple technology.. <S> but it works very well. <S> The noise immunity offered by RS485 (essentially, see here ) works great in an environment where there's lots of other electrical noise. <S> Whilst ethernet now seems like a simple solution with inexpensive parts it certainly wasn't 25 years ago. <S> If you can run your own wiring then ethernet has enough noise immunity for a theatre environment. <S> It's probably overkill, but if you have a microcontroller or modules that support ethernet and a higher layer protocol it should work fine. <S> Note that many modern lighting desks will also support ethernet but are likely to run their own proprietary protocols on top. <S> Unless you can figure out those details <S> it's unlikely you can interface with them. <A> Traditional DMX (not RDM) uses one master multiple slaves bus topology and is single direction. <S> The Ethernet versions, 10base-2 and 10base-5, that use bus topology are obsolete. <S> So Ethernet needs switches. <S> DMX is simple. <S> With Ethernet, you probably want to use the IP stack. <S> The stack is widely available for any popular platform, but it does add complexity. <S> Alternatively you can choose to "farm-out" the network interface by using some pre-packaged solution like Ethernet to UART bridge. <S> DMX has slow data rate and is good for controlling objects with simple numerical values only. <S> There is no bidirectional communication. <S> DMX equipments may or may not be isolated and <S> non-isolated DMX implementation can potentially be a source of ground loop issues. <S> Ethernet from 10Base-T on up are isolated. <S> While DMX still dominates in professional lighting applications, IoT and smart devices based on general networking are getting popular and will be more and more so.
| Practically any microcontroller with an UART interface can easily be made into a DMX receiver. This depends if you want interoperability with the equipment and wiring likely to be installed.
|
Transformers at full power? I have completed a lab studying using an AC step down transformer. However, I am wondering under normal operating condition, when would a transformer be operating at full power? It seems that during the lab all power ratings were very low, which makes sense to me. However, is there a case in which full power would be reached? perhaps testing? <Q> I would expect transformers in domestic appliances, and most other equipment, to be operated at near their rated power. <S> A manufacturer doesn't want to pay for a 100 VA transformer when he knows it will only be required to handle 25 VA. <A> If you have a load of 10MW (let's say you're a steel rolling mill), then you have a choice of buying a 10MW transformer, which will run at full load, or a 20MW transformer, which will run at half load. <S> The 20MW transformer will cost nearly twice as much as the 10MW transformer. <S> Which do you choose? <S> What percentage of full power is it most cost effective to run at? <A> Full power perhaps can be reached if a transformer which is connected to the rated primary voltage, is loaded as such where maximum rated secondary current is flowing. <A> Depends amortized cost and future demand for power and if you need redundancy. <S> 10MVA also depends on if PF caps are added e.g. 1MVAR 3phase PP rack mounted caps and <S> kVAR-h costs vs KWh costs based on peak demand. <S> then derate further probably costs $7(usd)/kVA out and down time <S> $10k/hr so dual units preferred at 75% load factor. <S> Efficiency and material costs are related for thin Si-coated CRGOS laminate <A> Transformers are actually more efficient when KVA secondary output is increased (mutual induction increased, counter emf is decreased). <S> However, like anything else, output should not exceed 80% of Xformer KVA, MVA, or VA rating.
| For instance, a transformer rated 1000VA @ 100 volt secondary and 240 volt primary is considered as fully loaded when it is being supplied by 240 volt power source and its secondary is supplying 10 amps of current to a connected load.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.