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Determine the amp hour rating of a home made battery pack I'm thinking of making my own electric bicycle, and I need help determining what the amp hour rating of my battery pack would be. I'm going for 36v and 8 or 9 amp hours. So, with 10 batteries in series, I can get a single, low capacity pack with 36v. If I were to make three of these, and put those in parallel, I'd have three times the capacity. But, with each battery providing 2500mah, that's 75 amp hours, and I know that's not correct. What am I forgetting? image according to my knowledge and to the picture, series doubles the voltage and the capacity stays the same <Q> If you series connect the batteries the amp-hour rating stays the same. <S> The voltage has increased though, so the power available has increased. <S> That's why using watt-hours is more useful than amp-hours when comparing battery packs of different voltages. <S> So ... If you're using 10 x 3.6 V batteries in a string you have a 36 V, 2.5 Ah battery. <S> That's a 36 V x 2.5 <S> Ah = 90 <S> Wh (watt-hour) battery. <S> Put three of those packs in parallel and you get a 36 V, 7.5 Ah battery. <S> That's a 36 V x 7.5 <S> A = 270 <S> Wh battery. <S> Now here's where watt-hours are so much more useful. <S> Let's say you're going to need 135 W average power for your motor. <S> Your battery duration will be 270 Wh / 135 <S> W = 2 hours duration. <S> My electric bike uses about 150 W in normal cycling mode at almost 30 kph. <S> I reckon my legs are putting in another 150 W. Little technical point: In the SI system of units it's capital letters for abbreviated units named after someone but lower-case when spelled out: V for volt, A for amp, W for watt, etc. <S> Yes, there will be exceptions ... <A> Showing my work: 2.5 amp hours (one cell) at 36 volts (10 cells in series) is 90 watt-hours. <S> Times 3 of these strings in parallel is still 36 volts (you knew that) and 7.5 amp hours or 270 watt hours. <S> So I think you forgot a decimal point. <S> :-) <S> Side note about connecting batteries in parallel: If you can avoid it, don't! <S> Use higher capacity batteries instead. <S> The reason appears when one cell gets weak or fails short-circuit while the others are still good: If you have one series string and that happens, the pack voltage drops and you have to replace that one cell. <S> Inconvenient, but still the best of all scenarios. <S> If you have multiple strings in parallel, current can flow backwards through the string that contains the bad cell, potentially damaging the entire string. <S> If you tie across all the strings at each level, making one series string where each "cell" is actually a parallel combination of cells, then you're pretty much guaranteed to lose an entire "rung" at once because the bad cell will drag the rest down with it. <S> And there's the issue of how much current actually flows in each scenario. <S> For example, when you first connect the batteries in parallel, whether single cells or strings of cells, they will be charged to slightly different voltages (nothing is ever equal, including end-of-charge from the same charger at different times), which will be imposed across the wires that connect the strings. <S> How much current is that? <A> The overall energy in Wh is increased because the voltage of the battery pack increases. <S> If you want more capacity in Ah, you need to put several 36V stacks in parallel; 3 of them in parallel will give you 7.5Ah. <S> That's purely to answer your question, though you should watch out for cell failure modes and charge balancing.
| The same current goes through batteries in series, therefore the capacity in Ah is not the sum of the individual batteries, it's the same.
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Is there a clever way to use a small variable capacitor where a larger capacitance is required? I am designing an activity where students should band pass a song to find a hidden message at 1000Hz. Problem I have considered using a notch filter and tank circuit , but using the formula $$f=\frac{1}{2\pi\sqrt{LC}}$$I cannot reach the range of 1000Hz with my variable capacitor. Is there a clever way to achieve a circuit to filter 1000Hz and use my variable capacitor for tuning? At my disposal: A variable capacitor (20-350pF) Various capacitor values ranging from 100pF to 220uF. (I can find more if need be) Inductors ranging from 250uH to 3mH The drive to work on the weekend because I find this project so interesting. <Q> You could try an op-amp capacitance multiplier circuit: (Courtesy of Wikipedia) <S> C = <S> C1 * R1 / R2 , <S> so adjust R1 / R2 to provide the level of multiplication desired. <S> Note that some op-amps have trouble driving capacitive loads . <S> So if you connect a significant-enough capacitance to the op-amp output, it may oscillate. <S> Edit: A variant to try is this one . <S> It uses the LM102 to buffer the input current, mostly eliminating the "resistance factor" from the previous circuit. <S> I have not tested it, but would be skeptical of it's performance with very small capacitances. <S> Also note that since C1 is being driven by the LM101A, this is "capacitive loading" and some op-amps could oscillate in this configuration. <S> Some experimentation would be required. <S> (Courtesy Texas Instruments) <A> Given your variable cap and your need for a fairly wide tuning range, I'm afraid you're out of luck. <S> Consider: from your formula $$f=\frac{1}{2\pi\sqrt{LC}}$$$$LC=\frac{1}{{(2\pi f)}^2} = 2.5\times 10^{-8}$$ so $$C=\frac{1}{L{(2\pi f)}^2} = <S> \frac{2.5\times 10^{-8}}{L}$$ and for a 3 mH inductor $$C= 8.45\times 10^{-6}$$ <S> If you pad your tuner with 8.5 uF, and assume 0 pF <S> on the tuner, you'll get a nominal frequency of $$f=\frac{1}{2\pi\sqrt{LC}} = <S> 1.0028356 <S> kHz$$ Cranking up tuner to 350 pF will give $$ <S> f=\frac{1}{2\pi\sqrt{L(C+\Delta C)}}=1.0028562 <S> \text{kHz}$$ <S> Basically, unless you can get much bigger inductors, your capacitance range is just too small to be useful. <S> A capacitance multiplier may sound like a good idea, but as presented it is not applicable. <S> The circuit shown does not produce a pure capacitance. <S> Instead, it is the equivalent of simulate this circuit – <S> Schematic created using CircuitLab where C is the multiplied value. <S> You will find that this is not exactly a useful circuit for what you want to do. <A> <A> You could use your variable caps and fixed inductors to make a LC oscillator .Of <S> course the frequency will be far too high but you could use it to clock a switched capacitor bandpass filter which these days would come in a simple 8 pin chip <S> .Switched <S> cap filters do have snags and the old MF4 is featured in www.badbeetles.com . <S> I am assuming that you will pick a more modern chip thats more suited to be a high Q filter. <S> If you dont want to do this you could at a pinch make a active bandpass filter out of high impedence elements like TLo74 MPF102 using your variable caps with some fixed cap in paralell and high value resistors .. <S> In fact the old Hatfield LMS measuring set has a variable cap in it <S> and it runs down to 20Hz using a wien bridge.
| Simply stick some fixed capacitors in parallel with the variable capacitor until you get to the capacitance you need -- your tuning range will be limited, but the resistor analog of this is a common way of implementing a "fine trim" type of control that operates on some sort of control voltage.
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Make a transistor act as a potentiometer So, how would one get a transistor (probably BJT) to act somewhat like a potentiometer? What I trying to do is have xl6009 buck boost regulator be electronically controlled to charge a li-ion battery pack i.e. using a micro controller to implement CV and CC. Ideally I would insert some kind of transistor into the feedback loop (instead of the normal 10k potentiometer) and thus vary the output voltage. The problem is most BJT's I've seen don't seem to have a large active (linear) region AND the resistances in those regions are too small to allow a broad range of output. <Q> The feedback input pin on the device will normally be sitting at 1.25 volts when the output is at the correct level. <S> R1 and R2 and the output level of 18.5 volts produce exactly 1.25 volts on this pin. <S> However, if you inject a DC current into the junction of R1 and R2 <S> you are telling the chip that it is creating too much output voltage and therefore the chip will alter its duty cycle accordingly and produce a lower output voltage. <S> This is the way to control the device - use a DAC and current source. <S> You can compromise this to a DAC, an op-amp gain stage and a biggish valued resistor. <S> This can be further compromised using a DAC and a medium value resistor. <S> To avoid changing the base value you need to ensure that your DAC can produce 1.25 volts. <S> If you want the output level to fall inject current into the node. <S> If you want the output to rise above the "nominal" take current from the node but be careful because you don't need to take much before the output voltage possibly doubles. <S> So, don't try and pursue getting a transistor to act as a pot. <S> About the only similarity is that it has three terminals. <S> Injecting/extracting current is the clean way of doing it. <A> You can use a digital pot to adjust the output voltage of this regulator. <S> The only tricky bit is that typically a digital pot can't withstand a higher voltage than the logic supply voltage on any of its pins. <S> So, assuming a 5V logic supply, we can do something like this (assume a digital pot such as the MCP4018 , but there are many other types). simulate this circuit – <S> Schematic created using CircuitLab <S> We need the voltage on the 'high' side of RV1 to be <= 5V under all conditions, so that constrains R1 to be no less than 3.33K. <S> That results in a 4:1 range in output voltage (5V/1.25V) for the XL6009 with its 1.25V reference. <S> R1(min) = <S> \$\text <S> R_{V1}\over(\frac {V_{supply}}{V_{REF} }-1)\$ <S> You can easily calculate the value of Rx from the maximum desired output voltage to be: <S> Rx = <S> \$\frac {V_{MAX} <S> R_1}{V_{REF}}-(R_{V1}+R_1)\$ <S> So, suppose you wanted 12V maximum output voltage and you used 3.33K for R1, <S> then Rx would be 18.6K, and the output voltage range would be 3V to 12V (4:1). <S> Note <S> : I've deliberately ignored tolerances and <S> such like- but you shouldn't. <S> Often the digital pot (like mechanical pots) has a poor tolerance for the total resistance. <S> That must be taken into account when doing the calculations. <S> In practice that means you won't get quite 4:1 range in the situation described. <A> The transistor allows you to control the current through this resistor, from 0 to about (1.25-0.3)/R1, by controlling its base current - <S> Ic is approximately Ib <S> * hfe - effectively making it look as though R1 is variable. <S> You can then change the fixed values of R1 and R2 to give you the range of voltages you like. <S> You now have another question - <S> how do you control the current into the base of the BJT to give you the output voltage you require? <S> The answer from @AndyAka is really the solution you are looking for.
| The way to use a transistor to control the voltage output of your regulator would be to put an NPN BJT in series with the resistor R1, with emitter to ground.
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Automation: How can I dispense thousands of paper sheets? I am wondering if anyone has a good idea for a theoretical problem. Let's assume I have 3000 different designs printed on paper. I have about 100 pieces of each design. Each design is not a regular print you could print with a laser printer or inkjet printer.Every design is stored in a little shelf so we have 3000 shelves with each 100 paper sheets in it.Each day we have to pick designs from different shelves. We run from shelf 150 to 356 and then to 788 (just an example). This task uses a lot of time so I was wondering if there might be a clever possibility to automate it. My thoughts so far: I use a linear system with 3 axis. A little mount with suction pads grabs the paper sheet from the shelves and dispenses it to a certain place. Maybe this is not clever, maybe there is a better solution. I appreciate your ideas and thoughts on this issue. Thanks Jo <Q> Amazon has had very good success implementing physical memory caching algorithms in their warehouses. <S> An operating system can map commonly accessed data in fast to access locations(near the centre of the platter) and near each other to reduce travel time. <S> You could organize your stock in such a way. <S> This optimization is difficult to apply after the fact, it is primarily based on where you choose to store things. <S> While this is not "automation" it is an optimization that might help. <S> I think you will find actual mechanical handling of paper from shelves unrealistic cost wise unless you have a massive scale to justify it. <A> Sounds like an ASRS (automatic Storage and retrieval system) you are describing. <S> I haven't seen one for such a small product like a paper, but you could do something like it with a pneumatic tube system like the ones banks and stores use. <A> You might look at how sheet-fed offset printers pick sheets. <S> From this website, here is a photo of the mechanism. <S> The stack of paper was also elevated as the sheets were used. <S> Very fast and very, very reliable. <S> A modern implementation with electronics would be a bit different, of course.
| Amazon found that sorting their warehouses so that commonly access items are closer and near each other has a significant increase in efficiency. The one I used (the Multi 1250, which used to be a classic in print shops before digital printing) used a horizontal flow of compressed air to ruffle the sheets, making sure that they were not stuck together, and the suction to lift the sheet and feed it into the transport mechanism.
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What is the name of the sensor that most inkjet printers use to keep track of the print head's position? I've been trying to build a cheap X-Y-Z table using mainly bits of old CD drives and printers I've got lying around. Since the printers already have a good single axis I would like to be able to, rather than having to buy a stepper motor and rebuild it, just repurpose the optical tabe the printer uses to keep track of the print head's position and keep the same DC motor. I've been able to extract the sensor the printer uses without damaging it or any part of the track, but I've been unable to find what the proper name of this sensor is, let alone documentation describing how to use it. The sensor in question is in a black, plastic housing with a slit that the tape fits into. There are six pins on its underside that solder to the board, two of which come from something on top of the slit and four from something underneath. The black housing is labeled '6536' on top and '15' and 'DSO' underneath. My first guess is that the top is an LED while the bottom is a light sensor of some kind, but beyond basic checks (applying a weak current through the possible LED in either direction did not cause it to noticeably light up) I'm not sure how to get any more information out of it safely. <Q> The tape has very small stripes on it, through which light cannot pass. <S> When this stripe blocks the IR light from the emitter, the receiver does not see it. <S> When the tape continues to move, the clear portion of the tape eventually passes through. <S> The light is then seen by the IR receiver, and the circuit can tell that the tape has moved. <S> I would expect the two-lead part to be the emitter and the 4-lead part to be the receiver. <A> It's an optical encoder just like in the old mice with the steel ball. <S> It will not only detect movement but also the direction. <S> That detection is done with some silicon. <S> And that is the reason for the six pins You have discovered. <S> Two pins belong to an IR-led (more or less easy to be identified with a multimeter), four pins are for the receiver, Vcc, Vss, left and right info. <S> Please try http://reprap.org/wiki/Optical_encoders_01 <S> There is info about how to identify which pin is what. <A> I see this is an old question, but I recently took apart an Avago optical encoder from an HP printer and can provide a photo of what is connected to the four pins mentioned in the question: <S> My first guess is that [...] <S> the bottom is a light sensor of some kind." <S> From the Avago AEDS-962x datasheet , here is the description: <S> As seen in Figure 1 ["AEDS-962x SERIES BLOCK DIAGRAM" below], the module contains a single Light Emitting Diode (LED) as its light source. <S> The light is collimated into a parallel beam by means of a single lens located directly over the LED. <S> Opposite the emitter is the integrated detector circuit. <S> This IC consists of multiple sets of photodetectors and the signal processing circuitry necessary to produce the digital waveforms. <S> The codewheel/codestrip moves between the emitter and detector, causing the light beam to be interrupted by the pattern of spaces and bars on the codewheel/codestrip. <S> The photodiodes which detect these interruptions are arranged in a pattern that corresponds to the radius and count density of the codewheel/codestrip. <S> These detectors are also spaced such that a light period on one pair of detectors corresponds to a dark period on the adjacent pair of detectors. <S> The photodiode outputs are fed through the signal processing circuitry. <S> Two comparators receive these signals and produce the final outputs for channels A and <S> B. Due to this integrated phasing technique, the output of channel A is in quadrature with channel B (90 degrees out of phase).
| It sounds to me like an optical encoder, which looks something like this: One side is an infrared emitter (which is why you can't see it--Infrared light is not visible to the naked eye) and the other side is the infrared receiver.
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How can a smartphone's extended battery be same size as the original battery? I know this is kind of a basic question, there might be a few ways to answer it... Recently I found out there is a slim extended battery which has the same size as OEM battery (I mean really same size height and width, not using a different battery cover,) how is that possible? In simply physics, if the OEM Battery already max out the capacity of storing mAh in such a size, how is the Extended Battery able to have more capacity with the same size? For example, I found original battery capacity is 2800mAh But here the extended battery can be up to 3800mAh, 4400mAh , that's a huge difference... Are those extended batteries just fake products? <Q> Usually just marketing <S> Most aftermarket batteries are labelled in way that is extremely optimistic (if not an outright lie). <S> It is possible to change chemistries to go to one with a higher energy density, but this is usually very unsafe as the phone's internal battery charge regulation circuitry is not tuned for the new chemistry and can cause a fire or explosion as a result. <S> It is also possible to gain capacity by utilizing the volume better. <S> Most cell-phone battery packs have integrated safety electronics (current-limit, thermal fuse, short-circuit protection, and/or similar safety features). <S> Using more advanced/smaller packaging for these electronics can yield more volume for active battery material. <S> Similarly, you can use more advanced/thinner substrates to get more active material into the same volume -- but these improvements are typically small (maybe no more than 10% in most cases). <S> So... <S> Yeah, it's probably just a fake product built on the (probably correct) assumption that nearly all buyers will not test the battery's capacity to verify the claim (e.g. they won't get caught). <A> This can also be more or less true and here's how. <S> First, battery technology slowly advances. <S> One major power tools producer launched its 10.8 volts line of professional power tools in perhaps 2004 and they shipped with batteries assembled of 1Ah Li-Ion cells. <S> In perhaps 2007 they switched to 1.3 Ah cells of the same size. <S> Later they somehow found 1.5 Ah cells and 2.0 Ah cells still of the same size. <S> Nowadays tools from that line come with 1.5 Ah and 2.0 Ah batteries. <S> Newer batteries can be used on older tools and older batteries can be used on newer tools. <S> I have no independent data to back the claim that batteries capacity indeed increases as claimed but batteries are key elements to power tools <S> so I assume a major tools producer would not fool us like that. <S> Second, it depends on how you measure. <S> You can charge the battery to different states. <S> One major power tools producer claims certain Li-Ion batteries are charged fully in 35 minutes. <S> Answers to this question explain that it's simply impossible and the most likely scenario is that the battery is actually charged to about 80% which isn't that bad. <S> So if you know that the target device only charges a battery to 80% you can charge your battery to 100% when measuring capacity. <S> Customers will be unable to use that full capacity <S> but you can simply disclose your measuring procedure in the finest print available. <S> Fourth, you could find a battery that uses more compact safety features or lacks some of them. <S> Major battery producers make some effort to ensure that batteries don't explode or burn under heavy loads. <S> That also takes some volume of the battery. <S> Make those less reliable or remove them - and you can fit more chemicals into the battery and get larger capacity. <A> Hypothetically, battery technology could have improved between the release of your phone and the release of the third-party battery. <S> My first mobile phone though, came with an NiCd battery. <S> A later released battery, from the original manufacturer, was an NiMH battery, and was indeed both slimmer and had higher capacity than the original one <S> (NiMH has up to three times the energy density of NiCd). <S> In the end I bought an NiMH battery even thicker than the original, and never had to charge my phone outside weekends. <S> Those were the days. :-) <A> A larger capacity battery is always heavier ,thicker than the lower capacity battery you are replacing usually . <S> If they are li-Ion as I am sure they are there is only one type approved for cellphones and that is the Lithium Cobalt Oxide(LiCoO2) because of their relatively long life and safety .Normal type are <S> 150–200Wh/kg. <S> but these higher capacity ones you are looking at maybe specialty cells which can provide up to 240Wh <S> /kg ( 37% more ) which matches your 2800mAh to 3800mAh increase quoted , a 36% increase . <A> In my experience, getting higher capacity cheap batteries is not that hard. <S> However, they are rarely worth the trouble since they lose 10% of the advertised capacity already on the shelf, are down to the nominal capacity of the original battery within a month, to the actual remaining capacity of the original battery within another month, and dead within another. <S> I guess that they don't have enough or proper material inside to withstand and contain the onslaught of the rather aggressive chemicals storing the energy. <S> It would be nice if batteries had to advertise the charge they are expected to provide over their lifetime rather than merely a one-time charge capacity. <S> So I try steering clear of cheap high-density offerings. <S> Particularly if they don't even fall short all that much of the original capacity claims, they are likely to degenerate far too fast for my liking. <A> If you see a Li-Ion battery for your phone that states more mAh <S> it is a lie. <S> The only way to increase mAh is to make the battery larger. <S> They usually supply a different back panel for the phone that allows for the thicker battery. <S> With today's smart phones I'm not seeing that as much as I did 10 to 15 years ago. <S> I bought a couple of the so called 4600mAh batts for my Note 8 that had 3300mAh <S> batts originally. <S> I saw no difference in battery time. <S> These larger mAh claims are fake specs!
| Third, you could find a battery with slightly thinner case which simply holds more chemicals and thus has larger capacity. This can be false marketing. It's not very likely though, as the other commenters have said, given that battery technology hasn't moved much in the last few years. A modern new stock Li-Ion battery of the same size cannot have more mAh .
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Sensor to track people sitting still in the room I'm currently using a PIR sensor to turn on the light when it detect my hand's motion inside the box. But when my hand stop moving, the light turn off immediately. Is there any sensor I can add to keep the light on after my hand stop moving? <Q> How about a thermal IR sensor ? <S> The description says "the D6T sensor is ideal for human detection; even for stationary objects a normal PIR structure would struggle to detect." <A> It is implicit in the design of a PIR sensor that there must be motion, because the electrical signal inside the sensor is AC-coupled. <S> Therefore, if the thing you're trying to sense is not moving, you need to move the sensor. <S> Scanning it back and forth through an angle that's at least as large as the angle between the Fresnel zones in its lens assembly should provide a continuous signal for any stationary target. <A> Provided the computational resources are there, you could use a cheap web camera and object detection algorithm. <S> For example, OpenCV's human detection algorithms are quite good and completely free (as in beer as well as in speech). <A> Assuming you are sticking your hand in a box, that your arm is blocking the way, you can use a simple IR LED/Photo Diode Pair. <S> It goes by many names, an IR obstacle sensor, a IR Proximity Detector, an IR Line Sensor or Line Follower circuit, IR Breakbeam sensor, or even an IR Distance Sensor. <S> The IR LED is used to illuminate the IR Photo Diode. <S> This makes it conduct. <S> This can be read in a digital or analog manner. <S> The Basic Circuit: <S> The Make version would have both diodes side by side, with one shielded from the other. <S> Anything in front of it will reflect the light and turn the sensor on. <S> The Break version would have the diodes facing each other, and anything that comes in between would turn the sensor off. <S> Just inverted logic. <S> You can get raw diodes, or complete modules. <S> The rest is handled by your microcontroller.
| Compared to a PIR sensor, this will keep the light on as long as the object is in the way, even if it is not moving. For your purposes, in either a make, or break, manner, someone sticking their hand in the path of the IR LED's light, will be sensed.
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How can i extract transfer function an unknown nonlinear system? I am trying to do PID control for my electroservo motor system by using nichols ziegler tuning method. My system has SSI encoder output for motor feedback mechanism. I will use this knowledge for control. According to nichols ziegler method i must know transfer function of my system. But i can not find its equation exactly. So how can i extract its transfer function? I need a methodology for this. Can i extract T.F. by using Matlab/Simulink or LAbview? <Q> You can definitely try System Identification Toolbox of Matlab. <S> Official page says You can use time-domain and frequency-domain input-output data toidentify continuous-time and discrete-time transfer functions, processmodels, and state-space models. <S> Which is what you are looking for. <A> Ziegler-Nicholls tuning does not require the TF to be known - that's the whole point of the method. <A> Your really have two choices. <S> The first, like @LvW suggested, is to use the motor specifications such as in the illustration. <S> This may be bore accurate but also more difficult because not all the specks will be available. <S> The second involves what you suggested. <S> By applying a voltage to the motor and recording how its speed behaves. <S> In effect this creates a step response that can be analyzed to find the specs for either a first order or second order system. <S> If the system oscillates then a second order is what you want. <S> If the system does not oscillate then you may be able to use a first order. <A> Nonlinear system models rather stick to time domain descriptions as nonlinear differential equations rather than frequency domain descriptions. <S> But in terms of current-in, speed out, your motor-encoder system is close enough to a linear system that you really don't need to concern yourself with nonlinear aspects (unless you are trying to control shaft angle to micro-radian precision!). <S> Perhaps the easiest way to obtain a linear model is to apply a simple proportional feedback control tuned to get the loop stable, then record input-output data to a step response. <S> Then fit the data to the closed loop transfer function. <S> From the closed loop transfer function you can calculate the open loop transfer function, factor out the proportional gain and voila - your motor model! <S> A simple linear DC motor model looks like: $$\frac{\omega}{i}=\frac{K_T}{Js+B}$$where <S> $$K_T$$ is the torque constant of the motor,$$J$$ is <S> the motro shaft and load inertiaand $$B$$is the linear viscous damping of the motor bearings <S> Perhaps your motor supplier already specifies these parameters in which case you don't have to test - you can write the model directly. <S> Note that even if you are using a permanent magnet synchronous motor, in feedback with a stiff current controller, the model approaches the model of the DC (brush) motor. <A> There are two very good methods for estimating transfer functions. <S> Look up moen4 and fitfrd. <S> To use moen4 you need basically input and an output of a test. <S> The algorithm then computes the transfer function that best fits the data. <S> The results tend to be pretty good for some systems, less so for systems that have significant non-linear behavior (for which a linear transfer function does not exist). <S> Here is code that you can use for both frd fit and moen4 fit. <S> You can plot the data of freq_response_frd (frd object) directly using the bode <S> () function to get a bode plot of your input data. <S> Your input data must have sufficient frequency coverage so use a chirp signal that increases with frequency in time and collect the resulting response in another array. <S> Then pass both arrays into the id_model_moen and you will get your transfer function back. <S> I typically limit the frequencies that are analyzed because if you plot the full range returned by fft you will get a lot of noise outside of the range for which you even have test data - <S> so that's useless part of results. <S> function [mag, phase, f] = <S> freq_response_mag_phase(out, in, t, freqlim) <S> dt = <S> (t(end) - t(1)) / (length(t) - 1); NFFT = length(t); Fs = 1.0 / dt; fb = fft(out, NFFT) <S> ; fa = fft(in, NFFT); f = [0:NFFT-1]*Fs/NFFT * 2 * pi; % find first bin after our test range. <S> We will discard bins after it. <S> ix = ceil(NFFT/2); <S> if(exist("freqlim <S> ", "var")) ix = find(f>freqlim,1); end f = <S> f(1 <S> :ix); mag = abs(fb(1: <S> ix)) ./ <S> abs(fa(1:ix)); phase = unwrap(angle(fb(1:ix)) <S> ) <S> - unwrap(angle(fa(1: <S> ix)));endfunction response = <S> freq_response_frd(out, in, t, freqlim) if(exist("freqlim", "var")) <S> [mag, phase, f] = <S> freq_response_mag_phase(out, in, t, freqlim); else [mag, phase, f] = <S> freq_response_mag_phase(out, in, t); end response <S> = frd(mag . <S> * exp(1i . <S> * phase) <S> , f);endfunction sys_tf = id_model_frd(out, in, t, nr) <S> resp = <S> freq_response_frd(out, in, t); sys = fitfrd(resp, nr); [b, a] = ss2tf(minreal(sys)); sys_tf = <S> tf(b, a); endfunction sys_tf = id_model_moen(out, in, t, nr) <S> dt = <S> (t(end) - t(1)) / (length(t) - 1); sys = moen4(iddata(out, in, dt), nr); [b,a] = ss2tf(d2c(sys)); sys_tf = tf(b, a); end
| The concept of Transfer Function is only defined for linear time invariant systems.
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Full bridge HVDC to modfied square sine - random half side MOSFET burn I am trying to make a step-down 350V to 220V voltage converter. I decided to go with rather simple circuit of full bridge MOSFET inverter since my loads are only resistive or switched. I had built it and tested today, however, without any success. It runs well with intended loads then, randomly, it blows the fuse. Further investigation shows that two fets (upper & lower) on one side dead shorted. I install another pair, after some time (mostly from minutes to no more than one hour) it burn. Then again, but opposite side. In all cases, both high and low fets burn. When I lucky to run it for a long time without blows, I remove power then check fets temperature and they are OK. My fets are IRFP460. I use two IR2110 unified into full bridge, the load is purely resistive (a set of series lamps plus one single bulb) and draws 450mA @ 220VAC. This load is not the intended load, I plan to power more resistive loads with this circuit. IR2110 are driven by TL494 as signal generator which has a onboard trimmer to adjust duty cycle. . This picture is reference only showing how my bridge is being built TL494 and IR2110 are powered by small onboard flyback converter which is NOT isolated from network common. Whole circuit gets power directly, no diode bridges. The power is 350VDC line which is another SMPS driven by lead acid battery pack. For two years I use exactly same circuit, but low voltage version: fets are IRF1404 (40V), and it gets power from separate 12V 1A low voltage flyback. Confirmed operation up to 30V on input. I probably miss something obvious, but can't figure out exactly what. Snubbers? If needed, I can post my development PCB picture. UPDATE: This is my old scope showing signal between TL494 and IR2110 (IR2110's input). Those IR2110 are working and there are no shorts on board after them. I can vary duty cycle from 0 to 97%. This is exactly signal waveform I want to see at device output. <Q> Furher there is no need both to switch both HI an LO side with PWM frequency, one can be ON/OFF and the other half is doing hard switching, you will get less loses. <A> I cant see any fast cycle by cycle current limiting . <S> Imagine if you put the square wave invertor into a load containing some capacitance .Many <S> mains loads have low ESR metal film capacitance between P & N for EMC or PFC reasons. <S> What now limits the prospective peak currents ? <S> Wiring ? <S> RDs on ? <S> ESR of HVDC buss capacitance? <S> So on your circuit you could get a prospective fault current of hundreds of amps .Chinese <S> invertors that operate this way complete with TL494 are popular in my country and are marketed as "modified sine wave "Placing inductive reactance in series with the AC output has been shown to stop the fets blowing up on a case by case basis. <S> Remember that George Ohm stated that resistance was proportional to Absolute temperature .Not always true for everything but <S> filament lightbulbs can have one tenth of thier resistance when cold ,meaning that the surge current at cold turn on could be 10 times the load current <S> so unprotected fet goes Bang again. <S> simple invertor can be really efficient but its only good for resistive loads .I <S> advise people to wire them in permanently to the known load ie hotwater element .This <S> stops people plugging in stuff that will blow the FETs .In <S> terms of $ per watt your invertor is one of the best if not the best . <A> Two things i can recommend: refuce the 47R to even 0, try catching both high and low side conducting. <S> It's almost certainly bad dead time management. <S> Also make sue the circuit is disabled on powerup. <A> Since my load does not require AC, I switched to DC "dimmer" circuit which requires only one MOSFET and it's robust and simple. <S> Just flyback supply + NE555 with common ground. <S> The left thing unknown is only that I connect any type of load to it and mosfet <S> does not blow at all, even if I short it when it is being supplied from limited power source. <S> I still think that my full bridge attempt has control issues, but without proper scope I cannot test it.
| You would need something more sophisticated than TL494, a complete H bridge wit four outputs and dead time setting, to prevent conduction at once. Also with bootstrapping there is a possibility that you can't have 100% duty cycle (not sure on that). This type of invertor is OK for powering a hotwater cylinder element as part of an off grid power system .Your
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College Computer/Electrical Engineering Circuits - Modulating Signal DC offset based of modulating index question Here is an image of the question. I am stuck at part a, which asks to find the DC Offset based of a modulating index of 0.7, or 70%. So essentially, you will transform the maximum and minimum, yes? How do you know the minimum and maximum values? I know that m (modulating index) = (max-min)/(max+min). Should the Vpp value be preserved and stay the same as original after the transformation? How do you do this? Would you do trial and error? Or am I using the wrong formula? <Q> Here's what 50% AM looks like: - Now imagine the slowish sinewave shape (in red) is your signal so roughly when your signal bottoms out at about -0.033 "units", this has to correspond to the -70% level as implied in my picture. <S> Yes I know my pictures shows +50% and -50% <S> but you can see that there will also be a point that equals + and -70%. <S> Your signal has to fit in those limits exactly and, because AM can be regarded as simple 2 quadrant multiplication, there needs to be a DC offset applied to your signal to get it to fit as I've prescribed i.e. the "centre line" of your signal needs to be moved to the nominal level in my picture. <S> Note that the above sentence previously said "the average level of your signal needs to be moved,,,," <S> BUT, because the signal is asymmetric about the average level this was misleading. <S> Now I have no idea what the graph in your question is actually shwoing in terms of volts because the axis aren't labelled and believe you me <S> I'm not going to start plotting out the formula <S> BUT, I can see that the graph does not bear much resemblance to the formula in terms of amplitude so you are on your own on this. <A> you have the correct formula -- look like your extremes are +0.045 and -0.032. <S> When you add a DC offset, it will add to each of these values. <S> Select the offset that meets the modulation index requirement. <A> I have this same question in my homework (we are probably in the same class). <S> You have the right formula, (modulating index) = <S> (max-min)/(max+min). <S> What I did is first find the max and min of the signal. <S> This is hard to do from the graph, so I used Mathematica to find the actual values from the given equation. <S> I get Vmax = 45 mV and Vmin = -32.6 mV, but you should check that to make sure. <S> Then I used the equation, adding in Vdc because that is what you are adding on and solving for. <S> For my calculations, I put all my values in Volts. <S> So we have <S> .7 = <S> [(Vdc + .045) - <S> (Vdc - .0326 <S> )] / [(Vdc + .045) <S> + (Vdc - .0326) ]. <S> Solve for Vdc.
| You can use MATLAB, your calculator, whatever, just find the max and min.
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Can 12V car windshield wiper motor be powered using PC power supply? As title says, can 12V car wiper motor be powered using PC power supply?According to amperage of typical car wiper motor I would say yes, but I want to check :)Thanks in advance <Q> Probably; assuming you use the 12 V line. <S> Some PC power supplies don't start unless there is a load (20 mA ?) <S> on the 5 V line. <S> You need to watch for the motor inrush (stalled) current, not the operating current. <A> It should work. <S> (Considering you checked the voltage&erage on your car and power supply) <S> Here is a link that has more details: <S> http://www.scary-terry.com/wipmtr/wipmtr.htm <S> and more about the power supply: <S> http://www.scary-terry.com/atxps/atxps.htm <S> Hope it helps. <A> You are probably better off using a somewhat older power supply. <S> At IBM XT times, floppy drives tended to be run on the 12V power supply (at PC times, even the RAMs needed a 12V rail) so the 12V rail is quite capable. <S> Older power supplies lack any of the modern sensing interlock controls. <S> That means that if you flip the power switch, the power supply will be on. <S> Modern power supplies have a complicated startup procedure including a soft-button, presence detection, minimum load requirements on the 5V rail, et. <S> al. <S> So using an older power supply will make your life easier. <S> It might also be a good idea to check that your circuitry contains a fly-back diode to keep the motor from trying to feed back power into the circuitry or power supply when switching it off. <A> That should probably work. <S> All matter is correct voltage and the amperage. <S> As usual a PC power supply of 250W wattage can output maximum current around 14A amperes. <S> The maximum output ranges are marked on the power supply according to the voltage range. <S> You can switch ON the power supply by connecting the “ Green ” color wire in the mother board socket with any black color ground wire with a jumper cable.-Make sure to use only “4 pin Molex connectors” or “floppy connectors” to obtain the voltages. <S> Don’t use any mother board sockets pins for that because they have not built for heavy amperage uses. <S> I also agree that use of a fly back diode for better safety of the power supply.
| It is safe for the power supply until the motors’ maximum input amperage is below than the maximum output of the power supply.
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FPGA Frequency divider with linear regulation I'm currently developing an FPGA gateware requiring a regulated frequency divider and I wonder if there is any tricky way to get a linear output frequency regulation? What I mean is that if I use a simple preloaded counter I get a f~1/x relevance, where x is a counter initial value. As I need to regulate frequency in a relatively wide range (say 0.5Hz-10kHz) I'd like it to be a linear function of some argument (of course, not necessarily a counter value-it can be something more complex ;-) ). <Q> At the moment, you are considering using a preloaded counter. <S> This means that each clock cycle you get a fixed increment, in a variable accumulator length that you control, hence the frequency is reciprocal to your control value. <S> If instead you use a controllable variable increment in a fixed accumulator length, your frequency will be linearly dependent on the control value. <S> This is how a DDS works. <S> A fixed length accumulator, typically a power of 2, for instance an 18 bit accumulator that counts up to 2^18 (250k-ish long) is incremented each clock cycle odf the system clock. <S> In pigHDL, you would write int count [17 downto 0]; process(on sys_clock rising): <S> _ count <= <S> count+freq; _ output := count[17]; <S> This may, or may not, give you what you need. <S> The MSB of the counter will give you the output, but unless freq is an exact power of 2, the output cycles will not be exactly the same length, they will vary by one count. <S> The average output frequency = <S> freq * fs / accumulator length. <S> You can get increased resolution for the frequency by increasing the length of accumulator and frequency word, to the right. <S> There is no way to remove this one cycle jitter, if you are going to use the clock as a digital source. <S> Bear in mind that for an FPGA, it is bad form to take the MSB output and use it as the clock line to other elements. <S> Much better to turn it into a one cycle ClockEnable, and use that to condition the clock into downstream elements. <A> I will suggest using a binary rate multiplier. <S> I have attached a reference sheet that shows what I mean. <S> You can accurately use the binary multiplication to get a good range of output frequencies. <S> Not exactly linear, but easily controlled. <S> http://www.ti.com/lit/ds/symlink/cd4089b.pdf <A> There are several things to consider. <S> 1) <S> The higher the main FPGA frequency you use the better. <S> This gives you more resolution at the higher frequencies. <S> This translated to more bits in your divider. <S> 2) <S> If you were to limit your frequency range you would have a more practical ability to realize a Freq -> Divisor relationship that could be considered more straight line.
| If it's for an external source, you can take the top several bits of the accumulator into a DAC, filter the waveform, and use a comparator, this will reduce the jitter considerably.
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Time delay dpdt relay using only pot, relay, and cap (discreet components and no transistors) I am trying to design a simple time delayed relay circuit that uses only a potentiometer, 5V dpdt EM relay, a single sufficiently large capacitor, additional resistors (if needed), and buttons/switches, powered by a 9V battery. The schematic below works to simply turn on an LED while a button is pressed and keeps it on for a moment. I have only been able to extend the time by adding more caps in parallel. I believe there is a simple solution to this logic problem that has the relay triggering itself almost as an oscillator, but when setting up as an oscillator I still can only vary the time by adding caps. How can I arrange and wire the components mentioned to allow a time delay of a few seconds to minutes, dependent upon a single potentiometer? I thought a toaster's circuit would work, but most I see use a 555 timer or a pair of transistors and other components. Is there a simple way to use an RC circuit alone as a timing circuit without the solid state components? I know how to do this using transistors and ICs, but that is not the point. It is an experiment to show a possible use of capacitors and RC circuits without first teaching about solid state components, ICs, etc. The time delay can be delay ON or delay OFF. Thanks simulate this circuit – Schematic created using CircuitLab <Q> I'd say forget it. <S> There is a reason all those circuits you see use a 555. <S> You might get it to work with a very very low current relay, maybe a 230V relay, with an idem capacitor. <S> But that is exactly the circuit you don't want a novice to play with. <S> (At least, if you want him to live long enough to rise above that level.) <A> The hold-on time will be in the order of the time constant given by the formula T = R x C <S> So the value of C required will be given by the formula (approximately) C = T / R <S> So, figure out your relay coil value and calculate C. <S> The answer will be in farads so multiply by 1,000,000 to convert to μF. One of the problems with this approach is that the voltage on the coil gradually drops off and the contacts slowly open. <S> If there was any significant current flowing then the contacts will start to arc and heat and lifespan would be reduced. <A> one way to extend the time is put a resistor parallel to the switch However: the resulting circuit is more vibration and supply voltage sentitive simple circuit start with 4 times the resistance of the relay coil. <S> lower resistances give longer extensions.
| A practical capacitor does not old enough energy to power a normal relay for the period of time you are aiming at.
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How to verify the reliability of torque rating of DC motors given on online sites? I have to decide the motors to be used for my robotics project, but while searching for the DC motors I have found out a rather unusual fact that I can find several motors who provide the same torque while there is a significant difference in weights. I am curious to know why? For eg. I can find 12 kg cm torque motor weighing just 125 gm whereas I can also find 17.5 kg cm torque motor weighing 600 gm. I really don't understand where does the difference lie? And is it reliable to use the lower torque motors. <Q> Its weight has nothing to do with its Torque. <S> These two numbers are not mutually exclusive. <S> While some heavier motors may have larger torque to weight ratios, there is no single rule of thumb to gauge them by. <A> Power is in proprotion of cube of dimensions, as well its weigth. <S> For example we have a motor with dimensions A x A x A with power of P1 = <S> 1W, then what will be the power rating of similar motor with dimensions 2A x 2A x 2A? <S> The answer is P1*2^3 = <S> P1 <S> *8. <S> Why is so? <S> Because the weight of machine is growing with x^3 of its dimensions (more steel, more copper), because the volume grows. <S> Power is related to torque and speed, P = M * omega, therefore the same motor can have large torque, but small final speed or high speed and small torque. <S> You should look both parameters: speed and torque. <A> It's all about efficiency. <S> Basically if you put more winding copper into the motor it will be able to produce a given amount of torque more efficiently than one with a small amount of windings. <S> However more windings will also make the motor much larger physically. <S> Which of these is important to <S> you really depends on the application. <S> For example, an air conditioner motor will have more windings because efficiency is more important than weight, while a model aircraft motor will have less windings and trade efficiency for weight. <S> In terms of reliability the main issue is heating in the motor. <S> The smaller motor will simply get a lot hotter and also have less mass to absorb the heat. <S> If you don't deal with this heat properly then it could fail. <S> The other aspect is the motor RPM. <S> If you can run a motor faster it is generally more efficient. <S> However at some point the motor will either fly apart, or core losses will start to dominate, reducing efficiency again. <S> In a well designed motor these points will all converge at the max operating speed (hopefully with a bit of safety margin on the flying apart RPM). <S> In terms of assessing this information you'll need to look at the rpm/voltage or torque/current constant and the winding resistance. <S> Using these two numbers you can work out the efficiency of your system and decide what sort of efficiency/size tradeoff is appropriate for your application.
| Motor Torque ratings are a factor of its electrical and electro-magnetic construction. You cannot verify the torque by weight alone.
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Switch (Relatively) High Voltage from Logic Level The Situation: Basically, I would like to switch a load that requires 30v using a switch that cannot have more than 5v across it. More descriptively, I am trying to drive the anode of a VFD segment from the output of a shift register (probably a 74HC595). What I Have So Far: Because the segments are essentially common-cathode, the anode has to be switched. You can't really switch that with an N-Channel FET, as when current is flowing drain to source, the source would instantly be at ~30v, which being higher than the gate voltage (5v), would instantly turn off (I guess maybe it would actually reach equilibrium at some very low conductance?). If a P-Channel FET were used, the gate can be pulled up to the 30v rail for off, but then you have 30v at the logic output, which is far, far outside of spec. Short of shifting the logic-level ground up to 25v, which seems like a major pain, the best that I can think of is to add an N-Channel FET to drive the main P-Channel FET, and pull down the gate of that to ground, like this: (I showed it with just a simple switch and load rather than an IC output and a floating anode so that this will be hopefully more applicable and/or easier to understand for people with other applications...) The Question: The problem that I have with this is that it seems like an excessive number of components for such a simple task. There has to be a simpler way to switch the higher voltage, right? I feel really stupid having so much trouble with something so simple, but I've spent all day on it, and this (or shifting up the logic ground) is all that I've been able to think of. I guess in the grand scheme, it's only one extra transistor and one extra resistor, but it kind of adds up when you're driving a large number of lines... <Q> There has to be a simpler way to switch the higher voltage, right? <S> No, there isn't. <S> Not from the high side. <S> You can put a single N-channel FET on the low side of the load though. <S> If you use an open-drain or open-collector device where you have your 5V switch, you can reduce the parts count though by eliminating Q2. <S> But conceptually it's the same thing. <S> Also, there exist integrated high-side switches that incorporate Q2. <S> These can be frustrating to find for your application because a lot of them only go up to 8V on the high side (they shift from really low levels like 1.8V), <S> e.g. TPS27081A , NTJD1155L , Si3865DDV . <S> and I don't know if you can still buy it. <S> Shop around, there are surely more. <S> There exist some high-side switch/driver arrays of this kind; <S> TPD2005F would work and has parallel inputs. <S> This one "cheats" by actually having a charge pump and N-MOSFETS as outputs. <S> There are a lot more with serial inputs, e.g. AMIS-39101 . <S> You could of course just buy one such array dedicated for VFDs at this point, e.g. MAX6922 , or even an integrated array+controller like STM86312 . <S> Which one of these to chose depends how much of a DIY exercise you want to make of this... <A> You can indeed get rid of Q2, although a normal open-drain part will have trouble with high voltage on the drain as the output FET will break down -- normal HCMOS output FETs are not rated to drive 30V! <S> A part like the TPIC6B595 will do the job, though -- that right there will save significantly on parts count. <S> Once you have the level-shifting handled, you can then use a high-side switch array to take care of the rest of the work. <S> (Oh, and the pullup on the output of the TPIC6B595 can be a resistor-pack, if that helps you any.) <A> This circuit has a problem -- when Q2 turns on, it will put 30 V across R1 -- which is also the VGS of Q1-- and this will exceed its specification significantly. <S> There are a few ways to fix this, but the most robust is to put a 5 V zener in parallel with R1 to keep the voltage reasonable. <S> Now, to control the total current, you have to put a new R in the source of Q2. <S> Assuming your speed requirements are simple, here are some good values to use: R2: <S> Not critical -- 10k or 100k would work. <S> R1 -- controls the turn off rate of Q1 -- 1k or 10k might be suitable (depends on FET and on speed requirements). <S> Select 10k <S> Rnew -- needs to drive 5V/10k <S> = 0.5 <S> mA. <S> When you drive the gate of Q2 with (say) 5 V, and VTH is 2 V, there is 3 V across this R == <S> > R = <S> 3V/0.5 <S> mA = 6k ohm -- use 4.7k.
| Or, you can simply use an integrated high-side driver/level shifter array. IPS511 would work for you voltage-wise, but it has more stuff in it than you probably want [over-temperature protection etc.]
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Why do multimeters have an aluminium foil at the back? I was wondering what is the purpose of aluminium foil at the back of some multimeters? I did a continuity test and found that it is connected to the ground (COM) terminal. My initial guess was that it was some kind of shield but since it wasn't fully enclosed and multimeters don't work in that high RF range it didn't seem plausible. Also, it wasn't some kind of ground plane as it is only connected to the board at one point via the spring. What is it exactly? <Q> It's a shield, but to keep RF out, not in. <S> The multimeter does not use high frequencies internally, but it does have high-impedance internal nodes that could be affected by external sources of RF. <A> Dave Tweed's answer has certainly got merit but my first thought is to prevent "hand" effects altering readings. <S> Clearly, if the DMM is sat on a wooden bench or maybe sat upright on it's stand it will produce the same reading but, if you are cupping it in your hand, all that extra capacitance could affect AC measurements given that some of the circuit nodes are (as Dave says) very high impedance. <S> The same argument for sitting the DMM on a conductive bench. <S> Similar sort of argument for stopping some digital areas <S> /signals coupling to some sensitive measurement areas via external capacitance such as hand or a metal bench. <S> I'm putting forward that there are significant reasons for using an "earthed" shield to keep measurement repeatability good. <A> In operation, the PCB of the meter may be a high voltage, perhaps AC, with respect to surrounding conductive objects (a metal bench, a wire, your hand). <S> That will cause increased noise on a dual slope A/D converter, and will cause a reading error when rectified by the AC-DC converter circuit used for AC volts and AC current. <S> When I designed a line of DC panel meters we did not find a shield necessary but for AC current meters (50mV full scale) a shield within the metal enclosure was necessary with the shield connected to the 'ground' of the circuit and the metal case earthed. <A> As said below, its indeed a partial faraday cage, it is necessary to protect high impedance nodes from rf, also however it protects sensible ICs namely in this case the ADC, which can be affected by EMI regardless of the input and supply configuration
| It's a partial Faraday cage to shield internal high impedance nodes against coupling.
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Can one tri-state pin drive two SS pins? This is how i see it; State H: SS out is High, Deselecting SS 2, turning off Q2 and on Q1, which grounds SS 1, turning it on. State L: SS out is Low, turning off Q1 and grounds SS 2 through Q2. SS 1 is deselected by R1 State Z: SS out is high impedance. R1&3 keep the SSs high. ARGH EDIT: R4 keeps Q1 off. R6 may keep Q2 off. IDK simulate this circuit – Schematic created using CircuitLab Am I wrong? It's purpose is for when gpio pins are a premium. It's a 3V3 system. <Q> My concern is that the low level produced by the PNP will be just under \$0.7\mathrm{V}\$ because of the required \$V_{be}\$ for the transistor to be on. <S> For \$5\mathrm{V}\$ CMOS, the low level input voltage is usually at most \$0.8\mathrm{V}\$ <S> (for \$3.3\mathrm{V}\$ CMOS it will be a little lower), which means you are right on the edge. <S> Personally I wouldn't be happy to run that close to the edge. <S> In fact, now that I have had a go simulating the circuit, I'm not convinced it will work as planned when floating. <S> Q2 won't turn off properly unless you add a pull up resistor - but in order to be strong enough to work, it would cause Q1 to turn on. <S> There is a way I have done this in the past, but it requires two comparators. <S> This isn't too bad as a dual-comparator package <S> is only 8 pins and would take up about the same as your discreet transistors. <S> Basically the approach is to turn the input into ternary - you have equal pull-up and pull-down resistors so that when floating the input voltage will be about half the power rail. <S> You then have a comparator for each output. <S> For the first device the comparator is set up so that it outputs a low only when the voltage is less than one third of the power supply. <S> For the second device you output a low when the voltage is above two-thirds of the power supply. <S> It will require 5 resistors and 1 dual-comparator. <S> The circuit is as follows: <S> The above can be simulated here . <S> It's simulated for \$5\mathrm{V}\$, but the circuit would be identical for \$3.3\mathrm{V}\$. <S> Essentially the top comparator will be low only when the input is driven high. <S> The bottom comparator will be low only when the input is driven low. <S> If the input floats, both comparator outputs will be high. <S> This is <S> almost a Ternary to Binary converter circuit, it's not strictly speaking as you need the outputs 01,10,11, whereas ternary would be 00,10(or 01),11, but its essentially the same thing just with one bit inverted (hence the comparator is the other way up). <S> This shouldn't cause a problem as you will get a nice strong logic 0. <A> As shown you concept will not work as when SSoutput is hi-Z, R1 + R2 will bias Q2base lowish and SS2 out will be at an intermediate voltage. <S> While whether this is seen as high or low will depend on the logic family used, it is not a "clean" solution. <S> Adding a diode or few at key locations may help. <S> FWIW: <S> The provision of a Hi-Z output state is not usually seen primarily as a means of expending pin count. <S> BUT any feature can be used for whatever function that the designer's ingenuity may decide :-). <A> It transpires that it's possible with just one PNP transistor. <S> It preys on the larger headroom given for logic-high than logic low. <S> R1 , R3 and R4 are quite delicately balanced to leave SS2 (which is directly connected to SSout) as high as possible while making sure Q1 is turned on when SSout is HiZ. Experimentation in the real world might be necessary, but you may also find that the threshold for logic-high is lower than stated in the datasheet. <S> With SSout driven high or low, the outputs are driven to 3.3V or <0.02V. <S> I'll leave you to work out which one is which.
| If the comparators are open-drain, which many are, you will also need a pull-up resistor on each output. Main use is to allow "wired OR" arrangements where various pins can be combined logically.
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How are very long component lifetime claims ascertained/evaluated? Its a common specification for household devices such as lightbulbs. However, I can't figure out how you'd truly evaluate/prove such a claim without running the device for the specified amount of time. Consider a lightbulb that is said to have a lifetime of 9000 hours. If I were to test this, the only way I can think of truly measuring this is to let the lightbulb run for 9000 hours, which is approximately a year! If a year is not long enough, consider certain LED bulbs rated at 50,000 hours! Clearly its not feasible to run a test for this long. So I guess I'm asking; on what basis are these claims made? Perhaps one way to test it is to stress the component at higher than normal operating conditions, so that it burns out faster, and then somehow create a prediction based on measurements. Or perhaps run the component for some (short) time and measure the deterioration/ageing and use that to create a prediction. <Q> One of the methods is, probably as you've predicted, that of accelerated aging . <S> This is used when the lifespan of a product is such that it would be impractical to run a normal lifespan test (such as LEDs, which have MTBFs of possibly 100,000 hours plus). <S> Here, one stresses the test item past what it will ever receive "in the field" to achieve a shortened lifespan, from which the data can be extrapolated. <S> A major consideration in the use of accelerated aging is the non-linear effect of operating outside of a part's recommended operating range. <S> This can be illustrated with mechanical systems, such as running a gearbox at 45,000 rpm instead of 15,000, and extrapolating the data by three. <S> However, suppose you're testing your lightbulb. <S> Common sense would say that running it at twice the current would make it run half as long; however due to those non-linear effects you may find that lifetime is only 1/4th of that at the proper current because of the additional stress of the overrating. <S> An important consideration is that the test device/subject/item should be well-characterized in both its intended and unintended operating range behavior prior to performing the test. <S> Because of that there are any number of studies ( a LOT of them) on the various refinements to the accelerated aging tests in various fields. <S> LEDs come to mind; photovoltaics are another . <A> As you intuited, lifetime rating usually involves testing at operating conditions more severe than the specifications allow. <S> Mathematical models - which can be empirical or theoretically derived - are then used to map the tested time to failure to a practical set of conditions. <S> In semiconductor devices, one well-known such 'law' is Black's equation and the general technique is called HTOL testing . <S> As you might imagine, it can be difficult to establish the validity of accelerated tests, and some engineers would recommend taking the numbers that result with a grain of salt. <S> In the semiconductor industry, many standards have been created and continue to develop. <A> I think you basically answered your question. <S> I found this regarding the LED life-time estimations that you mentioned: http://apps1.eere.energy.gov/buildings/publications/pdfs/ssl/lifetime_white_leds.pdf <S> Hope this helps.
| Depending on the technology the test cases and protocols to estimate the life-time vary of course, but overall they measure the degradation to a certain percentage and estimate the total life-time based on the protocols specific to that technology.
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What's the meaning of a Bode Plot when talking about nonperiodic inputs? The bode plot itself is built based on a transfer function, regardless of the input signal that will excite the circuit. Based on what I've learned, the bode plot show us how the response amplitude and phase will react when we change the input's frequency. However, I can only understand the meaning of the plot when the input is a sinusoidal signal. When the input is a non-periodic function, I can't see what would be the meaning of the bode plot, since I can't understand how would be possible to change the "frequency" of a non-periodic function. So, my question is: Is the bode plot only meaningful when we are analysing a circuit that is excited by a periodic signal? If not, what would be the meaning of the "frequency" (x-axis) of the plot when the input is non-periodic? I'm a bit confused because I always see bode plot examples when the instructor draw the bode plot of any transfer function, without caring if the circuit will be excited by a period function or not. Thank you ! <Q> The transfer function of a system, is valid for any signal. <S> Bode Diagram is another way to visualize the transfer function of a system. <S> An aperiodic signal, also has representation in the frequency domain . <S> An aperiodic signal, can also be considered as the sum of sinusoidal components, although its spectrum is not "constant" as in the case of a periodic signal. <S> We can consider that an aperiodic signal has a spectrum that varies every moment in its composition. <S> At one time, this spectrum has a specific distribution of components; at that moment, for the aperiodic signal applied to the system in question, the spectrum will be modified according to the frequency response of the system. <S> Accordingly, the aperiodic signal will be affected by the frequency response of the system to a greater or lesser extent, according to their instantaneous spectral composition. <S> For a basic relationship between the frequency response and an aperiodic signal, look at the step response and its relationship to the frequency response . <S> Example Consider the following aperiodic signal <S> wich has this spectrum <S> Of course, this spectrum is continuous, as for an aperiodic signal. <S> This signal is the input signal for a system with this transfer function <S> \$H(z <S> ) = \dfrac{0.2\,z + 1}{0.5\,z^2+1.5\,z}\$ and this Bode Plot As you can see, it is a high-pass filter. <S> The output signal looks like and has this spectrum Note that high frequencies have greater amplitude than the same frequencies in the input signal, as befits a high-pass filter. <A> A Bode plot is a means of crystallising system dynamics in a single graphic. <S> It can be obtained, practically, by a series of steady-state measurements that yield dynamic information when plotted. <S> The measurement process normally includes averaging, which provides significant noise filtering. <S> It gives prominence to high frequency effects, such as troublesome resonances, which may be hidden away, close to the origin in, say, a transient (e.g. step) response. <S> It gives information on relative stability, and allows the application of design tools such phase-lead, phase-lag, lead-lag. <S> To obtain the same level of information using time domain methods is often more tedious. <S> It is easily obtained analytically from the TF (for LTI systems). <S> Not an exhaustive list, by any means. <A> The transfer function H of a system (2-port) is the output-to-input ratio. <S> Because it makes no sense to find such a ratio if both ports (input, output) have different waveforms, this transfer function is defined for sinusoidal waveforms only (H=H(jw). <S> Only in this case and if the system has linear transfer properties both signals have the same waveform. <S> They differ only in magnitude and phase. <S> The BODE plot - a graphical representation of the magnitude and phase response versus frequency - is applicable to sinusoidal excitations only. <S> The BODE plot contains the same information as the Nyquist plot, which shows gain and phase in one single diagram. <S> You can construct a rough approximation of the BODE diagram (asymptotic lines) based on information about the the pole and zero locations only. <S> And - poles and zeros are defined for sinusoidal waveforms only.
| Bode Diagram characterizes a system, regardless of the excitation signal.
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Whats does "hr" mean among "V" and "Ah" in batteries? I have a battery which indicates 12V25Ah/10hr.What does 10hr mean here? I guess "hr" stands for "hour".But 25Ah already means if battery is fully charged and if I load 12V 25A (=300W) to it, it should last for an hour (correct me if I am wrong). What does "/10hr" mean here? How many watts I can load to use the battery for an hour? <Q> 25 <S> Ah is 25 Amps for 1 Hour which is equivalent to 2.5 Amps in 10 Hours. <S> So if you load the battery with 2.5 Amps it will last 10 Hours. <S> If loaded with a higher current usually battery capacity decreases so that is why the 10 Hours is mentioned, it results in a higher battery capacity making the battery look "better". <S> How many watts: simple 2.5 Amps <S> x 12 V = 30 Watts and that <S> for 10 Hours. <A> I concur with what others have said above. <S> Battery capacity is somewhat dependent on discharge current <S> At higher discharge current, the battery capacity decreases. <S> Here are discharge curves that illustrates this. <S> The same battery was discharged at different rates. <S> ( source ) <S> This chart is for a Li-ion battery. <S> A chart for a Lead-acid battery would also show the decrease in capacity. <S> Note that in this case, C doesn't stand for °Celsius. <S> In the chart above, C stands for discharge rate normalized by battery capacity. <S> 1C means such rate that will discharge a battery in 1 hour. <S> 0.5C and 4C correspond to 2 hours and 15 minutes, respectively. <S> Consider a 12Ah battery. <S> For such battery 0.5C, 1C, 4C correspond to 6A, 12A, 48A, respectively. <S> This kind of normalization helps to abstract away the size of the actual battery, which makes it easier to look at other aspects. <S> (More here .) <S> update: <S> Here's another similar chart. <A> The capacity of a battery depends partly on how fast you discharge it. <S> A specification like "25Ah/10hr" says that the battery has a capacity of 25 amp-hours when it is discharged at a rate that takes it from fully charged to fully discharged in 10 hours — in other words, at \$\frac{25Ah}{10 hr} = 2.5 A\$. <A> It means it will be 25Ah if discharged over 10 hours. <S> Battery capacity reduces if they are discharged fast. <S> Most battery capacities are quoted on a 20 hour discharge rate. <S> You would have to look at the battery's discharge curve to see what capacity it has at a 1 hour discharge rate to find the maximum current you can draw for 1 hour. <A> A lead battery has problems with load due to internal resistance. <S> That's why the capacity needs to be given dor a duration. <S> The capacity left is as follows: <S> From 10 to 6 hour the reduction is almost a straight line from 100% to 94%after that it gets bad.... <S> 5 hour discharge <S> you have only 92% <S> 4 hour makes 88% 1 hour has only 58% 250AH hours vs Ah <S> As a rule of thumb: a 25kg lead can charge 1kWh. <S> So a battery of i.e. 8kg can deliver around 280Wh (guessing a bit .... <S> 0.5kg housing, 0.5kg acid)this is almost independent of what type lead battery <S> it is since its based on the amount of lead used. <S> This can be useful if you don't know if you have a traction or a starter battery.
| It means that the battery has a capacity of 25 Ah when discharged in 10 hours.
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Can I use a Zener diode to convert 14 V into 5 V in a small-current circuit? I'm working on a microcontroller-based device, and I need it to detect an input voltage (distinguish between zero and non-zero). When it's not zero, it can be anything between 10 V DC and 14.6V with some added noise of various frequencies. When it is zero, it will either be at GND or floating, I'm not sure yet. Can I use a 5V Zener diode to stabilize this voltage and connect to the 5V-tolerant MCU's input? The current consumed will be however much the MCU requires, I expect it to be in the range of 1-3 mA (but I should probably see if I can find input current in the MCU's datasheet, which I couldn't so far but I'm sure it's there). <Q> I would personally use an 5V LDO regulator but as OP requested it is possible using a Zener diode, in my example I used a 1N751 5.1V Zener diode. <S> As @mkeith said you will need to limit the current. <S> Max current is calculated as Power Dissipation/Voltage, in our case it is 500mW max/5V <S> giving the maximum current as 100mA although the datasheet also states that the maximum current is 70mA so we will stick to that. <S> Rs = <S> (Vs - Vz)/Il <S> = (14.6v - 5.1v)/70mA = 135 Ohms E12 series makes that 150 Ohms Assuming a 1K loadIl = <S> Vz <S> /Rl = 5.1v/1k = <S> 5.1mA <S> Max zener <S> currentIz = <S> Is - Il = <S> 70mA - <S> 5.1mA = <S> 64.9mA <S> R1 is RsR2 is the Load <A> Because you describe the MCU input as "5 volt tolerant", this suggests that the MCU is operating at a lower power supply voltage, probably 3.3 volts. <S> In that case, simulate this circuit – Schematic created using CircuitLab will work. <S> When the source is 10 volts, the GPIO pin will be 3.3, and for larger voltages the diode will clamp the pin to no more than 4 volts, with a current drain of less than 1 mA. <S> This assumes, of course, that your low source voltage is low enough to be reliably detected as a zero after being divided by 3. <A> I think the best thing would be to use NMOS with gate connected to 14V. <S> The drain would be connected to 3.3V (the same one powering the IO for your microproceessor), and the source to your IO pin. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> With the Zener circuit, if the 14V is present when your micro power is not present, current will flow into the microprocessor by way of the clamp diode internal to your microprocessor IO pin. <S> Even if the current is limited, this could cause a partial or complete power-on of your microprocessor. <S> Best to avoid this situation. <S> This circuit will only turn on when VIN_14V is well above 3.3V. <S> But when it is on, all it does is connect the 3.3V VCC to the IO pin. <S> So if 3.3V is not present, then there is no harm, and no current will flow into the IO pin. <S> You might want to add a series resistor on the gate to prevent ringing. <S> The 100k resistor (R1) is just to make sure the node does not float high after 14V goes away. <S> 1 <S> M may be enough if you want to reduce sleep current. <S> Note: I am assuming that there is enough load on VIN_14V that it will quickly decay to below VCC_3.3V. <S> But if not, you might want to add a resistor from VIN_14V to GND to provide a path for decay. <A> Opto-isolator! <S> This has the optional advantage of completely isolating your noisy signal from your micro and the transistor output can turn hard on and hard off. <S> If you ever over-voltage it you only lose the opto-isolator. <S> simulate this circuit – <S> Schematic created using CircuitLab Bear in mind that with this circuit the input pin will pull low when the LED turns on. <S> If that's a problem then put the transistor on the high side and R2 pulling low to ground. <A> But you should consider the mcu's high level and low level voltages. <S> For example some 5v mcus accept voltages over 3.8v as high and below 0.8v as low. <S> Between these are fuzzy zones. <S> Which mcu can act unstable. <S> Feed resistor-zener output to gate and connect drain to mcu with a pullup resistor. <A> Why not just use a voltage divider? <S> If you're expecting a value between 10V and 14.6V but want to scale this down, add a 10K resistor and a 4.7K resistor in series. <S> 14.6V+ <S> ---- <S> [10K] --- [4.7K] --- GND | MCU <S> The noise issues can be dealt with by placing a capacitor across the 4K7 resistor. <S> A zener can also be connected across the 4K7 resistor to protect the MCU input from overvoltages.
| As long as you obey the datasheet of zener, this would be possible. Maybe you can add an additional logic-level-gate n-channel fet and use a lower voltage zener according to Vgs.
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convert 2.5VAC to 5V DC wave I have converted mains from 240V RMS (330V) AC @ 50Hz to 2.5V AC using a transformer and am now trying to put the wave into an ADC that operates at 0-5V. I am looking to convert a 1.76V RMS (2.5V) AC @ 50Hz (5V pp) sin wave into a 5V DC sin wave (5V pp), i know i require an op amp although i'm pulling blanks from my mind. Thanks for all the help. The reason is is i need to input the wave into an ADC that operated between 0-5V. At this point i guess i need a summing amplifier that acts as a DC level control. <Q> I recommend you AC couple the signal to the ADC. <S> This will create a high-pass effect, which, hopefully, will be acceptable. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The high-pass frequency is controlled by R1, R2 and C1. <S> You have to use the parallel combination of R1 and R2 (5k) for the RC calculation. <S> In this case, the cutoff frequency of the high-pass filter is around 3 Hz. <S> So 50 Hz will pass with very little attenuation. <S> Another thing to consider is that the source of the 50 Hz signal will see this as a 5k load. <S> I do not know if that will be a problem because you have not mentioned where the 50 Hz signal is coming from. <S> Assuming it can supply 1mA pp without sagging, there should be no problem. <S> You could make R1 and R2 larger (47k, or 100k) but then the effect of the 10 uA max leakage current might become a problem. <S> I chose 10k to make sure that effect would be negligible. <S> Addendum about Mains Power: Mains voltage is not well regulated. <S> Also, for safety, make sure your step-down transformer provides good isolation. <S> You should also probably have a fuse on the secondary side of the transformer. <S> I am not an expert on mains powered equipment, but there are many safety hazards associated with it. <A> So basically you have a sin wave that goes from +2.5V to -2.5V <S> and you want to convert it to go from 0V to +5V? <S> How about this...? <S> simulate this circuit – <S> Schematic created using CircuitLab Produces this output... <A> This circuit creates a 5V peak DC sine wave. <S> Can someone check this circuit please. <S> All my simulations lead me to believe that this is the solution. <S> Thanks.
| I would suggest you attenuate the signal further before sending it to the ADC, or add robust over-voltage protection to the ADC input.
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Which is faster, Executing a program from ROM or RAM? Most of us who are from electronics background knows that SRAM is faster than DRAM. But when it comes to comparing RAM with ROM, i am unsure. My question is related to micro-controller :"If a code is executing directly from RAM/ROM, whose performance will be better ?? 1) execution from RAM or 2) execution from ROM or 3) both will perform equal" Also considering the fact that ROM are designed to have higher READ speeds. whereas for RAM, there is a trade off of read speed for having write capabilities. <Q> The datasheet should tell you how long each instruction takes, and what differences there are, if any, between executing from RAM or ROM. <S> For microcontroller that offer the option of executing from RAM, that is probably faster, likely being the main point of using additional RAM space to execute code from. <S> There may also be some fetch overlap issues. <S> Again, the only way to know for any particular micro is to READ THE DATASHEET . <A> It depends entirely on the memory and CPU architecture. <S> As a rule of thumb, SRAM is faster than flash, particularly on higher-speed MCUs (>100 MHz). <S> SRAM bit cells produce a (more or less) logic-level output, while flash memory has to go through a slower current sensing process. <S> How much faster (if any) again depends on the architecture -- the word size of the memories, the number of wait states on each, the presence of caching, the size of the CPU instructions, etc. <S> If you're running at a low enough frequency, you could have zero wait states on flash and RAM, so they might run at the same speed. <S> The code also matters. <S> If your code is strictly linear (no branching), the flash could prefetch instructions fast enough to keep the CPU saturated even at higher frequencies. <S> As Olin said, a Harvard architecture CPU with separate program and data read paths could perform differently when code and data are in different memories. <S> Metal ROMs (and other nonvolatile memories such as FRAM) have their own characteristics, and may or may not be as fast as SRAM. <S> The ability to write doesn't necessarily make a difference; it's more about the characteristics of the bit cell output and sensing circuits. <S> The datasheet will give you a rough idea of the speed difference, but the only way to know for sure is to profile your code. <A> "Running a program" requires a CPU with a synchronous clock. <S> Slow memory can be accommodated by either running the entire system at a slow enough clock, or by inserting wait states (extra do-nothing clock cycles between the fetch and decode phases), active only for certain address ranges (see the ancient 8085 for example).The <S> CPU instruction fetch doesn't know or care exactly when the data is settled to its final value, just as long as it does not change during the setup/hold interval. <S> A microcontroller usually has all of its memory on-chip, so unless stated otherwise I'd assume the memory system is all zero-wait-state. <S> (but read datasheet to confirm). <S> Typical microcontrollers are meant to be simpler, single-chip solutions compared to a desktop, so wait states are unlikely in a microcontroller. <S> So it's unlikely that a microcontroller would have mismatched on-chip memory speeds. <S> Faster memory generally costs a premium (higher voltage, lower capacitance, more demand). <S> An 80xx86 has fast SRAM in L2 cache and even faster SRAM in L1 cache, and lots of slower DRAM off-chip attached to a memory controller. <S> This kind of system is a lot more complicated than a microcontroller, and is beyond the scope of the question. <S> (But of great interest to a computer engineer!)
| In some cases it might be faster to execute from ROM because it is a separate memory and RAM access can be going on concurrently.
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Is a linear voltage regulator necessary in AC to DC? I was doing research on AC to DC power supplies and I came across a lot of schematics that use regulators like the one below: credit: http://www.instructables.com/id/Make-a-simple-12-volt-power-supply/ Initially I drew out what I thought an AC to DC schematic would be and I basically did everything except including the LM7812 regulator. Is it necessary or could someone get an equally good power supply by just using an effective transformer, bridge rectifier, and capacitors? From the ( Wiki ) page it says the regulator is, "... conceptually an op amp with a relatively high output current capacity." Why this might be important, I do not know. It seems to be the de facto standard in the AC to DC converter world and I would really like to know the importance of it. <Q> The rectifier alone produces very rough DC; it is not smooth at all and contains much of the AC signal called "ripple." <S> A filtering capacitor helps to smooth this out some, but the output is still not perfectly invariant. <S> The addition of the 7812 in this schematic, regulates that DC down to 12v, and eliminates almost all of the remaining ripple, producing a very nice and clean DC output. <S> Such as if the power supply were just lighting a lamp or operating a fan. <S> Those "loads" do not require precise power control. <S> But for more sensitive electronics such as digital IC's and analog chips, their supply voltage must remain very steady for correct operation. <S> So the answer is, whether or not to use a regulator depends on the application. :) <A> Some applications will tolerate unregulated noisy supplies. <S> Motors, for example. <S> So no, you don't always need a regulator. <S> And if you do, it doesn't have to be a linear regulator : a switching regulator is more efficient and runs cooler, but it can be harder to keep the switching noise under control. <S> So a switching regulator is fine for digital logic or a computer, but a linear might be easier to use (but not the only option) for sensitive audio or radio circuits. <A> Regulators are required to provide a known, stable, ripple-free output voltage. <S> Without the regulator, the output voltage of the supply would vary with load, and the output would also have some "ripple voltage" which would also vary with the load on the supply. <S> The output voltage of a simple rectifier/capacitor power supply would also vary with the AC input voltage. <A> Switched mode power supplies are now the de-facto standard. <S> linear power-supplies with an iron transformer driven directly from the mains are now a specialty item.
| The regulator is needed if the fluctuations due to the pulsed current from the rectifier and variations in the line voltage give unsatisfactory results. There is nothing stating that you must use a regulator; and in fact, they are sometimes not used.
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Selecting a capacitor for use in feedback loop I'm designing a DC Lab Power Supply; the voltage regulation circuit looks like this: I'm developing the BOM, component by component, and next in line is specifying the capacitor in the feedback loop for the op amp, C1 in this schematic. As I'm going through the selection of specific components, I've developed a new appreciation for the diversity of available capacitors and the relative complexity of what I originally thought was a pretty straightforward component type. So my dwelling on this particular component is as much for the learning opportunity it presents as the desire to pick the right item for this specific case. My PCB is SMD wherever possible, and I'm inclined to believe a garden-variety 0805 X7R would do the job just fine. However, I've learned they can have surprising behaviors depending on, for example, the voltage applied, so wanted to get the perspective of more experienced designers. The design of the feedback loop itself was by far the biggest time investment in the circuit overall. I had to refresh my foggy recollection of Body plots, transfer functions, op amp particulars, etc. And it took me quite a few tries to get it this right (and understand why it worked then :). So I'm inclined to think if there's anywhere in the overall where the capacitance in critical that this would be it. So my question is: Is a regular surface-mount ceramic cap the right choice for using in this feedback loop? Or should I be thinking something a little fancier, like perhaps a film capacitor of some type? <Q> The hazard with some ceramic capacitors are that they lose capacitance as the voltage goes up, and as the temperature goes up. <S> Some under conditions within the data sheet T and V ratings can drop to 25% of their nominal capacitance. <S> The worst are the very high value, so > <S> 1uF caps in small packages where you look at them and think 'how the mike did they cram all that capacity into that ? <S> The answer is, by making other compromises on the dielectric material. <S> Used in that C1 position, losing some C would result in the break frequency of that RC moving up from your design of 340Hz. <S> If your stability margin is so small that you need high precision on that time constant, then I suggest you improve your margins. <S> I don't recall which of temperature and voltage spec designations like X7R leave underspecified, but as you have asked this question for your education, that's fine. <S> I will simply warn you, and you can go digging for the information. <S> You will need to dig deep. <S> I won't say the manufacturers try to hide this stuff, but they sure don't make it easy to find. <S> You will probably need to look in the material, rather than the capacitor, data sheets. <S> You might find a 1nF in that. <S> X7R - expect +/- <S> 15%, only goes to mid values X5R and Y5R - possibly -75% under some conditions, read the data sheets for the specific capacitor value, package (yes, even the package), voltage , manufacturer and material very carefully. <S> Keep your max voltage well below the rated voltage. <S> At 1nF X7R, you should not have any problems. <A> DC bias effects have already been noted (there is an excellent application note from Murata on this). <S> This link appears to be broken; this FAQ page may be of use. <S> C0G, although marginally more expensive, bring other things to the table, and in a feedback loop such as yours (I am designing some right now in an interesting application switching a few hundred volts that requires a linear ramp), I want to have a part that will remain at its rated capacitance across bias, time and temperature. <S> C0G: <S> No DC bias effect from most manufacturers (this has to do with the material used). <S> This is definitely true of AVX, Murata and Johnson. <S> No capacitor ageing Tempco <= <S> 30ppm: This will be important if the power supply box heats up significantly. <S> Compare that to X7R and you will find that in a control loop, C0G is the best choice in a ceramic. <S> I would not normally need to use a better part than that. <S> I am actually using a 1nF C0G, 50V, 5%, 0603 part from AVX (but all the usual suspects have them). <A> The type isn't critical, as long as it meets your tolerance needs. <S> Be aware that ceramics (not sure they are available so small) lose >> 20 % (to 50 % or more ) of their capacitance as the applied voltage reaches their rating. <S> To avoid this, you can use a cap that is rated at a higher voltage than you need. <S> On your circuit -- if your + and - 15 V supplies don't come up properly, is there any possibility that your output will become unregulated ? <S> The 1k will turn on Q1, but requires the opamp to turn it off...
| Amongst the main choices for materials are NP0 - very stable, but only very low values.
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How to conclude if logic circuit with feedback is combinational or sequential? In the below figure, the the inputs are a,b,c and d and the outputs are Zo and Z1.There is a loop in the circuit. How can i tell if the circuit is combinational or sequential? edit:The outputs of interest here are z1 and z2 which are given by: z1 = (a.b'.d) + (c.b.d) + (b.c.d) z2 = c + a.b'd . I have the outputs depending on inputs but can i conclude that it's a combinational ckt ? Is there any concept i am missing to understand? at the extreme left is the inputs a,b and c in order.The output of 1st level OR gate is z0 and "d" is the input to the 4th level AND in centre. The output of the same AND gate is taken as z1. The 5 th level lower AND gate has inputs "b and c". <Q> https://books.google.com/books?id=1QZEawDm9uAC&pg=PA176 <S> You have to make sure there's real state preserved there <S> , i.e. the state saved actually affects the output(s). <S> Otherwise it's a bogus/inconsequential "sequential" circuit, so really just combinatorial. <S> In real life nobody would design that, but this being an academic exercise... <S> you need to check. <S> You need to write (and simplify) <S> the expression for z2 [if that's what I've circled below is called] and see if it actually depends on the old value of any inputs or not. <S> Rafiquzzaman's textbook (to which I've linked) shows how to do this on an example (which actually has two state bits, but fewer inputs). <S> Basically you treat old and new values as different variables (in this case z2) and apply k-maps etc. <S> to simplify the expression to obtain a transition table for z2. <S> Regarding your equations, they seem wrong because there's no dependency of any on any [prior value of z]. <S> Unless you've already simplified them... <S> in which case, why are you still asking this question? <S> But then your image so blurry I can't read practically any letters in it. <A> Sequential logic implies that previous inputs effect current outputs, which given the feedback you have in the circuit you do have. <S> On the other hand when most people (including me) think of sequential logic they picture logic that is registered with state and clocked (e.g. you have flip flops). <S> What you have here is sequential logic, but it's asynchronous <S> (you don't have a clock input). <S> At first glance I would have wrongly called this combinational since I just looked for a register/flip flop somewhere, but because you have feedback your output depends on both current inputs and previous inputs. <S> To see this assume for a second that a=1, b=0, d=1, c= <S> x. <S> This means that your feedback path (the second input to the or gate with c as input) is driven high. <S> Z0 is now 1. <S> Assume now that you toggle a low. <S> Z0 will still be high. <S> That means that Z0 is dependent both on current inputs as well as previous inputs. <S> Now for fun set d=0. <S> Now you've cleared Z0. <S> From wikipedia : "In digital circuit theory, sequential logic is a type of logic circuit whose output depends not only on the present value of its input signals but on the sequence of past inputs, the input history. <S> 1 <S> [2][3][4] <S> This is in contrast to combinational logic, whose output is a function of only the present input. <S> That is, sequential logic has state (memory) while combinational logic does not. <S> Or, in other words, sequential logic is combinational logic with memory." <A> Does it store state? <S> If there exists a coding for the inputs, for which the internal nodes can be in a choice of states, then it's sequential. <S> For instance a cross-coupled pair of NAND gates, the classic SR latch, is sequential, because an input of (11) does not define the state of the gate outputs, they can be (01) or <S> (10) depending on the history of the inputs. <A> To be honest, this is a combinational feedback and combinational feedback is a SIN. <S> You should never use an output which you produced as an input to yourself without registering it. <S> I'm sure this was given to you as an exercise, but these kinds of circuits are never used in the industry as the output has a very good chance of going into a forever changing loop and will never settle down. <S> You should always use a flipflop to register the output first. <S> When the clockedge arrives, the output is used to do the new computation and it will be ready at the input of that flipflop before(a setup time) <S> the next clock edge.
| If you have combinatorial circuit with a feedback path from output to input then it's [by definition] an asynchronous sequential circuit.
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Eagle-how to change footprint? I know its basic but google just doesn't give me anything . I have Eagle on a mac, i have a design with a few resistors, i would like to change their footprint to 0805 . The menu is just not showing me this option to see the part footprint. thank you. <Q> Right click on a component and go to open device. <S> This will show the symbol and the package. <S> This only works with library parts that are in the lbr folder of the root eagle directory. <S> If you have your own custom libraries you have to open the library directly. <S> Also when placing components using the add tool, it shows you a preview of both the symbol and the package. <A> If you are using resistors from, say, the resistors or rcl libraries, the resistor part has multiple footprints already of all different shapes and size. <S> This will bring up a window of all the available packages to choose from. <S> The context menu looks something like this: <S> And this is the window that appears: <S> So in my case I was using an 0402, but can change it to one of many other sizes. <S> Simply select the one you want. <A> I understand that you want to change the footprint of multiple components together. <S> Starting in Eagle 9 , there are new tools available that simplify this task: Design Manager and Inspector <S> If they are not visible, enable them using the "View" menu. <S> Now follow these steps in "Browser" tab of Design Manager : <S> In "View" drop down menu, select "Devices". <S> In "Device Sets", make sure that "<All Devices>" is selected. <S> In "Devices" section, type in the footprint that you want to change. <S> This will filter the device list so devices with matching footprint are shown. <S> In the device list, select all devices (click the first device and shift-click the last device in the list) or only a subset of them (ctrl-click on a list entry). <S> Now in Inspector : <S> In "Quick Actions", <S> click on the second icon (from the left).This will open the "REPLACE" window where you can select the new footprint to assign to the components you selected before. <S> See the screenshot for a graphical view of the workflow. <S> Hope that helps.
| In this case all you need to do is right click on the part (either in layout or in the schematic), and click the option named package .
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Connect a wire to a circuit board I'm really new in designing PCB(it's my first one) and i'm using Eagle Cad do it, however i have no idea on how to connect an outside wire to the PCB(either in the schematic or in a real circuit board). In this case the wire only needs to be connected to GND don't know if its easier or not. thanks <Q> If it's earth you want, you may be able to use a mounting hole. <S> Show the mounting hole(s) on the schematic like any other component with a footprint and connect them where they should go (if anywhere). <S> If it's a wire to be soldered into a hole into the board, again, make a component footprint (preferably with strain relief). <S> You can also crimp a terminal onto the wire and solder the terminal into the board, use a board-to-wire connector pair (the 0.156" pitch series is good for high currents, whereas 2.54/2.50/2.0 mm pitch is good for finer individual wires. <S> Or a screw terminal strip if connections are to be made in the field <S> (there are even versions that use springs suitable for some applications that require no tools). <S> Another possibility is to use Faston (tm) spades soldered into the board and female spade connectors crimped to the ends of the wires. <S> The latter is very popular in appliance and commercial electronics that has to handle mains voltage at relatively high current. <S> Again, find or make a component symbol and a footprint that suits the part you want to use. <A> You can get them quickly, cheaply, and easily on Amazon, Digikey, and everywhere else. <S> You can find the footprints in Eagle con- libraries based on the number of terminals and the spacing you end up using . <A> I tend to use Pins, to just solder the wire directly onto the board, a lot of the time I use screw connectors like the answer above, i recommend Upverter, eagle cad is a load utter crap. <S> Upverter is free, easy to use and has a web app, but thats just my view. <S> Edited my question: For grounding your board you usually set a point where when the board is put into what ever your going to be putting it into it touches a metal part somewhere <S> , that will ground it, what is it your designing?
| For physically connecting wires to boards, I am a big fan of screw terminal blocks like these....
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Why 220V and 110V? Why did the industry choose these seemingly unintuitive values of 110V and 220V for the mains AC? Is there a physical reason why they aren't simply 100V and 200V? <Q> 'It's the history of how we got here' is a rather unsatisfactory answer, but it is true, if you mean 'why those exact figures?' <S> But we can explain why those figures are so close together. <S> There's a rather limited range of usable voltages for domestic power distribution. <S> Low enough to be safe to touch with bare hands, <50v, would entail massive copper conductors to shift any practical power at all, a kettle would need a 60A outlet at 40v. <S> When the voltage is lower than 350v or thereabouts, it's relatively safe to handle with casual insulation practices. <S> The voltage will not jump through cracks and pinholes in insulation, any thickness of dry clothing will protect from casual contact, and dry skin will often prevent a fatal shock. <S> So at this level, few enough people die accidentally that people accept the system as 'safe enough', rather like cars! <S> You need to be an expert to handle higher voltage, and the world is not full of experts. <S> If you read the history of the development of electrical systems around the world, you will find a whole slew of distribution voltages and frequencies. <S> Gradually, the minority choices died out, and today we are mostly left with just the 'big two'. <S> Why did Thomas Edison plump for 110v? <S> Who knows, but more than 100 probably had a nice ring to it, appealing to both his showmanship and his fledgling market. <A> Thomas Edison selected 110 V DC and AC systems were designed to match the existing standard. <S> You can probably find quite a bit of history details if you search. <A> They did choose nice round metric values of 100V, 200V and 500V. <S> And then, they wanted to increase transmission capacity without really adding copper to the system, so they made a series of sub-5% "bumps". <S> (this also made overload-induced voltage drop seem less severe.) <S> The gentle bumps were mild enough that people's light bulbs (the primary load of the day) didn't burn out that much faster, and of course light bulb makers tuned their products quickly with each bump. <S> By the time Edison was trying to popularize DC, the bumps were up to 110V -- DC, mind you. <S> And this was aggressively marketed to the general public, which is why "110V" stuck in the same way "Xerox copy" stuck. <S> 500V streetcar and subway voltage similarly got several bumps to 600V by that time. <S> All of them have since had several more bumps. <S> Streetcar and subway voltage has bumped to 750VDC without deleterious effect on motor commutators. <S> BART made a leap to 1000VDC but that was too much for the commutators and they had to begrudgingly back off to 900V. <S> North American AC power long ago bumped to 115V, some 117.5 and finally 120/240V. <S> There's talk of 125V and everything in the system is insulated for 125V. <S> Europe did the same thing, UK bumped a little more than the mainland and is now un-bumping to harmonise.
| There is no physical reason for any of standard voltages used around the world. When Edison threw in the towel on DC, they chose an AC voltage which would still work with the same DC light bulbs, which worked out to be the same as RMS, so 110 V AC.
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Detecting a disconnected analog input I am working with ADC on MCU SH7147 (Renesas). I have a problem with a sensor as outlined below. How do I detect a disconnect of the sensor cable on the analog input pin on the MCU? I tried to check it by cutting the the wire and seeing the analog data display in LCD, but it is random data, it is difficult to make a rule to detect disconnect. <Q> Add a pull down or pull up resistor on the A2D input - it will then always read full scale or zero when the sensor is disconnected. <S> A value of about 100K should do the trick, A2D input impedance is typically 1M ohm <S> or so. <S> and sensor output is typically less than 100 ohm. <A> Set the pin to digital output, set it to 1, set it to ADC input, capture value. <S> If the pin is not connected to anything, you should get a high/low ADC value, since the pin has a small capacitance, which stores the voltage level. <S> A connected sensor will set the voltage level back to 'his' value. <S> You may have to play around with delays between setting and measurement to get reliable results. <S> Of course, this only works when the ADC input itself is floating and not, when there's e.g. an opamp connected to the input. <A> What type of sensor is it? <S> One approach is to switch the MCU pin to be a digital output, then drive a 0 or a 1 on it for a while. <S> Then switch it to be an ADC input and read the value. <S> If it's still close to 0 or Vcc, the input could be disconnected. <S> However you need to be careful that driving a 0 or 1 output when the sensor is connected won't damage it, or the SH7147. <S> This depends a lot on what the sensor circuit looks like.
| Set the pin to digital output, set it to 0, set it to ADC input, capture value. This works because there is usually enough stray capacitance on your board to hold 0V or 3.3V/5V on the pin for a few microseconds while you switch over from digital output to analog in. And you have to check if a connected sensor tolerates this at its output .
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How can a powersupply have a large input volt range I see that computer power supplies that can take a voltage input of anything between 90V and 260V at a frequency between 47Hz and 63Hz. Meanwhile it can output power at a very precise voltage. How does that work? <Q> It is done by magic ;-) <S> Just kidding, it is not. <S> It may look like a big deal <S> but it's not actually. <S> See the other answers, they are also correct. <S> The power from the mains is actually converted into MAGnetIC energy (do you see what I did there <S> ;-) ) <S> and then back to electrical energy. <S> This conversion to magnetic energy also has the advantage that the output can be isolated from the mains supply so that you do NOT get an electrical shock when touching the output, I consider this is a very nice feature ! <S> A reference voltage is made internally in the power supply's chip(s). <S> Generating such a stable reference voltage is done using a bandgap circuit. <S> The output voltage is compared to this reference voltage and adjusted so that it will be the correct voltage, this is called feedback. <S> This feedback controls the circuit on the mains (high) voltage side and tells it to put more or less magnetic energy in the high frequency transformer depending on what the output voltage is doing. <S> If the output voltage is too low: more ! <S> If it is too high: less ! <S> Simple as that. <S> This feedback signal usually travels through an optocoupler (so using light) so that it does not need a direct wired connection to the mains side electronics and so keeping the isolation from the mains voltage side. <S> Due to this feedback the input voltage can vary over a wide range while the output voltage remains constant. <S> Brilliant isn't it ? ;-) <A> Computer power supplies are switch mode power supplies . <S> In input voltage is chopped up and switched on and off rapidy (usually in the tens of kilohertz range). <S> There is control circuitry that measures the output voltage and adjusts the amount of switching to compensate. <S> Computer power supplies are designed to accommodate a large range and frequency so that they can work anywhere in the world. <A> Typically the input AC is converted to a DC voltage, and the DC voltage is then converted to the desired output voltages by DC-DC converters.
| Using very fast switches at the high voltage side the amount of energy that is transferred to the low voltage side can be precisely controlled.
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Switch to stay ON until power removed (24V) In my bathroom, I would like to install some 24v LED tape, some in a warm-white and some in a cool-white, that I can switch between at will. IE: I will turn on the circuit from the wall-mounted dimmer outside the bathroom and then select either the warm or cool LEDs via another switch. I then want the ability to switch from one to the other at will. This, I realise will be easy - Install a switch after the 24v transformer to alternate between the two. However, I am curious whether a switch exists that will default to the warm-white circuit. So, for example, when I turn on the circuit at the wall the warm-white LEDs will be powered by default. I can then switch to the cool white however, when I turn off the power to the whole circuit at the wall the switch will automatically switch back to the default warm-white circuit so that next time I turn on the lights they are on the warm-white. Maybe some kind of self-powered relay? Electro-magnetic switch? To add more complexity this will be after the 24v transformer that will be dimmable by PWM. If I have to I suppose it could be done on the mains supply with one transformer for each of the circuits. So, to summarise: Default to circuit 'A' at power ON, switch to circuit 'B' after switch press, revert to circuit 'A' after power OFF. Thank you in advance to anyone who may be able to help! <Q> How about this? <S> : <S> simulate this circuit – <S> Your post mentions transformer <S> but it must be more than that to power the LEDs. <S> I'm assuming it's a PWM supply of some sort. <S> If that's the case each pulse will be full voltage and we can 'steal' some of that to charge a capacitor to make our power supply. <S> C1 is charged up by the PWM <S> pulses via D3 (to prevent discharge) and R1 (to limit current draw). <S> You'll need to calculate the values of these components to suit your relay coil. <S> Pressing S1 will energise the relay and latch itself on while turning on the cool LEDs. <S> When the PWM source is switched off C1 will discharge through the relay coil and reset the circuit. <A> The following circuit should do everything you asked for, and here's how it works: <S> U1 is a "D" type flip-flop, and when the 5 volt supply is turned on, R2 and C2 integrate the supply's rising edge and present it as a momentary signal starting at zero volts, to U1-S, which forces U1Q high, to +5 volts, and U1Qbar and U1D low, to 0 volts (ground). <S> U1Q going high turns on Q1 which energizes K1, causing the LED supply to be connected to the warm-white LED string, turning that LED string ON. <S> S1 is a normally-open momentary switch and, in conjunction with R1C1 (a switch debouncer), is used to send clock pulses to U1> <S> A "D" type flip-flop works by transferring whatever's on its D input to its Q output as soon as it sees a high-going edge on its clock input, so since D is low, Q will go low when S1 is pressed. <S> That'll cause the transistor to turn OFF which, which will de-energize K1, disconnecting the LED supply from the warm-white string and and connecting it to the cool white string through the relay's normally-closed contact. <S> D is now high because it's connected to Qbar, so if S1 is pressed again, the transistor will will turn ON again and... <A> The circuit below makes no assumptions about whether a separate switch is used to turn off the LED PWM (meaning one could run the +5 supply off the same switch), or it is part of the dimmer circuit itself, as it uses a missing pulse detector to determine when the PWM signal has quit rather than using a loss of power. <S> It also doesn't try to "harvest" power from the PWM source, <S> since at very low light levels (a few % on), I don't think there would be enough to pull in a relay coil. <S> The PWM is fed into a voltage divider that cuts the 24V down to 5V (signal conditioning or other circuitry be be needed, since I don't know what that signal really looks like). <S> The output of the retriggerable one-shot 74LS123 <S> stays high as long as there is a PWM pulse train. <S> The keeps the \$\small \overline{\text{CL}}\$ input of the 74LS74 flip-flop (being used as a simple latch) high, so <S> the flip-flop is not being cleared. <S> The period of the one-shot just needs to be longer than 1/frequency of the PWM; I set it arbitrarily to 500 ms since I have no idea what it looks like. <S> The formula for the time is T (in <S> ms) = 0.33 <S> * R (in K) <S> * C (in µF). <S> In this case, T = <S> 0.33 <S> * 150K <S> * 10 µF = 495 <S> ms. <S> If the pushbutton is pushed, this places a low on the \$\small\overline{\text{PR}}\$ input of the 74LS74, setting it and turning on the relay, switching the output from the warm LEDs to the cool ones. <S> The relay will not turn off until the one-shot times out meaning the PWM train of pulses has ended. <S> To see a larger version, right-click and click on View Image.
| Schematic created using CircuitLab Thus, on power-up, the default state will be warm-white and, after that, every time the switch is pressed, the LED strings will change states. If the external switch is a regular dimmer switch then this circuit is unlikely to work as the LED power supply is, more than likely, a simple transformer and rectifier.
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bi-directional flyback diode for relay spike protection A flyback diode is used to protect circuits against the reverse voltage generated by removing power from a relay coil. Like this: simulate this circuit – Schematic created using CircuitLab But what to do if the polarity of my power supply isn't known? placing two diodes in parallel in opposite won't work (short circuit)Any idea for bi-directional flyback diode? <Q> You are looking for a "bidirectional TVS diode" which is two Zeners back-to-back. <S> You can get them in many different voltages, so you can pick one that is above your supply. <S> That way only the inductor current will break through it. <S> Alternately you could use a diode bridge to set the polarity of the input no matter what you get fed. <A> Usually the polarity will be known because something like a transistor will not work if the voltage is reversed. <S> If you put it across the load, and the voltage surges above the breakdown voltage the diode will conduct and the switch and/or TVS could be destroyed, so it's sometimes better to put it across the switch. <S> Here are some example types. <S> You must make sure that the TVS can safely absorb the energy stored in the coil (less of a problem with the rectifier diode because most of the energy is dissipated in the coil if the operating voltage is much higher than the diode drop). <S> If the inductance is constant, that's simply the operating current squared multiplied by the inductance. <A> You don't need to use a diode; a capacitor will limit the open-circuit energy from the relay coil to a voltage usually not greater than twice 9V. <S> Lets say your inductance is 1 henry and the coil resistance is normally 750 ohm. <S> The current into the 1 henry is limited to 12 mA and therefore the coil stores an energy of 1 x 0.012 <S> x 0.012 / 2 = 72 <S> uJ. <S> If all of this is transferred to a 10uF capacitor the voltage increase above 9V on the capacitor is: - V = \$\sqrt{\dfrac{72uJ <S> \times 2}{10 <S> uF}}\$ = <S> 3.8 volts <S> i.e. the 9V, when disconnected by the switch, will create a peak voltage of 12.8 volts on the capacitor and it will likely but much less than this because of the series resistance of the coil burning off energy in the transfer of micro joules to the capacitor. <S> To address Spehro's comment a small values resistor in series with the capacitor can reduce inrush when activating the relay.
| If you have a bidirectional switch such as a mechanical contact or series back-to-back MOSFETs you can use a bidirectional TVS with the TVS voltage higher than the supply voltage.
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Programmed IO vs interrupt for devices Although I can understand the difference between programmed IO (PIO) and interrupt (INT) transfers, still there is something vague. In PIO, the processor repeatedly checks READY pin to see if the device is ready. However, in interrupt mode, the processor checks the INT pin at the end of the instruction cycle. My question is, in interrupt mode, the processor still has to repeatedly checks the INT pin. I know that in this mode, the external device send the interrupt signal, but how processor becomes aware of that interrupt? The processor has to repeatedly check the INT pin on every cycle (or instruction cycle). Doesn't it? References: {1} http://www.louiewong.com/archives/137 {2} https://www.quora.com/What-is-the-difference-between-programmed-driven-I-O-and-interrupt-driven-I-O-What-is-one-advantage-and-one-disadvantage-of-each {3} http://homepage.cs.uiowa.edu/~ghosh/Chapter3.ppt <Q> In PIO mode the processor is executing instructions that check the IO. <S> That means it's not executing other instructions during that time. <S> In IRQ mode the check is performed by the same part of the hardware that normally increments the instruction pointer, or handles jumps. <S> The check itself doesn't "cost" anything other than a tiny bit of processor chip area. <A> Typically the hardware interrupt saves the current PC on the stack and then jams the interrupt vector address into it, causing the CPU to jump to the interrupt handler code. <S> This may only take a few cycles, but since the main program is not aware that it has happened the interrupt handler must also save and restore any registers that it changes. <S> As a result the time required to process an interrupt may be much longer than using PIO. <A> It helps to look at a processor architecture. <S> Look up Mips pipeline with hazards. <S> It can get a little complex, but what is being pointed out is that the PC(program clock) of processors is in itself the pointer. <S> When something like an interrupt happens, it goes to the hazard unit, which also acts as the interrupt vector table. <S> This vector table takes care of all the mapping and controls a few pieces of the processor to stall, jump, branch (ISR things). <S> In many cases, it says to do something in Kernal mode, or syscall as MIPS calls it. <S> This means a software interrupt. <S> In the case of PIO, everytime you want to check a pin, you must process instructions to point to that pin, call a peripheral to do that, report back, and check the results with another instruction.
| The difference is that with PIO the CPU has to periodically execute instructions which read the READY pin and decide what to do about it, whereas in interrupt mode the CPU just keeps executing the main code and is 'interrupted' by extra hardware which is monitoring the INT pin.
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how to increase current in this given circuit I am am a beginner to this not having background from this feild n i I have a transformer less circuit which gives out 12vdc and 30 mamp how to increase the current to 2 amp <Q> First, and most important, this circuit is not safe. <S> A mistake putting it together can kill you or someone else. <S> Do not use it for anything. <S> It definitely will not work well at 2 amps. <S> Zener diode regulators work by converting the extra current capacity of the power supply to heat. <S> 2 amps at 12 volts is a huge amount of power to waste. <S> Passerby's answer is the right one. <S> Buy an AC-DC converter. <S> Don't try to build your own if you don't have the knowledge. <A> Its not feasible to modify the existing circuit. <A> Not using that circuit! <S> At the moment it is the 0.47 uF capacitor that is dropping about 220V AC across it (due to its reactance). <S> Using a bit of handwaving you could argue that the current taken from the AC is ~30mA AC therefore the reactance of the capacitor is 220/0.03 = 7333 ohms. <S> This would be a capacitance of 0.434 uF at 50 Hz. <S> To take 2A, you need a capacitor that has a value that is 2/0.03 times higher i.e. about 30uF and rated at 450V. <S> How big is one of these? <S> Here's one: <S> - I suggest you go researching on on ebay to get it's large dimensions. <S> Please not that it cannot be an electrolytic type. <S> Then you need to consider that off-load the 2A is flowing into the zener diode <S> - you'll need a zener diode rated at about 30 watts. <S> The one below is rated at 25 watts and will require a sizable heatsink: -
| Easiest solution, cut that circuit out, and use a commercial 120/240V mains to 12V 2 Amp power supply.
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Scanning/photocopying vs. photographing the material Why do scanners and photocopies scan the material? Couldn't they just have a digital camera + light behind the glass? A lens with fixed focus would suffice, since the sheet of paper would always be at the same distance. There would be no moving parts. Wouldn't than reduce costs? Would such a device be more robust? I fail to see whether the quality of the produced image would be worse. <Q> Consistent even lighting moving with scanner will give even illumination across the full scan. <S> No vignette effect. <S> Lens distortion - particularly at the edges. <S> Scanner head is a line-scan camera with each pixel vertically aligned with the subject. <S> Physical depth. <S> Scanner is very shallow. <S> Try it yourself with a camera and a flat printed page. <S> Look for evenness of tone across the page. <S> Try it with squared paper and see if the lines are straight in your photo. <A> Using a fixed focus lens wouldn't be the same. <S> The entire page is not at the same distance when you consider that there is a viewing cone on a camera. <S> The edges are farther from the lens than the center is. <S> That won't only affect focus, but lighting as well. <S> Consider that some inks have different apparent colors depending on angle of view and reflection from lights. <A> But they are expensive compared to getting a similar image quality with a moving-sensor 1 scanner. <S> Needs nice optics etc., also big sensor. <S> That one uses a 35Mpixel sensor per color (105Mpixel total). <S> Here's another one like that ; this one uses a glass press, so it's more like you envisioned it. <S> Also, a version of that that also turns pages . <S> 1 <S> In fact, you could have it the other away around, i.e. move the item/book as in the Google Books project , but to keep sensor costs down [to line sensor], you need to have relative movement of the sensor vs. material being imaged. <S> Actually if you envision a book scanner that automatically turns pages, you need to have some moving elements anyway... <S> so you might as well design the scanner around the mechanics of that movement. <S> In fact I've discovered a research project <S> I didn't know about in which they flip a book really fast and just photograph it with regular cameras and then apply sophisticated deskewing/restoration algorithms. <S> So it is possible to have your cake and eat it. <S> I'm not sure of the image quality in this system. <S> There's a paper about it at recent (2014) conference; in fact it got the best paper award there. <S> They scan only at 500 dpi. <S> To get that from the distance where the camera is at, they use ~25Mpixel (6.5Kx4.3K) camera. <S> From this you can see the issues: to minimize optical aberrations you need to have some distance to the camera, which in turn requires increasing its resolution much more than if you scan with a sensor up close. <S> By the way there exist some more consumer and DIY projects of this kind; something was showcased at CES 2011 ; I don't know if it made it into a mass market product. <S> Also something DIY .
| Actually there exist [book] scanners that are just that, a big sensor that images the whole page with no moving element .
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How does a logic gate behave with an input changing faster than its propagation delay? I am attempting to create a simple logic circuit simulator. I am having a hard time figuring out how does a logic gate behave with an input changing faster than its propagation delay. When attempting to connect a high frequency clock to an invertor logic gate with a 1 second delay in a simulator, the output stayed the same for 1 second and then started following the pattern of the clock. It seems as if the output was a following function: output(t) = NOT(input(t - delay)). Is this true? If so, why? Also, how would this work with different rise and fall delays? <Q> It really depends on how the gate is constructed. <S> For a completely accurate simulation, you have to do a transistor-level analog simulation. <S> However, it is possible to extract timing parameters from a transistor-level simulation and abstract them out a bit. <S> The output rise and fall times and propagation delays will depend on the input rise and fall times, output load capacitance, power supply voltage, temperature, and the state of the inputs. <S> Yes, it is possible for the same input transitioning to have a propagation delay that depends on the state of the other inputs. <S> These techniques are used in the timing models used in ASIC and FPGA design in both static timing analysis as well as timing-driven place and route. <S> For a single two transistor CMOS inverter, the propagation delay is determined by the analog electrical characteristics of the transistors and their parasitic capacitance. <S> The input will slew at some rate, then once the threshold is reached the output will start slewing. <S> If the input changes before the output finishes slewing, then the output will start to slew back the other way and you will end up with a highly distorted output. <S> So for a single inverter, the output for a change faster than the propagation delay would be an invalid logic level (i.e. x). <S> However, "gates" can be far more complicated than a single inverter. <S> For example, if you have a "gate" that is built from a string of 100,000 inverters, then the propagation delay of the whole unit will be 100,000 times the propagation delay of a single inverter and it is certainly possible to have multiple transitions 'in flight' at the same time, so long as these transitions are not faster than each individual inverter can handle. <A> What you are describing is called "transport delay" in VHDL, as opposed to "inertial delay" in which events shorter than 1 second would silently disappear. <S> "Transport delay" is useful for modelling delay lines, very long cables, and so on, but "inertial delay" is a more accurate simulation of logic. <S> More information here <A> The propagation delay is simply "the amount of time it takes for the head of the signal to travel from the sender to the receiver" (from here ), it is not the bandwidth, which is how fast the system can respond to the input. <S> The output will behave as expected, but delayed by the propagation delay time. <S> With different rise and fall propagation delays it will be more complicated, because different part of the signal will be delayed differently, and the behaviour will not be so obvious. <S> Of course, all of this is valid for simple systems (i.e. the inverter), increasing the complexity will lead to incresed chance of unpredictable behaviour. <S> (i.e. the race condition ) <S> There are even circuits based on the propagation delay, i.e. the ring oscillator <A> Roughly speaking there are two types of delay, "inertial" and "transport" (and combinations of both). <S> "inertial" delays represent capacitances (usually parasitic) that take time to charge and discharge. <S> Most basic logic gates would fall into this category, as charging and discharging capacitances is the main cause of delay. <S> Delays may not be symmetrical and may depend on the states of other inputs. <S> If you look at the output on a scope you would see it ramp up/down to it's final level. <S> If the input changes too often fast then it will never reach the levels it's supposed to reach and you will get an output that wanders up and down without actually reaching it's targets. <S> "transport" delays represent things that delay a signal without destorying it's shape. <S> One example is a properly terminated transmission line. <S> A long chain of gates can also look much like a transport delay for some switching rates as the individual gates have sufficient time to reach their target voltages but it takes some time for information to work it's way down the chain.
| Fundamentally, the propagation delay is determined by how long it takes for the output to transition in response to a change at the input. This depends on exactly how the gate is built at a transistor level.
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is soldering necessary for a robotics and electronics hobbyist? I know it is VERY common but I am worried about the exposure. I do understand it is minimal if an extraction device is used to get rid of the gasses and if a lead-free solder is used but I still have worries. I was wondering if it is essential since I am very interested in robotics and electronics. <Q> No soldering is not required. <S> Robotics and electronics is a huge area. <S> I know people that haven't touched an soldering iron in years because their job does not require them to do so because the field of robotics and electronics is so large. <S> However, depending on what your interests are and what your current level is, you may have to get your hands dirty and get into the low level stuff, such as soldering. <S> If this is of no interest to you, then you are looking at paying someone else to make custom cables and any other assembly work you require. <S> If time and money are of no concern to you, pay someone else to do it. <S> Problem solved. <S> Though, in my opinion, learn to solder. <S> Read up on whatever it is that is worrying you, and read about proper techniques and tools to ease yourself. <S> It's a basic skill in this field, and you are only robbing yourself of the basic skill by avoiding it. <A> If you are really interested in the details of robotics and electronics, then soldering probably can't be completely avoided. <S> Having said that, the dangers of soldering are not that high. <S> Of course lead free solders will eliminate any exposure to lead. <S> The biggest problem is the fumes and skin contact with the rosins used in most soldering operations. <S> These rosins can cause a variety of medical conditions. <S> However, if you only solder occasionally and in a well ventilated area, the risks are minimal. <S> If you google "solder safety", a lot of useful information will become available which you can study and decide for yourself whether you want to be involved with soldering. <A> You will occasionally break some wires here and there that's for sure. <S> Sometime the wires and sockets get detached (unintentionally) from the main PCB and will require little soldering. <S> You would not want to run to a technician to get it soldered right??? <S> ... <S> and that too when you already know what is broken. <S> Even though you may not require soldering in the beginning but you will eventually need to get yourself familiarize with the basics of soldering. <S> IT WILL BE REQUIRED.
| The temperatures used in normal soldering are not high enough to vaporize the metals used such as lead, tin and copper so metallic fumes are not a problem. It is not essential if you don't get down to actually building your own circuits but instead assemble purchased components.
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Voltage drops not adding up - beginner question I'm fairly new to electronics and playing around with a very simple series circuit - 2 resistors of ~220k Ohm and a 3v power supply, which I'm measuring with a digital multimeter. If I measure the voltage across both resistors (or direct across the power supply terminals) I get a reading of 3.37 volts, but if I take a reading across the individual resistors I get 1.656 and 1.676 V (adding up to 3.332). I'm wondering what would account for the difference in total voltage? Even if the multimeter wasn't calibrated properly I'd expect it to be consistent? Thanks! <Q> A typical voltage meter has about 10 MOhm input resistance, which you have to account for in your calculation. <S> The measurement error is about 2%. <A> I'll assume this is the circuit you`re testing <S> Usually a voltmeter ( PMMC ) type: <S> Is a galvanometer with a HUGE series resistance ( Mega ohm ), such that a very small amount of current pass through it whenever this voltmeter is connected in parallel with a resistance For the circuit shown <S> Voltage source is 3.3 V Current = 0.0075 <S> A Load resistances = <S> 220 <S> Ω <S> each <S> The voltage drop across each of the 220 Ω is 1.65 V = <S> (0.0075 <S> * 220) <S> BUT <S> when you connect your voltmeter now Assuming that: R : <S> The voltmeter resistance = <S> 10 MΩ <S> And you want to measure the drop across one of your resistances <S> so you'll now connect your voltmeter in PARALLEL with one of these 220 Ω resistances <S> Now you`re adding a parallel resistance with your load of value 10 MΩ <S> So the current is now not 0.0075 A since the circuit equivalent resistance is no longer the two series <S> 220 Ω resistances only its a combination of two parallel resistance 220 Ω , 10 MΩ and a series resistance <S> 220 Ω such that the equivalent resistance is now 439.9951 Ω (Almost less than 440) <S> so you'll not have the same amount of current (0.0075 <S> A) <S> but a slightly more amount of current in this case (0.00750008249 <S> A) Almost 0.0075! <S> this in turn affects the voltage drop across each resistance by a very very small value that we usually neglect <S> So you'd expect a slightly smaller reading than the calculated value <S> Other factors also affects your reading such as The source output resistance Wire losses <A> One is the input resistance of the meter. <S> Another is the full scale calibration error. <S> We also have offset error and nonlinearity errors. <S> Offset is normally negligible with modern meters and nonlinearity is pretty low for most (not all) meters. <S> Nonlinearity for a 7106-based 3.5 digit meter is typically + <S> /-0.2 count. <S> I'll assume you have such a meter. <S> So, a good first order approximation of the error in reading a voltage with source impedance RS is Vr = <S> Vx <S> * (Rin/(Rin+Rs)*(1+a) where |a| << 1 is the full scale error for a particular range and Rin <S> similarly is the input resistance for a particular range. <S> In your case, two equal resistors of value R connected to a stiff voltage source, the source resistance is R/2 for the measurements across each resistor or 110K\$\Omega\$. Looking at the total of the two measurements of closely equal resistors we have: <S> Vt ~= <S> Vx <S> * Rin/(Rin+Rs)*(1+a1). <S> The measurement of the total is just Vx = <S> Vx <S> * (1 + a2) <S> So Vt/Vm = Rin/(Rin+Rs)*(1+a1)/(1+a2). <S> To put some numbers on these errors, if the input resistance is 10M and the error on the 2V scale is -0.2% and the error on the 20V scale is +0.2%, the ratio would be 0.985. <S> If measured on the same scale the error would be 0.989, ignoring the +/- <S> 0.5 count uncertainty in each measurement. <S> Your number was 3.37/3.332 =0.989, so if the assumptions were correct you got lucky in using the two different ranges- <S> the error could have been much higher.
| There are several sources of error in your measurements.
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Is there a sensor for counting the number of cards in a flipping scorecard? I have a stack of cards and want to know how many cards are in the stack at any given time. Any thoughts on how I might be able to do this? Would it be possible for me to put a tag on every card and then use some type of sensor below the cards which could count the tags above it?The cards are suspended from rings like this: While there was a prior question that was similar, it does not appear there were any answers that came out of that question. <Q> Magnets? <S> Put a tiny magnet on each leaf, in a different position. <S> Put two hall sensors behind, attached to the wall. <S> When someone flips a leaf, the magnetic field will change detectably. <A> Mark a binary code on the bottom of each card and use photosensors to read them. <S> Scorecards showing '8' and '6'. <S> Binary code is 8 4 2 <S> 1- <S> - - - = 0- - - X = <S> 1- - X - = <S> 2- - X X = 3etc. <S> You would need some logic to decode the binary code and get it into whatever format you want (but your question doesn't explain that). <S> Code could be placed on rear if that suits the sensors. <A> Here's a slightly crazy idea. <S> You may be able to create a little pocket for the body of the resistor to help the stack lie flat. <S> In the schematic below SW0 + R0 represents card '0', etc. <S> Insulate the back half of the rings (or just one of them) so that they don't contact the resistors on the fliped-over cards. <S> Connect 12 V negative to one ring. <S> Connect 12 V positive to the other ring through a 10k resistor. <S> The voltage across your rings will be the score. <S> 0 <S> -> <S> 0V, 1 -> 1V, etc. <S> You may need to add weights to each card to pull them into contact with the rings. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> How it works <S> When '9' is showing it is the only resistor left in a circuit consisting of the 10kΩ RS and the 30kΩ pull-down, R9. <S> The result is 3/4 of supply voltage = <S> 9 <S> V. <S> When '8' is there <S> as well we want 8/12 of the 12 V <S> so we need to add enough resistance in parallel with the 30kΩ to bring the combination down to 20kΩ. <S> By parallel resistance formula we can show that 60kΩ is required. <S> The exercise continues for each step resulting in the following table for a 12 V supply and 10kΩ RS. <S> Digit <S> Rp <S> * <S> (k) <S> Rc <S> ** <S> (k)0 0.00 0.001 <S> 0.91 1.672 <S> 2.00 <S> 5.003 <S> 3.33 <S> 10.004 <S> 5.00 <S> 16.675 <S> 7.14 <S> 25.006 <S> 10.00 <S> 35.007 <S> 14.00 <S> 46.678 <S> 20.00 <S> 60.09 <S> 30.00 30.00 <S> Rp is the parallel combination of all the resistors left in contact with the rings. <S> Rc is the resistor on this card (digit). <S> If you want to run the simulator on this then double-click each switch in turn starting from the left and change the contact to 'open'. <S> A spreadsheet is your friend when working out some of this stuff. <S> You now have the choice of using a great big analog meter or feeding into an ADC. <S> Now where can we find a 3 foot diameter analogue 10 V voltmeter? <A> The photosensor option: Photo 1. <S> Card '0' and card '6'. <S> Blue dots represent the photosensors. <S> Position nine photosensors behind the cards to line up with the 0 to 8 cutouts shown in the photo. <S> The photosensors may need a short black tube in front to prevent stray light affecting operation. <S> On each card cut out the number of tabs to match the number on the card. <S> Card '0' has no tabs removed. <S> Card '6' has six tabs removed. <S> Tab '9' is never removed. <S> The logic for decoding the card is trivial although it does require more inputs than, for example, a binary coded system. <S> Configure the circuit to give reliable operation in various lighting conditions. <S> If you can get this to work reliably the big advantage is that the photosensors all go behind the cards, out of harm's way. <S> You may need to give some thought to protecting the bottom edge of the cards as any dog-earing through use may affect operation. <A> At 1/4" thick, ultrasonic measurement springs right to mind - 5 seconds of web search brought up sensors claiming 1mm resolution from over 1000mm away, which is about 6 times what you need to tell when one is added or removed... <S> Then again, a bar code reader would do nicely. <S> If the number & order of cards is fixed, you could do both of these things from the backside since you don't want anything out front - read the back of the next card to be flipped, or read the thickness of the stack of cards that have not yet been flipped.
| Glue a resistor to each card, threading the ends through the holes in a manner to ensure that when the card hangs on the rings the resistor touches both rings. Some open-sided lead fishing weights might do the trick.
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Using NPN-Transistor as switch not working I have following schematic: simulate this circuit – Schematic created using CircuitLab But the motor is not working, the NPN-Transistor does not seem to let any current through. What am I doing wrong? the transistor does work, I checked, but there may be something very basic wrong, electronics is not really my strong suit. <Q> You're very close. <S> instead of having the NPN in its present location, it need to be between the motor and GND. <S> So disconnect the NPN's emitter from the motor, its collector from the 9 V supply, and connect: Emitter -> GND Collector -> Motor Base - <S> > <S> 1k & 5 V as you have it. <S> The motor will go between the 9 V supply and the NPN's collector. <S> Whether or not 1k will work depends on the motor current. <S> 5 V and 1k will give about 4.3 mA of base current -- probably about 100 mA of collector current. <S> This sounds low for even a small motor. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Also, it is possibly you have (partly) damaged your NPN -- the way you have the circuit now, the base-emitter junction is reverse biased. <S> These usually break down around 6 V, and when they do, the beta (gain) of the transistor gets degraded. <A> As in my comment, BAT2 and diode are connected in reverse. <S> (as shown in your schematic). <S> However, you might have the schematic drawn incorrectly as it is more common for the second voltage source to be on the transistor's collector line (at the far right, with + facing up in this case). <S> The bottom lead of the motor would then be connected to the common - terminals, the diode on the motor still needs to be reversed. <A> What am I doing wrong? <S> Pretty much everything. <S> but you have this reversed. <S> Try putting the emitter to 0V and putting the motor in series with the collector up to 9V. keep the diode connected the same way round (about the only bit you got correct). <A> As another alternative (and close to what Andy has said), keep your original schematic but just swap the Emitter and Collector pins. <S> That gives you the correct polarity for the batteries and the diode.
| You are reverse biasing the collect-emitter polarity - an npn wants to see positive on the collector relative to emitter
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the relationship of amps to rpms in an induction motor when searching for a 2000 watt induction motor online the rpms seems to range from 3450-3600 Why? if you wanted to double rpms to say 7000 ,is that possible and if you are using 3 phase 220 V line, how would amps be effected. where can I study about this? I am a novice <Q> Speed: <S> The speed is measured in RPM. <S> 60 Hz excitation voltage corresponds to 3600 RPM. <S> The mechanical speed can be further reduced by using a motor with higher magnetic pole count. <S> Two pole motor has the same mechanical frequency as the excitation voltage. <S> Four pole motor will have half the frequency, six-pole motor will have a third of the frequency. <S> Induction machines slip behind the excitation frequency by definition. <S> Typical slips are 0-5%. <S> The higher the mechanical loading, the higher the slip. <S> 7000 RPM is not possible without either a variable frequency drive (VFD) or a gearbox. <S> Power <S> For simplicity, one can assume that the mechanical power out is about 90% of the electrical power in. <S> 2 kW motor can therefore draw ~2.2 kW of real power. <S> With 3-phase 220V line voltage the current would be: $$V_{phase} = <S> \frac{V_{line}}{\sqrt3 <S> } = 127 Vrms$$ <S> $$I_{phase <S> } = \frac{P}{3V} = \frac{2200 <S> W}{127 V*3} = <S> 5.8 Arms$$ <A> Induction motor speed is a matter of poles and frequency - on <S> 50 Hz you'll get 3000, 1500, 1000 as you go from 2-pole to 4 pole to 6 pole. <S> On 60 Hz its 3600, 1800 and 1200 - on 400 Hz its 24,000, 12,000 and 8000. <S> A variable frequency drive can provide a wide range of speeds, but it can also provide more speed than a motor may be designed for - if a motor purely intended to run on 60 HZ is attached to a VFD and wound up to 120 Hz, you might find the mechanical limits of strength in the rotor and destroy it (loud noises, sudden stops, great expense.) <S> There are motors specifically made for VFD operation which are rated for higher operation speeds - while expensive as compared to a normal motor, they are cheap as compared to what can happen when a normal motor comes apart in use. <S> The other common option for higher-speed motors is not an induction motor - rather a brushed type DC or universal motor. <S> On average these are fairly loud as compared to an induction motor, and there will be sparks from the brushes (and the brushes wear and need to be replaced) but they are a common approach to getting high speed operation, as seen in most portable saws, drills and routers, or small hand-held rotary tools. <A> Please comment on the following including if possible potential commercial applications. <S> We have just completed a working prototype that is comprised of two 1000W (1.34 hp each) induction type motors, which based on an efficiency of 0.885, equals to 1.18hp net for each motor. <S> The technology is modular. <S> The prototype contains two modules for proof of concept. <S> The RPM from the first motor (3,450) is transferred to the second motor, and then aggregates with the RPM from the second motor. <S> The combined RPM of about 7,000 is evidenced on the rotor of the second motor. <S> This increase in RPM is not due to either modulation of frequency or by using different sized pulleys/gears. <S> In addition to the accumulated RPM, the hp from both motors has also been aggregated. <S> This has been made possible due to way in which the RPM has been increased and accumulated, however combining the motors, the efficiency goes down somewhat to 0.78 (0.885 x 0.885), or a combined hp of 2.1hp (0.78 <S> x 2.68hp).The torque for the first motor is easily calculated using the formula ( hp x 5250)/RPM …. <S> (1.18 x 5250) /3450 or 0.7754 ft-lb. <S> The torque for the combined motors (2.1 x 5250) / 7000 equals 1.575 ft-lb, however the RPM has increased from 3,450 to about 7000 RPM. <S> In other words, the prototype demonstrates an increase in the second motor of both RPM, hp and torque.
| For an application requiring a fixed 7000 or 7200 RPM speed, a simple belt & pulley arrangement to double the motor shaft speed is likely to be the most cost-effective and power efficient approach (gearboxes are relatively expensive and can also be surprisingly inefficient.)
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What type of paper is good for the toner transfer PCB process? I've been using magazines and it works about half of the time. Thick sheets work better but they are hard to find and I must cut them to size. I'd rather just have something that works so I bought glossy yellow paper meant for toner transfer on Amazon. It didn't work at all, my printer can hardly get the toner to stick. Magazines are better, but they still are not good. Is there a type of paper that works consistently? <Q> Coated paper <S> Its the paper that I use, after the iron in the pcb, its easy to remove when you put the pcb in the water. <S> Just remember to use a lot of toner. <A> And as Rodion Gorkovenko mentioned you should make sure that the copper surface is clean. <S> And there are so many things that can go wrong. <S> You should apply heat to the paper atleast for 3-5 minutes to be sure. <S> And if some tracks are missing then you can use a good permanent marker to join them. <S> I use whitener for this purpose. <S> You should always take the print in a laser printer. <S> An inkjet printer wont work at all. <S> And make sure that the iron you are using is not a "steam iron", if possible use one with a complete flat surface. <S> Hope <S> this helps you. <A> It didn't work at all, my printer can hardly get the toner to stick. <S> Well, it probably will work all right but the problem is this specific printer (and my as well) could not heat the sheet well enough :( <S> So probably you will not be satisfied with any paper. <S> Though another thing to think about is the preparation of the PCB. <S> It should be cleaned very well to allow toner to stick better. <S> I use toothpaste and brush and some liquids for cleaning the bathtub usually :) <S> But I suggest you can use other approach instead, with photoresist. <S> It is bit more costly, of course compared to paper and printer, but I suspect the paper from amazon was not cheap also :) <S> For home use I suggest liquid photoresist rather than film. <S> Once you get all necessary paraphernalia and make one or two attempts you may find it more convenient... <A> One day I just done with normal paper,and ended up with failure. <S> Then I chose glossy sheet for toner transfer method. <S> It works fine for me for tssop package. <S> My advise is to use glossy sheet or OHP sheet. <S> While transferring the print to board with hot iron,dont scrub in it. <S> A small shake during transfer will make thinks bad. <S> Just apply heat over it without shaking. <S> During dry Iron,use maximum heat.
| Try to use glossy sheet,OHP sheet for toner transfer method. I personally use trace paper for the transfer.
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Is there an easy way to measure the thickness of the foil on FR-4 Sometimes I find myself trying to guess or recollect what foil thickness was specified for the given piece of FR-4. Shops do not always readily give it - and it appeared that what they write is not always reliable. And sometimes I, regretfully, simply could not remember what was specified. I do not care about much precision, but it would be good to be able to roughly distinguish between 70, 35 and 18 um foils (to select more suitable for intended traces width). I have few ideas in mind but none deserves to be called "simple"... <Q> Suddenly (reading loosely related "related questions" on the right of the page) I came to the idea which after testing seems better than others. <S> I've used kitchen scales to measure the weight of the sheet. <S> Then I used the ruler to calculate the area and vernier caliper <S> (if it is proper translation for what we locally call "shtangentzirkule") to check the overall sheet thinkness roughly. <S> Then I calculate the expected weight of laminate and subtract it from the total weight. <S> Now I have the weight of two copper sheets (in my case it was about 10g for area of about 160 cm2 ) <S> so I get the answer (0.0069 cm for two sheets total) looking very close to 0.035 mm per sheet. <S> e.g. S - area of the sheet (163 cm3 in my case)W - total weight of the sheet (55 g)d - thickness of the sheet (0.14 cm)p0 - laminate density (1.85 g / cm3) - googled, of coursep1 - copper density (9 g / cm3)w0 = <S> S <S> * d <S> * p0 (I get about 45.6 <S> g)w1 = <S> W - w0 <S> (i.e. 9g)d0= w1 / (S * p1) / 2 layers <S> This should be bit less effective for single-side sheets, but I hope it will still work well, since the copper is almost 5 times more dense compared to the laminate... <S> Other ideas I have were either try to measure resistance or the time of solving in the FeCl or something like this. <S> But I have no idea how to make them easy or reliable... <A> I'm not sure how interested you are/what your budget is, but what you're looking for is a simple profilometer. <S> I have done lots of work in cleanrooms and these are one of the basic tools of metrology. <S> If you have a budget, something like this is really what you should look at getting. <S> However, a much more inexpensive way I think would likely work would be to purchase a basic microscope ($200 - the ones you likely had in high school biology class) and get a few foil or plastic pieces as "standards". <S> Then you could place them on the board next to the foils and focus on the standards. <S> If you choose a microscope with a narrow depth of field lens (see link ), you can find the height by seeing if the foils are in focus next to the standard. <S> If the standard and the foil is in focus then the heights are within the depth of field of the lens. <S> This the closest simple equivalent to doing 3D microscopy I can think of. <S> Of course you'd have to calibrate this method quite a bit, but with practice and the right foil "standards" I think you could get within a few microns and make very quick measurements as well. <A> It is possible to measure 0.1um accurately with a good one, and even about 3um accuracy is not too difficult to get .
| If you have access to both sides of the foil, then the easiest way might be to directly measure the thickness using a micrometer .
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What are all the multiple pins on PSU connectors for? How does a 24-pin MB connector work exactly? I'm looking to learn more about power requirements and such that a Power Supply/Computer work with. I see this image and it has a lot of different pin layouts for various things and such. For a few projects of mine I'm looking to connect up to the power supply. I've heard I only need 1 12V+ and 1 ground, but I also need a 5V+ and a ground for another part of the project. I'm curious if it matters where this is taken from on the whole? I'm also curious, looking at that 24-pin connector, why the MB needs all of those values? 3.3v, 5v, a bunch of 12 volts.... Does that all get split into individual things or...? I see the VGA connector is 3 12V, and 5 grounds, so I'm curious if individual parts of the GPU require 12V+ or if it requires 36v total, or whatnot? Essentially I'm just curious how power works, how you know how much voltage, and amps and such you need to power your devices, and all that stuff with connectors and such, to make sure you have the proper power inputs and such? Any advice is appreciated, thanks a lot! <Q> As Passerby says, the main reason why there are several wires for any given voltage is that a computer may require tens of amps, so one wire by itself isn't enough. <S> The reason for all the voltages is largely historical. <S> For many years, the standard voltage for digital logic circuits was 5V. <S> Many circuits still run at 5V. <S> However, later chip designs could be made to work at lower voltages, and would run cooler when powered at a lower voltage. <S> So 3.3V was added to the power supply. <S> +12V is mainly used for powering motorized things, like disk drives and fans. <S> It's also needed for RS232 serial ports. <S> RS232 also requires -12V, so power supplies often have a low-current -12V line as well. <S> The 5Vsb line is a "standby" power line, to keep some parts of the computer powered on while the computer is in "sleep" mode. <A> An ATX power supply uses multiple wires for the same voltage, to increase its current delivery capacity. <S> And because using multiple if the a smaller gauge of wore can be cheaper than a single higher gauge of wire. <S> The same is done for the VGA connector. <S> The Layout is chosen to allow easier routing of power on a circuit board, minimizing the need to jump over or bridge traces from one side to another. <S> No, the voltage needed is not combined for 36 Volts. <A> Every wire is essentially a really small resistor, so if you put allot of current (tens of Amps) through it you will get losses in the wires. <S> It is easier to use multiple wires to reduce the combined resistance of the wires than one big one since those are usually less flexible and harder to route in your PC. <S> If you just want to power something little then just pick any 12V and GND Pin. <S> If you start to draw serious power from it, say more then 10A then connect to multiple 12V and Ground. <S> Sidenote: you have to turn on the PSU by connecting the PS_ON pin with ground to get 12V. <S> For 5V you don't even need to turn on the PSU, you can get standby power from 5V_SB, usually up to 5W.
| The PCIe plug might have extra Ground Pins because a GPU can also receive extra power over the PCIe bus (3.3V and 12V) from the mainboard.
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Determining resistance of 3 resistors connected in delta configuration I have 3 resistors of varying values connected in a delta configuration. I have access to all 3 nodes for measurements. I know that if I measure resistance across 2 points, I'm measuring 1 resistor in parallel with the sum of the other 2 resistors.Is there a way to determine each of the resistor values as if the circuit were opened (not connected in parallel) without any guarding and without destructively opening the circuit? I'm trying to verify my manufactured resistors in-house prior to sending them out for laser trim. <Q> The equations are easy to write down, but cumbersome to solve. <S> When you have a triangle of 3 resistors (a, b, c), and nodes A, B, C. Resistor <S> a is connected between nodes B & C; resistor b between A & C, and c between A & B. Make 3 measurements -- AC, BC, and AB between nodes A&C, B&C and A&B. <S> Mathematica gives this: $$a->\frac{(-AB^2+2\cdot AB \cdot AC-AC^2+ <S> 2\cdot AB\cdot BC+2\cdot AC\cdot <S> BC-BC^2)}{2 \left(AB+AC-BC\right)}$$$$b->\frac{(-AB^2+2\cdot <S> AB\cdot AC-AC^2+ <S> 2\cdot <S> AB\cdot BC+2\cdot AC\cdot <S> BC-BC^2)}{2 <S> (AB-AC+BC)}$$$$c->\frac{(AB^2-2\cdot AB\cdot AC+AC^2-2\cdot <S> AB\cdot <S> BC-2\cdot AC\cdot BC+BC^2)}{2 <S> (AB-AC-BC)}$$ <A> You have three nodes two measure from, so you can make two independent measurements to start from. <S> For example, you can measure R1 || (R2 + R3) and R2 || (R1 + R3). <S> To get a third independent measurement you can <S> (for example) short out R3 and measure R1 || R2. <A> First you can find the equivalent star. <S> Since we have a simple equation to solve (eg \$ R_1+R_2= <S> x, R_2+R_3 = <S> y, <S> R_3+R_1=z \$) from the results of \$R_1, R_2 and R_3 \$we can find the equivalent delta it is one of the best way without the need to remember all formulas
| With some algebra, you can now calculate each of the individual resistances.
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Christmas lights - how many strands can I connect? I am putting lights around my house this season, and am finding myself concerned about safety and what the company who makes the bulbs says is safe. I think they are being way over-cautious, because they say we can only connect two strands. But I'm reading different things on the web, and I'm not sure which is correct. I want to connect six strands of incandescant C9 bulbs, which have 25 bulbs each (150 bulbs total). I've carefully read all the safety instructions, but the only recommendation for how many strands to connect is on the box itself, and it says only two. This is very inconvenient because there will be three extension cords going across our yard and walkway. I found this equation via Google:Number of total bulbs = Volts from your outlet X amps on your circuit X 80%/Wattage from a single lamp So for a 120V outlet, and a 15A circuit, and each bulb having 7W, we should be ok with 205 bulbs. And so, is it ok to connect all six of these strands, with the above equation in mind? Is the company that makes the lights just being super-cautious so that they don't get sued? What would be the major concern (blowing a fuse, or something else)? And to clarify previous questions, my lights would in this case be connected in series (if it were possible to safely do so). Each strand is wired in series. <Q> If the strands are connected "in series" - strand <S> A plugs into an outlet, strand B plugs into the end of strand A, strand C plugs into the end of strand B, etc. <S> then all current for all of the strands will have to flow through the wires of strand A, which are probably too small to carry the current of more than two or three strands. <A> Very definitely, Peter's answer is correct, if you string all these lights together, you'll be putting yourself at a risk for an electrical fire. <S> The big problem is the lamps you're using, those 7 watt incandescent bulbs are a thing of the past. <S> You need to go get some LED type christmas lights, these draw less than 1/2 watt each. <S> A typical string of 100 lights is ~ 40 watts, so <S> I'm guessing you could string up to 5-6 of these together, but definitely follow the recommendation on the package. <A> If you can wire one string to the AC power then you can wire two strings in parallel to the same AC mains providing you don't exceed the 15A. OK so wire two in series and then take another two wired in series and wire these in parallel to the original two series strings. <S> Keep going until you run out of terminal blocks or you start to create an energy shortage in your district. <S> That's a joke by the way <S> but I'd hate to see anyone's xmas lunch spoilt by a power outage. <S> Just do it safely!!
| If you have eight strands connected "in series", the wires in the first couple of strands will likely overheat and melt the insulation.
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Bulb not lighting up if resistor is used I have a 12v adapter and small bulb of about 3.8v . If i give power to bulb with 12 volt adapter and with 100ohm to 10k ohm it doesn't light up and if i directly give it power without the resistor it fuses. This is the bulb i have When i give power to bulb with 9v battery without resistors it light up perfectly fine. Adapter gives 1.5amp (mentioned on sticker) How can i light up this bulb with 12volt adapter without blowing(fuse) it up?. I am not much experienced with electronics. <Q> I found a reference to a #13 radio pilot lamp rated at 3.8 volts, 0.3 amp. <S> To operate that lamp from a 12 volt supply, you need a resistor that will drop 8.2 volts at 0.3 amp. <S> Ohm's Law says that a 27 ohm resistor would work. <A> Many adapters of the type you describe are unregulated giving a higher-than-nominal output when unloaded or even lightly loaded. <S> They tend use the load current of their intended powered device to pull the output voltage down towards the nominal value (12V in your case). <S> You say that you think your MES lamp is 5V and that the adapter has a capability of 1.5A. <S> Your lamp probably draws between 60 and 300mA, consequently your adaptor is only lightly loaded at perhaps 6% of its capability, causing a high output voltage to appear across your lamp and 'blowing' it! <S> The value of any series resistance is calculated from Ohm's Law using the formula: <S> R = V/ <S> I where V is the difference between the nominal supply voltage (12V) and the lamp's voltage, which in this case gives us the result of 7V. <S> The value for I is the nominal lamp current, say 100mA (or 0.01A). <S> This gives a resistor value of 7/0.01 = 70 ohms. <S> The nearest E12 resistor value is 68 ohms. <S> If the lamp lights with this value of resistor but is somewhat dim, it means your lamp requires more current (assuming the supply is as described). <S> Try fitting a lower value resistor, say 47 ohms until you reach a satisfactory brilliance. <A> The problem is that the resistance of the bulb is probably very low, and you need to use a resistor roughly equal to the resistance value of the bulb, to equalize the voltage drop across both components. <S> The 12v is too high for your 5v bulb, so that's why it burns out when you connect it directly to the 12v. <S> I'm assuming you're connecting it in a parallel?
| Your question raises a number of other issues inherent in this situation, such as the required power rating of the resistor and the very fact that whilst this solution is a simple one, it represents a low efficiency solution owing to the energy that is wasted as heat in the resistor. If your lamp draws a different current, the you would need a different resistor, but 27 ohms should be a good starting point.
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Why do we need a pull-up resistor if we can just make a short to the power source? I saw the term pull-up resistor being mentioned everywhere, but I haven't grasp the reasoning behind its use. For example, I saw one in the book Practical Electronics for Inventors, in this schematic for 555 timer arranged for monostable operation. . It is explained that ... pin 2 is normally held high by the 10-k pull up resistors. But why use a resistor? If we want to keep pin 2 high, can we just make a short from this pin to Vcc? If we just want to limit the current, I thought we can do it somewhere else, for example by adding a resistor between the output pin 3 and the ground? <Q> In your drawing, the pull-up resistor does pull pin 2 towards Vcc, but allows a trigger input to pull pin 2 low in order to trigger the 555. <S> If you connected pin 2 directly to Vcc, you could never trigger the 555. <S> (often "inactive") state, but allow the line to be pulled to an active state by other circuits when required. <S> If you want a signal to be held to Vcc or Gnd at all times, you can usually make a direct connection to Vcc or GND. <S> (There have been logic families where a direct connection of an input to Vcc is not recommended, so a pull-up resistor should be used there, even if the pin will never be pulled low.) <A> Because once the "Trigger Voltage" goes low, you create a dead short between VCC and the "Trigger Voltage" source. <S> Dead shorts are bad. <S> The Pull-Up sets the default state (high), and is weak enough to be easily pulled low while preventing a high current draw in that situation. <A> In my experience, the term 'pull-up' (or pull-down) resistors originated in digital logic circuits, particularly those employing TTL devices with 'open collector' gate outputs. <S> The pull-up resistor ensured that when a gate's output was 'off' (i.e. non-conducting) <S> the logic level was Logic 1 (or if you prefer, HIGH or TRUE). <S> With a supply voltage of just 5V, usually 1k resistors were used to minimise current consumption when the gate's output was driven LOW (Logic 0). <S> Although not a digital circuit, the 555 timer inputs can be sensitive to spurious signals if not held at some pre-determined voltage level when in the quiescent or inactive state, hence the use of the resistor. <S> If the input was connected directly to Vcc, and ignoring any other effects, the input would never be able to change!
| In general, pull-up and pull-down resistors are used to hold a signal line at a default
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how do i select a battery for a circuit which has multiple ICs? I am using INA125 amplifier, a PIC controller, a load cell and a ESP 8266 wifi module. how can I calculate the power consumption to select a battery?Do I just add up the currents of my 4 devices or is there something I am overlooking? <Q> Read the data sheets and calculate or measure the current. <S> Youranswer will be in mA. <S> Decide how long you want the battery to last. <S> Answer will be in hours. <S> Multiply the two together. <S> The answer will be in mAh. <S> This will be the minimum battery capacity you require. <S> Purchase the next larger size. <A> The INA125 can be powered from 2.7V to 36V, most PIC controllers have a power source range of 2V to 5.5V and the ESP8266 appears to require 3.3V (max). <S> The key driver here is the ESP 8266 which would appear to require the use of a regulator to ensure it is not powered from a voltage that is too high. <S> All the devices listed would happily run at 3.3V which greatly simplifies interfacing as they will all be on the same power rail (you will need to make sure your load cell can operate in this environment). <S> Now read the datasheets and calculate the required current. <S> If you use a linear regulator, this will also be the battery load current. <S> I am not going to go into switching regulators in any detail at this point, although they have higher efficiencies than linear regulators, depending on the specifics of the circuit. <S> At very low Vin to Vout differences, this advantage disappears - even switch mode devices have a minimum dropout voltage, although some exist that can have the performance of a good low dropout device. <S> You have not stated what type of battery, but I can give a little guidance for Li+ A typical Li+ battery <S> has a linear discharge curve and at 3.5V (the minimum input to a linear regulator for 3.3V, typically, although there are very low dropout voltage regulators; the dropout voltage is usually load dependent), the battery will be at about 40% of rated energy ( very manufacturer dependent, check the datasheet; this is typical of the devices I used some years ago); i.e. if it states 2000mAH, then you would have a useable energy of about 60%, or 1200mAH. <S> Note that for all battery chemistries, the discharge curve and useable energy is dependent on the load , with a low load typically yielding higher useable energy. <S> Take this useable energy, divide it by the load and you have the length of time the battery will last. <S> Alternatively, take the load and multiply it by the time you wish the battery to last and then multiply it by the useable energy factor (0.6 in this case). <S> You will need to have a useable energy factor whatever you do. <A> Here is my suggestion if this is intended for a hobby application. <S> Use a single Li cell, that makes it easier to charge, using something like this one . <S> Use a DC-DC step up like this <S> one to power the load cell and the amplifier. <S> The PIC can possibly be powered directly from the Li cell. <S> Use a LDO regulator for the ESP8266. <S> If you want to go cheap, you could use a diode to drop the voltage or even connect the ESP8266 directly (In my experience they have survived 4.2v but it is not recommended). <S> Again this recommendation is for people looking for a quick and cheap solution and not for commercial development. <S> The key to designing battery powered devices is efficient power control. <S> Make sure the PIC and its peripherals are powered down when not in use. <S> The amplifier and load cell should be powered on only when taking the measurement. <S> As for the battery capacity the other answers have all the information you need.
| In all such cases, you should carefully check the power requirements of the devices you intend to use.
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STM32 & ST-LINK - Cannot connect to MCU after successful programming I have built my own board with STM32F7-45VGT6. I have successfully programmed it with ST-LINK v2 (not the original one though) and now I cannot even connect with MCU. I use ST-Link Utility from ST and SWD interface. It can be the case that I use SWD pins as output and in my code I set them as GPIO output. Can it be the case? Nevertheless, I connect my reset pin to GND and set "Connect under reset" option in ST-Link Utility but it doesn't work... What can I do? On the Internet, I have found something about using BOOT0 Pin, but I don't know exactly... <Q> I managed to solve that problem. <S> If anybody encounters similar problem, here's what I've done: <S> I used ST-Link v2 and ST-Link Utility. <S> In setting, I set "Connect under reset" and SWD interface (I'm not sure about frequency). <S> Then I press the reset button on my board and clicked "Target" -> "Erase chip" and just after clicking I released the button - It erased the chip so I can now reprogram my MCU. <S> Anyway, if you need to use SWD pins as output, then add some delay at the beginning of the program or use some jumper to disable/enable setting these pins as outputs. <A> For connect under reset to work the ST-Link must have control over the reset pin, if you tie it to ground the ST-Link has no chance to get the target running and gain access to it. <S> If you pull the BOOT0 pin high during power up, the MCU will start into the internal bootloader and you can gain access using several serial protocols (see the reference manual for more details). <S> Inside the bootloader the SWD pins should be available to gain access, but I'm not 100% sure on this. <S> The ST Flash Loader Demonstrator is a tool which allows you to erase / program the micro using the UART interface. <S> If you can't access any of the UARTs of your micro, this solution won't work for you. <A> if you're using stmcubemx, u need to configure the serial wire on stmcube pinout tab. <S> on pinout tab, click SYS and change debug option to serial wire. <S> it fix my problem, and maybe your problem too. <A> I downloaded some code to my own STM32F427 board. <S> Then I can not connect to my board using ST-LINK Utility anymore. <S> I think my code messup the debug port pin configurations <S> (? can not confirm). <S> What I did is the following to make the connection and reprogram my board: <S> Open the ST-LINK Utility and get ready to "Connect" in the Target menu. <S> Power your board(in <S> my case, I use a USB cable) and AT THE SAME TIME click the "Connect" from the ST-LINK Utility. <S> I restored 2 boards with this trick. <S> Hope this helps.--Bob <A> Like dili said: if you're using stmcubemx, u need to configure the serial wire on stmcube pinout tab. <S> on pinout tab, click SYS and change debug option to serial wire. <S> it fix my problem, and maybe your problem too. <S> STM32CubeMx doesn't configure the debug port by default, consequently ST-Link will stop working once you flash your code. <S> You have to erase the chip with ST-link Utility for example. <S> Then go to Tarjet menu and Erase chip . <A> To re-program the MCU, hold the reset button and choose connect to device in ST-Link Utility or press download in your IDE (for example Keil) and then release reset button.
| To connect with the MCU I have had to pull the BOOT0 pin high during power up to activate the bootloader.
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Sensing power supply - sensing after the on/off switch I have a sensing power supply powering a custom build LED light for video. There is an on/off switch after the sensing lines and everything works well. The setup looks like this: However I'd like to sense the voltage drop after the switch. As what I have read in the PSU manuals, this configuration will cause current to flow throught the sensing lines when the switch is open and damage the PSU's resistors. My question is, can the configuration work with a diode placed on the +SENS line to stop current when the switch is open? Like this: <Q> Why don't you put an on/off switch on the AC side of the power supply if you are worried about the sense lines being damaged. <A> Unless the power supply is specified to function correctly without the sense lines connected, this is a bad idea. <S> Even if your idea would work, you will cause the PS to raise its output voltage by the voltage drop through the diode. <S> This is because the diode will drop the voltage on the sense return which the supply will see as voltage drop to the target. <S> As @Andyaka suggested, the proper place to put a switch is on the AC input. <A> In short, a PSU sense line cannot include a diode in series; nor any other active/passive component, for what it matters. <S> Adding a diode as in the schematic you have depicted will prevent the main current blowing up the internal Rsense of the PSU, however, this will also make also impossible for the PSU to correctly measure the remote voltage at the load. <S> Sense cables can have very small cross-section area (diameter/gauge), as the sensing current will be very small and the voltage drop negligible. <S> This is always assuming that the PSU is being operated correctly . <S> Most remote sensing PSUs does not allow the main cables to be disconnected separately from the sensing cables. <S> This is precisely the type of PSUs the article from EDN refers to, http://www.edn.com/electronics-blogs/power-supply-notes/4418253/Power-supply--Remote-Sense--mistakes---remedies <S> In any case, delving a little bit more into your root needs, I cannot understand why you want "to sense the voltage drop after the switch". <S> Why would you like to break the power circuit but not the sensing lines? <S> That defeats the purpose of the sensing circuit and it's explicitly forbidden in many designs of sensing capable PSUs.
| The PS may, with the sense line disconnected, try to raise the output voltage Beyond its limit and shutdown or, worse, damage itself.
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Multiple brushes on a commutator The motor I saw has 14 brushes spaced around a commutator that is about two feet in diameter, and is used to move a large antenna. What is the advantage of having so many brushes and is the commutator wired any differently then normal. <Q> Multiple brushes are also needed to handle the current in the armature circuit. <S> It isn't just motors either, here on some high power slip rings, the slip ring is 18" diameter of brass, and each one has 168 2" carbon/metal composite brushes on them. <S> Neither carbon nor carbon/metal composites conduct anywhere nearly as well as copper. <A> Having more brushes and more solenoids in stator and/or rotor of a motor makes the torque more constant. <S> This is mostly useful if the torque needed to startup a machinery is rather high and low vibrations are desired. <S> If the commutator has only two contacts its wiring is not as special. <S> But the stator will have 14 or a multiple of 14 solenoids. <S> Those solenoids are connected in series with a brush connected to each node between two of them. <A> There are two types of windings. <S> Lap and wave. <S> Wave winding has only two parallel paths. <S> But lap winding has parallel paths equal to number of poles and hence it needs as many brushes. <S> The advantage of having more parallel paths is that you can handle more current. <S> And since torque produce is proportional to current hence more torque is produced.
| High power motors need multiple brushes just to handle the current and keep heating down to a reasonable level.
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How do I simplify microcontroller-based prototype into a sensible production-like circuit? First some context: I'm an IT person by trade, and I dabble in electronics as a hobby. I have rather limited experience - apart from some toy circuits I've only made a single "production" quality installation, a lighting system for a display case . On the toy front, I've done some simple Arduino and ARM-driven circuits. In general, I'm not that eager to resort to coding, as this is something I'm already familiar with, and I'd like to learn something new :) Now for the question itself: I'd like to do more practical circuits, similar to the light setup I've mentioned above. Solving practical problems is a great source of motivation for me. Problem is, while I can sort of work out how to implement what I'd like to achieve, I have tremendous problems with picking sensible physical components to put in the final assembly. For example, I have an RF-controlled (433MHz) socket. I got a simple circuit working, with a pushbutton triggering an Arduino to send the right command via RF transmitter to toggle the socket on and off. I'd like to make a permanent "production" version of this circuit now, one that doesn't include a whole big Arduino shield. Heck, probably the AVR itself is an overkill for "press button, pipe 24 bits over RF" solution. Another example: a magnet-triggered LEDs like these . Getting a prototype that uses a reed switch to go from "wave a magnet" to "toggle LED" is simple. Trying to achieve the same effect in a tiny surface of plastic model part sounds next to impossible for me. In short: drawing circuits is all fun and games, but how do I get from "that's a working prototype that has way too many cables and PCBs" to "that's a sensible implementation that's not an overkill and fits within the space allotted"? And without practical experience in this matter, it's hard to get that practical experience - I literally have no idea what kind of simple building blocks I have at my disposal, and what constitutes an overkill. On the other hand, perhaps going with a micro-based solution is cheaper/simple than trying to work out something without a microcontroller just for the sake of it? I guess this question can be condensed to: What do I read on to avoid slapping a microcontroller everywhere? :D <Q> Don't be afraid to use a MCU; just use the smallest MCU in the same family that you can get away with. <S> If you have it working with an Arduino then look at using a ATtiny25/45/85 or a ATtiny2313A/4313 . <S> If you have it working with AVR C then look at a ATtiny13 or a ATtiny4/5/9/10. <S> Don't hate on MCUs because they're "big and expensive", because in reality they're not. <A> You can absolutely attempt to move your design off the prototype platform onto dedicated logic or an analog implementation. <S> This will require redoing much of the work you have already done in a way that you are more unsure about. <S> An MCU does add some small cost (not much) and some headache dealing with programming and testing, but it does gives you a lot of flexibility. <S> Generally though, even small projects can justify the ever shrinking cost of the MCU (and supporting components). <S> You should start by deciding on what from your shields and arduinos that you actually need, combine their schematics and have a decent stab at laying out the PCB yourself. <S> Have it fabricated, test, rinse, repeat. <S> Many arduino shields and the arduino itself have freely available schematics and PCB layouts <S> so there is a lot of example material to go on. <S> This is a decent challenge in itself as well as the trying to figure out how to implement any design changes you may want. <A> This eliminates redundant components and all of the module-to-module wiring. <S> It also gives you the opportunity to shape the PCB to fit the packaging you have in mind for the final product. <S> Such a board, although it implements the exact same functionality in exactly the same way as the prototype, will be both more aesthetically pleasing and less costly to produce in any sort of volume.
| The direct way to repackage an existing prototype that's built from various subassemblies such as component evaluation boards and breadboards is to take the schematics of each subassembly, delete any circuitry not actually being used in your application, and combine them to produce a schematic for a single PCB that contains only what you need.
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Ways to estimate the rated current of an induction motor without nameplate I am looking at ways to find the rated current of the motor without the nameplate data. Methods for rated voltage has been found. Assuming that we have the rated voltage, what would be some ways or directions to find the rated current? I have looked into slip but different HP motors have different slip. Unless the motor HP is known, this method cannot be used. Are there any other methods? <Q> It's useful to measure the winding resistance. <S> If you use other methods, this method can be used to cross-check the results you get from other methods. <S> The voltage across a winding divided by that resistance gives you the stall/starting current in that winding. <S> Adjust for star/delta configuration as appropriate. <S> I(stall) = <S> V / R. <S> If you have only 2 windings of different resistance, you have a single phase motor : take the lower resistance winding as the "run" winding and use that figure; the other winding is just the start winding. <S> Now the stall current will typically be between 5x to 10x <S> the rated current, corresponding very roughly to motor efficiencies of 80% (usually small motors) to 90% (probably over 1 horsepower). <S> This is only a guide : if your motor is pulling a train, it may be designed for slightly higher efficiency. <S> I(rated) = <S> I(stall)/10 <S> up to I(stall)/5. <S> The same motor may be rated for both 80% efficiency (stall current/5) for (say) 20% duty cycle or 15 minutes, and 90% efficiency (stall current/10) for continuous duty. <S> These are approximate electrical efficiencies; magnetization losses, bearings, friction, windage etc will subtract from them. <S> To estimate those, measure current when running unloaded. <S> So if you're reverse engineering, with no better source of information, start in this range (say at stall current/10) and use additional parameters such as operating temperature to refine these estimates. <S> Note that to start the motor successfully, wiring and breakers must be rated to supply the full start current which may be in the 100A region : this does not imply a 100A breaker, but a "motor rated" (class C or D) breaker for your choice of rated current, which will momentarily supply starting current without tripping. <A> First measure the motor’s shaft diameter. <S> Compare the diameter with published data for similar motors. <S> That should provide a range of likely values for the rated current and torque. <S> Before taking data, get the motor near it’s rated operating temperature by running it for an hour or more at the minimum torque of the range that you estimate based on shaft diameter. <S> When running the motor above the maximum torque in the estimated range, take data as quickly as you can to avoid overheating the motor. <S> If possible, use automated data acquisition to get data from locked rotor through breakdown. <S> If you have access to automatic data acquisition equipment, you could take torque and speed data while accelerating a large inertia. <S> If you can record speed data but not torque data, you could calculate torque from the rate of speed change if you know the total motor plus load inertia. <S> Once you have that data, draw the curve and compare it with published curves or data for motors that might be similar. <S> You can use that method to more closely estimate a range for rated torque. <S> You might be able to narrow the range by running the motor at various value of torque. <S> At each torque value, run the motor until the temperature stabilizes and perform a temperature rise by resistance test. <S> Unfortunately, that still will not provide an exact value for rated torque and current unless you know the temperature rating of the insulation used for the windings. <S> However, higher temperature insulation is usually used to allow motors to operate in higher ambient temperatures rather than operate at heaver loads. <S> I believe that most motors are designed to operate at 80C temperature rise by resistance. <A> Induction motors are built for standard power: 0.25kw, 0.37, 0.5, 0.75, 1.1, 2.2, 3, 4, 5.5,...you can deduct the power rating from the size of the motor, then with known voltage you get also the current. <A> As you may be aware, induction motors are available in both single and three-phase versions and in the case of the former, a wide variety of types (e.g. split-phase; repulsion start; capacitor start; capacitor start & run; shaded pole, etc.) <S> Three-phase motors tend to be of the 'cage-rotor' design and nowadays are physically smaller than much older motors of the same power rating, however comparison of frame size with a known power of contemporary motor will give a good indication. <S> Assuming your question refers to a 3-phase motor, the connection of the windings (star or delta) also affects the current drawn. <S> Are you able to measure the phase currents, preferably under load (assuming the motor is already installed)? <S> The phase currents should be within 5% of each other for a healthy motor, ignoring control circuit consumption. <S> Typical values of phase currents for star and delta connected motors for 380-440V and 220-240V may be found on the Internet. <S> Alternatively, if you are able to provide some of the information referred to above, I may be able to give specific values. <S> Do you intend to run the motor from a conventional starter or an inverter drive?
| Run a dynamometer test to get data to draw curves for torque vs. speed and current.
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How to generate a short, slightly stepped up pulse? Given a 3v power source (2xAA batteries) and a microcontroller, what's the simplest cheapest way to generate a 3.6v 30ms 500mA pulse (to control a solenoid)? <Q> Update based on comments <S> So you have a solenoid with 9 Ohm resistance and need to push 500 mA through it. <S> At this peak moment you'll have 4.5 Volts (9 * 0.5) on this solenoid, so 3.6 do not look enough (if you are sure it is enough, then probably the required current is less). <S> I would take some step-up convertor chip (based on what I can get ready from the stores around - in my case it is MC33063 <S> which is available in convenient DIP or SOIC packages for far less than $1. <S> Then I google for "MC33063 datasheet pdf" and browse this document to find out the typical schematics. <S> I see I'll need few more components - resistors, capacitors, inductance. <S> They say the chip is able to provide up to 1.5 A <S> so the rest you need is to choose proper values according to their instruction, for, say, 5V output voltage. <S> Your batteries will be serving almost 3A at this moment <S> so it is important they are fresh. <S> I suspect here <S> is some misunderstanding there :) <S> You usually can control either voltage or current, not both. <S> I.e. you can apply given voltage to solenoid and see what current will flow <S> (it will not rise immediately). <S> The core idea of step-up conversion, uses inductance itself. <S> Imagine that your controller apply the voltage to the coil. <S> Coil starts rising the current and when the current becomes large enough, you switch it to capacitor. <S> Coil could not stop the current instantly so it charges capacitor until the current falls to zero. <S> At this moment capacitor has the maximum charge and maximum voltage. <S> Often diode prevents back flow from capacitor to the coil. <S> Look at this picture: http://paginas.fe.up.pt/~ee07229/images/classicDCDCconverter400px.gif <S> Imagine there is a MCU-controlled transistor (perhaps, MOSFET) instead of the key. <S> Now we only need to understand better what type of load <S> / solenoid you use. <S> perhaps it can get more than 500mA current with less than 3.6 Volts. <S> This depends on its resistance... <S> Anyway it may happen that cheapest / simplest way is to use the third battery. <A> It depends on the load in detail. <S> For a solenoid most, most will tolerate a higher voltage briefly without problems. <S> Most, once energised, will continue to hold in with a much lower voltage than was required to get them operated. <S> With those assumptions in mind, let's suggest a circuit that delivers a nominally 6v peak pulse, decaying exponentially with a time constant determined by the size of the capacitors you use, and the current drawn by the load. <S> It should be possible to get 30mS with reasonable value components, though you may find that less than 30mS will work OK with your load, especially with the higher voltage start. <S> Use a voltage doubler, consisting of two capacitors. <S> To deliver 30mS of 500mA, which is a charge of 15mC, would require a 1.5v drop in a 10mF (10000uF) capacitor. <S> At the low voltage you have, that's easily do-able. <S> Put one cap to ground, and use a fixed resistor to charge it to 3v. <S> This is your pedestal cap. <S> This is a permanent connection, make the resistor as large as you can so that it still meets your recharge time, given the duty cycle you want. <S> The second cap is your 'flying cap'. <S> Charge this with 2 resistors, one to ground, one to 3v, again the largest resistors you can for your duty cycle. <S> When you want to pulse, use a transistor (BJT, FET, or a relay even!) <S> to connect the 0v of the flying cap to the +ve of the pedestal cap. <S> The flying cap +ve goes up to 6v, and will deliver 500mA for 30mS while dropping back to 3v at the end of the pulse, when switched into your load. <S> A pedestal cap is to isolate its voltage droop from your +ve supply rail. <S> However, the resulting supply droop may disturb your micro, which is why I've suggested the slightly more complicated two cap approach first. <A> A micro power booster would do the trick. <S> This one can generate 5V from voltages as low as 1.5 volts: - Choose R1 and R2 to give you the desired output voltage of 3V6. <S> There are several devices like this from TI and LT. <S> Quiescent current in non-switching mode is 140 uA <S> and the device can be powered down to 10nA. <S> When not switching (or powered down) <S> the capacitor will discharge slowly so ensure R1 and R2 are as big as possible and that C2 is ceramic with low leakage. <S> D1 should also be chosen to be a low leakage type (probably silicon rather than a schottky diode. <A> simulate this circuit – Schematic created using CircuitLab <S> This is basically user44635's idea, except with the capacitor voltage going negative. <S> I drew it to see how to minimize the number of components. <S> When Q1 is off, C1 is charged through R1 and D1. <S> When Q1 is on, C1 is discharged into L1 through Q1 and Q3. <S> D3 dumps the energy of L1 back to C1 when turning off. <S> It may slow down the solenoid disengaging as the way it is. <S> The components are cheap, but there needs to have one relatively big capacitor and the time to charge it. <S> Although R1 can be easily modified to a transistor for fast charging. <S> Overall, I am not sure this is the way versus a simple booster such as the one suggested by Andy aka.
| If your battery supply is stiff enough, or you use a supply decoupling cap much larger than 10mF, you may get away with standing the flying cap directly onto the 3v supply.
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In production, is there a significant difference between QFN or TQFP components other than space? When producing a run of PCBs, does it matter if I choose QFN or TQFP components? Say, I want to have 1000 boards manufactured with an Atmel AVR that is available in QFN or TQFP packages. The QFN package is of course smaller, but I have enough space on the board, so that is not an issue. For prototypes, the TQFP is easier to handle because I could even hand-solder them, if a reflow oven and stencil is not available. But for automated production, this should't matter, or does it? For instance: would a PCB manufacturer recommend to use QFN because they are easier to handle by the pick-and-place machines, resulting in a price cut? The difference in component price appears to be small, too. <Q> I recommend QFN. <S> why? <S> smaller, cheaper in mass production. <S> it requires less solder ( because of small pins ) and therefore monetary efficient. <S> it is soldered faster than TQFP ( using an soldering machine like that pick and place that uses solder paste and hot air) because QFT packages stick to pad because of their GND tab. <S> The board will look niftier and neater :) and also more professional. <A> The BlueSky's answer isn't completely correct. <S> Price : There isn't big differnce or maybe it's better to say there is no differenc. <S> e.g. check out ATMEGA8A-MU and ATMEGA8A-AU . <S> Size : <S> in many cases QFN is half of TQFP but this factor doesn't decrease the price of manufacturing so much. <S> although it makes your PCB a bit compact. <S> Solder and Soldering : IMO solder consumption wouldn't be significant and most probably there wouldn't be any difference in soldering QFN and TQFP. <S> because I'm sure they just put components on PCB and soldering paste on pads. <S> NOW I'm sure you are asking yourself " <S> then What's the difference between these two packages? <S> well the big differences are in electrical characteristics. <S> for example QFN has smaller leg's length than TQFP and this means low capacitance and in some cases it's so important. <S> many RF ICs use these kind of packages. <S> or QFN packages include an exposed thermal pad to improve heat transfer out of the IC (into the PCB). <A> A QFP provides room under the chip to put vias, possibly making fan-out easier. <A> For anyone reading this down the line, there is one other major issue not yet mentioned regarding QFN vs QFP when it comes to commercial production of boards - and that is reliability. <S> If you have a product that deals with lots of thermal cycling or other vibrational/mechanical stress, leaded parts are the way to go. <S> Leadless parts look neater, but don't always take up less space since there's usually a big thermal pad that you can't route signals through. <S> If you're going to produce a thousand boards that will see a rough life (outdoors, automotive, etc), leaded parts generally last about 10x longer than leadless parts before you start to see issues with cracking solder joints, etc. <S> I work as a EE in oil and gas, formerly aerospace, and I swear against leadless parts in these applications for this reason. <S> Of course, you can always take additional measures to help even the playing field (primarily potting your leadless ICs), but this doesn't guarantee that you're giving your parts the same longevity as a leaded part. <S> Additionally, this will typically make your leadless implementations more expensive due to the additional manufacturing steps required.
| Again There is no big differnce indeed. Unless you need the heatsinking effect of the large middle ground pad, you may find the QFP easier to lay out the PCB for.
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LT3845 not giving 1.3V I am using the LT3845 to obtain 1.3V of output. From the formula given for calculating the resistors to obtain 1.3V, I conclude that R5(the one tapping into the Vout) should be approx5.69Ohms and the other resistor be 100KOhms. The test schematic simulation in LTSPICE is as shown below - . But, the Vout is almost 0V. The waveform is as shown below - Why so? <Q> In the datasheet, the minimum start voltage (see page 3, second parameter) is listed as 7.5V max. <S> There is also this note: <S> Note 4: <S> VIN voltages below the start-up threshold (7.5V) are only supported when the VCC is externally driven above 6.5V. <S> As you are not satisfying the startup voltage requirement with your 6V source, I do not expect your circuit to properly simulate (or operate, should you build it this way). <S> Either supply a minimum Vin of 7.5V, or externally drive Vcc (pin 13) with at least 6.5V. <S> You could always use a different converter , of course. <A> You ASSUME that it is the feedback network that is giving you an unexpected result. <S> But is that the problem ? <S> I think not. <S> Simulating a buck converter is not so simple. <S> Have you checked the basic operation of the switching converter ? <S> Do the outputs of the IC do anything ? <S> The model of the LT3845, does it even support a full switching mode transient simulation ? <S> Have a look inside the modelfile and see what is there. <S> You're trying to simulate 200ms, for a DCDC converter switching at 100 kHz that is a lot of switching. <S> That simulation could take hours. <A> I am using the LT3845 to obtain 1.3V of output. <S> From the formula given for calculating the resistors to obtain 1.3V, I conclude that R5(the one tapping into the Vout) should be <S> approx5.69Ohms and the other resistor be 100KOhms. <S> Incorrect. <S> Bizarrely you have R6 as 100k and R5 as 5.69 ohms - this is plain wrong. <A> Well, it all turned out to be a silly issue. <S> Now, when the Vin is > 20V all was fine and dandy. <S> The moment I kept reducing it (to a lowest of about 10V or so), it started to malfunction. <S> Finally found out that it was the SHDN pin that was playing spoil sport. <S> The buck was getting shut down as the Vin was reduced and there by the whole thing did not work. <S> The moment I ensured decent values of the voltage divider to make sure the SHDN was in the 5V range all worked fine.
| The Vfb voltage reference is 1.231 volts therefore, if R6 is a 1kohm resistor, then the current thru that resistor is 1.231 mA. That leaves 1.3 volts minus 1.231 volts across R5 (69 mV) and, at a current of 1.231 mA, the resistance should be 56 ohms.
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Splitting mains power into multiple relay-switched circuits Summary Introduction My goal is to split a 1-gang mains power extension lead into three, have my Raspberry Pi plugged into one of the outputs, and have the Pi able to switch the other two outputs using relays. Background I want to use my Raspberry Pi to switch on and off two mains-powered appliances using solid-state relays. I've found this useful video which demonstrates the concept on a mains-powered lamp. The video suggests cutting the lamp's power cord and wiring that up to the relay, but I don't want to cut the cords for my appliances. I am, though, happy to use an extension lead and cut that instead. However, I don't want to have to buy and cut an extension lead for each appliance; nor do I want to have to plug three things (the two appliances and the Pi) into my wall socket individually, for neatness' sake. So I thought maybe I can make a 3-gang extension cord and have the Pi run off one output and attach the relays to the other two. The Plan My idea is to get myself a nice long 1-gang extension chord and two spare extension chord sockets , so I'd have something like this (please excuse my crude illustrations!): I would then cut the extension cord say 3m from the plug end and cut two more times to give myself two pieces of headless chord, like so: Now I can wire those loose bits of cord into the spare sockets. Furthermore, I could cut those cords again to insert the relays, giving me this: The final piece of the plan would be to join all those loose ends in a junction box. The Question Sorry it's taken so long to get here, but here is the question (well, two actually): Is this plan feasible? And if so, what's the best (i.e., safest) way of connecting those loose ends? I thought of using a junction box and I've Googled endlessly but can't seem to find anything that takes one input and allows multiple outputs. The closest thing I can see is this but I think the cord might be too thick for it. Also, I'm a little concerned that shoving three wires under one screw in the terminal block of a junction box might be insecure and lead to them coming loose. Any advice and suggestions are most welcome! <Q> Option the first would be skip this if you are truly unfamiliar with mains power - it's not something you should attempt without having a very good idea what you are doing and how to do it right, especially if you are in a 220-240V mains region. <S> Use a box; One large enough to mount a normal wall-outlet type duplex receptacle in, and all your connections, SSRs and the RasPi and its power supply inside. <S> Use either a terminal strip (crimp-on spade connectors optional) or wire nuts to make the connections inside the box. <S> If the box is plastic, done. <S> If the box is metal, be sure to connect it to the grounding wire. <A> You'd need to find one that's openable. <S> Many are using a threaded 'nail' fastener now and they are not repairable. <S> Some of the MK brand might be worth a look. <S> These devices usually use metal strips to form a continuous connector and socket terminals. <S> If you can cut the live strip between socket 1 and 2, 2 and 3, etc., and wire each back to the first via your SSRs you should have a neat solution. <S> First socket is always powered, the remainder are switched. <S> Please keep adequate clearance between the power conductors and your signalling circuits. <S> Ensure that the SSRs are fully opto-isolated and adequately rated. <S> Figure out how much heat the SSRs are going to dissipate and calculate heatsink size bearing in mind that they won't have any ventilation. <S> If you don't internally fuse the individual SSR sockets then the 13A plug fuse should be replaced with one that will protect the weakest SSR. <S> e.g., If one of the circuits can only handle 3 A safely then a 3 A fuse should be fitted in the 13 A plug. <S> The feasibility of this all depends on finding a socket strip with enough space to take your electronics safely. <A> There are multi-way connectors for connecting several wires together. <S> If the wires aren't too thick, you should be able to get two into each end of a suitable terminal block strip (perhaps a 15A one). <S> Alternatively, Wago do a range of connectors for joining anything up to 5 wires at a time. <S> Make sure the whole lot is in a suitable insulated box. <S> I don't know if the solid state relays you're using provide adequate isolation between the mains and the drive circuit. <S> If not, act as if anything on the Raspberry Pi could be live when powered up.
| I suggest you find a decent quality plugboard / socket strip and mount everything inside that.
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How bad is it to places vias under a QFN thermal pad? I'm working on a PCB which has to be very small. There is a QFN IC which has a big thermal pad, although it doesn't really need to dissipate that much heat. So to save some space, I came up with the idea of reducing the footprint of the thermal pad so that I can place some vias and traces in there. It would be covered by solder resist so it doesn't sound unreasonable to believe that this might be okay. However, if it were such a good idea, everyone would be doing it, and I'm wondering how bad it is. I've seen this similar question but it's not exactly the same thing. For example: The traces are 0.2 mm (7.87 mil) wide, the via diameter is 0.7 mm and the drill is 0.3 mm. <Q> I would not go with this. <S> Small designs will not necessarily heat up that much, but running an exposes pad so close to your traces is risky. <S> Depending on volume, you will get some boards where part of that trace is exposed, or the via mask is scraped off. <S> Overall it is just not good design practice. <S> Use another layer. <A> It might work, but if you put soldermask over the traces under the pad, that will add extra height that will hold the chip above the pad. <S> You will definitely not have complete thermal contact, which may or may not be important to you. <S> I think this will probably generate a bewildered call from your assembly house. <A> If you want to comply with the ipc standard (standard for pcb design, production etc.). <S> You are not allowed to use the solder mask as an isolater. <S> Also experiences from the past lead me to choosing other options as the yield will drop and also dfm checks from production house are discarding this as the production pass yield will drop!
| The wear and tear on the board depending on use case might expose something, or maybe that part does heat up a slight amount and over time causes unexpected behavior.
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Higher current and power to an LED? I'm using an LED that is rated at (0.06W) current is 20mA and voltage is at 3V. I did some wrong calculations, and the resistors I'm using are not enough to limit the current to the diodes. simulate this circuit – Schematic created using CircuitLab The current is 0.024A now, can the LED work without blowing up? I'm maxing it with higher power would it manage? <Q> Unless you are grossly overloading the device, it will not blow immediately. <S> However, as with all devices too much heat will kill it eventually. <S> If you can add a bit of heatsinking (usually not possible with common LEDs) then it will be fine, otherwise derate the MTBF of the device accordingly. <A> If the LED current is 24 mA, and the recommended maximum current is 20 mA, you can probably expect a somewhat reduced life for the LED, but it is unlikely to be destroyed instantly. <A> If the values you placed are correct \$ Vled=3V, R1=200\Omega, V1=4.8V \$ <S> your actual current is: \$ Iled=\frac{V1-Vled}{R1}=9mA \$ <S> Therefore, no overload. <S> Check your values and/or your circuit. <A> It seems your error is serendipetous and you're actually pushing less current through the LED than it's rated for. <S> :) <S> If you don't exceed the LED's recommended forward current, it'll last for a long, long time. <S> To determine the value of the ballast resistor, you can do this: $$ R1 = <S> \frac {V1-VfLED}{IfLED} = <S> \frac {4.8V - 3V}{0.02A} <S> = 90 \text{ ohms <S> } $$
| If you keep the junction within reasonable temperatures the LED will not die immediately but you will accelerate the aging process.
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ENC28j60 ethernet interfacing with PIC18F4520 help needed I am a newbie to ethernet controller. I need to use ethernet controller to send sensor data to our server using http get method. I went through ENC28j60 documentation and microchip's TCP/IP stack as well. But I couldn't find a proper example similar to my project. briefing my application below. PIC18F4520 will read some sensor data and send it to server every 3 minutes.for example sensor data are voltage,current,temperature etc. It should push these data using http GET method to an URL likeabc.com/index.php/add/add_data?volt=dc_volt¤t=dc_cur&temp=temp. Where dc_volt,dc_cur and temp are the measured values in microcontroller. How can I do this using ethernet interfacing? Atleast what are the things I should learn to do this? Can I do this with the Microchip TCP /IP stack? any help is appreciated. Thanks <Q> Yes this is possible with TCP/IP stack . <S> I think you should not directly jump into reading some sensor data and send it to server every 3 minutes, instead try to build the demo app provided by Microchip and access their webpage (or your own page). <S> Microchip provide TCP/IP library which is included in Microchip Application Libraries v2013-06-15(MAL) . <S> Download and install MAL on your system. <S> Build the the TCP/IP demo app and upload it on your MCU. <S> Connect the LAN cable between ENC28J60 and your PC. <S> If everything went ok, then you can open the TCP/IP Discoverer Tool which is found in C:\microchip_solutions_v2013-06-15\Microchip\TCPIP Stack\Utilities , to get the IP address. <S> Enter the IP address in your browser and you can access their default webpage. <S> Now next thing you have to do is build your own webpage and put all web files inside webpages folder. <S> So next time when you enter the IP address in browser, it will open your webpage. <S> In the code you will have to put some dynamic variables to get and post data to your webpage. <S> I cannot explain everything here. <S> But you can read below tutorials which include everything you want. <S> Part 1 <S> : Web based monitoring Part 2 <S> : Web based control Part 3: Advanced web based control <S> You can watch these video tutorials for step by step explanation <S> tutorial1 <S> tutorial2 <S> tutorial3 <A> HTTP is not a "push" protocol. <S> Normally, your Windows/Linux/OSx client would use a GET method to GET the data from the PIC18F4520. <S> The Microchip TCP/IP stack will respond to GET requests. <S> The Microchip TCP/IP stack does not include a web browser like IE or FireFox or Safari, that send GET requests. <S> There is nothing in the stack to send GET requests. <S> You can modify the stack to send anything at any time, and you can write a Windows/Linux/OSx client to listen for what your sensors are sending, but that requires more work than simply using the Microchip TCP/IP stack. <S> The Microchip stack can understand GET requests, and there are examples provided that showing it replying with different types of data. <S> you just need to connect up your sensor to a variable, so that the stack has a value to send in the reply. <A> Considering that you are a "newbie", I think you should keep it simple. <S> The ENC28J60 is a complex and low-level Ethernet controller. <S> If Microchip provides a driver for it, that's great, but I would not want to write one! <S> Consider using something like a WIZnet module that would allow you to send TCP packets directly. <S> There is a lot more done for you there, and writing an HTTP client is relatively simple. <S> Even consider using a Raspberry Pi for something like this. <S> That thing really is amazing. <S> Unless you're just trying to learn something about the Microchip parts, don't feel bad about using these amazing and cheap components that are available.
| A simple method would be to use something like CURL to send a GET request to the Microchip part. You can just write shell scripts to read the data and send it out to a web server!
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How to stretch ON time of a pulse train when keeping the periods unchanged? I'm receiving varying frequency pulse train with very short ON time and I want to convert these to pulses with longer ON time but without changing the pulse periods. On time is now around 100us. Pulse period varies between 1000us and 2500us. The flipflop halves the frequency and skips one rising edge which I dont want in this case. How can I achieve this Duty cycle doesn't have to be fixed just the frequency of the pulses should remain the same. <Q> You want something called a one shot . <S> Basically, you ignore everything except rising edges of your input signal. <S> You copy the rising edge to the output, but make up your own falling edge. <S> A one shot is a timing component that does exactly that. <S> Since your pulses vary from 1 to 2.5 ms, 500 µs is the optimal time for guaranteeing the longest minimum level time. <S> Without prediction or delay, you can't make a square wave output. <A> I believe what you describe is a monostable circuit. <S> You can check the monostable multivibrator for reference. <S> What it does is starting a fixed duration pulse at every rising (falling) edge of the input. <S> You can tune it to your liking, as long as the 'on' time does not exceed the period of the signal (consider also the transition time). <S> You are basically creating a pwm this way. <A> I would try using a CD4046B and use the type II phase detector. <S> Both phase detectors use the leading edge of the input, but as your signal spans more than an octave, the type I detector may lock at the wrong frequency. <S> In addition, a type II phase detector is insensitive to input duty cycle, provided the input pulse width is long enough. <S> This would, admittedly, be a bit more difficult to implement than Olin's suggestion (which has the advantage of simplicity), but there is an application note available. <S> Much depends on the rate of change of frequency of the signal, though. <S> If that works, you will have a 50% duty cycle. <A> Try passing the signal to a NOT gate. <S> The period of the pulse train will remain the same and you will end up with larger ON time. <A> One way to do this is using: An edge-triggered flip-flop with asynchronous reset A delay circuit (could just be an RC filter, possibly followed by a schmitt trigger). <S> configured as follows: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The delay time will define the output high pulse width, so it needs to be selected to be narrow enough to not interfere with the next period, but wide enough to be detected. <S> There are probably already ICs that contain this circuitry, but I'm more familiar with IC design, so this solution is the first one that came to my mind. <A> This solution uses a microcontroller, although it also could be done with hardwired logic as well (probably a good application for an FPGA). <S> Set up an interrupt <S> so it will trigger on the rising edge of the input signal. <S> The uncertainty of the timing will then be half of one instruction cycle, typically in the tens of ns or less. <S> An input capture module could also be used. <S> It is assumed that this interrupt, and the three timer interrupts discussed below, are the only ones in the system so the latency is fixed in each case. <S> Call this first interrupt time A. <S> In the interrupt routine, capture the time from a free-running clock, and set a timer to interrupt at A + 2500 µs. <S> Call that A'. <S> The 2500 µs figure is chosen to be at least as long the maximum distance between pulses. <S> The leading edge of the next 100 µs pulse B is then acquired in the same fashion. <S> (This also starts the cycle for the next pulse, overlapping the first, so two overlapping timers are required, which are used alternately.) <S> A second timer is set to interrupt at A' + <S> (B - A) / 2, i.e. half the previous period. <S> Call that a'. <S> When the interrupt for timer A' occurs, set the output high. <S> When the interrupt for timer a' occurs, set the output low. <S> Since both of these are being handled by interrupts, the overhead for both will be the same and the relative time between the two equal to exactly a' - A', half the time between B - A, or a 50% duty cycle.
| You can set one up so that when a rising edge comes along, it starts a timed pulse.
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AC SSR does not switch off - why? I have a circuit meant to manage a shutter (or a shutter motor to be more precise). The motor has two power lines - depending on which power line one connects power, it moves up or down. I designed the following circuit to operate the shutter: RY1 and RY2 are operated via a uC / Mosfets. I now have the problem that the SSR (RY2) does not turn off even though there is no more current flowing through its control pins (voltage ~20mV, current <1uA). The SSR only turns off when the shutter reaches a final position (fully open or fully closed) - then the motor stops automatically, hence draws no more current on the 230VAC line, and the SSR then stops conducting. I've already applied a varistor to protect the SSR (R2) because the motor produces some nasty voltage spikes when it reaches an end position. I also added a snubber circuit (R1 / C1) and tried different values (100/100n, 22/100n, 47/47n) to reduce the dV/dt rise across the SSR (as described here ), but so far I haven't been successful. The motor draws roughly 110W. Is there an "easy fix" for this circuit or should I replace the SSR with a mechanical relay? Thanks! <Q> You might find that a AQH3223A works better. <S> The AQH3213A is a "zero-crossing" type SSR, while the AQH3223A is a "random" version. <S> Zero crossing types can have issues with motor loads. <S> From https://www.panasonic-electric-works.com/pew/eu/downloads/ds_x61_en_ssr_technical_information.pdf 4) <S> When controlling loads using zero- cross voltage types in which the voltage and current phases differ, since the triac sometimes does not turn on regardless of the input state, please conduct sufficient tests using actual equipment. <A> I wonder is the snubber in the wrong place? <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Try snubbing the triac. <S> And have a look at Choosing TRIAC snubber resistor for multi-purpose switching . <A> Assuming that you're using the higher-current version, I'd suggest that you look into the motor characteristics. <S> If the motor current and voltage have significant phase shift, and you're using a zero-voltage cutoff unit, it's entirely possible that the SSR never sees zero voltage and current simultaneously, so it never turns off.
| As you have mentioned, dv/dt can be a problem with triacs and it may be causing a false triggering.
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Simulate Gyroscope through MEMS accelerometer Is possible to simulate the Gyroscope using Accelerometer values? I would like to make that in Android Context so using MEMS smartphone accelerometer. <Q> No. <S> An accelerometer measures linear acceleration. <S> A gyroscope measures angular velocity. <S> You can't assume one from the other in any general sense. <S> Your smart phone probably has both sensors though. <A> One accelerometer will not work. <S> There are six degrees of freedom in mechanical motion <S> so you need three accelerometers and three gyros to capture them all. <A> Actually you can, if you have 3-axis accelerometers. <S> You will not be calculating the angular velocity, yet you can get your position in the space and with the aid of proper reference and post-processing your measurements, your device can simulate the functionality of the gyroscope. <S> However, your "gyroscope" will have the worst bias-drift in history, since double-integrating the accelerometers' output signals to get the position in space will take a lot of time on which an error will be accumulating. <S> I believer that's why inertial measurement units (IMUs) are typically 6-axis <S> (3-acc + 3-gyro).
| You can simulate a gyro with two accelerometers per axis.
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Why changing the potentiometer affects the whole circuit? simulate this circuit – Schematic created using CircuitLab I have the following simple circuit where the potentiometer is used as a rheostat. 9V battery (-) --- 0.470K Resistor --- LED --- 1K Potentiometer --- LED --- (+) When I turn the shaft of the potentiometer both LEDs change their brightness. My problem arise from my own understanding that electric charge flows from (-) to (+) so whatever happens after the first LED should not affect it (should not matter). Where is the problem in my mental picture? <Q> Current flows through the whole circuit and each element in your circuit <S> gets an equal amount of current, meaning that identical LEDs in series will have an identical brightness, disregarding minor manufacturing process variations. <S> Now, the current through your complete circuit is determined (in this case) by the total resistance and the voltage of the battery, so when you change your rheostat, the total series resistance changes, and the total current changes. <A> Electric current flows in a circuit a loop if you like. <S> It moves because one or more elements of the circuit, such as batteries or generators create an 'electro motive force' (EMF) - measured in Volts that drives a current round the loop. <S> The same current will pass through each element in the circuit, and the amount of current will depend on the total resistance of the elements in the loop. <S> It is true that electrons in the circuit conductors drift from the negative terminal towards the positive terminal of the driving EMF <S> but this happens quite slowly, on the other hand the effect of the EMF creates a field that travels through the conductors at close to the speed of light. <S> There is no real reason why, but conventionally we say that current is flowing from the positive terminal to the negative terminal. <A> I think this circuit might exhibit a response more like you expected. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> As others have commented D1 has to pass all the current in the circuit. <S> In the modified circuit the voltage across D2 + R1 varies with the potentiometer and so the current through D2 does as well. <S> As the pot is turned down D2 will dim to the point where it turns off. <S> The circuit still won't be exactly what you expected because D1's current still depends on the pot setting as the pot wiper is moved up <S> D2 will glow brighter but so will D1 because all the circuit current passes through it. <S> Try it and see. <S> It will be instructive to connect your voltmeter between the wiper (+) and the bottom of the pot (-). <S> If your meter has a mA range then try connecting that in series above D1, below D1 (should be the same), at the bottom of the pot and in the D2 - R1 branch. <S> You can also learn some of the characteristics of an LED if you connect the meter across D2. <S> You should see the voltage rise a good bit before you see any light. <S> You will notice that the voltage maxes out at some point (about 1.8 V for a red LED). <S> All this should teach you that the LED's voltage and current relationship is like that of a diode and not like a resistor. <S> Have fun!
| The brightness of a LED is controlled by the current flowing through it.
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In microcontroller (AVR), the interrupt vector take address $0000 to $003A so the program is start after that address? I'm studying microcontroller using Atmel 8-bit microcontroller datasheet, According to the datasheet, there's 3 types of internal memory in microcontroller Flash memory = program memory SRAM: 32 general purpose register as well as I/O register is located here EEPROM So, when program counter (PC) fetchs instruction, it will fetch from flash memory where the program (instructions) is stored, starting from $0000 but then when I read interrupt chapter (ATmega 2560) , it says that interrupt vectors is stored at $0000 to $003A in program memory then does it mean when the microcontroller starts, PC will go through the interrupt vector list first and then instructions later? Moreover, the interrupt is enabled by instruction which come after $003A so when the PC start from $0000 to $003A, microcontroller do nothing because at that time, intteruption haven't enabled yet? I'm currently use two datasheet, Atmel ATmega 2560 and ATmega 16A <Q> The first vector in the interrupt vector table (located at 0x0000) is the "Reset Vector". <S> This is the first program memory address which is read by the CPU on power up 1 . <S> The location in memory is usually filled with a JMP or RJMP instruction where the jump address is the start of your program. <S> If the reset vector is not correctly programmed (e.g. with an RJMP instruction or whatever), the CPU will simply keep counting along executing instructions as they appear - e.g. executing other interrupt vectors if there. <S> If there are any other interrupt routines programmed into the vector table, these would all be executed in turn regardless of whether it's interrupt source was enabled or disabled. <S> When using something like avr-gcc , it is aware of the interrupt vector table and its structure, and will make sure that the reset vector points to the start of your program. <S> In this case, the start of the program is actually not your main() function, but a load of stuff the compiler adds to initialise variables and whatnot. <S> But after all of the initialisation stuff, your main() function would be called. <S> As an example, this is the disassembled vector table for one of my programs: 0: <S> 0c 94 <S> 72 00 <S> jmp 0xe4 ; <S> 0xe4 <__ctors_end <S> > 4 <S> : <S> 0c 94 8f 00 <S> jmp <S> 0x11e ; <S> 0x11e <__bad_interrupt <S> > 8 <S> : <S> 0c 94 8f 00 <S> jmp <S> 0x11e ; <S> 0x11e <__bad_interrupt> <S> c: <S> 0c 94 8f 00 <S> jmp <S> 0x11e ; <S> 0x11e <__bad_interrupt <S> > <S> 10: <S> 0c <S> 94 4f 03 <S> jmp <S> 0x69e ; 0x69e <__vector_4 <S> > <S> 14: <S> 0c 94 8f 00 <S> jmp <S> 0x11e ; <S> 0x11e <__bad_interrupt <S> > 18 <S> : <S> 0c 94 <S> ef 03 <S> jmp 0x7de <S> ; 0x7de <__vector_6 <S> > <S> 1c: <S> 0c 94 8f 00 <S> jmp <S> 0x11e ; <S> 0x11e <__bad_interrupt <S> > ... <S> 6c: <S> 0c 94 8f 00 <S> jmp <S> 0x11e ; <S> 0x11e <__bad_interrupt> <S> Notice how avr-gcc adds a jump at address 0x0000 <S> to what it calls __ctors_end which is basically a memory address after the end of the vector table where it's initialisation stuff starts. <S> All unused vectors jump to something called __bad_interrupt - located at the destination of that jump <S> is basically another jump instruction back to 0x0000 <S> such that any occurrence of an unhandled interrupt resets the processor. <S> 1. <S> Some AVRs have a bootloader space and can be programmed to have the interrupt vector table at a different address than 0x0000, but the same still applies. <A> When an interrupt is triggered during a program <S> execution(after you must have set and programmed it for interrupt of course) there are unique addresses in which your program counter will point to which is specific for each specific interrupts of your AVR microcontroller. <S> These addresses just happens to be designated at the beginning of your flash memory, like you said from 0x0000(reset interrupt) to 0x003A.Now <S> when you program your AVR uC and then start your program with .ORG 0 <S> what you have done is tell your assembler to start placing your program from address 0x0000, downwards.. <S> meanwhile you have actually written your programs on top of those interrupt addresses downwards. <S> Now what do you think will happen, if paraventure during the course of execution of your program you have set an interrupt?. <S> By design, your Program Counter(PC) is to point to the unique address of that particular interrupt that just triggered which is usually among those specified addresses at the beginning of your flash memory. <S> Then on pointing to that address it see an instruction and then starts to execute, that is the logic.so therefore, what programmers do is avoid those address and then just place a function call on those interrupt address if there is need for any interrupt in your program that you have written. <S> I hope this helps <A> If you use interrupts, you will want to at least steer clear of the interrupt vectors you use. <S> Unplanned interrupts, if they occur, will jump into your code and would likely be bad. <S> (!) <S> Best practice is to start your code after the whole block of interrupt vectors. <S> I typically start a project with all int vectors populated with a JMP or RJMP (depending on processor) to a trap for bad interrupts, where I set a breakpoint during development. <S> As I populate the ISRs, they get vectors to the appropriate locations. <S> When done, all unused ints are trapped, and then trigger a fault in my sanity check routine. <S> You could place the entry point anywhere past the int vector block. <S> You might want to place tables between the int vector block and the code, but unless I have a specific reason to do otherwise, I typically allocate tables starting past the end of the code.
| You can place your start point anywhere after the reset vector address.
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MP1584 DC/DC step down converter cause whine in audio, how to avoid? Short story I use the MP1584 to step down 5VDC to 1.5VDC to power an audio equalizer taken from a Panasonic walkman (RQ-P255) from the nineties. However this is working great, sound is great, it introduces a heavy whine and don't know how to avoid this. See also the schematic below of the device I have build. What can I do to remove the whine? Long Story I have build/want to build a Micro Portable Stereo Amplifier With Battery, Equalizer, BlueTooth And Line-In/Line-Out (in a small candy case) so you can connect any speaker you want (nice to use outside in the garden/party) and you can use it as a BlueTooth Walkman. I have added an equalizer taken from an old walkman of the nineties, the Panasonic RQ-P255. This because most bluetooth transmitters don't support an equalizer and, some bluetooth speakers supports only EQ with presets (hate it). Figure out how the equalizer works and it was very straight forward but requires only 1.5VDC input (the whole device, all other parts, operates on 5VDC). Because of positive reviews of the MP1584, I use this little device to step down 5VDC to 1.5VDC. This is working great except the whine it produces, very annoying. The device has an equalizer on/off switch and when I turn it off, the whine is gone. What can I do to remove the whine? The whine i'm talking about This is recorded with a microphone of a Tascam DR07 nearby speaker. The picture below is the waveform 1600% amplified. To hear the whine 800% amplified: https://drive.google.com/file/d/0B2l-eQoHefcVaVd3ZVNZMXJYcXM/view?usp=sharing To hear the whine - whine original: https://drive.google.com/file/d/0B2l-eQoHefcVb2NMS3dINDU3SFU/view?usp=sharing Notice: I don't have an oscilloscope. Schematic of the device Notice: I'm not a hardware designer or schematic professional, so maybe the schematic is not as it should but I think it is very clear what goal I want to achieve. The headphones connection is not in the schematic.EDIT: 200K must be 200 Ohms Setup of the device Here some drawings how the device must be set up What I got so far BlueTooth module and switch and headphone jack not implemented yet. What it should look like when finished ;-) Used parts/devices in current setup: Battery charger / DC converter taken from this product: http://www.mrhandsfree.com/en/shop/7318-portable-power-chargers/product/7320-portable-power-charger-2600-mah Step down converter: http://artofcircuits.com/product/mp1584-buck-step-down-3a-adjustable-regulator-module Equalizer taken from this product from the past: http://img19.staticclassifieds.com/images_slandokz/91858292_1_644x461_kassetnyy-audiopleyer-panasonic-rq-p255-pavlodar.jpg Amplifier: http://www.diodes.com/_files/datasheets/PAM8403.pdf <Q> Most DCDC converters run well above 20KHz at your power levels .You are getting Audible noise <S> and I think that you should put the oscilliscope. <S> You will probably see ripple artifacts on the 5V input and 1.5V output that are in the low KHZ range. <S> When you scope the mosfet drain waveform you will probably see shimmering on the rising or falling edges depending on how you have got your scope triggering. <S> Even when the DC input and load are constant the shimmering will persist. <S> This low frequency crud cuts through the orthodox SMPS filter components like a knife through butter because those components are designed to attenuate the switching frequency. <S> Experimentaly loading the converter may cure this because these low frequency instabilities tend to be worse at light load .If <S> the load experiment nails this <S> then you can increase the buck inducter and you will probably have silence or use a smaller buck converter which will in effect be more heavily loaded. <S> If things get too bad consider linear post reg using a 1.5V LDO and setting the buck up at say 2.5V . <A> Your first task here is debugging/finding the root cause of your problem. <S> First simple advice for you to try: Disconnect both DC/DC converters. <S> A simple low cost linear supply will do. <S> If the noise is gone, good news, now you can try to pinpoint which one of the converters is causing the problem. <S> After the debugging phase, you can move on to redesign part of your system in order to minimize or filter our the offending noise. <A> I would use just a linear LDO from the 5V or 3.7V (battery) like many suggested before, because the current consumption of the equalizer is too small to justify switching mode power supply. <S> High sensitivity and high impedance preamps are generaly sensitive to perturbations. <S> Event through the ground line noises may come and render the configuration useless... <S> you have to always design the audio chain in such way that the ground line (and also +/- <S> Vcc's) to be from the main supply to the speakers (if not BTL), than to the power amp, then to the potentiometer, then to the auxiliary supplies stabilisers and then to the preamps. <S> From power to sensitive circuits; that is always the order.
| It is most probably related to noise induced/coupled from one or both DC/DC converters. Power your circuit with a decently cleaan Power Supply Unit.
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How does the mobile 3.5mm audio port work, and Can I light a LED from it? I am trying to do this, light a LED from the headphone wire(mobile) but it seems it gives ac in there, I don't have much idea, please help. <Q> Perhaps something out of a dead transistor radio. <S> Otherwise you could build a diode pump that say quadrupled the voltage. <S> Use shottkey diodes for your diode pump because they waste less voltage .Use <S> a red led for your experiment because they need less volts than the other colors. <A> Yes, audio signals are AC. <S> Headphone voltages will be too low to light an LED. <A> I don't really know for sure, but I would be checking out the mic wire. <S> Easy enough to check - just measure the mic to GND voltage, and then to be sure put in 1k - 10k resistor there, and measure voltage again. <S> From these two measurements you should be able to determine open circuit voltage of that source, and internal resistance.
| It could have bias current designed to power electret mics. If you must light your led without an external supply you could consider a small audio stepup transformer.
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How can I view USB traffic on an oscilloscope? I'm trying to examine the communication of a USB mouse with an oscilloscope (PicoScope 3204). When I connect the scope's ground to one of the two data signal lines (Data+ or Data-) the mouse shuts down. The mouse is the 1995 Microsoft Home device, so it must be using USB 1.1, not some fancy high-speed version. I also tried the same on a USB memory stick and experienced the same problem. Is there an simple circuit I can build to overcome this problem? I'm aware there are sophisticated USB test fixtures for this purpose, but I'm trying this as a demonstration/experiment, so I'm looking for a minimal home-brew solution. <Q> Don't ground the data lines. <S> They are above ground. <S> This is encoded using NRZI and is bit stuffed to ensure adequate transitions in the data stream. <S> On low and full speed devices, a differential ‘1’ is transmitted by pulling D+ over 2.8V with a 15K ohm resistor pulled to ground and D- under 0.3V with a 1.5K ohm resistor pulled to 3.6V. A differential ‘0’ on the other hand is a D- greater than 2.8V and a D+ less than 0.3V with the same appropriate pull down/up resistors. <S> You need to connect scope ground to USB ground. <S> Otherwise you're killing one of the signals and the differential receiver will not detect the required phase reversal. <S> With scope connected to ground a single channel scope can monitor either D+ or D-. <S> A dual channel scope can monitor both and you should see the phase inversion when data is transmitted. <A> Please see this very useful explanation why USB D- is not a GND. <S> If you want to probe USB data pairs; you need to connect your scope between GND and USB D- and/or D+. <S> Also remember that in most PCs the "GND" is earthed; and so is your scope GND lead. <S> This is likely the behaviour of shutting down because you are effectively connecting USB D- to GND (via earth) via your oscilloscope. <A> To add a little bit more of information and references, related to the safety operation of your equipment during testing. <S> You must be always very careful before connecting the ground or chassis of your test equipment to your DUT (Device Under Test) . <S> Some very bad things can happen , including: Putting you or other persons under risk of death. <S> Damaging permanently your test equipment. <S> Damaging permanently your tested circuit or DUT. <S> Some interesting references about floating vs. grounded test equipment and safety. <S> I cannot summarize them here as the topic is hugely broad: How not to blow up your oscilloscope, http://www.eevblog.com/2012/05/18/eevblog-279-how-not-to-blow-up-your-oscilloscope/ <S> Oscilloscope measurements safety, http://www.circuitspecialists.com/blog/floating-oscilloscope-measurements-safety/ <S> Why on earth are o-scopes earth referenced? <S> Why on earth are o-scopes earth referenced? <A> scope ground goes to usb 0V <S> [try the white wire. <S> test with a DC multimeter to the computer chassis that it is 0V and low ohms which you connect to] if your scope allows two channel difference A-B then use A for usb +data and B for usb -data. <S> Always test traffic between PC and an appliance such as a printer or webcam. <S> You need the appliance because usb employs a handshake before it gets going. <S> Scrape insulation off to expose wires and try a high impedance (10 Meg) scope probe. <S> It might self-adjust in order to work with ugly exposed cables. <S> Do not expect anything > 10MHz to work like this as scope probes are ugly and 240 ohm terminated probes would conflict with the handshake from the proper termination at the appliance. <S> USB-C is unknown to me; it might have a whole lot of extra switches in your way to decide the direction. <S> I do not know what the 100 Watt power lines might do. <S> Avoid USB-C <A> You need a 2-channel oscilloscope. <S> Then you connect one channel to D+ and the other one to D- . <S> The ground clips of the two probes you connect to each other. <S> Then you subtract the two channels: <S> X-Y, because it is a differential signal.
| Beyond Logic says USB uses a differential transmission pair for data.
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What can I replace copper with? I and my friends were trying to make a salt water battery but we cannot find copper strips in our town so is there any other electrode that I can use to make our battery. We have aluminum foil (which is one of our electrode) and what else should we use? <Q> What you need to look for is two metals with a large "Galvanic Potential" between them. <S> This is also known as a metals Anodic Index. <S> The following is a partial chart of Anodic Indices of various materials: Image source here . <S> As you can see, Copper is quite Cathodic at \$-0.35\mathrm{V}\$, while Zinc is quite Anodic at \$-1.25\mathrm{V}\$. You would expect a terminal voltage of about: $$V= <S> V_{Cu}-V_{Zn}=(-0.35)-(-1.25)=0.9\mathrm{V}$$ <S> Alternatively you could use something like Graphite - like from a pencil - which is at \$0.3\mathrm{V}\$, so would actually give you a much higher voltage. <S> Basically pick two materials, one which is Cathodic, and one which is Anodic. <A> If you really want copper, try looking in a home improvement store for roof flashing. <S> This is often available in copper and aluminum. <S> However, there are a number of other materials that will work. <S> There must be a chart somewhere that gives the electrochemical potential of various materials. <S> For example, early commercial batteries were made with carbon and zinc electrodes. <A> Try Stainless steel crubbers, while stainless steel is an alloy of iron/nickel and chrome, but it's standard reduction potential is the same or even more positive the Copper and the scrubbers have a large surface area
| To replace Copper and get a similar voltage, you could use something like Nickel which is similar to copper at \$-0.3\mathrm{V}\$. Gunmetal (Red Brass) would also work, giving a similar potential to copper. Copper and Zinc is just one of many options.
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Will this battery pack for an LED strip need a resistor? I want to use a small battery pack to power a strip of LEDs from Radioshack . They didn't have the power supply but list the requirements as 12VDC/1.5A. I know 8 AA batteries will provide 12 volts but I'm less certain how they will react to the 1.5 amp draw. We put 8 new alkaline batteries into a battery holder and did a quick non-solder test and the strip did light up nicely but the battery pack started to get warm even though we only had it connected for less than a minute or so. The plan is to solder a 9-volt type connector to the strip so it will be easy to connect/disconnect the battery pack instead of some type of switch. I know that I can add a current-limiting resistor inline with the wiring but I'm not sure what value to use. The strip looks like it has built in resistors so I'm at a loss as to what size resistor to try.The project is really just a short-term way to decorate a french horn for a school concert. My 15-year-old daughter can just connect the pack when the performance starts and then disconnect it when it is over. My concern is I don't want her to have the battery pack overheat during the performance. I don't care about how long the batteries last, this is a one-time use. Maybe another way to ask this is, what is causing the battery pack to heat up, the LED strip drawing excessive current or some other reason? I'm not trying to use rechargable batteries or build a charging circuit, just simply light up the instrument during a concert but safely. [EDIT]We did a test using the battery pack and after 30 minutes we couldn't detect any noticable dimming and the battery pack was still around 90-92 degrees F, tested with an IR thermometer. I told her she can just leave the pack in her lap and if it gets too hot to unhook the connector or remove a battery. She plans on taking an extra set of batteries with her as replacements for when the first set run down. My guess is that we previously had shorted the battery pack accidently which caused the temp to spike but now that the wiring has been connected properly we are seeing more typical behavior heat-wise. <Q> You could choose to use a larger sized battery. <S> Try C or D sized cells. <S> You will still get your 12V or so from the same number of cells. <S> The larger batteries have lower internal resistance and thus generate less internal heat at the same amount of load current as compared to the AA cells. <S> They are a bit heavier but at the same time will last for multiple concerts! <A> 1.5 Amps is too much for the type of AA cells which I use and they'd go dim before 1/4 of the 2000mAh are used up. <S> That they get warm indicates that they are being somewhat overused. <S> I'd prefer to go up a size to a Pb 'gel' battery or a scrap motorcycle battery for "12 V" <S> Those are too heavy for a pocket at a concert <S> so you should allow enough extra wire to place the battery on the floor. <S> Another way would be to replicate your AA set so that two sets in parallel each making about 12V do about half the work. <S> I'm not greatly in favour of adding series resistors as the LED already have some resistance. <A> The strip is an array of groups of 3 led's and a ballast resistor. <S> As long as you remain in the realm of 12VDC ie. <S> 10.5 <S> to 14.4VDC you won't need to add a ballast device. <A> I do not expect that LED strip to draw 1.5 A, more like 60 x 20 mA = 1.2 A. Still, that is a large current to draw from AA batteries. <S> Most Alkaline AAs are around 2000mAh <S> so they should last about 1.5 hours. <S> As the batteries deplete the brightness will be lower of course. <S> As long as the batteries are not too hot to hold in your hand <S> then I would not worry. <S> Make sure you do NOT isolate them thermally (put them in a small box for example) but make sure that the heat can escape. <A> You could forgo use of nickel based batteries, RC Lipo's are relatively cheap and have very high current sink capabilities. <S> A 3S 2200mAH will provide you with a nominal 11.1V <S> throughout most of its discharge cycle.
| But good Alkaline type batteries will be able to handle that current. You do not need a dropper resistor, there are resistors already on the LED strip and that can handle 12V which is exactly what you get from 8 x 1.5 V AA.
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What is the difference between PDIP and MDIP? What is the difference between these opamps? MDIP: http://dk.rs-online.com/web/p/komparatorer/5338186/ PDIP: http://dk.rs-online.com/web/p/komparatorer/0428458/ PDIP: http://dk.rs-online.com/web/p/komparatorer/7140789/ I need an LM311 which I can use on bread board as a comparator with fast switching. Some has MDIP some PDIP. What is the difference? ?Are all these the same except their switching times and voltage ranges? Will they be treated the same way? <Q> The LM311 was invented, if memory serves, by National Semiconductor, as were most parts beginning with 'LM'. <S> Since it was introduced (perhaps in the 1970s) it has been second-sourced by other manufacturers including Motorola (now On Semi) and Texas Instruments. <S> In some cases the specifications may be somewhat different, and the packaging will vary by manufacturer and, over time, even within a given manufacturer's line as they make changes. <S> In 2011, Texas Instruments purchased National Semiconductor and folded the line into their own, updating the data sheets over the next couple of years to elide the National Semiconductor name and replace it with their own. <S> In this case, the LM311N/NOPB was the National Semiconductor part (the NOPB was added when lead was eliminated), and LM311P was the TI part. <S> Now they are both TI parts, but they are almost surely made in different facilities and have slightly different specifications. <S> The packages also have slight differences in the plating, and the plastic shipping tubes are different sizes (40 vs. 50 to a tube) <S> etc. <S> Also, the TI LM311P datasheet , as of right now, has a rather glaring error- referring to the part as a quad! <S> Bottom line, as a hobbyist-user, it makes no difference to you. <S> Pick whichever you like - just as if they were from two different manufacturers. <S> Possibly at some time in the future one of them will be obsoleted- chances <S> are it will be the one that costs more since they are discouraging use of it. <S> To a real manufacturer, they are most certainly not the same, and purchasing department may require special dispensation to be able to substitute one for the other, or it may be prohibited in some cases where the substitute has not been qualified or has been found to cause problems. <S> The packages are different as are the chips inside. <A> The opamp is the same, it is only the packaging which is (slightly) different: PDIP = <S> Plastic <S> Dual-In-line Package <S> MDIP = <S> Molded Dual-In-Line Package <S> Since both are DIP (Dual-In-line Package) <S> you can use them on a breadboard. <S> The difference between Plastic and Molded is completely irrelevant for you ! <S> Just get the cheapest. <S> The silicon circuit inside is IDENTICAL (assuming they are from the same manufacturer). <S> So switching times and voltage ranges are also IDENTICAL. <A> This refers to the package. <S> PDIP = <S> Plastic Dual Inline Package, MDIP = <S> Moulded Dual Inline Package. <S> In practice, they are the usually the same; electronics is littered with multiple acronyms and descriptions for the same thing. <A> There is also another difference in the summary specifications, the voltage range is 5-28v, or 9-28v. <S> Never trust a summary spec, especially typical figures in it, without going back to the data sheet and seeing what measurement conditions are being imposed. <S> You may find that if a part is intended for a military or higher performance market, specifications may be quoted over a wider range of temperatures, or supply voltages, or for heavier capacitive loading on the output. <S> In the good old days, 50pF was frequently used as an output load, though these days much lower might be quoted. <S> Little differences like this can change the specs.
| PDIP and MDIP are just names used by the two companies before the purchase- to refer to very similar things.
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Vce too high. Help? I have been working on a high-powered LED light bar.The design powers them almost directly from the mains, but limits current using a 555 and beefier NPN. Below is a simplified schematic. Obviously the LED resistance changes with several factors, and I have blown out a few, but that was due to soldering issues on my part. Essentially I know the circuit works. The only hold-up is that I have tried two different NPN transistors (salvaged from old CRTs) and both end up with Vce voltages around 35V causing them to heat up rather quickly since the current is around 1 amp. I will link the datasheets for the two in below. I have measured the output of the timer at 2V RMS and set the resistor values for each transistor according to their Ib vs Vce graphs (around 0.2A) since neither include a Beta value. My question is: how do I bring the Vce down to around 5V or lower? Fuji C3866 | Toshiba 2SC5386 https://drive.google.com/folderview?id=0B4n242xWd2TBOUVBZF9uQ1BDc2M&usp=sharing EDIT: Removed terrible CAD, added hand drawn circuitThese are 10W, 12V LEDs <Q> You can't bring Vce down - you are operating it with pulse width modulation therefore the transistor is either fully-on of fully-off. <S> What you may measure is the average Vce. <S> And now to the main issue - you are not controlling the current it seems from your circuit diagram - there is nothing to stabilize the PWM duty cycle - there is no current feedback hence it is open-loop voltage control. <S> The next issue is you need an inductor and capacitor to filter the "strong" square wave delivered by your switching transistor - the peak voltage across the LEDs will be basically rectified AC and this could kill them unless there is internal current limiting but, if there is then why both with this ornate solution? <S> Bad, bad circuit it seems. <S> Take a look at this: - Note <S> the inductor L1 in series with the LED string and note Rsense to detect current and control it. <A> You must not treat (a string of) LEDs as having a resistance ! <S> Treating LEDs this way is a recipe for disaster (as you have noticed burning out those poor LEDs). <S> LEDs have a very sharp voltage-current relation. <S> You must choose at what current you want to operate them, make an estimate of the voltage (usually around 2.5 - 3 V per LED) and then add one or more resistors in series or make a proper current source using a transistor. <S> Your LED current is unlimited, no wonder they burn out. <S> For high power LED drivers like this you really must know what you're doing. <S> Be able to make power calculations etc. <S> Also 165 Volts is dangerous, 1 A through the transistor. <S> Are you sure you know what you're doing ? <S> I doubt it. <S> "My question is: how do I bring the Vce down to around 5V or lower?"This Vce is NOT your problem ! <S> Your problem is much more fundamental to your design. <S> Please educate yourself first before proceeding. <S> What you need to learn first cannot be explained here in a few lines. <A> As Andy aka has pointed out, you are not measuring the Vce(on) of the transistor. <S> You are measuring the average Vce in the presence of the PWM, and you should expect that to be much higher than Vce(on). <S> However, since you seem to be having heat problems, I'll assume that your transistor is not being properly driven. <S> The rule of thumb is that, when a transistor is in saturation, it has a gain of about 10 to 20. <S> This means that, for a load current of 1 amp, you'll need about 100 mA of base current, which should be well within the capabilities of a 555. <S> Actually, from the data sheet, at 1 amp collector current, a 100 mA base drive will only drop your Vce to 5 volts using the C3866, so I don't recommend using it. <S> Note that hfe is the same as beta. <S> Use the other one. <S> You need to keep in mind that, with a 5 volt supply, a 555 putting out .2 <S> amps will only produce about 2.5 volts. <S> Furthermore, the Vbe will be about 0.7 volts, leaving only about 1.8 volts across the resistor. <S> So the resistor should be about 10 ohms. <S> I'll disagree with Andy about your filtering. <S> I believe that your cap is adequate. <S> With a 50% duty cycle your average current is about 0.5 amps, and for a capacitor $$dV/dt = \frac{i}{C} = \frac{0.5}{.001} = 500 \text{volt/sec}$$ and with a full-wave rectifier the charge period is 1/120 second, so the variation in maximum peak voltage $$\Delta V = \frac{500}{120} <S> = \text{~}4\text{volts} $$ <S> If your PWM frequency is high, a large capacitor may have some difficulties with it, but I don't see how this is likely to produce your transistor problem. <S> I'd recommend that you double-check the LED current. <S> Do this by replacing the transistor with a 1 ohm, 2 watt resistor, and check the voltage across it. <S> If the voltage is more than 1 volt, you know you're trying to pull more than an amp.
| Your problem is most likely that you are not able to provide enough base current to turn the transistor fully on.
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Is a microwave's output power proportional to the mass of its contents? My friend and I are having a heated debate. On the one hand, he thinks that a microwave oven that is empty, consumes almost no power (not considering lights, lcd, etc). He says that once you put an item in the oven, such as a glass of water, the oven will begin to consume more power as the magnetron must output more energy to heat the contents. Basically, he says that the power consumption of the magnetron in a microwave oven is directly proportional to the mass of heat-absorbing molecules inside the oven. On the other hand, there is me. I think that the magnetron is always outputting what it's rated for (in an ideal world). I believe that an empty microwave simply dissipates its energy as heat through the chassis. I made the analogy of a radio tower that when transmitting, is always doing so at the same power regardless of the amount of listeners. Both of us have come up with some interesting arguments, but neither of us are engineers and lack the knowledge to prove our theories. So we turn to you! Thanks! <Q> Simple thought or practical experiment: <S> If he's right then the heating time to bring water to boiling point is independent of the quantity of water. <S> One cup will take as long as two. <S> If you're right two cups of water will take twice as long to boil. <A> Is a microwave oven's output power proportional to the mass of its contents? <S> No. <S> The magnetron develops a certain electromagnetic field strength (volts per millimeter or however it is measured) just like any radio transmitter. <S> I tried the experiment, measuring power draw for 2 cups water, 4 and 8 <S> , it was identical: <S> 1380 Watts for a 700 Watt rated unit. <S> This is about what we would expect taking in to account losses (most radio transmitters are about 50% efficient). <S> When operating a transmitter, there is a specification called standing wave ratio, which determines how well matched the source is to the load. <S> If the load is perfectly matched, it absorbs the entire output, regardless of how much power that might be. <S> If it is poorly matched, some of the power is "reflected back" and makes an in-phase voltage at the output terminal of the device. <S> If this reflected power causes the maximum safe voltage of the output device to be exceeded, it will arc over. <S> It is also possible for a mismatched load to draw too much current, so the device will self-destruct by overheating. <S> In essence, you have X amount of watts coming out, which will either be properly absorbed by a load, or will stress the device (magnetron in this case) and probably damage or destroy it. <S> The output power is unchanged, and the input draw is unchanged. <S> It is like connecting an electric motor to a load: Stall the motor and it might burn out, unload it <S> and it might overspeed and damage itself. <S> This is true for all forms of radio wave emission. <S> Addition: <S> All devices have loss as well, so even if "spinning the wheels" it will still draw and waste some energy. <S> In the case of a Class A audio amplifier, this is 50% of the input power. <S> In some systems it is more, in others less. <S> Since a magnetron is not ideal, it is simply going to draw some power no matter what. <A> So I have finally gotten around to testing this. <S> Using <S> this watt meter <S> I tested 0, 1, 2 and 4 cups of water in a plastic bowl. <S> Overall, there was a difference of roughly 1 watt between any of the quantities of water, including empty. <A> A magnetron that heats food produces a frequency of about 2.5 GHz. <S> An antenna is formed within the oven and as per antenna theories, beyond about one wavelength <S> a true electromagnetic wave is generated and real power is produced. <S> One wavelength is about 120mm. <S> This power is done and gone forever so unless there is some kind of reflection up close to the antenna plate <S> Don't try it at home. <S> The stirrer distributes the microwaves to a bigger area inside the oven by reflection. <S> Once they have exited this point there is no coming back for any energy reconciliation. <A> When the oven is turned on, the magnetron emits a transversal elecromagnetic wave into the cavity. <S> As soon the EM wave fills the cavity it begins to hit the walls an reflects back, now the standing wave is formed. <S> In standing wave the E-field and H-field is out of phase by 90 degrees, that means no real power is transfering it just bounces from wall to wall and to the magnetron, where the E-field is exactly in phase and magnitude as the field produced by magnetron, this implies that there is no potential difference and emmiting is blocked. <S> When there is an item in the cavity, then it bends the standing wave, such that magnetron "sees" the load and additional TEM wave is superimposed to the existing standing wave. <S> Practicaly the standing wave acts like a conducting channel and "brings" the load connected to the source without wires. <A> A simple way to get the answer is to place a current reader between the plug and your microwave <S> (example: https://en.wikipedia.org/wiki/Kill_A_Watt#/media/File:P3-Kill-a-watt.jpg ) <S> Then run it empty, then with something in it. <A> The output of the magnetron must be constant for it to work. <S> However, two cups of water take sqrt(2) times as long to boil- square law. <S> This is due to the surface area.
| I reckon the power taken by the magnetron will be pretty steady whether there's food inside or not.
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Can I use a TRIAC on a secondary isolated AC winding? Can a TRIAC be used on the secondary winding of an AC isolation transformer? Say I have 230 V on the primary, but I want to use the 12 V AC secondary, will a TRIAC work on the secondary? What I want to achieve is this: Use the TRIAC on a secondary winding (say 12-24 VAC) to use on a spot welder. The control pulse to the gate of the TRIAC will be delivered via two optocouplers (one for the positive cycle and one for the negative cycle) of the AC wave. The optocouplers will be controlled by a single-shot (mono-stable) pulse from a 555 . <Q> A simple answer is "yes you can". <S> The better answer is "you don't have to", because in your situation, it is better to use a triac control on the primary side. <S> I have seen many special heating devices that use this approach, the current is smaller compared to the secondary, so you will not have trouble with short circuits where the major resistance would be your triac, thus the spot welder will heat mostly your triac and not the spot. <S> Forget about using a zero cross detector circuit, the worst case for a transformer is to turn it on at 0V and don't use the 1.9T toroidal transformer <S> (this is what you'll get in a shop for hallogen lamp), a use transformer designed with max. <S> flux 1.6 Tesla or maybe less in order to keep dI/dt manageable. <A> Yes. <S> Remember that there is a voltage drop across the triac (check the data sheets) and that is a more significant fraction of 12 V than it is of 230 V. <A> Some details missing in your question, but I will try to answer in general. <S> You must not forget that to synchronize the switching of the TRIAC , you must take into account the zero crossing of the winding that feeds the circuit of the TRIAC. <S> If you can ensure this operations conditions, the TRIAC will function in the secondary of an isolation transformer. <S> Edit: <S> zero-crossing & pulse Synchronization with the zero crossings of the line, allows you to control the power applied to the load. <S> Basically it is turning on the TRIAC after a certain time after the line voltage zero crossing. <S> There are many methods to detect the zero crossing and to generate the shot. <S> It would be very extensive to list them all here. <S> Here you can access information about phase control using a TRIAC. <S> Here you will find information to turn a TRIAC completely in sync, without power control.
| A TRIAC operates in an AC circuit, provided that the particular conditions of a thyristor are met, that is, not exceed the maximum working voltage and ensure minimal holding-current .
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Do I need to make a DC path at the input of an op amp? I've heard that a DC path is needed at the input of an op amp,or the op amp won't be functional because there is no path for the input bias current.(in the image below,R1 provides a dc path for input bias current) But if I use MOSFETs to design an op amp,is there still input bias current(the gate of MOSFETs have no current)?Or do I still need to add R1 to make op amp functional? if R1 is needed,can it connected to a DC voltage source or it must be connected to gorund? <Q> Yes, you generally want the resistor there. <S> It forms a high-pass filter with C1, and you would normally want the time constant short enough that the level at the non-inverting input stabilizes reasonably fast, but long enough that the cutoff frequency of the high-pass filter lets through the desired signal. <S> The output voltage of the op-amp (with no input and power applied) will stabilize near (1+R3/R2) times the voltage on the lower end of R1. <S> So if you connect R1 to ground, the output voltage should be near zero (ignoring the bias current multiplied by R1 and op-amp input <S> offset voltage). <S> If you connect it to something other than 0V, the output voltage may be different from 0V and may even saturate. <S> If you omitted R1, and used a very low leakage op-amp and a low-leakage capacitor then it might appear to work, perhaps even for quite some time, but eventually the output average voltage will likely drift towards one rail or the other. <S> If you used a relatively leaky capacitor such as a bipolar electrolytic, then it might work fine, though under some conditions the capacitor could find itself charged (apply a voltage beyond +/-Vs to the input, for example) and might take many minutes to recover. <S> If the gain is high, you might want to add another capacitor in series with R2, which will reduce the maximum output offset voltage since the input offset voltage will no longer be amplified. <A> While part of the function of the resistor is to provide a path for the input bias current is is also needed to establish a DC level there. <S> With a FET input connected to a capacitor there is essentially no DC path and therefore the input could remain charged to an arbitrary DC voltage pretty much indefinitely. <S> So yes you do want the resistor on a FET amplifier. <S> Having said that you should be able to use a much higher resistor value on a FET amplifier. <A> To answer the second part of the question : R1 can be connected to a DC source other than GND, provided that you take its effect on the output voltage into consideration. <S> In other words, if you connect it to 1V, and the gain of the circuit at DC is 10, the output will settle at 10V DC. <S> If Vs is +5V that can't happen, so the output will settle somewhere close to 5V and stay there as long as Vin (AC) <S> > <S> -1V. <S> This DC bias on the input can be used for various purposes : for example, offsetting an AC audio input (+/-2V say) to a range of 0.5 to 4.5V for conversion in an ADC with an input range of 0 to 5V. <S> You'd connect R1 to 2.5V and set the opamp gain to 1 (i.e. remove R2)
| As the other answers say, you need R1 to define the DC input voltage.
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AVR-ATmega256 What is the use of PORTxn? In AVR ATmega2560 microcontroller, there're 3 register for I/O port: DDxn, PORTxn, and PINxn DDxn decide the direction of each pin while other two, according to the datasheet PINxn toggle the value of PORTxn by written 1 into PINxn If pin is input (DDxn is written 1), written 1 to PORTxn turn on pull-up resistor and written zero turn it off If pin is output (DDxn is written 0), port pin is driven high (one) if PORTxn is written 1 and low (zero) if PORTxn is written zero I don't understand what these term means: pull-up resistor, driven high and low. What is the difference between 2 terms. What do they do with I/O port and what is the use of them? Can anybody give me explanation and example. I also don't know what is register used for store address for I/O. I means for input and output there must be some address tell where input data is stored or address tell where data is ouput. Furthermore, if 8 pin of one port is configured to be mixed between output and input then how the data of each bit of each pin is organized to be store in a register (which contain 8 bit) <Q> Driven High - <S> The signal is driven to your positive voltage (VCC) Driven Low - The signal is driven down to ground (GND) <S> Pull-up resistor :In short, this means that there is an internal resistor that is connecting the pin to your positive voltage, thus driving it high unless a connection to GND happens. <S> Sparkfun has an amazing tutorial <S> Should your PORT register have a mix of pins defined as inputs and outputs, you can still read the values just fine. <S> To read the pins defined as inputs, you'd read the PIN register like so: <S> uint8_t inputs = PINx; As for the outputs though, you can just set the PORTx register and you won't have to worry about it affecting the inputs: <S> Say you have pin 2 defined as an input and the others as outputs, writing a value like this, PORTx = 0b10001111 , won't affect the value for the input pin and will set all the others to the respective values. <S> Arduino has a good article on this <A> Figure 13-2 in the datasheet (which shows a single bit of an I/O port) should make it clear. <S> I have highlighted some parts to be discussed:- <S> The Data Direction Register (red) controls a Tri-State buffer which connects the output of the PORT flipflop (green) to the pin. <S> When the DDR bit set to input the buffer is disabled and the PORT output is disconnected from the pin. <S> The triple input AND gate (blue) enables the pullup resistor only when the DDR bit is set to input, the PORT bit is a 1, and PUD (pullup disable) is not active. <S> The output PORT (green) is a register which you can write to and read. <S> When you read it you are seeing the value you put there, which may not be the same as the level on the pin. <S> Reading the PIN 'register' (purple) will always give you the current level on the pin. <S> The value may be synchronized to the I/ <S> O clock <S> but it is not stored - there is no flipflop to hold its value as the PORT and DDR registers have. <S> Writing the PIN 'register' actually toggles the PORT bit. <S> It does this by feeding the output of the PORT flipflop back into its input and clocking it to create a divide by 2 counter (yellow). <A> If a GPIO pin is configured as an output, it reflects the value of PORTxn <S> (0=GND, 1 <S> =Vcc). <S> When a port is mixed input and output, bit manipulation operations must be used when reading or writing the DDRx, PORTx, and PINx registers.
| If a GPIO pin is configured as an input, PORTxn determines whether the internal pullup is enabled (1) or if the input is floating (0). The current value on a pin is always available as PINxn.
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Having a hard time Identifying this part I think its a female berg pin. I have a requirement of a similar part But I am not able to find it. What is it called? Thanks <Q> Those are machine tooled single inline sockets . <A> Note that the ones you show were cut from longer strips, which is why the ends are all ragged. <S> A couple of generic descriptions via Digikey: <S> Mill-Max: "Connector Receptacle x Position 0.100" (2.54mm) <S> Gold <S> Through Hole"Preci-Dip: "Connector Socket x Position 0.100" (2.54mm) <S> Gold Through Hole" <S> There are also similar parts with stamped contacts rather than the (relatively expensive) <S> screw machine parts ('turned', 'machined') <S> that are in your photo. <S> The height above the board varies as well as the number of pins. <A> You can find them at certain distributors such as Digikey . <S> Also look on eBay for lower prices with correspondingly lower quality. <S> Manufacturers to look for include MillMax and Samtec.
| That is a breakaway female header strip with machined sockets and solder tails. Different manufacturers use slightly different terms.
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Is there any pure digital phase shifter circuit? It is possible to make analog phase shifter circuits: But In digital electronics (logic circuits), is it possible to shift the phase of an input digital wave? The frequency of the input wave is 1MHz at most. <Q> Yes of course it's possible - any analogue filter can be taken into the Z domain and a digital equivalent constructed. <S> This happens all the time. <S> So if you have an analogue circuit that does what your need there are several methodologies that produce an equivalent digital "circuit" that can be implemented in MCUs, FPGAs etc.. <S> Try looking up: - Impulse invariance method Bilinear transformation <S> As to whether your original analogue circuit is a pure delay, that is another matter. <S> Anyway, I forgot that I wrote a paper on it and the front sheet should help you understand: - It begins with a simple RC and converts it to an op-amp equivalent in position (2). <S> Position (3) re-shapes it slightly and position (4) arrives at the finished digital filter. <A> A shift register with a clock can be used as pure digital delay, but that will result in the output pulse being modified, such that the output pulse will be synchronised with the clock. <S> To minimise the synchronisation errors, you would use a high clock rate, but that has the disadvantage of requiring a larger shift register. <S> An alternative would be to use an analogue delay line (such as posted in your question), with a pulse shaper (Schmidt trigger) on the output, but again this will have some effect on your pulse shape. <A> There are components available to delay digital signals by specified amounts. <S> Many of these components, called delay lines, offer tapped delays in steps. <S> In days gone by there were lumped constant delay lines that offered from 1 to 10 delay taps with each delay element being an L/C component pair. <S> Since the L/C type delay lines did have the effect of changing the rise and fall time of the signal some various manufacturers then introduced buffered lumped constant delay lines. <S> These would have an internal high speed inverter chip such as an SN74S04 that would buffer the input and five outputs. <S> These were specified so that the delay of the inverters was included in the tap to tap delay. <S> You can find tapped delay lines and programmable delay lines. <S> Some manufacturers to consider are Linear Technology and Maxim.
| These days various manufacturers have produced components called "silicon delay lines" that offer various features.
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LED typical and forward voltages So, I was going through the spec sheet of super bright 5mm orange LED, and it was listed that the forward voltage is between 1.8V and 2.5V. Below that there was a property listed as 'typical voltage' and had a value of 3.2V. I'm confused as to what value of resistance I must use. <Q> The important thing to realize is that CURRENT is what MATTERS to an LED. <S> The voltage is given as a range because the voltage varies, depending on manufacturing tolerances, temperature, etc. <S> If you design well for LEDs, you feed them with a current source, and the voltage simply needs to be adequate for the number of LEDs you have in series on each current source. <S> In figuring the voltage compliance your current source should have, look at the highest voltage quoted. <S> This is sub-optimal .vs. <S> a current source, as it will generally be providing less than the maximum safe current in order not to provide more than the maximum safe current under conditions (or for parts) where the forward voltage is particularly low. <A> Quick solution. <S> If you want to know the forward voltage hook your LED to a power supply turned down to 0 volts and slowly crank up the voltage. <S> You LED will turn on <S> once forward voltage is reached. <S> However , you might blow an LED up doing this so proceed with caution. <S> To help answer your question. <S> Use forward current to decide on what resistor to use and take the worst case forward voltage. <A> If you have doubts on the stated specs, you can make a simple Constant Current regulator to test an LED out with a common adjustable LDO. <S> R1 can be a fixed resistor, or more conveniently, a variable resistor or trimpot. <S> For 20 mA, that's 62.5 Ω. <S> (R = V/I, in this case VRef / IOut). <S> Simply measure the voltage across the Load/LED <S> and you know how much it's <S> Forward Voltage Drop at 20mA is.
| If you use resistors for current-limiting LEDs, use either the LOWEST value of forward voltage provided, or a bit less than that (safety margin) when choosing a resistor to set the current.
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Voltage Drop Across Diode in Off State I was reading this sparkfun article to learn more about how transistors work, but I am confused by one of their diagrams. Here it is: If there is no current flowing, why is there a 1.3V drop across the LED? I looked at the VI characteristic graph for LEDs and it should be 0v when I=0, which is the case here. I think I am missing something, can someone point out exactly what? Thanks! <Q> They are probably trying to show you the voltage the LED would drop when some current is going thru it. <S> I agree that is inconsistent with how they are showing the voltages across other components. <S> The only current will be the transistor leakage, which quite small. <S> There is a case where what they show could actually be true. <S> LEDs also work as photocells in reverse, although rather poorly. <S> With no load on it and in reasonable light, the LED will develop a voltage close to its normal forward operating voltage. <S> However, the impedance of that will be so high that even a ordinary voltmeter can load it. <S> Whoever made that diagram may have probed around the circuit with a voltmeter, and at that illumination and that voltmeter, that's what was reported across the LED. <S> Whatever part of the supply voltage that doesn't appear across the LED would then be across the transistors, since no current is flowing. <S> Another possibility is that when they probed across the transistor with a voltmeter, the meter caused enough current to flow for the LED to develop 1.3 V across it, so the meter read 3.7 V. <S> They then subtracted 5 V from 3.7 V to say the LED had 1.3 V across it. <S> If the LED was measured directly, it would have less voltage across it. <A> You are right. <S> When the current is zero, the voltage will be zero. <S> However, when simulating 0mA is not necessarily absolutely zero. <S> What I speculate their simulation tool to make that image does is also calculate a small leakage current through the transistor. <S> To be honest I find it fairly shoddy they do not explain about that (I'm assuming, since I am too lazy to read all of it), because it is confusing, as you have experienced first hand. <S> A transistor may still conduct a tiny bit if you do not force a current into the base, due to small bits of unperfectness. <S> When you express that leakage as mA it will say "0mA". <S> But a simulation tool may estimate up to several dozens of μA (which rounds down to 0mA in the view) as leakage, depending on the type of model it uses to simulate. <S> This means the LED will see a small current, which will put it, apparently, at the 1.3V point of its estimated curve. <S> Of course, this is speculation, not knowing how they made the picture, but I think it's likely enough the simulator "estimated" all of those leakage numbers at the points you see. <S> You should be aware that a simulation is just a simulation and never absolute truth. <S> I think in the real world you would not see such a significant voltage across the LED, although with old fashion transistors, the first generations basically, it's possible. <S> If in the real world you see that happening, this may help a little bit: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Here, if you open the switch R2 will help pull the transistor closed. <S> With modern transistors (like newly made 2N3904's and such) controlling only a simple LED or Relay these resistors are usually not at all that important, but when you want to be sure there is as little off-leakage as possible, they do still help. <A> That article is seriously flawed. <S> For instance, in the picture above your LED picture it shows this: - At first glance is seems OK <S> then you notice the control voltages and they say that 5V turns the motor off. <S> What utter rubbish - the motor will still be on - <S> it needs above 11 volts to turn it off and this is a pretty serious error in my book. <S> They do redeem themselves with an explanation about this but the article is pretty unprofessional and could lead to someone building that circuit and never being able to turn the motor off. <S> Then they show what they say is a two input AND gate: <S> - However, what they fail to recognize is that with the top transistor base turned-off, the emitter of the lower transistor will be at a logic level of 1 when its base is high. <S> Bad stuff really. <S> Then they show this H bridge (yuk): - <S> There is no excuse for not showing base resistors given as they made a big deal about it early on in the article. <S> My advice is don't believe that article.
| In reality, the LED will have very close to 0 V across it when the transistor is off.
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Why should you use two resistors in parallel on an LED? So I was looking over the Arduino R3 schematics and noticed this little design choice: What is the reason for doing something like this? I mean it's hard to know what the designers were thinking, but maybe it was done this way to save space. Do you get any other benefits? <Q> Don't look to the arduino designs as examples of stellar electrical engineering. <S> However, there can be a legitimate case for doing this. <S> This part contains 4 resistors. <S> If it was already there for another reason, especially if several more of them are used on the same board, then using two of the resistors that would otherwise go unused in parallel to make a 500 Ω resistor is a reasonable thing to do. <S> It can often save more money overall to use fewer different parts, than a smaller number of total parts but more different ones. <S> For cheap parts like resistors, the dominant cost is not the price of the part, but the cost of purchasing, stocking, setting up the pick and place machine, etc. <A> For the same reasons as Olin mentioned, using two distinctly seperate resistors in parallel can be a saving if those resistors are used elsewhere on the PCB. <S> Line items in the build need to be stocked and counted and there is a real annual cost for this. <A> Additional to all the valid point given already, another reason could be a better heat dissipation. <S> Since these resitors are limiting current on a LED they could probably get relatively hot, given their small size. <S> In this case the power they absorb is around 20mW. <A> Historically, using two resistors in parallel can give you a resistance which is not available in a standard package. <S> In this case, two 1K resistors gives 500 ohms. <S> A the nearest 'standard' single resistor is 470 ohms. <S> This may not be as relevant today as it once was, but was a way to obtain very specific resistances. <A> I asked this myself an engineer doing layout reviews. <S> He gave me three answers, from which two were already mentioned here: Distribution of heat dissipation Reuse of known, proven, qualified, used, ... <S> components New: Increase of reliability for the case one component fails. <S> However, I do not know if this applies in this case.
| Having two items instead of one allows for better heat dissipation.
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Why is it that we use capacitors in Differentiator & Integrator circuits (comprising Op Amps) & not inductors? One of the possible reasonsI know is that capacitors are available in wide range of values and can be made more accurate than inductors. What other factor leads to the use of capacitors over inductors? <Q> A lot of circuits require medium/high impedance values for even moderately low frequencies such as audio and a capacitor of say <S> 10 nF at 1 kHz has an impedance of 15.9 kOhm. <S> An inductor having this impedance at 1 kHz would have a value of 2.53 henries. <S> Now that isn't a small value to fit in a space for a surface mount component. <S> That's one reason and the next is cost - try finding a 2.5 henry coil in farnell, digikey or mouser and see how much change you have out of a dollar or a pound. <S> The 10nF will cost you about a couple of pennies maximum. <S> So, it's not small and it's not cheap and <S> a side effect of it not being small is that it will possess significant parasitic capacitance (several pico farads if not more) and this makes it less than perfect. <S> I'm certainly not saying caps are perfect but they are a couple of orders of magnitude more perfect than an inductor. <S> Also it will have a significant DC resistance as it is made as small as it can be. <S> It won't be good at handling currents like capacitor is and the core will probably saturate. <S> Bad idea - use capacitors. <A> Generally speaking inductors are much more lossy than capacitors. <S> They depart much more from the ideal models that people learn at college, and in a poorly specified fashion. <S> In other words a circuit having inductors instead of caps is more likely to need tweaking. <S> Also inductors, depending on how they are constructed, are more prone to picking up stray fields, which means more PCB layout constraints. <S> The whole process from hand sketched circuit diagram to working PCB is more predictable when inductors are not there. <S> Finally, if you need to substitute a part, you are more likely to have problems with an inductor than a cap, as many inductors use non-standard footprints. <A> Both coils (inductors) and capacitors differentiate or integrate something <S> w.r.t. time. <S> It takes more to manufacture a coil than it takes to manufacture a capacitor. <A> It would be possible to build integrator or differentiator circuits using inductors, but these would integrate current instead of voltage. <S> You would need a current amplifier (low input impedance, high output impedance) instead of a voltage amplifier in this case as well. <S> These are somewhat more impractical to build.
| I think, in almost any context, an active circuit (one with op-amps or some kind of transistor amplifier) with inductors will cost more than one with the equivalent number of capacitors.
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Why are reed switches made in glass tubes? Why are reed switches made in glass tubes? Reed switches sense magnetic fields, and glass is not the only material not affected by a magnetic field. They can use plastic for example. Why are most of them in glass? This question came into my mind when one of them easily cracked and then broke. It could have been in a plastic tube; then it wouldn't have broken. Also, plastic would be cheaper than glass, and easier to melt and form. <Q> Glass is clean, dimensionally stable and very strong, doesn't outgas at the operating pressures in the interior of the capsule, won't react with the fill gas in the capsule, and doesn't soften under soldering temperatures. <S> Here's a beautiful link. <A> A cylinder of plastic of sufficient strength would be too thick for common magnets to switch. <S> While the glass may snap, the plastic would be worse. <S> Keep in mind, the outside is only meant to protect the easily bent metal arms. <S> Once they are bent out of shape there is no easy fix. <A> The reeds have to be held very strongly in relation to each other, because they are usually spaced very closely together and any bending of the package, usually due to assembly and soldering stresses in the normal manufacturing process and environmental changes, would damage the ability of the reed to work within it specifications. <S> Glass is a nice brittle, strong enclosure, and further you can find metals that are magnetic and have a similar temperature expansion rate so specifications can be maintained across the specified temperature range. <S> Plastics and many other materials could be used, but they wouldn't be as reliable without enlarging them (more material to make up for the lack of strength), and dealing with temperature expansion mismatches. <A> With glass you can see inside. <S> That's useful since reed switches have orientations. <S> With a opaque package, the orientation would have to be marked, but there isn't much room on something the size of a reed switch. <S> That would leave cryptic markings. <S> With glass you just don't need that. <S> Glass is also more hermetic and inert than other materials of the same thickness and strength that are candidates for reed switch casings. <A> I imagine you're asking the question "why are reed switches still made of glass?" <S> Reed switches find rarer use every day since there are Hall-effect switches and other devices that are smaller, cheaper, and more reliable. <S> Reed switches were invented in 1936, according to Wikipedia. <S> Nobody retools a buggy whip factory, just as nobody will retool a reed switch factory. <S> If all reed switch manufacturing disappeared tomorrow, it's unlikely a new site would appear to replace the lost capacity. <S> All of the other mentioned reasons (imperviousness to gas, rigidity, and so forth) can be addressed using plastics. <S> Microchips are currently manufactured in plastics when at one time they came in steel packinging (TO-packaged LM711 comes to mind) as well as ceramic (old intel 4004's, even up to the 8086), but the usefulness of these items and the expense of that technology drove retooling to plastics.
| Plastic of the same thickness of the glass would bend without much force at all. It's likely due to the strength vs thinkness needed for the reed switch. Glass is traditionally used.
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Suggestions for moldable ferromagnetic material? I'm trying to build a sort of three-axis stepper motor that consists of a sphere with a permanent bar magnet embedded in it surrounded by a cube (slightly smaller than the sphere, so that the sides of the sphere protrude) of ferromagnetic material. On each pair of opposite sides to the cube is a coil of wire, surrounding the hole in the cube through which the sphere protrudes. By forcing current through a combination of the wire loops, the sphere can be magnetically induced to align itself along any axis. I've tried to build a prototype without using the ferromagnetic cube, but the field from loops of wire alone is too small to easily and safely cause the sphere to rotate. My first few attempts have been to make a mold of the proper shape and fill it with iron filings, then pour some liquid that can dry or otherwise cure to a fixative (preferably non-conductive so eddy currents aren't a problem) to hold the filings in place- it is then easy to laminate the surfaces of the material so that corrosion of the iron, if not induced by the fixative itself, is not an issue. However, the only material I have on hand to make the mold out of is PLA plastic, which has a low tolerance for heat. I've tried a few different varieties of wax, but all that I've been able to lay hands on melt at too high a temperature for the PLA to stand or else are too cohesive and don't seep in between the iron filings, and I've had similar problems with various glues. Can anyone suggest either a possible fixative or an alternative moldable ferromagnetic material to use? <Q> It sounds as if you are needing moldable flux concentrator for hobby purposes. <S> They are quite high in price, but performance is very good. <S> Search Flux Concentrators such as Fluxtrol or Ferrotron. <A> | Increasing temperature will make epoxy more flowable and also accelerate setting times. <S> By reducing setting compound amounts in 2 pot mixes you may be able to combine flowability and acceptable setting times. <S> Polyurethane plastics (one version is seen as clear varnish spray) are atmospheric water setting (relatives of cyanoacrylate glues) and can be very 'runny'. <S> Cyanoacrylates may do what you want (at a cost) and can be of very low viscosity. <S> Not flowable <S> but if you want to hasten setting of silicone rubber then mixing it with cornflour <S> greatly increases setting rate - this is because the CF provides intimate access to the small amount of water needed for setting distributed throughout the mix rather than it having to permeate through the SiR at mm's per day rates. <A> Marine epoxy ... good stuff. <S> (Doesn't have to be the very expensive West Systems either). <S> It's a 2-part compound, mixed as per instructions, usually 2 parts resin to 1 part hardener. <S> (Usual warnings : can cause allergy, use rubber gloves. <S> Mix too much, say >150ml, too warm, or too much hardener, and you'll see an exothermic reaction and almost instant setting!) <S> Unlike hardware store epoxy, it's a viscous liquid (like varnish) rather than a paste in its just-mixed state, for 20 minutes or more at lower temperatures. <S> It's common to use chopped glass, fumed silica powder, microspheres etc as filler material, to thicken the consistency from a liquid anywhere up to a stiff paste. <S> But for a ferromagnetic material you could use any of: iron filings, iron powder, powdered ferrite, or 1mm ball bearings (not stainless steel, which is usually non-magnetic). <S> Pour into a mould to set overnight. <S> After about a week you can machine, grind, polish to final size. <S> You didn't say if you needed to charge it as a permanent magnet : that would be more difficult, though you could embed permanent magnets in the component. <A> Two part urethane molding compounds have quite low viscosity and cure quickly. <S> They are thermoset polymers so I would expect reasonable temperature performance. <S> There are suppliers such as Smooth-on in the US who have a large array of products for casting.
| Professionals in induction heating have several moldable flux concentrators available. You can get two pot epoxy embedding compounts which may be flowable enough.
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Reason for 1/2 watt resistors marked as 2 watt resistors I recently took apart a combination VCR/DVD player. Along with other questionable quality inside including a green LED display viewed through a red acrylic front panel and a DVD player assembly using only 3 brushed DC motors, I found that there were 3 mystery components marked as 27KΩ 2W resistors in a form factor that I have never seen resistors in. As you can see in the picture, I decided to cut one open and found that it was a 1/2 watt carbon composition resistor potted in a plaster-like material. It had also failed short for an unknown reason. What is the reason for potting the resistor in a larger package like this? Update: I cut up another one of these resistors in the name of science and it appears to have a metal film or wire wound resistor inside. The core is too hard to cut with diagonal cutters. <Q> You can see the white casing as a heat sink that provides better heat transport to the surrounding air. <S> That is possible because it has much higher surface area than the original resistor (from the picture I guess at least factor 5) and it is made of a material with high thermal conductivity. <S> So that really increases the wattage of the resistor. <A> For higher wattage, wire wound resistors are used in a similar but larger casing. <S> This type of construction is common as you can see here , here and here . <S> The surface area give higher wattage since it can dissipate more power. <S> The increased thermal mass gives it a higher surge rating. <S> From the pictures, it seems that the surface area has increased by much more than 4 times, making it possible to be marked as a 2W resistor. <S> Here's an image from the Xicon datasheet (2nd link): <S> And one from the Uchi datasheet (3rd link): <A> cement resistors are common design in power board. <S> Most engineer will use metel oxide film resistors under 5 watts, cement resitors above 10 watts. <S> To use a 2 watts cement resistos is not common, becuse it can not be in auto insertion process.
| The ceramic casing increases the surface area as well as the thermal mass. This is possibly a metal film resistor encased in a ceramic case, filled with high alumina cement.
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What happens if you put a capacitor in the wrong direction for a short time? I have a 6300uF capacitor rated for 50V. When I connected it to my circuit I put it in the wrong way. After a few seconds I realized my mistake and switch off the circuit. I want to know if any damage have been done to the capacitor. It is not bulging and it did not explode. Edit it is applied to a 13v 10 amp max supply. It is a aluminum Electrolytic capacitor Panasonic. Datasheet: http://industrial.panasonic.com/cdbs/www-data/pdf/ABA0000/ABA0000C121.pdf <Q> You didn't say what technology the capacitor is, but considering it's 6.3 mF, <S> 50 V, and polarized, I'll assume it's a aluminum electrolytic. <S> How badly damaged, and how irreversible the damage depends on what voltage was applied for how long. <S> A 50 V capacitor can probably take 5 V in reverse for a few seconds, and probably mostly recover when promptly forward biased. <S> The prognosis gets worse at higher voltage and longer time. <S> The insulating layer formed on the surface of the aluminum gets eaten away, so eventually there is a short. <S> You have already damaged the insulating layer somewhat. <S> It can actually heal somewhat when forward voltage is applied, so it's hard to say how bad the damage is. <S> If this is a hobby application where you can tolerate the capacitor blowing up or leaking in the future, then continue on and be more careful next time. <S> To paraphrase Dirty Harry: "Do you feel lucky? <S> Well do you, punk?" <A> I wouldn't use it even in a hobby application. <S> I had a near miss accident with a smaller cap blowing up on the table, perhaps some ten inches from where my head was. <S> The case of the capacitor flew off and got stuck in a plastic ceiling panel. <S> That cap you may have damaged is nearly not as expensive as to justify the risk. <A> If it's in a protected environment (Your application has some kind of protective case) I'd definitely use it. <S> Otherwise, I wouldn't. <S> I know it's a thrilling experience being beside a cap that can blow up in any moment, but it won't be that fun if it ends up in your eye socket :P.
| Yes, the capacitor has gotten damaged, at least somewhat. If this is a commercial application, toss the capacitor, replace it with a new one, and don't look back.
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What kind of oscilloscopes are qualified for measuring USB 3 Super Speed Signals? I understand that they probably will be very expensive. But I have little idea of what kind would qualify. How much bandwidth? At what sampling rate? Any other parameters? <Q> USB 3.0 runs the extra lanes at 5Gbps, which equates to a clock of 2.5GHz. <S> So you will probably need at least 3GHz bandwidth, at an absolute minimum . <S> Quite pricey! <S> To see anything clearly, you'll want even more bandwidth as the signals have multiple harmonics - you would need at least the 3rd at 7.5GHz and preferably the 5th harmonic at 12.5GHz to see anything even remotely square. <S> But the thing that people forget is, a scope is only as good as the probes that it connects to. <S> So not only do you need a scope with enough bandwidth, but also a probe which has the bandwidth too. <S> The signals are also differential, so a differential probe is likely to be required. <S> At high frequencies like USB 3.0 runs at, electrical signals in wires are basically EM waves trapped in a waveguide (the cable). <S> These signals are incredibly sensitive to impedance mismatches, so sticking any old probe with a long cable on it is just going to distort the signals. <S> Keeping the cable short essentially calls for an active differential probe. <S> Expect such a probe to be in the region of $7k+ just on its own! <A> For every oscilloscope manufacturer you will find some rules of thumb, like "5 times the signal rate" or "You want to see at least the 5th harmonic" <S> and it all really also depends on what you want to do with it. <S> Do you just want to try decoding it? <S> Do you want to judge the eye diagram quality with and without equalization? <S> As one of the examples, for signal quality examinations LeCroy recommends 2.5-3 times the bitrate (remember, they are kinda square waveish), which means for USB3 Super Speed which is at 4.8Gbps, an analogue bandwidth of 12-14.4 GHz which then translates to lots more sampling rate and really big chunks of money. <S> A corresponding LeCroy presentation can be found here <S> Now of course this is for measuring the signal quality, to see anything at all, you will probably get away with half the bandwidth. <A> Here is the chart. <S> Required scope <S> BW 13 Ghz
| You would need a probe which has very low capacitance and connected to the signals with as short of a run of extra cable as possible.
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Wireless power transfer through stainless steel The SS in question is 316, and is only weakly magnetic.Does anyone know whether this is possible? <Q> It's not its magnetic properties that are the problem - it will have induced into it AC eddy currents that will pretty much kill off your magnetic field. <S> (that's hertz not kilo hertz). <S> EDIT (not related to the question) - you'd be surprised about the odd magnetic properties of "mm sizes" of 316 SS at between 50 kHz and 1 MHz, specifically 300 kHz. <S> Just stuff I've observed when designing food and pharmaceutical metal detectors. <A> I think you have a reasonable chance of doing this at mains frequency. <S> 316 SS is quite resistive and almost non-magnetic at room temperature <S> so a 60Hz field should easily penetrate 12mm- skin depth at 50 or 60Hz <S> is more than 5cm. <S> There would be a lot of leakage inductance so the resulting transformer would have a lot of effective series impedance. <S> I'm visualizing something like an 'pot core' shape turned from iron or magnetically soft steel with the SS wall between the two halves. <A> This is certainly possible. <S> Since you do not list any efficiency requirements, one solution might be to drive an AC current across (not though) the stainless steel, and use a pickup inside the vessel to inductively generate power inside the vessel. <S> The coil does not touch the stainless steel wall. <S> Here I am showing the half of the AC where the current if flowing from left to right across a wall of the vessel (the arrow pointing left). <S> This generates the circular magnetic field lines around the path of the current though the stainless steel wall. <S> You would position the interior pickup coil so that these circular field lines cut though it as the get bigger and smaller (the coil is parallel with the path of the current in the wall). <S> You would pick the number of turns for the interior coil based on the volts/amps you need inside the vessel. <S> Think of it as a transformer where the primary (the stainless) has 1 turn. <S> If you need DC inside the vacuum, then you'd need to rectify and filter the power coming off the interior coil. <S> There are many ways you could potentially make this arrangement more practical, but they are dependent on the geometry of your setup. <A> There is no way you are going to move 1W of power through 12mm of SS, thin sheets are used for EMC shielding for a reason, it is decent at absorbing magnetic fields as well as electric ones. <S> Can you give more details about the device, maybe there are other methods of power transfer... <A> Would it be acceptable to replace the connector with a nickel-plated, ferrite/mild-steel 'core?' <S> If so, you could rabbet-joint & NASA low-gas-epoxy in a small diameter transformer core so that a portion of it sits flush with (or even behind, say, 1-2mm of un-removed SS) the 'device-side' surface, to closely couple the magnetic flux from your exterior coil to the base of your device, where your secondary windings are placed, to receive the power. <S> If that's not an option, then I guess there's always the "put a 300W electro-magnet on the outside of the SS & deal with the losses" approach. <S> However, I'm guessing that you most likely aren't trying to design an induction cooker that heats your whole SS surface with 299W of power. <S> Also, while not generally a 'preferred' option, it may be 'feasible' to transfer a small amount of power through 12mm of SS using ultrasound (or other physical vibrations), rather than <S> E/M coupling. <S> This would be a "far from efficient" approach, and 1W would likely be approaching an upper-limit for plausibility, but if all else fails, it could be worth consideration.
| I don't think you'd get 1 watt through 1 mm SS let alone 12 mm unless you were operating at 10 Hz
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What determines the number of pixels in front porch and back porch of VGA display? Besides this, is pixel clock dependant on the VGA resolution and referesh rate or independant of them? What does one know what duration each pixel should have? <Q> Pixel Clock is related to resolution / refresh rate. <S> You need to clock it so that for a given horizontal line, the pixels plus porch is clocked out... <S> As for the porch pixels, that is dependent on the monitor, the reason for their existence is so that while the electron beam is moving back to the beginning of a row or frame, it does not show up on the screen. <S> Basically as long as it is approximately correct it will work fine. <S> There are charts that show the timing information based on clk and resolution. <S> Here is one: http://www.epanorama.net/documents/pc/vga_timing.html <A> For example 480x640 <S> (VGA) has circa 25 MHz. <S> If you increase the resolution, you need to transfer more pixels per second to the screen -> higher frequency / pixel clock. <S> The frequency depends on other parameters, too: CVT mode <S> Reduced blanking mode Refresh rate (50/60 Hz) <S> The porches are related to pixels per line and lines per screen. <S> There is a Excel sheet from VESA to calculate VGA timings for all known resolutions - I just need to find the link again .... <S> http://www.vesa.org/vesa-standards/free-standards/ <S> Related SE questions: VESA <S> CVT Standard - How to calculate video timings? <A> The VGA system is derived from analogue monitor design. <S> In a pure analogue system there are two fixed-frequency sawtooth generators, one for the horizontal and one for the vertical. <S> These would phase-lock to the hsync and vsync signals, with a small fixed phase offset at the start. <S> That defines the "front porch" period, which is a time period rather than a pixel count. <S> On an analogue monitor there are a number of tuning knobs which let the user move the picture to fit the screen correctly: these adjust the front porch and frequency of the sawtooth generators. <S> Later monitors were capable of multiple sync frequencies and would pick one from a table by looking at the sync frequency, allowing the use of 600-visible-line modes as well as the original VGA 480-visible-line mode. <S> Fully digital flat panel monitors add more modes and an ADC stage in the process. <S> Monitors with EDID will tell you what their mode table is. <S> There's usually a few percentage variance in what will be accepted, so it need not be a precise count of pixels for the front porch and back porch. <S> Autoadjust will usually compensate. <S> You can have your own video modes, (eg PC video "mode X"). <S> If you're outside VGA timings the monitor may not bother to display it. <A> As the discussions below the other answers show, the term "pixel clock" is interpreted differently by the authors. <S> There is actually no clock transmitted via the VGA cable. <S> But, for a resolution of 640x480 <S> @ 60 Hz and using the old Safe Mode Timing (i.e. the one used before Coordinated Video Timings ), the controller must output each pixel color value for around 40 ns (see below). <S> This can be achieved with a controller clock of 25 MHz; then the controller must output a new pixel color value every clock cycle. <S> But, the controller can also use a 50 MHz clock, then it must output a new value every second clock cycle, and so on. <S> Beside the pixel color, the VGA signal also comprises of the HSYNC and VSYNC pulses. <S> The signals must be driven low (or high) for a certain amount of time, to indicate the start/end of a frame (VSYNC) and a scan line (HSYNC) within a frame. <S> Before and after these sync pulses, there is a safe margin (called porch) which separate the sync pulses from the image content. <S> During the sync and porch time, the pixel color lines should be driven low (black) to achieve a proper synchronization. <S> Catode ray tubes need these extra times, to bring the catode ray back to the left/top side. <S> But, even LCDs require at least the sync pulses and even some time to activate the next row in the pixel matrix. <S> The new Coordinated Video Timings allow a reduced vertical blanking for LCDs. <S> Let's take a look at the Safe Mode Timing for 640x480 <S> @ 60 Hz. <S> The total time of each scan line is 31.778 µs consisting of: <S> 25.442 µs for the 640 pixels in each row, 0.636 µs for the front porch, 3.813 µs for the HSYNC pulse and 1.907 µs for the back porch. <S> Thus, the controller has to output each pixel color for $$\frac{25.442 \,\mbox{µs}}{640} = <S> 39.75\,\mbox{ns}$$ <S> Thus, the controller would require a clock of at least \$1 / 39.75\,\mbox{ns} = <S> 25.157\,\mbox{MHz}\$. <S> This value is also refered as the pixel clock frequency int the VESA documents. <S> But, the controller could also use an internal clock twice as fast as mentioned above. <A> The exact timings vary somewhat and as long as you get them approximately right and try to avoid black backgrounds (see below) <S> you should generally be ok. <S> The pixel clock will obviously depend on the resoloution, higher resoloution means both more lines and more pixels to fit into each line. <S> On old CRT monitors pixel perfect match wasn't needed. <S> There were manual controls to put the picture in the right place and these would have to be adjusted for each resolution. <S> Some monitors had the ability to remember two or more sets of settings for different resoloutions. <S> Modern LCDs need to turn the VGA signal back into a digital signal. <S> To do this without creating a blurry mess they need to determine discrete pixel positions and they will contain auto-adjustment systems to try and work out the timing of the incoming signal. <S> Sometimes they guess wrong, especially if fed with a black screen or a screen with black borers.
| Pixel clock is mainly dependent on the resolution.
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Rectification of Micro DC voltage levels? I am building a payload for a high-altitude Balloon launch in the club I started which will be measuring DC voltages (electric field) at different altitudes using a wire dipole or two metal plate formation. I obviously need a low noise amplifier to even register any data in the logger, and we are not worried about polarity, only magnitude. To do this, I am thinking an instrumentation amplifier fed from the metal plates to the data-logging device, but I am facing a problem, how can I assure the DC amplifier is only measuring positive values? Even some good diodes have a few hundred millivolt drop, which would not be suitable for the experiment. A transformer would not work because this is only DC we are working with (even if it was AC, the high input impedance requirement would not be fulfilled). Any ideas on rectification or ways to amplify the small DC signals which we can't be 100 percent sure the polarity is constant. <Q> That won't work. <S> It may detect ion flows and AC fields above a certain frequency, but it can't detect DC fields. <S> After all, the plates constitute a coupling capacitor in series with the fields you want to detect. <S> It's a high-pass filter. <S> The usual instrument for this is called a "field mill," where the detector plate is placed close behind a grounded metal propeller or rotating "sector disk." <S> The rotating disk chops any DC e-fields into low-freq AC, which is easily amplified/rectified/A-toD'd etc. <S> Perhaps even use low-pass filtering and a synchronous detector to reject any unwanted AC. <S> DIY e-field sensor field mill from Scientific American: <S> Jul '99 <S> The Amateur Scientist <S> More electrometer and Field Mill links Besides "field mill," also search for "Electrometer" Antenna without the field mill: if your detector plate is 10 picofarad, and the amplifier's input is MOSFET with 100 giga-ohm Z-inp, then you've formed a high-pass filter which rejects all DC below ~1Hz (since RC is 1e11ohm <S> * 1e-11farad) <S> It's possible to use a much higher-Z amplifier, plus a periodic shorting relay that "resets" the detector's input before taking data. <S> But then usually you'll run into unknown humidity leakage across plastic insulators, and a (drifting) high-pass period of ? <S> tens? of seconds. <S> Better to just add a field mill, and measure true DC fields. <S> (edit) PS. <S> Here's something I've not tried: just build a non-fieldmill e-field antenna with well-known RC time constant of perhaps a few seconds, use high-res A/D input, then massage the numbers to cancel out the RC and restore the DC value. <S> Of course this includes a stage of integration, so any slight errors in zero-adjust would produce a constantly increasing drift. <S> The amplifier's DC-zero would need to be extremely stable across the environmental temperature changes involved. <A> I don't know anything about field measurement, but in case someone stumbles here looking to rectify and measure peak of small voltages, here's my suggestion: <S> simulate this circuit – <S> Schematic created using CircuitLab Note 1: Peak detection speed of response will be limited by R3 and R8 and also depend on the size of C1. <S> Note 2: Detected peak will decay with characteristic time (tau) = <S> 5050(ohm) <S> x 100(uF) = 0.505(seconds) in this example. <S> (5050 is parallel combination of serial combinations of (R1+R2) || (R6+R7)), 100uF is C1.Hope <S> this helps non field measuring folk. <S> BTW V1 is also amplified 100 X. If you want to minimize error due to the input bias current add a 99 ohm resistor in series to the positive inputs of OA1 and OA4. <S> (or for different amplifications add R1||R2). <A> One problem you will have is charging of the payload. <S> This means that your payload chassis will not have a net charge of 0. <S> You can either, measure the charging or develop a way to keep the charge of the payload at 0.
| Use an INA116, it's a great electrometer if you make sure your dynamic range of the sensor is within the rails.
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LED: should I smooth the current with a capacitor To control the brightness of a LED often pwm is directly used as input to the LED. Does this on/off turning by the pwm has any negative effect on the live expectation of the LED? Would it be better for the MTTF (mean time to failure) to smoothen the current by adding a capacitor? <Q> Its brightness is not controlled by voltage, it is controlled by current, and the amount of time it is switched on (i.e. the PWM duty cycle). <S> The capacitor might reduce the 'smoothed' PWM voltage below the minimum forward voltage, so the LED would no longer be visible, even though it would be visible using exactly the same PWM signal directly (without the capacitor). <S> So it would reduce the brightness range over which the LED can be controlled. <S> AFAIK, the bigger killer of LEDs is heat leading to a significant temperature rise, and not switching. <S> Typically we want to drive an LED with a constant current (or something near, e.g. a resistor), so that it is protected from too much heat leading to temperature rise and permanent damage. <S> Edit: <S> Depending on how the capacitor is connected, a capacitor may actually reduce the effectiveness of the constant current circuitry. <A> Would it be better for the MTTF (mean time to failure) to smoothen the current by adding a capacitor? <S> This is usually done with an inductor: <S> - The inductor will smooth the current into the LED. <S> Also note that average current is maintained by the sense resistor attached to the chip in the diagram. <S> A capacitor will not smooth the current into an LED unless there is a series component like a resistor or inductor. <S> If you put raw PWM voltages across an LED it is likely to destroy it. <S> Observe the LED spec regards current. <S> Here's other examples: - This one doesn't use an inductor but relies on the transistor arrrangement to regulate max current into the LED: - LED forward conduction characteristics: - <S> At 2V across this "typical" LED <S> the forward current is 20 mA. <S> With only 1.7 volts applied there's hardly any current and the LED will be very dim. <S> If you applied 2.5 volts, the current is off the scale and the LED is damaged. <S> But some LEDs are designed for 1A <S> I hear someone say and that is true but applying a few tens of millivolts more than the recommended value will kill it nonetheless. <S> The LED brightness should always be controlled by current. <A> It is unnecessary to smooth the current. <S> If you are PWMming at a rate faster than the eye can follow, any heat cycling effects will be negligible. <S> Smoothing the current might give a very small (but unnoticeable) increase in brightness because an LEDs emissivity vs. current is not perfectly linear, especially at higher currents. <S> If you are adjusting the PWM to control brightness, then it's best NOT to filter it because a) effective brightness will not be linear with PWM duty cycle, and b) LED colour changes slightly with current, so you would get a colour shift also. <S> This would be most noticeable in white LEDs. <S> TVs and displays that use LEDs all PWMthem without filtering at 50/60 Hz or faster. <S> LEDs have some internal series resistance, and overall power efficiency would be higher by filtering the current, but if you are regulating current with an external resistor or current source, there is no difference (same average current consumed from supply) <S> If you are controlling a single LED (or single string) from a DC/DC converter, the DC/DC will smooth the current anyway. <S> This is optimal for efficiency. <S> TV screens use DC/DC converters to generate a sufficient voltage, but then PWM the current to maintain colour and brightness accuracy.
| One potential problem with using a capacitor to 'smooth' a PWM voltage for the LED, is the LED has a minimum forward voltage before it turns on sufficient to be visible.
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Why would the transistor not switch? I was reading an example from a text book. And for this circuit above the author claims when R3 is less than 100 ohm Q3 will not switch. I couldn't figure out the "reason" why. But I verified with LTSpice the author is right. He just doesn't explain the reason. If lets say R3 is close to zero when Q2 on, why wouldn't Q3 also switch on? <Q> For Q3 to switch on, the voltage drop between its base and emitter must be about 0.6 V, which means that the same voltage must be dropped over R3, which means that the current flowing through R3 must be at least I3 = 0.6V / R3. <S> When there is less current flowing through R3, the voltage drop over R3 is smaller than Q3's minimal voltage drop, and Q3 will stay off. <S> For R3 = 100 Ω, the required current I3 would be 6 mA.However, in this circuit, the current through both R3 and Q3 is also limited by R2: a current of 6 mA would result in a voltage drop of 19.8 V over R2, which is not possible with a 15 V supply. <S> The largest possible voltage drop over R2 happens when Q2 is saturated, and is about 14 V, which results in a maximum possible current of about 14V/3.3kΩ = 4.2 mA. <A> PNP transistors turn on when \$V_{EB}\$ is large enough. <S> Intuitively, \$V_{EB}\$ is the same as the voltage across \$R_3\$. Since <S> \$R_2\$ and \$R_3\$ are roughly a voltage divider (there's very little base current in \$Q_3\$), the voltage is$$V_{EB} \approx \frac{R_3}{R_2 + R_3} \cdot \text{15 V} \approx <S> \frac{R_3}{R_2} \cdot \text{15 V}$$if <S> \$R_3 << R_2\$. <S> Clearly, when the fraction \$R_3 / R_2\$ is too small, the transistor can't turn on. <A> Since you're confused about the turn-on behavior of Q3 relative to R3 consider the equivalent circuit consisting only of the essential resistor divider (R3 and R2) and the base-emitter junction of Q3: <S> I'm varying here R3 over time from 0 to 1K. <S> The BE diode turns at about 0.65V which corresponds to 150 ohms for R3. <S> This is easily verified as 15V*150/(3300+ <S> 150)=0.65V. <S> Since the current through a diode that's turned on has an exponential variation with the voltage across it (Shockley's equation), and since the current here is limited by R2, the BE voltage will be roughly constant once the diode is on. <S> Once the junction is on, Vbe actually varies logarithmically with a diode current that has an upper bound (imposed by R2)... which is to say not much. <S> Note that the V(BE) curve (red trace) has a sharper turn than the I(BE) current (magenta)... because of the logarithmic relationship it has with the diode current. <S> Before the diode turns on, the BE voltage is a linear function of R3 since it's just a resitive divider with R2. <S> Also I(R2) doesn't vary a whole lot even before the diode turns on because the turn-on point is only at about R3=4.5% of the value of R2. <S> But on a separate plot of I(R2) [in the lower pane] you can see that's "even more constant" past the turn-on point of the diode. <S> So this verifies the usual assumption that Vbe is constant (and consequently so is I(R2) here) <S> once the BE junction is actually on. <S> Before that there's no restriction on what Vbe it can be as you can see; it only depends on the value of R3 when the diode is off. <A> Consider the voltage across a diode and the current that flows. <S> Below are the curves for an old germanium diode (1N34A) and a silicon diode (1N914): - Concentrate on the silicon diode (1N914). <S> With 0.6 volts across it, the current is about 0.6mA. <S> Now drop that voltage to 0.4 volts. <S> The current falls to 10 uA and, with 0.2 volts across it <S> the current is about 100 nA. <S> Now, the base-emitter junction in a BJT is a forward biased diode. <S> The forward biasing comes from the voltage you place across it <S> and this is usually via a biasing resistor. <S> In your circuit, R2 and the power supply voltage define the current that can jointly flow into the base and into R3. <S> When R2 supplies a decent amount of current, most of it flows thru the base emitter junction because you are on that part of the diode curve and that part of the diode curve has a dynamic resistance that is much smaller than R3. <S> As base-emitter voltage lowers, its dynamic resistance gets higher and R3 starts to become the "path" to which most of the current from R2 flows. <S> Dynamic resistance is the small change in applied voltage divided by the change in current. <S> You could look at the diode graph above and pick some points: - <S> At 0.60 volts the current is possibly 600 uA <S> At 0.62 volts the current is about 1000 uA <S> Dynamic resistance would be 20mV/200uA = 100 ohms <S> At 0.40 volts the current is about 10 uA <S> At 0.42 volts the current is about 11 uA <S> Dynamic resistance would be 20mV/1uA = 20 kohms. <S> So, when R3 lowers it becomes more dominant that the base emitter junction and rapidly the junction current falls away. <S> Given that we can approximate transistor action to a device with current gain, lowering R3 beyond a certain point means a rapidly falling collector current and, in effect, the transistor is regarded as turned-off. <A> A transistor needs about 0.7v VBE to start conducting. <S> As you have the benefit of a simulator there, experiment with different R2/R3 values and look at the voltage developed across R3, and whether the transistor turns on. <S> As to <S> why it's 0.7v, you need semiconductor physics! <A> Well, I think all the complicated answers have been given, but for my two cents: anything below 150 ohms "shorts out" the base to emitter junction
| When you make \$R_3\$ too small, there isn't enough voltage across the transistor's EB junction for it to turn on.
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Plane pull-back and board keep-out I create one 4-layer board using Altium's PCB Wizard . By default, it set the plane pullback to 0.5mm, the keep-out area is 1.3mm from the board edge. That is, the top/bottom layer keep-out region is less then the power/ground plane, as below, the red layer are the power plane, and the blue one is the top layer pour. Is it a good design considering EMC? <Q> There is no problem with power and ground planes extending a little further to the edge of the board than other traces. <S> The important thing is that all copper is far enough from edges for whatever process you are using. <S> The ground plane should extent at least as far as any signal, but the small distances you are talking about wouldn't make a significant difference either way. <A> The default values sound reasonable to me unless you have some kind of special requirements. <S> The 0.5mm pullback on the planes keeps the copper from showing up on the edges where the boards are routed (as in cut with a router bit) or perhaps pushed up against a metal part (though in the latter case I would add a bit more clearance for luck in case <S> the PCB manufacturer doesn't get the cutting registration perfect). <A> The frequency of a quarter-wavelength of 1.5mm (5.0mm) is \$\lambda=\frac{v}{f}\$ <S> \$5mm=\frac{3\times 10 <S> ^8}{f}\$ \$f <S> = <S> \frac{3\times <S> 10^8}{5mm}\$ <S> \$f = <S> 60GHz\$ <S> so for 1.5mm of exposed power plane, it will couple electrically (behave like an antenna) at 60GHz.
| You don't want the signal lines to come too close to the edge of the ground/power planes and you definitely don't want them to cross over or they will emit EMI.
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how would inductor react if I inject current impulse into parallel LC tank? What would happen if I inject current impulse into parallel LC tank? How would current flowing through inductor look like over time? If we look at the delta function in the S-domain (laplace transform), energy is uniformly distributed over s-domain. This means, delta function cannot be just treated as high frequency signal. This means, some of current impulse will flow through capacitor, and some will flow through inductor. Since they are LC tanks (let's assume they are ideal LC tanks), then oscillation will occur. This case, which will react to this oscillation first: inductor? or capacitor? would inductor start dumping current into capacitor first? or would capacitor start dumping current to inductor first? I want to know what you guys think about this problem Thanks, <Q> Injecting a current impulse onto a L-C parallel tank is like hitting a bell with a hammer. <S> If nothing was going on before (voltages and currents were 0), then this will start the tank ringing. <S> The inductor can't change it's current instantly, but the capacitor can <S> (I'm assuming we're talking theoretical ideal components here). <S> The impulse will have no immediate affect on the inductor's state. <S> A current impulse on a capacitor causes a step change in voltage. <S> Depending on whether this step adds or subtracts from the existing voltage, it either adds or subtracts energy from the system. <S> For example, if you happen to hit the system with +1 V when the capacitor has -1 V on it and the current is 0, then you've just removed all the energy from the system and now everything will just sit there at 0. <S> On the other hand, if the voltage was +1 V and the current 0, you've now doubled the voltage and quadrupled the energy of the system. <S> The amplitudes of the voltage and current sines will be twice what they were before. <A> If it's an impulse, then it will all flow in the capacitor. <S> The indctor cannot change its current instantaneously. <S> Best way to characterize a current impulse is the total charge dumped (integral of current.dt). <S> Basically, you end up the same as if you connected a charged capacitor (ith Q = dumped charge) across an inductor -- that starts the ringing. <A> What would happen if I inject current impulse into parallel LC tank? <S> How would current flowing through inductor look like over time? <S> At this point (and based on the current being an impulse) I can assume that following that impulse there will be an open circuit from the driving source. <S> I can make this assumption because you said "inject current impulse" and this assumes that the only source is that impulse of current and thereafter there exists an open circuit from the source. <S> This means, some of current impulse will flow through capacitor, and some will flow through inductor. <S> No, absolutely wrong. <S> Current will only flow into the capacitor because an inductor will reject changes in current in those few sub-fempto seconds and the capacitor will take ALL the current and form a voltage across its plates. <S> So, the cap will take the energy from the impulse and become charged to some voltage. <S> The inductor will barely have noticed anything had happened if you want to put humanesque viewpoint on things. <S> Shortly afterwards (a few more fempto seconds) <S> the source is out of the equation and the charge in the capacitor becomes sapped by the inductor and forever, until the end of time, there will be a perfect oscillating intechange of energy between capacitor and inductor forming a sinewave of frequency \$\dfrac{1}{2\pi\sqrt{LC}}\$ <A> Assuming a unit impulse current is applied at \$t=0\$, Laplace transform analysis gives: \$I_C=-\omega \:sin(\omega t)\$ \$I_L=\omega \:sin(\omega <S> t)\$ and voltage across combination: <S> \$V=\frac{1}{C}\:cos(\omega t)\$ <S> where \$\omega = <S> \frac{1}{\sqrt{LC}}\$ <S> L and C currents are sinusoidal and have 180deg phase difference, so there's zero overall current flowing into the combination, but there's a sinusoidal current of \$\omega\:sin(\omega t)\$ circulating through L and C. Also <S> , there's a (co-)sinusoidal voltage across the combination. <S> At \$t=0\$, the capacitor is charged instantaneously to \$V=\frac{1}{C}\$ by the unit impulse current, hence the cosine voltage function.
| If you injected a current impulse into a parallel capacitor and inductor then all that impulse of current will flow through the capacitor. None will flow thru the inductor but, gradually things will change.
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Why is XNOR gate not referred to as an "Equality" gate? I am a programmer and I was thinking about some logic I was just writing and the logical statement started out as a negated exclusive or, but this simplifies to a logical equality check, e.g. var x = a == b . That is to say, we can express this as x <- a XNOR b . Since I was looking at the XOR page i found that there is an XNOR gate which is just an XOR gate negated. It got me to wondering because XNOR seems like such a convoluted way to refer to the concept of what this gate does. I wonder if this is just how history played out, or if there is something subtle that I'm missing. <Q> I think that's a valid observation. <S> The reason it's called an XNOR is that all logic gates are based on boolean algebra. <S> The boolean operations of conjunction, disjunction and negation map isometrically to AND, OR and NOT. <S> Combining negation with conjunction and disjunction gives you NAND and NOR and exclusive disjunction gives you XOR. <S> Naturally then, when you add negation, you get XNOR (which is simply easier to say than NXOR). <S> The fact that XOR performs an inequality operation and XNOR performs equality operation is a by-product, but a valid one nonetheless. <S> Note that it is no more remarkable <S> then the fact that AND is modulo two multiplication and XOR is modulo two addition. <S> You're free to use the representation that is most convenient to the task at hand. <S> If you're programming, then converting your result into arithmetic operators makes sense. <A> I disagree with the accepted answer that "XNOR performs equality operation is a by-product". <S> Although you cannot expect more than uninformed upvotes for this statement on a EE site (you should have really asked on Math. <S> SE <S> if you expected that), in math or logic contexts it is more likely to be called <S> a biconditional (that's because equality in a logic context is equivalent with if and only if ); You can find that in several math or CS textbooks which don't even mention xnor: https://books.google.com/books?id=M5dBBAAAQBAJ&pg=PA73 https://books.google.com/books?id=yJIMx9nXB6kC&pg=PA13 https://books.google.com/books?id=6cMSAAAAQBAJ&pg=PA40 https://books.google.com/books?id=FS-sCQAAQBAJ&pg=PA15 https://books.google.com/books?id=jgJQce_GRyEC&pg=PA48 Others call it just equivalence : https://books.google.com/books?id=TQ1n03kEBOkC&pg=PA8 https://books.google.com/books?id=UQ7NSn4UOAsC&pg=PA160 <S> It's usually only when you get to circuit engineering books that you usually start to encounter the xnor <S> terminology: <S> https://books.google.com/books?id=3zcgIKPl8L0C&pg=PA130 <S> https://books.google.com/books?id=XQjVBQAAQBAJ&pg=PA102 <S> https://books.google.com/books?id=-ZAccwyQeXMC&pg=PA81 <S> https://books.google.com/books?id=rguQ-SNgkNIC&pg=PA93 <S> Some of these engineering books call it concidence [gate] as well (or say it implements the coincidence function), although they have a preference for xnor, no doubt. <S> https://books.google.com/books?id=sZYJAAAAQBAJ&pg=PT204 <S> https://books.google.com/books?id=o7enSwSVvgYC&pg=PA131 <S> https://books.google.com/books?id=o7enSwSVvgYC&pg=PA97 <S> And some engineering books call it equivalence in addition to xnor <S> https://books.google.com/books?id=eQrlBwAAQBAJ&pg=PA225 <S> https://books.google.com/books?id=QypINJ4oRI8C&pg=PA102 <S> https://books.google.com/books?id=1msXLZ360m0C&pg=PA67 <S> So it depends who (or where) you ask. <S> Furthermore, there are introductory logic texts that don't even mention xor [thus calling something xnor would be a big huh for the students]; for example Suppes explicitly refutes the needto introduce a symbol for xor in his introductory logic textbook . <S> But it's hard to discuss logic without ever getting to equivalence (iff aka biconditional). <S> As an aside, perhaps if Latin were Suppes' [or other logician's] first language, he/they would be more inclined to accept [something like] xor, because (quoting from Copi's textbook): "Although disjunctions are expressed ambiguously in English, they are unambiguous in Latin. ... <S> The Latin word "vel" expresses weak or inclusive disjunction, and the Latin word "aut" corresponds to the word "or" in its strong or exclusive sense. <S> " This uniform interpretation of Latin is disputed by others though because aut in negated sentences like neo timebat tribunos aut plebes <S> "no one feared the magistrates [x]or the mob" doesn't sound genuine with aut interpreted as xor instead of or because then you can read that as allowing for "everyone feared both the magistrates and the mob" as being possibly true. <A> It would make more sense to have the biconditional gate be the normal one and the XOR be the negated one. <S> That way we could have IFF and NIFF gates. <S> However, what I wonder is why we go with XNOR instead of NXOR since the latter is much easier to say (ex-nor vs. nexor) and more descriptive of the situation. <S> Exclusive nor by this description is equivalent to an exclusive or, which would imply that the gate isn't any different. <A> Fundamentally, the answer to your question is a historical one and not a mathematical one. <S> If George Boole thought the concept of "fundamental logical operations" was best described by "negation", "conjunction" and "If and only if" (equality) and everything else could be described as combinations of those fundamentals then the term for XNOR would be NIFF in modern Boolean algebra (since that is a functionally complete set). <S> Because, instead, George Boole felt that "negation", "conjunction" and "disjunction" were a fundamental set we derive terms from that set. <S> But another issue is the confusion caused by the name XNOR being used instead of NXOR since XNOR is calculated as NOT(a XOR b). <S> People often say it's <S> because XNOR is easier to say, I'm not sure, "nexor" is just fine for me. <S> But, more confusingly, people also sometimes say that it is more congruent with XOR and that we can think of NOR as fundamental, meaning "neither" in English (people don't always make this association). <S> And XNOR means "exclusive neither". <S> I have no idea how you apply the word "exclusive" to the word "neither" using either dictionary definitions of the adjective form of "exclusive". <S> This is an unfortunate confusion and I wish we used NXOR or just updated Boolean Algebra to add IFF and NIFF so that the derivation of the terms were congruent with NAND and NOR.
| I haven't found this written explicitly somewhere, but I think the established symbol for the xnor gate being generally used only in circuit contexts and being absent in more abstract math/logic contexts facilitates this terminology divergence.
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P-ch Mosfet behavior During Drain Load What will happen if the drain voltage in the below image increases up to 200V? (the Mosfet could be IRF9640 or could be p-ch mosfet in IRF7105) I want to know if the Mosfet temperature goes up and also if the Drain current is almost zero. <Q> It will heat up, and if you indeed manage to get it up to 200V <S> it will probably violently blow. <S> The internal body diode is forward based and causes excessive current to flow. <S> It'll probably be destroyed long before you actually reach 200V. <A> As others have said, for positive voltage, you cannot get to 200V without total destruction. <S> But if you meant for V Drain to be -200V, that should be OK for the IRF9640, which is rated to -200V. FET will stay off. <S> But the IRF7105 is only rated for 25V. <S> So it will fail due to Vds breakdown <S> if you try to go to -200V. Also, the IRF7105 is a dual FET with a P-channel and N-channel in the same package. <S> It is kind of strange that you say it could be either one when they are so very different in specifications and package. <A> This is an insane configuration, and the FET will die at almost any voltage. <S> The body diode is forward biased and will always conduct. <S> Depending on the heat sink, the lethal voltage will probably be in the range of 1 to 5 volts. <S> If the voltage on the drain is negative, and the gate is positive, the FET will not conduct at all.
| In theory, the drain current will be extremely small until the breakdown voltage of the FET is reached.
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Why can't I measure resistance when there is current on the potentiometer? Probably a very basic question but I have no idea why this happens. I have a 1K potentiometer. When I measure the resistance over its legs when it is not connected to anything, the results are as expected and alter expectedly when I move the knob. However, when I connect the pot to a 9V battery and try to measure the pot's resistance, I don't get any readings. What is the reason for this behavior? <Q> Because your multimeter can't measure resistance. <S> So it applies a known current, measures the resulting voltage, and computes the resistance from that. <S> 1 <S> So when you're applying an external current to the potentiometer you are upsetting the meter's procedure, and the resulting voltage is probably outside the measurement range. <S> 1 <S> Unless it's really old . <S> In which case it applies a fixed voltage, measures the current, and lets you read the resistance off an inverse scale. <A> Your meter measures resistance by injecting a small [voltage or current] and measures the resulting [current or voltage.] <S> That is fundamentally incompatible with having a voltage applied across the component by something else, such as your battery. <S> You could measure the current supplied by your 9V battery and deduce the resistance as R = <S> V / <S> I <A> simulate this circuit – Schematic created using CircuitLab <S> Switch your meter to mA range and connect in series with the battery. <S> I recommend that you add a series resistor to limit the current to protect the battery and meter. <S> 33 Ω will limit the current to about 1/4 A with the pot turned to zero and the LED failed short-circuit. <S> Be careful with your potentiometer. <S> A standard carbon-film pot might be rated at 0.25 W but that's the rating for the whole track. <S> In the case of your 1 kΩ pot max current would be $$R = \frac{0.25 W}{1000 Ω} = <S> 0.25 <S> mA$$ <S> You can see from this that a 0.25 W potentiometer might not last too long. <S> Be careful with your multimeter. <S> As others have pointed out, when measuring resistance your meter applies a small current from the internal battery to the resistor being tested. <S> The voltmeter is usually on a sensitive range because the maximum voltage will be limited to a volt or so. <S> simulate this circuit Note that the voltmeter has a very high internal resistance and <S> 'all' of the current will go to Rtest. <S> A little goes to the meter but in the case of a digital meter the input impedance is usually about 10 MΩ <S> so it's tiny.
| If you were to connect the meter, on resistance range, to a resistor with more than a few volts across it you might damage the meter.
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Connecting a RJ45 MagJack I've made a few things now using a RJ45 jack and a separate pulse transformer. To make the design smaller i'm looking to use a magjak or a RJ45 connector with the pulse transformer built in. The jack has the center tap pins tied together to a single pin. Does this mean that i need to tie the center pins together on the device and hook them to the single pin on the jack or is there some other connection scheme i'm not aware of? <Q> Connect the CT pin to the system power (e.g. 3V3) and place a 100nF cap as close to the CT pin as you can. <S> Here's a sample design. <S> On a different connector, you may find two CT pins, so you'll need to use one cap for each pin. <A> Now that it has become clear that you are using a Lantronix module , as opposed to a PHY chip , the easiest solution is to find a MagJack that has both CT pins broken out, for example: ARJC02-111008B <S> (just one of the cheaper connectors on Digikey). <S> I also mentioned in the comments that you could contact Lantronix support or see where those lines are connected on the module. <S> While this will not help you to get the job done quicker, it may give you some good experience for your future projects. <A> The common mode level for the differential TX/RX pairs may be different on some phy's. <S> I would not risk such an issue and find another ethernet jack with separate CT pins, so you can support separate common mode levels for TX and RX. <S> However, you may find that they will tell you the same conclusion. <S> After all, they broke out the pins separately, presumably so they can swap phy's on their modules while staying compatible with client's hardware.
| You could also contact Lantronix and ask if you can connect 1 CT to both TXCT and RXCT.
|
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