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You can measure a bunch of your resistors and find some that are close to each other.
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# ? Nov 2, 2012 04:02 |
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# ? May 9, 2024 00:38 |
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So I've started drawing European resistors instead of American ones because it's just so much easier. Especially for these resistor-solving questions in my electric circuits class. Seriously why would anyone build these nutty circuits???
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# ? Nov 2, 2012 07:26 |
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BattleMaster posted:So I've started drawing European resistors instead of American ones because it's just so much easier. Especially for these resistor-solving questions in my electric circuits class. Seriously why would anyone build these nutty circuits??? The American resistor drawing is literally just squiggling your pencil, you can do it without even lifting your hand. Laziest symbol ever. To answer your question though, it's to get you good at the process, which is useful because complex impedances combine in pretty much the same way. And if you're doing frequency-dependent stuff, you'll need to take stray impedances into account. Basically, every circuit you build is also accidentally a filter and sometimes you need to account for that.
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# ? Nov 2, 2012 07:34 |
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I don't feel good about doing squiggles. I'm a neat freak.
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# ? Nov 2, 2012 07:45 |
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Has anyone done any Blackfin development? I saw that it was discussed several years ago (this thread is old!). I've been working with FPGAs all this time, and I see that it would be cheaper and use up less board space to go with a Blackfin than an FPGA in my project. I would like to play around with one, but I don't know where to start.
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# ? Nov 2, 2012 08:18 |
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Arcsech posted:The American resistor drawing is literally just squiggling your pencil, you can do it without even lifting your hand. Laziest symbol ever. But the symbol for inductor is also just squiggling your pencil. This way you can't confuse them.
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# ? Nov 2, 2012 11:41 |
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tonberrytoby posted:But the symbol for inductor is also just squiggling your pencil. This way you can't confuse them. Well it all depends on your definition of "squiggling" and how good you are at it I actually don't think I've ever seen a student with such awful handwriting that I confused a resistor for an inductor, but I don't doubt it's possible. Krenzo posted:Has anyone done any Blackfin development? I saw that it was discussed several years ago (this thread is old!). I've been working with FPGAs all this time, and I see that it would be cheaper and use up less board space to go with a Blackfin than an FPGA in my project. I would like to play around with one, but I don't know where to start. I've played with their predecessor, the SHARC. Have you just been using DSP slices / MACs / etc in FPGA fabric up to this point for your projects? I forget if Blackfin is a MCU strapped to a DSP or vice-versa; they might be equally emphasized, now that I think about it. Either way, I think ADI makes some relatively affordable starter kits for them.
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# ? Nov 2, 2012 15:17 |
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movax posted:I've played with their predecessor, the SHARC. Have you just been using DSP slices / MACs / etc in FPGA fabric up to this point for your projects? I forget if Blackfin is a MCU strapped to a DSP or vice-versa; they might be equally emphasized, now that I think about it. Right now, I'm just capturing data with my FPGA and doing my DSP calculations on my PC. The calculations will eventually be moved to the FPGA, but I haven't done it yet. I'm currently designing a PCB to integrate all of my parts and the FPGA on to one board. The cheapest board I saw is this, but I'm not sure if it works with the open source tools that are available for the Blackfin: http://www.analog.com/en/evaluation/sdp-b/eb.html
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# ? Nov 2, 2012 20:10 |
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Base Emitter posted:Sorry if I was unclear. You should be able to make this work with the bits you've got... Assuming the method proves viable, the eventual goal for this circuit is to install some strain gauges under brew pots to determine how much fluid is in them (based on the specific gravity of the fluid and tare weights). Also because I think strain gauges are cool and I've been wanting to learn how to use op amps so it seemed like a good project to learn with. I find having a practical application to work towards makes it much more interesting to learn.
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# ? Nov 3, 2012 00:37 |
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CapnBry posted:Assuming the method proves viable, the eventual goal for this circuit is to install some strain gauges under brew pots to determine how much fluid is in them (based on the specific gravity of the fluid and tare weights). Also because I think strain gauges are cool and I've been wanting to learn how to use op amps so it seemed like a good project to learn with. I find having a practical application to work towards makes it much more interesting to learn. What you really want is an instrumentation amp. Something like an LTC1100, AD8230, or INA333. They are essentially perfect for Wheatstone bridge type sensors (like a strain gage). You can make do with multiple opamps and discrete components, but trying to beat IC designers in performance is a fool's game. They can control more parameters than the DIYer could ever hope to.
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# ? Nov 3, 2012 01:15 |
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Alright goons, I need your help. I was disassembling a phone charger (raspberry pi related project) and I nicked a 470uF capacitor with the dremel. I took off a bit of the outside rubberized coating and put a tiny gouge into the metal housing. I'm certain I did not penetrate into the capacitor proper. Can I still use this or is the cap compromised?
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# ? Nov 3, 2012 20:25 |
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I think it'll be alright as long as you didn't pierce the metal housing, at least if it's on the low voltage side which I imagine it is based on the size. If it's a 450V cap I can't say.
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# ? Nov 3, 2012 20:36 |
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yeah seems to be on the low voltage side. I'll solder everything up and let it run, I guess. Hopefully doesn't take out anything else if it pops. edit: Seems to be working so far. Have to run full-load tests yet. autism ZX spectrum fucked around with this message at 21:45 on Nov 3, 2012 |
# ? Nov 3, 2012 20:45 |
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I'm doing a lab design project for school and I think I have a working solution but there is one thing about my solution that just seems weird to me. This is my amplifier design: And this is the waveform for the input and output: Why is my output sinusoid so much "wider" than my input? No matter what kind of scaling I use it always seems to be wider at the top and narrow at the bottom of the waveform. My only two theories are that it's distortion from taking a 10mV input and blowing it up to over 1v and the gain is just really high. The second is that it is something to do with the frequency response and capacitors? We haven't talked about transistor frequency response and the only thing I really know about the capacitors in the amplifier is to act as DC blockers for the output and to change the AC analysis and the gain in the case of the emitter resistor.
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# ? Nov 4, 2012 03:49 |
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Eyeballing it and not doing any circuit analysis, I'd say you've got distortion, basically soft clipping, at the upper half of your output voltage range. It looks wide because the distortion isn't symmetric for the upper and lower half of the waveform. Try reducing the gain to see if you get less distortion. You might even try increasing the supply voltage from 8 to say 12V just to see if the output eyeballs better, just because it will give you more headroom. If you have a sinusoidal input, then rolloff from capacitors (which are linear) should only affect the gain; the waveform would only be distorted for nonsinusoidal inputs. Transistors do have frequency limits, but 200uS period is only 5KHz, nowhere near the limit (which is somewhere in MHz). (Also it looks like your input is 40 mV peak to peak, or 14 mV RMS, in case you were really expecting it to be lower.)
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# ? Nov 4, 2012 04:09 |
Base Emitter posted:Eyeballing it and not doing any circuit analysis, I'd say you've got distortion, basically soft clipping, at the upper half of your output voltage range. It looks wide because the distortion isn't symmetric for the upper and lower half of the waveform. It looks like classic clipping to me as well. Build it, run a guitar through it, come back with a trip report. EDIT: VVVVV Grrr, doing the math by pen and paper, it's been a while since I've analyzed something this complicated. I suspected the AC gain was too high with the whole resistance bypassed but I didn't want to say anything until I had finished. I need to do these sorts of circuits more often. Delta-Wye fucked around with this message at 05:45 on Nov 4, 2012 |
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# ? Nov 4, 2012 04:42 |
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Add resistance in series with the 100uF capacitor for the 100:1 ratio you want (160 ohms, basically). It's still distorted as hell but it starts looking sinusoidal, at least. I can't really explain why your model distorts like that, but basically running a CE amplifier wide open doesn't work well in my experience. I managed to bring distortion down even further by upping quiescent current.
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# ? Nov 4, 2012 05:33 |
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As others have noted, your amplifier is distorting. I believe this is due to too much gain in the input stage. Check the values of the emitter resistor and the compensation capacitor C2. The 100uF is a very high value and at the frequency of interest (5kHz) is essentially shorting out Re. With no emitter degeneration, the gain of the circuit will be very high. (Not to mention unpredictable since it's heavily dependent on the beta of the transistor.) One reason you have the emitter resistor is to make the circuit insensitive to the transistor characteristics at the expense of overall gain. Also I have to wonder if you might be better served by an op-amp solution. What goal are you trying to achieve?
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# ? Nov 4, 2012 14:45 |
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Boy With Stick posted:As others have noted, your amplifier is distorting. I believe this is due to too much gain in the input stage. Check the values of the emitter resistor and the compensation capacitor C2. The 100uF is a very high value and at the frequency of interest (5kHz) is essentially shorting out Re. With no emitter degeneration, the gain of the circuit will be very high. (Not to mention unpredictable since it's heavily dependent on the beta of the transistor.) One reason you have the emitter resistor is to make the circuit insensitive to the transistor characteristics at the expense of overall gain. Well, the original project was that our professor gave us a few goals to achieve in designing the amplifier. It needs to have a gain >= 50, input resistance >= 15kohm, output resistance < 100 ohm and Vcc no greater than 9v. We could use a 2 or 3 stage so I chose the two stage and derived all the equations for the gain, input resistance and output resistance as well as the transistor characteristics to make sure it is on and the DC load line. My only design choice wrt the DC load line was to center the Q point on the line to reduce the amount of clipping, but I didn't go further and get the AC and all the rest, I didn't think that would be necessary. He didn't give us any values at all except the maximum 9v for Vcc so I wrote a program to cycle through using all my equations and the characteristics of a BJT and output all combinations that exceed my specifications. I think in that specific case I told it to give me values with an input resistance greater than 20k and output less than 75 ohm and that was one of the highest gains I could get with those conditions. One thing he did note is that our input signal is probably going to be small, but 10mV seems really small and I can't go much above that point on any of the combinations I've tried without experiencing visual clipping, not just widening of the sinusoid. The power supply on this particular circuit is actually supposed to be set to 9v as it was in my calculations and it was one of the highest gains I was able to achieve (calculated at 84.3 @ 9v). I only have to meet 50 so I can probably try values for gains in the 60s and see if that reduces my distortion.
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# ? Nov 4, 2012 16:16 |
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IratelyBlank posted:Well, the original project was that our professor gave us a few goals to achieve in designing the amplifier. It needs to have a gain >= 50, input resistance >= 15kohm, output resistance < 100 ohm and Vcc no greater than 9v. We could use a 2 or 3 stage so I chose the two stage and derived all the equations for the gain, input resistance and output resistance as well as the transistor characteristics to make sure it is on and the DC load line. My only design choice wrt the DC load line was to center the Q point on the line to reduce the amount of clipping, but I didn't go further and get the AC and all the rest, I didn't think that would be necessary. Even a voltage gain of 50 is pretty high for a single stage, which is really what you have (I'm not counting the output emitter follower, because it doesn't add any gain). This would be equivalent to a 33 dB gain stage, and requires your transistor to be extremely linear to prevent distortion. While you can try to bias the hell out of it to increase linearity, that will diminish your gain because you're setting a different Q point. I think you'll have much better luck designing a two stage gain section with the emitter follower, or a class AB output stage, than trying to do it all in one stage. Since this is a class project, do you get any freedom to pick transistors? And 10 mV with a x50 voltage gain will still only be 500 mV on the output - not all that extreme if you have a 9v rail.
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# ? Nov 5, 2012 04:04 |
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does anyone know where I could find some of those silicone LED/Button covers like on monomes/bliptronics?
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# ? Nov 5, 2012 10:37 |
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Zaxxon posted:does anyone know where I could find some of those silicone LED/Button covers like on monomes/bliptronics? If 4x4 or 2x2 sizes work, Sparkfun has some: https://www.sparkfun.com/products/7835
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# ? Nov 5, 2012 13:20 |
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Zaxxon posted:does anyone know where I could find some of those silicone LED/Button covers like on monomes/bliptronics? You can get a larger one (8x8) at lividinstruments
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# ? Nov 5, 2012 14:01 |
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I'm tuning a MOSFET driver circuit that I'm using to PWM a heating element at 31 Hz. I'm wondering if I have way too high a gate resistor. It's a 12V system, using an IRFZ48N MOSFET, and I'm driving it with a BC327 through a 1K resistor. Can't quite assemble all the math needed to calculate my gate resistor. Suspecting it should be much smaller. Our target current flow is ~8 amps through the MOSFET. (paging movax)
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# ? Nov 5, 2012 16:17 |
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armorer posted:You can get a larger one (8x8) at lividinstruments ugh, all of them are so expensive. and not quite right sized. I know the livid guys custom make these things, maybe I Just have to figure out how they do it. I figured since you see a fair amount of this stuff there would be a company that makes a few lines of them for standard button sizes. I have a big bag of green LED buttons I got from china and I want to make some cool drum machine/monome like interfaces for them, but getting stuff to build the boxes is just a complete pain in the rear end.
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# ? Nov 6, 2012 03:58 |
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Zaxxon posted:ugh, all of them are so expensive. and not quite right sized. I know the livid guys custom make these things, maybe I Just have to figure out how they do it. I figured since you see a fair amount of this stuff there would be a company that makes a few lines of them for standard button sizes. There is probably a source for them from someone in China, but good luck finding it.
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# ? Nov 6, 2012 04:16 |
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Jonny 290 posted:I'm tuning a MOSFET driver circuit that I'm using to PWM a heating element at 31 Hz. I'm wondering if I have way too high a gate resistor. It's a 12V system, using an IRFZ48N MOSFET, and I'm driving it with a BC327 through a 1K resistor. Can't quite assemble all the math needed to calculate my gate resistor. Suspecting it should be much smaller. Our target current flow is ~8 amps through the MOSFET. (paging movax) For a MOSFET, the gate resistor is basically limiting the current that is drawn when the FET is switched on (charging the gate cap). The higher you make this, the slower your MOSFET switches. At high-frequencies, you do this to reduce ringing / oscillations on the gate signal. As the FET transitions through the amplification region to the saturation region, noise on the gate input can couple onto the output; bad for such applications like switching power supplies since that introduces ripple onto the output. In my designs where I've got FETs driven by a switching power supply controller, I usually start at 0 ohms for top/bottom gate, and then size gate resistors accordingly after I get boards built and can throw a scope on there (I usually end up with 0-10 ohms on the top, and then I'll adjust the bottom to make sure FET timing remains safe). In this case, I think throwing a 100ohm resistor there will be fine, especially since you have that BJT stage in there.
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# ? Nov 6, 2012 04:40 |
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This is probably a seriously stupid question but it goes to show how much I don't remember from college electromagnetics I'm working with an MSP430 chip that is delivering 3.3V on its IO pins. I'm looking to do something with sound and my first thought is PWM with a piezo buzzer. While I was looking around for resources I came across this little speaker: http://www.pololu.com/catalog/product/1261/resources quote:This automotive-quality 0.15W speaker is a compact way to add sound to your project. The 100Ω impedance makes it easy to drive this straight from a microcontroller I/O pin. I'm not quite sure if I know what the bolded part means. Here's my stab at it: I know V = IR and according to the Wikipedia page on impedance V = IZ too. Since VCC in my application is 3.3V and the impedance is 100ohms I think that means that the speaker would only draw 33 milliamps. Is that correct? Is that what they mean when the description says it's easy to drive from a microcontroller pin? Where does the 0.15W come into play here?
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# ? Nov 6, 2012 05:28 |
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csammis posted:This is probably a seriously stupid question but it goes to show how much I don't remember from college electromagnetics If it was a traditional 8ohm impedance speaker, you'd definitely need an amp or something. With 100ohm though, your math is correct and it's 33mA...which IMHO, is still a lot for only a MCU pin unless they're rated to source that much current. I don't recall waht MSP430 can do.
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# ? Nov 6, 2012 05:45 |
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Thanks for confirming my math was right! Moving right along, looking up how much the MSP430G2231 (the older Launchpad chip) could source leads me to another question. According to the datasheet if I did pull 33mA from one of the output pins then I would see somewhere between a 1V and 1.5V drop in the output voltage. Okay, I didn't realize that voltage drop would occur but I accept that this is a thing that happens...but then, wouldn't the lower voltage mean that the current consumption is lower too? 2V/100ohm = 22mA which would mean less voltage drop so the current consumption would go up and oh gently caress I'm lost again
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# ? Nov 6, 2012 06:07 |
I would round to 1.25V and model the internal resistance of the IO pin. 1.25V/33mA = 37.9R. Put that in series with your 100 ohm load, and you should draw (3.3V / 137.9R = 23.9mA) about 24mA and see 2.4V across the piezo buzzer. Honestly I would get a transistor and not drive it directly, but I have a ton of 2N2904s and 2N7000s etc just laying around already.
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# ? Nov 6, 2012 06:38 |
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Can someone explain this setup to me? How can the voltage potential reverse on the inductor, and yet the current flows the same direction? Current is going from negative to positive through the inductor? from http://en.wikipedia.org/wiki/Flyback_diode
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# ? Nov 6, 2012 08:12 |
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peepsalot posted:How can the voltage potential reverse on the inductor, and yet the current flows the same direction? It's physics. The magnetic field created by a current flowing through a conductor stores energy. When the current is turned off, that energy has to go somewhere; it induces a current back onto the same conductor. In effect, inductors "don't like" their current changing. A current through an inductor will keep going for as long as it can get energy back out of the magnetic field. This is why a series inductor filters out high frequency AC. The reversal in voltage is related. If current is being pushed along the inductor by the changing magnetic field, the moving charge has to come from somewhere, and there's no complete path for it to follow. The movement of charge creates a large potential difference between the two ends of the inductor. Adding the protection diode provides a path for the current to return after the circuit is turned off without interfering with normal operation.
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# ? Nov 6, 2012 08:24 |
peepsalot posted:How can the voltage potential reverse on the inductor, and yet the current flows the same direction? Current is going from negative to positive through the inductor? When using passive sign convention, a negative power (such as the inductor) indicates an element is a source. The battery also goes negative to positive, indicating it is a source!
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# ? Nov 6, 2012 09:38 |
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Delta-Wye posted:When using passive sign convention, a negative power (such as the inductor) indicates an element is a source. The battery also goes negative to positive, indicating it is a source! Yeah I was starting to come to that conclusion after considering the voltage potential across a battery. It just never occurred to me that a source flows opposite to the rest of the circuit, it makes sense though, since it's doing the "pushing". So in the case of a simple battery/switch/inductor, if you open the switch then the voltage rises and crosses the spark gap of the switch and draws current from the battery for a short time, right? Does this put extra stress on the battery or is it the same current that would have flowed through it in steady state with switch closed.
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# ? Nov 6, 2012 10:32 |
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peepsalot posted:So in the case of a simple battery/switch/inductor, if you open the switch then the voltage rises and crosses the spark gap of the switch and draws current from the battery for a short time, right? Does this put extra stress on the battery or is it the same current that would have flowed through it in steady state with switch closed. Nope, there's usually a transient or "inrush" current that happens. Capacitance helps combat voltage drooping when this happens. Bulk caps (>100uF) are good for this and smoothing out low frequency noise; smaller caps (1uF/0.1uF/etc) are placed close to IC power supply pins and combat high-frequency noise. Inter-planar capacitance (from a ground plane + power plane) helps remedy very high frequency noise as well. (At a certain point, a discrete capacitor can only do so much at higher frequency due to the inductance of the capacitor package itself).
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# ? Nov 6, 2012 16:28 |
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Ugh, I think my TDS420A digiscope is finally going back to the trash heap from whence it emerged. Now it just plain refuses to trigger properly leaving me with a useless nearly unreadable signal display. Checked prices on replacing it with something. Whoops. Guess I didn't really need a digital scope right now anyway
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# ? Nov 6, 2012 17:47 |
I have a laptop battery that decided to go all pear-shaped on me, so I ripped it open and pulled out the little circuit board. Aside from a host of ridicu-small resistors and capacitors and diodes and such, it has some things I'd like to further identify. I think this is a transformer or something? The closest I could find was something like a PoE transformer module. Starring in King of the Hill, I'm not sure how to correctly read the value of this resistor(?). I think this is an over-charge/over-heat fuse? Either a gas gauge or 1-channel power supply management. Of course, I'd guess the latter, but the text doesn't quite match up then. Who knows, with these fancy things! It probably has a flux capacitor in here somewhere. Battery protection of some sort. 3181 ?099 I guess? QZ...That Q1 on the board is certainly the label for this piece, if that helps. They seem to have labeled all resistors as Rx, caps as Cx, etc.
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# ? Nov 12, 2012 19:19 |
Bad Munki posted:I have a laptop battery that decided to go all pear-shaped on me, so I ripped it open and pulled out the little circuit board. Aside from a host of ridicu-small resistors and capacitors and diodes and such, it has some things I'd like to further identify. Bad Munki posted:Starring in King of the Hill, I'm not sure how to correctly read the value of this resistor(?). Bad Munki posted:I think this is an over-charge/over-heat fuse? Bad Munki posted:Either a gas gauge or 1-channel power supply management. Of course, I'd guess the latter, but the text doesn't quite match up then. Who knows, with these fancy things! It probably has a flux capacitor in here somewhere. Bad Munki posted:Battery protection of some sort. Bad Munki posted:QZ...That Q1 on the board is certainly the label for this piece, if that helps. They seem to have labeled all resistors as Rx, caps as Cx, etc.
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# ? Nov 12, 2012 20:05 |
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# ? May 9, 2024 00:38 |
Delta-Wye posted:Probably a large package mosfet? I'm guessing TI PowerPAD series. I would expect to see at least one if not more in a battery protection circuit. quote:I'm pretty sure it's a high power, high precision .005 ohm resistor. Maybe a Vishay WSH28185L000FEA resistor? So I guess the last question is, is it worth bothering to pull any of these parts? And if so, how best to do so? Hot plate and tweezers?
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# ? Nov 12, 2012 20:49 |