Discussing audio amplifier design -- BJT, discrete
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From: pimpom on 18 Feb 2010 17:47 Jon Kirwan wrote: > On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom" > <pimpom(a)invalid.invalid> wrote: > >> Jon Kirwan wrote: > >>> I'm expecting to use caps on the order of perhaps 2.2mF 50V, >>> to be secure about the rails. But I expect to want to play >>> with that, once everything is working, to see just how bad I >>> can make it while seeing what that means for the output. And >>> then see if I can calculate a prediction that isn't too far >>> from those results, on paper. >> >> I have my own rule of thumb here for acceptable levels of >> ripple >> and load regulation. I divide the full supply dc voltage with >> the >> current at maximum output. This gives the equivalent dc load >> as >> seen by the power supply. In the sample design under >> consideration, that's roughly 30 ohms on each side of the >> split >> supply. Calculate the reactance of the filter capacitor at the >> pulsating dc frequency which is twice the mains frequency for >> full-wave. My rule of thumb is to get an R/Xc ratio of the >> order >> of 50 for a medium quality amp. Your choice of 2200uF agrees >> well >> with this. > > Thanks for your thinking on this. I used more mathy stuff to > get there, but I like your practical slice through all that. > It is easy to follow. Rules of thumb are often based on previous mathematical derivations, as it was in this case. However, after having done umpteen calculations where absolute precision is not needed, the novelty wears off after some time and one tends to be satisfied with being able to intuitively predict the outcome within a per cent or so without actually putting anything on paper. It's been firmly etched in my mind for 40 years that 1000uF has a reactance of 1.5815 ohms (usually taken as 1.6) at 100Hz (twice the mains frequency here) and I quickly derive Xc for other values from that within a second. Then I mentally divide the equivalent DC resistance of a load (not necessarily an audio amplifier) with that reactance and have a good idea of what to expect in terms of ripple voltage amplitude, regulation, DC voltage, peak diode current, rms transformer current, etc. Heck, it's past 4 am over here. Time for bed. Bye.
From: Paul E. Schoen on 18 Feb 2010 18:24 "pimpom" <pimpom(a)invalid.invalid> wrote in message news:hlk96n$6g8$1(a)news.albasani.net... > Jon Kirwan wrote: >> On Wed, 17 Feb 2010 16:20:40 -0500, "Paul E. Schoen" >> <paul(a)peschoen.com> wrote: >> >>> "Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message >>> news:kvlmn5dhciv4c51a0au32qi52rjnl67sro(a)4ax.com... >>>> >>>> Okay. I'm back to the power supply, again. (I'm convinced >>>> that my junkbox unit will work fine -- I think it can hold >>>> maybe 18V minimum under load on each rail. Which seems more >>>> than enough headroom for 12.7V, plus output stage overhead. >>>> >> .... > ............<snip>............. >> >> Now, I want axial leaded diodes for the bridge. From >> simulating a load of 8 ohms, 1kHz, average power of 10W, and >> my secondary winding resistance of 2.6 ohms, I'm finding that >> each diode suffers under a quarter watt of dissipation. So, >> any recommendations about diodes? Obviously, for a one-off, >> cost is not really an issue. How important is 'fast >> recovery'? (Outside of its impact on dissipation.) Seems >> that anything with 100V or better for reverse voltage >> standoff, 1/4 watt or better, should work. Leakage probably >> isn't that important (except against as it may add to >> dissipation.) >> > > Fast recovery is not important here since the diodes work at mains > frequency. > > 10W sinusoidal into 8 ohms = 1.58A peak = 0.503A dc average for Class B. > Add some mAs for the driver stages. That's slightly more than 0.25A each > for diodes in full-wave rectification. The ubiquitous 1N4002 to 1N4007 > rated for 1 Amp diodes will do fine. They differ only in the maximum > reverse voltage ratings and cost almost the same. As a matter of > convenience, I stock only the 1000-volt 1N4007. At less than 2 cents US > each retail, I buy them in batches of hundreds at a time. At one time I had two reels of 5000 each of 1N4004 that I got surplus, but I sold most of them. I also have a bag of about 1000 pieces of 1N4003. So I have pretty much a lifetime supply. Either one is OK for 120 VAC mains and perfect for lower voltage applications. But now my new designs are mostly SMT. I was going to keep the thru holes and use the "free" parts I had, but I figured that the labor cost of inserting, soldering, and clipping leads on 6 diodes on 40 boards might be more than the $0.06 each for the S1G SMT diodes. Once a commitment is made to SMT it is usually cost-effective to use as many such parts as possible. I never fully analyzed it, though. I figure about 2 minutes for the six diodes. At $60/hr, or $1/minute, I spend $2/board for the leaded parts. The SMT assembly is probably $0.05 per part, so I spend a total of $0.66 per board. > >> I'm expecting to use caps on the order of perhaps 2.2mF 50V, >> to be secure about the rails. But I expect to want to play >> with that, once everything is working, to see just how bad I >> can make it while seeing what that means for the output. And >> then see if I can calculate a prediction that isn't too far >> from those results, on paper. >> > I have my own rule of thumb here for acceptable levels of ripple and load > regulation. I divide the full supply dc voltage with the current at > maximum output. This gives the equivalent dc load as seen by the power > supply. In the sample design under consideration, that's roughly 30 ohms > on each side of the split supply. Calculate the reactance of the filter > capacitor at the pulsating dc frequency which is twice the mains > frequency for full-wave. My rule of thumb is to get an R/Xc ratio of the > order of 50 for a medium quality amp. Your choice of 2200uF agrees well > with this. Some time ago I came up with a rule of thumb of 1000 uF per amp, and I revised that to 2000 uF per amp. I used an RC time constant of 8 mSec between peaks for a 37% discharge from peak which holds the approximate RMS value, and for a typical 8 VDC power supply at 1 amp R=8 ohms. So C = ..008/8 = 1000 uF. But two time constants gives only 13% discharge so 2000 uF is much better. For a 16 VDC supply, 1000 uF is OK, and as the voltage doubles the required capacitance is halved. So for most low voltage applications, 1000 to 2000 uF per amp is reasonable, and easy to remember. Of course, if you enjoy mathematical analysis, you can spend time working out effects of winding resistance and capacitor ESR and acceptable ripple. Or you can just use LTSpice. But if I need a quick and dirty junkbox power supply, 1000 uF/amp is good enough to grab and go. For example, using LTSpice, I find a 12.6 V transformer and I want to make a 12 VDC power supply at 1 amp. Using a 1000 uF capacitor and a 12 ohm load, my output is 13.3 V which has a peak of 16.1 V and drops to 10.4 V, which is a 35% drop as predicted. With 2000 uF it drops to 12.6 VDC so my output is high enough to provide the 12 VDC I wanted with a regulator. Of course there are line variations and transformer regulation, but not bad for a quick estimate. If I wanted 24 VDC, and I had a 25.2 V transformer, a 1000 uF capacitor gives me a minimum of 26 VDC for a regulator with a little bit of headroom. Now I actually add a simple emitter follower voltage regulator with a 2N3055 and two 12 V zeners and a diode in series, with 220 ohms and a 100 uF cap. I get an output of 24.18 VDC which varies from 23.99 VDC to 24.30 VDC. Adding the regulator improves the minimum voltage excursion on the 1000 uF main filter capacitor to 27.6 VDC. Since I was originally designing for just such a regulated power supply, my "grab-and-go" estimates for main filter capacitors seems to work out quite well. And I found it more fun to build and test the circuit using LTSpice rather than with math. Filter capacitors of this size are typically -20% / +80% tolerance, so chances are the results will be even better than expected. Paul
From: Jon Kirwan on 19 Feb 2010 01:19 On Fri, 19 Feb 2010 04:17:22 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >Jon Kirwan wrote: >> On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom" >> <pimpom(a)invalid.invalid> wrote: >> >>> Jon Kirwan wrote: >> >>>> I'm expecting to use caps on the order of perhaps 2.2mF 50V, >>>> to be secure about the rails. But I expect to want to play >>>> with that, once everything is working, to see just how bad I >>>> can make it while seeing what that means for the output. And >>>> then see if I can calculate a prediction that isn't too far >>>> from those results, on paper. >>> >>> I have my own rule of thumb here for acceptable levels of >>> ripple >>> and load regulation. I divide the full supply dc voltage with >>> the >>> current at maximum output. This gives the equivalent dc load >>> as >>> seen by the power supply. In the sample design under >>> consideration, that's roughly 30 ohms on each side of the >>> split >>> supply. Calculate the reactance of the filter capacitor at the >>> pulsating dc frequency which is twice the mains frequency for >>> full-wave. My rule of thumb is to get an R/Xc ratio of the >>> order >>> of 50 for a medium quality amp. Your choice of 2200uF agrees >>> well >>> with this. >> >> Thanks for your thinking on this. I used more mathy stuff to >> get there, but I like your practical slice through all that. >> It is easy to follow. > >Rules of thumb are often based on previous mathematical >derivations, as it was in this case. Don't mistake me. All I meant to say is that I _am_ new and therefore took a slower approach, not having developed the well worn ruts from good experience as you have done. And that I enjoyed seeing your way of cutting through it. >However, after having done >umpteen calculations where absolute precision is not needed, the >novelty wears off after some time and one tends to be satisfied >with being able to intuitively predict the outcome within a per >cent or so without actually putting anything on paper. I think I clearly understood exactly that from your writing. >It's been >firmly etched in my mind for 40 years that 1000uF has a reactance >of 1.5815 ohms (usually taken as 1.6) at 100Hz (twice the mains >frequency here) and I quickly derive Xc for other values from >that within a second. Then I mentally divide the equivalent DC >resistance of a load (not necessarily an audio amplifier) with >that reactance and have a good idea of what to expect in terms of >ripple voltage amplitude, regulation, DC voltage, peak diode >current, rms transformer current, etc. > >Heck, it's past 4 am over here. Time for bed. Bye. I didn't expect this and it all looks as though I may have unintentionally implied something. If so, I hope you will re-read what I wrote and understand that I'm merely commenting upon my own painstaking processes, which are at this point in time important steps for me to take, and in no way commenting about anything you are saying (except perhaps that I agree and otherwise like the way you thought about it.) That's all there was there. Thanks very much again, Jon
From: Jon Kirwan on 19 Feb 2010 04:30 On Thu, 18 Feb 2010 18:24:21 -0500, "Paul E. Schoen" <paul(a)peschoen.com> wrote: >"pimpom" <pimpom(a)invalid.invalid> wrote in message >news:hlk96n$6g8$1(a)news.albasani.net... >> Jon Kirwan wrote: >>> On Wed, 17 Feb 2010 16:20:40 -0500, "Paul E. Schoen" >>> <paul(a)peschoen.com> wrote: >>> >>>> "Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message >>>> news:kvlmn5dhciv4c51a0au32qi52rjnl67sro(a)4ax.com... >>>>> >>>>> Okay. I'm back to the power supply, again. (I'm convinced >>>>> that my junkbox unit will work fine -- I think it can hold >>>>> maybe 18V minimum under load on each rail. Which seems more >>>>> than enough headroom for 12.7V, plus output stage overhead. >>>>> >>> .... >> ............<snip>............. >>> >>> Now, I want axial leaded diodes for the bridge. From >>> simulating a load of 8 ohms, 1kHz, average power of 10W, and >>> my secondary winding resistance of 2.6 ohms, I'm finding that >>> each diode suffers under a quarter watt of dissipation. So, >>> any recommendations about diodes? Obviously, for a one-off, >>> cost is not really an issue. How important is 'fast >>> recovery'? (Outside of its impact on dissipation.) Seems >>> that anything with 100V or better for reverse voltage >>> standoff, 1/4 watt or better, should work. Leakage probably >>> isn't that important (except against as it may add to >>> dissipation.) >>> >> >> Fast recovery is not important here since the diodes work at mains >> frequency. >> >> 10W sinusoidal into 8 ohms = 1.58A peak = 0.503A dc average for Class B. >> Add some mAs for the driver stages. That's slightly more than 0.25A each >> for diodes in full-wave rectification. The ubiquitous 1N4002 to 1N4007 >> rated for 1 Amp diodes will do fine. They differ only in the maximum >> reverse voltage ratings and cost almost the same. As a matter of >> convenience, I stock only the 1000-volt 1N4007. At less than 2 cents US >> each retail, I buy them in batches of hundreds at a time. > >At one time I had two reels of 5000 each of 1N4004 that I got surplus, but >I sold most of them. I also have a bag of about 1000 pieces of 1N4003. So I >have pretty much a lifetime supply. Either one is OK for 120 VAC mains and >perfect for lower voltage applications. But now my new designs are mostly >SMT. I was going to keep the thru holes and use the "free" parts I had, but >I figured that the labor cost of inserting, soldering, and clipping leads >on 6 diodes on 40 boards might be more than the $0.06 each for the S1G SMT >diodes. Once a commitment is made to SMT it is usually cost-effective to >use as many such parts as possible. I never fully analyzed it, though. I >figure about 2 minutes for the six diodes. At $60/hr, or $1/minute, I spend >$2/board for the leaded parts. The SMT assembly is probably $0.05 per part, >so I spend a total of $0.66 per board. > >> >>> I'm expecting to use caps on the order of perhaps 2.2mF 50V, >>> to be secure about the rails. But I expect to want to play >>> with that, once everything is working, to see just how bad I >>> can make it while seeing what that means for the output. And >>> then see if I can calculate a prediction that isn't too far >>> from those results, on paper. >>> >> I have my own rule of thumb here for acceptable levels of ripple and load >> regulation. I divide the full supply dc voltage with the current at >> maximum output. This gives the equivalent dc load as seen by the power >> supply. In the sample design under consideration, that's roughly 30 ohms >> on each side of the split supply. Calculate the reactance of the filter >> capacitor at the pulsating dc frequency which is twice the mains >> frequency for full-wave. My rule of thumb is to get an R/Xc ratio of the >> order of 50 for a medium quality amp. Your choice of 2200uF agrees well >> with this. > >Some time ago I came up with a rule of thumb of 1000 uF per amp, and I >revised that to 2000 uF per amp. I used an RC time constant of 8 mSec >between peaks for a 37% discharge from peak which holds the approximate RMS >value, and for a typical 8 VDC power supply at 1 amp R=8 ohms. So C = >.008/8 = 1000 uF. But two time constants gives only 13% discharge so 2000 >uF is much better. For a 16 VDC supply, 1000 uF is OK, and as the voltage >doubles the required capacitance is halved. So for most low voltage >applications, 1000 to 2000 uF per amp is reasonable, and easy to remember. Okay. >Of course, if you enjoy mathematical analysis, you can spend time working >out effects of winding resistance and capacitor ESR and acceptable ripple. I suppose what I don't enjoy is taking on faith "rules" or "prepared charts" I'm handed. So I do the math once or twice, just to verify and make sure I have a small sense of understanding about the whys and wherefores. (And I enjoy the math practice, from time to time.) >Or you can just use LTSpice. Once I feel I grasp the theory I will use LTspice a lot and not give it that much thought later on. But isn't it better to make sure, at least once >But if I need a quick and dirty junkbox power >supply, 1000 uF/amp is good enough to grab and go. And I think I understand the details why. (Unless someone expresses an interest, I won't dump it out here.) >For example, using LTSpice, I find a 12.6 V transformer and I want to make >a 12 VDC power supply at 1 amp. Using a 1000 uF capacitor and a 12 ohm >load, my output is 13.3 V which has a peak of 16.1 V and drops to 10.4 V, >which is a 35% drop as predicted. With 2000 uF it drops to 12.6 VDC so my >output is high enough to provide the 12 VDC I wanted with a regulator. Of >course there are line variations and transformer regulation, but not bad >for a quick estimate. As I just wrote, once I've done it in the "forward direction" and feel I understand the details well enough, once or twice, just selecting rules to follow after that make sense. If something doesn't feel right in the simulation, you can always return to the fundamentals on paper to double-check. But I don't like using a tool in a fashion where I have no clue whatsoever how to check the work on my own, should I decide to do so. Doesn't feel right. (You are past that point, of course, so no problem there.) >If I wanted 24 VDC, and I had a 25.2 V transformer, a 1000 uF capacitor >gives me a minimum of 26 VDC for a regulator with a little bit of headroom. > >Now I actually add a simple emitter follower voltage regulator with a >2N3055 and two 12 V zeners and a diode in series, with 220 ohms and a 100 >uF cap. I get an output of 24.18 VDC which varies from 23.99 VDC to 24.30 >VDC. Adding the regulator improves the minimum voltage excursion on the >1000 uF main filter capacitor to 27.6 VDC. > >Since I was originally designing for just such a regulated power supply, my >"grab-and-go" estimates for main filter capacitors seems to work out quite >well. And I found it more fun to build and test the circuit using LTSpice >rather than with math. Filter capacitors of this size are typically -20% / >+80% tolerance, so chances are the results will be even better than >expected. Thanks. And I got it. Jon
From: pimpom on 19 Feb 2010 13:13
Jon Kirwan wrote: > On Fri, 19 Feb 2010 04:17:22 +0530, "pimpom" > <pimpom(a)invalid.invalid> wrote: > >> Jon Kirwan wrote: >>> On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom" >>> <pimpom(a)invalid.invalid> wrote: >>> >>>> Jon Kirwan wrote: >>> >>>>> I'm expecting to use caps on the order of perhaps 2.2mF >>>>> 50V, >>>>> to be secure about the rails. But I expect to want to play >>>>> with that, once everything is working, to see just how bad >>>>> I >>>>> can make it while seeing what that means for the output. >>>>> And >>>>> then see if I can calculate a prediction that isn't too far >>>>> from those results, on paper. >>>> >>>> I have my own rule of thumb here for acceptable levels of >>>> ripple >>>> and load regulation. I divide the full supply dc voltage >>>> with >>>> the >>>> current at maximum output. This gives the equivalent dc load >>>> as >>>> seen by the power supply. In the sample design under >>>> consideration, that's roughly 30 ohms on each side of the >>>> split >>>> supply. Calculate the reactance of the filter capacitor at >>>> the >>>> pulsating dc frequency which is twice the mains frequency >>>> for >>>> full-wave. My rule of thumb is to get an R/Xc ratio of the >>>> order >>>> of 50 for a medium quality amp. Your choice of 2200uF agrees >>>> well >>>> with this. >>> >>> Thanks for your thinking on this. I used more mathy stuff to >>> get there, but I like your practical slice through all that. >>> It is easy to follow. >> >> Rules of thumb are often based on previous mathematical >> derivations, as it was in this case. > > Don't mistake me. All I meant to say is that I _am_ new and > therefore took a slower approach, not having developed the > well worn ruts from good experience as you have done. And > that I enjoyed seeing your way of cutting through it. > >> However, after having done >> umpteen calculations where absolute precision is not needed, >> the >> novelty wears off after some time and one tends to be >> satisfied >> with being able to intuitively predict the outcome within a >> per >> cent or so without actually putting anything on paper. > > I think I clearly understood exactly that from your writing. > >> It's been >> firmly etched in my mind for 40 years that 1000uF has a >> reactance >> of 1.5815 ohms (usually taken as 1.6) at 100Hz (twice the >> mains >> frequency here) and I quickly derive Xc for other values from >> that within a second. Then I mentally divide the equivalent DC >> resistance of a load (not necessarily an audio amplifier) with >> that reactance and have a good idea of what to expect in terms >> of >> ripple voltage amplitude, regulation, DC voltage, peak diode >> current, rms transformer current, etc. >> >> Heck, it's past 4 am over here. Time for bed. Bye. > > I didn't expect this and it all looks as though I may have > unintentionally implied something. If so, I hope you will > re-read what I wrote and understand that I'm merely > commenting upon my own painstaking processes, which are at > this point in time important steps for me to take, and in no > way commenting about anything you are saying (except perhaps > that I agree and otherwise like the way you thought about > it.) That's all there was there. > > Thanks very much again, > Jon If I seemed to be snapping at you, I apologise. I'd be less than honest if I didn't admit that I was just a little bit irritated at the time, but you weren't really the cause. I'd had a frustrating day - no, make that a frustrating 2 weeks plus - from being given the runaround by a component manufacturer. That and the fact that I'm not a native user of English may have made me sound more abrupt than I meant to be. Again, I apologise. You and I are very similar in that I also like to dissect all the little bits and pieces of any new ground I'm venturing into. But I suspect that your keen desire to analyse everything in minute detail, and the fact that you're probably better at that than a lot of people here with more experience in applied electronics, has put off more than one of the regulars in this NG. |