Discussing audio amplifier design -- BJT, discrete
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From: pimpom on 19 Feb 2010 13:58 Paul E. Schoen wrote: > "pimpom" <pimpom(a)invalid.invalid> wrote in message > news:hlk96n$6g8$1(a)news.albasani.net... >> 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. > > 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. I have an even simpler rule of thumb for audio amps using the standard series push-pull configuration. The ratio of rated speaker impedance to the equivalent load resistance at full power as seen by the power supply stays approximately the same no matter what level of power and supply voltage. So, that rule of thumb is 1000uF rail-to-rail for an amp with an 8-ohm load, medium quality. 500uF where quality is not an important desing criterion, and 2000uF where it is. Jon may be interested in how those figures came about, but I think he would rather deduce them himself instead of having them handed to him on a platter. Hint: Calculate the voltage and current swings needed for a given power output. Add a few volts for transistor saturation and drops across emitter resistors. Derive the average DC current from the peak swing. That gives the equivalent load resistance as seen by the power supply, and that resistance remains approximately the same for a given speaker impedance, for a wide range of designed power level. In fact, if it weren't for the effects of transistor imperfection and the need for emitter resistors and other compensation techniques, the ratio of equivalent load resistance to speaker impedance would be exactly the same for any power level.
From: Jon Kirwan on 19 Feb 2010 15:36 On Fri, 19 Feb 2010 23:43:04 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >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. No problem. I worried a little and just wanted to make sure that I hadn't come across wrong. I had only meant that I was merely taking extra pains myself and that I liked the way you'd written, too. And I reserved out the possibility that there were language/concept issues. However, I had no clear idea either way, if English was your native tongue. If my wife had asked, I'd have only said "I don't know." You are just that good at it. Makes me wish I weren't so provincial, myself. >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. I appreciate that. >But >I suspect that your keen desire to analyse everything in minute >detail, I like to at least try my hand, at times. It's hard for me to gain some creative insight without that work. >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. I cannot say much there. They have to speak for themselves, I suppose. Jon
From: Paul E. Schoen on 20 Feb 2010 03:17 "pimpom" <pimpom(a)invalid.invalid> wrote in message news:hlmn0i$aks$1(a)news.albasani.net... > > I have an even simpler rule of thumb for audio amps using the standard > series push-pull configuration. The ratio of rated speaker impedance to > the equivalent load resistance at full power as seen by the power supply > stays approximately the same no matter what level of power and supply > voltage. So, that rule of thumb is 1000uF rail-to-rail for an amp with an > 8-ohm load, medium quality. 500uF where quality is not an important > desing criterion, and 2000uF where it is. > > Jon may be interested in how those figures came about, but I think he > would rather deduce them himself instead of having them handed to him on > a platter. > > Hint: Calculate the voltage and current swings needed for a given power > output. Add a few volts for transistor saturation and drops across > emitter resistors. Derive the average DC current from the peak swing. > That gives the equivalent load resistance as seen by the power supply, > and that resistance remains approximately the same for a given speaker > impedance, for a wide range of designed power level. > > In fact, if it weren't for the effects of transistor imperfection and the > need for emitter resistors and other compensation techniques, the ratio > of equivalent load resistance to speaker impedance would be exactly the > same for any power level. Without going into my quick calculations, I came up with an impedance of 10.6 ohms for each rail, or about 21 ohms total. I think a perfect amplifier would have an impedance that matched the 8 ohm output, so considering possibly 40% efficiency that sounds about right. But for your 50:1 R/Xc ratio it seems like you need about 2500 uF. I would expect the Xc/R ratio to be roughly the amount of ripple, and 2% seems unnecessary. 1000 uF gives a ratio of 6.3%, which is still a reasonable level of ripple. But then one must consider RMS versus P-P which is a 3:1 ratio, and perhaps that is where the 2% comes from. If the ratio of amplifier impedance to speaker impedance is roughly constant, then it seems that one may only need to consider the ratio of Xc to speaker impedance. And these approximations are based on acceptable power line ripple, whereas the amplifier needs to be able to draw on the power supply at audio frequencies to 20 Hz, which may imply another factor of three for the capacitance needed. My rule-of-thumb values are based more on the requirements of a general purpose power supply, and I was also assuming some sort of active regulation. Still, the values come out within a couple multiples of each other, so that shows validity of either method. Paul
From: Jon Kirwan on 20 Feb 2010 05:48 On Sat, 20 Feb 2010 00:28:22 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >Paul E. Schoen wrote: >> >> 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. > >I have an even simpler rule of thumb for audio amps using the >standard series push-pull configuration. Hehe. What bothers me about having "simpler rules" is that in the end it requires me to keep so many rules in mind. Luckily, for a hobbyist type there aren't that many output configurations. (I've read about perhaps 20 sufficiently different types that professionals might consider... but that is happily not my problem.) >The ratio of rated >speaker impedance to the equivalent load resistance at full power >as seen by the power supply stays approximately the same no >matter what level of power and supply voltage. This kind of writing confuses me a little. You start out saying "at full power" then conclude "no matter what level of power." But if I read this to mean that if the computation is taken for the full power situation, that it will hold regardless of the volume control setting or supply rail. But if so, I have perhaps at least the following problem: Let's say "rated speaker impedance" is 8 ohms. Fixed. The equivalent load resistance at full power as seen by the V+ power supply rail will be about the same as seen by the V- rail, so I don't need to consider both in the ratio. For the V+ rail, it's V(V+)/Ic_rms. The ratio is then: 8*Ic_rms/V(V+) Which you claim is about the same regardless of power setting, for example. However, while the power setting does NOT affect V(V+) or the 8 ohm speaker, it does impact Ic_rms. So the ratio does not seem to remain the same on that score. In the case where V(V+) rises or falls, I'd imagine that Ic_rms would similarly do so. So there, I think I follow your point. >So, that rule of >thumb is 1000uF rail-to-rail for an amp with an 8-ohm load, >medium quality. 500uF where quality is not an important desing >criterion, and 2000uF where it is. Let's hold on that point, for a moment. >Jon may be interested in how those figures came about, but I >think he would rather deduce them himself instead of having them >handed to him on a platter. Exactly! >Hint: Calculate the voltage and current swings needed for a given >power output. Add a few volts for transistor saturation and drops >across emitter resistors. Derive the average DC current from the >peak swing. That gives the equivalent load resistance as seen by >the power supply, and that resistance remains approximately the >same for a given speaker impedance, for a wide range of designed >power level. Okay, let's do that. Let's use P_out=10W average. We've already concluded that this means V_peak=SQRT(2*P*R)=12.65V. However, let's not assume even that much. In case anyone needs to follow the details why, that equation comes from: V0 = V_peak w = 2*pi*f V_t = V0 * SINE( 2*pi*f * t ) = V0 * SINE( w * t ) I_t = V_t / R R = speaker impedance (say, 8 ohms?) P = (Integral(V_t * I_t, dt) from 0 to t0) / t0 choose t0 where w*t0=2*pi, so t0=2*pi/w=1/f = (Integral(V0^2*SINE^2(w*t)/R, dt) from 0 to 1/f) * f move V0^2/R out: = (Integral(SINE^2(w*t), dt) from 0 to 1/f) * f*V0^2/R change of variable, x = w*t, gives dx = w*dt and dt = dx/w = (Integral(SINE^2(x)/w, dx) from 0 to 2*pi) * f*V0^2/R move 1/w out: = (Integral(SINE^2(x), dx) from 0 to 2*pi) * f*V0^2/(w*R) combine f/w at the end of the expression: = (Integral(SINE^2(x), dx) from 0 to 2*pi) * V0^2/(2*pi*R) solve integral: = ((1/2x-1/4*SINE(2*x)) from 0 to 2*pi) * V0^2/(2*pi*R) apply domain: = (pi) * V0^2/(2*pi*R) final solution: = V0^2/(2*R) Just for the record. Since P=V_peak^2/(2*R), it follows that V_peak=SQRT(2*P*R). And this documents the fact that we are talking about V_peak and not V_p-p or V_rms or some other term. It's V_peak. So for the p-p value, it's twice that. That's nailed down, now. So I_peak will either be V_peak/R or else SQRT(2*P/R), depending on how you like it stated. V_peak = SQRT(2*P*R) I_peak = SQRT(2*P/R) Add a few volts transistors and so on and we get: V_rail = V_peak + 2V Derive average current from the peak swing? Hmm. This is where I may get stuck. We can compute peak values, no problem. But now we need to consider class of operation and more. For example, in class-B, each rail is used only _half the time_, so to speak. So for half the time, it's zero. The rest of the time looks like this: I_avg = (Integral(I_t, dt) from 0 to t0) / t0 choose t0 where w*t0=pi, so t0=pi/w=1/(2*f) = (Integral(V0*SIN(w*t)/R, dt) from 0 to 1/(2*f)) * 2*f move V0/R out: = (Integral(SIN(w*t), dt) from 0 to 1/(2*f)) * 2*f*V0/R change of variable, x = w*t, gives dx = w*dt and dt = dx/w = (Integral(SIN(x)/w, dx) from 0 to pi) * 2*f*V0/R move 1/w out: = (Integral(SIN(x), dx) from 0 to pi) * V0/(pi*R) solve integral: = (-COS(x)) from 0 to pi) * V0/(pi*R) apply domain: = (2) * V0/(pi*R) final solution: = 2*V0/(pi*R) So I_avg so far is (2/pi) * V_peak/R. But that is only for half the time. The other half, it is zero. So the actual I_avg over the entire period of a full cycle from one rail, in class-B, is half that or (1/pi) * V_peak/R. The equivalent load resistance as seen by one rail is then: R_equiv = pi * R * (V_rail/V_peak) = pi * R * (V_rail/(R*I_peak)) = pi * V_rail / I_peak = pi * V_rail / SQRT(2*P/R) For a given R and V_rail and P, that does remain constant. Let's pick some values, again. Just to make it concrete. R=8, P=10, V_rail=18V, leads to just under 36 ohms for the class-B case. That's from each rail and obviously includes the speaker's R, itself. I'd guess that class-A would lead to different results due to the fact that the half-cycle I set to zero would no longer be zero. >In fact, if it weren't for the effects of transistor imperfection >and the need for emitter resistors and other compensation >techniques, the ratio of equivalent load resistance to speaker >impedance would be exactly the same for any power level. Hopefully, the above is about right. Let me know where I screwed up. I'm now also thinking about something else. The issue is peak power transfer, which usually is considered to take place when R_src and R_load are equal. In the above example, with 36 ohms total, R_src would be about 36-8 or 28 ohms (R_load is 8 ohms.) The ratio of power delivered to the load, from that point of view, would be poor wouldn't it? Bad on the order of 28/8 or 3.5:1 instead of an optimal 1:1. Wouldn't that suggest that if I dissipate 10W in R, that even in class-B with 18V rails I'll be dissipating 35W across both power BJTs? (Cutting it closer by using 15V rails would yield more like 22/8 or 2.75:1 for 27.5W across both power BJTs. But nothing very good there, either.) Does this suggest a desire for an output transformer for impedance matching and better transfer of power? Thanks, Jon
From: pimpom on 20 Feb 2010 06:33
Paul E. Schoen wrote: > "pimpom" <pimpom(a)invalid.invalid> wrote in message > news:hlmn0i$aks$1(a)news.albasani.net... >> >> I have an even simpler rule of thumb for audio amps using the >> standard series push-pull configuration. The ratio of rated >> speaker >> impedance to the equivalent load resistance at full power as >> seen by >> the power supply stays approximately the same no matter what >> level >> of power and supply voltage. So, that rule of thumb is 1000uF >> rail-to-rail for an amp with an 8-ohm load, medium quality. >> 500uF >> where quality is not an important desing criterion, and 2000uF >> where >> it is. Jon may be interested in how those figures came about, >> but I think he >> would rather deduce them himself instead of having them handed >> to >> him on a platter. >> >> Hint: Calculate the voltage and current swings needed for a >> given >> power output. Add a few volts for transistor saturation and >> drops >> across emitter resistors. Derive the average DC current from >> the >> peak swing. That gives the equivalent load resistance as seen >> by the >> power supply, and that resistance remains approximately the >> same for >> a given speaker impedance, for a wide range of designed power >> level. >> >> In fact, if it weren't for the effects of transistor >> imperfection >> and the need for emitter resistors and other compensation >> techniques, the ratio of equivalent load resistance to speaker >> impedance would be exactly the same for any power level. > > Without going into my quick calculations, I came up with an > impedance > of 10.6 ohms for each rail, or about 21 ohms total. I think a > perfect > amplifier would have an impedance that matched the 8 ohm > output, so > considering possibly 40% efficiency that sounds about right. > But for > your 50:1 R/Xc ratio it seems like you need about 2500 uF. This is how I go about it: Take the case of Jon's amp as an example. 10W into 8 ohms is 12.65V, 1.581A (both sinusoidal peak). That's 0.5A dc average. To get 12.65V swing, we'll need about +/-16V Vcc. 16V/0.5A = 32 ohms for each rail, 64 ohms total. 1000uF is 1.6 ohms at 100Hz, 1.326 at 120Hz. This gives R/Xc of 40 and 48 at 100Hz and 120Hz respectively. (I'm sure you knew that the 50:1 was a round figure). To get 1000uF rail to rail, we'd need 2000uF for each rail, or 2200uF in practice. > I would expect the Xc/R ratio to be roughly the amount of > ripple, and > 2% seems unnecessary. 1000 uF gives a ratio of 6.3%, which is > still a > reasonable level of ripple. But then one must consider RMS > versus P-P > which is a 3:1 ratio, and perhaps that is where the 2% comes > from. I also assume reasonable values of equivalent source resistances, contributed mainly by transformer losses. This equivalent source resistance has a big influence on ripple and power supply current and voltage ratios. These days I rely mostly on charts for those ratios. > > If the ratio of amplifier impedance to speaker impedance is > roughly > constant, then it seems that one may only need to consider the > ratio > of Xc to speaker impedance. That's true, and is really just another way of looking at the rule of thumb of 1000uF for an amp with an 8-ohm speaker. > > And these approximations are based on acceptable power line > ripple, > whereas the amplifier needs to be able to draw on the power > supply at > audio frequencies to 20 Hz, which may imply another factor of > three > for the capacitance needed. Which is why I said earlier that I double my rule-of-thumb value to 2000uF per 1/8ohms for a good quality amp, say 4700uF on each supply rail. And 10,000uF where cost and size are not important considerations. And double those values yet again for a 4-ohm load. > > My rule-of-thumb values are based more on the requirements of a > general purpose power supply, and I was also assuming some sort > of > active regulation. Still, the values come out within a couple > multiples of each other, so that shows validity of either > method. > > Paul |