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From: pimpom on 20 Feb 2010 10:16 Jon Kirwan wrote: > 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 been in electronics for 4 decades, but I still can't decide whether I'm a hobbyist or a profressional! :-) I earn a living with it, but I also approach my work as if it was a hobby. I waste a lot of time and earn only a fraction of what I could. I do a lot of things for others for free or for a nominal fee, but I also have much more fun. > (I've read about perhaps 20 sufficiently > different types that professionals might consider... but that > is happily not my problem.) IMO, not even a budding pro should try to master everything at once. I suggest you concentrate on the configuration presented in its basic form in Wikipedia. What now look like very different types may well turn out to be variants of a very few basic types. > >> 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: I may have been a bit careless with my wording. What I meant was the ratio at full power output no matter what the designed maximum power is. That is, the ratio holds equally (approximately) for a 1W amp and a 100-watt amp. > > 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 guess you enjoy reinventing the wheel all the time :-) Most people just use the long-established relation between peak, rms and average values for known waveforms - sine in this case. Your mathematical derivation was fundamentally correct, but where you went astray was in that we are dealing with two half waves in series because of the configuration. Therefore, the effective average current drawn from each rail is Ipeak/pi, not 2Ipeak/pi. If the two halves were drawn from the same rail, the total would indeed be 2Ipeak/pi, but current for the two halves are drawn from two separate rails in series, so the overall average is still Ipeak/pi. There's a once-common configuration (which you've probably come across) in which two transistors take their power from the same rail, and each transistor conducts on alternate half cycles. In that configuration, the total average current drawn from the power supply is indeed 2Ipeak/pi. That configuration has dropped out of favor except for certain cases, mainly because it requires an output transformer to combine the two output half waves. But for the time being, let's not complicate things by going further in that direction. > > 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 Class A, the output stage would be biased to a steady dc current at least equal to the peak swing. At full drive in an ideal Class A, the current will swing between 0 and twice the steady value, and the average current will still be equal to that without any output. Even with less-than-full drive, the swing is still +/- around the bias value, and the average stays the same. > >> 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_ screwed up again here because I failed to make it clear that by "any power level", I meant amplifiers designed for any level of full power output, NOT any level of output for the same amplifier. > > 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. This is a common misinterpretation of a law that's fundamentally correct. The law applies when we try to draw as much power as possible from a source and the limit is imposed by a series resistance that we have no control over. That's not the case here. Let's take a hypothetical 6V battery with an internal resistance of 1 ohm. Maximum power transfer to an external load occurs when that external load is also 1 ohm. Half of that power will be wasted inside the battery and efficiency is 50%. But that's often not what we want. More often, we use the battery as a power source and draw what we need from it. A load of 10 ohms results in 3.27 watts of power, out of which 2.975W goes to the external load, while the rest is wasted inside the battery. That's a transfer effciency of about 91%. If I managed to put that across clearly, I think you'll want to revise the approach you use below. > > 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: Jon Kirwan on 21 Feb 2010 15:43 On Sat, 20 Feb 2010 20:46:00 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >Jon Kirwan wrote: >> 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 been in electronics for 4 decades, but I still can't decide >whether I'm a hobbyist or a profressional! :-) I earn a living >with it, but I also approach my work as if it was a hobby. I >waste a lot of time and earn only a fraction of what I could. I >do a lot of things for others for free or for a nominal fee, but >I also have much more fun. It's great when work and fun coincide. That's the way it is with my profession, too, and I count myself very lucky for that fact. >> (I've read about perhaps 20 sufficiently >> different types that professionals might consider... but that >> is happily not my problem.) > >IMO, not even a budding pro should try to master everything at >once. I suggest you concentrate on the configuration presented in >its basic form in Wikipedia. What now look like very different >types may well turn out to be variants of a very few basic types. It's nice to do a survey to get a feel for the variations before drilling down on one or two. Partly, it's just about being able to _read_ your comments and those of Paul's and David's, for example. Sometimes, one of you might bring up something and if I had no idea of the scope of possibilities I might not have any chance to (a) interpret what I read or (b) connect what I read with some other topologies to ask further questions or (c) ask why still others aren't used. It's not vital I know them, in detail. But it does help to at least be able to "recognize" them when I see them. >>> 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: > >I may have been a bit careless with my wording. What I meant was >the ratio at full power output no matter what the designed >maximum power is. That is, the ratio holds equally >(approximately) for a 1W amp and a 100-watt amp. Okay. I think I'm following much better now. >> 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 guess you enjoy reinventing the wheel all the time :-) What I don't like is not knowing _exact_ theory where certain numbers/quantities come from. I think you already understood this when you wrote: "I think he [me] would rather deduce them himself instead of having them handed to him on a platter." I guess I don't mind being told summaries, but I also want to know how to get there entirely on my own, too. I need to work on developing skills necessary to think for myself, when faced with a novel situation. >Most people just use the long-established relation between peak, >rms and average values for known waveforms - sine in this case. Since I'm at the front end of the learning curve, I guess then I'm not most people. >Your mathematical derivation was fundamentally correct, but where >you went astray was in that we are dealing with two half waves in >series because of the configuration. Therefore, the effective >average current drawn from each rail is Ipeak/pi, not 2Ipeak/pi. Perhaps you missed something in my writing. I wrote this (also note above here, too): >: 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. If you read that last equation, you see what amounts to Ipeak/pi. Hopefully, anyway. I think I _did_ follow and didn't make a mistake. >If the two halves were drawn from the same rail, the total would >indeed be 2Ipeak/pi, but current for the two halves are drawn >from two separate rails in series, so the overall average is >still Ipeak/pi. Well, I came at the solution from about the same angle you do here and come up with the same answer, too. So it must be right. ;) >There's a once-common configuration (which you've probably come >across) in which two transistors take their power from the same >rail, and each transistor conducts on alternate half cycles. In >that configuration, the total average current drawn from the >power supply is indeed 2Ipeak/pi. That configuration has dropped >out of favor except for certain cases, mainly because it requires >an output transformer to combine the two output half waves. But >for the time being, let's not complicate things by going further >in that direction. Okay. >> 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 Class A, the output stage would be biased to a steady dc >current at least equal to the peak swing. At full drive in an >ideal Class A, the current will swing between 0 and twice the >steady value, and the average current will still be equal to that >without any output. Even with less-than-full drive, the swing is >still +/- around the bias value, and the average stays the same. Okay. That makes sense. >>> 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_ screwed up again here because I failed to make it clear that >by "any power level", I meant amplifiers designed for any level >of full power output, NOT any level of output for the same >amplifier. I think I got your point. >> 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. > >This is a common misinterpretation of a law that's fundamentally >correct. The law applies when we try to draw as much power as >possible from a source and the limit is imposed by a series >resistance that we have no control over. That's not the case >here. Okay. I was just taking a silly thought, entirely out of context, and "running it up the flag pole." >Let's take a hypothetical 6V battery with an internal resistance >of 1 ohm. Maximum power transfer to an external load occurs when >that external load is also 1 ohm. Half of that power will be >wasted inside the battery and efficiency is 50%. But that's often >not what we want. More often, we use the battery as a power >source and draw what we need from it. A load of 10 ohms results >in 3.27 watts of power, out of which 2.975W goes to the external >load, while the rest is wasted inside the battery. That's a >transfer effciency of about 91%. So the "maximum power" metric would make sense if and only if I were trying to get the most power possible out of some given rail voltage and could adjust the load to any value I wanted to achieve that. In other words, with your 9V battery with 1 ohm internal resistance the external 1 ohm gets the most out of the internal voltage source -- namely 9^2/2 or 40.5 watts, half of that or 20.25 watts occuring in the external load (of course, I'm sure the 1 ohm internal resistance model for a 9V battery would have long since failed to be useful with that kind of external load.) So maximum extractable power is 20.25 watts, math-wise. Your example with 10 ohms requires quite a lot less and simply doesn't meet the maximum power purpose. Which then allows it to be more efficient, of course. Conversely, if the external resistor were 0.2 ohms then total power would be 9^2/1.2 or 67.5 watts while the external power delivered to the load would be 11.25 watts. Down from the maximum power criteria. Applied to the audio amplifier case -- and let's assume for a moment that the computed resistance comes from using the average current and the voltage rail magnitude as already discussed -- the "maximum power analogy" using a calculated 28 ohms in the driver section (removing the 8 ohms in the speaker for a moment) would then suggest the best case would occur when the external impedance was 28 ohms. With 18V rails, maximum power with DC drive to the load would be 18^2/56 or half of 5.785 watts or about 2.9 watts. With 8 ohms in the load and _if_ the audio amplifier continued to "look like" 28 ohms, this would then be 2 watts into the 8 ohms. Which is less than 2.9 watts. However, the assumption that the amplifier is always presenting 28 ohms fails in that case. So that is why the whole idea of using that resistance calculation isn't any good for the purpose to which I applied it. It does make sense, broadly speaking, regarding capacitive loading _because_ the integral I took, it's area, shows _charge_ that must be delivered by the cap (time was included to compute a current.. but the reality is that it was based upon finding the total charge distributed, first.) Sizing a cap based upon necessary charge storage at some voltage makes sense. It's what they do and how they work in a circuit. It fails for the purpose I then forced onto that idea because the drivers don't present that resistance -- it's an artifact of trying to deal with charge on the caps, not the reality of the output driver BJTs. >If I managed to put that across clearly, I think you'll want to >revise the approach you use below. Yes. Thanks, Jon >> 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 21 Feb 2010 16:28 Jon Kirwan wrote: .......<snipped for now>......... Jon, it's getting very close to 3 am here and I've got to sign off for now because I have to get up earlier than my usual 11-12 noon tomorrow. Maybe I glanced through your calculation of Iav too cursorily and wrongly concluded that you made a mistake where you didn't. I'll give it a closer look and come back when I can.
From: Jon Kirwan on 21 Feb 2010 17:24 On Mon, 22 Feb 2010 02:58:40 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >Jon Kirwan wrote: > >......<snipped for now>......... > >Jon, it's getting very close to 3 am here and I've got to sign >off for now because I have to get up earlier than my usual 11-12 >noon tomorrow. I see that you are +530 and I am -800, so we are either +1330 or -1030 apart (using my time as 0000) from each other -- roughly speaking at opposing corners of the day. (Not sure how to apply the chemical analogy of para- to this. Para-dies?) 11AM your time is 11+10.5 or 9:30PM my time. Which suggests you are dead to me from 1PM to 10PM, my time. ;) >Maybe I glanced through your calculation of Iav too cursorily and >wrongly concluded that you made a mistake where you didn't. I'll >give it a closer look and come back when I can. I very much appreciate any thoughts. So a lot of thanks go to you for even taking a moment, at all. Whenever you feel able and willing is nothing less than a fantastic boon to me. Jon
From: David Eather on 22 Feb 2010 07:56
On 17/02/2010 2:28 PM, Jon Kirwan wrote: > On Thu, 11 Feb 2010 17:51:27 +1000, David Eather > <eather(a)tpg.com.au> wrote: > >> <snip> >> It is not a big stress. You can always use the junk-box transformer and >> if it really isn't suitable replace it latter. For your consideration - >> the RMS power of even compressed samples of music is only about 20% of >> the peak. >> >> There are a few variations on that figure. RCA did a lot of research in >> the area and found that Radio broadcasts of compressed FM signals of >> "Rock Music" - an undefined term, was the most demanding at 15%. Some >> companies are trying to redefine that. IRF who call the same figure 1/8 >> of max power (12.5%) - which just happens to make their newest audio >> mosfets look really good. It might be the other way around. They may >> really believe it, and designed the mosfets to match. I forget where but >> some group stated the 20% figure with respect to new modern music >> styles. IIRC they were regarded as technically competent in the area and >> had no axe to grind or wheelbarrow to push - so I filed the info away. >> In any case an overestimate leads to a more conservative design and 5% >> is not much. I'd be wary of definition of "modern music" too - badly >> played organ music can be a stream of full amplitude waveforms that only >> change in frequency at random intervals. >> >> I'd use the junk box transformer and forget about allowing for the >> electricity company slackness and just choose good sized caps that are a >> reasonable price. I think a learning experience allows for a little >> compromise. > > 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. > > I take a little issue with your use of terms in this phrase, > "RMS power of even compressed samples of music is only about > 20% of the peak." Power is average and I don't think RMS > applies to power. Volts-to-power is a squared-phenomenon. So > are amps-to-power. RMS makes sense for those two. But power > is an average (integrated Joules divided by time.) OK. My bad. > > So I believe I have to interpret your meaning as suggesting > that the short-term power required (also an average of some > ill-defined kind, I suppose) when playing music can be a > factor of 5 times more than its long-term average power. You > also mentioned a figure as low as 12.5%, which would suggest > a factor of 8 used as a margin instead of 5. Yes. With a max of a ten watt sine wave output, the average long term output would be between 1.25 watts and 2 watts. BUT that does depend on what you intend the amp to be. The figures would be correct for domestic, most PA and semi pro applications. IF the amp was intended for musicians you would have to allow for the amp to be driven hard into clipping - an output model more like a 12.7 volt square wave with short circuit protection. In serious pro audio setups you might even feel it is appropriate to design the amp to drive full time into short circuits (not that it would be used that way, but during setup and the enevitable modifications to the setup outputs can be shorted for long periods of time). > > But a requirement to support short-term power levels is > really just a compliance requirement on the power supply > rails, isn't it? Yes. > > So put another way, if I wanted a long-term average of 10W > output and I wanted the extra margins required to support the > worst case estimate of a factor of 8 for short-term power > bursts, then I'd need to design rails that support a voltage > compliance level substantially higher. The other ways around. The design will deliver ten watts maximum (disregarding clipping) but the average power output will actually be much lower - hence you can "skimp" a bit on the supply transformer and heatsinks - which wrt overheating have very long time constants relative the the peak output demands. The parts would need > to withstand it, too. And because of the much higher rail > voltages that need to be dropped most of the time, the output > BJTs would need to have just that much more capacity to > dissipate. > > Or put still another way, assuming that my output swing at > the output stage emitters cannot exceed a magnitude of 15V > and that everything is sized for dissipating 10W, does this > mean the amplifier is a 10W amplifier that can support a peak > of 14W=(15^2/(2*8))? (Which isn't so good, considering your > comments above regarding "music?") > > What is meant when one says, '10 watts?' A ten watt amp delivers a sine wave producing 10 watts of output power into a specified load. Ideally this would be 10 watts for an infinite period of time but for audio amps, due to the nature of the signal, an "infinite period of time" in practical terms may be as short as a few seconds. > > This gets worse when I consider the class of operation, > doesn't it? I mean, class-B might be specified as 10W into 8 > ohms, but wouldn't that be 20W into 4 ohms? 40W output. I**2 x R. The power supply voltage is approximately constant. But if class-A, > it's pretty much 10W no matter what? If class A, power is 5 watts out with 4 ohms. Current is held constant. > > I'm beginning to imagine amplifiers should be specified as to > their peak output voltage compliance into 8, 6, and 4 ohms; > instantaneous and sustained without damage to the unit. For > example, 35V into 8 ohms instantaneous, 15V sustained. Or > 80W instantaneous, 15W sustained. That way, someone might > have some knowledge about how well it might handle _their_ > music at, say, 15W average power. And could compare that > against another unit specified as 20V into 8 ohms, 15V > sustained. Your argument here is reasonable but ..... it is also the beginning of the PMPO fiasco - since no advertising department could agree on what constitutes "music" they used what ever figures looked best - and that led to the PFPO (peak fantasy power output) fiasco where you just put anything you like on the box. For a short time some (better) manufactures used a figure they called "headroom" which was the maximum possible instantaneous power output when the power caps are fully charged divided by the long term power output (10 watts in this case). It was always expressed in db - but was confusing to the customer - so it disappeared. > > How does one know what they are buying? What a headache. Wait till you start talking about speakers! > > Jon |