Discussing audio amplifier design -- BJT, discrete
in [Basics]
|
Prev: Watch replacemnt (button) batteries?
Next: Low Power Satellite Based Laser / Imaging System Could Easily Track Activity Disturbing Cheap Reflecting Fibers
From: David Eather on 24 Feb 2010 06:20 On 24/02/2010 6:06 PM, Jon Kirwan wrote: > On Wed, 24 Feb 2010 15:03:18 +1000, David Eather > <eather(a)tpg.com.au> wrote: > >> I'm leaving all the sniping to you because your the one who knows what >> you want to investigate further. > > "Sniping" in the US has a negative connotation, which I'm not > sure you intended. I hope I'm not coming across in some > negative way. If so, I do apologize and will try for better. > I really do appreciate the time you've offered me. > >> Also, for today, I am using mostly using "perfect components" and >> theoretical efficiencies etc to make my life easier - they will show the >> point and I am talking about an amp spec'd for consumer audio. > > Okay. > >> On 23/02/2010 2:41 AM, Jon Kirwan wrote: >>> On Mon, 22 Feb 2010 22:56:22 +1000, David Eather >>> <eather(a)tpg.com.au> wrote: >>> >>>> 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> >>>>> >>>>> But a requirement to support short-term power levels is >>>>> really just a compliance requirement on the power supply >>>>> rails, isn't it? >>>> >>>> Yes. >>> >>> Okay. These concepts are slowly settling into my brain. >>> >>>>> 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. >>> >>> Cripes! Really? So a 10W amplifier isn't designed to >>> actually deliver a full 10W steadily into a load? That's the >>> peak power capability? Cripes. >> >> No, that's the power of a 12.7 volt (peak) sine wave into an 8 ohm load. >> An instantaneous peak power figure would be (Vmax**2)/R or 20watts. > > Clearly understood. I just wasn't thinking well at the > moment. I'm exactly with you on this. > >>> Let me put this another way. I design a class-B output stage >>> with rails capable of 10W compliance into 8 ohms (roughly 13V >>> peak, so rails at maybe +/-17V or so?) With 10W into the 8 >>> ohm load, let's say this means the upper power BJT is >>> handling about 4-5W and the lower BJT is handling 4-5W, as >>> well. Call it 10W total dissipation inside the amp while 10W >>> are dissipated in the speaker. >>> >>> But I don't have to go find BJTs able to dissipate 4-5W, >>> because the 10W spec is just a max-unsustained case and the >>> real situation is more like 2W into the load, continuous? In >>> short, I need to find a BJT that only needs to dissipate 1W >>> for the high side and 1W for the low side? I could use two >>> PN2222As in parallel to do that! >>> >>> I can _cheat_ like that and call it a 10W design? It doesn't >>> actually _have_ to sustain 10W without burning up? >> >> Think of it this way: >> >> You build an amplifier that puts out a 10 watt sine wave into 8 ohms >> 100% of the time. For a power transformer you will need something like a >> 30 volt CT rated at 40VA for this design (this one is a realistic not >> theoretical estimate). >> >> Use this amp in a consumer environment and the customer is happy about >> everything except the cost. >> >> Can you lower the cost without damaging the output quality? >> >> Yes. As mentioned earlier the long term average output power of the >> amplifier will be about 2 watts and any transformer will have a very >> long thermal time constant compared to any other component in the >> amplifier, so there is no danger of overheating during a peak in the >> music output. 40VA x 0.2 = 8VA. An 8VA transformer is big enough (in >> real life you would use a little bigger because of the increased I**2 x >> R losses 10VA would probably be a good choice, (if you had more >> information you could make a better choice but the result would be very >> close to 10VA) >> >> A 10VA transformer costs a whole lot less than 40VA and all you have >> done is removed an unnecessary over-specification of a component and >> that will have zero effect to the consumer. >> >> You can do the same thing with the heatsink, but it is not so dramatic a >> change, and it needs more care. On a small amp like this where the cost >> of a heatsink is low I wouldn't bother - except as an exercise or if you >> were making hundreds of them. You won't be able to proceed here until >> you have a more finalized design. > > I think I follow all of this. I guess my earlier writing was > about my own realizations and nothing else. I wrote more > strongly then because I'm just "seeing" a little better, is > all. > >>> Okay, now I'm depressed. I go buy a 50W amplifier, stick a >>> sine wave signal generator on it and watch the thing toast >>> itself, bursting in fire soon enough? >>> >>>>> 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. >>> >>> Yeah. A few seconds. So... now I can go back with a >>> quasi-comp output stage and use a pair of those PN2222As for >>> it, without heat sinking! Nice little TO92 packages, too. ;) >>> >>>>> 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. >> >>> I was looking at some actual measurements taken by Mr. Self >>> on an actual class-B amplifier when I wrote that. I didn't >>> do a theory-based analysis. Just read off the figures when >>> he was comparing a class-A with a class-B into different >>> loads. >>> >>> Now I'll do that. >> >> recheck everyone's figures > > Tentatively, I'm lumping his tabled results into effects I'm > less aware of, for now. Context will become clearer, later. > >>> I had then imagine it came from V^2/R and knowing that V^2 >>> remains the same for a given amplifier and only the R changed >>> from 8 to 4. Which makes sense then that it would double, >>> not quadruple, the power output. From an I^2*R perspective, >>> I get the same estimate because a smaller load does double >>> the current, but the R divides in half, so the combination is >>> still just twice, not four-times. >>> >>> Can you explain this 40W statement better for me? >> >> 20 Watts - you were right. My figure wrong. > > Thanks. > >>>>> 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. >>> >>> Again, looking at Self's chart (page 322 on his 5th edition) >>> I see a slight degregation into 4 ohms, going from about 20W >>> into 8 ohms to 15W into 4 ohms. I'm not entirely sure of >>> 'theory' here, but I took this to suggest that at the higher >>> currents the drive circuitry's compliance coupled with the >>> likely somewhat lower gain caused by somewhat higher currents >>> now needed accounted for the droop. >>> >>> But his chart certainly doesn't suggest 1/2 rated power. >>> >>> I guess I need to delve into this a bit more to make sure I >>> understand. The class-B case seems easier for me to follow >>> (assuming I'm right, above, which I of course may not be.) >> >> You seem to be doing OK. Maybe we should brush up and compare notes on >> the meanings of average and RMS but that's about it I think. >> >> A class A amp say at 10 watts into 8 ohms will have an output stage with >> a constant current sink (or source) set at 1.59 amps. > > I know there are a number of structures, but I like to think > in terms of two BJTs, one NPN and one PNP, in a push-pull > arrangement to the rails -- for class-A -- with a Vbe > multiplier set to cause both BJTs to have at least some > emitter current at all times. > > I know that there is also a single-ended arrangement. But I > never give that one more than a very slight nod. It's way, > way too inefficient to care about. I'm wondering if that is > what you are talking about here. > > If so, then you'd indeed set Iq to be 1.59 amps, either to > the (+) or (-) rails, as I read you saying here. Because > then the single-ended BJT can either source/sink nothing > causing the -1.59 (or +1.59) amps to flow from speaker to > rail or the single-ended BJT can source/sink 3.18 amps, only > 1.59 of which gets wasted via Iq and the rest going to the > speaker. > > However, I don't think much about that arrangement and I'm > sure that Self wasn't talking about an amplifier designed > that way. I'm pretty sure he was discussing a push-pull > class-A amplifier. > >> If the speaker >> load changes to 4 ohms the maximum current into and out of the speaker >> is still 1.59 amps. > > In the single-ended case, _very_ generally, yes. But let me > walk you through my single-ended thoughts. (I hate single > ended designs, so I hope I don't have to think about them > again!) > > Let's assume we have (+), (-), and ground rails. Let's > arrange it so that the Iq current is a sink, as you stated. > So it goes from (-) to the speaker pin. There is a BJT that > goes from the speaker pin and up to (+). It's base is driven > by the VAS and the emitter simply follows that. The speaker > load goes to ground. That's what I'm imagining you are > talking about. > > At 8 ohms and Iq = -1.59A, lets say that the emitter can rise > up to the point where Iload = +1.59A. (So about +12.7V at > the emitter, as discussed regularly.) This means the BJT's > emitter must be sourcing 3.18 amps, enough to supply both the > speaker and Iq. If the BJT effectively turns off and its > emitter current goes to about 0, then Iq causes Iload = > -1.59A. > > A 4 ohm load would still "see" no lower than -1.59A, since > that is all that Iq can do. However, when the emitter rises > again to it's +12.7V (driven by the same exact signal at its > base, by assumption), then we will have 12.7V across 4 ohms > and +3.18A into the speaker. The BJT will not only have to > supply that, but also the 1.59A required by Iq. That's 4.77A > total. So it will operate in class-AB, now. > > So I'm pretty sure I'm _not_ following you when you say the > output current is the same. Which I take to mean there is > something very wrong with the way I'm seeing this. I am/was thinking of a single supply class A amp. It has a great big output capacitor. > >> How's the power now? > > Best to wait for your knock on my head about this. I would > have first preferred to talk about a push-pull class-A case, > which seems fundamentally different. But I'm still learning > and have to assume I am getting all of this wrong. > >> Self is talking about his practical results and if you dig around you >> will see/find he believes in over-biasing the output stage current >> source by 50% - 100% hence the apparent anomaly. >> <snip> > > I think he was NOT talking about single-ended class-A > designs, but instead push-pull class-A. I get your point, I > think, about over-biasing -- it is something I already feel > I'd want to do, too... though I'm not sure 50% is right and > even then I don't think he spent much of any time at all > talking about single-ended designs (for reasons I think I > agree with.) > > I feel like I probably failed to get some point I should have > and so I'll stop here and wait. > > Very much appreciated, > Jon
From: Jon Kirwan on 24 Feb 2010 13:40 On Wed, 24 Feb 2010 21:20:45 +1000, David Eather <eather(a)tpg.com.au> wrote: ><snip> >I am/was thinking of a single supply class A amp. It has a great big >output capacitor. Then I think class-AB remains the mode of operation when the 8 ohm is replaced with the 4 ohm output, per your question to me about that. A single-sided class-A with a 1.59A sink and an appropriately sized rails would barely work class-A with 8 ohms. And would move into AB, driving 4. I think. I talked a bit about the topology I was considering, earlier, so hopefully I didn't get that part wrong even if I did fail to add the output cap to the description. Just to be clear, here is what I'm imagining right now: >: V+ >: | >: | >: |/c Q1 >: VAS ----| TIP3055 >: |>e >: | C1 >: | || BIG >: +----||----, >: | || | >: | \ >: | / R1 >: / \ \ 8 or 4 >: | I1 / >: v 1.59A | >: \ / | >: | gnd >: | >: gnd Just an emitter follower feeding a sink and the speaker via a cap. Maybe I'm getting that wrong, though. Jon
From: Paul E. Schoen on 24 Feb 2010 19:50 "Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message news:3rrao55usonbnvt12kv6iqg89sdeacc9no(a)4ax.com... > On Wed, 24 Feb 2010 21:20:45 +1000, David Eather > <eather(a)tpg.com.au> wrote: > >><snip> >>I am/was thinking of a single supply class A amp. It has a great big >>output capacitor. > > Then I think class-AB remains the mode of operation when the > 8 ohm is replaced with the 4 ohm output, per your question to > me about that. A single-sided class-A with a 1.59A sink and > an appropriately sized rails would barely work class-A with 8 > ohms. And would move into AB, driving 4. I think. > > I talked a bit about the topology I was considering, earlier, > so hopefully I didn't get that part wrong even if I did fail > to add the output cap to the description. Just to be clear, > here is what I'm imagining right now: > >>: V+ >>: | >>: | >>: |/c Q1 >>: VAS ----| TIP3055 >>: |>e >>: | C1 >>: | || BIG >>: +----||----, >>: | || | >>: | \ >>: | / R1 >>: / \ \ 8 or 4 >>: | I1 / >>: v 1.59A | >>: \ / | >>: | gnd >>: | >>: gnd > > Just an emitter follower feeding a sink and the speaker via a > cap. Maybe I'm getting that wrong, though. My conception of a class A amplifier is one where instead of an active current sink, there is just a resistor. It may be in the form of an emitter follower as in this case with a unity gain, or the resistor may be in the collector to obtain a voltage gain greater than one. But in these cases, large signal linearity is not realized. So such a configuration is used for very small signals that are at least an order of magnitude smaller than the supply rails, and power levels in the milliwatt range. With a resistor load, the maximum power output is where the output impedance equals the resistor, and the maximum voltage that can be achieved is about half the supply rails. When one adds an active current source or sink, it involves another transistor and the circuit becomes essentially a half-bridge. For the circuit shown above, with I1 = 0.75V and V+ = 12 VDC, a sine wave of 6V amplitude will be reproduced across R1 = 8 ohms. With R1 = 4 ohms, the current source must be set to 1.5A. The efficiency under these conditions is 25%, and 4.5 W output. But this assumes that the current source can pull the output below ground, which is not possible with any practical component. And I did not factor in the power provided by the current source, so actual efficiency will be lower. If I use a 4 ohm resistor as the emitter load, and bias Q1 so that there is equal clipping at the output, I can get an output of about 7.6 volts P-P, or 1.8 W. The efficiency is about 8%. If I bias the resistor for 1/2 the supply rail (6 V), I can get at most about 2.2 VRMS into 4 ohms, or 1.2 W. Efficiency is 6.6%. With a realizable current source made from a 2N3055 and a 0.2 ohm emitter resistor, set at 1.55A, I can get about 3.9 watts into 4 ohms, and an efficiency of 20%. Now, I decided to see if I could get better efficiency by adding a variable current sink. Essentially I am now making a push-pull circuit where the lower half is not pulling so much when the upper half is pushing, and then it pulls harder when it needs to do so during the negative excursions of the signal. It simply required two additional components. My LTSpice simulation shows an output of about 3.9 W into 4 ohms, with an efficiency of about 27%. This design is similar to class A in that it burns up about 15 watts with low level signals. As such, maybe it is not so much and amplifier as an "Apple-fryer" :) And under those conditions the output stage is running 1.2A. But that is better than running 1.5 A as was the case with the original design. There does not appear to be any crossover distortion, and at high signal levels you just get clipping, and that occurs within 1 volt of the supply rails. As Scotty might say, "Cap'n, she just caint give ye no more!" OK, I've played around enough. I've attached the ASC file if anyone wants to play with it or criticize it. I just used a "shotgun" approach with simplicity in mind. Maybe it's worth building, but I'm happy enough with the usual Class B or AB amplifiers that don't function as space heaters when they're just sitting there. And a class A power amplifier will never win an Energy Star! Go green! Use PWM! Paul ============================================================ Version 4 SHEET 1 880 680 WIRE 16 -80 -160 -80 WIRE 240 -80 16 -80 WIRE 240 -64 240 -80 WIRE 160 -16 -80 -16 WIRE 176 -16 160 -16 WIRE 16 0 16 -80 WIRE 16 96 16 80 WIRE -160 112 -160 -80 WIRE 240 128 240 32 WIRE 288 128 240 128 WIRE 416 128 352 128 WIRE 160 144 160 -16 WIRE 240 160 240 128 WIRE -80 208 -80 -16 WIRE 176 208 112 208 WIRE 416 208 416 128 WIRE 16 240 16 192 WIRE 112 240 112 208 WIRE 112 240 16 240 WIRE 112 256 112 240 WIRE 240 272 240 256 WIRE -160 352 -160 192 WIRE -80 352 -80 288 WIRE -80 352 -160 352 WIRE 112 352 112 336 WIRE 112 352 -80 352 WIRE 240 352 112 352 WIRE 416 352 416 288 WIRE 416 352 240 352 WIRE 416 368 416 352 FLAG 416 368 0 SYMBOL npn 176 -64 R0 SYMATTR InstName Q1 SYMATTR Value 2N3055 SYMBOL cap 352 112 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName C1 SYMATTR Value 5000� SYMBOL res 400 192 R0 SYMATTR InstName R1 SYMATTR Value 4 SYMBOL voltage -80 192 R0 WINDOW 3 -71 180 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value SINE(6.75 5.6 400 0 0 0 1000) SYMBOL voltage -160 96 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V2 SYMATTR Value 12 SYMBOL res 224 256 R0 SYMATTR InstName R2 SYMATTR Value .2 SYMBOL npn 176 160 R0 SYMATTR InstName Q2 SYMATTR Value 2N3055 SYMBOL res 96 240 R0 SYMATTR InstName R3 SYMATTR Value 15 SYMBOL pnp 80 192 R180 WINDOW 0 52 29 Left 0 WINDOW 3 60 68 Left 0 SYMATTR InstName Q3 SYMATTR Value 2N3906 SYMBOL res 0 -16 R0 SYMATTR InstName R4 SYMATTR Value 50 SYMBOL res 176 128 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R5 SYMATTR Value 100 TEXT -66 392 Left 0 !.tran .1 startup
From: Jon Kirwan on 24 Feb 2010 21:50 On Wed, 24 Feb 2010 19:50:15 -0500, "Paul E. Schoen" <paul(a)peschoen.com> wrote: >"Jon Kirwan" <jonk(a)infinitefactors.org> wrote in message >news:3rrao55usonbnvt12kv6iqg89sdeacc9no(a)4ax.com... >> On Wed, 24 Feb 2010 21:20:45 +1000, David Eather >> <eather(a)tpg.com.au> wrote: >> >>><snip> >>>I am/was thinking of a single supply class A amp. It has a great big >>>output capacitor. >> >> Then I think class-AB remains the mode of operation when the >> 8 ohm is replaced with the 4 ohm output, per your question to >> me about that. A single-sided class-A with a 1.59A sink and >> an appropriately sized rails would barely work class-A with 8 >> ohms. And would move into AB, driving 4. I think. >> >> I talked a bit about the topology I was considering, earlier, >> so hopefully I didn't get that part wrong even if I did fail >> to add the output cap to the description. Just to be clear, >> here is what I'm imagining right now: >> >>>: V+ >>>: | >>>: | >>>: |/c Q1 >>>: VAS ----| TIP3055 >>>: |>e >>>: | C1 >>>: | || BIG >>>: +----||----, >>>: | || | >>>: | \ >>>: | / R1 >>>: / \ \ 8 or 4 >>>: | I1 / >>>: v 1.59A | >>>: \ / | >>>: | gnd >>>: | >>>: gnd >> >> Just an emitter follower feeding a sink and the speaker via a >> cap. Maybe I'm getting that wrong, though. > >My conception of a class A amplifier is one where instead of an active >current sink, there is just a resistor. I first thought of that, as well. But for the purposes at hand, it seemed a lot easier to plop a current sink in there. David had written this to me: "A class A amp say at 10 watts into 8 ohms will have an output stage with a constant current sink (or source) set at 1.59 amps. If the speaker load changes to 4 ohms the maximum current into and out of the speaker is still 1.59 amps. How's the power now?" I didn't want to wind up "getting corrected" for failing to read well what he wrote, introducing some pre-conception of mine. That's why you see my schematic which _uses_ a current sink. I'm trying to read David as accurately as I can and construct from his words what I think he may be talking about. To do otherwise would be to _change_ the subject on him and talk at cross-purposes. >It may be in the form of an emitter >follower as in this case with a unity gain, Yes, that's clear -- now that we are quickly moving to change the subject. :) >or the resistor may be in the >collector to obtain a voltage gain greater than one. Um. After moving the speaker/cap connection up there, too? Right? >But in these cases, >large signal linearity is not realized. You mean in the case where a resistor is used in the emitter and where a collector resistor may (or may not) be used. Right? In the case of the current sink I attempted, when trying to follow David, it seems 'large signal linear' -ish to me. (Speaking loosely. Except for Vbe variations on Ic and maybe also the Early effect, anyway.) >So such a configuration is used for >very small signals that are at least an order of magnitude smaller than the >supply rails, and power levels in the milliwatt range. With a resistor >load, the maximum power output is where the output impedance equals the >resistor, and the maximum voltage that can be achieved is about half the >supply rails. The configurations you now brought up? Or the one that I was talking about, earlier, when trying to deal with David's question to me? >When one adds an active current source or sink, it involves another >transistor Yes, that's a given of sorts. And that is why I'd almost certainly prefer to go with a push-pull style class-A of some kind. It seems crazy to go single-ended under the circumstances. >and the circuit becomes essentially a half-bridge. A term I need to follow a little better, I suppose. I would use it in the case of two diodes instead of four in a full-wave, center-tapped PS with CT to ground and only one other rail. You are using it differently than that, here. Which makes me feel behind the terms-curve, still. >For the >circuit shown above, with I1 = 0.75V and V+ = 12 VDC, Um... I1=.75A? Not 'V', right? (I assume we are getting back to my ASCII schematic, now.) >a sine wave of 6V >amplitude will be reproduced across R1 = 8 ohms. Yes, assuming as I know you must be that the drive is nicely centered on +6V so that it goes from 0V to +12V -- which is what I take you to mean here. Actually, maybe 5.2V or so would be better, so that the emitter can follow up and down well. >With R1 = 4 ohms, the >current source must be set to 1.5A. Yes, this much I understand. The question that David was asking me, if I understood him accurately, didn't permit me to arbitrarily change the current sink value. As such, the example case you are bringing up would be a more accurate analogy to his question if you kept the current source at .75A and changed R1 to 4 ohms. My reply, at least, was made on that basis. >The efficiency under these conditions >is 25%, and 4.5 W output. But this assumes that the current source can pull >the output below ground, Two issues here. One with and one without the output cap that David wisely mentioned in his response to me. In the case without the output cap, the current sink needs access to a rail _below_ that used by the speaker load's other end. Otherwise, if they are common to each other, then there is a DC bias current flowing through the speaker and that's not really a good thing. In the case with it, the cap provides the necessary 'most- negative' side for the speaker and allows, after a few cycles to pump up an equilibrium voltage on it, a DC center of 0A for the speaker. > which is not possible with any practical >component. Without the cap. With the cap, you are still right in that a 0V on the base of Q1 does not mean that the emitter can sink to -0.8V or whatever, since there is no rail there for it. (Unless some extra windings are added to the transformer to get it, of course.) >And I did not factor in the power provided by the current >source, so actual efficiency will be lower. Yup. Understood. I think Self says 12.5% is the best to be hoped. I've not done my own double-check. But with your estimate and adding in an equal amount for the sink, that seems to get to about there. >If I use a 4 ohm resistor as the emitter load, and bias Q1 so that there is >equal clipping at the output, I can get an output of about 7.6 volts P-P, >or 1.8 W. The efficiency is about 8%. If I bias the resistor for 1/2 the >supply rail (6 V), I can get at most about 2.2 VRMS into 4 ohms, or 1.2 W. >Efficiency is 6.6%. > >With a realizable current source made from a 2N3055 and a 0.2 ohm emitter >resistor, set at 1.55A, I can get about 3.9 watts into 4 ohms, and an >efficiency of 20%. > >Now, I decided to see if I could get better efficiency by adding a variable >current sink. Essentially I am now making a push-pull circuit where the >lower half is not pulling so much when the upper half is pushing, and then >it pulls harder when it needs to do so during the negative excursions of >the signal. It simply required two additional components. My LTSpice >simulation shows an output of about 3.9 W into 4 ohms, with an efficiency >of about 27%. Thanks for the circuit. I also ran it under LTspice. Selecting from 100ms to 500ms as a range by which things have settled out well, the resistor shows about 3.9 watts and 14.1 watts from the rail supply. Which gets to your number. As you hoped, most of the 14 watts is in Q1, at about 6W. Q2 shows about 2.9W. >This design is similar to class A in that it burns up about 15 watts with >low level signals. As such, maybe it is not so much and amplifier as an >"Apple-fryer" :) :) >And under those conditions the output stage is running 1.2A. But that is >better than running 1.5 A as was the case with the original design. There >does not appear to be any crossover distortion, and at high signal levels >you just get clipping, and that occurs within 1 volt of the supply rails. >As Scotty might say, "Cap'n, she just caint give ye no more!" > >OK, I've played around enough. I've attached the ASC file if anyone wants >to play with it or criticize it. I just used a "shotgun" approach with >simplicity in mind. Maybe it's worth building, but I'm happy enough with >the usual Class B or AB amplifiers that don't function as space heaters >when they're just sitting there. And a class A power amplifier will never >win an Energy Star! Go green! Use PWM! Thanks, Paul. All discussion is most welcome to me. I appreciate it very much. Jon
From: Paul E. Schoen on 24 Feb 2010 21:55
"Paul E. Schoen" <paul(a)peschoen.com> wrote in message news:pQjhn.13494$Ab2.6638(a)newsfe23.iad... > > When one adds an active current source or sink, it involves another > transistor and the circuit becomes essentially a half-bridge. For the > circuit shown above, with I1 = 0.75V and V+ = 12 VDC, a sine wave of 6V > amplitude will be reproduced across R1 = 8 ohms. With R1 = 4 ohms, the > current source must be set to 1.5A. The efficiency under these conditions > is 25%, and 4.5 W output. But this assumes that the current source can > pull the output below ground, which is not possible with any practical > component. And I did not factor in the power provided by the current > source, so actual efficiency will be lower. Like 13%. The magic active current source was contributing about half the power. Paul |