Discussing audio amplifier design -- BJT, discrete
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From: Jon Kirwan on 8 Feb 2010 17:11 On Mon, 08 Feb 2010 11:54:42 -0800, I wrote: ><snip> >(Also, I suppose, the Early effect will add yet another >slight modification, since the Vce is slightly changed so is >the Ic for the same Vbe. The higher required Ic (assuming >the current source or resistor is supplying more current, >instead of less) requires a higher Vbe, as stated. So the >multiplied voltage at Vce is higher. But that multipled >voltage also slides over on the Vce axis for whatever Vbe >that has become and that suggests still more Ic due to Early >effect, so it is a positive feedback contributing to the >already existing problem, I think. I haven't tried to work >out just what percent it contributes, though.) ><snip> I'm rethinking this. I think the Early effect counters the effect, slightly. The higher Ic does increase Vbe. The increased Vbe does increase Vce. So Vce is higher and we are also on a different Vbe curve in the Vce vs Ic graph where the Early effect shows up clearly. But Ic is 'given' because it is being divided by the structure, so I should have rotated the chart in my mind and using Ic as the independent variable. Had I done that, I would have 'seen' that a given rise in Ic would have suggested (holding Vbe constant for a moment) a certain change in Vce. But that a now slightly higher Vbe would have chosen a new Vbe curve that (before rotating the chart) is _above_ the earlier Vbe curve, which will cut the new Vbe curve where that same Ic intersects it a little sooner on the Vce axis. Thus, it acts against. My gut was telling me that nature works to oppose change and I should have gone with that instinct, I think. The effect is small. On the order of about 0.1%, roughly. Jon
From: Jon Kirwan on 8 Feb 2010 17:21 On Mon, 08 Feb 2010 11:54:42 -0800, Jon Kirwan <jonk(a)infinitefactors.org> wrote: >On Tue, 2 Feb 2010 22:30:31 +0530, "pimpom" ><pimpom(a)invalid.invalid> wrote: > >>Jon Kirwan wrote: >>> On Sat, 30 Jan 2010 01:11:28 +0530, "pimpom" wrote: >>> >>>> <snip> >>> I've seen this as a modification. In ASCII form: >>> >>>>> A >>>>> | >>>>> ,---+---, >>>>> | | >>>>> | \ >>>>> | / R3 >>>>> \ \ >>>>> / R2 / >>>>> \ | >>>>> / +--- C >>>>> | | >>>>> | | >>>>> | |/c Q1 >>>>> +-----| >>>>> | |>e >>>>> \ | >>>>> / R1 | >>>>> \ | >>>>> / | >>>>> | | >>>>> '---+---' >>>>> | >>>>> B >>> >>> We've already decided that R1 might be both a simple resistor >>> plus a variable pot to allow adjustment. The usual case I >>> see on the web does NOT include R3, though. However, I've >>> seen a few examples where R3 (small-valued) exists and one of >>> the two output BJTs' base is connected at C and not at A. >>> >>> The above circuit is a somewhat different version of the Vbe >>> multiplier/rubber diode thing. The difference being R3, >>> which I'm still grappling with. >> >>I've seen R3 used in that position too, but never gave it much >>thought until you brought it up. Offhand I still can't see a >>reason for it either. Maybe for stability against a local >>oscillation? Perhaps taking some time to think about it will >>bring some revelation. Or someone else can save us the trouble >>and enlighten us. >><snip> > >I had earlier said I thought you might be right about this R3 >value. Now, I don't. I think it deals with something else >-- unregulated rail voltage variations. > >In this thread, you've posted circuits with a resistor on one >side and the VAS on the other of this structure. The VAS >yanks one end around while the other side mostly follows it >around. However, the current through the resistor varies, of >course. Even if one of those BJTs+2 diodes thingies is used >for a current source instead of the resistor, which does >improve things, it still isn't very constant. Using the 2 >BJT structure doesn't change that fact, though it does impact >variations. No matter how you arrange it, resistor or >current source, the fact is that the current into the Vbe >multiplier device changes around as the VAS yanks around one >side of it. > >This variation means that Q1's Ic varies. To accomodate that >variation, Vbe varies. Since Vbe varies, so does the >multiplied value. And for no _other_ reason than variations >in signal. That changes the bias. Changing the bias changes >the quiescent current. Etc. > >(Also, I suppose, the Early effect will add yet another >slight modification, since the Vce is slightly changed so is >the Ic for the same Vbe. The higher required Ic (assuming >the current source or resistor is supplying more current, >instead of less) requires a higher Vbe, as stated. So the >multiplied voltage at Vce is higher. But that multipled >voltage also slides over on the Vce axis for whatever Vbe >that has become and that suggests still more Ic due to Early >effect, so it is a positive feedback contributing to the >already existing problem, I think. I haven't tried to work >out just what percent it contributes, though.) > >So a cludge fix for this is to insert a resistor in the >collector, which will act in the opposite direction to some >degreee. I'd imagine this would create a second degree poly >curve, with a maximum somewhere but gentle 'arms' outward, >which means less variation of the Vbe-multiplied value with a >tweakable peak point based upon a nominal Ic. > >I imagine this is NOT nearly as important for class-A >operation, though, since it is already "biased up" and >variation at that point of operation probably isn't so >important. It _would_ matter, I think, in other classes of >operation. > >Thinking as I am that I don't want to go with class-A, I am >trying to think of still better ways of replacing R3 with (or >adding) an active device to further improve it. Anyone have >a suggestion there? > >Jon Side bar: The small signal analysis of the Vbe multiplier, if I got it right, is based squarely upon the small-re of the BJT which is, itself: (kT/q)/Ic. There is, of course, also the value of R2, but since its effect is only affected by the change in base current, it's contribution is divided by beta. So the actual equation is something like: R = (1/Ic)*(kT/q)*(1+R2/R1) + R2/beta For a 2X multiplier where R2=R1, this is: R = (2/Ic)*(kT/q) + R2/beta Note the variation on Ic! (We _want_ the variation on T but do NOT want the variation on Ic.) With R2=1k, for example, that part contributes 5 ohms. But with Ic=5ma, for example, the other part of it is sitting at 10.4 ohms -- a total of 15.4 ohms. So a variation of half an mA in Ic suggests a 7.7mV change in the bias point, ignoring any residual Early effect on it. I mentioned the Early effect being on the order of 0.1%. It took me a moment to think about it, but the figure works out to something like this: R_early = dV/dI = -Ic/VA*R^2 It's a negative resistance that adds to R. If R is 15.4 ohms and Ic is around 5mA and VA=100V, for example, you get about -10mOhms. Which is roughly a factor of 1000 less than 15.4 Ohms. Which is where I get the 0.1% as a ballpark estimate. The fuller equation would be: R = (2/Ic)*(kT/q) + R2/beta - Ic/Va*R^2 Which requires a quadratic solution to solve for R. I'm comfortable, for now, that I can ignore the Early effect and just focus on the broader R figure for the Vbe multiplier, applying changes in Ic to it to see how the voltage shifts around. So, in the diagram, R3 now reduces the effect by that change in Ic*R3. If R3 is on the order of the above computed R, that generally sets things so that right at that Ic used to compute R the effect of R3 will be just in the right amount to compensate for nearby changes in Ic. I think. Jon
From: pimpom on 9 Feb 2010 02:07 John Larkin wrote: > On Sun, 7 Feb 2010 16:10:14 +0530, "pimpom" > <pimpom(a)invalid.invalid> > wrote: > >> Jon Kirwan wrote: >>> On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin >>> <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>> >>>> >>>> A few percent distortion at power levels is essentially >>>> inaudible. >>>> Speakers do that already. Low-level crossover distortion is >>>> obvious >>>> and obnoxious. >>> >>> Yes, I think that's now much clearer to me now than it was >>> say two weeks ago -- without needing my ears to say so. Just >>> on understanding better _what_ crossover distortion is and >>> does. >>> >> Jon, to see a graphical illustration of JL's point, see this >> screenshot of some simple simulations I just ran: >> http://img694.imageshack.us/img694/8967/crossoverdistortion.png >> >> On the right, the complementary output stage is driven without >> any bias, . The upper trace shows the output when the input >> amplitude is +/-1V peak. The transistors are operating in >> Class C >> and manage to conduct for less than half of each half cycle. >> Now >> that's going to sound awful by any standard. I _know_ it >> sounds >> awful because, when I was doing a lot of repairing work on >> consumer products in the 80s, I came across some amps whose >> biasing circuit had developed a fault. >> >> Do a Fourier analysis and you'll get lots of harmonics. Reduce >> the input amplitude even further and there won't be any output >> at >> all below a certain amplitude. >> >> The lower trace shows the output with +/-9V input. Crossover >> distortion is much reduced, though still evident. This may or >> may >> not be acceptable depending on the application. For anything >> that >> needs good audio quality, any waveform distortion that can be >> clearly seen in graphical form is still too high, especially >> in a >> low-resolution bitmap trace like this. >> >> On the left, we have the same amp with diode biasing added. >> Visible distortion of the waveshape has disappeared. The >> slight >> irregularities in the sinusoidal shape are due to limitations >> of >> the low-res, non-antialiased bitmap image. >> > > In your "unbiased" circuit, try adding a 1K resistor from the > bases to > the output. Now the flat spots in the output waveform become > slopes. > But the transfer function is now continuous, so negative > feedback can > reduce distortion without ugly slewing problems. And with zero > bias, > there's no idle power dissipation and no possibility of thermal > runaway. > > The second circuit, with the bias diodes, is a likely firebomb. > > John I thought it would be obvious that those were simplified arrangements purely for illustration. No one in his right mind would think of using them in a practical audio amp.
From: pimpom on 9 Feb 2010 03:43 John Larkin wrote: > > Mag tapes had inherently gross crossover distortion. The fix > was to > add a pretty high-level "bias" oscillator to the record path to > smear > it out. The bias voltage might be 20 volts at 60 KHz, way > bigger than > the record signal going into the head. > DC bias of the recording head was also used in many cheap portables. The erase head was a permanent magnet that swiveled out of the way on playback. On recording, the erase head magnetised the tape to saturation in one direction. The dc bias is polarised in the opposite direction, with enough strength to place the operating point in the linear region. The noise level is higher than with AC erase and bias, but it works.
From: Paul E. Schoen on 9 Feb 2010 16:01
"pimpom" <pimpom(a)invalid.invalid> wrote in message news:hkr77v$lvl$1(a)news.albasani.net... > John Larkin wrote: >> >> Mag tapes had inherently gross crossover distortion. The fix was to >> add a pretty high-level "bias" oscillator to the record path to smear >> it out. The bias voltage might be 20 volts at 60 KHz, way bigger than >> the record signal going into the head. >> > > DC bias of the recording head was also used in many cheap portables. The > erase head was a permanent magnet that swiveled out of the way on > playback. On recording, the erase head magnetised the tape to saturation > in one direction. The dc bias is polarised in the opposite direction, > with enough strength to place the operating point in the linear region. > The noise level is higher than with AC erase and bias, but it works. I had one of those cheap cassette recorders, and it worked OK. But I had an 8-track tape player in my car and I wanted to be able to record, so I built an AC bias circuit (I think mine was 40 kHz), using a circuit that I found in an old databook. It incorporated the RIAA non-linear amplitude curve as well. I also made a device which used a cheap turntable and crystal pickup, with two J-FET (2N3819) linear amplifiers and VU meters. http://en.wikipedia.org/wiki/RIAA_equalization http://freecircuitdiagram.com/2008/11/27/gramophone-pre-amp-a-pre-amplifier-with-riaa-response-curve/ This was in 1970, and the only decent piece of test equipment I had was a refurbished HP140A scope, with 100 kHz bandwidth and a fast blue phosphor CRT made for photography. I still have everything except the 8-track tape player for the car. It was stolen, along with most of my tapes, when I foolishly neglected to put them in the trunk when I parked in a marginally bad neighborhood in Washington, DC in 1972. Paul |