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: pimpom on 28 Jan 2010 08:49 Jon Kirwan wrote: > On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom" > <pimpom(a)invalid.invalid> wrote: > > >> Talking about RCA's 70W amp got me nostalgic about those days. >> Here's a 1W amp using _germanium_ transistors: >> http://img716.imageshack.us/img716/2583/1wamp.png >> This is one of my early solid-state designs based, of course, >> on >> topologies I'd learned by studying others' designs. It's no >> hi-fi >> by any stretch of imagination, but I actually constructed a >> few >> of these in the early 70s for myself and for friends. One of >> them >> fed the input from an early Sony Walkman to drive an 8-inch >> Philips dual-cone "Hi-Q" speaker and gushed over how good it >> sounded! > > Okay. So lets talk about some aspects. It'll expose my > terrible ignorance, but what the heck. I'm no expert myself, but I'm willing to help where I can. > Input loading. I think I can ignore the R2 feedback as it is > 10k. At least, for now. R2 provides both dc and ac feedback. DC for bias stabilisation and setting the emitters of the output transistors to about half of Vcc (more about that later). For ac, it may be easier to think of it as current feedback. Q2 needs about +/-50uA peak of base current at full drive. At signal frequencies, R2 (plus the much smaller input impedance of Q1) is effectively in parallel with the output. The output swings by about 4V peak at max power, which has 400uA of negative feedback current going back through R2. The input current requirement goes up by a factor of 9. IOW, a negative feedback of 19db. This is substantially better than nothing and should significantly reduce distortion and improve frequency response. > C1 will present about Z=800 at > 20Hz, Z=160 at 100Hz, and Z goes down from there. R1 is 1k, > obviously in series with C1. The already low input impedance of Q1 is further reduced by the negative feedback, so R1 represents practically the whole input impedance of the amp. The -3db cutoff frequency is 1/2*pi*C1*R1 which is about 16Hz. > Then there is R3=1k in parallel > with Q1's impedance, which maybe I can approximate as R4 > times beta, or call it 50*33 or about 1500 ohms? No. R4 is bypassed by C3 and has little effect on input impedance except at very low frequencies. > So about > 600 ohms counting that and R3 in parallel, that itself in > series with 1k and whatever C1 presents? So call it around > 2k ohms loading, or so? (Which adds to the idea that the R2 > feedback can be mostly ignored as a load.) Would that be an > okay, off-the-hip guess? Or how would you go about it? > It's mostly the internal dynamic emitter resistance that determines Q1's input impedance. That resistance is 26/Ie at 20 deg C. Q2 is biased at about 7.7mA emitter current, giving about 3.4 ohms. Multiply that by hfe, add the ohmic base resistance and you get Q1's basic input Z. I don't have my old data book handy, but I think the AC126 had a typical hfe of about 150 and rbb of maybe 100 ohms. This gives an input Z of about 600 ohms. > D1 is, I guess, silicon and given that you said _germanium_, > I'll take that to suggest that the Vbe on those are about > half that of a silicon BJT. Which is why only one DR25 was > needed there. > The output transistors need only about 0.1V each of Vbe to bias them at a few mAs of Ic. D1 is germanium, but at the dc current level flowing through it, two of them in series will have too much voltage drop (I measured several samples). Ge transistors have a more rounded knee than their Si counterparts in the Vbe vs. Ic curve. So I felt that a single diode would present less chance of thermal runaway for the output Trs and still cause a reasonably low crossover distortion. Oops. Have to go out for a while. Will take up the rest later.
From: pimpom on 28 Jan 2010 13:50 Jon Kirwan wrote: > On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom" > <pimpom(a)invalid.invalid> wrote: > > >> R6 should also be split and the split point >> bootstrapped with a capacitor to the mid point of the output >> stage. > > That one I really need to think about. This is what I wanted > to happen here. Throwing out things (I'm assuming correct > things, of course) that force me to consider and think. > Thanks. > This is what happens without a booststrap: When it's Q5's turn to conduct on the negative half-cycle of the signal, the base drive current has to come via R6. At the same time, Q5's current is pulling its emitter - and therefore the base - down towards ground, decreasing the voltage drop across R6. This decreases the base drive current available just when it's needed. Now look at it modified with a boostrap: http://img715.imageshack.us/img715/4259/boostrap.png For simplicity, let R6 = R7. At steady-state, C will be charged to about a quarter of Vcc. When Q5 pulls its emitter (and therefore the positive electrode of C) towards ground, the voltage across C cannot change instantaneously and will push its negative terminal down too. Beyond a certain level of drive, the junction of C, R6 and R7 will even go down past oV and become negative with respect to ground. This maintains the voltage across R6 at an approximately constant level. Oops again. Guests this time. Will be back when I can.
From: pimpom on 28 Jan 2010 15:39 Jon Kirwan wrote: > > > D1 is, I guess, silicon and given that you said _germanium_, > I'll take that to suggest that the Vbe on those are about > half that of a silicon BJT. Which is why only one DR25 was > needed there. > I think this is where I left off earlier. > DC bias point of Q1... hmm. Well, assuming no signal, SPK1 > is roughly a dead short, so R5 is tied one side to a rail. > The other side moves Q2's base and Q2's emitter follows. As > Q2's emitter rises with it, R2 and R3 act to split that as > 1/11th to the Q1 base. Q1's emitter follows up for a ways, > allowing DC current via R4 which must go through R5, dropping > Q2's base and thus Q2's emitter, lowering Q1's base voltage > in opposition. So there will be a middle point found. > > Assuming Q1's Vbe should be something on the order of 300mV > (random guess), and I(R4) roughly equals I(R5), let's > establish where Q1's base will wind up. Call it Vb. The > value at Q2's emitter (which is also the other side of R2 > from the Q1 base) will be 11 times higher because R2 and R3 > split things that way. And Q2's base will be 300mV (same > random guess, again) higher than that. The difference > between there and the 9V battery voltage sets the current in > R5 and, by implication, in R4 as well. Of course, Q1's > emitter is 300mV away from that Vb value we are fussing over. > The equation looks like: > > I(R5) = (9V - Vb*11 - 300mV) / 560 > I(R4) = (Vb - 300mV) / 33 > I(R4) = I(R5) > So, > (9V - Vb*11 - 300mV) / 560 = (Vb - 300mV) / 33 > 33/560 * (9V - Vb*11 - 300mV) = (Vb - 300mV) > Vb = 33/560 * (9V - Vb*11 - 300mV) + 300mV > Vb = 33/560*9V - 33/560*Vb*11 - 33/560*300mV + 300mV > Vb + 33/560*Vb*11 = 33/560*9V - 33/560*300mV + 300mV > Vb * (1 + 33/560*11) = (33/560*9V - 33/560*300mV + 300mV) > Vb = (33/560*9V - 33/560*300mV + 300mV) / (1 + 33/560*11) > or, > Vb = 493mV > and thus the current routing through R5, D1, Q1, and R4 is > about 193mV/33 or 5.85mA. That's not the total quiescent > current because D1 uses that 5.85mA to develop a voltage > across it that is probably on the order of 700mV. With that > between the Q2 and Q3 bases, both Q2 and Q3 are passing > collector currents, rail to rail. Hard to know how much > without data sheets, I suppose. But something. Their shared > emitter node would be on the order of 11*490mV or about 5.4V. > > That neglected the base current for Q1 flowing via R2. As > I'm now guessing almost 6mA as Ic, and since we are talking > germanium here, I will pick a beta of about 60 and figure > about 100uA base current, then. That's about another 1V > across R2, less than that a little because that lowers Vb a > bit which lowers the 5.85mA figure a bit, which probably then > gets things very darned close to the midpoint of 4.5V one > might wish there. > > Not too bad given I have no idea about the BJTs and am using > a lot of random guesses as I go. > Your reasoning is correct. However, the output transistors Q2 and Q3 need only about 100mV each of Vbe for Class AB bias. IIRC, beta of Q1 is about 150 and Vbe at that level of current is about 0.12V. I don't know about others, but with low voltage circuits, I usually try to fix the quiescent voltage at the output mid-point at slightly more than half of Vcc. This is because Q1's Ve plus its Vcesat reduces the available downward swing of Q3's base. For this design, I tentatively chose a target of 4.6V at Q3's emitter. Add Q2's Vbe and that leaves 4.3V for R5 plus the speaker's dc resistance. The speaker's resistance has only a minor effect but, just for the heck of it, let's take it as 6 ohms. So Q1's Ic = 4.3/566 = 7.6mA. Q1's dc beta = 150, so Ib is about 50uA, and Ie = 7.65mA = I(R4). 7.65*33 places Q1's emitter at 252.45mV above ground. That plus Vbe of 0.12V gives Vb = 372.45mV. I(R3) = 372.45uA I(R2) = I(R3) + Ib = 422.45uA I(R2)*R2 = 4.2245V V(R2) + Vb = (4.2245 + 0.37245)V = 4.59695V. It just so happens that, in this case, common resistor values produce almost exactly the desired quiescent bias level. If they didn't, a slight departure from the target voltages would be acceptable. In any case, tolerances on resistor values and transistor characteristics could throw off actual values a bit. > R2 is not only a DC divider but also NFB, I think. Can you > talk a little about how you figure on calculating both the > NFB you want _and_ the DC biasing of this thing, both of > which affect R2's value, I think? For such a simple design without a high level of audio quality as the target, I wasn't too particular about the amount of signal feedback as long as it's a reasonable amount. I chose a compromise value for R3 first - low enough for bias stability so that the current through it would be several times Ib, but not too low to avoid excessive shunting of the signal input current. Then I let the value of R2 be what it needs to be for correct bias. Then I calculate the amount of NFB as outlined in one of my earlier replies and accept it if it's within reason. If I really wanted more NFB, I'd parallel R2 with another resistor, but with a capacitor in series to avoid upsetting the dc levels. BTW, that can be used to provide some bass boost by choosing the proper values of cap and resistor. > And although I've _seen_ > miller feedbacks in the small nF range, could you talk a > little about how that was set at 2.2nF? That's a guesstimated value, partly empirical and partly based on observation of other people's designs. No PCs and simulation software 40 years ago. For such a simple circuit, I didn't bother with complex calculations for loops and phase shifts that wouldn't be precise anyway due to wide tolerances in component characteristics. The reason for the relatively high capacitance is that this was a low-Z low-gain circuit. But I might have made a mistake in showing it now as 2.2nF. I might have used something like 1nF. > Also, I think I > _almost_ get the idea of hooking one side of R5 to SPK1 > instead of to the (-) side of 9V... but not quite sure. Can > you talk about that choice, as well? > This is a variation of the bootstrap circuit I described in my other post. R5 and the speaker serve the same functions as R6 and R7 respectively in the other circuit.
From: Jon Kirwan on 28 Jan 2010 21:16 On Thu, 28 Jan 2010 19:19:06 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >Jon Kirwan wrote: >> On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom" >> <pimpom(a)invalid.invalid> wrote: >> >>> Talking about RCA's 70W amp got me nostalgic about those days. >>> Here's a 1W amp using _germanium_ transistors: >>> http://img716.imageshack.us/img716/2583/1wamp.png >>> This is one of my early solid-state designs based, of course, >>> on >>> topologies I'd learned by studying others' designs. It's no >>> hi-fi >>> by any stretch of imagination, but I actually constructed a >>> few >>> of these in the early 70s for myself and for friends. One of >>> them >>> fed the input from an early Sony Walkman to drive an 8-inch >>> Philips dual-cone "Hi-Q" speaker and gushed over how good it >>> sounded! >> >> Okay. So lets talk about some aspects. It'll expose my >> terrible ignorance, but what the heck. > >I'm no expert myself, but I'm willing to help where I can. > >> Input loading. I think I can ignore the R2 feedback as it is >> 10k. At least, for now. > >R2 provides both dc and ac feedback. DC for bias stabilisation >and setting the emitters of the output transistors to about half >of Vcc (more about that later). This much I can see, immediately. And thanks for dealing with more, later on. >For ac, it may be easier to think of it as current feedback. Okay. >Q2 needs about +/-50uA peak of base >current at full drive. At signal frequencies, R2 (plus the much >smaller input impedance of Q1) is effectively in parallel with >the output. R2 is connected from the output to an input, which effectively doesn't move much after arriving at it's DC bias point. As you later point out, the _AC_ input impedance is lowish (near 600 ohms), so the 10k is pretty close to one of the rails at AC, anyway. Is that a different way of saying what you just said? Or would you modify it? >The output swings by about 4V peak at max power, >which has 400uA of negative feedback current going back through >R2. The input current requirement goes up by a factor of 9. IOW, >a negative feedback of 19db. This is substantially better than >nothing and should significantly reduce distortion and improve >frequency response. Okay. This goes past me a little (as if maybe the earlier point didn't.) I'd like to try and get a handle on it. Let's start with the 4V peak swing at max power. Since you are discussing AC and converting it 400uA current via the 10k, I would normally take this to mean 4Vrms AC. Which in Vp-p terms would be 2*SQRT(2) larger, or 11.3V which I know is impossible without accounting for the BJTs, given the 9V supply. So this forces me to think in terms of something else. But what? Did you mean 4Vpeak, which would be 8Vp-p? If so, that would be about 2.8Vrms. In that case, wouldn't a better "understanding" come from then saying that the negative feedback is closer to 280uA? The next point is on your use of "goes up by a factor of 9." Can you elaborate more on this topic? Where the 9 comes from? For volts, not power, I think I can gather the point that 20*log(9) = 19.085), so I'm not talking about that conventional formula. I'm asking about the 9, itself, and also your thinking along the lines of concluding that it significantly reduces distortion. How does one decide how much is enough? > > C1 will present about Z=800 at >> 20Hz, Z=160 at 100Hz, and Z goes down from there. R1 is 1k, >> obviously in series with C1. > >The already low input impedance of Q1 is further reduced by the >negative feedback, so R1 represents practically the whole input >impedance of the amp. The -3db cutoff frequency is 1/2*pi*C1*R1 >which is about 16Hz. I think I follow. Leading towards a comment you make soon below (about C3), C3's very low impedance even after beta multiplication almost bypasses R3 completely so the impedance is as you say, mostly depending upon C1 and R1. I think. >> Then there is R3=1k in parallel >> with Q1's impedance, which maybe I can approximate as R4 >> times beta, or call it 50*33 or about 1500 ohms? > >No. R4 is bypassed by C3 and has little effect on input impedance >except at very low frequencies. Thanks for the knock in the head there. I had been ignoring C3 for DC bais-point thinking and forgot to put it back in when talking about AC loading. Your point is made. >> So about >> 600 ohms counting that and R3 in parallel, that itself in >> series with 1k and whatever C1 presents? So call it around >> 2k ohms loading, or so? (Which adds to the idea that the R2 >> feedback can be mostly ignored as a load.) Would that be an >> okay, off-the-hip guess? Or how would you go about it? > >It's mostly the internal dynamic emitter resistance that >determines Q1's input impedance. That resistance is 26/Ie at 20 >deg C. Q2 is biased at about 7.7mA emitter current, giving about >3.4 ohms. Multiply that by hfe, add the ohmic base resistance and >you get Q1's basic input Z. I don't have my old data book handy, >but I think the AC126 had a typical hfe of about 150 and rbb of >maybe 100 ohms. This gives an input Z of about 600 ohms. Okay. I get your point about small-case 're' based upon kT/q and Ie. I'm mostly following here. >> D1 is, I guess, silicon and given that you said _germanium_, >> I'll take that to suggest that the Vbe on those are about >> half that of a silicon BJT. Which is why only one DR25 was >> needed there. > >The output transistors need only about 0.1V each of Vbe to bias >them at a few mAs of Ic. D1 is germanium, but at the dc current >level flowing through it, two of them in series will have too >much voltage drop (I measured several samples). Ge transistors >have a more rounded knee than their Si counterparts in the Vbe >vs. Ic curve. So I felt that a single diode would present less >chance of thermal runaway for the output Trs and still cause a >reasonably low crossover distortion. Thanks. >Oops. Have to go out for a while. Will take up the rest later. Well, I think I'm roughly following so far. Please kick me where I'm still off-track, if you feel you can afford the moment to do it. I appreciate it very much. Jon
From: Jon Kirwan on 28 Jan 2010 21:21
On Fri, 29 Jan 2010 00:20:15 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >Jon Kirwan wrote: >> On Thu, 28 Jan 2010 14:00:18 +0530, "pimpom" >> <pimpom(a)invalid.invalid> wrote: >> >> >>> R6 should also be split and the split point >>> bootstrapped with a capacitor to the mid point of the output >>> stage. >> >> That one I really need to think about. This is what I wanted >> to happen here. Throwing out things (I'm assuming correct >> things, of course) that force me to consider and think. >> Thanks. >> >This is what happens without a booststrap: When it's Q5's turn to >conduct on the negative half-cycle of the signal, the base drive >current has to come via R6. At the same time, Q5's current is >pulling its emitter - and therefore the base - down towards >ground, decreasing the voltage drop across R6. This decreases the >base drive current available just when it's needed. > >Now look at it modified with a boostrap: >http://img715.imageshack.us/img715/4259/boostrap.png > >For simplicity, let R6 = R7. At steady-state, C will be charged >to about a quarter of Vcc. When Q5 pulls its emitter (and >therefore the positive electrode of C) towards ground, the >voltage across C cannot change instantaneously and will push its >negative terminal down too. Beyond a certain level of drive, the >junction of C, R6 and R7 will even go down past oV and become >negative with respect to ground. This maintains the voltage >across R6 at an approximately constant level. > >Oops again. Guests this time. Will be back when I can. I'm appreciating this very much. Printed the two and am looking at both. I think I'm following, but need to sit down and do a little paper calcs to make sure I burn it in a little more. If I think of something useful, I'll post a question or two. For now, I feel like I'm following you. Jon |