Discussing audio amplifier design -- BJT, discrete
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From: Paul E. Schoen on 30 Jan 2010 13:08 "John Larkin" <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message news:4ch8m516qf4n46fqm1uefefcc48hpb1i90(a)4ax.com... > > Discrete-transistor audio design, being such an ancient practice, > tends to refer to history and authority rather than design from > engineering fundamentals. > > If I were designing an audio amp nowadays (which I certainly aren't) > I'd use mosfets with an opamp gate driver per fet. That turns the fets > into almost-perfect, temperature-independent, absolutely identical > gain elements. That's what I do in my MRI gradient amps, whose noise > and distortion are measured in PPMs. > > ftp://jjlarkin.lmi.net/Amp.jpg > > Why keep repeating a 50-year-old topology when you could have a little > fun? Nice photo. But how about a schematic to show how you implemented the design? I made a simulation of an amplifier design where I used a MOSFET output stage similar to my other post, and I closed the loop with a single op-amp and appropriate negative feedback. I think the OP is looking for a basic learning experience using the simplest components. It may be argued that a MOSFET is simpler than a BJT, but experience with both is a good idea. If the object were just to make an audio amp, there are single package designs and kits that will do the job nicely. I prefer making the circuit using LTSpice, but it can be a thrill to build something with real components. There are usually some gotchas that cause unwanted behavior not indicated in the simulation. But I need a real reason to build something, other than practice with soldering and handling components, so I rarely commit these designs to copper and silicon. Paul
From: Jon Kirwan on 30 Jan 2010 15:41 On Fri, 29 Jan 2010 20:01:53 -0800, John Larkin wrote: >On Fri, 29 Jan 2010 10:34:49 -0800 (PST), George Herold ><ggherold(a)gmail.com> wrote: >> >>"I'd probably replace the two diodes with >>one of those BJT and a few resistor constructions I can't >>remember the name of (which allows me to adjust the drop.)" > >"Vbe multiplier." Got it. Since that time, I've found "rubber diode" as another term mentioned on wiki. ;) >The classic output stage biasing scheme uses small emitter resistors >and biases the output transistors to idle current using a couple of >junction drops between the bases, or a Vbe multiplier with a pot. Both >are good ways to have a poorly defined idle current and maybe fry >transistors. I've already expressed my concern about that. >Two alternates are: > >1. Use zero bias. Connect the complementary output transistors >base-to-base, emitter-to-emitter. Add a resistor from their bases to >their emitters, namely the output. At low levels, the driver stage >drives the load through this resistor. At high levels, the output >transistors turn on and take over. > >2. Do the clasic diode or Vbe multiplier bias, but use big emitter >resistors. Parallel the emitter resistors with diodes. > >In both cses, the thing will be absolutely free frfom thermal runaway >issues and won't need adjustments. Bothe need negative feeback to kill >crossover distortion. I need to get some basics down before I return to these. I'm not there, yet. But I _do_ see an issue with the Vbe multiplier if it isn't crafted carefully for the situation. >Or... > >3. Use mosfets No FETs. Jon
From: Jon Kirwan on 30 Jan 2010 15:37 On Fri, 29 Jan 2010 22:00:03 +0530, "pimpom" <pimpom(a)invalid.invalid> wrote: >Jon Kirwan wrote: >> On Thu, 28 Jan 2010 19:19:06 +0530, "pimpom" >> <pimpom(a)invalid.invalid> wrote: >> >> >>> 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? >> >That's another way of putting it, yes. Okay. Thanks. >>> 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. > >Yes. It's 4Vp, 8Vp-p and 2.8Vrms. I wanted to give you a mental >picture of how much the output voltage can swing. Each output Q >has about 4.4V of Vce available, and about 4V before hard >saturation is reached (these are all round figure values). That's >4V peak for a sinusoidal wave form. Got it. >> In that case, >> wouldn't a better "understanding" come from then saying that >> the negative feedback is closer to 280uA? > >Yes, it's 280uA rms. But I was talking in terms of the maximum >amplitude of instantaneous change, which is why I used the terms >"swing" and "peak". Understood. >> 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 > >Without feedback, the input transistor Q1 needs 50uA of AC input >signal to drive the output Qs to full power output (still talking >in terms of peak to avoid confusion). With NFB, we need an >additional 400uA to overcome the current fed back from the >output. That's a total of 450uA peak, which is 9 times the >original 50uA. Okay. Let me put it in my own words. Since you chose to stick a 10k in for the NFB, and since it supplies a peak of 400uA into the input node and since Q1 itself requires its own 50uA (corrected below), then the signal source itself must "comply" by supplying 450uA peak into the input node. And that is where you got your 9. If that is it, I've got it. >Actually, I made an error when I cited the 50uA figure. Q1 is >biased at Ic = 7.6mA, Ib = 50uA. But only 5mA peak is needed from >Q1's collector to drive the output transistors. Divide that by >Q1's hfe of 150 and you get 33uA (peak) of AC signal current >needed into the base of Q1. The corrected total needed from the >signal source is now 433uA. The gain reduction factor due to NFB >is now 13 instead of 9. That's 22db (feedback is usually given in >db). Okay. I can compute the numbers. I think the problem I'm having with this, as a mental concept, is that the feedback is _from_ the complementary BJT outputs at their emitters _backwards_ into the node at the base of Q1. Not outwards from that node _towards_ the emitters. Let me think a little about this from an AC point of view, not DC. The output is supposedly running with about 4Vpeak into an AC divider made up of what you've earlier described as about 600 ohms to ground via Q1 and about 1k for R1. So given that, we are talking about 1k from input to base node, 600 ohms from base node to ground, and 10k from 4V peak output into base node. The 4V peak has the opposite sign as the input because of the relationship of Q1's collector to its base voltage. I'm just spouting things here without a lot of understanding, so bear with me. The change in Ic of Q1 for a change in the base voltage (computing a transconductance of Q1) is 1 divided by re, which you computed as 3.4 ohms earlier. So gm=294mSeimens, assuming the emitter (small signal wise) follows the base exactly. A 1mV input change at the base yields 1mV*294mS or 294uA change in the collector, ignoring the 100 ohm rbb you earlier mentioned for now. Multiplied by the collector load of about 566 ohms (your figure) produces 166mV change at the collector. A voltage gain (base node to output) of -166. This 166mV change is fed back via the 10k into an existing divider composed of the 1k and the beta-multiplied 3.4 ohms to ground (if I'm following you.) So with the rbb of 100, about 600 ohms as you mentioned (still AC-minded.) So back to the 1k from input, 600 to ground, 10k in from that -166mV change in response to a +1mV change at the base node. The 1k/600 divider means the input had to vary by about 2.66mV to achieve that node change. That means our voltage gain wasn't really -166, but more like 62.5 -- audio input to Q1 collector and then to output drive. Am I going around the barn about right, so far? Here's what worries me now. The postulated thevenin base node change of 1mV is through a thevenin of about 375 ohms (the 1k and 600 ohm splitter.) The 10k feeds into this from the other side with -166mV there. If I imagine 1mV on one side via 375 ohms and -166mV on the other side via 10k ohms, what is the node itself at? Well, (1mV*10k+-166mv*375)/(10k+375) is -5mV or so. This is a lot, isn't it? And it is more than enough to oppose the postulated thevinin change at the input node of +1mV. So I follow the calculation of 22db. The problem I'm having is with what kind of signal will result at Q1's collector. Okay. Granted. I am sure my reasoning fails on some points you will make clearer. I'm just trying to see this in a variety of ways rather than just let you tell me stuff without running through different thinking to see if I get to the same place. So what did I do wrong here? I can't argue with success and I know I'm not doing this right. But there it is. >> also your thinking along the lines of concluding that it >> significantly reduces distortion. > >The basic principle of NFB is that it reduces THD and extends >frequency response by a factor equal to the feedback ratio. So, >in our example, if you have 10% THD without feedback, it will >drop to 0.77% with the feedback factor of 13. But there are >caveats. E.g., phase shifts can cause undesireable effects, >especially with large amounts of feedback. I'm afraid a detailed >treatment of such things is really outside the scope of this >discussion - unless someone else is willing to take it up. I can do phase shift calcs given simple cases and if I take into account all the necessary parts (which, being ignorant about all this, I'm unlikely to do without more thought experiments to clarify my thinking.) So when I get to that point where I can actually walk myself along better, I'll be able to handle that (I hope.) >> How does one decide how much is enough? > >For one thing, how much distortion one is willing to put up with. >Another factor is input sensitivity, or IOW, how much gain is >needed. E.g., to drive the 1W amp to full output, we need 433uA >peak (306uA rms) from the signal source into 1k. That's 306mV >rms, plus some millivolts at the b-e junction. Say about 0.32V >rms total input voltage into about 1k input impedance. > >To present the basic concepts, I've made several approximations. >E.g., I neglected the shunting effect of R2. Besides, the input >resistance of Q1 is constant at 600 ohms only for very small >signal amplitudes relative to the quiescent dc levels. This >dynamic input resistance changes significantly with large signal >swings and adds distortion while also complicating precise >calculations. Okay. I'm going to leave things with the above "issue" laid out for you. It's bugging me right now. I am refusing, by the way, to attempt any simulation. I don't want to be "told" something by a simulator or handed things on some platter. I want to try and work through my thinking, find the flaws, slap myself for them, get back and try again, until I'm good from a paper-and-brain point of view. _Then_ I'll go and check it out to see where the chips fall in the simulator. Might bring up a good question at that point. But until I can get the gross aspects down, that won't really matter. Thanks so much for your efforts so far. It's been helpful to me, at least. Jon
From: John Larkin on 30 Jan 2010 17:28 On Sat, 30 Jan 2010 12:41:16 -0800, Jon Kirwan <jonk(a)infinitefactors.org> wrote: >On Fri, 29 Jan 2010 20:01:53 -0800, John Larkin wrote: > >>On Fri, 29 Jan 2010 10:34:49 -0800 (PST), George Herold >><ggherold(a)gmail.com> wrote: >>> >>>"I'd probably replace the two diodes with >>>one of those BJT and a few resistor constructions I can't >>>remember the name of (which allows me to adjust the drop.)" >> >>"Vbe multiplier." > >Got it. Since that time, I've found "rubber diode" as >another term mentioned on wiki. ;) > >>The classic output stage biasing scheme uses small emitter resistors >>and biases the output transistors to idle current using a couple of >>junction drops between the bases, or a Vbe multiplier with a pot. Both >>are good ways to have a poorly defined idle current and maybe fry >>transistors. > >I've already expressed my concern about that. The basic tradeoff is to use big emitter resistors to prevent thermal runaway, but that wastes power at large signal swings. Another tradeoff is to use a small Vbe multiplier voltage (ie, small quiescent DC drops across the emitter resistors) to reduce idle power and heatsink temp at the cost of more crossover distortion. Or change the rules. Semiconductors are cheap, heatsinks are expensive. Take a crack at calculating the thermal runaway situation of a typical class AB output stage. It's interesting. John
From: George Herold on 31 Jan 2010 00:31
On Jan 29, 3:11 pm, Jon Kirwan <j...(a)infinitefactors.org> wrote: > On Fri, 29 Jan 2010 09:19:31 -0800 (PST), George Herold > > <ggher...(a)gmail.com> wrote: > >Hi Jon, I'm enjoying your posts. > > Thanks. I feel like I'm way behind some curves, but it's fun > taking a moment to think about things and it is fantastic > that anyone else is willing to help talk about things with > me. That is priceless. So the real thanks go to those who > are sharing their knowledge and experience here. > > >What's a pin driver? > > Hmm. I think I first heard the idea when talking about > testing ICs, to be honest. But imagine instead a micro with > software to test some discrete part (could be an IC, too, > that that's more complex.) For example, to automatically > derive some modeling parameters for a BJT. > > Take a look at this datasheet, for an example of the features > one might support: > > http://www.analog.com/static/imported-files/Data_Sheets/AD53040.pdf > > >I made a nice > >switchable current source (10nA to 1mA) from a voltage reference, > >opamp and switchable resistors. (circuit cribbed from AoE.) > > I'd require at least one that can either sink _or_ source to > the pin. And that would be only one of the pin driver's > required features. I think the datasheet mentioned above > provides some more. But that part is expensive and not > readily available to us hobbyist types and doesn't teach me > anything about various trade-offs I might want to make or how > to design it at all, besides. > > Jon Wow, that's some chip. Thanks, George H. |