Maintaining a Vbe Multiplier's bias value
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From: George Herold on 8 Feb 2010 22:16 On Feb 8, 8:39 pm, Jon Kirwan <j...(a)infinitefactors.org> wrote: > I think this fits in sci.electronics.design, not .basics. > > I'd like to consider the Vbe multiplier often used in audio > amplifiers to maintain a bias voltage for the output stage. > The purpose is to better mitigate against ripple in the > unregulated power supply rails and against the the VAS > voltage output resulting from amplified signal voltages. > > (The only active device under consideration is a BJT, though. > No JFETs or MOSFETS or opamps or other ICs.) > > The basic starting form for a Vbe multiplier is shown in Fig. > 1 and the bias voltage output is indicated there. Assume Q1 > is thermally coupled in some magic way, for now, in just the > right way so that if the current through the Vbe multiplier > were perfectly stable, that the bias voltage would track just > as needed (The 'Eg' of Q1 is exactly what's needed for the > output stage's temperature tracking in some nice way and the > values of R1 and R2 are set correctly and the thermal > coupling and location is somehow where it needs to be.) The > focus is on the Vbe multiplier's variation of bias in the > face of changes in sourcing current at the top of Fig. 1. > > > > > > >: +V > >: | > >: resistor or > >: current source > >: | > >: ,---+---, > >: | | > >: \ | > >: / R2 | > >: \ +-----> upper quadrant > >: / | ^ > >: | | | > >: | |/c Q1 BIAS > >: +-----| VOLTAGE > >: | |>e | > >: \ | | > >: / R1 | v > >: \ +-----> lower quadrant > >: / | > >: | | > >: '---+---' > >: | > >: VAS ---' > >: > >: FIGURE 1 > > If I use a resistor as the load for the VAS, it's obvious to > me that the Vbe multiplier will need to cope with varying > currents. But even if I use a BJT (or two) to make a current > source sitting above the Vbe multiplier, it's still not going > to hold entirely still with +V ripple and with varying VAS > drive voltages. That variation will ultimately manifest > itself in a varying Vbe bias voltage. That will change the > operating point for the output stage. > > If it is class-A, I suppose it doesn't matter that much. But > I don't want to be forced into class-A operation. Nor do I > want to be forced into regulated rails. So it becomes a > little more important, I think, to get this nailed down > better. > > There's the problem, anyway. > > To quantify how bad all this really is, I tried my hand at > figuring out the small signal analysis of the Vbe multiplier. > If I got a first approximation about right, it is based > squarely upon the small-re of the BJT. The very familiar > value for (kT/q)/Ic. > > There is also the value of R2 shown in Fig. 1, but since its > effect is only affected by the change in base current, I > believe it's contribution is divided by Q1's beta. So the > actual equation is something like: > > R_ac = (1/Ic)*(kT/q)*(1+R2/R1) + R2/beta > > For a 2X multiplier where R2 is about R1, this is: > > R_ac = (2/Ic)*(kT/q) + R2/beta > > The Vbe multipler value is: > > V_bias = Vbe*(1+(R2/R1)) + R2*Ic/beta > > (The latter term being a correction for base current.) > > Ignoring base current for now and assuming I had Ic set > around 5mA and placed R1=R2=1k for the 2X factor, this R_ac > value works out to about 15.4 ohms. > > A variation of half an mA in Ic yields about 7.7mV change in > the bias point. > > I decided to see if the Early effect made much of a > difference. The adjustment appears to be something like > this: > > R_early = dV/dI = -Ic/VA*R^2 > > If I'm interpreting it right, it really does show as negative > resistance added to R_ac. The fuller equation, then, > including the Early effect, would be: > > R_ac = (2/Ic)*(kT/q) + R2/beta - Ic/VA*R^2 > > (Which requires a quadratic solution to solve for R.) > > If R_ac is 15.4 ohms and Ic is around 5mA, a VA of 100V would > suggest about R_early=-10mOhms. Which is roughly a factor of > 1500 less than 15.4 Ohms. Since it now appears to be on the > order of 0.1% or so for typical Ic, VA, and, R_ac values, I > think I can ignore it for these considerations. > > So drop it, I will. > > I had scouted around a few weeks back (not for this reason) > and found what is shown in Fig. 2. I remembered it, but > didn't understand it then. > > > > > > >: +V > >: | > >: resistor or > >: current source > >: | > >: ,---+---, <-- node A > >: | | > >: | \ > >: | / R3 > >: \ \ > >: / R2 / > >: \ | > >: / +-----> upper quadrant > >: | | ^ > >: | |/c Q1 | > >: +-----| BIAS > >: | |>e VOLTAGE > >: \ | | > >: / R1 | v > >: \ +-----> lower quadrant > >: / | > >: | | > >: '---+---' > >: | > >: VAS ---' > >: > >: FIGURE 2 > > I think I now understand why R3 was there. Changes in Ic > create changes in Q1's collector voltage, per Ic*R3. The > result is that dV=dI*R3. If R3 is on the order of the above > computed R_ac, then variations at node A caused by changing > currents through the Vbe multipler (most of which are seen as > Ic changes) will be neatly compensated for the change in the > voltage drop caused by R3. > > However, that can only be set for some assumed Ic. Nearby > changes will work pretty well. But further deviations will > start to show problems again. Also, the Fig. 2 version will > use a slightly higher multiplier value to get node A up high > enough for the R3 drop to hit the right place required to > bias the output stage. That higher multiplier means that > while, let's say, the two (or four, if that's it) output > BJT's Vbe values vary over temp and the thermally coupled Q1 > above also varies it's own Vbe value, the multiplier other > than 2 (or 4) will mean the variation of the bias will match > at only one place -- if it ever did more than one spot. How > important that is, I've not considered yet. > > I'm wondering about additional topology changes to improve > the performance still more. Obviously, if they are crazy and > wild, I'm probably going to live with the above and be done > with it. But I think there's got to be something still > better. Another BJT as a bypass route across Q1 and R3? > > Getting this nailed down should help mitigate against both > unreg supply ripple (on one side, anyway) putting hum into > the output and also against large scale changes in the VAS > amplified signal voltage (which means distortion.) > > Jon- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - "I'm wondering about additional topology changes to improve the performance still more." Hi Jon, I've been 'sorta' following your thread on s.e.basics. I wonder if you abandoned class A operation too early? Why not keep things linear evreywhere and avoid the dead band? So what if you need a bigger heat sink. Its certainly a lot simpler. George H.
From: Jon Kirwan on 8 Feb 2010 23:43 On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >Hang a big capacitor across it. Nice try. Jon
From: John Larkin on 8 Feb 2010 23:49 On Mon, 08 Feb 2010 20:43:03 -0800, Jon Kirwan <jonk(a)infinitefactors.org> wrote: >On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin ><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: > >>Hang a big capacitor across it. > >Nice try. > >Jon No, seriously, that solves a bunch of problems. John
From: Jon Kirwan on 9 Feb 2010 01:06 On Mon, 08 Feb 2010 19:09:28 -0700, Jim Thompson <To-Email-Use-The-Envelope-Icon(a)My-Web-Site.com> wrote: >What's a "VAS"? Sorry. I read it somewhere regarding audio amplifiers and the term stuck in my mind, I suppose. It's short-hand for Voltage Amplifier Stage. It's almost so simple that no one would bother creating a term for it, except that it seems as though someone did and folks have used it in places where I've been reading. By the way, if you look at the semi-conceptual schematic at the top of this page: http://en.wikipedia.org/wiki/Electronic_amplifier You will see Q3 acting as the VAS. Together with R6 it converts the beta multiplied current into drive voltage. (The Vbe/Ic transfer nasties this up, but I think it may be survivable. Everything is important, but I'm leaving worrying about this till later.) That schematic isn't entirely realistic, either. R3/R4 are better replaced with a mirror, regular, Wilson, or otherwise. R5 is often itself a current source or sink (depending on which way you flip the schematic polarities) and may be a BJT and diodes or two BJTs, etc. >What exactly are you trying to do? If you look again at the schematic mentioned above, note the function of D1 and D2. They stack to create a bias voltage. That's used to set the point of operation for the output stage (two-quadrant emitter follower -- which may be just two BJTs as in that picture, or more.) Often, this is replaced with an adjustable BJT configured as a Vbe multiplier. That's what I'm trying to do. Except that I'd like to have the +V and -V supply rails (ground is also present in the system) be unregulated. Part of the function of the Vbe multiplier is to also track the Vbe requirements for the output stage as it heats up and cools down. The variation of Vbe is quite large, as you know, where the controlling Eg term in the Is(T) equation overwhelms the otherwise oppositely-signed dV/dT of the Shockley equation. Above -2mV/K. And with the exponential dependance of Ic on Vbe... well, it serves that function as well. So the Vbe value needs to track temperature in just such a way that it maintains the design operating point for the output stage, over temperature, while also ignoring variations in the current that sources through it. I'm trying to keep my options open, regarding the amplifier's class. If it were operating class-A all the time, my limited understanding suggests that some variation across the Vbe multiplier isn't nearly as important as it clearly would be for, say, class-B operation. I'm not exactly sure where I want to wind up biasing things. So I am slowly learning this stuff and, assuming the Vbe multiplier has some part within it thermally coupled as appropriate to some well-chosen part of the output stage, trying to gather how I'd: (1) stabilize the voltage at some fixed temperature T against variations in the current flowing through it, and (2) calibrate it's Vbe multiplication factor in just the right way so that it tracks well with the effective Eg found in the Is(T) function of the output stage needed to hold the operating point steady vs temperature. My question here was regarding (1), not (2). I'm not far enough along on that one to even begin on that one, yet. To be honest, I just started learning about audio amplifier design, including terms like VAS, starting around the 26th last month. So I may be far off the mark in a few places. I'm finding it a very interesting education, though, and I'm glad I started down the road a small bit. But "being exact" about what I want remains part of the learning process, itself. So what you see here is as far as I've gotten to. >My nickname, as a kid engineer at Motorola (48 years ago), was "Vbe" >Thompson, because I could pull so much magic with Vbe compensation >methods ;-) Well, I can believe it. And I mean that as a sincere compliment. If you can suggest something still better than what I've already posted, I'd like to look at it. >(Vbe multipliers generally are used just to create a smaller dead-band >that is temperature stable. In this case, I want it to track the output stage so I'm going to have to couple it thermally in some useful way. What I'm considering, right now, is how to make it immune to unregulated supply variations and VAS output voltage swings. >Class AB bias is an art form of which I >am expert, but cannot divulge publicly at this time :-) Well, I want to examine class-AB at some point. It may be where I want to settle, though class-B would be quite fine for my needs. If you can't help with class-AB, then you can't. I will have to struggle along. However, anywhere else you can send me a clue I'd certainly appreciate it. There is no interest other than personal. Certainly nothing commercial in mind. I'm just a hobbyist trying to learn. Jon
From: Jon Kirwan on 9 Feb 2010 01:11
On Mon, 08 Feb 2010 20:49:24 -0800, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Mon, 08 Feb 2010 20:43:03 -0800, Jon Kirwan ><jonk(a)infinitefactors.org> wrote: > >>On Mon, 08 Feb 2010 17:54:13 -0800, John Larkin >><jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >> >>>Hang a big capacitor across it. >> >>Nice try. >> >>Jon > >No, seriously, that solves a bunch of problems. > >John Which problems does a slew-dependent, C*dV/dt bypass current solve? Jon |