miller & co
There is an interesting article in EW March issue precisely on this. The author makes the case that a combination of input lead and lag compensation in the Vas stage gives the best results. The Vas compensation is modified from the usual cap from C to B, to circumvent the problems of slew limiting and decreasing Vas gain with frequency which would cause increasing distortion with frequency. Check it out.
Jan Didden
capslock said:
There is an interesting article in EW March issue precisely on this. The author makes the case that a combination of input lead and lag compensation in the Vas stage gives the best results. The Vas compensation is modified from the usual cap from C to B, to circumvent the problems of slew limiting and decreasing Vas gain with frequency which would cause increasing distortion with frequency. Check it out.
Jan Didden
I am somewhat surprised that nobody has brought up
two-pole compensation of the VAS. Proponents of this
technique, e.g. Randy Slone, seem to consider it superior
to a single c-to-b capacitor. It seems good to me fromt a
theoretical point of view, but maybe there is some obvious
drawback that I have failed to see. What do you guys think
of two-pole compensation.
two-pole compensation of the VAS. Proponents of this
technique, e.g. Randy Slone, seem to consider it superior
to a single c-to-b capacitor. It seems good to me fromt a
theoretical point of view, but maybe there is some obvious
drawback that I have failed to see. What do you guys think
of two-pole compensation.
The idea of two-pole compensation is to have two capacitors
in series between the base and collector of the VAS. Let C1
be closest to the collector and C2 closest to the base. Further
connect a resistor R from the common point of these capacitors
to ground (usually it is connected to the closest rail instead,
but that is equivalent from a theoretical point of view). C2 is
usually choosen much lower in value than C1, say 10 times
lower.
C1 and R act as a voltage divider giving a pole at frequency
f1. However, the reactance of C2 is much higher than the
resistance of R here, so there is very little feedback at f1.
Let f2 be the frequency where the reactance of C2 equals R,
where f2 is much higher than f1.
The combined effect of this is that the pole at f1 formed by C1
and R will have almost no effect for frequencies below f2. As
a net result we get a flat frequency response up to f2, where
it starts to fall by 12 dB/octave, not the usual 6 dB/octave.
Eventually, at a yet higher frequency the response will switch
over to a 6dB/octave behaviour. One might say that we try
to simulate a double pole at f2 by this method.
With a single Miller capacitor we will usually have to put the pole
at a rather low frequency to ensure stability. With two-pole
compensation, we can often have the simulated double-pole
above the audible band and still ensure stability, since we
roll off the response more sharply.
in series between the base and collector of the VAS. Let C1
be closest to the collector and C2 closest to the base. Further
connect a resistor R from the common point of these capacitors
to ground (usually it is connected to the closest rail instead,
but that is equivalent from a theoretical point of view). C2 is
usually choosen much lower in value than C1, say 10 times
lower.
C1 and R act as a voltage divider giving a pole at frequency
f1. However, the reactance of C2 is much higher than the
resistance of R here, so there is very little feedback at f1.
Let f2 be the frequency where the reactance of C2 equals R,
where f2 is much higher than f1.
The combined effect of this is that the pole at f1 formed by C1
and R will have almost no effect for frequencies below f2. As
a net result we get a flat frequency response up to f2, where
it starts to fall by 12 dB/octave, not the usual 6 dB/octave.
Eventually, at a yet higher frequency the response will switch
over to a 6dB/octave behaviour. One might say that we try
to simulate a double pole at f2 by this method.
With a single Miller capacitor we will usually have to put the pole
at a rather low frequency to ensure stability. With two-pole
compensation, we can often have the simulated double-pole
above the audible band and still ensure stability, since we
roll off the response more sharply.
Christer,
I am not familiar with Slone's explanation of this. I understand your explanation: you are asking why not start the O.L. roll-off at a f above 20kHz, then roll it off at 12dB/oct, then introduce a zero to return the slope to 6dB/oct ahead of the C.L. unity gain f. I suppose you can. You would need to look at the affects upon step response of this more complicated roll-off.
My first question is how do you calculate the pole frequency? How do you know whether it will be at 1kHz or 20kHz? What drives this frequency?
My second question is - why does it matter where the pole is?
I am not familiar with Slone's explanation of this. I understand your explanation: you are asking why not start the O.L. roll-off at a f above 20kHz, then roll it off at 12dB/oct, then introduce a zero to return the slope to 6dB/oct ahead of the C.L. unity gain f. I suppose you can. You would need to look at the affects upon step response of this more complicated roll-off.
My first question is how do you calculate the pole frequency? How do you know whether it will be at 1kHz or 20kHz? What drives this frequency?
My second question is - why does it matter where the pole is?
I have once done the maths for this one applied to op-amps. If someone is interested in the formulas I can supply these ones (which would take some time since I don't have it in electronic form).
The two-pole compensation was once described in an article series within EW+WW (I think it was a series by Douglas Self).
Regarding step response of the amp one has to pay attention to the phase marging then it shouldn't be a problem.
Regards
Charles
The two-pole compensation was once described in an article series within EW+WW (I think it was a series by Douglas Self).
Regarding step response of the amp one has to pay attention to the phase marging then it shouldn't be a problem.
Regards
Charles
traderbam said:
The first pole, at f1, is decided by C1 and R. This pole decides
the basic 6dB/octave roll-off and should be selected to
give stability, just as with a single Miller cap. The second pole
is decided by C2 and R, and "hides" the first pole for frequencies
below f2.
The point of this is that the first pole at f1 will have no effect
for frequencies below f2. That is, even though you place it at
a sufficiently low frequency to guarantee stability, it "does not
affect" the OLG, which will remain flat up to f2, which may often
be above the audible band. Whether it is an advantage to have
a flat response over the audible band is another question. Many
seem to believe this is important, and I think it was one of the
things brought up in the first postings.
Poles apart
Christer,
Can you relay what Slone says about the pole frequencies? It is just that I don't think the poles and zero are related in a simple way to C1, R and C2 due to the feedback around the transistor and due to the influence of the Z on the collector.
If you have some example values for C1, C2 and R and an assumed collector Z I'd like to have them so I can simulate it and confirm it.
Bam
Christer,
Can you relay what Slone says about the pole frequencies? It is just that I don't think the poles and zero are related in a simple way to C1, R and C2 due to the feedback around the transistor and due to the influence of the Z on the collector.
If you have some example values for C1, C2 and R and an assumed collector Z I'd like to have them so I can simulate it and confirm it.
Bam
Re: Poles apart
I see now that I made an error previously, mixing up C1 and C2,
which affected the whole explanation. I am sorry for that.
Maybe that is the reason for your scepticism. Actually I was a
little bit puzzled myself after rereading Slone and posting my
previous explanation. I sensed something was not quite right.
I am not sure if I had forgotten how it works, or if I had never
quite understood it before. Anyway, I'll make a new attempt at
explaining it.
In order not to confuse for those who happen to have Slones
book, I swap the capacitors to agree with him. That is, the
circuit is as before, but C1 is connected to the base and C2
to the collector. As for the choice of values, Slone says that
C1 is choosen in the same way as if we were doing single-pole compensation, ie. using a single Miller capacitor. He then suggests as a rule of thumb that R = 1kOhm and C2 = 10*C1.
If we were only using C1 to do single pole compensation, we
would get a pole at f1, decided by C1 and the input impedance
of the VAS (and/or the output impedance of the input stage?).
With two pole compensation, C2 and R form a voltage divider,
with a pole at f2, where f2 > f1. Below f2, C2 will have a high
impedance compared to R, so there is almost no feedback
signal reaching C1. Above f2, C1 will start to act as a short
circuit, so assuming the impedance of C1 < R here, the circuit
simplifies to the single capacitor C1.
I have made Spice simulations of some of Sloanes amps, and
it seems to work fine. You can find an example in the Optimos
amplifier at Slones web page:
http://www.sealelectronics.com/kits/index.htm
traderbam said:
I see now that I made an error previously, mixing up C1 and C2,
which affected the whole explanation. I am sorry for that.
Maybe that is the reason for your scepticism. Actually I was a
little bit puzzled myself after rereading Slone and posting my
previous explanation. I sensed something was not quite right.
I am not sure if I had forgotten how it works, or if I had never
quite understood it before. Anyway, I'll make a new attempt at
explaining it.
In order not to confuse for those who happen to have Slones
book, I swap the capacitors to agree with him. That is, the
circuit is as before, but C1 is connected to the base and C2
to the collector. As for the choice of values, Slone says that
C1 is choosen in the same way as if we were doing single-pole compensation, ie. using a single Miller capacitor. He then suggests as a rule of thumb that R = 1kOhm and C2 = 10*C1.
If we were only using C1 to do single pole compensation, we
would get a pole at f1, decided by C1 and the input impedance
of the VAS (and/or the output impedance of the input stage?).
With two pole compensation, C2 and R form a voltage divider,
with a pole at f2, where f2 > f1. Below f2, C2 will have a high
impedance compared to R, so there is almost no feedback
signal reaching C1. Above f2, C1 will start to act as a short
circuit, so assuming the impedance of C1 < R here, the circuit
simplifies to the single capacitor C1.
I have made Spice simulations of some of Sloanes amps, and
it seems to work fine. You can find an example in the Optimos
amplifier at Slones web page:
http://www.sealelectronics.com/kits/index.htm
Hi Christer,
Ok I see now. The Slone circuit does increase the pole f at the expense of phase margin in the interval before the zero kicks in.
I didn't get much benefit from using C1=20pF, C2=100pF, R=1k into a 100k collector load, the VAS fed from a current generator. The Slone increased the -3dB f from 1kHz (using only a 20pF miller cap) to 2.2kHz and 30deg of phase margin was lost at 500kHz. Very little phase gain (if any) was observed above 1MHz.
Bam
Ok I see now. The Slone circuit does increase the pole f at the expense of phase margin in the interval before the zero kicks in.
I didn't get much benefit from using C1=20pF, C2=100pF, R=1k into a 100k collector load, the VAS fed from a current generator. The Slone increased the -3dB f from 1kHz (using only a 20pF miller cap) to 2.2kHz and 30deg of phase margin was lost at 500kHz. Very little phase gain (if any) was observed above 1MHz.
Bam
traderbam said:
I guess his rule of thumb for selecting component values will
not work in all cases. I did a lot of Spice simulations on two or
three of Slones amps, playing around with the values in the
compensation circuits and it could make a lot of difference.
It was usually no problem to shift the roll-off corner one or
two decades up or down. I admit it was a bit tricky to find
suitable values sometimes, though, since I did no attempts
at calculating them, but just tried to change the values and
simulate.
Christer,
Nice description of two pole lag compensation.
Typically you'd use 100pF from collector to R, then 1nF from R to base. I've not examined the math; perhaps someone can elucidate the principles?
This technique was used on Russell Breden's amp publish in Electronics World some three or four years back, if memory serves.
A friend built it; some nasty instabilities as I recall, didn't sound that much better, he moved to other, simpler designs.
This approach should pull back gain under unity by the critical pole without as much effect as a single lag comp cap. However, if the collector cap is identical to a normal, the loading is increased by the value of R, but this is trivial, and so it would be reasonable to assume that the slew rate influence would be unchanged over the normal single pole compensation.
So, I have to ask, what is the real benefit here??
Cheers,
Hugh
Nice description of two pole lag compensation.
Typically you'd use 100pF from collector to R, then 1nF from R to base. I've not examined the math; perhaps someone can elucidate the principles?
This technique was used on Russell Breden's amp publish in Electronics World some three or four years back, if memory serves.
A friend built it; some nasty instabilities as I recall, didn't sound that much better, he moved to other, simpler designs.
This approach should pull back gain under unity by the critical pole without as much effect as a single lag comp cap. However, if the collector cap is identical to a normal, the loading is increased by the value of R, but this is trivial, and so it would be reasonable to assume that the slew rate influence would be unchanged over the normal single pole compensation.
So, I have to ask, what is the real benefit here??
Cheers,
Hugh
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