Hi, Mikeks,
I wanted to ask you your opinion.
Assume I've already use a good output stage (classA or EC).
What do I get and what I don't get if I use
1. High OL front stage (high feedback design)
2. Low OL front stage (low feedback design)
I wanted to ask you your opinion.
Assume I've already use a good output stage (classA or EC).
What do I get and what I don't get if I use
1. High OL front stage (high feedback design)
2. Low OL front stage (low feedback design)
As much forward-path gain as you can get, with as much intrinsic linearity as possible.
Low feedback designs are for the perpetually deluded.
Low feedback designs are for the perpetually deluded.
Hi Andrew T
R12 - just to balance
probably better without R12 and input side readjusted.
cheers
John
R12 - just to balance
probably better without R12 and input side readjusted.
cheers
John
mikeks said:
Hi Mike,
Mike,
I must disagree with many of the points you have made here. Let me respond to each main point in a separate post.
Your first point had to do with my concern about the two halves of a complementary VAS “fighting” each other when there was a Miller compensation capacitor returned from the VAS output to each of the pnp and npn bases. I'll try to explain it better here.
What I had in mind when I brought this up was the very popular dual differential input pair architecture, in which a pnp input differential pair provides single-ended drive to an npn VAS transistor, and in which a parallel-fed npn input differential pair provides a single-ended drive to a pnp VAS transistor. An example of this approach, with two Miller compensation capacitors, is the Leach amplifier.
Consider the back-of-the-envelope analysis below where I assume a generic version of an amplifier with this complementary topology. I first assume that the npn and pnp input differential pairs are perfectly matched and that the Miller compensation capacitors are identical. Under these conditions, the npn and pnp VAS transistors should each contribute the same amount of signal current to moving the output node of the VAS against the net load capacitance on that node. We thus have the desired symmetrical push-pull operation of the VAS. I assume a signal excursion of 40V peak at 20 kHz and analyze the circuit and look at the currents. I then assume slightly mis-matched Miller compensation capacitors, one 5% high, the other 5% low, and see what happens. Here we see the effects of the “fight” I was worried about. We are connecting together the outputs of two shunt-feedback amplifiers (the shunt feedback being provided by the Miller compensation capacitors).
Under the ideal conditions, each VAS transistor contributes 285 uA peak to the signal excursion, for a total contribution of 570 uA. Under the condition of slightly mis-matched Miller capacitors, the VAS transistor surrounded by the smaller capacitor contributes 460 uA peak, while the VAS transistor surrounded by the larger capacitor contributes 110 uA peak. Most of the symmetry advantages have thus been lost with only a small mismatch in Miller capacitors.
Generic Complementary Pair Example:
Assume each transistor in each input pair biased at 1 mA
Assume each emitter is degenerated with 225 ohms
Single-ended gm of each pair is then about 2 mmho
Assume the VAS is biased at 10 mA and each emitter is degenerated with 100 ohms
Assume the VAS is preceded with an EF buffer
DC voltage at VAS base is 1.0V+0.6+0.6 = 2.2V relative to the rail
We therefore set the Rload RL of diff pair = 2.2 K
Zin of VAS = 250,000 ohms (transistor betas of 50)
Assume amplifier closed loop gain = 20
Assume amplifier closed loop bandwidth = 500 kHz
Assume Cmiller = Cc around each VAS Darlinton
For 500 kHz closed loop bandwidth, Cc = 0.159 * gm / (CLG * Fc)
= 0.159 * 2e-3 /(2e1 * 5e5) = 0.0318 e-9 = 32 pF
Assume miscellaneous load on VAS collector is 50 pF
Total capacitive load on VAS collectors = 32 pF + 32 pF + 50 pF = 114 pF
Assume 40V peak at 20 kHz at VAS output
Impedance of 114 pF at 20 kHz = 0.159 / (0.114e-9 * 2e4) = j70k.
Total VAS signal current = 40V/70k = 570 uA peak
Each VAS transistor contributes half of this signal current
VAS signal current per side = 285 uA peak (per VAS transistor)
VAS signal input voltage = 100 ohms * 0.29 mA = 29 mV peak
Signal current in input pair load resistor RL = 29 mV/2.2k = 13 uA pk
Impedance of Cc at 20 kHz = 0.159 / (32e-12 * 2e4) = j250k
Current in Cc = 40V/250k = 0.16 mA pk = 160 uA pk.
Each input diff pair must be putting out 160+13 = 173 uA pk single-ended.
This is 17.3% of max peak input pair output current of +/- 1 mA
Error signal input = 173 uA/2 mmho = 173e-6/2e-3 = 86.5e-3 = 87 mV
NOW, assume Cc top is 5% high and Cc bottom is 5% low.
Cc top = 33.6 pF; Cc bottom = 30.5 pF
I Cc top = 168 uA
I Cc bottom = 152 uA
Each diff pair is still putting out 173 uA pk
The left-over current is what drives the 2.2k RL and forms the drive voltage for the VAS
I RL top = 173 – 168 = 5 uA net
V RL top = 5 uA * 2.2k = 11 mV peak
VAS signal current top = 11 mV / 100 ohms = 110 uA peak
I RL bot = 173 – 152 = 21 uA net
V RL bot = 21 uA * 2.2k = 46.2 mV peak
VAS signal current bot = 46mV / 100 ohms = 460 uA peak
Total VAS signal current = 110 + 460 = 570 uA peak.
A +/- 5% tolerance difference in Cc capacitors causes strong VAS imbalance.
The benefits of symmetrical balanced push-pull VAS are lost.
This is what I was concerned about.
Cheers,
Bob
Bob, you make a good point about cap mismatch, but I have always found that it is difficult to make current mirror second stages that are as good as complementary drive. I attribute this to thermal differences in the mirror devices. I usually assign a value of 2 to the difference in distortion between a typical current mirror second second stage and a complementatry driven second stage, and a value of 5 to a typical single drive stage with an active load, from measurements over the years.
Also, I must point out the value of 'pole splitting' that the Miller compensation gives. This is pointed out on page 616 of my copy of: 'Analysis and Design of Analog Integrated Circuits' by Gray and Meyer. I think that this can be important as well.
Also, I must point out the value of 'pole splitting' that the Miller compensation gives. This is pointed out on page 616 of my copy of: 'Analysis and Design of Analog Integrated Circuits' by Gray and Meyer. I think that this can be important as well.
john curl said:
Hi John,
I agree with what you have said here. In particular, the pole-splitting nature of Miller compensation is important and not to be under-estimated. In a later post, I'll point out that the form of compensation I use also provides the pole-splitting advantages.
I also agree that the current mirror approach to getting the complementary VAS drive needs to be executed carefully in order not to suffer higher distortion due to the difference in the paths that the two polarities of the signals take. None of these circuits is perfect, that is for shure. I have always said that one of the biggest influences on an engineer's choices is what he fears the most; because different engineers fear different things, they make different choices, all of which are usually sensible in the context of what they wish to avoid. In my case, my fear here was the "fight" between the two (upper and lower) halves of the complementary VAS.
Cheers,
Bob
Bob Cordell said:
In my youth I feared DC drift in this stage without a nice CCS.
I've gotten over it.
😎
mikeks said:
Hi Mike,
You expressed concern about my assertion in an earlier post that Miller compensation was sub-optimal with respect to slew rate, and asked for more specifics. Here I’ll try to do a better job of explaining what I meant.
Conventional Miller compensation creates a link among input pair bias current, input pair gm, closed loop gain, closed loop bandwidth and achievable slew rate. The slew rate is I/C, where I is the peak output signal current available from the input stage, and C is the value of the Miller compensation capacitor. The Miller capacitor, by being wrapped around the CE VAS, effectively makes the forward path into an integrator. Indeed, the term Miller Integrator is familiar to many. At the same time, the compensation capacitor C is set by the combination of input stage gm, closed loop gain, and closed loop bandwidth. Once you set these, with conventional Miller compensation you have set the slew rate.
Here is an example:
Assume a degenerated differential input pair with RE = 225 ohms per emitter
Assume this pair provides single-ended drive to the base of a CE VAS transistor
Assume 1 mA through each transistor
Net gm = 2 mmoh (single-ended output of diff pair); gm = 0.002
Maximum available current for slewing = +/- 1 mA (single-ended output of diff pair)
Now set the compensation assuming a 500 kHz gain crossover
(most power amplifiers will have a gain crossover between 200 kHz and 2 MHz)
Assume 26 dB closed loop gain = 20:1.
To achieve this gain crossover frequency, we need OLG of 20 at 500 kHz.
Xc = 20 / gm at 500 kHz = j10,000 ohms
Cm = 0.159/(10k * 500k) = 32 pf
Slew rate SR = I/C = 1e-3 / 0.032e-9 = 31 V/us
Setting value of Miller cap and determining slew rate:
Xc = CLG/gm at Fc where CLG is closed loop gain and gm is that of input stage
Cm = 0.159/Xc * Fc where Fc is the NFB gain crossover frequency
Cm = 0.159 * gm/(CLG * Fc)
SR = I/C = I * CLG * Fc /(0.159 * gm) = 6.28 * (I/gm) * CLG * Fc
Notice that I/gm is a key factor in this; degenerating the input pair increases I/gm.
In the case here, I/gm = 1 mA/2 mmho = 0.5
Notice that amount of NFB and closed loop bandwidth do not enter into this expression.
There are other compensation schemes, like input compensation, that break the relationship SR = 6.28 * (I/gm) * CLG * Fc and permit a higher slew rate for a given set of relevant parameters. This is what I mean by conventional Miller compensation being sub-optimal for slew rate.
Keep in mind that in some designs, the above slew rate cannot be achieved because of other circuit design issues that may limit it needlessly. Also keep in mind that high slew rate is only one of many parameters influencing the sonics of an amplifier.
Cheers,
Bob
Hi Bob,
You've, in essence, covered ground previously set out in Solomon and other works.
This, however, does not validate your blanket averment that Miller compensation gives ''sub-optimal slew rates'' in audio applications.
From your example above, a slew rate of 31V/uS is obtained, which, it would appear, you deem inadequate.
But, of great significance, you neglect to state the output voltage swing with respect to which this figure, and, therefore, implied inadequacy is referred.
If it is assumed that your design swings many kilovolts then there could be no rational disagreement.
If, on the other hand, your amplifier is merely required to swing, say, 30V peak, then your figure of 31V/uS is handsomely in excess of the absolute minimum by several orders of magnitude.
You've, in essence, covered ground previously set out in Solomon and other works.
This, however, does not validate your blanket averment that Miller compensation gives ''sub-optimal slew rates'' in audio applications.
From your example above, a slew rate of 31V/uS is obtained, which, it would appear, you deem inadequate.
But, of great significance, you neglect to state the output voltage swing with respect to which this figure, and, therefore, implied inadequacy is referred.
If it is assumed that your design swings many kilovolts then there could be no rational disagreement.
If, on the other hand, your amplifier is merely required to swing, say, 30V peak, then your figure of 31V/uS is handsomely in excess of the absolute minimum by several orders of magnitude.
mikeks said:
Yes, what I described is well-known stuff, and I did not claim it to be new. I just thought that you might be unaware of it.
I said that Miller compensation was sub-optimal in providing slew rate, meaning that other forms of compensation can do better in the same situation. I certainly did not say that the slew rate it provides is inadequate. That is dependent on the application and individual needs. That is why there was no need for me to state the voltage swing context.
If you insist, I'll tell you that I consider a slew rate of 31 V/us to be only modest for a 100 watt amplifier. On the other hand, the slew rate of 300 V/us that I achieved in my 50W MOSFET power amplifier was way more than necessary.
I think you belong in a debating society.
Bob
Bob Cordell said:
Agreed.
And - Miller compensated amps are not the best sounding ones, in my experience.
Bob, I agree that 31V/us is a little low. Walt Jung and everyone else agrees that a higner slew rate---at least 50V/us, is better.
john curl said:
John -
should not we speak about voltage swing (or output power) as well. I do agree that for 1OOW / 8 ohm amp we should have at least 50V/us.
PMA said:
I assume that you are applying the 10:1 rule. However, with
real music you aren't very likely to ever see the 5 V/uS implied
by the 100 watt figure.
Actual testing with real material shows that .5 V/uS is the
probable maximum slew rate, and the highest I have measured
for a 100 watt amp is 1.5 V/uS (Rim shots on the Sheffield Drum
record - Vinyl). The test setup was confirmed able to do 40 V/uS.
Walker and others have performed similar experiments, and
come up with even more conservative figures.
😎
Well - according to measurements shown here:
http://www.stereophile.com/features/404metrics/index2.html
the slew rate would reallly not be higher than 2V/us at 100W/8 ohm. When I made my own measurements, some electronic music showed higher value, just about 5V/us, the same number you are speaking about. You are right, I would like to see 10:1 reserve.
http://www.stereophile.com/features/404metrics/index2.html
the slew rate would reallly not be higher than 2V/us at 100W/8 ohm. When I made my own measurements, some electronic music showed higher value, just about 5V/us, the same number you are speaking about. You are right, I would like to see 10:1 reserve.
I probably measured the actual worst case slew rate with mistracking vinyl recordings using an Ortofon cartridge and no transformer (because it is bandwidth limited). It shows that a 10us risetme AFTER RIAA EQ is possible.
Nelson Pass said:
True!
Add P. Baxandall to your illustrious list.
Clearly, ill-informed speculation is no substitute for proper tests.