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Control Grid Feedback (as opposed to Cathode FB) and Miller Capacitance
Control Grid Feedback (as opposed to Cathode FB) and Miller Capacitance
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Old 17th November 2015, 01:32 AM   #1
ballpencil is offline ballpencil  Indonesia
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Default Control Grid Feedback (as opposed to Cathode FB) and Miller Capacitance

Assuming as grounded cathode stage where the output is fed back to the control grid. Rfb is the one that spans from the output to the control grid while Rg is the one on the control grid.

Rg pretty much determines the input impedance so we want this as high as possible without making Rfb so big that it is noisy. Am i correct in thinking that Rg still forms a 1st order LPF with the tube's Miller capacitance? Hence, let's say on a 12AX7 tube with typical OL gain of 60x, i can't just use 100k for Rg as that would make the 3dB cutoff sit below 20kHz. Or.. since i have introduced feedback which lowers gain, Miller multiplication is lower so that i can use higher than usual value for Rg? "Usual" means the value i'd use for a plain-jane grounded cathode stage.
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Old 17th November 2015, 02:08 AM   #2
Johan Potgieter is offline Johan Potgieter  South Africa
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Partly right: The determining factor being that the junction point of Rfb and Rg (at the grid) is a relatively low impedance point because of feedback. One can, in fact, equate Rfb to the Miller capacitance reactance (it is after all in parallel), only without any frequency effect, as a capacitor would have.

Also then, if I understand you correctly, a larger Rfb would have less effect on the noise, not more, as it is essentially swamped by other resistances. The noise will mostly result from Rg, which is the limitation of this topology. (Also then why the early simple RIAA circuits analogous to this - mostly Rg = 100K, - had more noise than a two-stage anode-cathode feedback, or a low noise designed input stage feeding a passive RIAA network. But this is a simplification; the story is more complex.)

So BUT: the Miller effect itself will not simply limit the h.f. response, as per a passive way. The feedback will counter that, moving the h.f. pole up again, in a manner of speaking. (Again very simplified, just to get in line with your question. One has two effects working in opposite direction, etc.)
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Old 17th November 2015, 03:47 AM   #3
cerrem is offline cerrem  United States
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Keep in mind this can be viewed as a simple OPAMP ... inverting gain stage...
The cathode is the non-inverting input and the grid is the inverting input...
Obviously the cathode is low Z input ....but keep this locked down with a bypass cap...
There are 3 resistors to this Gain stage... (Rin series), Rg (Grid to Ground), Rfb .....
Rfb and Rg make a divider to the grid.... Virtual GND at this node... The input resistor Rs in series from source to this node and is the dominant input Resistance...
This is a bit over simplified, due to not a true balances opamp model....
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Old 17th November 2015, 04:28 AM   #4
cerrem is offline cerrem  United States
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Correction to previous post... The grid node is not a "true" virtual GND but rather the equivalent Rg is reduced by Feedback Factor... Rin is simply added to increase and adjust the input Z ...
For example.... if Rg is 100K and the feedback factor reduces this node to roughly 20K, then Rin would be set to 27K if a 47K input Z is needed...
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Old 17th November 2015, 09:44 AM   #5
DF96 is offline DF96  England
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Quote:
Originally Posted by ballpencil
Am i correct in thinking that Rg still forms a 1st order LPF with the tube's Miller capacitance?
Yes.

Quote:
Or.. since i have introduced feedback which lowers gain, Miller multiplication is lower so that i can use higher than usual value for Rg?
No. The open-loop gain is unchanged. All that has happened is that you have put a resistor in parallel with the anode-grid capacitance so the reduction in input impedance happens at lower frequencies too. Miller capacitance is unchanged, but it will have less effect as it is now in a lower impedance circuit.

I think there may be some confusion in this thread. I think the OP meant Rg to be the grid 'stopper' resistor in series with the input signal, but more normally Rg means the grid leak resistor connected to ground. Better to use Rs - source or series resistor or Rin.

It is unclear to me what the OP is trying to achieve.
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Old 17th November 2015, 10:58 AM   #6
ballpencil is offline ballpencil  Indonesia
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Thank you all for the help.

Yes, i was stumbled when i named Rg and Rfb, not knowing if this was the conventional way. For further discussion, i will use Rg as grid leak and Rin as the series/grid stopper resistor.

I am trying to understand this:
Quote:
All that has happened is that you have put a resistor in parallel with the anode-grid capacitance so the reduction in input impedance happens at lower frequencies too. Miller capacitance is unchanged, but it will have less effect as it is now in a lower impedance circuit.
Trying to explain with my own words for the above quote:
Taking the 12AX7 example with 60x OL gain.. Miller cap is estimated as 200pF.
At 20kHz the Xmc reactance is 39k. This will yield a -3dB cutoff if i use equal value for Rin (39k). Now, if i use Rfb=100k, this will be in parallel with the 200pF miller cap.. At 20kHz, the parallel Rfb/Xmc impedance is about 37k.. This confuses me, along with Rin 39k, won't this mean even higher loss (about -3.1dB)? I must have misunderstood something.

What i am trying to understand is how to estimate the value of Rin for the highest input impedance possible without compromising the audio bandwidth too much. I have also been corrected that noise is mainly from Rin, not from Rfb.. thank you Johan for this.
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Old 17th November 2015, 01:00 PM   #7
DF96 is offline DF96  England
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Quote:
Originally Posted by ballpencil
Now, if i use Rfb=100k, this will be in parallel with the 200pF miller cap.
No. Rfb is in parallel with Cag (or Cpg - depending where in the world you are). This is the real capacitance which gives rise to Miller capacitance.

If you have significant amounts of feedback then the HF rolloff is determined by Rfb and Cag, not Rin and Cmiller. Rin just sees a virtual ground at the grid. Feedback is significant if Rfb/Rin < mu, which appears to be the case for your example.
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Old 17th November 2015, 01:26 PM   #8
ballpencil is offline ballpencil  Indonesia
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I think i got it.

Combining with what cerrem mentioned the stage as being analogous to an inverting op-amp.. I believe this diagram explains it.
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Cf on the diagram above would be Cga as you mentioned. In 12AX7 case, this is around 1.5pF. Rf is of course Rfb. With Rfb = 100k, the -3dB cutoff would then be a whopping 1.1Mhz. Is this correct?

So.. this raises another question: why did you say i was correct when i thought that Rg (now Rin) still forms a 1st order LPF with the tube's Miller capacitance? In other words, what is stopping me from raising Rin too high if i want as high input impedance as possible? Plain resistor noise?
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Old 17th November 2015, 01:27 PM   #9
Merlinb is offline Merlinb  United Kingdom
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Quote:
Originally Posted by ballpencil View Post
Trying to explain with my own words for the above quote:
Taking the 12AX7 example with 60x OL gain.. Miller cap is estimated as 200pF.
At 20kHz the Xc reactance is 39k. This will yield a -3dB cutoff (of 20kHz) if I use equal value for Rin (39k).
Correct. And since you said the OL gain was 60 and the Miller capacitance was 200pF, we know the 'cold' anode-grid capacitance must be 200pF/(60+1) = 3.3pF approx.

Quote:
Now, if i use Rfb=100k, this will be in parallel with the 200pF miller cap..
You have reduced the gain to approximately 100k/39k = 2.6. Therefore, the anode-grid capacitance is now multiplied by the Miller effect to 3.3pF*(2.6+1) = 11.9pF.* The total resistance 'seen' by this capacitance is the parallel combination of the feedback and input resistors, which is 100k||39k = 28k. The cut-off frequency is therefore 477kHz.

Notice that the bandwidth has been increased by a factor of 477kHz/20kHz = 23.9, which is the same as the feedback factor: 60/2.6 = 23 (OK close enough, since I rounded a lot of numbers off).

*(OK, the OL gain is still 60 "inside" the loop. But then the 200pF of Miller capacitance 'sees' the external resistances divided by the feedback factor, so the outcome is the same whichever way you look at it).

Quote:
since i have introduced feedback which lowers gain, Miller multiplication is lower so that i can use higher than usual value for Rg?
Yes, if you don't care about noise.

Last edited by Merlinb; 17th November 2015 at 01:45 PM.
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Old 17th November 2015, 01:47 PM   #10
DF96 is offline DF96  England
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Quote:
Originally Posted by ballpencil
So.. this raises another question: why did you say i was correct when i thought that Rg (now Rin) still forms a 1st order LPF with the tube's Miller capacitance?
Because I wasn't thinking clearly enough. However, there are often several ways to analyse a circuit.

In this case we can treat the circuit as a gain=60 voltage amp with a C and R in parallel for feedback. The result is an HF rolloff set by the feedback C and R.

Alternatively, we can treat the circuit as a gain=2.6 voltage amp with some Miller capacitance. The net result is the same (as it has to be) but it looks like we arrived at it by a different route.

The danger is to mix the two different analyses together wrongly, and in effect get double counting.

The best option is to construct a full analysis (or simulation) but that gets more complicated so most of the time we use approximations. Most of the time they work OK, but they don't issue a warning when they fail and give misleading results.
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