Hi,
I am currently getting trying to get to grips with the theory behind tube amps.
One thing that is puzzling me is negative feedback. I can understand how it does what it does but I am getting lost working out how it is applied to a real circuit.
For instance I have been looking at the Mullard 5-20 circuit.
How are the values of the feedback resistors calculated?
Can anybody point me to a good discussion of this topic?
Thanks,
James
I am currently getting trying to get to grips with the theory behind tube amps.
One thing that is puzzling me is negative feedback. I can understand how it does what it does but I am getting lost working out how it is applied to a real circuit.
For instance I have been looking at the Mullard 5-20 circuit.
How are the values of the feedback resistors calculated?
Can anybody point me to a good discussion of this topic?
Thanks,
James
Hi,
Well, I can well imagine some newcomer developing a headache by looking at circuits like that one...
The theory is the same, whether you look at Solid Sand or Liquid Tubes:
Imagine a circuit...
In order to be able to inject a FB loop, you'll need to have an out of phase signal.
This signal is what you inject back, to some extend, into the input in order for it to cancel non-linearities by superposition onto the main signal.
It's basically the same, much oversimplified, as saying two minuses make a plus really...In principle that is.
Look up Harold S. Black on the net for more detail.
The game you play is frequency pole, time lag and phase behaviour in order to reduce distortion.
The common rug and carpet trick, really...You don't see it, so it can't be there.
Cheers,
Well, I can well imagine some newcomer developing a headache by looking at circuits like that one...
The theory is the same, whether you look at Solid Sand or Liquid Tubes:
Imagine a circuit...
In order to be able to inject a FB loop, you'll need to have an out of phase signal.
This signal is what you inject back, to some extend, into the input in order for it to cancel non-linearities by superposition onto the main signal.
It's basically the same, much oversimplified, as saying two minuses make a plus really...In principle that is.
Look up Harold S. Black on the net for more detail.
The game you play is frequency pole, time lag and phase behaviour in order to reduce distortion.
The common rug and carpet trick, really...You don't see it, so it can't be there.
Cheers,
One good example of NFB is a cathode follower. Take a normal plate-loaded stage and all the conditions that entails, then put the load in the cathode. (Remember that electrically it's the exact same. Analysis takes grid vs. cathode voltage, and the load remains the same since Ia = Ik.)
Now. Apply a signal. For 11V input, it might make 10V output. The difference is what the tube is amplifying, that is, 11-10 = 1V appears between grid and cathode as signal drive to it. When analyzing this stage, you would find that 1V input creates 10V output on the plate, that is, its plate loaded gain is 10. But as a CF, it has 10/11ths gain, a reduction of 11 times (about 21dBV). By the nature of NFB (this can be proven by math), distortion and output impedance is reduced by the same amount, so if your plate loaded distortion was 1% THD, it will be reduced to .09%. The Zo of a plate loaded stage is Rp || Rsupply || RL (Rsupply goes from plate to +V, RL is the external load which contributes no DC supply to the tube). Say Zo used to be 5k, it will now be 450 ohms.
The reason for this is that the output voltage is connected in series with the input, so the tube can correct for errors (distortion, frequency response, gain, Zo..). That's the premise behind most NFB, but it can also be applied in other ways, current NFB for instance. That has the effect of reducing distortion but increasing Zo (due to correcting for the current in the output, rather than the voltage).
Oh yeah, NFB also tries to maintain a stable point. For voltage NFB, it tries to produce a constant-voltage output (i.e. reduced Zo), while current NFB tends to make a constant-current (high Zo) output.
...Long post... could you link to the 5-20 schematic you mentioned?
Tim
Now. Apply a signal. For 11V input, it might make 10V output. The difference is what the tube is amplifying, that is, 11-10 = 1V appears between grid and cathode as signal drive to it. When analyzing this stage, you would find that 1V input creates 10V output on the plate, that is, its plate loaded gain is 10. But as a CF, it has 10/11ths gain, a reduction of 11 times (about 21dBV). By the nature of NFB (this can be proven by math), distortion and output impedance is reduced by the same amount, so if your plate loaded distortion was 1% THD, it will be reduced to .09%. The Zo of a plate loaded stage is Rp || Rsupply || RL (Rsupply goes from plate to +V, RL is the external load which contributes no DC supply to the tube). Say Zo used to be 5k, it will now be 450 ohms.
The reason for this is that the output voltage is connected in series with the input, so the tube can correct for errors (distortion, frequency response, gain, Zo..). That's the premise behind most NFB, but it can also be applied in other ways, current NFB for instance. That has the effect of reducing distortion but increasing Zo (due to correcting for the current in the output, rather than the voltage).
Oh yeah, NFB also tries to maintain a stable point. For voltage NFB, it tries to produce a constant-voltage output (i.e. reduced Zo), while current NFB tends to make a constant-current (high Zo) output.
...Long post... could you link to the 5-20 schematic you mentioned?
Tim
Basically, the ratio of R13 to R5 sets the closed loop gain (in reality, it will be a bit lower because the open loop gain is far from infinity). The input tube's cathode acts as the inverting input.
Feedback theory, as Frank so correctly said, is the same for any kind of amp. The best tutorials on it focus on op amps, where it's easier to understand, but it's applicable to any technology.
Feedback theory, as Frank so correctly said, is the same for any kind of amp. The best tutorials on it focus on op amps, where it's easier to understand, but it's applicable to any technology.
Sy and others,
For Global NFB (or as I prefer, Inverted Feedback):
With high gain solid state amps, I would agree that the R2/R1 ratio sets gain, but it is not as simple as that for a low open loop gain tube amp.
I found somewhere (I do not know where I did or what the formula is off the top of my head) a formula for calculating the feedback resistor that also takes into consideration the output impedance. I think it was in "Valve Amplifiers" by Morgan Jones. The ratio then is different.
I mean, if it was R2/R1 then my amp with a 3.9K feedback resistor with a 390 ohm cathode resistor would only have a V gain of about 10. It has a V gain of about 20 (Vout/Vin, or 10/0.5, RMS, for about 13-14 watts into 8 ohms).
That, the output impedance, is one reason why the feedback resistor depends on the tap you use on the output transformer, whereas with a solid state amp it doesn't matter. The other reason is that the output taps produce different output voltages.
So... if I can dig up the book I will put the actual equation out here.
But I am sure someone else here has one. The Jones book is highly recommended among tube DIYers, BTW.
My
Gabe
For Global NFB (or as I prefer, Inverted Feedback):
With high gain solid state amps, I would agree that the R2/R1 ratio sets gain, but it is not as simple as that for a low open loop gain tube amp.
I found somewhere (I do not know where I did or what the formula is off the top of my head) a formula for calculating the feedback resistor that also takes into consideration the output impedance. I think it was in "Valve Amplifiers" by Morgan Jones. The ratio then is different.
I mean, if it was R2/R1 then my amp with a 3.9K feedback resistor with a 390 ohm cathode resistor would only have a V gain of about 10. It has a V gain of about 20 (Vout/Vin, or 10/0.5, RMS, for about 13-14 watts into 8 ohms).
That, the output impedance, is one reason why the feedback resistor depends on the tap you use on the output transformer, whereas with a solid state amp it doesn't matter. The other reason is that the output taps produce different output voltages.
So... if I can dig up the book I will put the actual equation out here.
But I am sure someone else here has one. The Jones book is highly recommended among tube DIYers, BTW.
My
Gabe
Well, the feedback resistor will indeed set the gain referenced to the tap it comes from. The other taps scale in output voltage as usual. I should have been clearer.
I'm not so sure about the output impedance thing- after all, open loop source Z in a tube amp might typically be on the order of magnitude of 10 ohms, whereas the feedback resistor is something like 1K-10K ohms.
I'm not so sure about the output impedance thing- after all, open loop source Z in a tube amp might typically be on the order of magnitude of 10 ohms, whereas the feedback resistor is something like 1K-10K ohms.
SY said:
The value of the two feedback resistors, set the gain of the amp and in you case you point the amp as a open loop output impedance of 10 Ohms (a reasonable value for a triode output stage ,not for a pentode stage) if when you connected the fedback loop the gain become 10 times smaler you are puting 20 dB of feedback in your amp.
So the output impedance will be 10 times lower and become 1 Ohm
Well, look what I found in the basement of my old brain 
If Ao=Vo/Vin is the differential voltage gain of the amplifier without feedback and B is the voltage gain of the feedback network the closed loop gain is:
Acl = Ao/(1+Ao*B)
B = Ri/(Rf+Ri) if Rf is the feedback resistor and Ri is the input impedance of the cathode of the input tube.
If Ao is very large, Acl can be approximated to 1+Rf/Ri, a formula we recognize as the gain for a noninverting opamp. Now, with our tube amp, Ao isn't large enough to do this approximation.
If Ao=Vo/Vin is the differential voltage gain of the amplifier without feedback and B is the voltage gain of the feedback network the closed loop gain is:
Acl = Ao/(1+Ao*B)
B = Ri/(Rf+Ri) if Rf is the feedback resistor and Ri is the input impedance of the cathode of the input tube.
If Ao is very large, Acl can be approximated to 1+Rf/Ri, a formula we recognize as the gain for a noninverting opamp. Now, with our tube amp, Ao isn't large enough to do this approximation.
Sch3mat1c said:
Actually, the 390 ohm resistor is in series with a 2.2K resistor to the cathode. There is a kind of voltage divider between the cathode and ground, where the 2.2K is connected to the cathode, the other end is connected to the 390 ohm resistor and the feedback resistor, and the other end of the 390 ohm is connected to ground.
So the 2.2K acts as a "snubber" of sorts to the cathode input. Cathode resistance is negligible in this case, IMHO. Like this:
(ASCII Art doesn't work to well)
TUBE
K
|
R4 2K2
|------FBR 3.9K----- |
R5 390 T1 Secondary
|----------------------|
Ground
I surmise that the linear DC resistance of the output transformer secondary tends to come into play somehow, since it conects to ground. Hadn't had the chance to rummage through my storage bin where my books are. I will this weekend.
And, even if the 390 ohms were directly connected to the cathode, it is the resultant voltage between the feedback resistor and the resistor to ground that is used as the feedback voltage, I believe.
According to the equation that I used, the resistor is not double for 16 ohms versus 8 ohms, as you seem to imply by your thought above, but the next step up from 3.9K is 4.7K, for 8 and 16 ohms, respectively.
My thoughts... whatever they are worth!
Gabe
Jax said:
Of course you are correct.
I apologize for the confusion, but I was referring to my amp, which is not bypassed, and explaining NFB in general.
In the context of the 5-20, absolutely, the 2.2K is as if it isn't there. But, and I may be wrong in my thinking, the cathode resistance does not come into play, since the voltage across the 390 ohm is what is being added to... by the inverted signal, that is. It is this sum that is injected as input to the cathode.
Gabe
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