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Old 19th January 2004, 09:49 PM   #1
psysjal is offline psysjal  United Kingdom
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Default Trying to understand negative feedback

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
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Old 19th January 2004, 11:07 PM   #2
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Hi,

Quote:
For instance I have been looking at the Mullard 5-20 circuit.
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,
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Old 19th January 2004, 11:43 PM   #3
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Hi,

Quote:
The common rug and carpet trick, really...
Should have said: Brush and carpet...Sorry.

Cheers,
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Old 20th January 2004, 02:41 AM   #4
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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
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Old 20th January 2004, 06:45 AM   #5
dhaen is offline dhaen  Europe
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Mullard 5-20 schematic
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Old 21st January 2004, 02:06 AM   #6
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Looks good to me, aside from the small filter caps. Pentode input, LTP splitter, UL class A output.

Tim
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Old 21st January 2004, 02:23 AM   #7
SY is offline SY  United States
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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.
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Old 22nd January 2004, 12:24 PM   #8
psysjal is offline psysjal  United Kingdom
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Thanks for your help people.

I understand now (I think ).
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Old 22nd January 2004, 05:38 PM   #9
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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

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Old 22nd January 2004, 06:14 PM   #10
SY is offline SY  United States
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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.
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