Calculating Zout for Pentode amp with NFB?

[Jeb-D.]

Gents,
Could some one tell me how to calculate output impedance of a Pentode amplifier that uses global negative feedback?

This is easy for triode amplifiers, but since Pentodes theoretically have an infinite open loop output impedance, I'm unsure how to go about it.

[Tubes4e4]

Hi Jeb-D.,

Could some one tell me how to calculate output impedance of a Pentode amplifier that uses global negative feedback?

This is easy for triode amplifiers, but since Pentodes theoretically have an infinite open loop output impedance, I'm unsure how to go about it.

This isn´t the right approach, I am afraid. In general, for class A SE operation:

1) Just take the rp of the pentode (or any tube type, for that matter) at the chosen operation point by either reading it from the spec sheet (not often given there) or by deriving it yourself from the plate curves, which is easily done and precisely enough.

2) Then divide this rp by the pri:sec impedance ratio of the OPT

3) Then multiply by feedback factor (i.e 6dB = 2)

4) Don´t fall from chair when comparing Zout (and DF) of triode based and pentode based power stages ...

For pure class A PP, just use 1/2 rp of a single tube for your computations.

For class A/B PP or class B PP, you have to consider that the reflected impedance each tube sees when operation strays into class B softly changes from 1/2 Raa to 1/4 Raa due to cutoff of one tube of a pair during each half-cycle. So you will have to consider and recompute the actual impedance ratio the OPT gives (or more precisely, what each tube sees) for class B, which is easy enough But keep in mind that there is no hard changeover point between class A PP and class B PP.

So, for class A/B PP amps, often it is easier to really measure open loop Zout at the OPT secondary and plot against output power (and maybe frequency, too), then just apply the feedback factor - given the chosen feedback type is of reasonable linear frequency character - which it usually ain´t considering deliberately introduced time constants (like "speedup caps, etc") being involved.

Regards,

Tom Schlangen

[ray_moth]

3) Then multiply by feedback factor (i.e 6dB = 2)

I think you mean divide, assuming negative feedback?

[Tubes4e4]

I think you mean divide, assuming negative feedback?

Yes, of course, it should be divide (or multiply with the inverse). Thank you for reading close :-)

Tom

[jane]

Originally posted by Tubes4e4
1) Just take the rp of the pentode (or any tube type, for that matter) at the chosen operation point by either reading it from the spec sheet (not often given there) or by deriving it yourself from the plate curves, which is easily done and precisely enough.

Let's take an example: Single ended pentode EL84 and OPT=4k.
EL84 rp = ~40k
Zo no feedback = 8 x 40k/4k = 80 ohm.
According to your calculations Zo with 20dB (10x) NFB should be 80 ohm / 10 = 8 ohm.
When measuring Zo in this simple circuit you will find it much lower, why?

If you look up a classic text book you'll find that Zo is more like rp/(1+(mu x B)). It is crucial to see that this will give you another result than the normal Aol/Acl. The tricky part may be to find a good figure for mu in a pentode circuit. An approximation is to measure Aopen with no load connected to the amplifier. And of course, since Aopen =~ Rl x gm, and Rl is infinite, Aopen is much higher than Aol measured the normal way with load.

Jan E Veiset

[Jeb-D.]

So, it seems that it is pretty much the same as triode.

The tricky part may be to find a good figure for mu in a pentode circuit.

Isn't a Pentodes mu= (gm x Rl) / Vin ?

rp/(1+(mu x B))

What is B in that equation? That looks more like the equation for a cathode follower Zo.

Thanks!

[jane]

B is the fraction of the signal that is fed back.
In a cathode follower B is close to 1.

Jan E.

[Jeb-D.]

B is the fraction of the signal that is fed back.

Ahh.. I see. Thats very useful. Thanks!

Now i'm thinking Pentodes mu = (gm x Rl)

So that equation can work like Zo = rp/(1+((gm x Rl) x B))

Where Rl is R load (primary impedance of the OPT).

[oldeurope]

Originally #8 posted by Jeb-D.


Ahh.. I see. Thats very useful. Thanks!
...

Originally #1 posted by Jeb-D.
Gents,
Could some one tell me how to calculate output impedance of a Pentode amplifier that uses global negative feedback?
...

Yes,


rout = 1 / (s x vu x FD)


rout => output resistance of a pentode amp, anode output
s => transconductance of the pentode
vu => gain of the stages from input of negative feedback to the grid of the output pentode

FD = Re / (Rg + re)

FD => feedback divider (see post#7)
re => input resistance of the stage where feedback is input
Rg => feedback resistor


Kind regards,
Darius

[Tweeker]

8ohms Zo with NFB sounds about right if the pentode cathode is not bypassed.