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 i3alan 20th May 2019 08:02 PM

How to measure the output impedance of a valve preamp?

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Hi everyone,
I was trying to measure the output impedance of a DIY preamp for the past couple of days without any luck. Each time, I got to the same result of around 3K3 ohm, which is unbelievably low.

I have a rather long monologue with a bunch of pictures so im attaching a PDF here with all the schematics and my solution attempts.

Was I on the right track? What have i done wrong?
If not, then how do I calculate the output impedance of a preamp?

Any help greatly appreciated.

 JonSnell Electronic 20th May 2019 08:14 PM

R3 is the anode load resistor and the value is 120k therefore the output impedance of that stage is 120k.
There are a few non standard values employed in your models so this is pure theory.
Where you get 3k3 from I am not sure unless you are measuring the 3k3 resistor effectively across the 1M0 pot. Seems odd as to why you would load a reasonably high impedance, (120k) with 3k3.

 6A3sUMMER 20th May 2019 08:20 PM

The output impedance of the 12AX7 plate resistance, rp, of 60k Ohm, is in parallel with the 120k Ohm plate load resistor. That is 120k/3 = 40k. (120k/2 = 60k, and that is in parallel with 120k, so 120k/3).

All that other stuff including the potentiometers and 3.3k resistor ought not to be in the circuit.

A 12AX7 plate circuit can not drive that 3.3k Ohm resistor.

 rayma 20th May 2019 08:59 PM

Quote:
 Originally Posted by i3alan (https://www.diyaudio.com/forums/instruments-and-amps/338165-measure-output-impedance-valve-preamp-post5797354.html#post5797354) I was trying to measure the output impedance of a DIY preamp for the past couple of days without any luck. Each time, I got to the same result of around 3K3 ohm, which is unbelievably low.
The idea is to compare the open circuit output voltage, to the output voltage when a load resistor is connected, and adjust the load for half voltage. Of course, the output impedance with the 3k resistor actually connected will be around 3k, since the circuit's 40k output impedance is much higher. The Thevenin equivalent output impedance (with the 3k connected) is the 3k in parallel with the circuit's 40k output impedance, which is still about 3k.

 jan.didden 20th May 2019 09:05 PM

I don't see anything in your write-up about output impedance measurement. Here's how to do it (Spice or test bench, same method, but I will describe Spice as it is much faster ;-).

Set the input source to zero or ground the input. Put an AC current source, let's call it Itest, between output and ground. Say 0.1mA. Then plot Vout/Itest. That is Zout. Ohms Law.

But maybe first clean up so that you get a well-functioning amp, without all that low impedance load stuff. That'll kill it.

Jan

 rayma 20th May 2019 09:31 PM

1 Attachment(s)
Quote:
 Originally Posted by i3alan (https://www.diyaudio.com/forums/instruments-and-amps/338165-measure-output-impedance-valve-preamp-post5797354.html#post5797354) I was trying to measure the output impedance of a DIY preamp for the past couple of days without any luck. Each time, I got to the same result of around 3K3 ohm, which is unbelievably low.
Here's the idea.

 i3alan 20th May 2019 09:33 PM

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By pulling the signal with 3K3 resistor to the ground I am trying to lower the output level of the preamp. I found this value just by trial and error and it brings the output to the acceptable level. Without that the preamp is way too loud. So I was wondering what output impedance I am getting after all? Please see attached the re-drawn version of the last 2 gain stages. When I am calculating the output impedance of the last gain stage alone, I am getting 40K which seems logical. But when I am trying to calculate the total impedance at the output of the circuit, I am getting around 3.25K which doesn't make any sense to me. How come Z out became 10 times smaller after all the passive components that create an extra load on the output signal?:confused:

 rayma 20th May 2019 09:37 PM

Quote:
 Originally Posted by i3alan (https://www.diyaudio.com/forums/instruments-and-amps/338165-measure-output-impedance-valve-preamp-post5797430.html#post5797430) By pulling the signal with 3K3 resistor to the ground I am trying to lower the output level of the preamp. When I am calculating the output impedance of the last gain stage alone, I am getting 40K which seems logical. But when I am trying to calculate the total impedance at the output of the circuit, I am getting around 3.25K
You need a volume control (or fixed attenuator) at the input to lower the level. A heavy load resistor
connected in shunt on the output will cause high distortion and poor bass response.

See posts #4 and #6 for the second answer.

 i3alan 20th May 2019 09:52 PM

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Right, but this is a guitar preamp, so I actually want to overdrive the gain stages. So I can't lower the input signal. I just need to quiet it down at the output. Perhaps, I am not getting it right, but to my understanding the output impedance is a sort of resistance in series to the output signal. As it is shown in Merlin's book. See the pic attached. So, based on my calculations, my Z out is getting as low as 3k3 and the output signal is meeting a very low resistance. While the logic says it is actually the opposite. The signal is heavily loaded. Just can't get my head around it.:spin:

 6A3sUMMER 20th May 2019 10:01 PM

The 270pf from plate to cathode makes a great low pass filter. Xc = 29k Ohms at 20kHz, and the 12AX7 plate rp and 120k load is 40k. response is -3 dB at less than 20kHz. Is that what you want? What is the intended purpose and function of your preamp? What is the signal source at the input? A microphone, phono cartridge, CD player, Tuner, etc.? What is the output of the preamp going to drive? A power amp or what?

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