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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.

This is the circuit I’m working with:

Starting Circuit

I’m using the book “Designing Valve Preamps,” by Merlin Blencowe as a reference. What I have tried:

  1. Connecting a resistor to LTSPICE circuit and finding its value at 50 % output gave the output impedance around 3250 Ω.

  2. The author mentioned that theoretically if you were to feed 1 V signal into the output, the ratio between the forced voltage and the current drawn will give you the output impedance. This gave me again around 3K3 Ω.

  3. I have replaced the potentiometers by fixed value resistances, and simplified the circuit as shown below:

Simplified Circuit

Then, I have used Merlin’s Thevenin equivalent for the gain stage (page 31), I tried both for bypassed and unbypassed capacitors, which resulted in:

Final Simplified Circuit

Now, coming to basic circuit analysis, I have tried 2 ways:

a) set all capacitors as short circuits
b) treat the capacitors as complex impedances.

Both of those ways resulted in a total impedance of around 3K3 Ω.

This is impossibly low since It means the signal transmitted almost ideally to the amp. However, when connecting it to a high Z input and attaching a cathode follower in between, the sound seems to have more “breathing space”.

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 is greatly appreciated.

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  • \$\begingroup\$ Are you measuring the DC output impedance? Or AC output impedance? \$\endgroup\$
    – Voltage Spike ♦
    Commented May 20, 2019 at 20:26
  • \$\begingroup\$ AC because im trying to measure the audio impedance \$\endgroup\$
    – i_alan
    Commented May 20, 2019 at 20:38
  • \$\begingroup\$ At AC the capacitors would look like shorts, this still doesn't explain how your seeing 3.3k, because I would expect that even with the capacitors being like AC shorts, R3 would be also limiting current. \$\endgroup\$
    – Voltage Spike ♦
    Commented May 20, 2019 at 20:53
  • \$\begingroup\$ R3 becomes Ra in the second circuit. This is the thevenin equivalent of the unbypassed gain stage. and if you take the time to do the math after replacing the capacitors with short circuits its indeed around 3K3 \$\endgroup\$
    – i_alan
    Commented May 20, 2019 at 22:03
  • \$\begingroup\$ The thevening model is from Fig.1.31 page 31: books.google.ca/… \$\endgroup\$
    – i_alan
    Commented May 20, 2019 at 23:16

1 Answer 1

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you've go this (ignoring C2)

schematic

simulate this circuit – Schematic created using CircuitLab

and this looks like 3.3Kohm

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  • \$\begingroup\$ Isn't R2 in the wrong place? \$\endgroup\$
    – Voltage Spike ♦
    Commented May 21, 2019 at 1:06
  • \$\begingroup\$ This is an incremental_voltage model, where we might input 1uV or 1mV of signal and examine the circuit behavior. Another name is "small signal model". In such modeling, we assume the supply voltages DO NOT MOVE, and we happily connect them to ground. Thus the 120K discrete resistor, in the plate path, is correctly drawn in parallel with the internal resistance of the 12AX7. \$\endgroup\$
    – analogsystemsrf
    Commented May 21, 2019 at 4:59
  • \$\begingroup\$ Do you have access to plots of the 12AX7 plate characteristic? \$\endgroup\$
    – analogsystemsrf
    Commented May 22, 2019 at 18:14

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