How to measure the output impedance of a valve preamp?
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 20th May 2019, 10:10 PM #11 i3alan   diyAudio Member   Join Date: May 2019 This a guitar preamp pedal. It will be used in line with the other guitar effects, plugged to the front end or to the power amp directly. Yes, the low pass filter is intended.
 20th May 2019, 10:20 PM #12 wintermute   just another diyAudio Moderator     Join Date: Aug 2003 Location: Sydney moved to instruments and amps __________________ Any intelligence I may appear to have is purely artificial! Photography (another hobby)
rayma
diyAudio Member

Join Date: Apr 2011
Quote:
 Originally Posted by i3alan 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 circuit's output impedance is 40k, and you are loading that with 3k to ground. The voltage divider formed by these two resistances reduces the output level a lot, more than a factor of ten.

However, the output impedance of the circuit with the 3k in place is changed from 40k to 3k. You would have to study Thevenin's theorem to have an intuitive understanding of it. Failing this, just accept that the output impedance of the circuit with the 3k load is the 40k in parallel with the 3k.

The Thevenin Theorem - A Simple Explanation - Build Electronic Circuits

 21st May 2019, 03:18 AM #14 PRR   diyAudio Member     Join Date: Jun 2003 Location: Maine USA If you truly need 3K on the output to get your signal level, then of course Zout will be <3K. With your backward-connected Master Volume pot in the latest plan, final Zout will vary from 250k to <3K. To my mind, _I_ would think to look at a normal-connection 5K pot (possibly with 10K across to make your 3K) which would give <5K at all settings and be much less susceptible to various loadings.
 21st May 2019, 03:43 AM #15 i3alan   diyAudio Member   Join Date: May 2019 Thank you, PRR. So, I got the concept of the output impedance all the way wrong. Could you tell, on the most basic level, the definition of the output impedance in such circuits and how it will interact with the different loads? I am just trying to get the basic simplified idea of it.
 21st May 2019, 10:02 AM #16 DF96   diyAudio Member   Join Date: May 2007 You can model the output of almost any circuit as a voltage source with a resistor in series with it. The voltage source will give whatever voltage it gives, however much you load it. The series resistor will drop some voltage if you take some output current. You can use Ohm's Law to calculate how much voltage drop. The series resistor in this model is the output impedance.
 22nd May 2019, 01:35 AM #17 djgibson51   diyAudio Member   Join Date: May 2010 Location: Hawkes Bay Side issue: Adding a tone stack of your choice prior to the volume pot will drop output level. Tone controls after distortion are good for guitar preamps. Last edited by djgibson51; 22nd May 2019 at 01:39 AM.
rayma
diyAudio Member

Join Date: Apr 2011
Quote:
 Originally Posted by i3alan the definition of the output impedance in such circuits and how it will interact with the different loads?
See post #6.

 22nd May 2019, 04:14 AM #19 PRR   diyAudio Member     Join Date: Jun 2003 Location: Maine USA > output impedance Why do you even care?? If you do, I suggest you study Voltage Dividers. EVERYTHING can be analyzed as a voltage divider. It is a powerful concept.
 22nd May 2019, 01:23 PM #20 i3alan   diyAudio Member   Join Date: May 2019 The post #6 shows one of the ways to calculate the output Z. Which I can totally understand. I care about the output impedance topic just because I would like to understand the concept of the correct signal matching. To my understanding the lower the output impedance is - the better. The ideal z out is 0. But I am confused by the fact that in real circuit Zout is going down when the output signal is pulled to ground. Which is not a good thing for a signal, right? I guess I am missing something between the lines.

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