Passive Preamp output impedance calculation

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Hi all, I am puzzled about the output impedance calculation for passive preamp such as the Khozmo one with 10k Ohm series attenuator like this https://www.khozmo.com/series_khozmo_attenuator.html

from Arek it seems the input impedance of the preamp will always at 10k Ohm constant, but the Output impedance in a quick query from Arek will be 1k Ohm at -20db attenuation and 20-30 ohm at -60db attenuation, for calculation part I got puzzled and no idea how the output impedance will be, can any expert here shed some light on how I should plug in the resistor values used and get the output impedance at any given level?

 

[Blumlein 88]

The information is correct.
Let me see if I can do some simple illustrations and explanations in the next couple posts.

 

[Blumlein 88]

Okay in the illustration the green line shows all the resistance in line from input to ground. If you put a multi-meter between input and ground it will read 10 kohm.

Now if the output which is tapped somewhere in the middle connects to a very low input impedance power amp that wouldn't be true. But as long as the power amp input impedance is high (say 100 kohms or more) you can basically ignore it. So volume setting has no bearing on the input impedance. It will equal essentially the impedance of the string of resistors between input and ground.

 

[Blumlein 88]

Now for output impedance.

If the source component is a low output impedance, which it normally will be, you could ignore it. And treat it as in the illustration with a ground on both ends of the resistor string. So if you connected a multimeter between the output connector and either ground you'll read a resistance number made up of two paths to ground in parallel.

In this illustration I am showing a setting where the output is halfway so you get 5 kohm in one direction, and 5 kohm in parallel in the other direction. These in parallel will read an output impedance of 2500 ohms. This btw is the highest reading you'll get, and will be the - 6 db setting.

 

[Blumlein 88]

Now I change the volume in this case I lower it quite a bit. A setting of -20 db.
From input to output most of the voltage is used up by 9 k ohm leaving only 1 k ohm above ground and a lower voltage. 1/10th of the total. The output impedance is the value from paralleling 1 k ohm and 9 k ohm or about 900 ohms output impedance.

Here I have raised the volume up almost to maximum. It would be - .45 db from maximum.
And the two paths to ground again are paralleled to determine the output impedance. In this case 475 ohms output impedance.

So the input impedance is always the same value in this type volume control circuit. The output impedance varies and is never more than 1/4th the value of the volume control, in this case 10 k ohms yielding a maximum output Z of 2500 ohms. At all other settings output Z is lower and near either max volume or minimum volume it is the least.

There are other volume circuits which act a little differently, but this one is most common and is also the way volume potentiometers work.

I hope this helps rather than confuses, if it doesn't help I've done a poor job explaining. Ask further questions.

 

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Okay in the illustration the green line shows all the resistance in line from input to ground. If you put a multi-meter between input and ground it will read 10 kohm.
View attachment 95903
Now if the output which is tapped somewhere in the middle connects to a very low input impedance power amp that wouldn't be true. But as long as the power amp input impedance is high (say 100 kohms or more) you can basically ignore it. So volume setting has no bearing on the input impedance. It will equal essentially the impedance of the string of resistors between input and ground.

thanks for the explaining, I think I basically understands everything with my high school physics failed brain, only question is that if I understand correctly, for a 10k Ohms pot the input impedance of the power amp >=10 k Ohms will basically makes the least resistance path at 10k Ohms either way, so the input impedance will be kept at 10k Ohms, and for output impedance except the -6db 5k Ohm per side setting the output impedance is at 2500 Ohms and others near either max or min will be lower progressively.

 

[Blumlein 88]

thanks for the explaining, I think I basically understands everything with my high school physics failed brain, only question is that if I understand correctly, for a 10k Ohms pot the input impedance of the power amp >=10 k Ohms will basically makes the least resistance path at 10k Ohms either way, so the input impedance will be kept at 10k Ohms, and for output impedance except the -6db 5k Ohm per side setting the output impedance is at 2500 Ohms and others near either max or min will be lower progressively.

Yes, I think you have it .

 

[RayDunzl]

Buy a couple of 10K audio taper pots for $5 and wire it up.

Why two? Easy channel balance.

I used one for years - a block of wood, some curtain hanger hardware to hold the pots, cut a couple of RCA cables in half, solder - between a CD player and an amp.

 

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Yes, I think you have it .

Thanks for the advise and teaching man, now not only I can sort out my preference and start saving up for a sexy Khozmo passive, but also I can give advise to my friends when they wanna get a pair of Genelecs after listening to mine