[Fred_fred2004]

I'm trying to build a passive preamp using this circuit, but the range of control of the pot is very small it's either off or full on which is still heavily attenuated, just wondering if anyone has experience of this type of circuit

 

[FallenAngel]

Well, with that resistor divider, you won't get huge range.

Why not follow the "traditional" voltage divider schema? 1K inline, 470K to ground, 50K pot.

 

[amb]

Quote:

Originally Posted by FallenAngel /img/forum/go_quote.gif Why not follow the "traditional" voltage divider schema? 1K inline, 470K to ground, 50K pot.

Problem with that is, when the pot is at minimum position, the source would "see" only a 1K ohm load, which is too low in some cases. You need the series resistor to be at least 10x the output impedance of the source, and also take into account the high-pass filter characteristics (which would change with the volume knob) if the source output is capacitively-coupled.

But that might also mean a high-ish output impedance after the "passive preamp" to the next stage, which would become susceptible to interference, stereo crosstalk, and cable capacitance induced high frequency rolloff.

 

[nikongod]

Quote:

Originally Posted by FallenAngel /img/forum/go_quote.gif Why not follow the "traditional" voltage divider schema? 1K inline, 470K to ground, 50K pot.

The signal dosnt go through the pot (as much) in this design. In the traditional configuration the pot is 100% in the signal path. This design makes the hi-fi use of cheap pots possible, and nice pots OMG nice.

Anyways, to the OP: that is very weird. with a 26K ohm shunt resistor and a 100K ohm pot the circuit should not give very much attenuation at all. IE: off and full volume.

Is the 18K ohm resistor on the input side of the 26K ohm one? I can see some problems if its on the "pot" side. It still should not be as bad as you describe.

Can you measure the input and output ac voltages? It may just be that whatever you are running off of this actually needs a "proper" preamp with some gain to go as loud as you would like it.

 

[Fred_fred2004]

This may be my fault would driving off a low impedence source (cmoy output) be the problem?

 

[amb]

Actually, a cmoy should have no trouble driving a 1K ohm load. After all, it could drive headphone loads that have even lower impedance. But, the resistances of the series resistor and the pot will establish an attenuator that, even at maximum volume position, has some loss (dependent on the values). You can work out the control range by "mapping" the attenuation vs. pot position and translating that into dBs. Just use an audio (log) taper pot curve when you do this.