Author Topic: lowering effects output  (Read 1186 times)

royzic

lowering effects output
« on: December 12, 2006, 12:47:12 PM »
theoretically speaking,if one wants to lower the output of an effect,
will a ressistor in series with the output signal be OK? :-\

Roy

Seljer

Re: lowering effects output
« Reply #1 on: December 12, 2006, 01:00:21 PM »
use 2 resistors, as a voltage divider (just like a volume pot, but in a fixed postion)


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the level of the output = R of the bottom one / (R of both of them together)
« Last Edit: December 12, 2006, 01:10:10 PM by Seljer »

royzic

Re: lowering effects output
« Reply #2 on: December 12, 2006, 01:43:48 PM »
hi Seljer !
thank you for your reply ;)

there is an obvious question ,why is it better done with such voltage divider rather then a  ressistor in series? :-[


Ben N

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Re: lowering effects output
« Reply #3 on: December 12, 2006, 02:29:37 PM »
First, the disclaimer:  I am no expert.

But, I think the reason is that you would need a lot of resistance in series to get much attenuation, generating a lot of heat and introducing a lot of noise.   With a voltage divder, very low values can be used, so long as they are in the correct proportion, so there is much less heat and noise.  Also, a big series resistance on the output means your output impedance is increased, which would require you to have a muc higher input impedance on the next device in the chain to avoid signal loss and tone suck.  Also, for many devices I am guessing you need a path to ground on the output for correct bias, but here I am on very thin ice.

Is that right?

R.G.

Re: lowering effects output
« Reply #4 on: December 12, 2006, 02:56:37 PM »
Quote
there is an obvious question ,why is it better done with such voltage divider rather then a  ressistor in series?
The obvious answer is that there is always a second resistor there - it's the input resistance of whatever you are driving with the output, even if that's only an oscilloscope probe.

Anything you connect up has an input resistance. So there is a voltage divider composed of the "single" series resistor and that input resistance. If the input resistance is about 1M like most tube amps, then you have to have a 1M resistor in series to cut the output by half. If you want to cut it by 20db - which is about half the volume - then you need a 10M resistor in series. And you are, as Ben N supposes well into the area of objectionable noise.

If you use two smaller resistors, say 10K and 1K, then the voltage at the top of the 1K is 1/11th of the voltage at the top of the two in series, and introduces little noise. When you stick that into your 1M tube amp, the amp sees the smaller voltage but its 1M is so large compared to the resistors supplying it that it has little effect. So you get the reduction in level without the problems.
R.G.

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Sir H C

Re: lowering effects output
« Reply #5 on: December 12, 2006, 02:56:58 PM »
because you don't know what the other resistance will be at the far end.  Since you are driving a voltage output and the next stage is looking for a voltage, the voltage it sees is based on the output impedance and input impedance acting as a voltage divider.  The series resistor will cause the output to vary on how the input load of the next stage is.

royzic

Re: lowering effects output
« Reply #6 on: December 12, 2006, 03:18:51 PM »
now i understand it,,

thank you all very much!
 :icon_smile:

so, I'll just put a second standard volume pot to check the values i need and replace with ressistors