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# noob question : value of potentiometer for lower volume?

noob question : value of potentiometer for lower volume?

I currently use a 100k pot in my cmoy, if I want to reduce the volume by default , should I reduce pot to 20k? or increase?

help a noob

The value of the pot makes absolutely no difference to volume, because they are voltage dividers.

Turning the knob on your pot changes the ratio of voltage from your source that reaches the amp or goes to ground. At similar pot positions, bigger value and lower value pots all provide the same ratio.

If you want to reduce the volume, reducing the gain might be more appropriate.
uhmm how do I reduce the gain?
That depends. Is it a Cmoy you put together yourself? Somebody else did it for you? A commercial product?

In any case, this should get you started: Tweaks
In a CMoy, the op-amps are used in a very common non-inverting amplifier configuration. This means that the input is connected to the non-inverting input (IN+) and the feedback loop is the output (OUT) fed through a voltage divider attached to the inverting input (IN-).

The voltage divider is comprised of two resistors. The formula for gain is 1+R2/R1, where R1 is the resistor between IN- and ground and R2 is the resistor between OUT and IN-.
You can easily adjust gain by replacing R1 or R2.

You can locate these resistors by determining which pins they are attached to. The pinouts depend on whether the amp uses one double op-amp or two single op-amps.

If there's one double op-amp, these are the pinouts:
Code:
```    ┌────┐
OUTL┤1  8├ V+
IN-L┤2  7├ OUTR
IN+L┤3  6├ IN-R
V-  ┤4  5├ IN+R
└────┘
```
If there's two chips, one for each channel, these are the pinouts:
Code:
```    ┌────┐
NC  ┤1  8├ NC
IN- ┤2  7├ V+
IN+ ┤3  6├ OUT
V-  ┤4  5├ NC
└────┘
Pins 1, 5, and 8 might be used for offset adjustment,
but they can usually be left disconnected.
```
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• noob question : value of potentiometer for lower volume?