Jon said:

The circuit shown here:

http://www.benchmarkmedia.com/pdf/panpot.pdf
says that the total gain to one of the outputs will be .707 times the

normal gain when the pot is in the extreme position. Is this correct?

Something wrong, here. In one extreme position, the output goes to

zero. In the other the output must be greater than when the pot is

centered. This cannot be a fractional multiple of the centered

output.

According to my calculations, the 4.87k, 10k, and 2k make a voltage

divider (because the input impedance of the inverting amplifier is

2k), which is (|| = in parallel with):

(10k || 2k)

------------------- = .255

(10k || 2k) + 4.87k

for the center position, it is:

(5k || 2k)

------------------- = .226

(5k || 2k) + 4.87k

since .255 * .707 = .180, the gain ratio of .707 (-3 dB) seems

incorrect.

the gain of the inverting amplifier is 11.8k/2k = 5.9, which,

multiplied by the voltage divider is 1.5 and 1.339, for the two

configurations. right?

If you ignore the input impedance of the amplifier, the values are

then:

10k

----------- = .672

10k + 4.87k

and

5k

---------- = .506

5k + 4.87k

and the total gains are 3 and 4, which are a ratio of .75 apart, which

is much closer. but does this mean that they forgot to include the

input impedance, or that i am in the wrong by including it?

Remember that the input of an inverting opamp is a virtual ground, so

you can figure the input signal as a current entering that ground.

The output is then, that current passing through the feedback

resistor. If you figure an input of 1 volt, then the gain is the

drop across the feedback resistor.

So, for the pot at 10k, and a 1 volt input, the current arriving at

the - input is:

Solving for the voltage across the 2K resistor:

1 volt * (2k || 10k) / (4.87k + (2k || 10k) = .255 volt

So the current to the inverting node is .25 volt / 2k = .1275 ma.

This causes a drop in the feedback resistor of:

..127 ma * 11.8K = 1.504 volts

For a gain of 1.504

Repeating for the pot at center:

Solving for the voltage across the 2K resistor:

1 volt * (2k || 5k) / (4.87k + (2k || 5k) = .227 volt

So the current to the inverting node is .227 volt / 2k = .1134 ma

This causes a drop in the feedback resistor of:

..1134 ma * 11.8K = 1.338 volts

For a gain of 1.338

So the gain ratio from full to center is 1.504/1.338=1.124

If they got the ratio upside down it would be 1.338/1/504=.89

I think they screwed up.