No you can't. With the wiper all the way to the bottom, the gain is -1. As the wiper is moved up toward the mid point, the gain gets progressively higher, but is always negative. With the wiper at the mid point or higher, the gain is -∞, so the opamp works like a comparator. The gain is independent of the value of the pot.
Hi Mike,
Sorry, but i have to disagree. With the wiper all the way up the gain is equal to 1, and as it moves toward center the gain progressively increases until it reaches center where the gain is infinite, then as it just passes though center the gain goes highly negative and then progressively gets less negative until at the bottom end the gain reaches -1. Thus, the gain varies from -1 to 1, but it goes to very high extremes as the pot is turned toward center. This is with the other two resistors equal to 10k each and the pot is 10k too.
Here is the gain formula:
A=(R2*R4+R2*R3)/(R2*R4-R1*R3)
where
R1 is the input resistor,
R2 is the upper negative feedback resistor,
R3 is the upper part of the pot,
R4 is the lower part of the pot.
and note that when R2*R4=R1*R3 the gain is infinite, but with R2*R4>R1*R3 the gain is positive, and with R2*R4<R1*R3 the gain is negative.
If we keep the pot value at 10k we can simplify the formula (working in K ohms):
A=(10*R2)/(10*R2-(R2+R1)*R3)
and note the restriction 10*R2=(R2+R1)*R3 can not be allowed or the gain is infinite.
What this means is with R1=10k and R2=10k and the pot=10k, the gain never gets between -1 and +1, but it gets to every other possible value. Of course in real life the output will be limited.
With all resistors equal to 10k the formula simplifies to:
A=5/(R4-5)
where R4 is again the bottom part of the pot and from this we can see that the gain goes to infinity when R4=5.
We can also express the gain as:
A=R2/(R2-x*R2-x*R1)
where x is the ratio of R3 to the total pot resistance, so x=1/2 means the pot is at the center.