# Op amp design need advice

1. Jan 16, 2007

### Number2Pencil

1. The problem statement, all variables and given/known data
Design an op-amp that uses a variable 50k ohm resistor. when this variable resistor is at one extreme, the gain is 13, and at the other extreme, the gain is 3. you may use up to 2 op-amps, and up to 7 resistors (including the variable resistor)

3. The attempt at a solution

here is the circuit i came up with:

I made it so all I would have to find is Rf to satisfy the design.

here is how I solved for Rf

first extreme:

$$(1 + \frac{R_f}{60k-ohms}) = 13$$

next extreme:

$$(1 + \frac{R_f}{110k-ohms}) = 3$$

combine the two equations:

$$11 + \frac{R_f}{110k-ohms} = 1 + \frac{R_f}{60k-ohms}$$

$$10 = R_f (\frac{1}{60k-ohms} - \frac{1}{110k-ohms})$$

$$R_f = 1.32M-ohms$$

this answer gives a gain of 13 on one extreme of the variable resistor and 23 on the other. whoops. any help?

Last edited: Jan 16, 2007
2. Jan 16, 2007

### Gokul43201

Staff Emeritus
Two equations in one variable? Why should they have the same solution? As expected, if you solve them individually, you find they do not.

How would you fix this?

3. Jan 16, 2007

### Number2Pencil

you're right, I get a different resistor values for each equation, yet it needs to be the same value.

Mathematically, I should have one equation for each variable. There's only one but I forget how to solve an equation that has an answer of 3 OR 13...

either i need more variables or i need less equations

yeah i need another pointer

Last edited: Jan 16, 2007
4. Jan 16, 2007

### Number2Pencil

What I make the 60kohm resistor R1 (or another unknown)...

that way I could have two equations with two unknowns. Is this the correct way to approach this problem?

5. Jan 16, 2007

### Gokul43201

Staff Emeritus
Yes, that would be one way to fix it. To solve a problem with n independent boundary values, you need n independent parameters.

6. Jan 16, 2007

### Number2Pencil

Good, cause it worked. I got R1 = 10k ohm and Rf = 120k ohm which give me the correct gain(s). Thank you so much I appreciate all the help this forum gives me.

But out of curiousity, you said that's "one" way to solve it. got another trick up your sleeve?

Last edited: Jan 16, 2007
7. Jan 17, 2007

### AlephZero

There are a huge number of ways you could design the circuit within the constraints you are given. For example this is a completely different idea which doesn't need any algebra to figure it out.

(Except its deliberately wrong because it has gains of -3 to -13, so its not a complete solution to your question!)

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Last edited: Jan 17, 2007
8. Jan 17, 2007

### Staff: Mentor

Getting kind of sneaky there, are we?

9. Jan 17, 2007

### Gokul43201

Staff Emeritus
I would have put the trimpot on the feedback loop, instead of as an input resistor.

10. Jan 17, 2007

### Number2Pencil

yeah, i realized it today that by putting the trimmer on the feedback, i'd get the low, 3 gain when it's shorted and 13 when it's the full 50k, which would just make it less confusing to use