# finding gain of differentiating amp

Discussion in 'Homework Help' started by no12thward, Oct 24, 2011.

1. ### no12thward Thread Starter Member

Jan 10, 2011
30
0
How would i go about creating this amp with a gain of -30. Resistors are no prob, but how would i find the gain with the capacitor and resistor in this differentiating amp?

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Oct 22, 2011
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3. ### crutschow Expert

Mar 14, 2008
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3,229
What is the frequency at which you need a gain of -30?

The gain is simply the capacitor reactance at the design frequency divided by the input resistance.

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4. ### steveb Senior Member

Jul 3, 2008
2,433
469

I agree, that is a bit of a confusing question. The way I interpret the question is to say that a true derivative is assigned a gain of one at all frequencies, even though it is frequency dependent. The gain is then the overall constant which scales the pure derivative.

For example, let's say we have the following relation.

$y(t)={{dx(t)}\over{dt}}$

Here, we can think of x as an input and y as the output. The function is a pure derivative, and in the Laplace domain it is simply

$Y(s)=sX(s)$

Now lets amplify the above equation by -30, and we get the following.

$y(t)= -30 {{dx(t)}\over{dt}}$

$Y(s)= -30sX(s)$

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5. ### no12thward Thread Starter Member

Jan 10, 2011
30
0

So assuming there is no frequency as a dependant variable here, my gain would just be capacitive reactance/ value of input resistor?

6. ### steveb Senior Member

Jul 3, 2008
2,433
469

I think the gain would simply be -RC because the output function formula is the following.

$v_o=-RC{{dv_i}\over{dt}}$

So, you need to find the best combination of standard resistor and capacitor values that gives you RC=30, as closely as possible.

Last edited: Oct 24, 2011
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7. ### no12thward Thread Starter Member

Jan 10, 2011
30
0

Then that means I'd have a capacitor value of 3.3F and resistor value of 9.1Ω. How would the input and output voltage be found now, if neither are known to begin with?

8. ### steveb Senior Member

Jul 3, 2008
2,433
469
I'm not sure I follow you. You need to know the input to calculate the output.