# amplifier with 2 capacitors

Discussion in 'Homework Help' started by mzsyed85, Jan 7, 2009.

1. ### mzsyed85 Thread Starter Member

Jan 1, 2009
10
0
Hey guys,
I am having some problems making the correct assumptions involving an inverting amplifier (attached)...
Assumptions:
I1 = I2
V- = 0

I1 = C1 (dVin/dt)
I2 = C2 (-dVout/dt)

Question: Prove Vout = -Vin(C1/C2)

I have arrived at this solution, but I am not sure I did it the right way or I got my assumptions right by coincidence - specifically for I2. Even then, I may have jumped a step which seems a bit dodgy to me now...

Since I1 = I2,

C1(dVin/dt) = C2(-dVout/dt)
C1(dVin) = C2(-dVout)
dVout = -dVin(C1/C2)

therefore Vout = -Vin(C1/C2)

Could somebody tell me whether I did this correctly? For some reason I feel I am not allowed to get rid of the 'd' that easily...
Thank you

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2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
To remove the 'd' you have to integrate both sides of your equation

dVout = -dVin(C1/C2)

integral of dVout=Vout

integral of dVin=Vin

and thus

Vout=-Vin(C1/C2)

3. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
Of course, in the real world the op amp would promptly saturate on one or the other power supply rail. Interesting concept though, I don't think they ever hit us with that one in school.

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
You are right Bill,this is the problem with schools and universities, they don't explain to the students what practical problems arise from the circuits.

5. ### mzsyed85 Thread Starter Member

Jan 1, 2009
10
0
oh yeah... i forgot about integration lol! Thanks guys, much appreciated