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following up my op-amp question - still confused

P

Peter

Jan 1, 1970
0
I designed the circuit listed below in the simulator using an ideal op-amp.

Rf = 20k
R1 = 2k
C1 = 4uF

the way it was designed: The Rf is the feedback resistor and R1 and C1 are
in series to ground. The input is 20Hz 2Vp-p on the non-inverting input. I
calculated a gain of 8.08 so the output should be 16.2Vp-p. But the
simulator shows an output of 15.6Vp-p.

Now this isn't a huge difference, but since I used exact numbers in the
calculator (i.e. didn't round off) and I used an ideal op-amp, shouldn't I
have the exact numbers???? I even calculated a 45 degree phase shift and
the simulator has a 38 degree phase shift.

If anyone can assist me, I'd appreciate it.

Thanks
 
P

Peter

Jan 1, 1970
0
The lowest gain is unity gain because at low frequencies the capacitor is
open and at high frequencies the capacitor is a short to ground so your
gain is 1+Rf/R1.

But my calculated and the simulation are different. Does anyone know
why????


Thanks
 
P

Paul

Jan 1, 1970
0
Wow!!!!! for 20k, 2k, 4uF and 20Hz I get exactly 7.8305 for a gain
(dividing my output by the input on the oscilloscope). This entire time I
was trying to find out why the exact gain can't be calculated and now it
looks like you solved the gain by calculationing and got exact (ideal)
answers. Remember, I'm using an ideal op-amp and exact value components, I
don't expect this to be true in the real world, but for purposes of
understanding the math behind it, I can use ideal components to verify my
math is correct.

I am not sure of your method though. It looks like you didn't use polar
cordinates at all and came up with an exact gain. Why using polar
cordinates (as everyone else did) does my gain not come out exact and
whichever method you used it came out correct???


If this is easier in an email, you can reach me at [email protected]


Thanks
 
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