Following up on the Op-Amp Question

Wow!! there were a lot of replies. But none of which actually answered my
question as to why the simulator got a different gain than what I
I'm using an ideal op-amp and when I calculate it, I'm not rounding off, so
my answers should be just about identical.
Can someone explain this to me and/or explain if there is something else I
should be taking into account?
Reply to
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My guess:
In your calculation for gain, you have added
1 + Zf/Z1 = 1 + 7.09 and came up with 8.09. You should have 1 + Zf/Z1 = 1 + 7.09 /_45 degrees, and come up with 7.82 /_ ??, where I haven't bothered to find angle, but the gain is correct per your simulation.
You can use the formulae for gain but make sure you use complex values. Better still, understand the opamp and do not even use canned formulae, just solve the circuit.
Reply to
operator jay
Oops I might have answered a different op amp question. What is it, exam time?
Reply to
operator jay
--Is this the question you mean??
If I have a non-inverting amplifier, the gain is normally 1+Rf/R1.
So if I had 9k for Rf and 1k for R1, the gain would be 10.
Now how do I calculate it if I put a capacitor with an equal Xc value to R1 in series with R1????
I thought I just calculate Xc, add it to R1. Then I would divide Rf by that answer. But it doesn't seem to be the case, my gain is higher. I'm using 18k for Rf, 2k for R1, 5uF for C1 and 17Hz input. At 17Hz, my gain is 7.3.
Can someone shed some light on this?? I've spent a few hours on this problem and I don't know what to search for because I don't know what this type of circuit is called.
thanks in advance!
--Because if it is, here is the answer you were given
The gain of the non-inverting op amp is 1 + Zf / Zi. Now, Xc = - j 1872 so Zi = 2000 - j 1872 and Zf = 18000. The magnitude of the gain becomes 7.57.
--But that doesn't make sense because it appears you already saw that
I'm running a computer simulation. I got the answer in the next message, but I'm trying to understand polar cordinates right now, so the answer is a bit confusing for me. But the gain he calculated is correct.
--What's up.
Reply to
operator jay
No, it's not exam time. In fact, I am just working on understanding some op-amp circuits. Here is the website that actually got me interested in this question:
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The answer that was given to me was I should have a gain of 7.5 but the computer simulation came up with a bit higher gain. This is using ideal resistors, ideal op-amp (no offset voltage, input current or anything) and I'm not rounding off. So I'm confused as to where the extra gain is coming from.
Reply to
oops, I made two mistakes. I did not see the reply for 1+Zf/Z1 =1+7.09 = 7.82
The second is I said in my previous message the simulator got higher than this. the computer simulation got a gain of 7.33 where as my calculated value (according to the message) 7.82.
It's still not the 7.33 the simulator got. so I am wondering why I can't get exact values.
Reply to
The circuit was analyzed at the cutoff frequency of your R1 and C1 network.
Reply to
Jeff Thon
I get pretty much exactly 7.33 for your opamp problem. My answer to this opamp problem is identical to my answer to the other, identical opamp problem, just a change in the numbers.
1 + Zf/Z1 1 + (18000) / (2000 - j1872.4) 1 + (18000) / (2739.7/_-42) 1 + 6.57/_42 1 + 4.88 + j4.40 5.88 + j4.40 7.34/_whatever
I just WAGged 42 degrees, no calculator.
You are correct that complex numbers is your difficulty. Find a text with complex numbers and work through the conversion between a+jb form to polar form and vice versa. Do some problems on adding and subtracting, then on multiplying and dividing complex numbers. Then rattle off a couple Z=jwL and Z=1/jwC practice calculations. Then your formula for gain will work. Of course, it is still better to understand how an opamp works and not rely on a formula.
Reply to
operator jay

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