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Op Amp

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JasonMcG

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Hi there

I need help analysing this circuit.

I have taken the op amp on the left to be amp A and the one on the right to be amp B.
I calculated the following:
The gain for A is (1 + 10k/1k = 11).
The gain for B is (1 + 56k/1k = 57).
3n3 capacitor : Xc = 1/2pi f c = 3215.25 ohms.

It says the input signal is a 15kHz sine wave with a voltage of 0.1 Vp-p.
I know it is an AC circuit and need to know what else to do.

I also need to find the resonant frequency of the LCR circuit.

https://i1362.photobucket.com/albums/r687/JasonMcG1/OpAmp_zps70d48f9a.jpg

Regards
Jason McG
 
You need to work out the 'gain' of the LCR circuit in the middle if you want to know what signal will be present at the output. From wikipedia, the impedance of an inductor is 4ae79a5db34995327db1c245828b859c.png and the impedance of a capacitor is 33e2a105a5db00ac8867a7c8a0569a0b.png, where 'j' is [LATEX]\sqrt(-1)[/LATEX]. If you don't know how to calculate the 'gain', see https://en.wikipedia.org/wiki/Voltage_divider

Resonant freq and impedance info can be found here https://en.wikipedia.org/wiki/LC_circuit
 
The series resonant LC shorts the output of opamp #1 at a very high frequency. But the lousy old 741 opamp does not work at high frequencies and works poorly even at 15kHz.

Google has links that calculate the frequency for you.
 
Hi,

The resonant frequency for the LCR circuit in the center is 1/sqrt(L*C) which comes out to around 27705Hz with the values shown in the drawing:
R=10 Ohms
L=0.010 H
C=3.3e-9 Farads

The -3db points are found with:
w1=sqrt(-(R*sqrt(C^2*R^2+4*C*L))/C+R^2+(2*L)/C)/(sqrt(2)*L)
w2=sqrt((R*sqrt(C^2*R^2+4*C*L))/C+R^2+(2*L)/C)/(sqrt(2)*L)

and of course f1=w1/(2pi) and f2=w2/(2pi),

so the 3db points are about plus and minus 80Hz away from the center frequency.

The amplitude across the 10 ohm resistor might be useful:
Vampl=Vin*(w*C*R)/sqrt(w^2*C^2*R^2+(1-w^2*C*L)^2)
 
The lousy old 741 opamp works poorly above 9kHz.
Most modern inexpensive opamps work well up to 100kHz.
 
I have since been able to calculate the answer.
I calculated the gain for the first op amp using the 0.1V and got 1.1 V coming out the first amp. Using the inductive reactance and capacitive reactance I found the impedance to be 2.27 k ohms. Using this and (I=V/Z) the current is 0.484mA. V10ohm = Vamp2= (0.484 * 10) = 4.84mV. Using the gain on the second amp i got the output to be 0.28 Vp-p.
 
Hi again,

Very good Jason. And if you use the formula i gave you get 0.27587 vpp for a 0.1 vpp input, so the results agree. That's also assuming ideal op amps.

One extra small note here:
If the frequency gets near the resonate frequency it could overload the output of the op amp because at the resonate frequency the reactance of the cap and inductor cancel out making those two zero ohms, and so we then have a 10 ohm resistor connected from the output of the first op amp to ground. With 1.1v output that would mean 110ma of current to ground and that would be too high for the op amp.
 
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