Cutoff Frequency Not What I Expected

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mwazowzki 20 years ago

Hello All, I've run into a circuit that I'm simulating but the results are not what I expected. The circuit is a negative feedback op-amp circuit with a 100 Ohm resistor at the inverting input, a 100 Ohm resistor from the non-inverting input to ground (for bias currents probably) and a parallel resistor-capacitor in the feedback path (8.45kOhm and 47 pF, respectively). According to my calculations, the gain equation should be:

A = Rf/Rin( 1/ (1+j2*pi*f*Rf*C) )

Using the values Rin = 100 Ohm, Rf = 8.45kOhm and C = 47 pF, the cutoff frequencty (-3 dB) should be about 400 kHz.

When I run the simulation, the gain falls 3 dB at 12.4 kHz.

What am I forgetting?

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J

John Larkin 20 years ago

Opamp gain-bandwidth?

John

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Joerg 20 years ago

Hello,

That would mean the opamp must have a GBW product in excess of 30MHz. Does it?

Regards, Joerg

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Richard Hosking 20 years ago

You are trying to get a very high gain - nearly 40dB. The opamp will run out of gain at some freq, depending on it's performance. It will effectively have a capacitance in parallel with your circuit capacitance . The higher the gain the earlier this effect will cut in.

R

snipped-for-privacy@yahoo.com wrote:

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theJackal 20 years ago

Your analysis is based on the simplified /idealized and well known /used model where the input resistance is Rin. A more accurate model shows that this is actually an approximation that works at low frequencies, high open loop gains . After some Math its easy to show that the actual expression at dc is Ri = Rin + Rf +ro/ 1 +a + (Rf +ro)/rd

a = open loop gain ro= output impedance in opamp rd= differential resistance in opamp

at signal frequencies replace the above with their impedances.

theJackal

"Go easy on the Whiskey"

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T

theJackal 20 years ago

Your analysis is based on the simplified /idealized and well known /used model where the input resistance is Rin. A more accurate model shows that this is actually an approximation that works at low frequencies, high open loop gains . After some Math its easy to show that the actual expression at dc is Ri = Rin + Rf +ro/ 1 +a + (Rf +ro)/rd

a = open loop gain ro= output impedance in opamp rd= differential resistance in opamp

at signal frequencies replace the above with their impedances.

theJackal

"Go easy on the Whiskey"

Vote

T

the Jackal 20 years ago

Your analysis is based on the simplified /idealized and well known /used model where the input resistance is Rin. A more accurate model shows that this is actually an approximation that works at low frequencies, high open loop gains . After some Math its easy to show that the actual expression at dc is Ri = Rin + Rf +ro/ 1 +a + (Rf +ro)/rd

a = open loop gain ro= output impedance in opamp rd= differential resistance in opamp

at signal frequencies replace the above with their impedances.

theJackal

"Go easy on the whiskey"

Vote

M

mwazowzki 20 years ago

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Cutoff Frequency Not What I Expected

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