Cutoff Frequency Not What I Expected

[[email protected]]

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?

 

[John Larkin]

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?

Opamp gain-bandwidth?

John

 

[Joerg]

Hello,

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

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

Regards, Joerg

 

[Richard Hosking]

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

 

[theJackal]

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

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"

 

[theJackal]

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?

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"

 

[the Jackal]

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?

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"

 

[[email protected]]

Thanks to all of for your responses. You've given me some things to
think about.