 ### Network # Problems of building integrator circuit using current feedback op-amps

S

#### sunpeak

Jan 1, 1970
0
Hi,

I am wondering if anyone can give me some suggestion on building an
integrator circuit using current feedback op-amps. It is a simple
circuit with R and C, while C is connected for the negtive feedback.
Since the current feeback op-amps like some specific feedback
resistance, I connect a 1k Ohm at the inverting input and then the
summing point for connecting R and C is the other side of this 1k Ohm
resistor. However, the test result is not resonable. If the input
signal is +/- 2.5V square wave, I was supposed to get a triangler wave
at the output if the non-inverting input of the opamp is grounded, but
what I saw at the op-amp output was a severe distorted triangler wave
with some offset volatge as high as 13V (the supply voltages are +/-
15V for the op-amp). Anyone knows why's this large offset and the
distortion?

Thanks,

Spunky

R

#### Rene Tschaggelar

Jan 1, 1970
0
sunpeak said:
Hi,

I am wondering if anyone can give me some suggestion on building an
integrator circuit using current feedback op-amps. It is a simple
circuit with R and C, while C is connected for the negtive feedback.
Since the current feeback op-amps like some specific feedback
resistance, I connect a 1k Ohm at the inverting input and then the
summing point for connecting R and C is the other side of this 1k Ohm
resistor. However, the test result is not resonable. If the input
signal is +/- 2.5V square wave, I was supposed to get a triangler wave
at the output if the non-inverting input of the opamp is grounded, but
what I saw at the op-amp output was a severe distorted triangler wave
with some offset volatge as high as 13V (the supply voltages are +/-
15V for the op-amp). Anyone knows why's this large offset and the
distortion?

CFAs have a low impedance on the negative input and
are therefore not suitable for integrators.

Rene

B

#### [email protected]

Jan 1, 1970
0
DEAR FRIEND.
R ASSUMED TO BE 1K OHM
F IS THE FREQUENCY OF APPLIED SQUERE WAVE
USE THE FORMULA F=1/(2piRC) TO CALCULATE THE VALUE OF C
choose C a virable capacitor to calpration
ANY QUSATION WRITE TO MY EMAIL [email protected]

S

#### sunpeak

Jan 1, 1970
0
I have read some document on CFAs and they suggested to connect some
resistance at the inverting input and then connect the RC network for
integrator. That is the purpose for puting that 1K Ohm in my circuit.
Will that do any good?

K

#### Ken Smith

Jan 1, 1970
0
I have read some document on CFAs and they suggested to connect some
resistance at the inverting input and then connect the RC network for
integrator. That is the purpose for puting that 1K Ohm in my circuit.
Will that do any good?

The inverting input of a CFA amplifier cares a lot what impedance is
connected to it. A major oversimplification: The resistor you put there
determines the gain of the first stage of the amplifier. The lower the
resistance, the higher the gain. If you make the gain too high, the
circuit will oscillate. Hence, you need a resistor there to keep the gain
reasonable.

F

#### Fred Bartoli

Jan 1, 1970
0
sunpeak said:
Hi,

I am wondering if anyone can give me some suggestion on building an
integrator circuit using current feedback op-amps. It is a simple
circuit with R and C, while C is connected for the negtive feedback.
Since the current feeback op-amps like some specific feedback
resistance, I connect a 1k Ohm at the inverting input and then the
summing point for connecting R and C is the other side of this 1k Ohm
resistor. However, the test result is not resonable. If the input
signal is +/- 2.5V square wave, I was supposed to get a triangler wave
at the output if the non-inverting input of the opamp is grounded, but
what I saw at the op-amp output was a severe distorted triangler wave
with some offset volatge as high as 13V (the supply voltages are +/-
15V for the op-amp). Anyone knows why's this large offset and the
distortion?

That should work except that you've not provided a path for DC feedback.
That coupled with the fact that CFB opamps tend to have a rather minus input
high bias current (the AD8005 datasheet I have on my desk right now spec 5uA
typ, +/-10uA max) won't let any chance to your pure integrator to have a
correct bias point.

This high offset current adds 5-10mV input offset voltage, the 1K as you've
placed it, adds to input referred voltage noise (still for the 8005:
opamp=4nV/rtHz, 1K=4nV/rtHz, 1K*In=9nV/rtHz for a total 10.5nV/rtHz), all
that for performances that won't be better than a VFB opamp of the same BW.

All this makes your nice idea not so nice.

Why a CFB opamp?

C

#### [email protected]

Jan 1, 1970
0
sunpeak said:
Hi,

I am wondering if anyone can give me some suggestion on building an
integrator circuit using current feedback op-amps. It is a simple
circuit with R and C, while C is connected for the negtive feedback.
Since the current feeback op-amps like some specific feedback
resistance, I connect a 1k Ohm at the inverting input and then the
summing point for connecting R and C is the other side of this 1k Ohm
resistor. However, the test result is not resonable. If the input
signal is +/- 2.5V square wave, I was supposed to get a triangler wave
at the output if the non-inverting input of the opamp is grounded, but
what I saw at the op-amp output was a severe distorted triangler wave
with some offset volatge as high as 13V (the supply voltages are +/-
15V for the op-amp). Anyone knows why's this large offset and the
distortion?

Thanks,

Spunky

Feeding a squarewave into an integrator will not give you a triangle
wave out, at least not for very long. Your integrator will "integrate"
any offset over time, thats what it's supposed to do, and send the
output to one of the rails.

J

#### Jim Thompson

Jan 1, 1970
0
Feeding a squarewave into an integrator will not give you a triangle
wave out, at least not for very long. Your integrator will "integrate"
any offset over time, thats what it's supposed to do, and send the
output to one of the rails.

The OP hasn't specified frequency or symmetry of the "square" wave.

Could be as simple as adding a large R across the integrating cap, OR
as complex as an overall DC restoration loop.

The easiest way to generate triangular waves is to use a pair of
comparators that switch current sources... then the end points are
always bounded.

...Jim Thompson

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