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Problems of building integrator circuit using current feedback op-amps

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

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

Rene

Reply to
Rene Tschaggelar

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 snipped-for-privacy@gmail.com

Reply to
bughuffa

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?

Reply to
sunpeak

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.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

"sunpeak" a écrit dans le message de news: snipped-for-privacy@z34g2000cwc.googlegroups.com...

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?

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Thanks,
Fred.
Reply to
Fred Bartoli

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.

Reply to
cbarn24050

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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|  James E.Thompson, P.E.                           |    mens     |
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Reply to
Jim Thompson

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