Why Isn't My Gm-C Integrator Design Working as Expected?

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The Gm-C integrator design is not functioning as expected because it behaves more like a low-pass filter rather than an integrator. The circuit requires a large series resistor between the amplifier and the capacitor to achieve proper integration, ensuring that the RC time constant is significantly longer than the integration time. For 1% linearity, the RC time constant should be at least 100 times the square wave integration time. The mathematical relationship indicates that to maintain a constant output rate, the exponential decay must remain close to 1 for all time. Proper modifications to the circuit are essential for it to perform as an effective integrator.
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Gm-C Integrator design...

Dear all,

I have been asked to design Gm-C continuous time integrator with no rigid specifications. I have used a two stage OTA with 46dB gain and applied a capacitor at the end as integrating element. But the circuit is merely acting as an amplifier and not integrator. It does not convert square wave into triangular... What modifications do i need to bring? Circuit schematic attached...

http://images.elektroda.net/32_1284460009.jpg
the circuit quite well acts as a low pass filter after installation of capacitor at the end...I believe just a capacitor at the end can't transform an amplifier into integrator...
 
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I believe you need a large series resistor R between the amplifier and capacitor C, such that RC is long compared to the integration time. For 1% linearity, RC>=100 x square wave integration time.

Bob S

[added]
Suppose after t=0 for square wave input, Vout(t) = V0[1-e-t/RC]

Then dVout(t)/dt = [V0/RC]e-t/RC

To achieve integration linearity of 1% (meaning dVout(t)/dt = constant) , e-t/RC >= 0.99 for all t
 
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