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While doing this project I ran into a problem. We are using an opamp in a integrator circuit. However we noticed the existence of a DC component that only apears when the input frequency is high. I think it is somewhat related to the capacitor being charged, and the opamp offset, but i'm not sure how it occurs. Does anyone know why this happens? circuit

square wave 100hz 5V amp

sine wave 1kHz 1V amp

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This is an answer to your duplicate question asked later.

At low frequency the output is saturating and so the capacitor is forced to charge to the same voltage in either direction. If you reduce the amplitude of the input signal so that the output is no longer saturating I would expect the DC offset to return.

An integrator acts like a low pass filter. As the frequency increases, the output amplitude reduces which brings the output out of saturation enabling the DC offset to return.

At very high frequency the low pass filter effect has reduced the output to zero, but I would expect the output to slowly drift downwards. Try running the simulation for longer.

The output offset comes from a combination of the input offset voltage and the input bias current. These two cause a larger voltage drop across the 10k resistor when the input swings in one direction then when it swings the other way causing more current to flow "through" the capacitor one way then the other.

A parallel resistor ensures that if there is a DC offset then there will be less capacitor charge current in one direction than the other due to the higher voltage across the resistor. This eventually evens up the output swings.

As a matter of interest, the output amplitude will equal the input amplitude at f = 1 / (2 * pi * 10k *47nF) = 338Hz

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As DC charge builds up, the baseline will shift. Place a 100,000 ohm resistor across the cap.

Any slight imbalance in the input waveform, compared to the OpAmp's Voffset, will cause this buildup.

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  • \$\begingroup\$ The thing is I cant change the circuit, only explain it. Sorry but I dont understand what you mean by "builds up", you mean in the capacitor? \$\endgroup\$ – Catarina Apr 23 at 18:54
  • \$\begingroup\$ So, this means that you have no access to the opamps output and input terminals? Note that each working opamp circuit needs a DC negative feedback path for a stable operating point. However, if your integrator stage is used as a block (and not as a stand-alone circuit) within an overall control loop, it will work as desired. \$\endgroup\$ – LvW Apr 24 at 8:37

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