# Understanding Integrators and Input Offset Voltage

Discussion in 'Homework Help' started by SkiBum326, Jul 8, 2014.

1. ### SkiBum326 Thread Starter Member

May 16, 2014
33
0
Hi Everyone,

I have another question from Electronic Principles page 864 regarding integrators. I attached an image for clarification.

The book discusses the use of a resistor in parallel with the capacitor to account for input offset voltage. If I understand this correctly, the resistor limits the voltage gain, thereby limiting how much this offset voltage charges the cap. However, why doesn't the resistor have the same effect on the valid input voltage? The book mentions that the offset signal is DC and implies the valid voltage is not, but I'm not exactly sure how this is relevant..

The book then introduces the idea of using a JFET instead of the resistor. The JFET resets the cap before the input pulse. However, does input offset no longer have any effect?

Thanks everyone for the help!

Austin

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2. ### blah2222 Well-Known Member

May 3, 2010
565
33
Think about the circuit with only the capacitor connected in feedback.

To a DC input (offset voltage between the op-amp terminals), the capacitor is essentially an open circuit. A small DC perturbation will saturate the op-amp in either direction depending on its polarity.

To an AC input (valid source) the capacitor provides a closed feedback path, with impedance dependent on operation frequency and capacitance.

Since AC and DC are superimposed, the circuit needs to have paths for both of them in order to keep the op-amp from saturating. A simple way to do this is to give the DC inputs a closed feedback path to the output through a resistor.

The JFET acts as a low-resistance path for DC signals and also shorts the capacitor voltage to ground.

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3. ### MrAl Distinguished Member

Jun 17, 2014
2,556
515
Hi,

The time function for the circuit with only the capacitor in the feedback path is:
Vout=Vin*t/(Rin*C)

From this we can see that it is a ramp that ramps up to infinity for t approaching infinity, or just some long time relative to Rin*C. This is a perfect integrator.

The time function for the circuit with the resistor Rfb also in the feedback path is:
Vout=Vin*(Rfb/Rin)*(1-e^(-t/(Rfb*C))

From this second one we see that it is an exponential with a limit:
Lim[t-->inf](Vout)=Vin*Rfb/Rin

So the first one has no limit (will saturate the op amp) and the second one does have a limit but still depends on Vin. If Vin is too large it will still saturate, so Rfb/Rin is chosen to avoid that.

The input offset becomes a problem with the first one because for Vin*t/(Rin*C) and t a long time, the output still gets very very high and takes the op amp out of range. Even a small Vin could cause this depending on the time frame in question.

Using the second one and including the input offset, we get:
(Vin-Voffset)*(Rfb/Rin)*(1-e^(-t/(Rfb*C)))

so yes the input offset affects the reading, but the total response is now limited because of that second resistor.

The result of the integration is not a true integration anymore. The true integration of a constant in time with zero initial condition is:
a*t

where a is some constant. What we have with the second resistor is:
a*(1-e^(-b*t))

which is clearly not the same thing.

Look at both forms, we find that they are nearly the same for smaller time periods, and as the time period gets larger the error goes up. It looks like the error becomes as high as 5 percent for time periods of length approximately equal to Rfb*C/10, but you should check the two yourself for verification of the error due to Rfb.