Single Supply Integrator Questions

[eev_carl]

Hi,

I built this single-supply integrator amplifier but am not getting the output on the scope I expected.  From the specification in the lower-right of the book image (Vin=+-2.5V, 2000Hz, Vout=+=1.3V), I was expecting a triangle wave at +1.3V/-1.3V.  However, I got a curved triangle at a much smaller amplitude.   See the first scope image which shows a lower amplitude than what the book asked for.  It shows a +-1.25V Vin, but I got similar results with the +2.5V/-2.5V input.  My V+ is 9V and I've tried 5V and 15V as well.

Also, there is a lot of overshoot at the square wave transitions and was wondering if this was in spec.  See the second image.  I'm using a Rigol 1054Z scope and DG1022 function generator.

Thanks in advance,
Carl

[Zero999]

Try increasing R2 to 470k.

[orin]

The scope trace looks fine to me.  See attached ltspice run (Using LT1013 as the op-amp.  I scaled and offset the traces; the rectangular wave is actually 2.5V pk-pk).

The curved triangle is due to the R1-C2 time constant of 470us which is significant considering the pulse width is 250uS.  The R1-C2 time constant must be much greater than the pulse width, or C2 charges, reducing the current into the integrator.  Remember that this circuit is integrating the current into the inverting input node, so as C2 charges, there is less voltage across R1, less current into the integrator and the triangle becomes an exponential curve.

If you exclude the effect of C2, then 2.5V for 250uS will put 2.5*250E-6/4700 charge on the integrating capacitor; applying Q=CV, that's about 1.3V as originally claimed.

Edit: corrected component designation (C2, not C1).

[Zero999]

The scope trace looks fine to me.  See attached ltspice run (Using LT1013 as the op-amp.  I scaled and offset the traces; the rectangular wave is actually 2.5V pk-pk).

The curved triangle is due to the R1-C2 time constant of 470us which is significant considering the pulse width is 250uS.  The R1-C2 time constant must be much greater than the pulse width, or C2 charges, reducing the current into the integrator.  Remember that this circuit is integrating the current into the inverting input node, so as C2 charges, there is less voltage across R1, less current into the integrator and the triangle becomes an exponential curve.

If you exclude the effect of C2, then 2.5V for 250uS will put 2.5*250E-6/4700 charge on the integrating capacitor; applying Q=CV, that's about 1.3V as originally claimed.

Edit: corrected component designation (C2, not C1).

You're right. I overlooked C2. I erroneously assumed the circuit looked like a resistive load at the input, but it doesn't, it's a constant current load. The circuit is both a differentiator and integrator, depending on which capacitor is larger. To make it more of an integrator, C2 could be a 10µF electrolytic capacitor, with the anode connected to R1.