0
\$\begingroup\$

I am trying to implement an RC integrator circuit in my design. The purpose of this circuit is to convert input pulse signals swinging between 0 to 2.5V into a ramp signal. The integrator circuit was designed with a time constant of 2ms. The circuit is as shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

According to the simulation results and the theoretical calculations for output voltage, for a 10Hz input pulsed signal, the output ramp signal should rise up to the level of 2.48V. Whereas in the PCB, I am finding the output voltage rising only till 2.06V.
I tried one more experiment with R1 = 20kohm and C1 = 100nF (time constant still at 2ms). With this combination, I am seeing the output ramp signal rising till 2.5V (approx.). Both capacitors used were of ceramic type with a voltage rating of 16V.

Does this behavior have anything to do with the capacitor's charging current? What could be the reasons for this behavior?

\$\endgroup\$
7
  • \$\begingroup\$ Your circuit is not an integrator, it's an LPF. You have to feed your input as current in order for it to integrate. \$\endgroup\$
    – Mike
    Feb 13, 2015 at 7:47
  • \$\begingroup\$ Perhaps the probe is 1 M\$\Omega\$ \$\endgroup\$
    – HKOB
    Feb 13, 2015 at 7:54
  • \$\begingroup\$ @HKOB what will happen if the probe is 1Mohm? \$\endgroup\$
    – Avin
    Feb 13, 2015 at 8:18
  • \$\begingroup\$ Yes,it is a lowpass filter - however, it can be used (with some restrictions) as an integrator for frequencies far above the corner frequency (3 dB) . By the way - in principle, the same restrictioins apply also to opamp integrators, but mostly with a much lower 3dB frequency. \$\endgroup\$
    – LvW
    Feb 13, 2015 at 8:34
  • \$\begingroup\$ For "proper" integration and a time constant T=2ms the input frequency should be at least some hundreds of Hz. For a 10 Hz input, the time constant must be much larger. \$\endgroup\$
    – LvW
    Feb 13, 2015 at 8:43

1 Answer 1

2
\$\begingroup\$

If your oscilloscope probe input loaded the circuit with 1 Mohm (normal for oscilloscopes, the peak voltage seen would be: -

2.5 volts \$\times\dfrac{1,000,000}{1,000,000 + 200,000} = \$ 2.08 volts

If you have a x10 facility on your probe (10 Mohm input resistance , the voltage would rise to about 2.45 volts.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.