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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?

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  • \$\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
    Commented Feb 13, 2015 at 7:47
  • \$\begingroup\$ Perhaps the probe is 1 M\$\Omega\$ \$\endgroup\$
    – HKOB
    Commented Feb 13, 2015 at 7:54
  • \$\begingroup\$ @HKOB what will happen if the probe is 1Mohm? \$\endgroup\$
    – Avin
    Commented 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
    Commented 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
    Commented Feb 13, 2015 at 8:43

1 Answer 1

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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.

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