# RC integrator behavior

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:

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?

• Your circuit is not an integrator, it's an LPF. You have to feed your input as current in order for it to integrate.
– Mike
Commented Feb 13, 2015 at 7:47
• Perhaps the probe is 1 M$\Omega$
– HKOB
Commented Feb 13, 2015 at 7:54
• @HKOB what will happen if the probe is 1Mohm?
– Avin
Commented Feb 13, 2015 at 8:18
• 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.
– LvW
Commented Feb 13, 2015 at 8:34
• 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.
– LvW
Commented Feb 13, 2015 at 8:43

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