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IVC102 integration time 1us

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Dear community,

I use IVC102 that is powered with +/-15V in order to measure low currents.

Internal capacitors are all connected together and at pin 3. So total capacitance is 100pF.

pin 2 is input,

pin1 is directly connected to analog ground.

With the formula of slew rate in the datasheet and taking 1V/us as maximum, Cint = 100pF then the maximum current to measure is 100uA.

Anyway...I use 3.3V and 10MOhm resistor to produce 0.33uA at pin2. So I tried to integrate for 1 us and I expect the output of IVC102 at pin 10 to be a voltage ramp with a maximum of about -3.3mV (gain = -10,000).

Well...I get a more squarish form with a maximum of approximately 200mV.

If I increase integration time to more than 100 us I get a voltage ramp with the expected voltage value (I mean now g = -1,000,000 so I get about 330mV)

Why does this happen? Any explanation?

Thank you a priori.

Gabriel


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