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Original question:

OPA380: Designing with low impedance input signal

OPA380: Functioning of the IC

Swapniil Kanwar
Prodigy 160 points
Part Number: OPA380

Hi, 

I was studying about transimpedance amp and came accross this IC. In the datasheet, there is a circuit which is intriguing me. 

On the first page I came across this circuit which, as per my knowledge, is a simple transimpedance circuit with a small bias voltage at non-inverting pin of top opamp (to prevent opamp going into saturation)
(prior knowledge is based on video by TI on youtube: www.youtube.com/watch

But with this knowledge I am unable to infer what voltage will be developed at non-inverting pin of top opamp. The 1Mohm and 67pF forms an integrator circuit, which is basically a low pass filter with cutoff 2.3KHz and then followed by 100kohm and 75pF which again forms a low pass filter with cutoff 21.2Khz. 

I am not able to understand the fact that how does taking voltage at photodiode and passing it with two low pass filter and giving it as a Bias voltage helping? Could anyone give me the proper explanation, it will be very helpful!! 

Thanks 

  • 0
    Guru 281620 points

    This circuit does not measure a voltage. A photodiode generates a current, and the transimpedance amplifier converts this into a voltage.

    All opamp inputs are kept at GND. (Shorting the photodiode minimizes the dark current.) The photocurrent flows through RF; the output voltage is Iphoto × RF.

    The second amplifier does offset correction (see p. 9 of the datasheet); I think the circuit does not show everything.