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My goal is to measure the UV light using this UV diode and a trans-impedance amplifier, but I need help with the understanding of similar circuits.

I have checked out the following image and I don’t understand the function of the 51k resistor.

enter image description here

The 4.7M resistor is for the conversion factor. I have to change the value of this resistor to 38k when I want to measure 2500 nA with an output voltage of 950 mV. But what is the correct value for the 51k resistor? The second opamp is for the impedance conversion to achieve a high output impedance. Is this correct?

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I have checked out the following image and I don’t understand the function of the 51k resistor.

Normally, designers put a resistor in series with the non-inverting input to balance the impedance seen by both input pins. This then largely prevents offset voltages errors due to bias currents flowing in or out of the inputs.

However, on the CA3130, the input bias current is a paltry 50 pA maximum and so, the worst case error correction voltage it can muster is 51 kΩ × 50 pA = 2.55 μV and, that is waaaaaay below the natural offset voltage produced by the CA3130 which is typically 8 mV.

So, my considered conclusion is that the 51 kΩ resistor is surplus to requirement. It (the appropriate value of course) may be needed on other op-amps but not on this one.

The second opamp is for the impedance conversion to achieve a high output impedance. Is this correct?

It produces a very low output impedance (circa ohms); not a high output impedance.

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    \$\begingroup\$ One addition: The schematic is designed for some other diode and not for my UV diode. So it´s not clear if the 51k resistor was necessary. But your explanation helps me! Thank you very much! \$\endgroup\$
    – Kampi
    Commented Aug 11, 2022 at 16:13

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