Electrical – Photodiode max. Output power


I am designing a transimpedance amplifer to run at 10mA in 100uS to detect proximity of an object (low power application)

I am intending to use the VSMB1940ITX01 IR LED and the VEMD6110X01 IR photodiode.

I am looking at using the standard transimpedance amplifier in either passive or active configuration. The output current when illuminated is in the uA range but what determines the max voltage I can get from this in choosing my load resistor?

Also does using an active over a passive configuration help? I've had a look at multiple application notes but to little avail.

Any help greatly appreciated!

Best Answer

In active configuration, the diode is shorted, so the voltage it produces is 0. In this case, you are measuring its short-circuit current. There is therefore no issue of "max voltage".

In passive configuration, you apply a reverse voltage across the diode. Received light makes the diode leak more, and you measure this leakage current. In that case, you decide what voltage to hold the diode at. The reverse leakage current is usually quite insensitive to the reverse voltage over a reasonable range.