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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!

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    \$\begingroup\$ The bigger the feedback R the bigger the voltage. (for a fixed current.) Bigger R is also slower. If you are not getting near the saturation level of the PD then biasing does not make much difference. (If you are near saturation then reverse biasing helps a lot... sweeps carriers out of the junction faster.) \$\endgroup\$ – George Herold Jun 5 '17 at 16:38
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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.

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  • \$\begingroup\$ Okay, maybe I should re-phrase it. If i put a 1Giga-ohm resistor in the active or passive configuration. The active will only be limited by the op-amps available voltage rails for gain? The 1G resistor is a hyperbole but it may illustrate my issue. But surely the passive configuration would not be able to create such a proportionally huge voltage? Is there a way to calculate the max voltage for a given photo current? Also can you not put it in photovoltaic mode, as this was the mode I was intending to use? \$\endgroup\$ – ConfusedCheese Jun 5 '17 at 17:02
  • \$\begingroup\$ @ConfusedCheese What is your dynamic range for the current? There are ways to change the gain or to use a log-amp, if you fell trapped between exceeding the voltage rails on one end, or by ADC resolution on the other, using a single gain resistor (and associated BW limiting capacitor, usually.) \$\endgroup\$ – jonk Jun 5 '17 at 17:27
  • \$\begingroup\$ It can output up to 10uA. But i intend to run it around 1uA. What is the max resistor I can put across this that will create linear voltage? \$\endgroup\$ – ConfusedCheese Jun 5 '17 at 17:38
  • \$\begingroup\$ @Conf: You don't put a resistor across it. The resistor goes in the feedback path of the opamp. \$\endgroup\$ – Olin Lathrop Jun 5 '17 at 18:34
  • \$\begingroup\$ Sorry, yes, I know how to create the set up, i mis worded it. Apologies! \$\endgroup\$ – ConfusedCheese Jun 5 '17 at 18:35

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