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I'm designing a readout circuit for a photodiode consisting of a transimpedance amplifier. I've noticed that the photodiode has a relatively large capacitance (320 pF). Are there any tricks/adjustments I would need to compensate for this capacitance (in terms of speed and noise)?

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Yes, there are many things beyond the textbook transimpedance amplifier configuration.

For example, you can use a cascoded transimpedance amplifier and bootstrap it to reduce the effect of the PD capacitance. Dr. Phil Hobbs is an expert in this subject, and I would recommend his book on Electro-optical Systems. Here's an article on the subject that is freely downloadable, and below is a schematic for such a PD front end. The BFG25 acts as part of the cascode, and the MPSA18 provides the bootstrapping.

enter image description here

This is by no means the final word on PD front ends, but the major ideas are present. When the FB resistor has to be very high value (G ohms), another set of tricks comes into play.

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  • \$\begingroup\$ Could this cricuit be used with a single +12V/+5V supply? \$\endgroup\$
    – joaocandre
    May 12, 2014 at 14:42
  • \$\begingroup\$ It would be better to generate the negative bias for the PD from the +12. \$\endgroup\$ May 12, 2014 at 14:48
  • \$\begingroup\$ But how would I control exactly the reverse bias applied to the PD? I'm trying to use an avalanche photodiode (APD) which would require high voltage bias (~70V) \$\endgroup\$
    – joaocandre
    May 12, 2014 at 16:05
  • \$\begingroup\$ I've noticed that several manufacturers provide several transimpedance ICs supposedly low-noise and with large bandwidth - do they implement these "tweaks" or is a custom amplifier like this still a better option? \$\endgroup\$
    – joaocandre
    May 13, 2014 at 9:57
  • \$\begingroup\$ what exactly would I need to change (regarding component ratings) if I wanted to change PD bias? \$\endgroup\$
    – joaocandre
    May 15, 2014 at 10:18
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You say the photodiode current is going into a transimpedance amplifier. That means ideally the voltage accross the diode isn't going to change, only the current thru it. The capacitance accross the diode (whether built into the diode or external) therefore doesn't matter. Since the voltage accross the diode doesn't change, there won't be any current thru the capacitor.

This is in fact the main reason to use a transimpedance amplifier. The input signal is current, not voltage. The voltage can therefore usually be held pretty constant, mitigating the effects of capacitance that would otherwise form a low pass filter with the impedance of the source.

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    \$\begingroup\$ Makes sense, however this TI document ti.com/lit/an/sboa122/sboa122.pdf seems to suggest otherwise. \$\endgroup\$
    – joaocandre
    May 8, 2014 at 13:16
  • \$\begingroup\$ @joaocandre especially since the input C affects the phase margin of the op-amp and will cause it to ring. www.ti.com/lit/an/sboa055a/sboa055a.pdf \$\endgroup\$ May 8, 2014 at 14:08
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    \$\begingroup\$ Olin, I'm going to have to disagree in a big way. The capacitance of the photodiode, hanging of the input of the transimpedance amp can produce serious instability. See any article about such amplifiers. There are two straightforward mitigations - increasing photodiode bias voltage (which reduces the diode's capacitance), and providing a feedback capacitor on the amplifier. See either of the two articles already referenced. \$\endgroup\$ May 8, 2014 at 17:12

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