Light Catcher

The problem

A photodiode’s current (a current source on the net “inn”, with 5 pF of capacitance to ground, both supplied by the test bench) must become a voltage. Build a transimpedance amplifier: a feedback resistor Rf from “out” to “inn” (the gain is Rf in ohms), with the non-inverting input tied to a 0 V source named “inp”. Choose Rf = 100 kΩ for a transimpedance of 100 dBΩ.

With only Rf and the input capacitance, the loop gain has a pole at 1/(2π·Rf·Cin) = 320 kHz, and the amplifier’s own pole near 10 MHz: the response peaks and may ring. A small capacitor Cf across Rf adds a zero in the loop that restores the phase margin. A good value is √(Cin/(2π·Rf·GBW)), about 1 pF here.

The test bench

• The test bench supplies the op amp as a ready-made block with terminals on the nets named “inp”, “inn” and “out”: DC gain 80 dB, one pole at 1 kHz (unity-gain frequency 10 MHz), output resistance 100 Ω. The photodiode is an AC current source of 1 A into “inn” in parallel with 5 pF to ground.
• “Gain” here is the transimpedance in dB relative to 1 Ω (100 kΩ = 100 dBΩ).

Numbers to hit

• ◆Transimpedance between 99.5 dBΩ and 100.5 dBΩ
• ◆Bandwidth (−3 dB) at least 1 MHz
• ◆Peaking at most 1.5 dB