Post Go back to editing

Photodiode transimpedance amplifier with ADA4627-1

Hi everyone,

I have to design a photodiode transimpedance amplifier for an instrumentation application.

I've used the Analog Device analog photodiode wizard with the following parameter:

- reverse bias : positive, 12V

- photodiode capacitance @reversebias : 6 pF (I can choose a smaller photodiode, it is not the problem)

- shunt resistance : 1 GigaOhm

- peak photo current : 100 µA

- target bandwidth : 1 MHz

- peak voltage : 10 V

- corresponding gain : 100 kOhm

- feedback capacitance : 1.8 pF

The wizard have found a single stage solution based on the ADA4627-1 operational amplifier. I've done an other modelisation using SPICE and the provided opamp spice model. No bad surprise.

What I dont understand is the obtained bandwidth of the circuit (1MHz) vs the provided 19MHz GBW of the ADA4627-1,as it come from the datasheet.

If I'm correct, the gain of the transimpedance amplifier is set only by the feedback resistor following G(db) = 20*log10(100 kOhm) = 100db. To achieve a bandwidth of 1MHz, the corresponding GBW of the opamp should be at least 100*1MHz = 100Mhz. So the 19MHz of the ADA4627-1 should not be sufficient, but simulation work perfectly...

I've suspected an error in the provided spice file, so here is what I've done:

I've tried to replace the ADA4627-1 by its decompensated twin, the ADA4637-1, in SPICE. The result show small difference, indicating that the two spice file are different. It also work perfectly. However, even the decompensated version with it's 79MHz GBW should not work.

Maybe there is something that I didn't get in all that calculation. Any help in understanding that would be greatly appreciated by the modest beginner I am.

Thank you in advance, and have a good day


  • Hi ZeusKnight,

    ADA4627-1 should work fine. The use of the feedback resistor is for the increase the output voltage from a current input. The one that will affect your bandwidth is your RC feedback. The calculation using the GBW is to determine what Cf to use. For example, if ((Rf/Rin)+1) ≥ 2 (GBW*Rf*Cs) then your Cf would be (Cs)/(2*((Rf/Rin)+1)) otherwise, Cf would be (Cs/(GBW*Rf)). In your case, since ((Rf/Rin)+1) ˂ 2 (GBW*Rf*Cs) which is  8.717E+10, your Cf will be 1.77pF 1.8pF as given by the photodiode wizard.



  • This question has been assumed as answered either offline via email or with a multi-part answer. This question has now been closed out. If you have an inquiry related to this topic please post a new question in the applicable product forum.

    Thank you,
    EZ Admin