[SOLVED] Questions about a TIA Product (TI OPA380) Data Sheet

[FvM]

Saying that the transimpedance value isn't directly related to voltage gain of the inner amplifier doesn't mean of course, that the voltage gain doesn't affect the overall behaviour. But other parameters like input capacitance and source impedance have to be added to the calculation.

 

[coxstreet]

No I meaned that voltage gain numbers (unit V/V) can't be compared to transimpedance numbers (unit V/A).

Hi FvM,

YES! You are right! Thank you!

Cox

 

[erikl]

Hi Erikl, I think we made a mistake at the beginning. The transimpedance gain of TIA should not be compared with its openloop gain by just so simple way. Smaller openloop gain still can have large TIA gain.

My TIA transfer function is wrong. It should be (for first order system) Vout/Iin=[-Ao/(Ao+1)] * [Rf/(1+Rf*Cdiode*s/(A+1)] (This is from Razavi's "Design of Integrated circuits for optical communication", 2th, Pg 87). So for this equation, at the low frequency, Vout/Iin= [-Ao/(Ao+1)] * Rf and thus A0>>1 then Vout/Iin=Rf. A0 can be no larger than Rf to achieve this.

For the TI data sheet, I use one OPAMP has dc_Gain=45dB to achieve TIA's Rf=100k ohms without no degrades.

Cox

Hi Cox,

seems I've made the same mix-up of the 2 gain values dB and dBΩ.
Thank you for your feedBack - and FvM for the hint!

erikl