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OPA320: A question about the "1 MHz, Single-Supply, Photodiode Amplifier Reference Design"

Yan Fan
Intellectual 360 points
Part Number: OPA320

Hi all, 

I am reading the "1 MHz, Single-Supply, Photodiode Amplifier Reference Design" application note and have a question about the pole mentioned on Page 5. Why the R in parallel with C gives a pole of 1/2pi*C*R? 

Thanks.

Transimpedance amplifier_TI.pdf

  • 0
    Guru 140200 points
    Hi B,

    below this frequency the parallel circuit of C1 and R1 looks mainly like a resistor (R1) and above this frequency this parallel circuit looks mainly like a capacitor (C1).

    A pole is only a certain position in the Bode plot, where something happens. No rocket science...

    Kai
  • 0 in reply to kai klaas69
    Intellectual 360 points
    Thanks Kai,

    I have figured out. There is a Cj in sereis with R1 // C1 and if you write the transfer function, there is a pole at 1/2piR1C1.

    Thanks
  • 0 in reply to Yan Fan
    Guru 140200 points
    Ah, now I see what you mean. But I think you don't need Cj. It's enough to take the photocurrent source into consideration. Then the pole exists even without any Cj.

    Kai
  • 0 in reply to kai klaas69
    TI__Mastermind 23050 points

    Hi Kai,

    Just for clarification, you should always account for the photodiode capacitance (Cj). The interaction of the feedback resistance with the parallel combination of your photodiode capacitance and op-amp input capacitance creates a zero in the feedback network (technically 1/beta). This is the real reason why C1 is added, because without it your transimpedance amplifier will very likely be unstable. Effectively Cj places a limit on the maximum bandwidth you can get out of the circuit, and consequently on the minimum value of C1 you can use and still keep the op-amp stable.

    Also, Go Green!

  • 0 in reply to Zak Kaye
    Guru 140200 points
    Dear Zak,

    I'm sorry if my bad english lead to confusion. It was never my intention to suggest that Cj can be neglected in the design of a TIA. I only wanted to mention, that for the calculation of the pole R1 // C1 no other component in the equivalent circuit is necessary. Anyway... I know, of course, that Cj leads to a phase lag and that this phase lag would erode the phase margin of operational amplifier. A phase lead capacitance C1 is needed to recover the phase margin, at least partially.

    From my experience I know, that for very high detector capacitances this phase lead compensation can be unsatisfactory, sometimes, especially if R1 is in the kOhm range...

    Kai
  • 0 in reply to kai klaas69
    Intellectual 360 points
    Hi Kai,

    Why only R1//C1 can give a pole. My comprehension is only R in series with C can give you a pole from the transfer function.

    Thanks,
  • 0 in reply to Zak Kaye
    Intellectual 360 points

    Hi Zak,

    Do you happen to know how the equation 9 was derived?  where can I find the theory of using 1/beta line to determine the stability?

    Thanks, Go White!

    Bin

  • 0 in reply to Yan Fan
    TI__Guru 67601 points

    Bin,

    Please watch the video series under TI Precision Labs covering the topic of stability - see link below:

  • 0 in reply to Yan Fan
    Guru 140200 points
    Hi B,

    you wrote: "Why only R1//C1 can give a pole. My comprehension is only R in series with C can give you a pole from the transfer function."

    Maybe this video helps:

    www.youtube.com/watch

    Kai
  • 0 in reply to Marek Lis
    Intellectual 360 points

    Hi Marek, 

    Thank you very much for your suggestion. I have studied the material you mentioned but still could not figure out why "in the region above the pole, the frequency of intersection is fI = (C1/Cin+C1)fGBW.

    Bin

  • 0 in reply to Yan Fan
    Guru 140200 points

    Hi Bin,

    equation 9 is just taken out of figure 6 of the application note you refer to:

    The blue graph is 1/beta = (Cin + C1)/C1. The red graph is a line with the slope of -20db/decade, which crosses the 0dB line at fGBW. And the point of intersection of the blue and red graph is just at

    fi = C1/(Cin +C1) x fGBW

    An example: If 1/beta is 20dB, then the intersection point is one decade under fGBW. If 1/beta is 40dB, then the intersection point is two decades under fGBW.

    Kai