I am now designing a transimpedance using ADA4062-2 to detect signal from a photodiode. The first image is the equivalent circuit.

The signal from PD contains 1kHz sine wave and 2kHz sine wave and their equation are:

1kHz: 1u*sin(a)cos(2*pi*1k*t);

2kHz: 10n*cos(a)cos(2*pi*2k*t+b);

(while a is a fixed value set manually and b means the phase of 2kHz signal)

My target is to measure the amplitude and the phase of 2kHz sine wave. As the signal is quite small, I decide to use T feedback network. However, I found that the phase of 2kHz would change a lot, especially when a is set to be near 90 or 270 degree.

I doubted that it may be due to the T feedback network. So I substitute the T network with equivalent resistance(the second image). And luckily the phase change is gone.

But I don`t know why would the T feedback network affect the phase. Distortion? or any other reason? Does any one have idea about it? Thank you very much!

enter image description here enter image description here

  • \$\begingroup\$ The T feedback network was bypassed by C1. So you have a direct feedback path at HF and a T path at LF, which may be relevant. \$\endgroup\$
    – user16324
    Commented Feb 13, 2015 at 11:48
  • \$\begingroup\$ @BrianDrummond: But could you tell me why would they affect the phase of my signal? \$\endgroup\$
    – billyzhao
    Commented Feb 13, 2015 at 13:18
  • \$\begingroup\$ I think LvW is doing a better job of that. \$\endgroup\$
    – user16324
    Commented Feb 13, 2015 at 13:20

1 Answer 1


Billyzhao, you cannot simply replace a capacitive bypassed T-network with a corresponding resistor (also bypassed) - and expect the same behaviour. This can be shown on a simple intuitive way: In the first circuit the left side of C1 works upon the input network in parallel to app. (40+1)=41 kohms. In the second circuit you have 860k in parallel.

Both circuits are equivalent only without the bypass capacitor.

  • \$\begingroup\$ Thank you for your reply! Do you mean the Y-∆ transform is incorrect when bypassing a cap? \$\endgroup\$
    – billyzhao
    Commented Feb 13, 2015 at 13:16
  • \$\begingroup\$ Yes - because the feedback factor is different in both cases. (Without C both are identical). \$\endgroup\$
    – LvW
    Commented Feb 13, 2015 at 14:11
  • \$\begingroup\$ Ok. Do you know why would the T feedback network + bypass cap affect the phase? \$\endgroup\$
    – billyzhao
    Commented Feb 14, 2015 at 2:56
  • \$\begingroup\$ The simple Y-∆ transform applies only in case the relevant network is not bypassed and/or loaded by other circuitry. \$\endgroup\$
    – LvW
    Commented Feb 14, 2015 at 10:42

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