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I am developing a transimpedance amplifier as an assignment for a faculty course and I am running into problems when trying to simulate my circuit in ADS Keysight, even though I am quite sure how my circuit works and I know what am I supposed to get as a result. My circuit with TIA+photodiode

I am sweeping for all photocurrents I am expecting to obtain from my diode, and while trying to do a Bode plot, I get the response of a simple integrator.Bode plot of the transimpedance gain

Since I am very closely following the schematic from my literature (in which they managed to obtain sensible simulation results for almost the same schematic), I am puzzled, why does my Bode plot of gain looks like this? I was expecting to get a constant gain until reaching cutoff frequency of 7.3 kHz which is determined by capacitor and resistor in the feedback loop of TIA. Also, I have tried several varied values of capacitor and resistor, but I am getting the same result all the time.

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    \$\begingroup\$ You are plotting the voltage gain which will just be the open-loop gain of the opamp. You need to plot the transfer impedance which will be Vout to input current ratio. \$\endgroup\$ – Kevin White May 30 at 19:56
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    \$\begingroup\$ What's the point of SRC2? With the voltage divider formed by R3 and R2 you're only getting ~0.1 V at the non-inverting input of the op-amp. \$\endgroup\$ – The Photon May 30 at 20:11
  • \$\begingroup\$ The two comments above are spot-on, and though it makes no difference for simulation your polarized caps are backwards. It looks wrong. And you probably can't find polarized caps that are 30pF or 2.2pF anyway. \$\endgroup\$ – John D May 30 at 20:26
  • \$\begingroup\$ @JohnD And you probably can't find polarized caps that are 30pF or 2.2pF anyway This is a circuit simulator and the caps are ideal, the symbol might look like it is a polarized cap but the model "under" that symbol isn't polarized. ADS can also be used for simulating caps on a chip or a substrate and while these aren't polarized they do have a "top" and "bottom" plate which is why the symbol is such that you can see which side is top and which side is bottom. \$\endgroup\$ – Bimpelrekkie May 30 at 20:57
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    \$\begingroup\$ @DaveTweed As Kevin wrote above the OP is just measuring the open loop response of the amplifier. You can see the dominant pole and the single-pole roll-off to the GBW point. The transimpedance response should be current in to Vout, which should not show the dominant pole. \$\endgroup\$ – John D May 30 at 22:01

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