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I want to calculate the maximum amplification that I can get out of my transimpedance amplifier and I'm struggling a little bit with how to do determine this value.

Is the amplification simply the value of the feedback resistor? Because that is what they are doing in this Texas Instruments tutorial: Transimpedance Amplifiers - What Op Amp Bandwidth do I Need? In the design example beginning at page 4 they choose a 1 Meg resistor and say that the amplifier has a gain of 1 Meg and a I/V bandwidth of more than 100 kHz. But when I check the datasheat of the used OPA316 on page 11 the Open Loop gain at 100 kHz is only 40 dB - and if I am not mistaken we would need 120 dB for an amplification of 1 Million.

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Is my assumption wrong? Do I mix up something here? Can the I/V amplification be higher than the Open Loop Gain?

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  • \$\begingroup\$ What's the open loop current gain of the op-amp? \$\endgroup\$ – Spehro Pefhany May 31 at 13:30
  • \$\begingroup\$ I'm not sure what an open loop current gain is. In open loop there is no current? \$\endgroup\$ – jusaca May 31 at 13:53
  • \$\begingroup\$ There's almost no input current in a FET input op-amp so the (DC) current gain is almost infinite. Input resistance is typically something like 1T ohm. \$\endgroup\$ – Spehro Pefhany May 31 at 13:54
  • \$\begingroup\$ Ohh, I see. So the Open Loop Gain for voltage does not matter in this circuit, because the input is a current. \$\endgroup\$ – jusaca May 31 at 14:04
  • \$\begingroup\$ It matters, but in a way I don't have time to properly answer. Hopefully someone else will provide a complete answer. The paper you referenced looked pretty good at a glance. \$\endgroup\$ – Spehro Pefhany May 31 at 14:06
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Since the internal compensation for guaranteed stability with any RF value is at 30Hz, the GBW is 100MHz from 100Hz and up. Therefore indeed 40dB gain at 100kHz and the gain in mV/mA is =Rf.

You can see the DC gain is less than 120dB and the integrator starts at 30 Hz.

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