[SOLVED] Transimpedance amplifier

[rrpilot]

Hello Guru's,

I have been trying to wrap my head around some transimpedance amplifier basics and there is some things that trouble me.

Referring to TI's OPA655 op-amp datasheet (**broken link removed**)

The first page shows a bode plot with transimpedance gain. Two things I would like to discuss; transimpedance gain is 120dB when op-amp open loop gain is only 57dB, bandwidth is 1MHz when GBP is only 240MHz.

I have been doing a significant amount of research around these amplifiers and it seems as though the design is done around noise voltage gain on the noninverting terminal. Compensation capacitor in parallel with the feedback resistor is set so that the op-amp is unconditionally stable in terms of the noise voltage not the transimpedance gain.

My Questions:

1) Why does it seem like we can violate GBP formula when operating in transimpedance mode?
2) Why can transimpedance gain (V/A) exceed open loop voltage gain (V/V)?

I am willing to purchase books to help me understand, whatever you guys think will help. I have already been reading "Photodiode Amplifiers - op amp solutions" by Jerald Graeme but I haven't been able to answer these questions from reading in there.

 

[keith1200rs]

Transimpedance is the ratio of output voltage to input current i.e. a resistance. If that is plotted on a dB scale then 1MΩ comes out as 120dB. It is really 1MΩ. It is not a 120dB voltage gain.

Keith.

 

[rrpilot]

Hi Keith,

Thank you for your reply. The thing that bothers me is if I replace the current source input with a voltage source in series with a 1-ohm resistor then I am now limited by the GBP of the op-amp. I don't understand when directly working with a current input that things all of a sudden change.

 

[keith1200rs]

If you replace the current source with a voltage source and a resistor then it is not a transimpedance amplifier any more (unless you use a very large resistor so it is like a current source). So, if you use 1 ohm you will be trying to get 120dB of voltage gain and it just cannot do that - as you know the open loop gain is not that high. If you look at the voltage on the inverting input when it is running as a transimpedance amplifier then you will see that the ratio between the output and input voltage is 58dB - which just happens to be the open loop gain. The amplifier is working as a unity VOLTAGE gain amplifier but that doesn't mean its current to voltage conversion ratio is zero.

Keith.

 

[rrpilot]

If you look at the voltage on the inverting input when it is running as a transimpedance amplifier then you will see that the ratio between the output and input voltage is 58dB - which just happens to be the open loop gain. The amplifier is working as a unity VOLTAGE gain amplifier but that doesn't mean its current to voltage conversion ratio is zero.

This is starting to make sense. Okay, let's say we have a current input of 1uA for the circuit in the datasheet (1Mohm gain). I would expect to see +-1V on the output (depending on input polarity). Now the inverting input terminal voltage should be 58dB down from the output as you mentioned which would be 1.26mV. Then since there is still the 1V drop across the 1Mohm feedback resistor this error carries forward to the output which would now be 1.00126V. Now back to the input (58dB down) would be 1.26mV again (I should have carried more precision initially). Awesome! so the output reaches some stable value with error due to open loop gain.

Would what I did be accurate?

Keith that simple statement I quoted from you was invaluable, thank you so much!

---------- Post added at 11:14 ---------- Previous post was at 10:45 ----------

I ran through the same example except replacing the 1uA source with a 1uV source and 1ohm resistor and everything is making sense now! When I ran through that example the output ended up reversing the expected output polarity and saturating the output.

Thank you so much, you have cleared up so much for me here!

 

[keith1200rs]

If you feed in 1uA to the inverting input it will end up 1.26mV or so ABOVE the non-inverting one and the output will be at -1V.

The finite gain of an opamp is also the reason for the finite input resistance of the transimpedance amplifier.

Keith.