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Output impedance of transimpedance amplifier

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rrpilot

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Hello Guru's,

The circuit in question is a basic op-amp current-to-voltage converter with feedback resistor placed from output back to inverting terminal. signal current also connected to the inverting terminal and non-inverting terminal is connected to ground.

I remember back in school that if I have a voltage amplifier then I can express the output impedance as:

Zout = Zout,opamp(s) / (1 + beta(s)*A(s))
where Zout,opamp(s) is the open loop output impedance of the amp from the datasheet vs frequency.
beta(s) is the feedback factor which may or may not be frequency dependant and
A(s) open loop op-amp gain vs frequency.

Now I'm assuming I can use something like this for my transimpedance amplifier but I guess I'm not sure how to go about figuring out what beta(s) is for this particular circuit configuration. Would the model used for calculating noise voltage gain be used?

Thank you for any assistance.

Regards,
Tim
 

It is about the same thing. If you assume that the input impedance looking into the negative input is infinite, you work at low frequencies, so input capacitances don't matter and that your input current signal source is ideal i.e. infinite source resistance, then it becomes quite easy. In this case the beta is 1/Rf, where Rf is your feedback resistor. As a matter of fact 1/beta=Rf is your gain assuming ideal opamap i.e. infinite gain one.
 
What's your problem in determining beta?

Beta of the transimpedance amplifier isn't different from other inverting OP circuits, usually the transfer function of a divider Z1/(Z1 + Z2), in fact it's related to noise gain. Z1 of a basic transimpedance converter is the source impedance, you have to know it to calculate beta.
 
Thanks to both of you for clearing this up.

I was getting confused as to whether having the input signal a current source changed anything in terms of beta.
 

Notice, that the impedance of the current source can considerably change beta, e.g. the capacitance of a photodiode.
 

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