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Subject: Re: input impedance of transimpedance amplifier
Date: Sat, 23 Nov 2002 23:08:14 +0100
Organization: Worldonline Belgium
References: <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
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Indeed in a real world circuit the input impendance will be low at dc
but not zero due to the finite gain of the opamp.
As frequency rises the input impedance will increase due to the
decreasing gain and the growing phase lag of the opamp.
The stray capacitance of the (-)input node of the opamp may cause a
maximum of the input impendance at some frequency, it looks like a
You may ahve a look at :
I hope this helps,
Asa Cannell wrote:
> I don't understand what you mean by first order. The way I see it, if
> the voltage change is zero, the input impedance is zero right? I can
> see how the AC impedance might change as frequency increases and the
> opamp can't keep up as fast, and I think thats what you and the
> previous poster with the Z=Rf/A was saying.
> James Meyer wrote in message news:<email@example.com>...
>>On 22 Nov 2002 10:45:30 -0800, firstname.lastname@example.org (Asa Cannell) wroth:
>>>How do I determine the input impedance of a transimpedance amplifier?
>>>I am using an opamp as the active element, with a 100k resistor from
>>>output to inverting input. Non inverting input is grounded. What is
>>>the input impedance at the 'virtual ground'? I think it might be zero
>>>because a change in current results in zero change in voltage (at the
>> For a first order approximation, you are correct.
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