Reply-To: "Kevin Aylward"
From: "Kevin Aylward"
References: <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
Subject: Re: input impedance of transimpedance amplifier
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Sat, 23 Nov 2002 18:17:03 -0000
NNTP-Posting-Date: Sat, 23 Nov 2002 18:17:11 GMT
Asa Cannell wrote:
> James Meyer wrote in message
>> On 22 Nov 2002 10:45:30 -0800, email@example.com (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 input node).
>> For a first order approximation, you are correct.
> 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.
First order means a first order *approximation*. If the gain is say,
100,000 and rf=100k, then Zin=1ohm, which is quite low.
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