From: Winfield Hill
Subject: Re: input impedance of transimpedance amplifier
Date: 23 Nov 2002 13:51:01 -0800
Organization: Rowland Institute
References: <firstname.lastname@example.org> <email@example.com> <firstname.lastname@example.org>
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> 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.
Actually the input impedance looks like an inductor, with the value
. L = Rf / 2 pi f_T
When used with a real signal source, which always has capacitance,
the input inductor and the node capacitance resonate at frequency
. f_r = 1 / 2pi sqrt (LC) = sqrt (2pi f_T / Rf C)
This is a rather ugly oscillator-capable LC resonance, which can be
eliminated by placing a small capacitor across the transresistance
amplifier's feedback resistor, with a reactance equal to Rf at f_r.
Thuis capacitor creates a loss resistance, rs = sqrt Rf / 2pi fT C
in series with the input inductor, nicely damping the resonance.
In practice one doesn't measure or assume the actual value of C,
but instead calculates a worse-case value, and compensates this.
The system is then stable for any input capacitance at or below
this value. Note, C must also include stray capacitance as well
as the opamp's internal capacitance.
For example, take an 8MHz opamp with a 1M feedback resistor. The
input looks like a 20mH inductor. Let's say we'll compensate it
for up to 300pF of input capacitance, allowing for 50pF of stray
and opamp capacitance, plus a 100pF photodiode on a 5-foot cable.
The resonate frequency is 65kHz. That means we want Cf = 2.5pF,
which will create a 8k loss resistance for the 20mH inductance.
A final note, we'll simplify and say the bandpass of this 1M-ohm
transresistance amplifier is about 65kHz.
The view I have presented above is not the customary one used.