From: "Gary Richardson"
Subject: Re: Voltage error caused by input bias current
Date: Thu, 2 Jan 2003 14:17:24 -0800
Organization: Posted via Supernews, http://www.supernews.com
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"Winfield Hill" wrote in message
> Bill Sloman wrote...
> > Stefan Salewski Salewski@PHYSnet.Uni-Hamburg.de wrote
> >> I try to understand the effect of input bias current
> >> to the output voltage of an inverting OpAmp amplifier.
> >> I have a simple noninverting OpAmp amplifier as shown
> >> on page 193 (Figure 4.30) in "The Art Of Electronics":
> >> Noninverting input is connected to ground, signal is
> >> connected with R1 to inverting input, and inverting
> >> input is connected to output with R2, so gain is -R2/R1.
> >> On page 194 Mr. Horowitz and Mr. Hill writes:
> >> In this circuit the inverting input sees a driving
> >> impedance of R1 || R2, so the bias current produces a
> >> voltage Vin = IB (R1 || R2), which is then amplified by
> >> the gain at dc, -R2/R1.
> > You've used the wrong gain - the "noise gain" of an inverting
> > op amp is (R2+R1)/R1 - the gain of the circuit seen as a
> > non-inverting amplifier, and this should be applied to all
> > voltage changes appearing at the op-amp inputs, as opposed
> > the free end of R1.
> Thanks Bill, looking at our sentence (written many years ago),
> I see it's not only misleading, but wrong. Properly applying
> the non-inverting gain yields Vo = Ib R2, the correct answer.
> >> I am not absolutely sure if I have understood this correct.
> >> In my simple mind the inverting input of the OpAmp should
> >> be at ground potential too because of the large differential-
> >> mode voltage gain of the OpAmp. So nearly all bias current
> >> have to flow through R2, and the voltage offset at the output
> >> should be just Vout=R2*IB. Indeed I have found a less popular
> >> german book with this expression.
> That's correct, and an easier way to view the issue.
> > This is an easier way of getting the right answer, but Win and
> > Paul Horowitz have other pedagogic points to make, and they need
> > the concept of "noise gain" for predicting other non-idealities.
> The use of noise gain wasn't a popular approach 20 years ago (it
> appears only once in our book) but it's become widely used in the
> last 10 years, especially for analyzing transimpedance amplifiers,
> and we'll introduce it earlier in the 3rd edition.
> - Win
Is there an errata for A of E on the 'net somewhere?