From: "Bill Sloman"
Subject: Re: Voltage error caused by input bias current
Date: Wed, 1 Jan 2003 15:35:18 +0100
Organization: Planet Internet
NNTP-Posting-Date: 1 Jan 2003 14:35:51 GMT
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
"Stefan Salewski" wrote in message
> 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.
> 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.
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.
Bill Sloman, Nijmegen