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Reply-To: "Kevin Aylward"
From: "Kevin Aylward"
References: <email@example.com> <firstname.lastname@example.org>
Subject: Re: Op-amp virtual earth not 0V
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Tue, 17 Sep 2002 11:16:01 +0100
NNTP-Posting-Date: Tue, 17 Sep 2002 11:16:04 BST
wrote in message news:email@example.com...
> "Nghia" a écrit dans le message news:
> > Hi all,
> > I've just done an experiment with an unknown op-amp where we put in
> > triangle wave (1 volt) at 100Hz into an inverting voltage gain
> > gain of -10. The op-amp saturates at +-15V.
> > Now when observing the virtual earth point I noticed that its not
> > but a square wave of very low amplitude, also at 100Hz. Can't
> > how much was zoomed in so can't remember the exact value.
> > I've searched everywhere on the web and haven't found anything that
> > could help me explained this. Why isn't it zero volts?
> Because opamps are not perfects.
> At frequencies above a few Hz (or 10s Hz, more for some high speed
> opamps behave like integrators (gain rolling off 6db/octave).
> Hence the differential input voltage (the one you observe at the
> ground node in an inverter scheme) has to be the derivative of the
> voltage divided by the opamp gain.
This might be a bit misleading. It depends on what you mean by its gain.
If the gain is taken to be what it really is, i.e. A(s), then dividing
by the gain automatically accounts for the derivative bit.
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