Reply-To: "Kevin Aylward"
From: "Kevin Aylward"
Subject: Re: Amplifying small DC signal with large DC bias...
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Tue, 12 Nov 2002 07:47:17 -0000
NNTP-Posting-Date: Tue, 12 Nov 2002 07:47:20 GMT
"Mike Deblis" wrote in message
> "Kevin Aylward" wrote in message
> > Off the bat, I would guess that your system is not set up optimally.
> > Where does the 4V come from?. If it were stable you would only have
> > subtract this from the wanted signal before amplification. However,
> > bet its not. If not, you need to extract its value in some way.
> Its a decaying signal about 500uS long - I need to take a window along
> the curve a few uS wide, and compare the average signal in that window
> with a normalised (average). The curve is an exponential decay and
> I'll be measuring it on the flatish portion when there may still be
> several volts on average in the signal.
I actually meant what is the *system*, not what the signal is. If we
know the system, there might be a better way to do things. What is the
actual problem you are trying to solve.
> The difference between the
> signal I measure and the average may be as little as 1uV, so I want to
> remove the DC bias so I can prevent the input op-amp clipping
Oh....So you think you want to extract 1uv dc out of a DC of 4V, in
500us. Ahmmm... Off the cuff, you aint going to do it. That is,
theoretically, it may be possible, but in engineering reality, its not
goanna happen. In some systems, e.g. strain gauges, one uses a bridge so
that DC terms can be cancelled. That is, two matched sensors are used,
one with the signal, one without. Achieving 10uv resolution when you are
integrating over at least one power line cycle integration (20ms) for
noise rejection is still usually quite tricky in such cases.
Now..., you say that the signal is a decaying exponential, this is in
conflict of your original spec of saying it was DC. You need to describe
the *problem*, that way it might be determined if you can say, extract
the DC from the signal itself, because the real signal is really AC.
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