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References: <3D92F99F.firstname.lastname@example.org> <email@example.com>
Subject: Re: Current Feedback Amp Questions
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
NNTP-Posting-Date: Fri, 27 Sep 2002 08:53:23 GMT
Organization: AT&T Broadband
Date: Fri, 27 Sep 2002 08:53:23 GMT
>Thanks for the reply. This looks like it won't work for me becasue it
>minimum gain of 10 (I need a gain of 2 max.) ...
Don't give up on using an amplifier just because it has a minimum stable
gain of 10.
You can easily use that amplifier in a gain of 2 with the addition of a
single resistor. Place the resistor directly across the +input and
the -input. Lets suppose that your gain of 2 is defined with a feedback
resistor (Rf) of 1K and a gain resistor (Rg) of 1K. The parallel
combination is 500 Ohms. Then your resistor across the inputs (Rn) should
be 100 Ohms. This combination of three resistors satisfies the amp
requirements of gain of 10 while giving an actual gain of 2. The formula to
use to calculate the value of the resistor across the inputs (Rn) is the
Rn = Rf / Minimum gain.
Use Rg to define the actual gain in the following way:
Actual Gain = Rf / Rg.
This is a simple way to work around the minimum gain requirement. Perhaps
somebody can make one of those neat text schematics to illustrate this fact.
The disadvantage of using this technique is that the amplifier noise is
amplified by the minimum gain requirement, in this case 10, no matter what
the actual gain of the circuit is. But if amplifier noise is not critical
for you then go ahead and use it.
Hope this helps.
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