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From: Fred Bloggs
User-Agent: Mozilla/5.0 (Windows; U; Win 9x 4.90; en-US; rv:1.0.1) Gecko/20020823 Netscape/7.0
X-Accept-Language: en-us, en
Subject: Re: Current Feedback Amp Questions
References: <3D92F99F.firstname.lastname@example.org> <email@example.com> <70Vk9.19089$wH.1079@sccrnsc01>
Date: Fri, 27 Sep 2002 13:24:28 GMT
NNTP-Posting-Date: Fri, 27 Sep 2002 06:24:28 PDT
Organization: EarthLink Inc. -- http://www.EarthLink.net
>>Thanks for the reply. This looks like it won't work for me becasue it
> needs a
>>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.
Something like this:
View in a fixed-width font such as Courier.
R1||R2||R3 | \
| | \
Vin----+ (G-1)R/(10-G) | / -----+--Vout=GVin
+---/\/\/\--+---|- / |
R1 | | / | G<10
| | / |
| (G-1)R |
R2 \ R
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