Yes.
If you are using an inverting configuration, you would need three
resistances. For example, choose a 5k resistor for the input
resistor. The feedback resistance would consist of a fixed 500 ohm
resistor in series with a 50k pot wired as a rheostat.

This would give a gain variable from 1/10x to 10.1x

If you absolutely want a gain range from 1/10x to 10x, then you
have to solve the set of simultaneous equations:

.1=Rf/Rin   and
10=(Rf+50k)/Rin

Rearranging the first equation we get:

Rf=.1*Rin

Substituting into the second equation we get:

10=(.1*Rin+50,000)/Rin

1=(.1*Rin+50,000)/(10*Rin)
1=(.01*Rin+5,000)/Rin
Rin=.01*Rin+5,000
Rin-.01*Rin=5,000
.99*Rin=5,000
Rin=5,000/.99
Rin=5050.5050

So, Rin would be 5,050.505 ohms

Rf=.1*Rin   Rf=.1*5,050.505  which is 505.05 ohms

When the 50k pot is added, the gain becomes:

gain=(Rf+50,000)/R1   which is  gain=(5,050.5+50,000)/5,050.505,
 which is 10.000

So, the gain is now variable from 1/10x to 10x

Have fun trying to get resistances (and the pot!) this accurate.

If you use 5.1k and 51k 1% resistors you would get a gain range from
.1x to 9.9x, which is within 1% of what you want.

Fr. Thomas McGahee






----- Original Message -----
From: "Richard Sloan" <rsloan@THEMINDFACTORY.COM>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Thursday, March 28, 2002 3:04 PM
Subject: [EE]: opamp info


Is it possible to make an opamp design with one variable 50K resistor and
have its gain variable from 1/10x to 10x?

Where on the web is good opamp info?

Richard.

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