PHYS345 Electricity and Electronics

Answer for S14.1

a. Rf/Ri = 100. For example, Ri = 1k and Rf = 100k.

b. Rf/Ri = 99.

c. For the noninverting configuration, the effective input impedance is extremely high and does not load the transducer. There is no change in the gain.

For the inverting configuration, the effective input impedance is the sum of Ri and RTH. If Ri = 10k (and Rf = 1.0M) then the effective gain is 16.7. As Ri decreases, the effective gain approaches 20.

To eliminate loading effects or changes in gain when using an inverting configuration, use a voltage follower between the transducer and the inverting stage.

Answer for P14.1

The ratio of Rf/Ri needs to vary between 1 and 20 for each input. This may be accomplished, for example, by using a feedback resistor of 100k and an input network of a 5k resistor in series with a variable resistance (potentiometer) that can vary from 0-95k.

You may be satisfied with using a 100k trimmer potentiometer which is typically available; full-range variation will yield a variation of c from 20 to 0.95 (only a 5% error on the low side).

The low-end stop of the range may be improved by putting 1% 1.91k resistor is parallel with the potentiometer...

(You may find many tables of "standard resistor values" on the web. Another one!)

Answer for P14.2

Availability of a -5 V supply makes is easy: Use Rf = Ri and tie v2 at -5 V.

To use a different voltage requires a voltage divider. Then we seek a voltage divider of -15 V to ground via R4 and R3 in series such that its Thevenin equivalent is the same as -5 V and Rf above.

If Rf = 10k above, then R4 = 30k and R3 = 15k works.

Differential amplifier with voltage divider

Answer for P14.3

Put a voltage follower between each input and associated transducer.


"http://www.physics.udel.edu/~watson/phys345/protected/exercises/answers/1117.html"
Last updated Dec. 4, 1998.
Copyright George Watson, Univ. of Delaware, 1998.