TC 9-62
Figure 7-25. Difference Amplifier
7-96. With the same inputs that were used for the subtractor (E1 = +3 V; E2 = +12 V)
the output of the difference amplifier should be five times the output of the subtractor
(EOUT = +45 V). Following the same steps used for the subtractor, first compute the value
of R2 plus R4:
R2 + R4 = 1 kΩ + 5 kΩ
R2 + R4 = 6 kΩ
With this value, compute the current through R2 (IR2):
E2
I R2 =
R2 = R4
+ 12V
=
I R2
6kΩ
= + 2 mA
I R2
Next, compute the voltage drop across R2 (ER2):
ER2 = (R2) x (IR2)
ER2 = (1 k Ω) x (+2 mA)
ER2 = +2 V
Then, compute the voltage at point B:
Voltage at point B = E2 - ER2
Voltage at point B = (+12 V) - (+2 V)
Voltage at point B = +10 V
Since point A and point B will be the same potential in an operational amplifier:
Voltage at point A = +10 V
Now compute the voltage developed by the R1 (ER1):
7-34
TC 9-62
23 June 2005
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