# 1.6 kΩ 200 Ω + 1 kΩ ```Example 8
1.6 kΩ
200 Ω
_
+
+
1 kΩ
12 V
+
_
2V
+
_
8 kΩ
vo
_
Find the voltage v o due to the difference amplifier.
In this problem, we want to determine the output voltage of the ideal operational
amplifier. To make it general, I will label the two input voltages v1 and v 2 . four resistors
as R1 , R1 ' R 2 , and R 2 ' . The voltages of the input terminals as v n and v p ,
The input currents of the ideal op amps should be zero. So all the current through
resistor R 2 flows through R 2 ' . That is i 
v2
. The voltage v p should be i multiplied
R2  R2 '
by R 2 ' , That is
vp 
v2
R2 ' 
R2  R2 '
1
v .
R2 2
1
R2 '
For the two input terminals, they should have the same voltages.
vn  v p 
1
v (1)
R 2
1 2
R2 '
Let us write a KCL equation at the inverting node. The entering current will be assigned
as a negative sign. summing the currents gives us

v n  v1 v o  v n

0.
R1
R1 '
Move the first term to the right side of the equation,
v o  v n v n  v1

R1 '
R1
The equation is multiplied by R1 ' at both sides
vo  v n 
vn  v1
R1 '
R1
vo  vn 
v n  v1
R1 '
R1
Collect the coefficients of v n
R1 '
R'
)v n  1 v1 (2)
R1
R1
v o  (1 
Substitute (1) into (2)
vo  (1 
R1 '
R'
)vn  1 v1
R1
R1
R 





 1 1 
R1 '  1 
R1 '
R1 ' 
R1 ' 
R'
v o  (1 
)
v2 
v1 
v 2  1 v1 (3)



R
R
R1
R1
R1
R1
 1  2 
 1  2 
R2 ' 
R2 ' 


R
R
If the resistance ratio 1  2 , equation (3) becomes
R1 ' R2 '
v o  v1 
R1 '
(v 2  v1 ) .
R1
This op amp circuit is a difference amplifier. If R1  R1 ' , the circuit is a subtractor.
vo 
R1 '
(v 2  v1 )  8  (2  12 )  80V
R1
```