Example 14
500 Ω
+
4.5 kΩ
_
500 Ω
3V
+
_
1.2 kΩ
6 kΩ
20 kΩ
i
5 kΩ
Determine the current i in the op amp circuit shown in the figure above.
For the op amp circuit, we are interested in the current through the 500Ω resistor. Our
target current i is associated with the output voltage of the op amp. So we need to find the
output voltage of the op amp first. Let us label it as v o . To find the output voltage, it is
very often that we try to associate the output voltage with the voltages at the input side.
The input voltages are label as v p and v n here.
The circuit can be simplified before we analyze it further. The 1.2kΩ resistor is in parallel
with the 6kΩ resistor so that they can be combined to one equivalent resistor. The
equivalent resistance for two parallel resistors is the product of the two individual
resistances divided by the sum of the two resistances. 1.2k times 6k divided by 7.2k gives
us 1k. Here is the simplified circuit.
Req 
1.2  6
 1k
1.2  6
For the ideal op amp, it doesn’t draw any current. So the same current flows through the
500Ω and 1kΩ resistors.
i1 
3
 2mA
0.5  1
The voltage across 1.2kΩ resistor should be the current times the resistance,
v p  2  1  2V .
The ideal op amp has the same voltages at both input terminals. So
v n  v p  2V .
The current through the 5kΩ resistor should be
i2 
vn  0 2

 0.4mA
5k
5k
Again there is no current that flows into the op amp. So the current through the 20kΩ
resistor is the same as that of the 5kΩ resistor. The output voltage of the op amp should
be the current multiplied by the total residence.
v o  i 2  (5k  20 k )  0.4  25  10V
The current i is the voltage divided by the resistance i 
vo
10

 20 mA .
0.5k 0.5k