Page 82 - Schaum's Outline of Theory and Problems of Electric Circuits

CHAP. 5]
2
p 1k
¼ v C =1000 ¼ 0:001 W ¼ 1000 mW
1k
: AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS 71
2
2k
: p 2k
¼ðv 2 v C Þ =2000 ¼ 0:00242 W ¼ 2420 mW
2
5k
: p 5k
¼ v 1 =5000 ¼ 0:00005 W ¼ 50 mW
2
8k
: p 8k
¼ v 2 =8000 ¼ 0:00128 W ¼ 1280 mW
2
10 k
: p 10 k
¼ v C =10 000 ¼ 0:0001 W ¼ 100 mW
The total power dissipated in the resistors is
p 2 ¼ p 1k
þ p 2k
þ p 5k
þ p 8k
þ p 10 k
¼ 1000 þ 2420 þ 50 þ 1280 þ 100 ¼ 4850 mW

5.5 INVERTING CIRCUIT
In an inverting circuit, the input signal is connected through R 1 to the inverting terminal of the op
amp and the output terminal is connected back through a feedback resistor R 2 to the inverting terminal.
The noninverting terminal of the op amp is grounded (see Fig. 5-13).

Fig. 5-13
To find the gain v 2 =v 1 , apply KCL to the currents arriving at node B:

v 1 v 2 v 2 R 2
þ ¼ 0 and ¼ ð10Þ
R 1 R 2 v 1 R 1
The gain is negative and is determined by the choice of resistors only. The input resistance of the circuit
is R 1 .

5.6 SUMMING CIRCUIT
The weighted sum of several voltages in a circuit can be obtained by using the circuit of Fig. 5-14.
This circuit, called a summing circuit, is an extension of the inverting circuit.
To find the output, apply KCL to the inverting node:
v 1 v 2 v n v o
þ þ þ þ ¼ 0
R 1 R 2 R n R f
from which

R f R f R f
v o ¼ v 1 þ v 2 þ þ v n ð11Þ
R 1 R 2 R n

1
1
1
EXAMPLE 5.10 Let the circuit of Fig. 5-14 have four input lines with R 1 ¼ 1; R 2 ¼ ; R 3 ¼ ; R 4 ¼ , and R f ¼ 1,
2 4 8
all values given in k
. The input lines are set either at 0 or 1 V. Find v o in terms of v 4 , v 3 , v 2 , v 1 , given the
following sets of inputs:
(a) v 4 ¼ 1V v 3 ¼ 0 v 2 ¼ 0 v 1 ¼ 1V