Page 106 - Schaum's Outline of Theory and Problems of Electric Circuits
P. 106

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
               CHAP. 5]
                     The circuit is a difference amplifier.                                             95


               5.23  Find v o in the circuit of Fig. 5-46.

















                                                        Fig. 5-46

                         Apply KCL at node B.  Note that v B ¼ v A ¼ v 2 .  Thus,
                                                      v 2   v 1  v 2   v o
                                                            þ       ¼ 0
                                                        R 1     R 2
                     Solving for v o , we get v o ¼ v 2 þðR 2 =R 1 Þðv 2   v 1 Þ.



               5.24  Find v o in the circuit of Fig. 5-47.
















                                                        Fig. 5-47

                         The left part of the circuit has a gain of ð1 þ R 1 =R 2 Þ.  Therefore, v 3 ¼ð1 þ R 1 =R 2 Þv 1 .  Using results of
                     Problem 5.23 and substituting for v 3 results in

                                         R 2             R 2    R 2    R 1        R 2
                                 v o ¼ v 2 þ  ðv 2   v 3 Þ¼ 1 þ  v 2    1 þ  v 1 ¼ 1 þ  ðv 2   v 1 Þ
                                         R 1             R 1    R 1    R 2        R 1


                                                                                    6
                                                                      6
               5.25  In Fig. 5-48 choose resistors for a differential gain of 10 so that v ¼ 10 ðv   v Þ.
                                                                               o
                                                                                           1
                                                                                       2
                         The two frontal op amps are voltage followers.
                                                             and
                                                    v A ¼ v 1       v B ¼ v 2
                     From (16), Sec. 5.9, we have
                                                      R 2         R 2
                                                  v o ¼  ðv B   v A Þ¼  ðv 2   v 1 Þ
                                                      R 1         R 1
                                                                   6
                         To obtain the required differential gain of R 2 =R 1 ¼ 10 , choose R 1 ¼ 100 
 and R 2 ¼ 100 M
.
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