Page 84 - Schaum's Outline of Theory and Problems of Electric Circuits
P. 84

AMPLIFIERS AND OPERATIONAL AMPLIFIER CIRCUITS
                                                                                                      73
               CHAP. 5]
               The gain v 2 =v 1 is positive and greater than or equal to one. The input resistance of the circuit is infinite
               as the op amp draws no current.

               EXAMPLE 5.11   Find v 2 =v 1 in the circuit shown in Fig. 5-16.


















                                                        Fig. 5-16

                   First find v A by dividing v 1 between the 10-k
 and 5-k
 resistors.
                                                         5       1
                                                   v A ¼     v 1 ¼  v 1
                                                       5 þ 10    3
               From (12) we get

                                           7     9     9 1                    v 2
                                   v 2 ¼ 1 þ  v A ¼  v A ¼  v 1  ¼ 1:5v 1  and  ¼ 1:5
                                           2     2     2 3                    v 1

               Another Method
                   Find v B by dividing v 2 between the 2-k
 and 7-k
 resistors and set v B ¼ v A .
                                              2      2     1            v 2
                                         v B ¼   v 2 ¼  v 2 ¼  v 1  and   ¼ 1:5
                                             2 þ 7   9     3            v 1


               EXAMPLE 5.12   Determine v o in Fig. 5-17 in terms of v 1 ; v 2 ; v 3 ; and the circuit elements.













                                                        Fig. 5-17

                   First, v A is found by applying KCL at node A.

                                   v 1   v A  v 2   v A  v 3   v A       1
                                         þ       þ       ¼ 0   or   v A ¼ ðv 1 þ v 2 þ v 3 Þ         ð13Þ
                                     R       R       R                   3
               From (12) and (13) we get

                                                  R 2     1    R 2
                                          v o ¼ 1 þ   v A ¼  1 þ   ðv 1 þ v 2 þ v 3 Þ                ð14Þ
                                                  R 1     3    R 1
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