So point A is at virtual ground in this circuit also. To check the results, compute the

voltage at V2 as follows:

V2 = (voltage at point A) - ER2

V2 = (0 V) - ( +10 V)

V2 = -10 V

You can compute the values for view (C) and prove that point A in that circuit is also at

virtual ground.

7-63. The whole point is that the inverting input to the operational amplifier shown in

Figure 7-12 (for all practical purposes) is at virtual ground since it is at 0 volts. Since the

inverting input is at 0 volts, there will be no current (for all practical purposes) flowing into

the operational amplifier from the connection point of R1 and R2.

7-64. Given these conditions, the characteristics of this circuit are determined almost

entirely by the values of R1 and R2. Figure 7-14 should help show how the values of R1

and R2 determine the circuit characteristics.

NOTE: It should be stressed at this point that for purpose of explanation, the

operational amplifier is a theoretically perfect amplifier. In actual practice we

are dealing with less than perfect. In the practical operational amplifier there

will be a slight input current with a resultant power loss. This small signal can

be measured at the theoretical point of virtual ground. This does not indicate

faulty operation.

7-65. As shown in Figure 7-14, the input signal causes current to flow through R1. Only

the positive half cycle of the input signal is shown and will be discussed. Since the voltage

at the inverting input of the operational amplifier is at 0 volts, the input current (IIN) is

computed by using the following formula:

IIN = EIN

R1

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