0195136047.pdf

5.4 APPLICATIONS OF OPERATIONAL AMPLIFIERS 245

vo=Ao(v 2 −v 1 ) (5.4.1)

Sincev 2 =0, because terminal 2 is grounded,

v 1 =−

vo

Ao

(5.4.2)

Thus one gets

i 1 =

vi−v 1

R 1

=

vi+(vo/Ao)

R 1

(5.4.3)

i 2 =

v 1 −vo

R 2

=

−(vo/Ao)−vo

R 2

(5.4.4)

Since no current is drawn by terminal 1 of the ideal op amp,

i 2 =i 1 (5.4.5)

Hence,

vo

vi

=−

R 2

R 1

1

1 +[(R 1 +R 2 )/AoR 1 ]

(5.4.6)

which is the expression for the circuit gain. Now, ifAo→∞, we have
vo
vi

=−

R 2

R 1

, ifAo→∞ (5.4.7)

which is independent of the op amp’s gainAo(see Example 5.2.1). The sign inversion associated
with Equation (5.4.7) makes the amplifier “inverting.” Generally ifAois at least 200 times the
magnitude ofR 2 /R 1 , then the actual gain given by Equation (5.4.6) will be within 1% of the ideal
gain given by Equation (5.4.7). Assuming infinite gain of the op amp, it follows that

vd=v 2 −v 1 =0orv 1 =v 2 (5.4.8)

Since terminal 2 is at ground potential (zero) in Figure 5.4.1, terminal 1 is said to be avirtual
ground. Notice that whenR 2 =R 1 , the inverting amplifier becomes avoltage followerwith inverted
sign and a gain magnitude of unity.

Noninverting Amplifier

It is called “noninverting” because there is no sign inversion. A typical circuit is shown in Figure
5.4.2. With a finite op-amp gainAo,

v 2 −v 1 =

vo

Ao

or v 1 =v 2 −

vo

Ao

(5.4.9)

−

+

vo

vi

i 2

i 1

v 1

v 2 = vi

3

2

1

R 1

R 2

+

−

Figure 5.4.2Noninverting amplifier.