5.4 APPLICATIONS OF OPERATIONAL AMPLIFIERS 251
which corresponds to the response of adifferential amplifier.Usually resistor values are so chosen
thatR 3 =R 1 andR 4 =R 2 for some practical reasons. Improved versions of differential amplifiers
are available commercially.
Integrators
Figure 5.4.10 shows anoninverting integrator,which can be seen to be a negative impedance
converter added with a resistor and a capacitor. Noting thatvo= 2 v 1 andi 3 =v 1 /R, the total
capacitor current is
i=iin+i 3 =
vin−v 1
R
+
v 1
R
=
vin
R
(5.4.36)
The capacitor voltage is given by
v 1 (t)=
1
C
∫t
−∞
i(ξ)dξ=
1
RC
∫t
−∞
vin(ξ)dξ (5.4.37)
Thus,
vo(t)=
2
RC
∫t
−∞
vin(ξ)dξ (5.4.38)
which shows that the circuit functions as an integrator.
ReplacingR 2 in the inverting amplifier of Figure 5.4.1 by a capacitanceCresults in the
somewhat simpler integrator circuit shown in Figure 5.4.11, known as an inverting integrator, or
Miller integrator. With ideal op-amp techniques,iC=iin=vin/R. The voltage acrossCis just
vo, so that
−
+
vo
R (^1) v
1
v 1
iin i 3
i
C
R 1
R
Negative impedance converter
R
vin
Figure 5.4.10Noninverting integrator.
−
- vo
1
2 3
C
iC
iin
R
vin
Figure 5.4.11Inverting integrator (Miller integrator).