0195136047.pdf

5.4 APPLICATIONS OF OPERATIONAL AMPLIFIERS 249

io

iS

= 1 +

Rf

R 1

(5.4.28)

Charge-to-Charge Amplifier

A circuit is shown in Figure 5.4.7 in which there is a capacitorC 1 in the−input line and a
capacitorCfin the feedback loop. KCL at nodeXgives

dq 1

dt

+

dqf

dt

= 0 (5.4.29)

whereq 1 andqfare charges on the input and feedback capacitors. Thus,

q 1 =−qf or C 1 vi=−Cfvo or

vo

vi

=−

C 1

Cf

(5.4.30)

Negative Impedance Converter

The op-amp circuit of Figure 5.4.8 causes a negative resistanceRinbetween the input terminal
and ground. In the more general case, whenRis replaced by an impedanceZ,the circuit gives a
negative impedance. Using ideal op-amp techniques, one has

v 1 =vin; i 1 =

v 1

R 1

=

vin

R 1

=i 2

vo=i 2 (R 1 +R 2 )=vin

(

1 +

R 2

R 1

)

; i 3 =

vo−vin

R

=vin

R 2

RR 1

=−iin

so that

Rin=

vin

iin

=−R

R 1

R 2

(5.4.31)

−

+

X vo

Cf

C 1

vi

Figure 5.4.7Charge-to-charge amplifier.

−

+

vo

i 2

iin i^3

i 1

v 1

Rin vin

2 3

1

R 1 R 2

+ R

−

Figure 5.4.8Negative impedance converter.