Applications Using
Operational Amplifiers
• Basic op-amp applications with negative feedback: inverting amplifiers,
noninverting amplifiers, differential amplifiers, summing amplifiers, etc.
• Others linear and non-linear applications:
 Integrators and differentiators
 Current sources with high output resistance
 Half-wave and full-wave precision rectifiers
 Precision peak detectors
 Logarithmic and exponential amplifiers
 Circuits for multiplication and division
1/9
Integrator
Time domain analysis
it  
vI t 
R
t
1
vO t   vC t     it dt  vC (0) 
C0
1 vI t 
 
dt  vC (0)
C0 R
t
t
1
vO t   
vI (t )dt  vC (0)

RC 0
RC – time constant,
integrating constant
Problem:
• There is no NF in dc because the equivalent impedance of the
capacitor is infinite,
• The op-amp can become saturated due to the dc offset voltage
and biasing currents
2/9
Integrator
with NF in dc
Frequency domain analyses
vO  j 
Z ech
Av  j  

v I  j 
R
Z ech
R1
1
 R1 ||

jC 1  jR1C
R1
1
Av  j   
R 1  jR1C
R1 large enough to be
neglected in
comparison with
equivalent impedance
of the capacitor at the
working frequency
Active lowpass filter
Example:
R=1KΩ;
R1=100KΩ
C=100pF
3/9
Differentiator
Thye circuit acts as a “noise
amplifier” because of the derivation
of the input signal.
In practical approaches a small
resistor has to be connected in series
with the capacitor.
dvI t 
it   C
dt
dv I t 
vO t    Ri   RC
d t 
vO  j  R
Av  j  

 jRC
v I  j  Z C
Av  j   RC
Active high-pass filter
f0=∞
1
f0 
2R1C
4/9
Precision (Active) Rectifiers
Half-Wave Rectifier
 Can not rectify small signals
 Some voltage (0.6V) is
wasted across the conducting
diode
Precision rectifier:
For the rectified half-wave we need: vO = vI
Superdiode – (almost) zero voltage drop across the
conducting superdiode
Op amp + NF + D
5/9
vO - cannot
became
negative
iD  0
 half-wave rectifier for positive half-wave
 rectification of the negative half-wave?
6/9
Shortcoming:
• vI <0, simple comparator
vO,oa=VOL - saturation
slows down the operation speed
• limits the operation frequency
Solution: avoid the saturation
VTC
7/9
Full-Wave Rectification
• The principle
vI >0, D1-(on), D2–(off)
vO=vI
vI <0, D1-(on), D2–(off)
vO=-vI
• precision rectifier
How does the
circuit look like ?
8/9
Precision Positive Peak Detector
Precision Positive Peak Detector that Holds the Voltage
D2 role?
R role ?
9/9