EE101: Op Amp circuits (Part 4)
M. B. Patil
mbpatil@ee.iitb.ac.in
www.ee.iitb.ac.in/~sequel
Department of Electrical Engineering
Indian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Half-wave rectifier
Consider a diode rectifier:
Vo
Vi
Vo
D
R
slope=1
VD
Vi
M. B. Patil, IIT Bombay
Half-wave rectifier
Consider a diode rectifier:
Vo
Vi
Vo
D
R
slope=1
VD
Vi
If Vi VD , the diode drop can be ignored.
However, if Vi is small, e.g., Vi = 0.2 sin ωt V , then the circuit does not rectify, and
Vo (t) = 0 V .
M. B. Patil, IIT Bombay
Half-wave rectifier
Consider a diode rectifier:
Vo
Vi
Vo
D
R
slope=1
VD
Vi
If Vi VD , the diode drop can be ignored.
However, if Vi is small, e.g., Vi = 0.2 sin ωt V , then the circuit does not rectify, and
Vo (t) = 0 V .
Precision rectifier circuits overcome this drawback.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
Vo
Vi
D
R
Half-wave precision rectifier
i−
Vo
Vi
D
R
Vi
iD
Vo1
Vo
VD
R
iR
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (except
for the diode drop) the buffer we have seen earlier.
Half-wave precision rectifier
i−
Vo
Vi
D
R
Vi
iD
Vo1
Vo
VD
R
iR
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (except
for the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =
Vo1
Vo + 0.7 V
=
≈ 0 V → Vo = Vi .
AV
AV
Half-wave precision rectifier
i−
Vo
Vi
D
R
Vi
iD
Vo1
Vo
VD
R
iR
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (except
for the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =
Vo1
Vo + 0.7 V
=
≈ 0 V → Vo = Vi .
AV
AV
This situation arises only if iD > 0 (since the diode can only conduct in the
forward direction), i.e., Vo > 0 → Vi = Vo > 0 V .
Half-wave precision rectifier
Vo
i−
Vo
Vi
D
R
Vi
iD
Vo1
Vo
VD
R
iR
slope=1
Vi
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (except
for the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =
Vo1
Vo + 0.7 V
=
≈ 0 V → Vo = Vi .
AV
AV
This situation arises only if iD > 0 (since the diode can only conduct in the
forward direction), i.e., Vo > 0 → Vi = Vo > 0 V .
M. B. Patil, IIT Bombay
Half-wave precision rectifier
Vo
Vi
D
R
Half-wave precision rectifier
Vo
Vi
D
R
Vi
Vo
Vo1
R
(ii) D is not conducting → Vo = 0 V .
Half-wave precision rectifier
Vo
Vi
D
R
Vi
Vo
Vo1
R
(ii) D is not conducting → Vo = 0 V .
What about Vo1 ?
Since the Op Amp is now in the open-loop configuration, a very small Vi is
enough to drive it to saturation.
Half-wave precision rectifier
Vo
Vi
D
R
Vi
Vo
Vo1
R
(ii) D is not conducting → Vo = 0 V .
What about Vo1 ?
Since the Op Amp is now in the open-loop configuration, a very small Vi is
enough to drive it to saturation.
Note that Case (ii) occurs when Vi < 0 V . Since V+ − V− = Vi − 0 = Vi is
negative, Vo1 is driven to −V sat.
Half-wave precision rectifier
Vo
Vo
Vi
D
R
Vi
Vo
Vo1
R
Vo = 0
Vi
(ii) D is not conducting → Vo = 0 V .
What about Vo1 ?
Since the Op Amp is now in the open-loop configuration, a very small Vi is
enough to drive it to saturation.
Note that Case (ii) occurs when Vi < 0 V . Since V+ − V− = Vi − 0 = Vi is
negative, Vo1 is driven to −V sat.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
2
0
Vo
D off
D
Vo1
Vi
Vi
−4
Vo = Vi
Vo
R
superdiode
Vo
−2
D on
−6
Vo = 0
−8
Vi
−10
−12
0
* The circuit is called a “superdiode” (i.e., a diode with zero Von ).
1
t (ms)
2
Half-wave precision rectifier
2
0
Vo
D off
D
Vo1
Vi
Vi
−4
Vo = Vi
Vo
R
superdiode
Vo
−2
D on
−6
Vo = 0
−8
Vi
−10
−12
Vo1
−Vsat
0
* The circuit is called a “superdiode” (i.e., a diode with zero Von ).
1
t (ms)
2
Half-wave precision rectifier
2
0
Vo
D off
D
Vo1
Vi
Vi
−4
Vo = Vi
Vo
R
superdiode
Vo
−2
D on
−6
Vo = 0
−8
Vi
−10
−12
Vo1
−Vsat
0
1
t (ms)
* The circuit is called a “superdiode” (i.e., a diode with zero Von ).
* Note that the Op Amp needs to come out of saturation when Vi changes from
negative to positive values. This is a relatively slow process, and it limits the
speed of this circuit.
2
Half-wave precision rectifier
2
0
Vo
D off
D
Vo1
Vi
Vo
−2
D on
Vi
−4
Vo = Vi
Vo
R
−6
Vo = 0
superdiode
−8
Vi
−10
−12
Vo1
−Vsat
0
1
t (ms)
2
* The circuit is called a “superdiode” (i.e., a diode with zero Von ).
* Note that the Op Amp needs to come out of saturation when Vi changes from
negative to positive values. This is a relatively slow process, and it limits the
speed of this circuit.
SEQUEL file: precision half wave 1.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
R2
iD1
iR1
Vi
D1
R1
iD2
Vo
Vo1
D2
R
iR
Improved half-wave precision rectifier
iR2
R2
iD1
iR1
Vi
D1
R1
iD2
Vo
Vo1
D2
R
iR
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
Improved half-wave precision rectifier
iR2
R2
iD1
iR1
Vi
D1
R1
iD2
Vo
Vo1
D2
R
iR
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo ).
→ iR2 = 0, Vo = V− = 0 V .
Improved half-wave precision rectifier
R2
iR2
R2
iD1
iR1
Vi
D1
R1
D1
iD2
Vo
Vo1
D2
R
Vi
R1
iR
Vo
Vo1
D2
R
Vi > 0
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo ).
→ iR2 = 0, Vo = V− = 0 V .
Improved half-wave precision rectifier
R2
iR2
R2
iD1
iR1
Vi
D1
R1
D1
iD2
Vo
Vo1
D2
R
Vi
R1
iR
Vo
Vo1
D2
R
Vi > 0
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo ).
→ iR2 = 0, Vo = V− = 0 V .
iR1 = iD1 which can only be positive ⇒ Vi > 0 V .
Improved half-wave precision rectifier
R2
iR2
R2
iD1
iR1
Vi
D1
R1
D1
iD2
Vo
Vo1
D2
R
Vi
R1
iR
Vo
Vo1
D2
R
Vi > 0
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo ).
→ iR2 = 0, Vo = V− = 0 V .
iR1 = iD1 which can only be positive ⇒ Vi > 0 V .
(ii) D1 is off; this will happen when Vi < 0 V .
Improved half-wave precision rectifier
R2
iR2
iR1
Vi
D1
R1
R2
R2
iD1
D1
iD2
Vo
Vo1
D2
R
Vi
R1
iR
D1
Vo
Vo1
D2
Vi
R1
R
Vi > 0
Vo
Vo1
D2
R
Vi < 0
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo ).
→ iR2 = 0, Vo = V− = 0 V .
iR1 = iD1 which can only be positive ⇒ Vi > 0 V .
(ii) D1 is off; this will happen when Vi < 0 V .
In this case, D2 conducts and closes the feedback loop through R2 .
Improved half-wave precision rectifier
R2
iR2
iR1
Vi
D1
R1
R2
R2
iD1
D1
iD2
Vo
Vo1
D2
R
Vi
R1
iR
D1
Vo
Vo1
D2
Vi
R1
R
Vi > 0
Vo
Vo1
D2
R
Vi < 0
(i) D1 conducts: V− = V+ = 0 V , Vo1 = −VD1 ≈ −0.7 V .
D2 cannot conduct (show that, if it did, KCL is not satisfied at Vo ).
→ iR2 = 0, Vo = V− = 0 V .
iR1 = iD1 which can only be positive ⇒ Vi > 0 V .
(ii) D1 is off; this will happen when Vi < 0 V .
In this case, D2 conducts and closes the feedback loop through R2 .
„
«
0 − Vi
R2
Vo = V− + iR2 R2 = 0 +
R2 = − Vi .
R1
R1
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
R2
Vo
1k
1k
Vi
R1
R2
− Vi
R1
D1
Vo1
Vo
D2
Vo = 0
R
Vi
Improved half-wave precision rectifier
2
R2
Vo
1k
1k
Vi
R1
−
D1
Vo1
R2
Vi
R1
1
Vi
Vo
D2
Vo = 0
R
Vo
0
Vi
−1
0
1
t (ms)
2
Improved half-wave precision rectifier
2
R2
1k
1k
Vi
R1
−
D1
Vo1
Vo1
Vo
R2
Vi
R1
1
Vi
Vo
D2
Vo = 0
R
Vo
0
Vi
−1
0
1
t (ms)
2
Improved half-wave precision rectifier
2
R2
1k
1k
Vi
R1
−
D1
Vo1
Vo1
Vo
R2
Vi
R1
1
Vi
Vo
D2
Vo = 0
R
Vo
0
Vi
−1
0
1
t (ms)
* Note that the Op Amp does not enter saturation since a feedback path is
available for Vi > 0 V and Vi < 0 V .
2
Improved half-wave precision rectifier
2
R2
1k
1k
Vi
R1
−
D1
Vo1
Vo1
Vo
R2
Vi
R1
1
Vi
Vo
D2
Vo = 0
R
Vo
0
Vi
−1
0
1
t (ms)
2
* Note that the Op Amp does not enter saturation since a feedback path is
available for Vi > 0 V and Vi < 0 V .
SEQUEL file: precision half wave.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
R2
Vo
Vo = 0
D1
Vi
R1
Vo1
Vi
Vo
D2
R
−
R2
Vi
R1
The diodes are now reversed.
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
R2
Vo
Vo = 0
D1
Vi
R1
Vo1
Vi
Vo
D2
R
−
R2
Vi
R1
The diodes are now reversed.
By considering two cases: (i) D1 on, (ii) D1 off, the Vo versus Vi relationship shown in
the figure is obtained (show this).
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
R2
Vo
Vo = 0
D1
Vi
R1
Vo1
Vi
Vo
D2
R
−
R2
Vi
R1
The diodes are now reversed.
By considering two cases: (i) D1 on, (ii) D1 off, the Vo versus Vi relationship shown in
the figure is obtained (show this).
SEQUEL file: precision half wave 2.sqproj
M. B. Patil, IIT Bombay
AM demodulation using a peak detector
0.15
Super
diode
AM
signal
Vo1
filter
Vi
Vo1
Vo2
0
−0.15
0
1
t (ms)
2
M. B. Patil, IIT Bombay
AM demodulation using a peak detector
0.15
Super
diode
AM
signal
Vo1
filter
Vi
Vo1
Vo2
0
−0.15
0
1
t (ms)
2
* charging through superdiode, discharging through resistor
M. B. Patil, IIT Bombay
AM demodulation using a peak detector
0.15
Super
diode
AM
signal
Vo1
filter
Vi
Vo1
Vo2
0
−0.15
0
1
t (ms)
2
* charging through superdiode, discharging through resistor
* The time constant (RC ) needs to be carefully selected.
M. B. Patil, IIT Bombay
AM demodulation using a peak detector
0.15
Super
diode
AM
signal
Vo1
filter
Vi
Vo1
Vo2
0
−0.15
0
1
t (ms)
2
* charging through superdiode, discharging through resistor
* The time constant (RC ) needs to be carefully selected.
SEQUEL file: super diode.sqproj
M. B. Patil, IIT Bombay
Full-wave precision rectifier
VB
Vi
x (−1)
Vi
Half−wave
rectifier
(inverting)
Vo1
VB
x (−2)
Vo1
Vo
VA
Vo
VA
Vi
Vi
Vi
Full-wave precision rectifier
VB
Vi
x (−1)
Vi
Half−wave
rectifier
(inverting)
Vo1
VB
Vo
VA
x (−2)
Vo1
Vo
VA
Vi
Vi
Vi
R
R1
R
D1
Vi
Vo
R/2
R1
D2
Vo1
inverting half−wave rectifier
inverting summer
(SEQUEL file: precision_full_wave.sqproj)
Full-wave precision rectifier
VB
2
Vi
1
x (−1)
Vi
Half−wave
rectifier
(inverting)
Vo1
VB
Vo
VA
x (−2)
Vo1
0
Vo
VA
Vi
Vi
−1
Vi
Vo
Vi
−2
0
R
R1
1
t (ms)
R
D1
Vi
Vo
R/2
R1
D2
2
Vo1
inverting half−wave rectifier
inverting summer
(SEQUEL file: precision_full_wave.sqproj)
M. B. Patil, IIT Bombay
Wave shaping with diodes
R0
i
0V
V
A
D
R′
R
−V0
Wave shaping with diodes
i
0V
R0
i
0V
R′
V
A
R′
V
A
D
R0
R
−V0
D off
R
−V0
Wave shaping with diodes
i
0V
R0
i
0V
R′
D off
R
−V0
When D is off, VA is (by superposition),
R0
R
− V0
.
R + R0
R + R0
VA = V
V
A
R′
V
A
D
R0
R
−V0
Wave shaping with diodes
i
0V
R0
i
0V
R′
V
A
R′
V
A
D
R0
D off
R
−V0
R
−V0
When D is off, VA is (by superposition),
R0
R
− V0
.
R + R0
R + R0
VA = V
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =
R
(V0 + Von ) + Von .
R0
Wave shaping with diodes
R0
i
0V
R0
i
0V
R′
V
A
D
R′
V
A
D off
R
−V0
−V0
R0
i
0V
V
A
Von
D on
R
R
R′
−V0
When D is off, VA is (by superposition),
R0
R
− V0
.
R + R0
R + R0
VA = V
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =
R
(V0 + Von ) + Von .
R0
Wave shaping with diodes
R0
i
0V
R0
i
0V
R′
V
A
D
R′
V
A
D off
R
−V0
−V0
R0
i
0V
V
A
Von
D on
R
R
R′
−V0
When D is off, VA is (by superposition),
R0
R
− V0
.
R + R0
R + R0
VA = V
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =
When D is on,
V
V − Von
−V0 − Von
i =
+
+
R0
R
R0
»
–
1
1
=V
+
+ (constant)
R0
R
R
(V0 + Von ) + Von .
R0
Wave shaping with diodes
R0
i
0V
R0
i
0V
R′
V
A
D
R′
V
A
D off
R
−V0
−V0
R0
i
0V
V
A
Von
D on
R
R
R′
−V0
When D is off, VA is (by superposition),
R0
R
− V0
.
R + R0
R + R0
VA = V
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =
When D is on,
V
V − Von
−V0 − Von
i =
+
+
R0
R
R0
»
–
1
1
=V
+
+ (constant)
R0
R
i.e., V = (R0 k R) i + (constant) .
R
(V0 + Von ) + Von .
R0
Wave shaping with diodes
R0
i
0V
R0
i
0V
R′
V
A
D
R′
V
A
D off
R
−V0
slope = R0
−V0
Vbreak
R0
i
0V
slope = R0 k R
V
A
Von
D on
V
R
R
R′
i
−V0
When D is off, VA is (by superposition),
R0
R
− V0
.
R + R0
R + R0
VA = V
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =
When D is on,
V
V − Von
−V0 − Von
i =
+
+
R0
R
R0
»
–
1
1
=V
+
+ (constant)
R0
R
R
(V0 + Von ) + Von .
R0
i.e., V = (R0 k R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
i
0V
R0
V
Wave shaping with diodes
V
i
0V
R0
V
i
Wave shaping with diodes
V
R0
i
D1B
R1B
R1B′
0V
−V0
V
i
Wave shaping with diodes
V
R0
i
D1B
R1B
R1B′
0V
−V0
V
i
Wave shaping with diodes
V
R0
i
R1B
D2B
R2B
R2B′
D1B
R1B′
0V
−V0
V
i
Wave shaping with diodes
V
R0
i
R1B
D2B
R2B
R2B′
D1B
R1B′
0V
−V0
V
i
Wave shaping with diodes
V0
R1A′
V
R1A
D1A
R0
i
R1B
D2B
R2B
R2B′
D1B
R1B′
0V
−V0
V
i
Wave shaping with diodes
V0
R1A′
V
R1A
D1A
R0
i
R1B
D2B
R2B
R2B′
D1B
R1B′
0V
−V0
V
i
Wave shaping with diodes
R2A
D2A
R1A
D1A
R0
i
R1B
D2B
R2B
R2B′
D1B
R1B′
0V
V
R2A′
R1A′
V0
−V0
V
i
Wave shaping with diodes
R2A
D2A
R1A
D1A
R0
i
R1B
D2B
R2B
R2B′
D1B
R1B′
0V
V
R2A′
R1A′
V0
−V0
V
i
Wave shaping with diodes
R2A
D2A
R1A
D1A
R0
i
Ra = 5 k
R1A
D1A
D2B
R2B
R1B′
D1B
R1B
−V0
Ra
R2B
−V0
R0
i
Vi
D2B
R2A
D2A
i
R1B
R2B′
R2A′ = R2B′ = 10 k
D1B
R1B′
R1A′ = R1B′ = 60 k
R2A′
R1A′
R2A = R2B = 5 k
R2B′
R1A = R1B = 15 k
V0 = 15 V
0V
V0
R0 = 20 k
V
R2A′
R1A′
V0
Vo
RL
V
i
Wave shaping with diodes
R2A
D2A
R1A
D1A
R0
i
Ra = 5 k
R1A
D1A
−V0
R0
D1B
R1B
D2B
R2B
R1B′
R2B
R2A
D2A
i
D2B
V
R2B′
R2A′ = R2B′ = 10 k
R1B
R1B′
R1A′ = R1B′ = 60 k
R2A′
R1A′
R2A = R2B = 5 k
D1B
R2B′
R1A = R1B = 15 k
V0 = 15 V
0V
V0
R0 = 20 k
V
R2A′
R1A′
V0
−V0
i
Vi
Ra
Vo
RL
Since Vi = −Ra i, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra ) factor.
i
Wave shaping with diodes
R2A
D2A
R1A
D1A
R0
i
Ra = 5 k
R2B
−V0
R0
D2B
R2B
R1B′
D1B
R1B
10
5
−V0
Ra
D2B
V
R2B′
R1A
D1A
i
i
Vi
R1B
R1B′
R2A′ = R2B′ = 10 k
D1B
R2A
D2A
Vo (V)
R1A′ = R1B′ = 60 k
R2A′
R1A′
R2A = R2B = 5 k
R2B′
R1A = R1B = 15 k
V0 = 15 V
0V
V0
R0 = 20 k
V
R2A′
R1A′
V0
Vo
RL
0
−5
−10
−5
0
Vi (V)
5
Since Vi = −Ra i, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra ) factor.
i
Wave shaping with diodes
R2A
D2A
R1A
D1A
R0
i
Ra = 5 k
R2B
−V0
R0
D2B
R2B
R1B′
D1B
R1B
10
10
Vo
5
i
Ra
D2B
i
V
R2B′
R1A
D1A
i
−V0
Vi
R1B
R1B′
R2A′ = R2B′ = 10 k
D1B
R2A
D2A
Vo (V)
R1A′ = R1B′ = 60 k
R2A′
R1A′
R2A = R2B = 5 k
R2B′
R1A = R1B = 15 k
V0 = 15 V
0V
V0
R0 = 20 k
V
R2A′
R1A′
V0
Vo
RL
Vi
0
0
−5
−10
−5
0
Vi (V)
5
−10
0
2
4
time (msec)
6
Since Vi = −Ra i, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra ) factor.
8
Wave shaping with diodes
R2A
D2A
R1A
D1A
R0
i
Ra = 5 k
R2B
−V0
R0
D2B
R2B
R1B′
D1B
R1B
10
10
Vo
5
i
Ra
D2B
i
V
R2B′
R1A
D1A
i
−V0
Vi
R1B
R1B′
R2A′ = R2B′ = 10 k
D1B
R2A
D2A
Vo (V)
R1A′ = R1B′ = 60 k
R2A′
R1A′
R2A = R2B = 5 k
R2B′
R1A = R1B = 15 k
V0 = 15 V
0V
V0
R0 = 20 k
V
R2A′
R1A′
V0
Vo
RL
Vi
0
0
−5
−10
−5
0
Vi (V)
5
−10
0
2
4
time (msec)
6
8
Since Vi = −Ra i, the Vo versus Vi plot is similar to the V versus i plot, except for the (−Ra ) factor.
SEQUEL file: ee101 wave shaper.sqproj
M. B. Patil, IIT Bombay
Wave shaping with diodes: spectrum
10
5
Vi
0
0
−10
Wave shaping with diodes: spectrum
10
5
Vi
0
0
−10
10
10
Vo
0
−10
0
2
4
time (msec)
6
8
0
0
2
6
4
8
10
N
M. B. Patil, IIT Bombay