EE101: Op Amp circuits (Part 4) - Indian Institute of ...sequel/ee101/ee101_opamp_4.pdf · EE101:...
66
EE101: Op Amp circuits (Part 4) M. B. Patil [email protected] www.ee.iitb.ac.in/~sequel Department of Electrical Engineering Indian Institute of Technology Bombay M. B. Patil, IIT Bombay
-
Upload
trinhhuong -
Category
Documents
-
view
272 -
download
7
Transcript of EE101: Op Amp circuits (Part 4) - Indian Institute of ...sequel/ee101/ee101_opamp_4.pdf · EE101:...
EE101: Op Amp circuits (Part 4)
M. B. [email protected]
www.ee.iitb.ac.in/~sequel
Department of Electrical EngineeringIndian Institute of Technology Bombay
M. B. Patil, IIT Bombay
Half-wave rectifier
Consider a diode rectifier:
slope=1
DR
VoVi
Vi
Vo
VD
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, andVo(t) = 0 V .
Precision rectifier circuits overcome this drawback.
M. B. Patil, IIT Bombay
Half-wave rectifier
Consider a diode rectifier:
slope=1
DR
VoVi
Vi
Vo
VD
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, andVo(t) = 0 V .
Precision rectifier circuits overcome this drawback.
M. B. Patil, IIT Bombay
Half-wave rectifier
Consider a diode rectifier:
slope=1
DR
VoVi
Vi
Vo
VD
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, andVo(t) = 0 V .
Precision rectifier circuits overcome this drawback.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
VDVi
Vo1
Vo
iR
i−iD
slope=1
Vi
Vo
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (exceptfor the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =Vo1
AV=
Vo + 0.7 V
AV≈ 0 V → Vo = Vi .
This situation arises only if iD > 0 (since the diode can only conduct in theforward direction), i.e., Vo > 0→ Vi = Vo > 0 V .
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
VDVi
Vo1
Vo
iR
i−iD
slope=1
Vi
Vo
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (exceptfor the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =Vo1
AV=
Vo + 0.7 V
AV≈ 0 V → Vo = Vi .
This situation arises only if iD > 0 (since the diode can only conduct in theforward direction), i.e., Vo > 0→ Vi = Vo > 0 V .
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
VDVi
Vo1
Vo
iR
i−iD
slope=1
Vi
Vo
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (exceptfor the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =Vo1
AV=
Vo + 0.7 V
AV≈ 0 V → Vo = Vi .
This situation arises only if iD > 0 (since the diode can only conduct in theforward direction), i.e., Vo > 0→ Vi = Vo > 0 V .
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
VDVi
Vo1
Vo
iR
i−iD
slope=1
Vi
Vo
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (exceptfor the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =Vo1
AV=
Vo + 0.7 V
AV≈ 0 V → Vo = Vi .
This situation arises only if iD > 0 (since the diode can only conduct in theforward direction), i.e., Vo > 0→ Vi = Vo > 0 V .
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
VDVi
Vo1
Vo
iR
i−iD
slope=1
Vi
Vo
Consider two cases:
(i) D is conducting: The feedback loop is closed, and the circuit looks like (exceptfor the diode drop) the buffer we have seen earlier.
Since the input current i− ≈ 0, iR = iD .
Further, V+ − V− =Vo1
AV=
Vo + 0.7 V
AV≈ 0 V → Vo = Vi .
This situation arises only if iD > 0 (since the diode can only conduct in theforward direction), i.e., Vo > 0→ Vi = Vo > 0 V .
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
ViVo1
Vo
Vi
Vo
Vo = 0
(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 isenough to drive it to saturation.
Note that Case (ii) occurs when Vi < 0 V . Since V+ − V− = Vi − 0 = Vi isnegative, Vo1 is driven to −V sat.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
ViVo1
Vo
Vi
Vo
Vo = 0
(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 isenough to drive it to saturation.
Note that Case (ii) occurs when Vi < 0 V . Since V+ − V− = Vi − 0 = Vi isnegative, Vo1 is driven to −V sat.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
ViVo1
Vo
Vi
Vo
Vo = 0
(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 isenough to drive it to saturation.
Note that Case (ii) occurs when Vi < 0 V . Since V+ − V− = Vi − 0 = Vi isnegative, Vo1 is driven to −V sat.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
ViVo1
Vo
Vi
Vo
Vo = 0
(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 isenough to drive it to saturation.
Note that Case (ii) occurs when Vi < 0 V . Since V+ − V− = Vi − 0 = Vi isnegative, Vo1 is driven to −V sat.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
DR
Vi
Vo
R
ViVo1
Vo
Vi
Vo
Vo = 0
(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 isenough to drive it to saturation.
Note that Case (ii) occurs when Vi < 0 V . Since V+ − V− = Vi − 0 = Vi isnegative, Vo1 is driven to −V sat.
M. B. Patil, IIT Bombay
Half-wave precision rectifier
superdiode
D
2
0
−2
−4
−6
−8
−10
−12
D onD off
t (ms) 1 2 0
R
Vo1 Vo
Vi
Vo
Vo = 0
Vo = Vi
Vi
Vo
Vi
−Vsat
Vo1
* 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 fromnegative to positive values. This is a relatively slow process, and it limits thespeed of this circuit.
SEQUEL file: precision half wave 1.sqproj
M. B. Patil, IIT Bombay
Half-wave precision rectifier
superdiode
D
2
0
−2
−4
−6
−8
−10
−12
D onD off
t (ms) 1 2 0
R
Vo1 Vo
Vi
Vo
Vo = 0
Vo = Vi
Vi
Vo
Vi
−Vsat
Vo1
* 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 fromnegative to positive values. This is a relatively slow process, and it limits thespeed of this circuit.
SEQUEL file: precision half wave 1.sqproj
M. B. Patil, IIT Bombay
Half-wave precision rectifier
superdiode
D
2
0
−2
−4
−6
−8
−10
−12
D onD off
t (ms) 1 2 0
R
Vo1 Vo
Vi
Vo
Vo = 0
Vo = Vi
Vi
Vo
Vi
−Vsat
Vo1
* 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 fromnegative to positive values. This is a relatively slow process, and it limits thespeed of this circuit.
SEQUEL file: precision half wave 1.sqproj
M. B. Patil, IIT Bombay
Half-wave precision rectifier
superdiode
D
2
0
−2
−4
−6
−8
−10
−12
D onD off
t (ms) 1 2 0
R
Vo1 Vo
Vi
Vo
Vo = 0
Vo = Vi
Vi
Vo
Vi
−Vsat
Vo1
* 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 fromnegative to positive values. This is a relatively slow process, and it limits thespeed of this circuit.
SEQUEL file: precision half wave 1.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
iR2
iR1iD2
iD1
R
Vo
iR
R1
R2
D2
D1
ViVo1
R
Vi > 0
VoR1
D2
D1
ViVo1
R2
R
Vi < 0
VoR1
D2
D1
ViVo1
R2
(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.
Vo = V− + iR2R2 = 0 +
„0− Vi
R1
«R2 = −R2
R1Vi .
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
1 k
1 k
R
VoR1
D2
D1
Vi
Vo1
R2
Vo
Vi
Vo = 0
−R2
R1Vi
2
1
0
−1
t (ms) 1 2 0
VoVi
Vo1
* Note that the Op Amp does not enter saturation since a feedback path isavailable for Vi > 0 V and Vi < 0 V .
SEQUEL file: precision half wave.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
1 k
1 k
R
VoR1
D2
D1
Vi
Vo1
R2
Vo
Vi
Vo = 0
−R2
R1Vi
2
1
0
−1
t (ms) 1 2 0
VoVi
Vo1
* Note that the Op Amp does not enter saturation since a feedback path isavailable for Vi > 0 V and Vi < 0 V .
SEQUEL file: precision half wave.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
1 k
1 k
R
VoR1
D2
D1
Vi
Vo1
R2
Vo
Vi
Vo = 0
−R2
R1Vi
2
1
0
−1
t (ms) 1 2 0
VoVi
Vo1
* Note that the Op Amp does not enter saturation since a feedback path isavailable for Vi > 0 V and Vi < 0 V .
SEQUEL file: precision half wave.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
1 k
1 k
R
VoR1
D2
D1
Vi
Vo1
R2
Vo
Vi
Vo = 0
−R2
R1Vi
2
1
0
−1
t (ms) 1 2 0
VoVi
Vo1
* Note that the Op Amp does not enter saturation since a feedback path isavailable for Vi > 0 V and Vi < 0 V .
SEQUEL file: precision half wave.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
1 k
1 k
R
VoR1
D2
D1
Vi
Vo1
R2
Vo
Vi
Vo = 0
−R2
R1Vi
2
1
0
−1
t (ms) 1 2 0
VoVi
Vo1
* Note that the Op Amp does not enter saturation since a feedback path isavailable for Vi > 0 V and Vi < 0 V .
SEQUEL file: precision half wave.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
R
VoR1
D2
D1
Vi
Vo1
R2
Vi
Vo = 0
Vo
−R2
R1Vi
The diodes are now reversed.
By considering two cases: (i) D1 on, (ii) D1 off, the Vo versus Vi relationship shown inthe figure is obtained (show this).
SEQUEL file: precision half wave 2.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
R
VoR1
D2
D1
Vi
Vo1
R2
Vi
Vo = 0
Vo
−R2
R1Vi
The diodes are now reversed.
By considering two cases: (i) D1 on, (ii) D1 off, the Vo versus Vi relationship shown inthe figure is obtained (show this).
SEQUEL file: precision half wave 2.sqproj
M. B. Patil, IIT Bombay
Improved half-wave precision rectifier
R
VoR1
D2
D1
Vi
Vo1
R2
Vi
Vo = 0
Vo
−R2
R1Vi
The diodes are now reversed.
By considering two cases: (i) D1 on, (ii) D1 off, the Vo versus Vi relationship shown inthe figure is obtained (show this).
SEQUEL file: precision half wave 2.sqproj
M. B. Patil, IIT Bombay
AM demodulation using a peak detector
AMsignal
Superdiode
filter
−0.15
0
0.15
t (ms) 1 2 0
Vo2
Vi
Vo1
Vo1
* 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
AM demodulation using a peak detector
AMsignal
Superdiode
filter
−0.15
0
0.15
t (ms) 1 2 0
Vo2
Vi
Vo1
Vo1
* 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
AM demodulation using a peak detector
AMsignal
Superdiode
filter
−0.15
0
0.15
t (ms) 1 2 0
Vo2
Vi
Vo1
Vo1
* 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
AM demodulation using a peak detector
AMsignal
Superdiode
filter
−0.15
0
0.15
t (ms) 1 2 0
Vo2
Vi
Vo1
Vo1
* 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
Half−waverectifier(inverting)
x (−1)
x (−2)Vo1 Vo
Vi
Vi
Vo1
Vi
VB
Vi
VA
Vi
Vo
VA
VB
(SEQUEL file: precision_full_wave.sqproj)
R
R/2
R
inverting summerinverting half−wave rectifier
VoR1
D2
D1
R1
ViVo1
−2
−1
0
1
2
t (ms) 1 2 0
ViVo
M. B. Patil, IIT Bombay
Full-wave precision rectifier
Half−waverectifier(inverting)
x (−1)
x (−2)Vo1 Vo
Vi
Vi
Vo1
Vi
VB
Vi
VA
Vi
Vo
VA
VB
(SEQUEL file: precision_full_wave.sqproj)
R
R/2
R
inverting summerinverting half−wave rectifier
VoR1
D2
D1
R1
ViVo1
−2
−1
0
1
2
t (ms) 1 2 0
ViVo
M. B. Patil, IIT Bombay
Full-wave precision rectifier
Half−waverectifier(inverting)
x (−1)
x (−2)Vo1 Vo
Vi
Vi
Vo1
Vi
VB
Vi
VA
Vi
Vo
VA
VB
(SEQUEL file: precision_full_wave.sqproj)
R
R/2
R
inverting summerinverting half−wave rectifier
VoR1
D2
D1
R1
ViVo1
−2
−1
0
1
2
t (ms) 1 2 0
ViVo
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V
D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
D
A
R
i
R′
R0
−V0
V0 V D off
A
R
i
R′
R0
−V0
V0 V
D on
A
R
i
R′
R0
−V0
V0 V
Von
i
V
Vbreakslope = R0 ‖ R
slope = R0
When D is off, VA is (by superposition),
VA = VR′
R + R′ − V0R
R + R′ .
For D to turn on, VA = Von ≈ 0.7 V → V ≡ Vbreak =R
R′ (V0 + Von) + Von .
When D is on,
i =V
R0+
V − Von
R+−V0 − Von
R′
= V
»1
R0+
1
R
–+ (constant)
i.e., V = (R0 ‖ R) i + (constant) .
M. B. Patil, IIT Bombay
Wave shaping with diodes
i
0 V
R0
V
i
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
V
i
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
i
0 V
R0
Vi
V
D1B R1B
R1B
′
i
−V0
0 V
R0
Vi
V
D2B R2B
R2B
′
i
V
D1A R1A
R1A
′
V0
i
V
D2A R2A
R2A
′
i
V
D1B
D2B
D2A
D1A
R1A
′
R2A
′
R1B
R2B
R1B
′
R2B
′
RL
Vo
i
R2A
R1A
i
V0
Ra
−V0
R0
Vi
Ra = 5 k
R0 = 20 k
R1A = R1B = 15 k
R2A = R2B = 5 k
R1A′ = R1B′ = 60 k
R2A′ = R2B′ = 10 k
V0 = 15 V
−100 5−5
−5
0
5
10
Vi (V)
Vo
(V)
20 4 6 8
10
−10
0
time (msec)
Vi
Vo
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
0
510
−10
0
Vi
20 4 6 8
10
−10
0
time (msec) N0 2 4 6 8 10
0
10
Vo
M. B. Patil, IIT Bombay
Wave shaping with diodes: spectrum
0
510
−10
0
Vi
20 4 6 8
10
−10
0
time (msec) N0 2 4 6 8 10
0
10
Vo
M. B. Patil, IIT Bombay
![Filter Design with Op-amptera.yonsei.ac.kr/class/2020_1_1/lecture/Lect 29 Project... · 2020. 7. 8. · • Filter Design with ideal Op-amp [10], Operational Amplifier [20] ,Filter](https://static.fdocument.org/doc/165x107/60a6ee382a29926a6905df36/filter-design-with-op-29-project-2020-7-8-a-filter-design-with-ideal.jpg)
![CIT3136 - ict. · PDF filederivation (Slide 8 was a . rightmost): (1) exp ⇒exp op exp [exp →exp op exp] ... Principle of Syntax-directed . Semantics. The parse tree will be used](https://static.fdocument.org/doc/165x107/5a9629457f8b9ab6188c88c2/cit3136-ict-slide-8-was-a-rightmost-1-exp-exp-op-exp-exp-exp-op.jpg)
