ee101 diodes 1
TRANSCRIPT
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EE101: Diode circuits
M. B. Patil [email protected]
Department of Electrical EngineeringIndian Institute of Technology Bombay
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Diodes
flow
pressureV
i
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Diodes
flow
pressureV
i
* A diode may be thought of as an electrical counterpart of a directional valve
(“check valve”).
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Diodes
flow
pressureV
i
* A diode may be thought of as an electrical counterpart of a directional valve
(“check valve”).* A check valve presents a small resistance if the pressure p > 0, but blocks the
flow (i.e., presents a large resistance) if p
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Diodes
flow
pressureV
i
* A diode may be thought of as an electrical counterpart of a directional valve
(“check valve”).* A check valve presents a small resistance if the pressure p > 0, but blocks the
flow (i.e., presents a large resistance) if p
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Diodes
flow
pressureV
i
* A diode may be thought of as an electrical counterpart of a directional valve
(“check valve”).* A check valve presents a small resistance if the pressure p > 0, but blocks the
flow (i.e., presents a large resistance) if p
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Diodes
flow
pressureV
i
* A diode may be thought of as an electrical counterpart of a directional valve
(“check valve”).* A check valve presents a small resistance if the pressure p > 0, but blocks the
flow (i.e., presents a large resistance) if p
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Simple models: R on/R off model
V V
i i R = Ron if V > 0
R = Roff if V < 0
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Simple models: R on/R off model
V V
i i R = Ron if V > 0
R = Roff if V < 0
* Since the resistance is different in the forward and reverse directions, the i − V relationship is not symmetric.
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Simple models: R on/R off model
V V
i i R = Ron if V > 0
R = Roff if V < 0
* Since the resistance is different in the forward and reverse directions, the i − V relationship is not symmetric.
* Examples:
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Simple models: R on/R off model
V V
i i R = Ron if V > 0
R = Roff if V < 0
* Since the resistance is different in the forward and reverse directions, the i − V relationship is not symmetric.
* Examples:
10
0
−5 −4 −3 −2 −1 0 1 −5 −4 −3 −2 −1 0 1
V (Volts) V (Volts)
i
( m A )
Ron = 5Ω
Roff = 500Ω
Ron = 0.1 Ω
Roff = 1 MΩ
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Simple models: ideal switch
S
i
i ii
V V V
i
V
S closed, V > 0
S open, V < 0
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Simple models: ideal switch
S
i
i ii
V V V
i
V
S closed, V > 0
S open, V < 0
* V > 0 Volts → S is closed (a perfect contact), and it can ideally carry anyamount of current. The voltage drop across the diode is 0 V .
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Simple models: ideal switch
S
i
i ii
V V V
i
V
S closed, V > 0
S open, V < 0
* V > 0 Volts → S is closed (a perfect contact), and it can ideally carry anyamount of current. The voltage drop across the diode is 0 V .
* V
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Simple models: ideal switch
S
i
i ii
V V V
i
V
S closed, V > 0
S open, V < 0
* V > 0 Volts → S is closed (a perfect contact), and it can ideally carry anyamount of current. The voltage drop across the diode is 0 V .
* V
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Shockley diode equation
p n
V
i i
V
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Shockley diode equation
p n
V
i i
V
i = I s exp V
V T − 1, where V T = k B T /q .
k B = Boltzmann’s constant = 1.38× 10−23 J /K .
q = electron charge = 1.602× 10−19 Coul.
T = temperature in ◦K .
V T ≈ 25 mV at room temperature (27 ◦C).
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Shockley diode equation
p n
V
i i
V
i = I s exp V
V T − 1, where V T = k B T /q .
k B = Boltzmann’s constant = 1.38× 10−23 J /K .
q = electron charge = 1.602× 10−19 Coul.
T = temperature in ◦K .
V T ≈ 25 mV at room temperature (27 ◦C).
* I s is called the “reverse saturation current.”
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Shockley diode equation
p n
V
i i
V
i = I s exp V
V T − 1, where V T = k B T /q .
k B = Boltzmann’s constant = 1.38× 10−23 J /K .
q = electron charge = 1.602× 10−19 Coul.
T = temperature in ◦K .
V T ≈ 25 mV at room temperature (27 ◦C).
* I s is called the “reverse saturation current.”
* For a typical low-power silicon diode, I s is of the order of 10−13 A.
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Shockley diode equation
p n
V
i i
V
i = I s exp V
V T − 1, where V T = k B T /q .
k B = Boltzmann’s constant = 1.38× 10−23 J /K .
q = electron charge = 1.602× 10−19 Coul.
T = temperature in ◦K .
V T ≈ 25 mV at room temperature (27 ◦C).
* I s is called the “reverse saturation current.”
* For a typical low-power silicon diode, I s is of the order of 10−13 A.
* Although I s is very small, it gets multiplied by a large exponential factor, givinga diode current of several mA for V ≈ 0.7 V .
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S
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Shockley diode equation
p n
V
i i
V
i = I s exp V
V T − 1, where V T = k B T /q .
k B = Boltzmann’s constant = 1.38× 10−23 J /K .
q = electron charge = 1.602× 10−19 Coul.
T = temperature in ◦K .
V T ≈ 25 mV at room temperature (27 ◦C).
* I s is called the “reverse saturation current.”
* For a typical low-power silicon diode, I s is of the order of 10−13 A.
* Although I s is very small, it gets multiplied by a large exponential factor, givinga diode current of several mA for V ≈ 0.7 V .
* The “turn-on” voltage (V on) of a diode depends on the value of I s . V on may be
defined as the voltage at which the diode starts carrying a substantial forwardcurrent (say, a few mA).For a silicon diode, V on ≈ 0.7 V .For a GaAs diode, V on ≈ 1.1 V .
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Sh kl di d i
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Shockley diode equation
p n
V
i i
V
i = I s
exp
V
V T
− 1
, where V T = k B T /q .
Example: I s = 1× 10−13 A, V T = 25 mV .
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Shockley diode equation
p n
V
i i
V
i = I s
exp
V
V T
− 1
, where V T = k B T /q .
Example: I s = 1× 10−13 A, V T = 25 mV .
V x = V /V T e x i (Amp)
0.1 3.87 0.479×102 0.469×10−11
0.2 7.74 0.229×104 0.229×10−9
0.3 11.6 0.110×106 0.110×10−7
0.4 15.5 0.525×107 0.525×10−6
0.5 19.3 0.251×109 0.251×10−4
0.6 23.2 0.120×1011 0.120×10−2
0.62 24.0 0.260×1011 0.260×10−2
0.64 24.8 0.565×1011 0.565×10−2
0.66 25.5 0.122×1012 0.122×10−1
0.68 26.3 0.265×1012 0.265×10−1
0.70 27.1 0.575×1012 0.575×10−1
0.72 27.8 0.125×1013 0.125
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Shockley diode equation
p n
V
i i
V
i = I s
exp
V
V T
− 1
, where V T = k B T /q .
Example: I s = 1× 10−13 A, V T = 25 mV .
V x = V /V T e x i (Amp)
0.1 3.87 0.479×102 0.469×10−11
0.2 7.74 0.229×104 0.229×10−9
0.3 11.6 0.110×106 0.110×10−7
0.4 15.5 0.525×107 0.525×10−6
0.5 19.3 0.251×109 0.251×10−4
0.6 23.2 0.120×1011 0.120×10−2
0.62 24.0 0.260×1011 0.260×10−2
0.64 24.8 0.565×1011 0.565×10−2
0.66 25.5 0.122×1012 0.122×10−1
0.68 26.3 0.265×1012 0.265×10−1
0.70 27.1 0.575×1012 0.575×10−1
0.72 27.8 0.125×1013 0.125V (Volts)
i ( A m p )
1
i ( m A )
20
40
60
80
100
0
log scale
linear scale
0 0.2 0.4 0.6 0.8
10-6
10-12
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Shockley equation and simple models
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Shockley equation and simple models
p n
V
i i
V
i = Is
e
V/VT − 1
, Is = 10−13 A , VT = 25 mV .
Shockley equation and simple models
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Shockley equation and simple models
p n
V
i i
V
i = Is
e
V/VT − 1
, Is = 10−13 A , VT = 25 mV .
Model 1:
(open circuit)
V (Volts)
0
20
60
80
100
i ( m A )
40
Model 1
equation
Shockley
0.2 0.4 0.6 0.8
0
V > Von
V < Von
i Von
Von
Von = 0.7 V
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Shockley equation and simple models
p n
V
i i
V
i = Is
e
V/VT − 1
, Is = 10−13 A , VT = 25 mV .
Model 1:
(open circuit)
V (Volts)
0
20
60
80
100
i ( m A )
40
Model 1
equation
Shockley
0.2 0.4 0.6 0.8
0
V > Von
V < Von
i Von
Von
Von = 0.7 V
Model 2:
(open circuit)
V (Volts)
Shockleyequation
Model 2
0.2 0.4 0.6 0.80
VonV > Von
V < Von
i Ron
slope = 1/Ron
Von
Von = 0.668 V
Ron = 0.5 Ω
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Shockley equation and simple models
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Shockley equation and simple models
p n
V
i i
V
i = Is
e
V/VT − 1
, Is = 10−13 A , VT = 25 mV .
Model 1:
(open circuit)
V (Volts)
0
20
60
80
100
i ( m A )
40
Model 1
equation
Shockley
0.2 0.4 0.6 0.8
0
V > Von
V < Von
i Von
Von
Von = 0.7 V
Model 2:
(open circuit)
V (Volts)
Shockleyequation
Model 2
0.2 0.4 0.6 0.80
VonV > Von
V < Von
i Ron
slope = 1/Ron
Von
Von = 0.668 V
Ron = 0.5 Ω
* For many circuits, Model 1 is adequate since R on is much smaller than otherresistances in the circuit.
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Shockley equation and simple models
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Shockley equation and simple models
p n
V
i i
V
i = Is
e
V/VT − 1
, Is = 10−13 A , VT = 25 mV .
Model 1:
(open circuit)
V (Volts)
0
20
60
80
100
i ( m A )
40
Model 1
equation
Shockley
0.2 0.4 0.6 0.8
0
V > Von
V < Von
i Von
Von
Von = 0.7 V
Model 2:
(open circuit)
V (Volts)
Shockleyequation
Model 2
0.2 0.4 0.6 0.80
VonV > Von
V < Von
i Ron
slope = 1/Ron
Von
Von = 0.668 V
Ron = 0.5 Ω
* For many circuits, Model 1 is adequate since R on is much smaller than otherresistances in the circuit.
* If V on is much smaller than other relevant voltages in the circuit, we can useV on ≈ 0 V , and the diode model reduces to the ideal diode model seen earlier.
M. B. Patil, IIT Bombay
Shockley equation and simple models
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Shockley equation and simple models
p n
V
i i
V
i = Is
e
V/VT − 1
, Is = 10−13 A , VT = 25 mV .
Model 1:
(open circuit)
V (Volts)
0
20
60
80
100
i ( m A )
40
Model 1
equation
Shockley
0.2 0.4 0.6 0.8
0
V > Von
V < Von
i Von
Von
Von = 0.7 V
Model 2:
(open circuit)
V (Volts)
Shockleyequation
Model 2
0.2 0.4 0.6 0.80
VonV > Von
V < Von
i Ron
slope = 1/Ron
Von
Von = 0.668 V
Ron = 0.5 Ω
* For many circuits, Model 1 is adequate since R on is much smaller than otherresistances in the circuit.
* If V on is much smaller than other relevant voltages in the circuit, we can useV on ≈ 0 V , and the diode model reduces to the ideal diode model seen earlier.
* Note that the “battery” shown in the above models is not a “source” of power!It can only absorb power (see the direction of the current), causing heatdissipation.
M. B. Patil, IIT Bombay
Reverse breakdown
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Reverse breakdown
0
−20
20
40
V (Volts)
i ( m A )
−6 −5 −4 −3 −2 −1 0 1
V
i
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Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
−6 −5 −4 −3 −2 −1 0 1
V
i
* In the reverse direction, an ideal diode presents a large resistance for any appliedvoltage.
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Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
−6 −5 −4 −3 −2 −1 0 1
V
i
* In the reverse direction, an ideal diode presents a large resistance for any appliedvoltage.
* A real diode cannot withstand indefinitely large reverse voltages and “breaksdown” at a certain voltage called the “breakdown voltage” (V BR).
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Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
−6 −5 −4 −3 −2 −1 0 1
V
i
* In the reverse direction, an ideal diode presents a large resistance for any appliedvoltage.
* A real diode cannot withstand indefinitely large reverse voltages and “breaksdown” at a certain voltage called the “breakdown voltage” (V BR).
* When the reverse bias V R > V BR, the diode allows a large amount of current.If the current is not constrained by the external circuit, the diode would getdamaged.
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Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
Symbol for a Zener diode
−6 −5 −4 −3 −2 −1 0 1
V
i
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Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
Symbol for a Zener diode
−6 −5 −4 −3 −2 −1 0 1
V
i
* A wide variety of diodes is available, with V BR ranging from a few Volts to a fewthousand Volts! Generally, higher the breakdown voltage, higher is the cost.
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Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
Symbol for a Zener diode
−6 −5 −4 −3 −2 −1 0 1
V
i
* A wide variety of diodes is available, with V BR ranging from a few Volts to a fewthousand Volts! Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high V BR are generally used in power electronics applications andare therefore also designed to carry a large forward current (tens or hundreds of Amps).
M. B. Patil, IIT Bombay
Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
Symbol for a Zener diode
−6 −5 −4 −3 −2 −1 0 1
V
i
* A wide variety of diodes is available, with V BR ranging from a few Volts to a fewthousand Volts! Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high V BR are generally used in power electronics applications andare therefore also designed to carry a large forward current (tens or hundreds of Amps).
* Typically, circuits are designed so that the reverse bias across any diode is lessthan the V BR rating for that diode.
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Reverse breakdown
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0
−20
20
40
V (Volts)
i ( m A )
Symbol for a Zener diode
−6 −5 −4 −3 −2 −1 0 1
V
i
* A wide variety of diodes is available, with V BR ranging from a few Volts to a fewthousand Volts! Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high V BR are generally used in power electronics applications andare therefore also designed to carry a large forward current (tens or hundreds of Amps).
* Typically, circuits are designed so that the reverse bias across any diode is lessthan the V BR rating for that diode.
* “Zener” diodes typically have V BR of a few Volts, which is denoted by V Z . Theyare often used to limit the voltage swing in electronic circuits.
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Diode types
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Apart from their use as switches, diodes are also used for several other purposes. Thechoice of materials used, fabrication techniques, and packaging depend on thefunctionality expected from the device.
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Diode types
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Apart from their use as switches, diodes are also used for several other purposes. Thechoice of materials used, fabrication techniques, and packaging depend on thefunctionality expected from the device.
* Light-emitting diodes (LEDs) emit light when a forward bias is applied.Typically, LEDs are made of III-V semiconductors.
An LED emits light of a specific wavelength (e.g., red, green, yellow, blue).
White LEDs combine individual LEDs that emit the three primary colors (red,green, blue) or use a phosphor material to convert monochromatic light from ablue or UV LED to broad-spectrum white light.
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Diode types
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Apart from their use as switches, diodes are also used for several other purposes. Thechoice of materials used, fabrication techniques, and packaging depend on thefunctionality expected from the device.
* Light-emitting diodes (LEDs) emit light when a forward bias is applied.Typically, LEDs are made of III-V semiconductors.
An LED emits light of a specific wavelength (e.g., red, green, yellow, blue).
White LEDs combine individual LEDs that emit the three primary colors (red,green, blue) or use a phosphor material to convert monochromatic light from ablue or UV LED to broad-spectrum white light.
* Semiconductor lasers are essentially light-emitting diodes with structuralmodifications that establish conditions for coherent light.
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Diode types
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Apart from their use as switches, diodes are also used for several other purposes. Thechoice of materials used, fabrication techniques, and packaging depend on thefunctionality expected from the device.
* Light-emitting diodes (LEDs) emit light when a forward bias is applied.Typically, LEDs are made of III-V semiconductors.
An LED emits light of a specific wavelength (e.g., red, green, yellow, blue).
White LEDs combine individual LEDs that emit the three primary colors (red,green, blue) or use a phosphor material to convert monochromatic light from ablue or UV LED to broad-spectrum white light.
* Semiconductor lasers are essentially light-emitting diodes with structuralmodifications that establish conditions for coherent light.
(source: wikipedia)
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Diode types
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* Solar cells are generally silicon diodes designed to generate current efficientlywhen solar radiation is incident on the device. A “solar panel” has a large
number of individual solar cells connected in series/parallel configuration.A solar cell can be modelled as a diode in parallel with a current source(representing the photocurrent). In addition, parasitic series and shuntresistances need to be considered.
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Diode types
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* Solar cells are generally silicon diodes designed to generate current efficientlywhen solar radiation is incident on the device. A “solar panel” has a large
number of individual solar cells connected in series/parallel configuration.A solar cell can be modelled as a diode in parallel with a current source(representing the photocurrent). In addition, parasitic series and shuntresistances need to be considered.
photo
current
p
n
Rseries
Rshunt
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Diode types
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* Solar cells are generally silicon diodes designed to generate current efficientlywhen solar radiation is incident on the device. A “solar panel” has a large
number of individual solar cells connected in series/parallel configuration.A solar cell can be modelled as a diode in parallel with a current source(representing the photocurrent). In addition, parasitic series and shuntresistances need to be considered.
photo
current
p
n
Rseries
Rshunt
* Photodiodes are used to detect optical signals (DC or time-varying) and toconvert them into electrical signals which can be subsequently processed by
electronic circuits. They are used in fibre-optic communication systems, imageprocessing, etc.
A photodiode works on the same principle as a solar cell, i.e., it converts lightinto a current. However, its design is optimized for high-sensitivity, low-noise, orhigh-frequency operation, depending on the application.
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Diode circuit analysis
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* In DC situations, for each diode in the circuit, we need to establish whether it ison or off, replace it with the corresponding equivalent circuit, and then obtainthe quantities of interest.
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Diode circuit analysis
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* In DC situations, for each diode in the circuit, we need to establish whether it ison or off, replace it with the corresponding equivalent circuit, and then obtainthe quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns onor off, and analyse the circuit in intervals between these time points.
M. B. Patil, IIT Bombay
Diode circuit analysis
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* In DC situations, for each diode in the circuit, we need to establish whether it ison or off, replace it with the corresponding equivalent circuit, and then obtainthe quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns onor off, and analyse the circuit in intervals between these time points.
* In AC (small-signal) situations, the diode can be replaced by its small-signalmodel, and phasor analysis is used. We will illustrate this procedure for a BJTamplifier later.
M. B. Patil, IIT Bombay
Diode circuit analysis
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* In DC situations, for each diode in the circuit, we need to establish whether it ison or off, replace it with the corresponding equivalent circuit, and then obtainthe quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns onor off, and analyse the circuit in intervals between these time points.
* In AC (small-signal) situations, the diode can be replaced by its small-signalmodel, and phasor analysis is used. We will illustrate this procedure for a BJTamplifier later.
* Note that there are diode circuits in which the exponential nature of the diodeI-V relationship is made use of. For these circuits, computer simulation would berequired to solve the resulting non-linear equations.
M. B. Patil, IIT Bombay
Diode circuit example
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36 V3 k
6 k
1 k
i=?D
A B
C
IDR1
R2 R3
0.7 VVD
Diode circuit example
6 k
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36 V3 k
6 k
1 k
i=?D
A B
C
IDR1
R2 R3
0.7 VVD
Case 1: D is off.
36 V3 k
6 k
1 k
i
A B
C
R1
R2 R3
Diode circuit example
6 k
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36 V3 k
6 k
1 k
i=?D
A B
C
IDR1
R2 R3
0.7 VVD
Case 1: D is off.
36 V3 k
6 k
1 k
i
A B
C
R1
R2 R3
VAB = VAC =3
9 × 36 = 12 V ,
which is not consistent with ourassumption of D being off.
Diode circuit example
6 k
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36 V3 k
6 k
1 k
i=?D
A B
C
IDR1
R2 R3
0.7 VVD
Case 1: D is off.
36 V3 k
6 k
1 k
i
A B
C
R1R2
R3
VAB = VAC =3
9 × 36 = 12 V ,
which is not consistent with ourassumption of D being off.
→D must be on.
Diode circuit example
6 kA B
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36 V3 k
6 k
1 k
i=?D
A B
C
IDR1
R2 R3
0.7 VVD
Case 1: D is off.
36 V3 k
6 k
1 k
i
A B
C
R1R2
R3
VAB = VAC =3
9 × 36 = 12 V ,
which is not consistent with ourassumption of D being off.
→D must be on.
36 V3 k
6 k
1 k
i
A B
C
Case 2: D is on.
R1R2
R3
0.7 V
Diode circuit example
6 kA B
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36 V3 k
6 k
1 k
i=?D
A B
C
IDR1
R2 R3
0.7 VVD
Case 1: D is off.
36 V3 k
6 k
1 k
i
A B
C
R1R2
R3
VAB = VAC =3
9 × 36 = 12 V ,
which is not consistent with ourassumption of D being off.
→D must be on.
36 V3 k
6 k
1 k
i
A B
C
Case 2: D is on.
R1R2
R3
0.7 V
Taking VC = 0 V,
VA − 36
6 k +
VA3 k
+VA − 0.7
1 k = 0 ,
→ VA = 4.47 V, i = 3.77 mA .
Diode circuit example
6 kA B
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36 V3 k
6
1 k
i=?D
A B
C
IDR1
R2 R3
0.7 VVD
Case 1: D is off.
36 V3 k
6 k
1 k
i
A B
C
R1R2
R3
VAB = VAC =3
9 × 36 = 12 V ,
which is not consistent with ourassumption of D being off.
→D must be on.
36 V3 k
6 k
1 k
i
A B
C
Case 2: D is on.
R1R2
R3
0.7 V
Taking VC = 0 V,
VA − 36
6 k +
VA3 k
+VA − 0.7
1 k = 0 ,
→ VA = 4.47 V, i = 3.77 mA .
Remark: Often, we can figure out by inspection if a diode is on or off.
M. B. Patil, IIT Bombay
Diode circuit example
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1 V
1 k
1.5 k
0.5 k
B
A
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
(b) Plot Vo(t) for a triangular input:
−5 V to +5 V, 500 Hz.
ID
Vi Vo
R1
R2
D2
R
i2i1
D1
0.7 VVD
M. B. Patil, IIT Bombay
Diode circuit example
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1 V
1 k
1.5 k
0.5 k
B
A
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
(b) Plot Vo(t) for a triangular input:
−5 V to +5 V, 500 Hz.
ID
Vi Vo
R1
R2
D2
R
i2i1
D1
0.7 VVD
First, let us show that D 1 on ⇒ D 2 off, and D 2 on ⇒ D 1 off.
M. B. Patil, IIT Bombay
Diode circuit example
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1 V
1 k
1.5 k
0.5 k
B
A
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
(b) Plot Vo(t) for a triangular input:
−5 V to +5 V, 500 Hz.
ID
Vi Vo
R1
R2
D2
R
i2i1
D1
0.7 VVD
First, let us show that D 1 on ⇒ D 2 off, and D 2 on ⇒ D 1 off.
Consider D 1 to be on → V AB = 0.7 + 1 + i 1R 1 .
M. B. Patil, IIT Bombay
Diode circuit example
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1 V
1 k
1.5 k
0.5 k
B
A
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
(b) Plot Vo(t) for a triangular input:
−5 V to +5 V, 500 Hz.
ID
Vi Vo
R1
R2
D2
R
i2i1
D1
0.7 VVD
First, let us show that D 1 on ⇒ D 2 off, and D 2 on ⇒ D 1 off.
Consider D 1 to be on → V AB = 0.7 + 1 + i 1R 1 .
Note that i 1 > 0, since D 1 can only conduct in the forward direction.
⇒ V AB > 1.7 V ⇒D 2 cannot conduct.
M. B. Patil, IIT Bombay
Diode circuit example
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1 V
1 k
1.5 k
0.5 k
B
A
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
(b) Plot Vo(t) for a triangular input:
−5 V to +5 V, 500 Hz.
ID
Vi Vo
R1
R2
D2
R
i2i1
D1
0.7 VVD
First, let us show that D 1 on ⇒ D 2 off, and D 2 on ⇒ D 1 off.
Consider D 1 to be on → V AB = 0.7 + 1 + i 1R 1 .
Note that i 1 > 0, since D 1 can only conduct in the forward direction.
⇒ V AB > 1.7 V ⇒D 2 cannot conduct.
Similarly, if D 2 is on, V BA > 0.7 V , i.e., V AB < −0.7 V ⇒D 1 cannot conduct.
M. B. Patil, IIT Bombay
Diode circuit example
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1 V
1 k
1.5 k
0.5 k
B
A
(a) Plot Vo versus Vi for −5 V < Vi < 5 V .
(b) Plot Vo(t) for a triangular input:
−5 V to +5 V, 500 Hz.
ID
Vi Vo
R1
R2
D2
R
i2i1
D1
0.7 VVD
First, let us show that D 1 on ⇒ D 2 off, and D 2 on ⇒ D 1 off.
Consider D 1 to be on → V AB = 0.7 + 1 + i 1R 1 .
Note that i 1 > 0, since D 1 can only conduct in the forward direction.
⇒ V AB > 1.7 V ⇒D 2 cannot conduct.
Similarly, if D 2 is on, V BA > 0.7 V , i.e., V AB < −0.7 V ⇒D 1 cannot conduct.
Clearly, D 1 on ⇒ D 2 off, and D 2 on ⇒ D 1 off.
M. B. Patil, IIT Bombay
Diode circuit example (continued)
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* For −0.7 V < V i < 1.7 V , both D 1 and D 2 are off.
→ no drop across R , and V o = V i . (1)
1 V
1 k
1.5 k
0.5 k
A
B
Vi Vo
R1
R2
D2
R
i2i1
D1
M. B. Patil, IIT Bombay
Diode circuit example (continued)
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* For −0.7 V < V i < 1.7 V , both D 1 and D 2 are off.
→ no drop across R , and V o = V i . (1)
* For V i < −0.7 V , D 2 conducts. → V o = −0.7 − i 2R 2 .
Use KVL to get i 2: V i + i 2R 2 + 0.7 + Ri 2 = 0.
→ i 2 = −V i + 0.7
R + R 2, and
V o = −0.7 − R 2i 2 = R 2
R + R 2V i − 0.7
R
R + R 2. (2)
1 V
1 k
1.5 k
0.5 k
A
B
Vi Vo
R1
R2
D2
R
i2i1
D1
M. B. Patil, IIT Bombay
Diode circuit example (continued)
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* For −0.7 V < V i < 1.7 V , both D 1 and D 2 are off.
→ no drop across R , and V o = V i . (1)
* For V i < −0.7 V , D 2 conducts. → V o = −0.7 − i 2R 2 .
Use KVL to get i 2: V i + i 2R 2 + 0.7 + Ri 2 = 0.
→ i 2 = −V i + 0.7
R + R 2, and
V o = −0.7 − R 2i 2 = R 2
R + R 2V i − 0.7
R
R + R 2. (2)
* For V i > 1.7 V , D 1 conducts. → V o = 0.7 + 1 + i 1R 1 .
Use KVL to get i 1: −V i + i 1R + 0.7 + 1 + i 1R 1 = 0.
→ i 1 = V i − 1.7
R + R 1, and
V o = 1.7 + R 1i 1 = R 1
R + R 1
V i + 1.7 R
R + R 1
. (3)
1 V
1 k
1.5 k
0.5 k
A
B
Vi Vo
R1
R2
D2
R
i2i1
D1
M. B. Patil, IIT Bombay
Diode circuit example (continued)
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* For −0.7 V < V i < 1.7 V , both D 1 and D 2 are off.
→ no drop across R , and V o = V i . (1)
* For V i < −0.7 V , D 2 conducts. → V o = −0.7 − i 2R 2 .
Use KVL to get i 2: V i + i 2R 2 + 0.7 + Ri 2 = 0.
→ i 2 = −V i + 0.7
R + R 2, and
V o = −0.7 − R 2i 2 = R 2
R + R 2V i − 0.7
R
R + R 2. (2)
* For V i > 1.7 V , D 1 conducts. → V o = 0.7 + 1 + i 1R 1 .
Use KVL to get i 1: −V i + i 1R + 0.7 + 1 + i 1R 1 = 0.
→ i 1 = V i − 1.7
R + R 1, and
V o = 1.7 + R 1i 1 = R 1
R + R 1
V i + 1.7 R
R + R 1
. (3)
* Using Eqs. (1)-(3), we plot V o versus V i .
(SEQUEL file: ee101 diode circuit 1.sqproj)
1 V
1 k
1.5 k
0.5 k
A
B
Vi Vo
R1
R2
D2
R
i2i1
D1
M. B. Patil, IIT Bombay
Diode circuit example (continued)
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* For −0.7 V < V i < 1.7 V , both D 1 and D 2 are off.
→ no drop across R , and V o = V i . (1)
* For V i < −0.7 V , D 2 conducts. → V o = −0.7 − i 2R 2 .
Use KVL to get i 2: V i + i 2R 2 + 0.7 + Ri 2 = 0.
→ i 2 = −V i + 0.7
R + R 2, and
V o = −0.7 − R 2i 2 = R 2
R + R 2V i − 0.7
R
R + R 2. (2)
* For V i > 1.7 V , D 1 conducts. → V o = 0.7 + 1 + i 1R 1 .
Use KVL to get i 1: −V i + i 1R + 0.7 + 1 + i 1R 1 = 0.
→ i 1 = V i − 1.7
R + R 1, and
V o = 1.7 + R 1i 1 = R 1
R + R 1V i + 1.7
R
R + R 1. (3)
* Using Eqs. (1)-(3), we plot V o versus V i .
(SEQUEL file: ee101 diode circuit 1.sqproj)
1 V
1 k
1.5 k
0.5 k
A
B
Vi Vo
R1
R2
D2
R
i2i1
D1
−5 0 5
−0.7 1.7
−5
0
5
Vo
Vi
D1 off
D2 on
D1 on
D2 off
D1 off
D2 off
M. B. Patil, IIT Bombay
Diode circuit example (continued)
A5 5
R
Vo Vo
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t (msec)
t (msec)
t ( m s
e c )
t ( m s
e c )
1 V
1 k
1.5 k
0.5 k
B
0
1
2
0 1 2
−5 0 5
0
0
0 . 5
1
1 . 5
2
0 1 2
0
−5−5
−5 0 5
Vo
R1
R2
D2
i2i1
D1
Vi
t2t1
Point-by-point construction of
are shown as examples.
Two time points, t1 and t2,
Vo versus t:
Vi
Vi
D1 off
D2 on
D1 on
D2 off
D1 off
D2 off
M. B. Patil, IIT Bombay
Diode circuit example
D
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Plot Vo versus Vi for −5 V < Vi < 5 V .
I
D
Vi VoR2
R1
D
0.7 VVD
1 k
0.5 k
Diode circuit example
D
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Plot Vo versus Vi for −5 V < Vi < 5 V .
I
D
Vi VoR2
R1
D
0.7 VVD
1 k
0.5 k
D on
0.7 V
Vi VoR2
R1
Vo = Vi − 0.7
Diode circuit example
D0 7 V VD
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Plot Vo versus Vi for −5 V < Vi < 5 V .
I
D
Vi VoR2
R1
D
0.7 VVD
1 k
0.5 k
D on
0.7 V
Vi VoR2
R1
Vo = Vi − 0.7
D off
Vi VoR2
R1
VD
Vo =R2
R1 + R2
Vi
M. B. Patil, IIT Bombay
Diode circuit example
D0 7 V VD
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Plot Vo versus Vi for −5 V < Vi < 5 V .
ID
Vi VoR2
R1
0.7 VVD
1 k
0.5 k
D on
0.7 V
Vi VoR2
R1
Vo = Vi − 0.7
D off
Vi VoR2
R1
VD
Vo =R2
R1 + R2
Vi
At what value of V i will the diode turn on?
M. B. Patil, IIT Bombay
Diode circuit example
D0 7 V VD
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Plot Vo versus Vi for −5 V < Vi < 5 V .
ID
Vi VoR2
R1
0.7 VVD
1 k
0.5 k
D on
0.7 V
Vi VoR2
R1
Vo = Vi − 0.7
D off
Vi VoR2
R1
VD
Vo =R2
R1 + R2
Vi
At what value of V i will the diode turn on?
In the off state, V D = R 1
R 1 + R 2V i .
M. B. Patil, IIT Bombay
Diode circuit example
D0 7 V VD
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Plot Vo versus Vi for −5 V < Vi < 5 V .
ID
Vi VoR2
R1
0.7 VVD
1 k
0.5 k
D on
0.7 V
Vi VoR2
R1
Vo = Vi − 0.7
D off
Vi VoR2
R1
D
Vo =R2
R1 + R2
Vi
At what value of V i will the diode turn on?
In the off state, V D = R 1
R 1 + R 2V i .
For D to change to the on state, V D = 0.7 V .
i.e., V i =
R 1 + R 2
R 1 × 0.7 = 1.05 V .
M. B. Patil, IIT Bombay
Diode circuit example
D0.7 V VD
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Plot Vo versus Vi for −5 V < Vi < 5 V .
ID
Vi VoR2
R1
0.7 VVD
1 k
0.5 k
D on
0.7 V
Vi VoR2
R1
Vo = Vi − 0.7
D off
Vi VoR2
R1
D
Vo =R2
R1 + R2
Vi
At what value of V i will the diode turn on?
In the off state, V D = R 1
R 1 + R 2V i .
For D to change to the on state, V D = 0.7 V .
i.e., V i =
R 1 + R 2
R 1 × 0.7 = 1.05 V .
(SEQUEL file: ee101 diode circuit 2.sqproj)
M. B. Patil, IIT Bombay
Diode circuit example
D0.7 V VD
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Plot Vo versus Vi for −5 V < Vi < 5 V .
ID
Vi VoR2
R1
0.7 VVD
1 k
0.5 k
D on
Vi VoR2
R1
Vo = Vi − 0.7
D off
Vi VoR2
R1
D
Vo =R2
R1 + R2
Vi
At what value of V i will the diode turn on?
In the off state, V D = R 1
R 1 + R 2V i .
For D to change to the on state, V D = 0.7 V .
i.e., V i =
R 1 + R 2
R 1 × 0.7 = 1.05 V .
(SEQUEL file: ee101 diode circuit 2.sqproj)
−5 0 5
1.05
0
5
−5
Vo
Vi
D onD off
M. B. Patil, IIT Bombay
Diode circuit example
D1 D2R1
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1 k
1 k
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
V
i
VoR2
Diode circuit example
D1 D2R1
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1 k
1 k
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
V
i
VoR2
5 V0.7 V1 k
1 k
Vo = i R2 =Vi − 5.7
R1 + R2R2
Since i > 0, this can happen only when Vi > 5.7 V.
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
i
Vi Vo
R1
R2
Diode circuit example
D1 D2R1
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1 k
1 k
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
V
i
VoR2
5 V0.7 V1 k
1 k
Vo = i R2 =Vi − 5.7
R1 + R2R2
Since i > 0, this can happen only when Vi > 5.7 V.
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
i
Vi Vo
R1
R2
5 V 0.7 V1 k
1 k
Vo = −i R2 =Vi + 5.7
R1 + R2R2
Since i > 0, this can happen only when Vi < -5.7 V.
D2 on (forward), D1 in reverse breakdown
i
Vi Vo
R1
R2
Diode circuit example
D1 D2R11.5
Vo
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1 k
1 k
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
V
i
VoR2
5 V0.7 V1 k
1 k
Vo = i R2 =Vi − 5.7
R1 + R2R2
Since i > 0, this can happen only when Vi > 5.7 V.
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
i
Vi Vo
R1
R2
5 V 0.7 V1 k
1 k
Vo = −i R2 =Vi + 5.7
R1 + R2R2
Since i > 0, this can happen only when Vi < -5.7 V.
D2 on (forward), D1 in reverse breakdown
i
Vi Vo
R1
R2
0
−1.5
−8 −4 4 80
D2 r.b.D1 r.b.D1 r.b. D2 r.b.Vi
(breakdown)(breakdown)
M. B. Patil, IIT Bombay
Diode circuit example
D1 D2R11.5
Vo
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1 k
1 k
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
Vi VoR2
5 V0.7 V1 k
1 k
Vo = i R2 =Vi − 5.7
R1 + R2R2
Since i > 0, this can happen only when Vi > 5.7 V.
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
i
Vi Vo
R1
R2
5 V 0.7 V1 k
1 k
Vo = −i R2 =Vi + 5.7
R1 + R2R2
Since i > 0, this can happen only when Vi < -5.7 V.
D2 on (forward), D1 in reverse breakdown
i
Vi Vo
R1
R2
0
−1.5
−8 −4 4 80
D2 r.b.D1 r.b.D1 r.b. D2 r.b.Vi
(breakdown)(breakdown)
(SEQUEL file: ee101 diode circuit 3.sqproj)
M. B. Patil, IIT Bombay
Diode circuit example (voltage limiter)
1 k
R
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Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
Vi
D2
D1
Vo
Diode circuit example (voltage limiter)
1 k
R
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Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
Vi
D2
D1
Vo
1 ki
0.7 V
5 V
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
Vi
R
Vo
D2
D1
Vi > 5.7 V
Vo = 5.7 V
C
Diode circuit example (voltage limiter)
1 k
R
D
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Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
Vi
D2
D1
Vo
1 ki
0.7 V
5 V
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
Vi
R
Vo
D2
D1
Vi > 5.7 V
Vo = 5.7 V
C
1 ki
0.7 V
5 V
D2 on (forward), D1 in reverse breakdown
Vi
R
Vo
D2
D1Vo = −5.7 V
Vi < −5.7 VA
Diode circuit example (voltage limiter)
1 k
R
D
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Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
Vi
D2
D1
Vo
1 ki
0.7 V
5 V
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
Vi
R
Vo
D2
D1
Vi > 5.7 V
Vo = 5.7 V
C
1 ki
0.7 V
5 V
D2 on (forward), D1 in reverse breakdown
Vi
R
Vo
D2
D1Vo = −5.7 V
Vi < −5.7 VA
In the range, −5.7 V < Vi < 5.7 V, no current flows, and Vo = Vi. B
Diode circuit example (voltage limiter)
1 k
R
D1 4
8Vo
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87/112
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi.
Vi
D2
D1
Vo
1 ki
0.7 V
5 V
D1 on (forward), D2 in reverse breakdown
For a current to flow, we have two possibilities:
Vi
R
Vo
D2
D1
Vi > 5.7 V
Vo = 5.7 V
C
1 ki
0.7 V
5 V
D2 on (forward), D1 in reverse breakdown
Vi
R
Vo
D2
D1Vo = −5.7 V
Vi < −5.7 VA
In the range, −5.7 V < Vi < 5.7 V, no current flows, and Vo = Vi. B
−8 −4 0 4 8
4
0
−4
−8
Vi
A B C
(SEQUEL file: ee101 diode circuit 4.sqproj)M. B. Patil, IIT Bombay
Peak detector
10
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time (msec) 0 1 2
VoVi
0C
D
10
-10
Vo (V)
Vi (V)
M. B. Patil, IIT Bombay
Peak detector
10
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time (msec) 0 1 2
VoVi
0C
D
10
-10
Vo (V)
Vi (V)
Let V o (t ) = 0 V at t = 0, and assume the diode to be ideal, with V on = 0 V .
M B Patil IIT Bombay
Peak detector
10
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time (msec) 0 1 2
VoVi
0C
D
10
-10
Vo (V)
Vi (V)
Let V o (t ) = 0 V at t = 0, and assume the diode to be ideal, with V on = 0 V .
For 0 < t < T /4, V i rises from 0 to V m. As a result, the capacitor charges.
M B Patil IIT Bombay
Peak detector
10
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time (msec) 0 1 2
VoVi
0C
D
10
-10
Vo (V)
Vi (V)
Let V o (t ) = 0 V at t = 0, and assume the diode to be ideal, with V on = 0 V .
For 0 < t < T /4, V i rises from 0 to V m. As a result, the capacitor charges.
Since the on resistance of the diode is small, time constant τ T /4; therefore thecharging process is instantaneous ⇒ V o (t ) = V i (t ).
M B Patil IIT Bombay
Peak detector
10
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time (msec) 0 1 2
VoVi
0C
D
10
-10
Vo (V)
Vi (V)
Let V o (t ) = 0 V at t = 0, and assume the diode to be ideal, with V on = 0 V .
For 0 < t < T /4, V i rises from 0 to V m. As a result, the capacitor charges.
Since the on resistance of the diode is small, time constant τ T /4; therefore thecharging process is instantaneous ⇒ V o (t ) = V i (t ).
For t > T /4, V i starts falling. The capacitor holds the charge it had at t = T /4 since
the diode prevents discharging.
M B Patil IIT Bombay
Peak detector
10
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time (msec) 0 1 2
VoVi
0C
D
0
-10
Vo (V)
Vi (V)
Let V o (t ) = 0 V at t = 0, and assume the diode to be ideal, with V on = 0 V .
For 0 < t < T /4, V i rises from 0 to V m. As a result, the capacitor charges.
Since the on resistance of the diode is small, time constant τ T /4; therefore thecharging process is instantaneous ⇒ V o (t ) = V i (t ).
For t > T /4, V i starts falling. The capacitor holds the charge it had at t = T /4 since
the diode prevents discharging.
SEQUEL file: ee101 diode circuit 5.sqproj
M B Patil IIT Bombay
Peak detector (continued)
10
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5 k
time (msec) 0 1 2
VoVi RC
D
1µF
10
-10
0
Vi (V)
Vo (V)
M B Patil IIT Bombay
Peak detector (continued)
10
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5 k
time (msec) 0 1 2
VoVi RC
D
1µF
-10
0
Vi (V)
Vo (V)
If a resistor is added in parallel, a discharging path is provided for the capacitor, andthe capacitor voltage falls after reaching the peak.
M B Patil IIT Bombay
Peak detector (continued)
10
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5 k
time (msec) 0 1 2
VoVi RC
D
1µF
-10
0
Vi (V)
Vo (V)
If a resistor is added in parallel, a discharging path is provided for the capacitor, andthe capacitor voltage falls after reaching the peak.
When V i > V o , the capacitor charges again. The time constant for the chargingprocess is τ = R ThC , where R Th = R R on is the Thevenin resistance seen by thecapacitor, R on being the on resistance of the diode.
M B Patil IIT Bombay
Peak detector (continued)
10
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5 k
time (msec) 0 1 2
VoVi RC
D
1µF
-10
0
Vi (V)
Vo (V)
If a resistor is added in parallel, a discharging path is provided for the capacitor, andthe capacitor voltage falls after reaching the peak.
When V i > V o , the capacitor charges again. The time constant for the chargingprocess is τ = R ThC , where R Th = R R on is the Thevenin resistance seen by thecapacitor, R on being the on resistance of the diode.
Since τ T , the charging process is instantaneous.
M B Patil IIT Bombay
Peak detector (continued)
10
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5 k
time (msec) 0 1 2
VoVi RC
D
1µF
-10
0
Vi (V)
Vo (V)
If a resistor is added in parallel, a discharging path is provided for the capacitor, andthe capacitor voltage falls after reaching the peak.
When V i > V o , the capacitor charges again. The time constant for the chargingprocess is τ = R ThC , where R Th = R R on is the Thevenin resistance seen by thecapacitor, R on being the on resistance of the diode.
Since τ T , the charging process is instantaneous.SEQUEL file: ee101 diode circuit 5a.sqproj
M B Patil IIT Bombay
Peak detector (with V on = 0.
7 V )
10
Vo (V)
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time (msec) 0 1 2
VoVi R
0
-10
0
-10
10C
D
1µF
Vi (V)
Vi (V)
Vo (V)
R = 1 M
R = 5 k
M B P til IIT B mb
Peak detector (with V on = 0.
7 V )
10
Vo (V)
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time (msec) 0 1 2
VoVi R
0
-10
0
-10
10C
D
1µF
Vi (V)
Vi (V)
Vo (V)
R = 1 M
R = 5 k
With V on = 0.7 V , the capacitor charges up to (V m − 0.7 V ).
M B P til IIT B b
Peak detector (with V on = 0.
7 V )
10
Vo (V)
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time (msec) 0 1 2
VoVi R
0
-10
0
-10
10C
D
1µF
Vi (V)
Vi (V)
Vo (V)
R = 1 M
R = 5 k
With V on = 0.7 V , the capacitor charges up to (V m − 0.7 V ).Apart from that, the circuit operation is similar.
M. B. Patil, IIT Bombay
Peak detector (with V on = 0.
7 V )
10
Vo (V)
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time (msec) 0 1 2
VoVi R
0
-10
0
-10
10C
D
1µF
Vi (V)
Vi (V)
Vo (V)
R = 1 M
R = 5 k
With V on = 0.7 V , the capacitor charges up to (V m − 0.7 V ).Apart from that, the circuit operation is similar.
SEQUEL file: ee101 diode circuit 5a.sqproj
M. B. Patil, IIT Bombay
Clamped capacitor
20VC Vo
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0
10
0 0.5 1 1.5 2
time (msec)
Vi Vo
iD
VC
Vi
C
D
* Assume V on = 0 V for the diode.When D conducts, V D = −V o = 0 ⇒ V C + V i = 0, i.e., V C = −V i .
M. B. Patil, IIT Bombay
Clamped capacitor
20VC Vo
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0
10
0 0.5 1 1.5 2
time (msec)
Vi Vo
iD
VC
Vi
C
D
* Assume V on = 0 V for the diode.
When D conducts, V D = −V o = 0 ⇒ V C + V i = 0, i.e., V C = −V i .
* V C can only increase with time (or remain constant) since i D can only bepositive.
M. B. Patil, IIT Bombay
Clamped capacitor
20VC Vo
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0
10
0 0.5 1 1.5 2
time (msec)
Vi Vo
iD
VC
Vi
C
D
* Assume V on = 0 V for the diode.
When D conducts, V D = −V o = 0 ⇒ V C + V i = 0, i.e., V C = −V i .
* V C can only increase with time (or remain constant) since i D can only bepositive.
* The net result is that the capacitor gets charged to a voltage V C = −V i ,corresponding to the maxmimum negative value of V i , and holds that voltagethereafter. Let us call this voltage V 0
C
(a constant).
M. B. Patil, IIT Bombay
Clamped capacitor
20VC Vo
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0
10
0 0.5 1 1.5 2
time (msec)
Vi Vo
iD
VC
Vi
C
D
* Assume V on = 0 V for the diode.
When D conducts, V D = −V o = 0 ⇒ V C + V i = 0, i.e., V C = −V i .
* V C can only increase with time (or remain constant) since i D can only bepositive.
* The net result is that the capacitor gets charged to a voltage V C = −V i ,corresponding to the maxmimum negative value of V i , and holds that voltagethereafter. Let us call this voltage V 0
C
(a constant).
* V o (t ) = V C (t ) + V i (t ) = V 0
C + V i (t ), which is a “level-shifted” version of V i .
M. B. Patil, IIT Bombay
Clamped capacitor
20VC Vo
C
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0
10
0 0.5 1 1.5 2
time (msec)
Vi Vo
iD
VC
Vi
C
D
* Assume V on = 0 V for the diode.
When D conducts, V D = −V o = 0 ⇒ V C + V i = 0, i.e., V C = −V i .
* V C can only increase with time (or remain constant) since i D can only bepositive.
* The net result is that the capacitor gets charged to a voltage V C = −V i ,corresponding to the maxmimum negative value of V i , and holds that voltagethereafter. Let us call this voltage V 0
C
(a constant).
* V o (t ) = V C (t ) + V i (t ) = V 0
C + V i (t ), which is a “level-shifted” version of V i .
(SEQUEL file: ee101 diode circuit 6.sqproj)
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp Peak detector
A B C
VC1
C1 VD2
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−10
0
10
20
−10
0
10
20
R100 k
0 1 2 3 4 5 6 7 8 9 10
time (msec)
0 1 2 3 4 5 6 7 8 9 10
time (msec)
Von = 0 V Von = 0.7 V
C1
C2
VA
VB
VC
D1
D2
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp Peak detector
A B C
VC1
C1 VD2
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−10
0
10
20
−10
0
10
20
R100 k
0 1 2 3 4 5 6 7 8 9 10
time (msec)
0 1 2 3 4 5 6 7 8 9 10
time (msec)
Von = 0 V Von = 0.7 V
C1
C2
VA
VB
VC
D1
D2
* The diode clamp shifts V A up by V m (the amplitude of the AC source), making V B go from0 to 2 V m.
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp Peak detector
A B C
VC1
C1 VAD2
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−10
0
10
20
−10
0
10
20
R100 k
0 1 2 3 4 5 6 7 8 9 10
time (msec)
0 1 2 3 4 5 6 7 8 9 10
time (msec)
Von = 0 V Von = 0.7 V
C1
C2
VA
VB
VC
D1
D2
* The diode clamp shifts V A up by V m (the amplitude of the AC source), making V B go from0 to 2 V m.
* The peak detector detects the peak of V B (2 V m w.r.t. ground), and holds it constant.
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp Peak detector
A B C
VC1
C1 VAD2
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−10
0
10
20
−10
0
10
20
R100 k
0 1 2 3 4 5 6 7 8 9 10
time (msec)
0 1 2 3 4 5 6 7 8 9 10
time (msec)
Von = 0 V Von = 0.7 V
C1
C2
VA
VB
VC
D1
* The diode clamp shifts V
A up by V
m (the amplitude of the AC source), making V
B go from0 to 2 V m.
* The peak detector detects the peak of V B (2 V m w.r.t. ground), and holds it constant.
* Note that it takes a few cycles to reach steady state. Plot V C 1, i D 1, i D 2 versus t and explainthe initial behaviour of the circuit.
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp Peak detector
A B C
VC1
C1 VAD2
http://find/
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112/112
−10
0
10
20
−10
0
10
20
R100 k
0 1 2 3 4 5 6 7 8 9 10
time (msec)
0 1 2 3 4 5 6 7 8 9 10
time (msec)
Von = 0 V Von = 0.7 V
C1
C2
VA
VB
VC
D1
* The diode clamp shifts
V A up by
V m (the amplitude of the AC source), making
V B go from0 to 2 V m.
* The peak detector detects the peak of V B (2 V m w.r.t. ground), and holds it constant.
* Note that it takes a few cycles to reach steady state. Plot V C 1, i D 1, i D 2 versus t and explainthe initial behaviour of the circuit.
(SEQUEL file: ee101 voltage doubler.sqproj)
M. B. Patil, IIT Bombay
http://find/