EE101: Diode circuits
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
Diodes
i
flow
V
pressure
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* A diode may be thought of as an electrical counterpart of a directional valve
(“check valve”).
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* 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 < 0.
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* 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 < 0.
* Similarly, a diode presents a small resistance in the forward direction and a large
resistance in the reverse direction.
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* 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 < 0.
* Similarly, a diode presents a small resistance in the forward direction and a large
resistance in the reverse direction.
* In the forward direction, the diode resistance RD = V /i would be a function of
V . However, it is often a good approximation to treat it as a constant (small)
resistance.
M. B. Patil, IIT Bombay
Diodes
i
flow
V
pressure
* 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 < 0.
* Similarly, a diode presents a small resistance in the forward direction and a large
resistance in the reverse direction.
* In the forward direction, the diode resistance RD = V /i would be a function of
V . However, it is often a good approximation to treat it as a constant (small)
resistance.
* In the reverse direction, the diode resistance is much larger and may often be
treated as infinite (i.e., the diode may be replaced by an open circuit).
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
V
R = Roff if V < 0
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
V
R = Roff if V < 0
* Since the resistance is different in the forward and reverse directions, the i − V
relationship is not symmetric.
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
V
R = Roff if V < 0
* Since the resistance is different in the forward and reverse directions, the i − V
relationship is not symmetric.
* Examples:
M. B. Patil, IIT Bombay
Simple models: Ron /Roff model
i
i
V
R = Ron if V > 0
R = Roff if V < 0
V
* Since the resistance is different in the forward and reverse directions, the i − V
relationship is not symmetric.
* Examples:
10
Ron = 0.1 Ω
Roff = 500 Ω
Roff = 1 MΩ
i (mA)
Ron = 5 Ω
0
−5
−4
−3
−2
V (Volts)
−1
0
1
−5
−4
−3
−2
−1
0
1
V (Volts)
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
S
i
S closed, V > 0
i
V
V
V
S open, V < 0
i
V
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
S
i
S closed, V > 0
i
V
V
V
S open, V < 0
i
V
* V > 0 Volts → S is closed (a perfect contact), and it can ideally carry any
amount of current. The voltage drop across the diode is 0 V .
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
S
i
S closed, V > 0
i
V
V
V
S open, V < 0
i
V
* V > 0 Volts → S is closed (a perfect contact), and it can ideally carry any
amount of current. The voltage drop across the diode is 0 V .
* V < 0 Volts → S is open (a perfect open circuit), and it can ideally block any
reverse voltage. The current through the diode is 0 A.
M. B. Patil, IIT Bombay
Simple models: ideal switch
i
i
S
i
S closed, V > 0
i
V
V
V
S open, V < 0
i
V
* V > 0 Volts → S is closed (a perfect contact), and it can ideally carry any
amount of current. The voltage drop across the diode is 0 V .
* V < 0 Volts → S is open (a perfect open circuit), and it can ideally block any
reverse voltage. The current through the diode is 0 A.
* The actual values of V and i for a diode in a circuit get determined by the i-V
relationship of the diode and the constraints on V and i imposed by the circuit.
M. B. Patil, IIT Bombay
Shockley diode equation
i
i
p
V
n
V
M. B. Patil, IIT Bombay
Shockley diode equation
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
i
p
V
n
V
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
M. B. Patil, IIT Bombay
Shockley diode equation
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
i
p
V
n
V
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
M. B. Patil, IIT Bombay
Shockley diode equation
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
i
p
V
n
V
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
* For a typical low-power silicon diode, Is is of the order of 10−13 A.
M. B. Patil, IIT Bombay
Shockley diode equation
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
i
p
V
n
V
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
* For a typical low-power silicon diode, Is is of the order of 10−13 A.
* Although Is is very small, it gets multiplied by a large exponential factor, giving
a diode current of several mA for V ≈ 0.7 V .
M. B. Patil, IIT Bombay
Shockley diode equation
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
i
i
p
V
n
V
kB = Boltzmann’s constant = 1.38 × 10−23 J/K .
q = electron charge = 1.602 × 10−19 Coul.
T = temperature in ◦ K .
VT ≈ 25 mV at room temperature (27 ◦ C).
* Is is called the “reverse saturation current.”
* For a typical low-power silicon diode, Is is of the order of 10−13 A.
* Although Is is very small, it gets multiplied by a large exponential factor, giving
a diode current of several mA for V ≈ 0.7 V .
* The “turn-on” voltage (Von ) of a diode depends on the value of Is . Von may be
defined as the voltage at which the diode starts carrying a substantial forward
current (say, a few mA).
For a silicon diode, Von ≈ 0.7 V .
For a GaAs diode, Von ≈ 1.1 V .
M. B. Patil, IIT Bombay
Shockley diode equation
i
i
p
V
n
V
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
Example: Is = 1 × 10−13 A, VT = 25 mV .
M. B. Patil, IIT Bombay
Shockley diode equation
i
i
p
V
n
V
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
Example: Is = 1 × 10−13 A, VT = 25 mV .
V
x = V /VT
ex
i (Amp)
0.1
3.87
0.479×102
0.469×10−11
7.74
0.229×104
0.229×10−9
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
24.8
0.565×1011
0.565×10−2
25.5
0.122×1012
0.122×10−1
26.3
0.265×1012
0.265×10−1
27.1
0.575×1012
0.575×10−1
27.8
0.125×1013
0.125
0.2
0.3
0.64
0.66
0.68
0.70
0.72
M. B. Patil, IIT Bombay
Shockley diode equation
i
i
p
V
n
V
«
–
»
„
V
i = Is exp
− 1 , where VT = kB T /q .
VT
Example: Is = 1 × 10−13 A, VT = 25 mV .
V
x = V /VT
ex
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
10−12
100
0.62
24.0
0.260×1011
0.260×10−2
80
0.64
24.8
0.565×1011
0.565×10−2
25.5
0.122×1012
0.122×10−1
26.3
0.265×1012
0.265×10−1
20
27.1
0.575×1012
0.575×10−1
0
27.8
0.125×1013
0.125
0.68
0.70
0.72
i (Amp)
log scale
i (mA)
0.66
1
10−6
linear scale
60
40
0
0.2
0.4
V (Volts)
0.6
0.8
M. B. Patil, IIT Bombay
Shockley equation and simple models
i
i
V
p
n
V
h
i
i = Is eV/VT − 1 , Is = 10−13 A , VT = 25 mV .
Shockley equation and simple models
i
i
p
V
i (mA)
100
V
Model 1:
80
V > Von
60
V < Von
h
i
Von
(open circuit)
40
Shockley
equation
20
0
Von = 0.7 V
0
0.2
Model 1
0.4
V (Volts)
0.6
i
i = Is eV/VT − 1 , Is = 10−13 A , VT = 25 mV .
n
0.8
Von
Shockley equation and simple models
i
i
p
V
i (mA)
100
V > Von
60
V < Von
i
V
Model 1:
80
h
i = Is eV/VT − 1 , Is = 10−13 A , VT = 25 mV .
n
i
Model 2:
Von
40
0
Von = 0.7 V
0.2
Ron
(open circuit)
Shockley
equation
0
Von
V < Von
(open circuit)
20
i
V > Von
Von = 0.668 V
Model 1
0.4
V (Volts)
Ron = 0.5 Ω
0.6
0.8
Von
0
0.2
slope = 1/Ron
Shockley
equation
Model 2
0.4
V (Volts)
0.6
0.8
Von
M. B. Patil, IIT Bombay
Shockley equation and simple models
i
i
p
V
i (mA)
100
V > Von
60
V < Von
i
V
Model 1:
80
h
i = Is eV/VT − 1 , Is = 10−13 A , VT = 25 mV .
n
i
Model 2:
Von
40
0
Von = 0.7 V
0.2
Ron
(open circuit)
Shockley
equation
0
Von
V < Von
(open circuit)
20
i
V > Von
Von = 0.668 V
Model 1
0.4
V (Volts)
Ron = 0.5 Ω
0.6
0.8
Von
0
0.2
slope = 1/Ron
Shockley
equation
Model 2
0.4
V (Volts)
0.6
0.8
Von
* For many circuits, Model 1 is adequate since Ron is much smaller than other
resistances in the circuit.
M. B. Patil, IIT Bombay
Shockley equation and simple models
i
i
p
V
i (mA)
100
V > Von
60
V < Von
i
V
Model 1:
80
h
i = Is eV/VT − 1 , Is = 10−13 A , VT = 25 mV .
n
i
Model 2:
Von
40
0
Von = 0.7 V
0.2
Ron
(open circuit)
Shockley
equation
0
Von
V < Von
(open circuit)
20
i
V > Von
Von = 0.668 V
Model 1
0.4
V (Volts)
Ron = 0.5 Ω
0.6
0.8
Von
0
0.2
slope = 1/Ron
Shockley
equation
Model 2
0.4
V (Volts)
0.6
0.8
Von
* For many circuits, Model 1 is adequate since Ron is much smaller than other
resistances in the circuit.
* If Von is much smaller than other relevant voltages in the circuit, we can use
Von ≈ 0 V , and the diode model reduces to the ideal diode model seen earlier.
M. B. Patil, IIT Bombay
Shockley equation and simple models
i
i
p
V
i (mA)
100
V > Von
60
V < Von
i
V
Model 1:
80
h
i = Is eV/VT − 1 , Is = 10−13 A , VT = 25 mV .
n
i
Model 2:
Von
40
0
Von = 0.7 V
0.2
Ron
(open circuit)
Shockley
equation
0
Von
V < Von
(open circuit)
20
i
V > Von
Von = 0.668 V
Model 1
0.4
V (Volts)
Ron = 0.5 Ω
0.6
0.8
Von
0
0.2
slope = 1/Ron
Shockley
equation
Model 2
0.4
V (Volts)
0.6
0.8
Von
* For many circuits, Model 1 is adequate since Ron is much smaller than other
resistances in the circuit.
* If Von is much smaller than other relevant voltages in the circuit, we can use
Von ≈ 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 heat
dissipation.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
* In the reverse direction, an ideal diode presents a large resistance for any applied
voltage.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
* In the reverse direction, an ideal diode presents a large resistance for any applied
voltage.
* A real diode cannot withstand indefinitely large reverse voltages and “breaks
down” at a certain voltage called the “breakdown voltage” (VBR ).
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
* In the reverse direction, an ideal diode presents a large resistance for any applied
voltage.
* A real diode cannot withstand indefinitely large reverse voltages and “breaks
down” at a certain voltage called the “breakdown voltage” (VBR ).
* When the reverse bias VR > VBR , the diode allows a large amount of current.
If the current is not constrained by the external circuit, the diode would get
damaged.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few
thousand Volts! Generally, higher the breakdown voltage, higher is the cost.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few
thousand Volts! Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high VBR are generally used in power electronics applications and
are therefore also designed to carry a large forward current (tens or hundreds of
Amps).
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few
thousand Volts! Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high VBR are generally used in power electronics applications and
are 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 less
than the VBR rating for that diode.
M. B. Patil, IIT Bombay
Reverse breakdown
40
i
20
i (mA)
V
0
Symbol for a Zener diode
−20
−6
−5
−4
−3
−2
V (Volts)
−1
0
1
* A wide variety of diodes is available, with VBR ranging from a few Volts to a few
thousand Volts! Generally, higher the breakdown voltage, higher is the cost.
* Diodes with high VBR are generally used in power electronics applications and
are 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 less
than the VBR rating for that diode.
* “Zener” diodes typically have VBR of a few Volts, which is denoted by VZ . They
are often used to limit the voltage swing in electronic circuits.
M. B. Patil, IIT Bombay
Diode types
Apart from their use as switches, diodes are also used for several other purposes. The
choice of materials used, fabrication techniques, and packaging depend on the
functionality expected from the device.
M. B. Patil, IIT Bombay
Diode types
Apart from their use as switches, diodes are also used for several other purposes. The
choice of materials used, fabrication techniques, and packaging depend on the
functionality 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 a
blue or UV LED to broad-spectrum white light.
M. B. Patil, IIT Bombay
Diode types
Apart from their use as switches, diodes are also used for several other purposes. The
choice of materials used, fabrication techniques, and packaging depend on the
functionality 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 a
blue or UV LED to broad-spectrum white light.
* Semiconductor lasers are essentially light-emitting diodes with structural
modifications that establish conditions for coherent light.
M. B. Patil, IIT Bombay
Diode types
Apart from their use as switches, diodes are also used for several other purposes. The
choice of materials used, fabrication techniques, and packaging depend on the
functionality 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 a
blue or UV LED to broad-spectrum white light.
* Semiconductor lasers are essentially light-emitting diodes with structural
modifications that establish conditions for coherent light.
(source: wikipedia)
M. B. Patil, IIT Bombay
Diode types
* Solar cells are generally silicon diodes designed to generate current efficiently
when 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 shunt
resistances need to be considered.
M. B. Patil, IIT Bombay
Diode types
* Solar cells are generally silicon diodes designed to generate current efficiently
when 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 shunt
resistances need to be considered.
Rseries
p
photo
current
Rshunt
n
M. B. Patil, IIT Bombay
Diode types
* Solar cells are generally silicon diodes designed to generate current efficiently
when 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 shunt
resistances need to be considered.
Rseries
p
photo
current
Rshunt
n
* Photodiodes are used to detect optical signals (DC or time-varying) and to
convert them into electrical signals which can be subsequently processed by
electronic circuits. They are used in fibre-optic communication systems, image
processing, etc.
A photodiode works on the same principle as a solar cell, i.e., it converts light
into a current. However, its design is optimized for high-sensitivity, low-noise, or
high-frequency operation, depending on the application.
M. B. Patil, IIT Bombay
Diode circuit analysis
* In DC situations, for each diode in the circuit, we need to establish whether it is
on or off, replace it with the corresponding equivalent circuit, and then obtain
the quantities of interest.
M. B. Patil, IIT Bombay
Diode circuit analysis
* In DC situations, for each diode in the circuit, we need to establish whether it is
on or off, replace it with the corresponding equivalent circuit, and then obtain
the quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns on
or off, and analyse the circuit in intervals between these time points.
M. B. Patil, IIT Bombay
Diode circuit analysis
* In DC situations, for each diode in the circuit, we need to establish whether it is
on or off, replace it with the corresponding equivalent circuit, and then obtain
the quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns on
or off, and analyse the circuit in intervals between these time points.
* In AC (small-signal) situations, the diode can be replaced by its small-signal
model, and phasor analysis is used. We will illustrate this procedure for a BJT
amplifier later.
M. B. Patil, IIT Bombay
Diode circuit analysis
* In DC situations, for each diode in the circuit, we need to establish whether it is
on or off, replace it with the corresponding equivalent circuit, and then obtain
the quantities of interest.
* In transient analysis, we need to find the time points at which a diode turns on
or off, and analyse the circuit in intervals between these time points.
* In AC (small-signal) situations, the diode can be replaced by its small-signal
model, and phasor analysis is used. We will illustrate this procedure for a BJT
amplifier later.
* Note that there are diode circuits in which the exponential nature of the diode
I-V relationship is made use of. For these circuits, computer simulation would be
required to solve the resulting non-linear equations.
M. B. Patil, IIT Bombay
Diode circuit example
6k
A
R1
B
R2
36 V
3k
C
D
i=?
R3
ID
1k
0.7 V
VD
Diode circuit example
6k
A
R1
B
R2
36 V
i=?
R3
D
1k
3k
0.7 V
C
Case 1: D is off.
6k
A
B
i
R1
R2
36 V
3k
C
ID
R3
1k
VD
Diode circuit example
6k
A
R1
B
R2
36 V
i=?
R3
D
1k
3k
0.7 V
C
Case 1: D is off.
6k
R1
36 V
A
B
i
R2
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
ID
VD
Diode circuit example
6k
A
R1
B
R2
36 V
i=?
R3
D
1k
3k
0.7 V
C
Case 1: D is off.
6k
A
R1
36 V
B
i
R2
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
ID
VD
Diode circuit example
6k
A
R1
B
R2
36 V
i=?
R3
D
1k
3k
Case 2: D is on.
Case 1: D is off.
A
R1
36 V
6k
B
i
R2
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
VD
0.7 V
C
6k
ID
0.7 V
A
B
i
R1
R2
36 V
3k
C
R3
1k
Diode circuit example
6k
A
R1
B
R2
36 V
i=?
R3
D
1k
3k
Case 2: D is on.
Case 1: D is off.
A
R1
36 V
6k
B
i
R2
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
VD
0.7 V
C
6k
ID
0.7 V
A
B
i
R1
R2
36 V
3k
R3
1k
C
Taking VC = 0 V,
VA − 0.7
VA − 36 VA
+
+
= 0,
6k
3k
1k
→ VA = 4.47 V, i = 3.77 mA .
Diode circuit example
6k
A
R1
B
R2
36 V
i=?
R3
D
1k
3k
Case 2: D is on.
Case 1: D is off.
A
R1
36 V
6k
B
i
R2
3k
R3
1k
C
3
× 36 = 12 V ,
9
which is not consistent with our
assumption of D being off.
VAB = VAC =
→ D must be on.
VD
0.7 V
C
6k
ID
0.7 V
A
B
i
R1
R2
36 V
3k
R3
1k
C
Taking VC = 0 V,
VA − 0.7
VA − 36 VA
+
+
= 0,
6k
3k
1k
→ 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
R
A
ID
1k
D1
D2
1V
Vi
R2
0.5 k
R1
1.5 k
i1
i2
Vo
0.7 V
VD
(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 .
B
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.5 k
R1
1.5 k
i1
i2
Vo
0.7 V
VD
(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 .
B
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.5 k
R1
1.5 k
i1
i2
0.7 V
Vo
VD
(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 .
B
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.5 k
R1
1.5 k
i1
0.7 V
Vo
i2
VD
(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 .
B
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
Note that i1 > 0, since D1 can only conduct in the forward direction.
⇒ VAB > 1.7 V ⇒ D2 cannot conduct.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.5 k
R1
1.5 k
i1
0.7 V
Vo
i2
VD
(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 .
B
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
Note that i1 > 0, since D1 can only conduct in the forward direction.
⇒ VAB > 1.7 V ⇒ D2 cannot conduct.
Similarly, if D2 is on, VBA > 0.7 V , i.e., VAB < −0.7 V ⇒ D1 cannot conduct.
M. B. Patil, IIT Bombay
Diode circuit example
R
A
ID
1k
D1
D2
1V
Vi
R2
0.5 k
R1
1.5 k
i1
0.7 V
Vo
i2
VD
(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 .
B
First, let us show that D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
Consider D1 to be on → VAB = 0.7 + 1 + i1 R1 .
Note that i1 > 0, since D1 can only conduct in the forward direction.
⇒ VAB > 1.7 V ⇒ D2 cannot conduct.
Similarly, if D2 is on, VBA > 0.7 V , i.e., VAB < −0.7 V ⇒ D1 cannot conduct.
Clearly, D1 on ⇒ D2 off, and D2 on ⇒ D1 off.
M. B. Patil, IIT Bombay
Diode circuit example (continued)
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
A
1k
D1
D2
1V
Vi
R2
Vo
0.5 k
R1
1.5 k
i1
i2
B
M. B. Patil, IIT Bombay
Diode circuit example (continued)
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
* For Vi < −0.7 V , D2 conducts. → Vo = −0.7 − i2 R2 .
Use KVL to get i2 : Vi + i2 R2 + 0.7 + Ri2 = 0.
Vi + 0.7
, and
→ i2 = −
R + R2
R2
R
Vo = −0.7 − R2 i2 =
Vi − 0.7
.
(2)
R + R2
R + R2
A
1k
D1
D2
1V
Vi
R2
Vo
0.5 k
R1
1.5 k
i1
i2
B
M. B. Patil, IIT Bombay
Diode circuit example (continued)
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
* For Vi < −0.7 V , D2 conducts. → Vo = −0.7 − i2 R2 .
Use KVL to get i2 : Vi + i2 R2 + 0.7 + Ri2 = 0.
Vi + 0.7
, and
→ i2 = −
R + R2
R2
R
Vo = −0.7 − R2 i2 =
Vi − 0.7
.
(2)
R + R2
R + R2
* For Vi > 1.7 V , D1 conducts. → Vo = 0.7 + 1 + i1 R1 .
Use KVL to get i1 : −Vi + i1 R + 0.7 + 1 + i1 R1 = 0.
Vi − 1.7
→ i1 =
, and
R + R1
R1
R
Vo = 1.7 + R1 i1 =
Vi + 1.7
.
(3)
R + R1
R + R1
A
1k
D1
D2
1V
Vi
R2
Vo
0.5 k
R1
1.5 k
i1
i2
B
M. B. Patil, IIT Bombay
Diode circuit example (continued)
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
* For Vi < −0.7 V , D2 conducts. → Vo = −0.7 − i2 R2 .
Use KVL to get i2 : Vi + i2 R2 + 0.7 + Ri2 = 0.
Vi + 0.7
, and
→ i2 = −
R + R2
R2
R
Vo = −0.7 − R2 i2 =
Vi − 0.7
.
(2)
R + R2
R + R2
* For Vi > 1.7 V , D1 conducts. → Vo = 0.7 + 1 + i1 R1 .
Use KVL to get i1 : −Vi + i1 R + 0.7 + 1 + i1 R1 = 0.
Vi − 1.7
→ i1 =
, and
R + R1
R1
R
Vo = 1.7 + R1 i1 =
Vi + 1.7
.
(3)
R + R1
R + R1
A
1k
D1
D2
1V
Vi
R2
Vo
0.5 k
R1
1.5 k
i1
i2
B
* Using Eqs. (1)-(3), we plot Vo versus Vi .
(SEQUEL file: ee101 diode circuit 1.sqproj)
M. B. Patil, IIT Bombay
Diode circuit example (continued)
R
* For −0.7 V < Vi < 1.7 V , both D1 and D2 are off.
→ no drop across R, and Vo = Vi .
(1)
D1
* For Vi < −0.7 V , D2 conducts. → Vo = −0.7 − i2 R2 .
Use KVL to get i2 : Vi + i2 R2 + 0.7 + Ri2 = 0.
Vi + 0.7
, and
→ i2 = −
R + R2
R2
R
Vo = −0.7 − R2 i2 =
Vi − 0.7
.
(2)
R + R2
R + R2
* For Vi > 1.7 V , D1 conducts. → Vo = 0.7 + 1 + i1 R1 .
Use KVL to get i1 : −Vi + i1 R + 0.7 + 1 + i1 R1 = 0.
Vi − 1.7
→ i1 =
, and
R + R1
R1
R
Vo = 1.7 + R1 i1 =
Vi + 1.7
.
(3)
R + R1
R + R1
* Using Eqs. (1)-(3), we plot Vo versus Vi .
A
1k
D2
1V
Vi
Vo
R2
0.5 k
R1
1.5 k
i2
i1
B
Vo
5
D1 off
D2 on
D1 off
D2 off
D1 on
D2 off
0
(SEQUEL file: ee101 diode circuit 1.sqproj)
−5
−5
0
−0.7
5
1.7
Vi
M. B. Patil, IIT Bombay
Diode circuit example (continued)
Vo
Vo
R
A
5
5
1k
D1
Vi
D2
1V
R2
Vo
D1 off
D1 off
D1 on
D2 on
D2 off
D2 off
0
0
0.5 k
R1
1.5 k
i1
i2
B
5
−5
0
5
Vi
−5
0
0
1 t2
2 t (msec)
1
2
0
Vi
t1
t (msec)
0.5
Two time points, t1 and t2,
are shown as examples.
0
0
Point-by-point construction of
Vo versus t:
−5
−5
1
1
1.5
2 t (msec)
2 t (msec)
M. B. Patil, IIT Bombay
Diode circuit example
D
1k
ID
R1
Vi
R2
0.5 k
Vo
0.7 V
Plot Vo versus Vi for −5 V < Vi < 5 V .
VD
Diode circuit example
D
0.7 V
1k
ID
R1
Vi
R2
0.5 k
R1
Vi
Vo
0.7 V
Plot Vo versus Vi for −5 V < Vi < 5 V .
R2
VD
D on
Vo = Vi − 0.7
Vo
Diode circuit example
D
1k
ID
R1
Vi
VD
0.7 V
R2
0.5 k
R1
Vi
Vo
0.7 V
Plot Vo versus Vi for −5 V < Vi < 5 V .
R1
R2
Vo
R2
Vi
Vo
VD
D on
Vo = Vi − 0.7
Vo =
D off
R2
Vi
R1 + R2
M. B. Patil, IIT Bombay
Diode circuit example
D
1k
ID
R1
Vi
VD
0.7 V
R2
0.5 k
R1
Vi
Vo
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D on
Vo = Vi − 0.7
Vo =
D off
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
M. B. Patil, IIT Bombay
Diode circuit example
D
1k
ID
R1
Vi
VD
0.7 V
R2
0.5 k
R1
Vi
Vo
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D on
Vo = Vi − 0.7
Vo =
D off
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
M. B. Patil, IIT Bombay
Diode circuit example
D
1k
ID
R1
Vi
VD
0.7 V
R2
0.5 k
R1
Vi
Vo
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D on
Vo = Vi − 0.7
Vo =
D off
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
For D to change to the on state, VD = 0.7 V .
R1 + R2
i.e., Vi =
× 0.7 = 1.05 V .
R1
M. B. Patil, IIT Bombay
Diode circuit example
D
1k
ID
R1
Vi
VD
0.7 V
R2
0.5 k
R1
Vi
Vo
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D on
Vo = Vi − 0.7
Vo =
D off
R2
Vi
R1 + R2
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
For D to change to the on state, VD = 0.7 V .
R1 + R2
i.e., Vi =
× 0.7 = 1.05 V .
R1
(SEQUEL file: ee101 diode circuit 2.sqproj)
M. B. Patil, IIT Bombay
Diode circuit example
D
1k
ID
R1
Vi
VD
0.7 V
R2
0.5 k
R1
Vi
Vo
0.7 V
R1
R2
Vo
R2
Vi
Vo
VD
Plot Vo versus Vi for −5 V < Vi < 5 V .
D on
Vo = Vi − 0.7
Vo =
D off
R2
Vi
R1 + R2
Vo
At what value of Vi will the diode turn on?
R1
In the off state, VD =
Vi .
R1 + R2
For D to change to the on state, VD = 0.7 V .
R1 + R2
i.e., Vi =
× 0.7 = 1.05 V .
R1
(SEQUEL file: ee101 diode circuit 2.sqproj)
5
0
D off
−5
−5
D on
0
5
Vi
1.05
M. B. Patil, IIT Bombay
Diode circuit example
R1
D1
D2
1k
R2
Vi
1k
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
Vo
Diode circuit example
R1
D1
D2
1k
R2
Vi
1k
Vo
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
For a current to flow, we have two possibilities:
D1 on (forward), D2 in reverse breakdown
R1
i
1k
0.7 V
5V
R2
Vi
Vo
1k
Vo = i R2 =
Vi − 5.7
R2
R1 + R2
Since i > 0, this can happen only when Vi > 5.7 V.
Diode circuit example
R1
D1
D2
1k
R2
Vi
1k
Vo
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
For a current to flow, we have two possibilities:
D2 on (forward), D1 in reverse breakdown
D1 on (forward), D2 in reverse breakdown
R1
R1
i
1k
0.7 V
i
5V
R2
Vi
Vo
1k
5V
0.7 V
R2
Vi
Vo = i R2 =
Vi − 5.7
R2
R1 + R2
Since i > 0, this can happen only when Vi > 5.7 V.
Vo
1k
1k
Vo = −i R2 =
Vi + 5.7
R2
R1 + R2
Since i > 0, this can happen only when Vi < -5.7 V.
Diode circuit example
Vo
R1
D1
1.5
D2
1k
R2
Vi
1k
Vo
0
−1.5
−8
Von = 0.7 V, VZ = 5 V.
D1 r.b.
Plot Vo versus Vi .
−4
D1 r.b.
0
4
8
D2 r.b.
D2 r.b.
(breakdown)
Vi
(breakdown)
For a current to flow, we have two possibilities:
D2 on (forward), D1 in reverse breakdown
D1 on (forward), D2 in reverse breakdown
R1
R1
i
1k
0.7 V
i
5V
R2
Vi
Vo
1k
5V
0.7 V
1k
1k
Vo = i R2 =
Vi − 5.7
R2
R1 + R2
Since i > 0, this can happen only when Vi > 5.7 V.
Vo
R2
Vi
Vo = −i R2 =
Vi + 5.7
R2
R1 + R2
Since i > 0, this can happen only when Vi < -5.7 V.
M. B. Patil, IIT Bombay
Diode circuit example
Vo
R1
D1
1.5
D2
1k
R2
Vi
1k
Vo
0
−1.5
−8
Von = 0.7 V, VZ = 5 V.
D1 r.b.
Plot Vo versus Vi .
−4
D1 r.b.
0
4
8
D2 r.b.
D2 r.b.
(breakdown)
Vi
(breakdown)
For a current to flow, we have two possibilities:
D2 on (forward), D1 in reverse breakdown
D1 on (forward), D2 in reverse breakdown
R1
R1
i
1k
0.7 V
i
5V
R2
Vi
Vo
1k
5V
0.7 V
1k
1k
Vo = i R2 =
Vi − 5.7
R2
R1 + R2
Since i > 0, this can happen only when Vi > 5.7 V.
Vo
R2
Vi
Vo = −i R2 =
Vi + 5.7
R2
R1 + R2
Since i > 0, this can happen only when Vi < -5.7 V.
M. B. Patil, IIT Bombay
Diode circuit example (voltage limiter)
R
1k
D1
Vo
Vi
D2
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
Diode circuit example (voltage limiter)
R
1k
D1
Vo
Vi
D2
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
For a current to flow, we have two possibilities:
D1 on (forward), D2 in reverse breakdown
R
1k
i
D1
0.7 V
Vi
Vo
C
D2
5V
Vo = 5.7 V
Vi > 5.7 V
Diode circuit example (voltage limiter)
R
1k
D1
Vo
Vi
D2
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
For a current to flow, we have two possibilities:
D1 on (forward), D2 in reverse breakdown
D2 on (forward), D1 in reverse breakdown
R
R
1k
i
1k
i
D1
D1
0.7 V
Vi
Vo
C
D2
5V
Vo = 5.7 V
Vi > 5.7 V
Vi
A
5V
D2
0.7 V
Vo
Vo = −5.7 V
Vi < −5.7 V
Diode circuit example (voltage limiter)
R
1k
D1
Vo
Vi
D2
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
For a current to flow, we have two possibilities:
D1 on (forward), D2 in reverse breakdown
D2 on (forward), D1 in reverse breakdown
R
R
1k
i
1k
i
D1
D1
0.7 V
Vi
Vo
C
D2
Vo = 5.7 V
Vi > 5.7 V
Vi
5V
A
5V
In the range, −5.7 V < Vi < 5.7 V, no current flows, and Vo = Vi .
D2
0.7 V
B
Vo
Vo = −5.7 V
Vi < −5.7 V
Diode circuit example (voltage limiter)
R
Vo 8
1k
D1
4
Vo
Vi
D2
0
−4
Von = 0.7 V, VZ = 5 V.
Plot Vo versus Vi .
−8
−8
A
B
−4
0
C
4
8
Vi
For a current to flow, we have two possibilities:
D1 on (forward), D2 in reverse breakdown
D2 on (forward), D1 in reverse breakdown
R
R
1k
i
1k
i
D1
D1
0.7 V
Vi
Vo
C
D2
Vo = 5.7 V
Vi > 5.7 V
Vi
5V
A
D2
Vo
Vo = −5.7 V
Vi < −5.7 V
0.7 V
5V
In the range, −5.7 V < Vi < 5.7 V, no current flows, and Vo = Vi .
B
(SEQUEL file: ee101 diode circuit 4.sqproj)
M. B. Patil, IIT Bombay
Peak detector
10
Vo (V)
D
Vi
C
0
Vo
Vi (V)
-10
0
1
time (msec)
2
M. B. Patil, IIT Bombay
Peak detector
10
Vo (V)
D
Vi
C
0
Vo
Vi (V)
-10
0
1
time (msec)
2
Let Vo (t) = 0 V at t = 0, and assume the diode to be ideal, with Von = 0 V .
M. B. Patil, IIT Bombay
Peak detector
10
Vo (V)
D
Vi
C
0
Vo
Vi (V)
-10
0
1
time (msec)
2
Let Vo (t) = 0 V at t = 0, and assume the diode to be ideal, with Von = 0 V .
For 0 < t < T /4, Vi rises from 0 to Vm . As a result, the capacitor charges.
M. B. Patil, IIT Bombay
Peak detector
10
Vo (V)
D
Vi
C
0
Vo
Vi (V)
-10
0
1
time (msec)
2
Let Vo (t) = 0 V at t = 0, and assume the diode to be ideal, with Von = 0 V .
For 0 < t < T /4, Vi rises from 0 to Vm . As a result, the capacitor charges.
Since the on resistance of the diode is small, time constant τ T /4; therefore the
charging process is instantaneous ⇒ Vo (t) = Vi (t).
M. B. Patil, IIT Bombay
Peak detector
10
Vo (V)
D
Vi
C
0
Vo
Vi (V)
-10
0
1
time (msec)
2
Let Vo (t) = 0 V at t = 0, and assume the diode to be ideal, with Von = 0 V .
For 0 < t < T /4, Vi rises from 0 to Vm . As a result, the capacitor charges.
Since the on resistance of the diode is small, time constant τ T /4; therefore the
charging process is instantaneous ⇒ Vo (t) = Vi (t).
For t > T /4, Vi 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
Vo (V)
D
Vi
C
0
Vo
Vi (V)
-10
0
1
time (msec)
2
Let Vo (t) = 0 V at t = 0, and assume the diode to be ideal, with Von = 0 V .
For 0 < t < T /4, Vi rises from 0 to Vm . As a result, the capacitor charges.
Since the on resistance of the diode is small, time constant τ T /4; therefore the
charging process is instantaneous ⇒ Vo (t) = Vi (t).
For t > T /4, Vi 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
Vo (V)
D
Vi
C
1 µF
R
5k
0
Vo
Vi (V)
-10
0
1
time (msec)
2
M. B. Patil, IIT Bombay
Peak detector (continued)
10
Vo (V)
D
Vi
C
1 µF
R
5k
0
Vo
Vi (V)
-10
0
1
time (msec)
2
If a resistor is added in parallel, a discharging path is provided for the capacitor, and
the capacitor voltage falls after reaching the peak.
M. B. Patil, IIT Bombay
Peak detector (continued)
10
Vo (V)
D
Vi
C
1 µF
R
5k
0
Vo
Vi (V)
-10
0
1
time (msec)
2
If a resistor is added in parallel, a discharging path is provided for the capacitor, and
the capacitor voltage falls after reaching the peak.
When Vi > Vo , the capacitor charges again. The time constant for the charging
process is τ = RTh C , where RTh = R k Ron is the Thevenin resistance seen by the
capacitor, Ron being the on resistance of the diode.
M. B. Patil, IIT Bombay
Peak detector (continued)
10
Vo (V)
D
Vi
C
1 µF
R
5k
0
Vo
Vi (V)
-10
0
1
time (msec)
2
If a resistor is added in parallel, a discharging path is provided for the capacitor, and
the capacitor voltage falls after reaching the peak.
When Vi > Vo , the capacitor charges again. The time constant for the charging
process is τ = RTh C , where RTh = R k Ron is the Thevenin resistance seen by the
capacitor, Ron being the on resistance of the diode.
Since τ T , the charging process is instantaneous.
M. B. Patil, IIT Bombay
Peak detector (continued)
10
Vo (V)
D
Vi
C
1 µF
R
5k
0
Vo
Vi (V)
-10
0
1
time (msec)
2
If a resistor is added in parallel, a discharging path is provided for the capacitor, and
the capacitor voltage falls after reaching the peak.
When Vi > Vo , the capacitor charges again. The time constant for the charging
process is τ = RTh C , where RTh = R k Ron is the Thevenin resistance seen by the
capacitor, Ron 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 Von = 0.7 V )
10
Vo (V)
0
D
Vi
C
1 µF
R
Vo
Vi (V)
R = 1M
-10
10
Vo (V)
0
R = 5k
-10
0
Vi (V)
1
time (msec)
2
M. B. Patil, IIT Bombay
Peak detector (with Von = 0.7 V )
10
Vo (V)
0
D
Vi
C
1 µF
R
Vo
Vi (V)
R = 1M
-10
10
Vo (V)
0
R = 5k
-10
0
Vi (V)
1
time (msec)
2
With Von = 0.7 V , the capacitor charges up to (Vm − 0.7 V ).
M. B. Patil, IIT Bombay
Peak detector (with Von = 0.7 V )
10
Vo (V)
0
D
Vi
C
1 µF
R
Vo
Vi (V)
R = 1M
-10
10
Vo (V)
0
R = 5k
-10
0
Vi (V)
1
time (msec)
2
With Von = 0.7 V , the capacitor charges up to (Vm − 0.7 V ).
Apart from that, the circuit operation is similar.
M. B. Patil, IIT Bombay
Peak detector (with Von = 0.7 V )
10
Vo (V)
0
D
Vi
C
1 µF
R
Vo
Vi (V)
R = 1M
-10
10
Vo (V)
0
R = 5k
-10
0
Vi (V)
1
time (msec)
2
With Von = 0.7 V , the capacitor charges up to (Vm − 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
VC
20
C
Vi
Vo
10
D
iD
VC
Vo
0
Vi
0
0.5
1
time (msec)
1.5
2
* Assume Von = 0 V for the diode.
When D conducts, VD = −Vo = 0 ⇒ VC + Vi = 0, i.e., VC = −Vi .
M. B. Patil, IIT Bombay
Clamped capacitor
VC
20
C
Vi
Vo
10
D
iD
VC
Vo
0
Vi
0
0.5
1
time (msec)
1.5
2
* Assume Von = 0 V for the diode.
When D conducts, VD = −Vo = 0 ⇒ VC + Vi = 0, i.e., VC = −Vi .
* VC can only increase with time (or remain constant) since iD can only be
positive.
M. B. Patil, IIT Bombay
Clamped capacitor
VC
20
C
Vi
Vo
10
D
iD
VC
Vo
0
Vi
0
0.5
1
time (msec)
1.5
2
* Assume Von = 0 V for the diode.
When D conducts, VD = −Vo = 0 ⇒ VC + Vi = 0, i.e., VC = −Vi .
* VC can only increase with time (or remain constant) since iD can only be
positive.
* The net result is that the capacitor gets charged to a voltage VC = −Vi ,
corresponding to the maxmimum negative value of Vi , and holds that voltage
thereafter. Let us call this voltage VC0 (a constant).
M. B. Patil, IIT Bombay
Clamped capacitor
VC
20
C
Vi
Vo
10
D
iD
VC
Vo
0
Vi
0
0.5
1
time (msec)
1.5
2
* Assume Von = 0 V for the diode.
When D conducts, VD = −Vo = 0 ⇒ VC + Vi = 0, i.e., VC = −Vi .
* VC can only increase with time (or remain constant) since iD can only be
positive.
* The net result is that the capacitor gets charged to a voltage VC = −Vi ,
corresponding to the maxmimum negative value of Vi , and holds that voltage
thereafter. Let us call this voltage VC0 (a constant).
* Vo (t) = VC (t) + Vi (t) = VC0 + Vi (t), which is a “level-shifted” version of Vi .
M. B. Patil, IIT Bombay
Clamped capacitor
VC
20
C
Vi
Vo
10
D
iD
VC
Vo
0
Vi
0
0.5
1
time (msec)
1.5
2
* Assume Von = 0 V for the diode.
When D conducts, VD = −Vo = 0 ⇒ VC + Vi = 0, i.e., VC = −Vi .
* VC can only increase with time (or remain constant) since iD can only be
positive.
* The net result is that the capacitor gets charged to a voltage VC = −Vi ,
corresponding to the maxmimum negative value of Vi , and holds that voltage
thereafter. Let us call this voltage VC0 (a constant).
* Vo (t) = VC (t) + Vi (t) = VC0 + Vi (t), which is a “level-shifted” version of Vi .
(SEQUEL file: ee101 diode circuit 6.sqproj)
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp
Peak detector
VC1
A
B
C
D2
C1
D1
VA
VB
R
C2
100 k
VC
Von = 0 V
Von = 0.7 V
20
20
10
10
0
0
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp
Peak detector
VC1
A
B
C
D2
C1
D1
VA
VB
R
C2
100 k
VC
Von = 0 V
Von = 0.7 V
20
20
10
10
0
0
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
* The diode clamp shifts VA up by Vm (the amplitude of the AC source), making VB go from
0 to 2 Vm .
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp
Peak detector
VC1
A
B
C
D2
C1
D1
VA
VB
R
C2
100 k
VC
Von = 0 V
Von = 0.7 V
20
20
10
10
0
0
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
* The diode clamp shifts VA up by Vm (the amplitude of the AC source), making VB go from
0 to 2 Vm .
* The peak detector detects the peak of VB (2 Vm w.r.t. ground), and holds it constant.
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp
Peak detector
VC1
A
B
C
D2
C1
D1
VA
VB
R
C2
100 k
VC
Von = 0 V
Von = 0.7 V
20
20
10
10
0
0
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
* The diode clamp shifts VA up by Vm (the amplitude of the AC source), making VB go from
0 to 2 Vm .
* The peak detector detects the peak of VB (2 Vm w.r.t. ground), and holds it constant.
* Note that it takes a few cycles to reach steady state. Plot VC 1 , iD1 , iD2 versus t and explain
the initial behaviour of the circuit.
M. B. Patil, IIT Bombay
Voltage doubler
Diode clamp
Peak detector
VC1
A
B
C
D2
C1
D1
VA
VB
R
C2
100 k
VC
Von = 0 V
Von = 0.7 V
20
20
10
10
0
0
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
−10
0
1
2
3
4
5
6
time (msec)
7
8
9
10
* The diode clamp shifts VA up by Vm (the amplitude of the AC source), making VB go from
0 to 2 Vm .
* The peak detector detects the peak of VB (2 Vm w.r.t. ground), and holds it constant.
* Note that it takes a few cycles to reach steady state. Plot VC 1 , iD1 , iD2 versus t and explain
the initial behaviour of the circuit.
(SEQUEL file: ee101 voltage doubler.sqproj)
M. B. Patil, IIT Bombay