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EE101: Diode circuits

M. B. Patil [email protected]

Department of Electrical EngineeringIndian Institute of Technology Bombay

M. B. Patil, IIT 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


(“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




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Diodes

flow

pressureV

i




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Diodes

flow

pressureV

i




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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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V V


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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V V


R = Roff if V < 0


* Examples:


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V V


R = Roff if V < 0


* 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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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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S

i

i ii

V V V

i

V

S closed, V > 0

S open, V < 0


* V

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S

i

i ii

V V V

i

V

S closed, V > 0

S open, V < 0


* V

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Shockley diode equation

p n

V

i i

V


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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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p n

V

i i

V

i = I s exp V






* I s is called the “reverse saturation current.”


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p n

V

i i

V

i = I s exp V







* For a typical low-power silicon diode, I s is of the order of 10−13 A.


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p n

V

i i

V

i = I s exp V








* 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 .


S

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p n

V

i i

V

i = I s exp V








* 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 .


Sh kl di d i

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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 .


Sh kl di d ti

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p n

V

i i

V

i = I s

exp

V

V T

− 1


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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p n

V

i i

V

i = I s

exp

V

V T

− 1


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


Shockley equation and simple models

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p n

V

i i

V

i = Is

e

V/VT − 1

, Is = 10−13 A , VT = 25 mV .


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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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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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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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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


* 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.



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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


* 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.


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


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.


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


* A real diode cannot withstand indefinitely large reverse voltages and “breaksdown” at a certain voltage called the “breakdown voltage” (V BR).


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


* 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.


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


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

* 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.


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


* 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).


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



* Typically, circuits are designed so that the reverse bias across any diode is lessthan the V BR rating for that diode.


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



* 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.


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.


Diode types

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* 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.


Diode types

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* Semiconductor lasers are essentially light-emitting diodes with structuralmodifications that establish conditions for coherent light.


Diode types

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* Semiconductor lasers are essentially light-emitting diodes with structuralmodifications that establish conditions for coherent light.

(source: wikipedia)


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.


Diode types

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photo

current

p

n

Rseries

Rshunt


Diode types

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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.


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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* 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.



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* 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.



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* 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.


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


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


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.


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 ,


→D must be on.


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 ,


→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


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 ,


→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 .


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 ,


→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.



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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



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1 V

1 k

1.5 k

0.5 k

B

A



−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.



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1 V

1 k

1.5 k

0.5 k

B

A



−5 V to +5 V, 500 Hz.

ID

Vi Vo

R1

R2

D2

R

i2i1

D1

0.7 VVD


Consider D 1 to be on → V AB = 0.7 + 1 + i 1R 1 .



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1 V

1 k

1.5 k

0.5 k

B

A



−5 V to +5 V, 500 Hz.

ID

Vi Vo

R1

R2

D2

R

i2i1

D1

0.7 VVD



Note that i 1 > 0, since D 1 can only conduct in the forward direction.

⇒ V AB > 1.7 V ⇒D 2 cannot conduct.



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1 V

1 k

1.5 k

0.5 k

B

A



−5 V to +5 V, 500 Hz.

ID

Vi Vo

R1

R2

D2

R

i2i1

D1

0.7 VVD





Similarly, if D 2 is on, V BA > 0.7 V , i.e., V AB < −0.7 V ⇒D 1 cannot conduct.



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1 V

1 k

1.5 k

0.5 k

B

A



−5 V to +5 V, 500 Hz.

ID

Vi Vo

R1

R2

D2

R

i2i1

D1

0.7 VVD





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.


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



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* 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



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→ 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



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→ 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)



→ 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



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→ 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)



→ 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 .


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



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



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

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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


D0 7 V VD

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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



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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?



D0 7 V VD

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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


In the off state, V D = R 1

R 1 + R 2V i .



D0 7 V VD

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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



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 .



D0.7 V VD

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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



R 1 + R 2V i .


i.e., V i =

R 1 + R 2

R 1 × 0.7 = 1.05 V .




D0.7 V VD

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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



R 1 + R 2V i .


i.e., V i =

R 1 + R 2

R 1 × 0.7 = 1.05 V .


−5 0 5

1.05

0

5

−5

Vo

Vi

D onD off



D1 D2R1

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1 k

1 k

Von = 0.7 V, VZ = 5 V.

Plot Vo versus Vi.

V

i

VoR2


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


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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




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.


i

Vi Vo

R1

R2


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




i

Vi Vo

R1

R2

5 V 0.7 V1 k

1 k

Vo = −i R2 =Vi + 5.7

R1 + R2R2



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)



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




i

Vi Vo

R1

R2

5 V 0.7 V1 k

1 k

Vo = −i R2 =Vi + 5.7

R1 + R2R2



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)



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 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



Vi

R

Vo

D2

D1

Vi > 5.7 V

Vo = 5.7 V

C


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



Vi

R

Vo

D2

D1

Vi > 5.7 V

Vo = 5.7 V

C

1 ki

0.7 V

5 V


Vi

R

Vo

D2

D1Vo = −5.7 V

Vi < −5.7 VA


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



Vi

R

Vo

D2

D1

Vi > 5.7 V

Vo = 5.7 V

C

1 ki

0.7 V

5 V


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


1 k

R

D1 4

8Vo

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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



Vi

R

Vo

D2

D1

Vi > 5.7 V

Vo = 5.7 V

C

1 ki

0.7 V

5 V


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)


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)


For 0 < t < T /4, V i rises from 0 to V m. As a result, the capacitor charges.


Peak detector

10

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time (msec) 0 1 2

VoVi

0C

D

10

-10

Vo (V)

Vi (V)



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 ).


Peak detector

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time (msec) 0 1 2

VoVi

0C

D

10

-10

Vo (V)

Vi (V)




For t > T /4, V i starts falling. The capacitor holds the charge it had at t = T /4 since

the diode prevents discharging.


Peak detector

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time (msec) 0 1 2

VoVi

0C

D

0

-10

Vo (V)

Vi (V)




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


Peak detector (continued)

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5 k

time (msec) 0 1 2

VoVi RC

D

1µF

10

-10

0

Vi (V)

Vo (V)



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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.



10

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5 k

time (msec) 0 1 2

VoVi RC

D

1µF

-10

0

Vi (V)

Vo (V)


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.



10

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5 k

time (msec) 0 1 2

VoVi RC

D

1µF

-10

0

Vi (V)

Vo (V)



Since τ T , the charging process is instantaneous.



10

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5 k

time (msec) 0 1 2

VoVi RC

D

1µF

-10

0

Vi (V)

Vo (V)



Since τ T , the charging process is instantaneous.SEQUEL file: ee101 diode circuit 5a.sqproj


Peak detector (with V on = 0.

7 V )

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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


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


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.



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


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 .


Clamped capacitor

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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.


Clamped capacitor

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0

10

0 0.5 1 1.5 2

time (msec)

Vi Vo

iD

VC

Vi

C

D




* 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).


Clamped capacitor

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0

10

0 0.5 1 1.5 2

time (msec)

Vi Vo

iD

VC

Vi

C

D





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 .


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





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 .



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


Voltage doubler


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.


Voltage doubler


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.


Voltage doubler


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.


* 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.


Voltage doubler


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.


* 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)

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Ee101 Diodes 1

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