L12.hegde.unlocked

Analysis  of  Diode  Models  

ESc201  :    Introduc:on  to  Electronics  L12

rhegde Dept. of Electrical Engineering

IIT Kanpur

1  

Diode Anode

Cathode

2  

PN Junction Diode

P N

Inside a PN junction at equilibrium (zero applied voltage), there is built-in voltage with N region being positive and P-region negative.

The built-in voltage (also called potential barrier) prevents electrons and holes to give rise to current.

-­‐

-­‐

-­‐

-­‐

-­‐ -­‐

+

+

+

+

+ +

P N

3  

Simplified Picture

4  

Extrinsic Semiconductors

Adding small amounts of suitable impurity atom can drastically alter number of electrons and holes in a semiconductor !

Addition of a group V element impurity to Silicon should increase electrons while addition of group III element impurity should increase number of holes

5  

Doping

6  

B

7  

N-type : n > p

A Semiconductor such as Silicon doped with a donor impurity such as Phosphorous or Arsenic from group V of periodic table. The donor impurity donates an electron to conduction band thereby increasing their concentration

P-type : p > n

A Semiconductor such as Silicon doped with a Acceptor impurity such as Boron from group III of periodic table. The acceptor impurity increases number of holes in valence band.

N and P-type Semiconductors

8  

No. of silicon atoms per unit volume 22 34 10 cm−×;

17 310AN cm−=Impurity concentration :

1 in 400,000 Silicon atoms is replaced by Boron

Very small amounts of impurity atoms can cause a drastic change in electrical property of a semiconductor.

9  

PN Junction Diode

P N

Inside a PN junction at equilibrium (zero applied voltage), there is built-in voltage with N region being positive and P-region negative.

The built-in voltage (also called potential barrier) prevents electrons and holes to give rise to current.

-­‐

-­‐

-­‐

-­‐

-­‐ -­‐

+

+

+

+

+ +

P N

10  

V

Forward and Reverse Bias P N

vd

Forward Bias: P is biased at a higher voltage compared to N. It lowers the built-in potential and allows current to flow.

V

P N

vdReverse Bias: N is biased at a higher voltage compared to P. This increases built-in potential and very little current flows. 11  

Forward Bias Reverse Bias

The p-n junction only conducts significant current in the forward-bias region.

-2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0.20

Cur

rent

(A)

Voltage (V)

Diode : 1N4001

VV

12  

Breakdown

13  

Diode : I-V Characteristics

{exp( ) 1}dD S

T

vi InV

= × −

Vd

n is called ideality factor and is equal to 1 for ideal diodes / 26 at T 300KTV kT q mV= ≅ =

Is : Reverse Saturation Current id:diode current; vd: applied voltage

14  

Forward Bias

{exp( ) 1}dD S

T

vi IV

= × −

26d Tv V mV>> =

exp( )dD ST

vi IV

≅ ×

P N

vd

15  

Reverse Bias

{exp( ) 1}dD S

T

vi IV

= × −

d Rv v= −

{exp( ) 1}RD S S

T

vi I IV

= × − − ≅−

P N

vd

R Tv V>>16  

17  

18  

Analysis using non-linear diode model is not easy

VS

R

D

VO = ? (1)S OV I R V= × +

{exp( ) 1} (2)OS

T

VI InV

= × −

S

I ln( 1) (4)IS TV IR nV⇒ = + × +

S

I ln( 1) (3)IO TV nV⇒ = × +

I

19  

Iterative Method:

Assume 0.6V =OV

S OV VIR−

=Calculate

1)IIln( S +×= TO nVVRe-calculate

ε≤Δ

IIConvergence:

(1) OS VIRV +=S

I ln( 1) (3)IO TV nV= × +

20  

Assume OV

S OV VIR−

=

1)IIln( S +×= TO nVV

D

VO = ?1K

2V

VO = 0.5

I = 1.5 x 10-3

VO = 0.711

VO = 0.707

I = 1.293 x 10-3

VO = 0.707

VO = 0.711

I = 1.289 x 10-3

VO = 0.707

CONVERGENCE

15

{exp( ) 1}

2 10/ 26

D ST

S

T

Vi IV

I AV kT q mV

−

= × −

= ×

= ≅at T 300K=

21  

15

{exp( ) 1}

2 10/ 26 at T 300K

D ST

S

T

Vi IV

I AV kT q mV

−

= × −

= ×

= ≅ =

Assume OV

S OV VIR−

=

1)IIln( S +×= TO nVV

D

VO = ?1K

2V

VO = 1.0

I = 1.0 x 10-3

VO = 0.7

VO = 0.707

I = 1.293 x 10-3

VO = 0.707

VO = 0.7

I = 1.3 x 10-3

VO = 0.707

CONVERGENCE to the same Result 22  

Graphical Method: Method of Load Line

S O

S O

V I R VV VIR

= × +

−⇒ =

VS

R

D

VO

{exp( ) 1}OS

T

VI InV

= × −

I

VO VS

VS/R solution

23  

How about something that is

simple & easy to work with

24  

Simple Models

DESIGN

SIMULATE

Accurate but complex Models

Design meets Specs.

Implement

Role of simple model in design cycle

25  

• Analysis using a non-linear diode model is relatively difficult and time consuming. • It also does not give a symbolic expression that can provide insight and help in the design of the circuit. Need SIMPLER and LINEAR Device Models

Vγ

open circuit

rf

V > Vγ

V < Vγ

26  

I

V Vγ

Slope = 1/rf

Vγ

open circuit

rf

V > Vγ

V < Vγ

Piece-Wise Linear Model

Vγ is called cut-in or turn-on voltage and depends on nature of diode and range of current considered

10f f

I r V V I rV V γ

γ

−−= ⇒ = + ×

−

For most of our analysis, we will take Vγ = 0.7V and rf ~10Ω

27  

Even Simpler Diode Models

open circuit

V > Vγ

V < Vγ

Vγ

Vγ

open circuit

rf

V > Vγ

V < Vγ

Constant voltage drop model

Vγ V

I

Vγ

I

V 28  

open circuit

V > 0

V < 0

Even Simpler Diode Models

Ideal diode model

Vγ=0

I

V

29  

Diode Models

open circuit

V > Vγ

V < Vγ

Vγ

open circuit

V > 0

V < 0

Vγ

open circuit

rf

V > Vγ

V < Vγ

+ vd -

I

{exp( ) 1}dD S

T

vi IV

= × −

Simplicity

Accuracy

30  

D

1K

10V

Analysis using ideal diode model

open circuit

V > 0

V < 0

1K

10V D

mAk

I 10110

==31  

Analysis with a constant voltage diode model

D

1K

10V open circuit

0.7V

V > 0.7

V < 0.7

1K

10V 0.7V

mAk

I 3.917.010=

−=

32  

Vγ

open circuit

rf

V > Vγ

V < Vγ

Analysis with a constant voltage plus resistor diode model

D

1K

10V

mAI 208.91010007.010=

+

−=

33  

D 5K

10V

5K2mA

Example

Find the current through the diode using ideal diode model

Is the diode forward biased? – Not Sure!! Assume that it is forward biased J Carry out analysis and then check if current through the diode is in appropriate direction. If not, diode is reverse biased and we carry out the analysis again!! 34  

Example

D 5K

10V

5K2mA

102 05 DmA iK

−− + + =

10V

5K2mAiD

4Di mA=

Current is positive, so our assumption is correct

Assume forward bias

35  

Example Find the current through the 5K resistor using ideal diode model

Assume forward bias

36  

D 5K

10V

2mA 4K5K

D 5K

10V

2mA 4K5K

D 5K

10V

2mA 4K5K

102 05 DmA iK

−− + − =

10V

2mA 4K iD

4Di mA= −

This is not possible. Therefore, our assumption is incorrect

37  

5K

D 5K

10V

2mA 4K

5K

10V

2mA 4K

Assume reverse bias

1 1

1

2

102 05 5

010

V VmAk k

VV V

+− + + =

=

=

V1 V2

38  

5K

5K