PN-JUNCTION DIODE AND DIODE CIRCUITSj.guntzel/ine5442/pn-junction.pdf · Electronics 2 PN-JUNCTION...

31
Electronics 1 PN-JUNCTION DIODE AND DIODE CIRCUITS Objectives Understand the basic principles of diode operation and the current- voltage exponential characteristic; Use equivalent models of diodes to analyze typical diode circuits Analyze and determine experimentally the characteristics of typical diode circuits; Use circuit simulator to analyze diode applications.

Transcript of PN-JUNCTION DIODE AND DIODE CIRCUITSj.guntzel/ine5442/pn-junction.pdf · Electronics 2 PN-JUNCTION...

Electronics1

PN-JUNCTION DIODE AND

DIODE CIRCUITS

Objectives

Understand the basic principles of diode operation and the current-

voltage exponential characteristic;

Use equivalent models of diodes to analyze typical diode circuits

Analyze and determine experimentally the characteristics of

typical diode circuits;

Use circuit simulator to analyze diode applications.

Electronics2

PN-JUNCTION DIODE AND

DIODE CIRCUITS

1 Main properties of semiconductors

2 The PN-junction in Equilibrium

3 The i – v Characteristics of the PN-Junction

4 Deviations from the Ideal Diode

5 Analysis of the Diode Circuits

6 Rectifier Circuits

Electronics3

Electronics4

Electronics5

Electronics6

Electronics7

Electronics8

Electronics9

Electronics10

PN-junction

Electronics11

1016

1014

1012

1010

106

108

104

102

nn = ND

pp = NA

np = ni2 / NA pn = ni

2/ ND

N A=1016

cm-3

PN D=10

15cm

-3

N

ohmic

contact

ohmic

contact

A B

-N A

N D

x(a) Step junction

(b) “initial” carrier concentrations

ni=1010cm-3

“Before contacting”, the two

regions are electrically neutral

( ρ = 0 ).

Electronics12

Electronics13

Diffusion current

- A spatial variation ( gradient ) in concentration causes a net

transport of particles from regions of higher concentration to

regions of lower concentration.

0 L x

Hole current

p(x)

A

0 L x

dp/dx < 0 electron

current

n(x)

0 x

dx

dp DqdiffJ pp −=

dx

dn DqdiffJ nn =

Dn(Dp): electron (hole) diffusion coefficient

Electronics14

NA=1016 cm-3 ND=1015 cm-3 P-type Si N-type Si

A B

1016

1012

108

104 n

Space charge region or

depletion region

quasi-neutral regions

ρ = E = 0

Ions (NA-)

Ions (ND+) +

_

VAB = 0

∗ electrons (holes) diffuse from

the n (p) side to the p (n) side,

leaving behind ND+ (NA

-) ionized

donor (acceptor) atoms and,

consequently, a net charge density

ρ ≠ 0, which gives rise to an

electric field ≠ 0

∗ pn = ni2 because VAB = 0

Electronics15

EE = 0 E = 0

-xp xn

-Emax

-xp xn

-qNA = -1.6 x10-3

c/cm3

x

x

1.6 x10-4

c/cm3qND

ρ (c/cm3)

+

-

P- NNA = 10

16 cm

-3 ND = 1015

cm-3A B

0

x

φJ

xn-xp 0

V

Electronics16

Electronics17

−=

kT

PEexpnn O

PE: potential energy

n=nO at sea level (h=0)

0 20 40 60 80

10-1

100

0.0

0.2

0.4

0.6

0.8

1.0

n/no (log scale)

height (km)

n/no

The Exponential Atmosphere and the Boltzmann’s Law

k = 1.38 x10-23 J/K = 86.2 x10-6 eV/K

T: absolute temperature

K: Boltzmann’s constant

Electronics18

Charge carriers in semiconductor: PE=-qV for electrons

PE= qV for holes

n = nO exp (-PE/kT ) nO is the electron concentration for V=0

pO is the hole concentration for V=0

n = nO exp (qV/RT) p = nO exp (-qV/RT)

k: Boltzmann’s constant

T: absolute temperature

q: electronic charge

φt: thermal voltage

φt ≈ 25 mV (@ 290 K)

26 mV (@ 301.5 K)

Electronics19

x

φ (x)

xn

-x p

v = 0D

v > 0D

v < 0D

- vD

φj

+ -P-side N-side

+ vD -

Electrostatic junction potential for different applied voltages

Barrier potential = ( φJ - vD )

is controlled by the applied

voltage.

Carrier concentration depends

exponentially on barrier potential

( as in the Boltzmann’s

atmosphere ).

Electronics20

I

VA

VA ≠ 0

I = IUP - IDO

IUP α NA exp [(-φJ+VA) / φt]

IUP = IS exp (VA / φt)

IDO = IS

= 1

VexpII A

SD

VA > 0 forward bias

VA < 0 reverse bias

IUP – current uphill α NA exp(-φJ / φt )

IDO – current downhill α p ( N-side )

Equilibrium: IUP = IDO = IS

where IS α NA exp(-φJ / φt )

P N

φJφJ -VA

-xp

VA=0

xn

Equilibrium (VA=0)

Electronics21

1.51.00.50.0-0.5-1.0-1.5-0.02

0.00

0.02

0.04

0.06

0.08

0.10

Diode Voltage (V)

Diode Current (A)

Figure 3.9

Graph of the i-v characteristics of a p-n junction diode

+ vD -

iD

= 1

vexpIi D

SD

tηφ

φt = kT/q

Electronics22

0.10.0-0.1-0.2-2.00e-15

0.00e+0

2.00e-15

4.00e-15

6.00e-15

8.00e-15

1.00e-14

1.20e-14

Diode Voltage (V)

Dio

de

Cu

rren

t (A

)

IS

Diode behavior near the origin with IS=10-15 A and η=1

φt = 25 mV

Electronics23

Diode Voltage (V)

1.00.90.80.70.60.50.40.30.20.10.010 -15

10 -14

10 -13

10 -12

10 -11

10 -10

10 -9

10 -8

10 -7

10 -6

10 -5

10 -4

10 -3

10 -2

10 -1

Diode Current (A)

φt = 0.025 V

Slope - 1 decade/60 mV

I = 10 AS

-15

Diode i -v characteristic on semilog scale

IS : saturation current

n : ideality factor ( 1 to 2 )

φt : kT/q

k : 1.38 x10-23 J/K

Typical values of IS:

10-18 A ≤ IS ≤ 10-9 A

= 1

vexpIi D

SD

tηφ

+ vD -

iD

Electronics24

Deviations from the Ideal Diode *

Reverse-biased deviations from ideal

Forward-biased deviations from the ideal

∗ G.W. Neudeck, The PN Junction Diode,

2nd edition, Prentice Hall, 1989

Electronics25

vD

iDIdeal Diode

Characteristic

Ideal Diode Symbol

iD

vD+ -

A

C

"On"

C

A

"Off"

"Short" Circuit

"Open" Circuit

+

-V

10 kΩ

VDID

+

-10 V

R

R

VVI D

D

−=

10k Ω

10VI D =

1mAID =

Ideal diode model

Electronics26

iD

iD

(a) (b)

vD Von

Von vD

iD Ideal Diode +

Source

Characteristic

(c)

- Constant voltage drop model for diode

(a) Actual diode

(b) Ideal diode plus voltage source Von

(c) Composite i-v characteristic

+

-10 V

10 k Ω

I

(a)

+

-10 V

10 k Ω

I D

-

(c)

V = 0.6 Von

I = ID

+

(b)

- Diode circuit analysis using constant voltage drop mode

(a) Original diode circuit

(b) Circuit whit ideal diode replaced by piece-wise linear model

I = 10 – 0.6

10 k Ω

I = 0.94 mA

• Constant-voltage-drop model

Electronics27

+

- R

iD

D1

v = VPsin ω t

S

Half-wave rectifier circuit

0.0200.0150.0100.0050.000-15

-10

-5

0

5

10

15

Time (sec)

Voltage (V)

InputVoltage

OutputVoltage

InputVoltage

OutputVoltage

Half Half-wave rectifier output voltage with VP = 10 V and Von = 0.7 V

Simplified analysis

VS ≤ Von → Diode is ON

VS > Von → Diode is OFF

Electronics28

vO

+

-

+

-

v = VPsin ωt

SC+

-v1

Rectifier with capacitor load (peak detector)

Time

Voltage (V)

V - VP on

0 T 2T

0

vo

Input and output wave forms for the peak detector circuit

0iD ≥ 0dt

dvC O ≥

• The half-wave rectifier: capacitive load

Electronics29

0s 10ms 20ms 30ms 40ms 50ms 60ms

Time

Initial Surge Current

Repetitive Diode Current

0V

Output Voltage

VS

0s

0A

100A

A

V

0V

25V

PSPICE simulation of the half-wave rectifier circuit

17.8V212.6Vp =×=

0.747RC

TVV pr =≈

% 4.4VV pr =

rad 0.29V

2Vω∆Tθ

p

rC ===

f = 60 Hz

Von = 1V

R = 15Ω

C = 25 mF

1.12A15

16.8

R

VI

p

DC ==≅

48.6A∆T

2TII DCp ==

168AωCVI pSC ==

35.6V2VPIV p =≈

Electronics30

VS D1

D2 VX

C1

C2

+

_

VO

•Voltage doubler

Clamping circuit & peak detector

2Vp

Vp

-Vp

2Vp

VO

VS

t

t

Vx

Electronics31

References

– EEL 7061 Eletrônica Básicahttp://www.eel.ufsc.br/electronics/index7061.htm

– Reid R. Harrison, “Analog Integrated Circuit Design”ECE/CS 5720/6720 Department of Electrical and Computer Engineering University of Utah

– Charles Sodini, “6.012 Microelectronic Devices and Circuits”,OpenCourseWarehttp://ocw.mit.edu