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Semiconductor

Diodes

Energy Band Gap in Materials

2

Extrinsic Semiconductors

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N-type P-type

Typical doping level 1 part per 10 million

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Semiconductor Diode

The semiconductor diode is formed by bringing p and n-type materials together (constructed from the same base Ge or Si)

Diode Symbol

P-N Junction (barrier) Voltage

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Diode equation

where is the reverse saturation current

is the applied forward-bias voltage across the diode

is an which is a function of operating conditions

( 1)

ideality factor

D T

s

D

V nVsD

I

V

n

I I e

-23

T

is Boltzman constant=1.38 10 J/K

is the absolute temperature in Ke

and physical construction; it has a range between 1 &2.

(n=1 will be assumed unless otherwise noted)

V

k

T

kT

q

-19

lvins

is the electronic charge=1.6 10

For 0, and fo 0, r D TD DV nV

s sD D

q C

V VI I e I I

VT 0.26 V at room temperature

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V-I characteristic of Diode

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Temperature Effect

The reverse saturation current Is approximately doubles for every 10oC rise in temperature. If Is = Is1 at T = T1, then at temperature T2, Is2 is given by, Is2 = Is1x 2

(T2 T1)/10

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PIV (Peak Inverse Voltage Rating)

The maximum reverse-bias potential that can be applied to the diode without damaging it or causing it to break down.

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Diode Equivalent Circuit

Ideal Diode:

Piecewise Linear Model

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Diode Equivalent Circuit

Simple Diode Model Typically Used:

VD

ID

VK

VK=0.7 V for Si

VK=0.3 V for Ge

VSVK I=(VS-VK)/R and VA= VS-VK

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A more complicated model for a diode (Do not use this

for your circuit analysis)

Follows the actual diode characteristics better

What will be the states

(ON/OFF) of the two

diodes, D1 and D2?

Assume the diodes to

be ideal with a forward

voltage drop of 0.7 V.

Find the currents ID1

and ID2

Answer:

D1 ON & D2 OFF

ID1=0.953 mA ID2=0

Example 1

10 V

D1

D2

R1

4 K

R2

5 K

V1I1

I2 I3

V2

R3

1 K

For the circuit shown, find the

voltages V1 and V2 and the

currents I1, I2 and I3. Assume

the diodes to be ideal with a

forward voltage drop of 0.7 V.

Answer:

V1=9.3 V V2=7.53 V

I1=1.51 mA I2 =1.07 mA

I3=0.44 mA

Example 2

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Zener Diode

Zener diodes are special diodes manufactured

with adequate power dissipation capabilities to

operate in the breakdown region.

Symbol:

Equivalent ckt. Approx. Eq. ckt.

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V-I Characteristics of Zener Diode

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Zener Regulator

Loaded Zener regulator

Vs>Vz for breakdown Rs is the current limiting resistance to limit the zener current to less than its maximum rating

Is=(Vs Vz)/Rs

Vth = voltage across zener when it is not in breakdown = Vs x RL/(Rs+RL) For breakdown, Vth>Vz

IL = VL/RL = VZ/RL Iz = Is IL Power dissipated by zener diode = VzIz Is=(Vs Vz)/Rs, Is = Iz + IL

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Zener Regulator contd.

Loaded Zener regulator

Typically, for a Zener Diode, one would specify the zener voltage VZ and the maximum power dissipation PZ,max in the zener diode.

Is=(Vs Vz)/Rs, Is = Iz + IL

Additionally, we may also

specify the minimum zener

current IZ,min that must flow

through the zener diode to

provide the zener action