Diode is the simplest semiconductor device. It’s a two-terminal device

Diode

Ideal diode conducts current in only one direction and acts like openin the opposite direction

Basic operation

Conduction region - the voltage across the diode is zero- the current near infinite- the diode acts like short

Characteristics of an ideal diode: Conduction region

Non-conduction region - all of the voltages is across the diode- the current is zero- the diode acts like open

Characteristics of an ideal diode: Non-conduction region

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Semiconductor materials

Two types of materials used in development of semiconductor areSilicon(Si) and Germanium(Ge)

Doping is a process to add impurity(such as Antimony(Sb) or Boron(B)) to Si or Ge to make two types of semiconductor materials:

n- type material: make Si(or Ge) more negative (has “free” electrons)p- type material: make Si(or Ge) more positive (has “holes”)

Joining n-type and p-type materials makes a p-n junction diode

P-n junction diode

“electrons” in n-types material migrate across the junction to p-typematerial and forms a so-called “depletion region” around the junction.

No bias condition

Reverse bias condition

Forward bias condition

Actual diode characteristics

Zener region

Resistance levels

•DC or Static resistance

•AC or Dynamic resistance

•Average AC resistance

DC or Static resistance

AC or Dynamic resistanceForward bias region:

• The resistance depends on the amount of current(ID) in the diode.

• The voltage across the diode is fairly constant(26 mA for 25 C).

• rB ranges from a typical 0.1 Ohms for high power device to 2 Ohms for low power, general purpose diodes.

Reverse bias region:

• The resistance is infinite. The diode acts like an open.

AC or Dynamic resistance(cont’d)

Average AC resistance

Diode specification sheets

Other types of diodes

•Zener diode

•Light emitting diode (LED)

•Diode arrays

Zener diode

Light emitting diode (LED)

Diode arrays

Diode equivalent circuits

•Piecewise linear model

•Simplified model

•Ideal device

Piecewise linear model

Simplified model

Ideal device

Determine VDQ, IDQ and VR if 20 1(a) E = V, R = kOhms kk kk k() =1 0 , =1 0

E = VD + VR = VD + RID

Try VD = 0.5ID = (E - VD)/R= 19.5/1000 195= . mA

kkk kkkkkk kk k, D k kk k195 D = 0 .9 VTry VD = 0.9

ID = (E - VD)/R= 19.1/1000 191= . mATherefore,kD 191= . mA, VD 09= . , VR 191= . V

Try VD = 0.5ID = (E - VD)/R= 9.5/10000 095= . mA

From the graph, at ID 095= . mA, VD 055= . VTry VD = 0.55

ID = (E - VD)/R= 9.45/10000 094= . mATherefore,kD = 094. mA, VD 055= . , VR 945= . V

(a) E=20V, R = 1k:

(b) E=10V, R = 10k:

Load-line analysis

Example: Simple diode circuit

Example: Simple diode circuit

Find VDQ, IDQ, and VR if

(a) E = 10 Volts, R = 1 kOhms(b) E = 10 Volts, R = 2 kOhms(c) Repeat (a) using the approximateequivalent model for the Si diode.(d) Repeat (b) using the approximateequivalent model for the Si diode.(e) Repeat (a) using the ideal diode model.(f) Repeat (b) using the ideal diode model.

Example: (a) E = 10 Volts, R = 1 kOhms

Y-intercept = E/R= 10 mA

Slope = -1/ R= 1 mA/V

From the load-line: IDQ = 9.15 mA VDQ = 0.82 V VR = IDQ x R

= 9.15 V

Example: (b) E = 10 Volts, R = 2 kOhms

Y-intercept = E/R= 5 mA

Slope = -1/ R= 0.5 mA/V

From the load-line: IDQ = 4.6 mA VDQ = 0.78 V VR = IDQ x R

= 9.2 V