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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
Www.cpe.kmutt.ac.th/~sanan
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.
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.
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:
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