Semiconductor Diodes

SEMICONDUCTOR, DIODE AND POWER SUPPLIES

CHAPTER TWO

SEMICONDUCTOR

Meaning of SemiconductorNeither a conductor nor an insulator but rather halfway in between the two.The resistive properties of a semiconductor can be varied between those a conductor and those of an insulator.

Three most commonly used semiconductor materials are silicon (Si), Germanium (Ge) and Carbon (C).

Si and Ge widely used in the production of solid state components.

Atomic structure of (a) silicon; (b) germanium; and (c) gallium and arsenic.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

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Energy Level

Intrinsic and Extrinsic

Intrinsic- pure material: semiconductor which has a very low level of impurities.

Pure Si and Ge are poor conductor due to partially to the number of valence electrons, covalent bonding and relatively large energy gap.

Extrinsic- those semiconductor that has been subjected to doping process and no longer pure

Doping- is the process of adding impurity atoms to intrinsic Si or Ge to improve the conductivity of the semiconductor.

N-Type material

N stand for negative charge of electron

N type is created by adding with five valence electron into a pure Si or Ge base.

P-Type material

P stand for positive charge of holes

P created by adding with three valence electron into pure Si and Ge base

PN junction

A p–n junction with no external bias. (a) An internal distribution of charge; (b) a diode symbol, with the defined polarity and the current direction; (c) demonstration that the net carrier flow is zero at the external terminal of the device

when VD = 0 V.

Diode

Forward-biased p–n junction. (a) Internal distribution of charge under forward-bias conditions; (b) forward-bias polarity and direction of resulting current.

Forward and Reverse bias

Series diode configuration.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Fig. 2.9 (a) Determining the state of the diode of Fig. 2.8; (b) substituting the equivalent model for the “on” diode of Fig. 2.9a.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Fig. 2.9 (continued) (a) Determining the state of the diode of Fig. 2.8; (b) substituting the equivalent model for the “on” diode of Fig. 2.9a.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Fig. 2.12 Substituting the equivalent model for the “off” diode of Fig. 2.10.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Fig. 2.11 Determining the state of the diode of Fig. 2.10.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Fig. 2.10 Reversing the diode of Fig. 2.8.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Fig. 2.13 Circuit for Example 2.4.

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Fig. 2.14 Determining the unknown quantities for Example 2.5.  

Robert L. BoylestadElectronic Devices and Circuit Theory, 9e

Copyright ©2006 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.

Load line Analysis

Diode Characteristic

Load line and operating point

Steps:

To find load line From equation V=Vd+IR,

Let Vd=0, so Id=V/R

Let Id = 0, so V=Vd Draw a straight line between these two values.

The intersection between the load line and characteristic curve is the operating point.

Example 1

Approximate Model

Ideal Diode

Find the Operating Point using graphical method or network

Parallel Configuration of Diode

Example 2

Example 3

Solution I1 = Vk2 / 3.3k

= 0.7 / 3.3k= 0.212mA

Find V2, using KVL-V+Vk1+Vk2+V2=0

V2=V-Vk1-Vk2 = 20-0.7-0.7 = 18.6V

Find I2 using Ohm’s LawI2=V2 / R2 = 18.6 / 5.6k

=3.32mA

To find IdId=I2-I1 = 3.32mA- 0.212mA = 3.11mA