Chapter 1: Semiconductor material

Semiconductor is a material that has a conductivity level somewhere between that of -----------

and that of -----------.

2 types of semiconductor that commonly used in industries are ---------- and------------. Between

these two, --------------is widely used. This is because of its low cost, stable at high temperature

and low reverse saturation current.

Electron at outermost orbit of atom is called-----------

Si and Ge are element from which group?-------------

Covalent bonding is---------------------------------------------------------------------------------------

2 charge carrier in semiconductor are------------and-----------.

Pure semiconductor is called-------------------. It has poor conductivity because--------------------.

In order to increase the conductivity of the material, semiconductor undergoes doping process.

Doping process is a process that---------------------------------------------------------------------------.

The doped semiconductor is called-------------------semiconductor. (opposite to the pure

semiconductor)

n-type semiconductor is formed by doping with---------------------------. The majority carrier is----

---. The minority carrier is----------------.

p-type semiconductor is formed by doping with---------------------------. The majority carrier is----

---. The minority carrier is----------------.

Chapter 2: PN junction

PN junction is formed by combining--------------with----------------.

The term bias means---------------------------------------------------------------------------------.

Complete the figure A, B and C by drawing

o the carriers (Notethat amount of majority is higher than that of minority carrier)

o The charged dopants (donor or accepter)

o The depletion region (show the difference when zero, forward and reverse bias)

o Direction of current follow.

P-type n-type

A. Zero bias

P-type n-type

B. forward bias

P-type n-type

C. reverse bias

Depletion region is a region where----------------------------------------------------------------.

The barrier potential is the potential difference of the electric field in the depletion region and is

the amount of energy required to move electron through the depletion region. The barrier

potential is also called as knee volatage. Knee voltage for Si and Ge are------------and ----------,

respectively.

Draw the symbol of diode and show the electrical connection for reverse bias.

In the reverse bias, a small current due to the movement of minority carrier can be observed. The

current is called-----------. The direction of the current is -------------------.

Draw the I-V (current-voltage) characteristic for

A. actual Si diode

B. Complete model for Si diode

C. Practical diode for Si diode

D. Ideal diode

At reverse bias, diode can be replace with--------circuit. The current is------.

Circuit analysis

Procedure when analyzing diode circuit:

i) Determine whether diode is ON or OFF. Condition so that diode is ON are------------------------

and -------------------------------------------------------

ii) Based on the result of step i), diode is replace with appropriate equivalent circuit.

iii) Use KVL, KCL and ohm’s law to solve the calculation

Find Vo and current through the diode for figure (a), (b), (c) and (d). Assume diode is ideal.

Find V1 and V2.

Vo

+ 3V

- 3V

(a)

10 kΩ

- 3V - 3V - 3V

10 kΩ

10 kΩ 10 kΩ

Vo Vo Vo

+ 3V + 3V + 3V

(b) (c) (d) Vo=+3V

ID=0.6mA Vo=-3V

ID=0 A

Vo=+3V

ID=0 A

Vo=-3V

ID=0.6mA

Find current through diode..

Find V1 and V2.

Find diode current and V1.

2 kΩ

5 kΩ

V2

V1 Si

Ge

-10 V V1= -9.7 V

V2=-6.93 V

Si

20 V

20 Ω

10 Ω 0.965 A

10 V

V1 V2

5kΩ Si Ge

V1= 9.3 V

V2=0.3 V

Find V1 and V2. Find again V1 and V2 if the polarity of battery is reversed.

Find V1, V2 and I

20 V

5.6 kΩ

2.7 kΩ

3.3 kΩ

Ge

Ge

Si

Si

V1

V2

V1=5.02V

V2=-5.28V

When

reverse,

V1=-4.81V

V2=5.52V

20 V

1 kΩ

1.5 kΩ

5 kΩ

5 kΩ

Si

V1 V1=13.3 V

ID=2.67mA

V2 V1

Si Ge 20 V

1 kΩ 0.47kΩ

I

V1=0.7V

V2=0.3V

I=18.45 mA

Chapter 3: Zener diode

Draw the symbol of zener diode and its I-V characteristics. Explain the difference in comparison

with normal diode.

(a) Determine VL, IL, IZ and IR if RL=180 Ω

(b) Repeat part (a) if RL=470 Ω

(c) Determine the minimum value for RL to ensure that zener diode is in “ON” state.

Referring to the following figure, the zener diode is able to regulate the output voltage for diode

current between 1 mA to 300mA and load current between 0 and 200 mA. Determine the input

voltage range that can be regulated by this circuit.

Vs: 9.62~11.6V

(a)VL=9V,IL=0.05A,IR=0.05A

(b)VL=10V, IL=0.021A, IR=0.04545A

(c) RL min=220 ohm

Chapter 4: Rectifier and power supply circuit

In each silicon diode circuit, vi is 1 kHz 10 V peak sine wave. Draw the the waveform of v0

Draw v0 and iR

halfwave, Vp=9.3V,T=1ms No output signal halfwave (negative cycle),

Vp=-8.6V, T=1ms

Draw vo and the required PIV rating for each diode.

Refer to the following figure. Assume all diodes are ideal, v1= tV sin20 where V0 = 240 volt,

ω = 2πf, f=50Hz and t = time in second (s)

a. Describe clearly the function of Block A, Block B and Block C

b. Find the peak value of v1, v2 and v3

c. Draw and completely label the waveforms of v1, v2 and v3

d. Calculate the average value of v3

e. What is the DC value of v4?

vo is full wave (negative value), Vpp=-100V

PIV=100V

Refer to the rectifier circuit in the following figure, assume that all diodes are ideal, perform the

following tasks:

f. Draw and label completely the output voltage waveform. State the peak value and the

frequency of the waveform.

g. Calculate the average value of the output voltage and the load current.

h. Find the Peak Inverse Voltages of D4 and D3 when Vi is positive

i. If D1 is open circuited, draw and completely label the output voltage waveform.

j. If D4 is open circuited, draw and label completely the output voltage waveform.

k. If the positions of all diodes in the circuit are reversed, draw and label completely the

output voltage waveform.

240V vi

vORL

50Hz

20 : 1

+

_

D1

D3D2

D4 iO

100W

(a)A: Rectifier: DC to AC, B: Filter (to smoothes out) C: Volatge regulator

(regulate output at Vz)

(b) v1=339.4V, v2=33.94V, v3=33.94V

(c) vi: sine wave, Vpeak=339.4 V, T=20ms

v2: sine wave, Vpeak=33.94V, T=20 ms

v3=full wave, Vpeak=33.94 V

(d) average voltage=21.6V

(e) DC value=9V

rms

Refer to the following circuit, assume the diodes are ideal.

(a) What is the name of this circuit?

(b) Draw the waveform of Vout if the input is 100 Hz 10 V peak sine wave.

(c) Determine the PIV

(d) Find the DC voltage across R.

2:1

D1

R

F

+

Vout

-

D2

Vsec/2

Vsec/2

AC

(f) full wave, Vpeak=16.968 V

(g) DC voltage=10.8 V

(h) PIV=16.968 V

(i) half wave (positive cycle “OFF”, negative cycle”ON’)

(j) half wave(positive “ON”, negative”OFF”)

(k) full wave(negative value), Vpeak=16.968 V

(a) Center tap full wave rectifier

(b) full wave, Vpeak=2.5 V

(c) PIV=5 V

(d) DC=1.59