CENTURION UNIVERSITY OF TECHNOLOGY AND MANAGEMENT

SCHOOL OF ENGINEERING & TECHNOLOGYDEPARTMENT

OF ELECTRONICS & COMMUNICATION ENGINEERING

ELECTRONIC DEVICES

Section: ECE SEM: II

PART-A

1.

a) In a N-type silicon sample, the electron concentration 2.15×𝟏𝟎𝟏𝟎/𝒄𝒎𝟑. If the intrinsic

carrier concentration is 1.5×𝟏𝟎𝟖/𝒄𝒎𝟑.Then calculate the hole concentration.

Given that electron concentration n= 2.15×1010/𝑐𝑚3

intrinsic carrier concentration 𝑛𝑖=1.5×108/𝑐𝑚3

according to the law of mass action n×p = 𝑛𝑖2

hole concentration p = 𝑛𝑖

2

n =

(1.5×108

𝑐𝑚3)2

2.15×1010/𝑐𝑚3

p = 1.046 ×106/𝑐𝑚3

b) The intrinsic carrier density at 300K is 2.5×𝟏𝟎𝟏𝟓/𝒄𝒎𝟑, in silicon. For p-type silicon

dopped to 1.25×𝟏𝟎𝟏𝟐 atoms/𝒄𝒎𝟑 then find out the equilibrium electron and hole

density.

The question mentioned is wrong because the given doping concentration

is less than intrinsic concentration which is practically impossible.

Mathematically ,if we calculate then the semiconductor will become N-type

c) Determine the value of load resistor.

𝑅𝐿= 12𝑣−0.3𝑣−0.7𝑣

2 mA

𝑅𝐿= 12𝑣−1𝑣

2 mA =

11𝑣

2 mA = 5.5kΩ.

d) What is the ripple, 3.5 𝑽𝑹𝑴𝑺 on average of 120 V?

Given that R.M.S value of ac component 𝑉𝑅𝑀𝑆 = 3.5𝑉

Average value of output voltage 𝑉𝑑𝑐 or 𝑉𝑎𝑣𝑔 =120V

Ripple factor ( ᵧ) =𝑉𝑅𝑀𝑆

𝑉𝑑𝑐 =

3.5 𝑉

120 V = 0.029

e) A dc voltage supply provides 36v when the output is unloaded. When connected to a

load the output drops to 30v. Calculate the value of voltage regulation.

No load voltage 𝑉𝑁𝐿 = 36v

Full load voltage 𝑉𝐹𝐿 = 30v

Voltage regulation = 𝑉𝑁𝐿 − 𝑉𝐹𝐿

𝑉𝑁𝐿

= 36𝑣−30𝑣

36v

= 6 𝑉

36 V

=0.1667

f) If ἀ is 0.98, 𝑰𝑩 is 100 μA ,𝑰𝑪𝑶 is 6 μA .then find out 𝑰𝑪 .

β = ἀ

( 1−ἀ ) =

0.98

( 1−0.98 ) =

98

2 =49.

β = 𝐼𝐶

𝐼𝐵

𝐼𝐶 = β 𝐼𝐵 = (49)(100 μA) = 4.9mA.

𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵 = (4900+100) μA = 5000 μA =5mA.

For common base transistor configuration:

𝑰𝑪𝒏𝒆𝒕 𝒄𝒖𝒓𝒓𝒆𝒏𝒕= 𝑰𝑪𝒎𝒂𝒋𝒐𝒓𝒊𝒕𝒚

+ 𝑰𝑪𝒎𝒊𝒏𝒐𝒓𝒊𝒕𝒚

Where as𝑰𝑪𝒎𝒂𝒋𝒐𝒓𝒊𝒕𝒚= ἀ𝐼𝐸 and 𝑰𝑪𝒎𝒊𝒏𝒐𝒓𝒊𝒕𝒚

= 𝐼𝐶𝐵𝑂

𝐼𝐶 = ἀ𝐼𝐸 + 𝐼𝐶𝐵𝑂 = (0.98×5mA)+(6 μA) =4.906mA.

g) Differentiate between fixed bias and self bias.

Fixed Bias Self Bias

1. S=β + 1.

It has poor stability due to large stability factor.

1. S<(β + 1).

Stability is improved i.e. better than fixed

biasing.

2.Absence of 𝑅𝐸 2. presence of 𝑅𝐸

3.Due to large value of s as β is greater ,circuit

is unstable.

3.Due to less value of s which increases the

stability.

h) “FET is known as voltage controlled device” justify the statement.

It is called as voltage controlled device because the output drain current (𝐼𝐷) is

controlled by input gate to source voltage (𝑉𝐺𝑆).

i) What do you mean by intrinsic standoff ratio of UJT?

The Greek letter η (eta) is called the intrinsic stand-off ratio of the device and is

defined by

η=𝑅𝐵1

𝑅𝐵1 + 𝑅𝐵2|𝐼𝐸 = 0 =

𝑅𝐵1

𝑅𝐵B

j) What the different modes of operation of BJT?

a. saturation mode

b.active mode

c. cut off mode

PART-B

2.

a) Where is 𝑬𝑭 located in the energy band of silicon, at 300K with p=𝟏𝟎𝟏𝟒𝒄𝒎−𝟑 and

𝑵𝑽 =1.40×𝟏𝟎𝟏𝟗.

p= 𝑁𝑉exp [−(𝐸𝐹 − 𝐸𝑉 )

𝐾𝑇]

p=𝑁𝑎 = 1014𝑐𝑚−3 and

𝑁𝑉 =1.40×1019

exp[(𝐸𝐹 − 𝐸𝑉 )

𝐾𝑇] =

𝑁𝑉

p

[(𝐸𝐹 − 𝐸𝑉 )

𝐾𝑇] =ln(

𝑁𝑉

p)

(𝐸𝐹 − 𝐸𝑉 ) = KT ln(𝑁𝑉

p)

(𝐸𝐹 − 𝐸𝑉 ) = (26𝑚𝑒𝑉)ln (1.40×1019

1014𝑐𝑚−3)

(𝐸𝐹 − 𝐸𝑉 ) = 0.31eV.

Hence 𝐸𝐹 is above 𝐸𝑉 by 0.31eV

b) Define mobility, conductivity and diffusion?

Mobility :It is defined as the ratio between the particle drift velocity (𝑽𝒅) per unit electric

field (E) .

Mobility = 𝑽𝒅

E

Conductivity :reciprocal of resistivity. It is the measure of semiconductor’s ability to

conduct current. Depends on mobility of charge carriers and dopant concentration.

Diffusion :It is a process of doping a semiconductor with impurities .It is the net movement

of a substance (ae., an atom, ion or molecule) from a region of high concentration to a region

of low concentration.

3.

a) A sample of N-type semiconductor has a resistivity, of 0.1 Ω-cm and hall

coefficient of 100 𝒄𝒎𝟑/𝒄𝒐𝒖𝒍𝒐𝒎𝒃 .assuming only electrons as carriers determine the

electron density and mobility.

Given that resistivity = 0.1 Ω-cm

hall coefficient 𝑅𝐻= 100 𝑐𝑚3/𝑐𝑜𝑢𝑙𝑜𝑚𝑏

electronic charge (e) = 1.6×10−19 𝑐𝑜𝑢𝑙𝑜𝑚𝑏.

Conductivity = 1

resistivity =

1

0.1 Ω−cm =10 S/cm.

Electron density = 1

𝑅𝐻

= 1

100 𝑐𝑚3/𝑐𝑜𝑢𝑙𝑜𝑚𝑏 = 0.01coulomb/𝑐𝑚3.

Mobility = (Conductivity)×(𝑅𝐻) =(10)(100) =(1000 S 𝑐𝑚2)/ coulomb.

b) Neatly draw and explain the V-I characteristics of a tunnel diode?

Here forward current (𝐼𝐹) increases sharply as applied voltage (𝑉𝐹) increases.

𝐼𝐹increases upto the point ‘A’ on the curve i.e, the peak voltage.

As the F/B is increases beyond this point ,𝐼𝐹 decreases and continuous to drop until the

point B is reached. This is known as “Valley Voltage 𝑉𝑣 ”.

At B, the current starts to increase once again and does so rapidly as bias is increases further.

Beyond this point, the Tunnel Diode will behaves as a P-N junction Diode.

4.

a) A 230v 50Hz ac voltage is applied to the primary of 5:1 step down transformer,

which is used in center tapped rectifier, having load resistance 800ohm assuming the

diodes to be an ideal then determine, ac power input , dc power output ,rectification

efficiency and ripple factor.

𝑽𝟏

𝑽𝟐

= 𝑵𝟏

𝑵𝟐

=>𝑽𝟐 = (𝑵𝟐

𝑵𝟏)𝑽𝟏 = (

1

5)(230)=46v = 𝑽𝒔

Hence the peak value of the secondary voltage (𝑽𝒔)𝒓𝒎𝒔 =(𝑽

𝒔)

𝒎𝒂𝒙

√2

=>(𝑽𝒔)𝒓𝒎𝒔 = (𝑽𝒔)𝒎𝒂𝒙

√2

=46

√2

(𝑽𝒔)𝒓𝒎𝒔 = 32.5269v

A.C input power = 𝐼𝑟𝑚𝑠2 × (𝑹𝑳+ 𝑹𝑭) =

𝐼𝑚𝑎𝑥2

𝟐× (𝑹𝑳 + 𝑹𝑭)

𝑷𝒊𝒏(𝒂.𝒄) = 𝐼𝑚𝑎𝑥

2

𝟐×𝑹𝑳 = (

(𝑽𝒔

)𝒎𝒂𝒙

𝑹𝑳

)2 × 𝑹𝑳

2

𝑷𝒊𝒏(𝒂.𝒄) = ( 𝟒𝟔𝐯

𝟖𝟎𝟎𝛀)2 ×

𝟖𝟎𝟎𝛀

2

=>𝑷𝒊𝒏(𝒂.𝒄) = 1.3225watt

Rectification efficiency (η) = 𝑷𝒅𝒄(𝒐/𝒑)

𝑷𝒂𝒄(𝒊/𝒑)

=𝐼𝑑𝑐

2 × 𝑹𝑳

𝐼𝑟𝑚𝑠2 × (𝑹𝑳 + 𝑹𝑭)

= (

2

3.14𝐼max )2 × 𝑹𝑳

𝐼𝑚𝑎𝑥2

𝟐× (𝑹𝑳 + 𝑹𝑭)

=>η =𝟖∗𝑹𝑳

(𝐩𝐢∗𝐩𝐢)×(𝑹𝑳 + 𝑹𝑭)

=>η =𝟖

(𝐩𝐢∗𝐩𝐢)×(𝟏 + 𝑹𝑭/𝑹𝑳)

Assuming diode is ideal 𝑹𝑳 ≫ 𝑹𝑭

=>η =0.𝟖𝟏𝟐

(𝟏 + 𝑹𝑭/𝑹𝑳) × 100% = 0.812 × 100% = 81.2%

We know Rectification efficiency (η) = 𝑷𝒅𝒄(𝒐/𝒑)

𝑷𝒂𝒄(𝒊/𝒑)

= 0.812

Dc power 𝑷𝒅𝒄(𝒐/𝒑) = 𝑷𝒂𝒄(𝒊/𝒑) × 0.812

= 0.812 × 𝟏. 𝟑𝟐𝟐𝟓𝐰𝐚𝐭𝐭

𝑷𝒅𝒄(𝒐/𝒑) = 1.0738watt

Form factor (𝐾𝐹) = 𝑰𝒓𝒎𝒔

𝑰𝒂𝒗𝒈

= (

𝑰𝒎𝒂𝒙

√2 )

(𝟐𝑰𝒎𝒂𝒙

pi ) =

𝐩𝐢

𝟐 √2 = 1.11

Ripple factor (ᵧ ) = √((𝐾𝐹)2) − 1 = √(1.11 ∗ 1.11) − 1 = 0.482

b) With a neat sketch, explain the zener regulator circuit.

Under reverse bias condition ,the voltage across the diode remains almost constant

although the current through the diode increases.

Hence, the voltage across the zener diode can be used as a reference voltage which

can be used as “ voltage regulator ” or “ zener regulator “.

It is required to provide a constant voltage across load resistancs (𝑅𝐿) where the

input voltage may be varying over a range.

Zener diode is reverse bias and as long as input voltage 𝑉𝑖𝑛not less than 𝑉𝑧 .

Voltage across the diode will be constant.

Load voltage will also constant.

5.

a) A full wave rectifier has a peak output voltage of 50 volts at 50 Hz and uses a

shunt capacitor filter with c= 45 μf .The connected load is of 8kΩ determine, 1) ripple

voltage and 2) form factor.

Given c= 45 μf ,

𝑅𝐿 = 8kΩ,

f = 50Hz

As per formula Ripple factor (ᵧ ) = 𝑉𝑎𝑐(𝑟𝑚𝑠)

𝑉𝑑𝑐

= 𝑉𝑟

2 √3𝐼𝑑𝑐𝑅𝐿

Ripple factor (ᵧ ) = 1

4√3f C𝑅𝐿

= 0.00801875

Maximum voltage across the load (𝑉𝐿)𝑚𝑎𝑥 = (𝑉𝑠)𝑚𝑎𝑥= 50v

We know that 𝑉𝑑𝑐 = 𝑉𝐿 𝑚𝑎𝑥 - (𝐼𝑑𝑐

4 f C)

𝐼𝑑𝑐𝑅𝐿 + (𝐼𝑑𝑐

4 f C) = 𝑉𝐿 𝑚𝑎𝑥

𝐼𝑑𝑐(8000 + 1

4∗50∗45∗ 10−6 ) = 50

𝐼𝑑𝑐 = 0.00616438 Amp = 6.164 mA

𝑉𝑑𝑐 = 𝐼𝑑𝑐 * 𝑅𝐿 = 49.31v

Ripple voltage 𝑉𝑟 = 𝐼𝑑𝑐

2 f C =1.3697v

(𝑉𝑎𝑐)𝑟𝑚𝑠 = 𝑉𝑟

2√3 = 0.3953

Ripple factor (ᵧ ) = 𝑉𝑎𝑐(𝑟𝑚𝑠)

𝑉𝑑𝑐

= 0.3953v

49.312v = 0.0080163

Ripple factor (ᵧ ) =√((form factor)2) − 1

ᵧ = √((𝐾𝐹)2) + 1

𝐾𝐹 = √((ᵧ)2) + 1

𝐾𝐹 = √((0.0080163)2) + 1

𝐾𝐹 = √1.00006426 = 1.

b) Differentiate between Half wave and full wave rectifier.

Half waverectifier Full wave rectifier

S.No Particulars Half waverectifier Center tapped full

wave rectifier

Bridge type full

wave rectifier

1. Number of

diodes

1 2 4

2. Transformer

required

Not essential Essential Not essential

3. 𝐼𝑑𝑐 𝐼𝑀𝐴𝑋

𝜋

2𝐼𝑀𝐴𝑋

𝜋

2𝐼𝑀𝐴𝑋

𝜋

4. 𝐼𝑟𝑚𝑠 𝐼𝑀𝐴𝑋

2

𝐼𝑀𝐴𝑋

√2

𝐼𝑀𝐴𝑋

√2

5. Peak Inverse

Voltage 𝑉𝑀𝐴𝑋 2𝑉𝑀𝐴𝑋 𝑉𝑀𝐴𝑋

6. η 40.6% 81.2% 81.2%

6.

a) In a voltage divider biasing circuit of BJT, R1 =47kΩ ,R2=82k Ω ,Rc=2.2kΩ

,Re=1.2kΩ ,and β =352,Vcc=12V.Draw the load line for the above biasing circuit and

locate the Q-point that is Ic and 𝑉𝐶𝐸.

Given that R1 =47kΩ ,

R2=82k Ω ,

Rc=2.2kΩ ,

Re=1.2kΩ ,and

β =352,

Vcc=12V.

Exact Analysis: on applying thevenins theorem, we have

𝑅𝑇𝐻 = (𝑅1 ||𝑅2 ) = (47kΩ ||82k Ω)

=29.875kΩ

𝑉𝑇𝐻 = 𝑉𝑐𝑐 ×𝑅2

(𝑅1+𝑅2)

= 7.627V

On applying the K.V.L to the input loop equation we have 𝑉𝑇𝐻 − 𝑅𝑇𝐻𝐼𝐵 − 𝑉𝐵𝐸 − 𝑅𝐸𝐼𝐸 = 0

𝑉𝑇𝐻 − 𝑅𝑇𝐻𝐼𝐵 − 𝑉𝐵𝐸 − 𝑅𝐸𝐼𝐵(1 + 𝛽) = 0

𝐼𝐵 = 𝑉𝑇𝐻−𝑉𝐵𝐸

𝑅𝑇𝐻+(1+𝛽)𝑅𝐸

= 0.0153mA

= 15.3159μA .

𝐼𝐶 = 𝐼𝐵𝛽 = 5.391mA

On applying the K.V.L to the output loop equation we have 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 − 𝑉𝐶𝐸 − 𝑅𝐸𝐼𝐸 = 0

𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶(𝑅𝐶 + 𝑅𝐸)

𝑉𝐶𝐸 = (−6.33)𝑉

As –ve signifies that collector to emitter is reverse bias .

Again from the output loop equation we know that 𝑉𝐶𝐶 − 𝑅𝐶𝐼𝐶 − 𝑉𝐶𝐸 − 𝑅𝐸𝐼𝐸 = 0

𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶(𝑅𝐶 + 𝑅𝐸)

𝐼𝐶(𝑠𝑎𝑡) = 𝑉𝐶𝐶

(𝑅𝐶+𝑅𝐸) |𝑉𝐶𝐸 = 0

𝑉𝐶𝐸 = 𝑉𝐶𝐶|𝐼𝐶 = 0

Point A (0,𝑉𝐶𝐶

(𝑅𝐶+𝑅𝐸) )

Therefore Point A = (0, 3.529mA)

Point B (𝑉𝐶𝐶 , 0)

Therefore Point B = (12v,0)

Therefore Point A and Point B provides the load line.

b) Draw the V-I characteristic of a UJT and explain how it could be use as an oscillator.

Upto the peak point P, the diode is reverse bias and hence, the region to the left of the

peak point is called “cutoff region”.

There is a negative resistance region from peak point to valley point.

After the valley point, the device is driven into saturation and behaves like a

conventional forward bias pn junction diode.

The region to the right of the valley point is called “saturation region”.

In the valley point, the resistance changes from negative to positive.

The resistance remains positive in the saturation region.

Due to negative resistance property a UJT can be used to produce sawtooth

waveform generator which is properly known as “UJT Relaxation Oscillator”.

It consists of a UJT and a capacitor which is charged through 𝑅𝐸 as the supply

voltage 𝑉𝐵𝐵 is switched on.

The voltage across the capacitor increases exponentially and 𝑉𝑐 = 𝑉𝑝 and UJT starts

conducting.

After the peak voltage of a UJT is reached, it provides negative resistance to the

discharge path which is useful in working of relaxation oscillator.

As the 𝑉𝑐 =0 , the device is cutoff and capacitor 𝐶𝐸starts to charge again.

This cycle is repeated continuously generating a sawtooth waveform across 𝐶𝐸.

7.

a) N-type semiconductor: A semiconductor in which electrical conduction is due chiefly

to the movement of electrons.

b)LED: Light Emitting Diode is nothing but a P-N junction diode which emits light when it

is forward bias. In all semiconductors or P-N junction diode some of energy is radiated as

heat and photons. Here light is generated by recombination of electrons and holes where by

the excess energy is transferred to a emitted photon. The brightness of emitted light is

transferred to an emitted photon.

c)Avalanche Breakdown:

It is due to thermally generated charge carriers.

It is lightly doped.

It occurs more than 5volts.

Deplection layer will be wider.

d) L-filter: The output of rectifier circuit contains dc and ac components. In order to get

the pure form of dc we are using one of the filter circuit i.e, L-filter circuit.An inductor filter

in which the rectified output signal will be the input for this filter circuit. The property of

inductor will allows to pass dc and blocks ac.

𝜈 =𝑅𝐿

𝜔𝐿3√2

PART-C

8.

a) Write down how to measure the average value of output voltage and ripple factor

by help of a digital multimeter in the laboratory.

Let us assume a rectifier circuit, here to find the output voltage first we are setting the

digital multimeter in voltage measurement mode and measuring the output of rectifier circuit

by placing the multimeter probes (red probe and black probe) across the load resistance

𝑅𝐿side of the rectifier circuit.

b)How to measure the knee voltage and identify the terminals of an isolated diode as

well as the terminals of an isolated BJT only by the help of a digital multimeter.

For Diode: The small diode symbol as the bottom option of the rotating dialWhen set in this

position and hooked up, the diode should be in the “on” state and the display will provide an

indication of the forward-bias voltage such as 0.67 V (for Si).

For Transistor:Digital multimeter provides the level of hFE using the lead sockets appearing

at the bottom left of the dial. The choice of pnp or npn and the availability of two emitter

connections to handle the sequence of leads as connected to the casing. In the diode testing

mode it can be used to check the p-n junctions of a transistor. With the collector open the

base-to-emitter junction should result in a low voltage of about 0.7 V with the red (positive)

lead connected to the base and the black (negative) lead connected to the emitter. A reversal

of the leads should result in to represent the reverse-biased junction. Similarly, with the

emitter open, the forward- and reverse-bias states of the base-to-collector junction can be

checked.

c) Draw the pictorial diagram of a digital multimeter with proper labelling. What are

the parameters that can be measures by help of the digital multimeter

Parameters that we can measure by using digital multimeter are:

Identifying the terminals of diode

Identifying the terminals of transistor

Measuring voltage

Finding knee voltage

Whether the transistor is PNP or NPN

β value of transistor

To find exact resistor values

Measuring current

Depending upon the resistance values we can identify the terminals of

transistor.