Download Given that electron concentration n= 2.15×10 / intrinsic carrier

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 = 𝑛𝑖 2
hole concentration p =
𝑛𝑖 2
n
=
108 2
)
𝑐𝑚3
2.15×1010 /𝑐𝑚3
(1.5×
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
ᵧ
𝑉𝑅𝑀𝑆
Ripple factor ( ) =
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
6𝑉
36 V
=0.1667
f) If ἀ is 0.98, 𝑰𝑩 is 100 μA ,𝑰𝑪𝑶 is 6 μA .then find out 𝑰𝑪 .
β=
β=
ἀ
( 1−ἀ )
=
0.98
( 1−0.98 )
=
98
2
=49.
𝐼𝐶
𝐼𝐵
 𝐼𝐶 = β 𝐼𝐵 = (49)(100 μA) = 4.9mA.
 𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵 = (4900+100) μA = 5000 μA =5mA.
For common base transistor configuration:
𝑰𝑪 𝒏𝒆𝒕 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 = 𝑰𝑪 𝒎𝒂𝒋𝒐𝒓𝒊𝒕𝒚 + 𝑰𝑪 𝒎𝒊𝒏𝒐𝒓𝒊𝒕𝒚
Where as𝑰𝑪 𝒎𝒂𝒋𝒐𝒓𝒊𝒕𝒚 = ἀ𝐼𝐸
and 𝑰𝑪 𝒎𝒊𝒏𝒐𝒓𝒊𝒕𝒚 = 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.
1.
S<(β + 1).
It has poor stability due to large stability factor. 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 3.Due to less value of s which increases the
is unstable.
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=𝑅
𝐵1
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=𝟏𝟎𝟏𝟒 𝒄𝒎−𝟑 and
𝑵𝑽 =1.40×𝟏𝟎𝟏𝟗 .
p= 𝑁𝑉 exp
[
]
−(𝐸𝐹 − 𝐸𝑉 )
𝐾𝑇
p=𝑁𝑎 = 1014 𝑐𝑚−3 and
𝑁𝑉 =1.40×1019

exp

[
[
]=
(𝐸𝐹 − 𝐸𝑉 )
𝐾𝑇
(𝐸𝐹 − 𝐸𝑉 )
𝐾𝑇
] =ln(
𝑁𝑉
p
𝑁𝑉
p
)
 (𝐸𝐹 − 𝐸𝑉 ) = KT ln(
𝑁𝑉
p
)
1.40×1019
 (𝐸𝐹 − 𝐸𝑉 ) = (26𝑚𝑒𝑉)ln( 14 −3 )
10 𝑐𝑚
 (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
 Electron density =
1
𝑅𝐻
=
=
1
0.1 Ω−cm
1
100 𝑐𝑚3 /𝑐𝑜𝑢𝑙𝑜𝑚𝑏
=10 S/cm.
= 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.
𝐼𝐹 increases 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 = (𝑵𝑵𝟐)𝑽𝟏 = (
𝟏
1
)(230)=46v = 𝑽𝒔
5
(𝑽𝒔 )
Hence the peak value of the secondary voltage (𝑽𝒔 )𝒓𝒎𝒔 =
𝒎𝒂𝒙
√2
(𝑽𝒔 )𝒎𝒂𝒙
=>(𝑽𝒔 )𝒓𝒎𝒔 =
√2
46
=
√2
(𝑽𝒔 )𝒓𝒎𝒔 = 32.5269v
2
A.C input power = 𝐼𝑟𝑚𝑠 × (𝑹𝑳 + 𝑹𝑭 ) =
𝑷𝒊𝒏(𝒂.𝒄) =
𝑷𝒊𝒏(𝒂.𝒄) =
𝐼𝑚𝑎𝑥 2
𝟐
(
×𝑹𝑳 = (
𝟒𝟔𝐯
𝟖𝟎𝟎𝛀
)2 ×
(𝑽𝒔 )
𝒎𝒂𝒙
𝑹𝑳
)2 ×
𝐼𝑚𝑎𝑥 2
𝟐
× (𝑹𝑳 + 𝑹𝑭)
𝑹𝑳
2
𝟖𝟎𝟎𝛀
2
=>𝑷𝒊𝒏(𝒂.𝒄) = 1.3225watt
Rectification efficiency (η) =
𝑷𝒅𝒄(𝒐/𝒑)
𝑷𝒂𝒄(𝒊/𝒑)
=
𝐼𝑑𝑐 2 × 𝑹𝑳
𝐼𝑟𝑚𝑠 2 × (𝑹𝑳 + 𝑹𝑭 )
=
(
2
𝐼
)2 × 𝑹𝑳
3.14 max
𝐼𝑚𝑎𝑥 2
× (𝑹𝑳 + 𝑹𝑭 )
𝟐
=>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) =
Dc power 𝑷𝒅𝒄(𝒐/𝒑) =
𝑷𝒅𝒄(𝒐/𝒑)
𝑷𝒂𝒄(𝒊/𝒑)
𝑷𝒂𝒄(𝒊/𝒑) ×
= 0.812
0.812
= 0.812 × 𝟏. 𝟑𝟐𝟐𝟓𝐰𝐚𝐭𝐭
𝑷𝒅𝒄(𝒐/𝒑)
= 1.0738watt
Form factor (𝐾𝐹 ) =
𝑰𝒓𝒎𝒔
𝑰𝒂𝒗𝒈
𝑰
( 𝒎𝒂𝒙 )
=
√2
𝟐𝑰𝒎𝒂𝒙
(
pi
)
= 𝟐 𝐩𝐢√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 𝑉𝑖𝑛 not 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
Ripple factor (ᵧ ) =
=
𝑉𝑟
2 √3𝐼𝑑𝑐 𝑅𝐿
1
4√3f C𝑅𝐿
= 0.00801875
Maximum voltage across the load
We know that 𝑉𝑑𝑐 = 𝑉𝐿 𝑚𝑎𝑥 - (
𝐼𝑑𝑐 𝑅𝐿 + (
𝐼𝑑𝑐 (8000 +
𝐼𝑑𝑐
4fC
(𝑉𝐿 )𝑚𝑎𝑥 = (𝑉𝑠 )𝑚𝑎𝑥 = 50v
𝐼𝑑𝑐
4fC
)
) = 𝑉𝐿 𝑚𝑎𝑥
1
4∗50∗45∗ 10−6
) = 50
𝐼𝑑𝑐 = 0.00616438 Amp = 6.164 mA
𝑉𝑑𝑐 = 𝐼𝑑𝑐 * 𝑅𝐿 = 49.31v
Ripple voltage 𝑉𝑟 =
(𝑉𝑎𝑐 )𝑟𝑚𝑠 =
𝑉𝑟
2√3
𝐼𝑑𝑐
2fC
=1.3697v
= 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
S.No Particulars
1.
3.
Number of
diodes
Transformer
required
𝐼𝑑𝑐
4.
𝐼𝑟𝑚𝑠
2.
5.
6.
Peak Inverse
Voltage
η
Half waverectifier
1
Not essential
Full wave rectifier
Center tapped full
wave rectifier
2
Essential
Bridge type full
wave rectifier
4
Not essential
𝐼𝑀𝐴𝑋
𝜋
2𝐼𝑀𝐴𝑋
𝜋
2𝐼𝑀𝐴𝑋
𝜋
𝐼𝑀𝐴𝑋
2
𝐼𝑀𝐴𝑋
𝐼𝑀𝐴𝑋
√2
√2
𝑉𝑀𝐴𝑋
2𝑉𝑀𝐴𝑋
𝑉𝑀𝐴𝑋
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 − 𝑅𝑇𝐻 𝐼𝐵 − 𝑉𝐵𝐸 − 𝑅𝐸 𝐼𝐸 = 0

𝑉𝑇𝐻 − 𝑅𝑇𝐻 𝐼𝐵 − 𝑉𝐵𝐸 − 𝑅𝐸 𝐼𝐵 (1 + 𝛽) = 0

𝐼𝐵 =
𝑉𝑇𝐻 −𝑉𝐵𝐸
𝑅𝑇𝐻 +(1+𝛽)𝑅𝐸
= 0.0153mA
= 15.3159μA .

𝐼𝐶 = 𝐼𝐵 𝛽 = 5.391mA
On applying the K.V.L to the output loop equation we have 𝑉𝐶𝐶 − 𝑅𝐶 𝐼𝐶 − 𝑉𝐶𝐸 − 𝑅𝐸 𝐼𝐸 = 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 − 𝑅𝐶 𝐼𝐶 − 𝑉𝐶𝐸 − 𝑅𝐸 𝐼𝐸 = 0
𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 (𝑅𝐶 + 𝑅𝐸 )
𝐼𝐶(𝑠𝑎𝑡) =
𝑉𝐶𝐶
(𝑅𝐶 +𝑅𝐸)
|
𝑉𝐶𝐸 = 0
|
𝑉𝐶𝐸 = 𝑉𝐶𝐶 𝐼𝐶 = 0
Point A (0,(𝑅
𝑉𝐶𝐶
𝐶 +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 𝐶𝐸 starts 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
𝑅𝐿 side 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:
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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.