Link to download- https://docs.google.com/document/d/12i1mAKKJbFj92g9wWyRGPC_unyjmo-lE/edit?usp=sharing&ouid=111175923012321593217&rtpof=true&sd=true
EXPERIMENT NO. 1
Object: To study the Hall
Effect and to determine the hall coefficient.
Observation:
Resistivity ‘ρ’ = 0.05 ohm-m.
Table I
Magnetic field, B = 2
Kilo Gauss = 3×10-1 T (1 Gauss = 10-4 T)
|
S.No. |
Current
I (mA) |
Hall voltage VH (mV) |
Hall coefficient (m3/C) |
|
1. |
1 |
-0.6 |
1.5×10-3 |
|
2. |
2 |
-1.1 |
1.375×10-3 |
|
3. |
3 |
-1.7 |
1.417×10-3 |
|
4. |
4 |
-2.3 |
1.438×10-3 |
|
5. |
5 |
-2.8 |
1.40×10-3 |
|
6. |
6 |
-3.3 |
1.375×10-3 |
|
7. |
7 |
-3.8 |
1.357×10-3 |
|
8. |
8 |
-4.3 |
1.344×10-3 |
|
9. |
9 |
-4.9 |
1.361×10-3 |
|
10. |
10 |
-5.5 |
1.375×10-3 |
Table II
Current I = 2 mA
|
S.No. |
Magnetic field |
Hall
voltage VH
(mV) |
Hall coefficient(109mm3/C) |
|
|
k
gauss |
Wb/m2 |
|||
|
1. |
0.5 |
0.05 |
-0.2 |
1.0×10-3 |
|
2. |
1 |
0.1 |
-0.5 |
1.25×10-3 |
|
3. |
1.5 |
0.15 |
-0.7 |
1.167×10-3 |
|
4. |
2 |
0.2 |
-1.0 |
1.25×10-3 |
|
5. |
2.5 |
0.25 |
-1.3 |
1.3×10-3 |
|
6. |
3 |
0.3 |
-1.6 |
1.333×10-3 |
|
7. |
3.5 |
0.35 |
-1.9 |
1.357×10-3 |
|
8. |
4 |
0.4 |
-2.2 |
1.375×10-3 |
|
9. |
4.5 |
0.45 |
-2.5 |
1.389×10-3 |
Calculations:
1. Slope 
= 
= -0.533 V/A
2. Slope 
= 
= -6 x 
V/A
RH1
= (Slope 
)×
= -1.332 x 
m3/C
RH2
= (Slope 
)×
= -1.5 x m3/C
[Note: VH, I, t and B are to be taken in volts,
amperes, meters and Wb/ m2 respectively].
Mean = RH = 
= 
= 
= 
m3/C.
Result:
1. The polarity of the
Hall Voltage is Negative. Therefore, the given semiconductor is of N
Type.
2. The value of Hall
coefficient is 
m3/C.
EXPERIMENT NO. 2
Object:
To determine wavelength of LASER light using
diffraction phenomena.
Observation
for Grating Diffraction:
Number
of lines on the diffraction grating = a = 300 per mm
N = a×10 = 3000 per cm
Grating element
(a+b) = 1/N = (1/3) ×10-3 cm
Distance
between grating and screen = r = 25.5 cm
|
S.No. |
Order of Spectrum (n) |
Distance between Bright Spots (cm) |
Angle of Diffraction nθ = (radian) |
θ = |
sinθ |
nλ = (a+b) sin θ |
||||
|
|
|
On one side (x1) |
On other side (x2) |
Mean x |
|
|
|
|
||
|
1. |
For 1st order n = 1 |
5 |
5.1 |
5.05 |
0.1980 |
11.35 |
0.196 |
0.6468 × 10-4 |
||
|
2. |
For 2nd order n = 2 |
10 |
10.5 |
10.25 |
0.4019 |
23.03 |
0.391 |
1.2903 × 10-4 |
||
Calculations:
For n = 1;
nλ = 0.6468 × 10-4 cm
= 0.6468
× 10-4 cm
For n = 2;
nλ = 1.2903 × 10-4 cm
= 1.2903
× 10-4 / 2 = 0.6451 × 10-4 cm
Mean λ = 
= 
cm
λ = 645.95 nm
Relative % error = 
× 100 = − 2.07%
Result:
By diffraction grating = λ ± δλ = 645.85 ± 2.07% nm
EXPERIMENT
NO. 3
Object: To sketch the following basic op-amp
circuits using ME 625 and verify then theoretically –
a. Inverting
amplifier
b. Non-inverting
amplifier
c. Summing
amplifier
d. Differential
amplifier
Inverting Amplifier:
Observation
Table 1:
Vin = 1 volt
|
|
Vout |
Vout |
Vout |
Vout |
||||
|
R1 |
RF
= 10 kΩ |
RF
= 22 kΩ |
RF
= 33kΩ |
RF
= 47 kΩ |
||||
|
|
Experimental |
Calculated |
Experimental |
Calculated |
Experimental |
Calculated |
Experimental |
Calculated |
|
10 |
-0.99 |
-1.0 |
-2.16 |
-2.2 |
-3.24 |
-3.3 |
-4.67 |
-4.7 |
|
22 |
-0.44 |
-0.45 |
-1.0 |
-1.0 |
-1.45 |
-1.5 |
-2.10 |
-2.13 |
|
33 |
-0.30 |
-0.30 |
-0.65 |
-0.66 |
-1.0 |
-1.0 |
-1.40 |
-1.42 |
|
47 |
-0.20 |
-0.21 |
-0.45 |
-0.46 |
0.67 |
-0.7 |
-1.0 |
-1.0 |
|
Avg |
0.4825 |
0.49 |
1.086 |
1.106 |
1.786 |
1.83 |
2.72 |
2.752 |
Non-Inverting
Amplifier:
Observation
Table 2:
|
|
Vout |
Vout |
Vout |
Vout |
||||
|
R1 |
RF
= 10 kΩ |
RF
= 22 Kω |
RF
= 33kΩ |
RF
= 47 kΩ |
||||
|
|
Experimental |
Calculated |
Experimental |
Calculated |
Experimental |
Calculated |
Experimental |
Calculated |
|
10 |
2.6 |
2.0 |
3.2 |
3.2 |
4.27 |
4.3 |
5.72 |
5.7 |
|
22 |
1.45 |
1.45 |
- |
2 |
2.46 |
2.5 |
3.12 |
3.13 |
|
33 |
1.28 |
1.30 |
1.65 |
1.66 |
- |
2 |
2.40 |
2.42 |
|
47 |
1.19 |
1.21 |
1.45 |
1.46 |
1.68 |
1.7 |
- |
2 |
|
Avg |
1.63 |
1.49 |
2.116 |
2.106 |
2.80 |
2.83 |
3.74 |
3.75 |
Observation Table 3
|
Vin1 |
Vin2 |
Vout |
|
|
Experimental |
Calculated |
||
|
1 V |
1 V |
-1.97 |
-2.0 |
|
2 V |
2 V |
-4.01 |
-4.0 |
|
3 V |
3 V |
-5.92 |
-6.0 |
|
4 V |
4 V |
-7.95 |
-8.0 |
Difference Amplifier:
Observation Table 4
|
Vin1 |
Vin2 |
Vout |
|
|
|
|
Experimental |
Calculated |
|
1 V |
1 V |
1.01 |
1 |
|
2 V |
2 V |
1.49 |
2 |
|
3 V |
3 V |
2.95 |
3 |
|
4 V |
4 V |
4.00 |
4 |
|
AVG |
2.3625 |
2.5 |
|
EXPERIMENT
NO. 4
Object: To study the input and output characteristic
of PNP transistor in CE mode and to calculate h parameters of a transistor.
Observation:
Input characteristics: At constant Vce = 0V, 1V,
2V, 3V
|
Sr. No. |
Vce = 00 V |
Vce = 0.5 V |
Vce = 1.0 V |
Vce = 1.5 V |
||||
|
Emitter Base voltage Vbe |
Base current Ib (μA) |
Emitter Base voltage Vbe |
Base current Ib (μA) |
Emitter Base voltage Vbe |
Base current Ib (μA) |
Emitter Base voltage Vbe |
Base current Ib (μA) |
|
|
1. |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
2. |
0.2 |
0 |
0.2 |
0 |
0.2 |
0 |
0.2 |
0 |
|
3. |
0.4 |
0 |
0.4 |
0 |
0.4 |
0 |
0.4 |
0 |
|
4. |
0.6 |
5 |
0.6 |
2.5 |
0.6 |
2.5 |
0.6 |
2 |
|
5. |
0.8 |
10 |
0.8 |
8 |
0.8 |
7 |
0.8 |
7 |
|
6. |
1.0 |
17.5 |
1.0 |
15 |
1.0 |
15 |
1.0 |
15 |
|
7. |
1.2 |
20 |
1.2 |
19 |
1.2 |
18 |
1.2 |
18 |
|
8. |
1.4 |
29 |
1.4 |
25 |
1.4 |
25 |
1.4 |
25 |
|
9. |
1.6 |
36 |
1.6 |
35 |
1.6 |
35 |
1.6 |
35 |
|
10. |
1.8 |
44 |
1.8 |
41 |
1.8 |
40 |
1.8 |
40 |
|
Sr. No. |
Ib =
25μA |
Ib =
50μA |
Ib =
100μA |
Ib =
150μA |
||||
|
|
Collector
Emitter Voltage (Vce) |
Collector
current Ic (mA) |
Collector
Emitter Voltage (Vce) |
Collector
current Ic (mA) |
Collector
Emitter Voltage (Vce) |
Collector
current Ic (mA) |
Collector Emitter
Voltage (Vce) |
Collector
current Ic (mA) |
|
1. |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
2. |
0.1 |
0.5 |
0.1 |
1 |
0.1 |
3.5 |
0.1 |
4 |
|
3. |
0.15 |
2 |
0.15 |
3.5 |
0.15 |
6 |
0.15 |
9 |
|
4. |
0.2 |
2.5 |
0.2 |
5 |
0.2 |
8.5 |
0.2 |
13.5 |
|
5. |
0.25 |
2.5 |
0.25 |
5.5 |
0.25 |
10 |
0.25 |
15 |
|
6. |
0.3 |
2.5 |
0.3 |
5.5 |
0.3 |
10.5 |
0.3 |
16 |
|
7. |
0.35 |
2.5 |
0.35 |
5.5 |
0.35 |
10.5 |
0.35 |
16 |
|
8. |
0.4 |
2.5 |
0.4 |
5.5 |
0.4 |
10.5 |
0.4 |
16 |
|
9. |
0.45 |
2.5 |
0.45 |
5.5 |
0.45 |
10.5 |
0.45 |
16 |
|
10. |
0.5 |
2.5 |
0.5 |
5.5 |
0.5 |
10.5 |
0.5 |
16 |
Calculations:
Input
Impedance:
For the
graph VCE = 0V;
Ri =
(1.0-0.8)/ (17.5-10) µA = 26.6×10-3Ω
Output
Impedance:
For the
graph Ib = 50µA;
Ro =
(0.20-0.15)/(5.5-3.5) mA = 0.025×103 Ω
Result:
h-parameters
of a transistor in CE configuration are:
Input
Impedance: h11 = hie = 26.6×10-3 Ω
Output
Impedance: h22 = hoe = 0.025×10-3 Ω
EXPERIMENT:
5 (A)
Object: To study characteristics of a Field
Effect Transistor (FET)
Observation Table:
·
Output Characteristic:
|
Sr. No. |
Vgs =
- 0. 2 Volts |
Vgs =
- 0.4 Volts |
Vgs =
- 0.8 Volts |
|||
|
Vds (volts) |
Id (mA) |
Vds (volts) |
Id (mA) |
Vds (volts) |
Id (mA) |
|
|
1. |
0.5 |
0.4 |
0.5 |
0.4 |
0.5 |
0.4 |
|
2. |
1.5 |
1.1 |
1.0 |
0.7 |
1.0 |
0.8 |
|
3. |
2.0 |
1.7 |
1.5 |
1.2 |
1.5 |
1.2 |
|
4. |
2.5 |
2.0 |
2.0 |
1.6 |
2.0 |
1.6 |
|
5. |
3.5 |
2.6 |
2.5 |
1.8 |
2.5 |
1.8 |
|
6. |
4.0 |
2.9 |
3.0 |
2.3 |
3.0 |
2.1 |
|
7. |
5.0 |
3.2 |
4.0 |
2.8 |
3.5 |
2.4 |
|
8. |
6.0 |
3.4 |
4.5 |
2.9 |
4.0 |
2.6 |
|
9. |
8.0 |
3.5 |
6.0 |
3.0 |
4.5 |
2.7 |
|
10. |
10.0 |
3.6 |
|
|
5.0 |
2.8 |
·
Transfer Characteristic:
|
S. No. |
Vds (volt) =
5 V |
Vds (volt) =
10 V |
||
|
Vgs (volt) |
Id (mA) |
Vgs (volt) |
Id (mA) |
|
|
1 |
0.2 |
3.0 |
0.2 |
3.4 |
|
2 |
0.4 |
2.9 |
0.4 |
3.2 |
|
3 |
0.8 |
2.7 |
0.8 |
3.0 |
|
4 |
1.0 |
2.6 |
1.0 |
2.8 |
|
5 |
1.2 |
2.4 |
1.2 |
2.7 |
|
6 |
1.4 |
2.3 |
1.4 |
2.5 |
|
7 |
1.8 |
2.0 |
1.8 |
2.3 |
|
8 |
2.0 |
1.8 |
2.0 |
2.0 |
|
9 |
2.2 |
1.7 |
2.2 |
1.9 |
|
10 |
2.4 |
1.6 |
2.4 |
1.7 |
Note: Only one transfer characteristics curve
is to be plotted on one graph paper.
Graphs: Output characteristics: Plot graphs of
Id against Vds on constant Vgs, one graph
paper.
Graphs: Transfer characteristics: Plot graphs
of Id against Vgs on a separate graph paper.
Results: The drain characteristics are similar
to that of a pentode value. The current is almost independent of drain voltage
above ‘PINCH OFF’. The input resistance is essentially that of a reverse biased
p-n (gate of source) junction, therefore is very high.
Transfer characteristic is similar to
that of a vacuum triode.
EXPERIMENT NO. 5 (B)
Object: To study output and transfer characteristics of a Metal Oxide Silicon
Field Effect Transistor (MOSFET) and to identify type of MOSFET.
Observation Table:
- Output
Characteristic:
|
S. No. |
Vgs
(volt)
= 3.2 Volts |
Vgs
(volt)
= 3.3 Volts |
Vgs
(volt)
= 3.4 Volts |
|||
|
Vds (volt) |
Id (mA) |
Vds (volt) |
Id (mA) |
Vds (volt) |
Id (mA) |
|
|
1. |
0.05 |
0.1 |
0.05 |
0.15 |
0.05 |
0.35 |
|
2. |
0.1 |
0.1 |
0.1 |
0.2 |
0.1 |
0.60 |
|
3. |
0.15 |
0.1 |
0.15 |
0.25 |
0.15 |
0.80 |
|
4. |
0.2 |
0.1 |
0.2 |
0.25 |
0.2 |
0.80 |
|
5. |
0.25 |
0.1 |
0.25 |
0.25 |
0.25 |
0.80 |
·
Transfer Characteristic:
|
S. No. |
Vds (volt) = 2 Volts |
Vds
(volt)
= 5 Volts |
||
|
|
Vgs (volt) |
Id (mA) |
Vgs (volt) |
Id (mA) |
|
1. |
0 |
0 |
0 |
0 |
|
2. |
0.2 |
0 |
1.5 |
0 |
|
3. |
0.4 |
0 |
2.5 |
0 |
|
4. |
2,5 |
0 |
3.0 |
0.3 |
|
5. |
3.0 |
0.1 |
3.4 |
1.0 |
|
6. |
3.4 |
1.0 |
3.6 |
3.6 |
|
7. |
3.6 |
1.3 |
4.0 |
4.0 |
|
8. |
3.9 |
1.3 |
4.2 |
4.0 |
|
9. |
5.0 |
1.3 |
4.4 |
4.0 |
|
10. |
_ |
_ |
4.6 |
4.2 |
Graphs: Output
Characteristic: Plot graph between Vds and Id keeping Vgs constant.
Graphs: Transfer
Characteristic: Plot drain current Id against Vgs on a separate
graph paper.
Results:
Cut-off region
is a region in which
the MOSFET will
be OFF as there will be no current flow through it. Ohmic or linear
region is a region where
in the current
IDS increases with an increase in the value of VDS. When
MOSFET's are made
to operate in this region,
they can be used as amplifiers.
In saturation region, the MOSFETs have their IDS constant in spite
of an increase in VDS and occurs
once VDS exceeds
the value of pinch-off voltage
VP.
EXPERIMENT
NO. 6
Object: To study the V-I characteristics of a
Uni-Junction Transistor (UJT).
Diagram:
UJT
Characteristics Apparatus & UJT as Relaxation Oscillator
Observation Table:
|
S.No. |
VB2B1=5V |
VB2B1=10 V |
VB2B1=15 V |
|||
|
VE(V) |
IE (mA) |
VE
(V) |
IE (mA) |
VE
(V) |
IE (mA) |
|
|
1 |
0
V |
0
mA |
0
V |
0
mA |
0
V |
0
mA |
|
2 |
0.5
V |
0
mA |
0.5
V |
0
mA |
0.5
V |
0.025
mA |
|
3 |
1.25
V |
0.1
mA |
1.25
V |
0.075
mA |
1.25
V |
0.075
mA |
|
4 |
2
V |
0.125
mA |
2
V |
0.125
mA |
2
V |
0.125
mA |
|
5 |
2.5
V |
0.15
mA |
2.5
V |
0.15
mA |
2.5
V |
0.150
mA |
|
6 |
3
V |
0.2
mA |
3
V |
0.175
mA |
3.75
V |
0.225
mA |
|
7 |
1.25
V |
10
mA |
3.75
V |
0.225
mA |
5
V |
0.3
mA |
|
8 |
1.25
V |
15
mA |
6.25
V |
0.375 mA |
6.25
V |
0.375
mA |
|
9 |
1.25
V |
20
mA |
1.25V |
5
mA |
7.5
V |
0.45
mA |
|
10 |
|
|
1.25
V |
10
mA |
8.75
V |
0.525
mA |
|
11 |
|
|
1.25
V |
15
mA |
10
V |
0.575
mA |
|
12 |
|
|
|
|
1.25
V |
5
mA |
|
13 |
|
|
|
|
1.25
V |
10
mA |
|
14 |
|
|
|
|
1.25
V |
15
mA |
|
15 |
|
|
|
|
1.25
V |
20
mA |
Characteristics of UJT:
In figure shows the curve between
emitter voltages (VE) and the emitter current (IE) of a UJT at a
given voltage VBB between the bases. This is known as the emitter
characteristic:
1.
Initially, in the cutoff region,
as VE increases from zero, slight leakage current flows from terminal B2 to the
emitter. This current is due to the minority carriers in the reverse biased diode.
2.
Above a certain value of VE,
forward current IE begins to flow, increasing until the peak voltage
VP and current IP are reached at point P.
3.
After the peak point P, at
attempt to increase VE is followed by a sudden increase in emitter
current IE with a corresponding decrease in VE. Then
negative portion of the curve because with increase in IE, VE
decreases.
4. The
negative portion of the curve lasts until the valley point V is reached with
valley point voltage Vv and valley point current Iv. After the valley point,
the device is to saturation.
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