# Physics II Practicals

EXPERIMENT NO. 1

Object: To study the Hall Effect and to determine the hall coefficient.

# 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 = × (degree) 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 Summing Amplifier:

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.