Government Polytechnic, Muzaffarpurgpmuz.bih.nic.in/docs/BEE III.pdf · Basic Electronics Lab...

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Government Polytechnic, Muzaffarpur

Basic Electronics Lab

Subject Code: 1620308

1. Aim of experiment

To plot Forward & Reverse biased characteristics of diode

2. Apparatus required

a) A diode

b) ADC voltage supplier

c) Bread board

d) 1000 Ω resistor

e) 2 multi meter for measuring current and voltage

f) Connecting wires

3. Theory of experiment

Donor impurities (pentavalent) are introduced into one-side and acceptor

impurities into the other side of a single crystal of an intrinsic semiconductor

to form a p-n diode with a junction called depletion region (this region is

depleted off the charge carriers). This region gives rise to a potential barrier

Vγ called Cut- in Voltage. This is the voltage across the diode at which it

starts conducting. The P-N junction can conduct beyond this Potential.

The P-N junction supports uni-directional current flow. If +ve terminal of

the input supply is connected to anode (P-side) and –ve terminal of the input

supply is connected to cathode (N- side), then diode is said to be forward

biased. In this condition the height of the potential barrier at the junction is

lowered by an amount equal to given forward biasing voltage. Both the

holes from p-side and electrons from n-side cross the junction

simultaneously and constitute a forward current ( injected minority current –

due to holes crossing the junction and entering N-side of the diode, due to

electrons crossing the junction and entering P-side of the diode). Assuming

current flowing through the diode to be very large, the diode can be

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approximated as short-circuited switch. If –ve terminal of the input supply

is connected to anode (p-side) and +ve terminal of the input supply is

connected to cathode (n-side) then the diode is said to be reverse biased. In

this condition an amount equal to reverse biasing voltage increases the

height of the potential barrier at the junction. Both the holes on p-side and

electrons on n-side tend to move away from the junction thereby increasing

the depleted region. However the process cannot continue indefinitely, thus

a small current called reverse saturation current continues to flow in the

diode. This small current is due to thermally generated carriers. Assuming

current flowing through the diode to be negligible, the diode can be

approximated as an open circuited switch.

The volt-ampere characteristics of a diode explained by following equation:

I = Io(Exp(V/ ηVT)-1)

I= current flowing in the diode

Io=reverse saturation current

V=voltage applied to the diode

VT=volt-equivalent of temperature=kT/q=T/11,600=26mV(@ room temp).

η=1 (for Ge) and 2 (for Si)

Circuit diagram:

Forward Bias:

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Reverse Bias:

Forward Biased Condition:

1. Connect the PN Junction diode in forward bias i.e Anode is connected to

positive of the power supply and cathode is connected to negative of the

power supply.

2. Use a Regulated power supply of range (0-30)V and a series resistance of

1kΏ.

3. For various values of forward voltage (Vf) note down the corresponding

values of forward current(If) .

Reverse biased condition:

1. Connect the PN Junction diode in Reverse bias i.e; anode is connected to

negative of the power supply and cathode is connected to positive of the

power supply.

2. For various values of reverse voltage (Vr ) note down the corresponding

values of reverse current ( Ir ).

4. Observation

Tabular column:

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Forward Bias:

S. No Vf (volts) If (mA)

Reverse Bias:

S.No Vr (volts) Ir (μA)

Graph ( instructions)

1. Take a graph sheet and divide it into 4 equal parts. Mark origin at the

center of the graph sheet.

2. Now mark +ve x-axis as Vf

-ve x-axis as Vr

+ve y-axis as If

-ve y-axis as Ir.

3. Mark the readings tabulated for diode forward biased condition in first

Quadrant and diode reverse biased condition in third Quadrant.

Graph:

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Precautions:

1. While doing the experiment do not exceed the ratings of the diode. This

may lead to damage of the diode.

2. Connect voltmeter and Ammeter in correct polarities as shown in the

circuit diagram.

3. Do not switch ON the power supply unless you have checked the circuit

connections as per the circuit diagram

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Name of the Lab: Basic Electronics Lab


AIM: To draw the drain and transfer characteristics of a given FET.

APPARATUS: FET (BFW-11)

Regulated power supply

Voltmeter (0-20V)

Ammeter (0-100mA)

Bread board

Connecting wires

THEORY:

A FET is a three terminal device, having the characteristics of high input

impedance and less noise, the Gate to Source junction of the FET s always reverse

biased. In response to small applied voltage from drain to source, the n-type bar

acts as sample resistor, and the drain current increases linearly with VDS. With

increase in ID the ohmic voltage drop between the source and the channel region

reverse biases the junction and the conducting position of the channel begins to

remain constant. The VDS at this instant is called “pinch of voltage”.

If the gate to source voltage (VGS) is applied in the direction to provide additional

reverse bias, the pinch off voltage ill is decreased.

In amplifier application, the FET is always used in the region beyond

the pinch-off.

IDS=IDSS(1-VGS/VP)^2

CIRCUIT DIAGRAM

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PROCEDURE:

1. All the connections are made as per the circuit diagram.

2. To plot the drain characteristics, keep VGS constant at 0V.

3. Vary the VDD and observe the values of VDS and ID.

4. Repeat the above steps 2, 3 for different values of VGS at 0.1V and 0.2V.

5. All the readings are tabulated.

6. To plot the transfer characteristics, keep VDS constant at 1V.

7. Vary VGG and observe the values of VGS and ID.

8. Repeat steps 6 and 7 for different values of VDS at 1.5 V and 2V.

9. The readings are tabulated.

10. From drain characteristics, calculate the values of dynamic resistance (rd) by using

the formula

rd = ∆VDS/∆ID

11. From transfer characteristics, calculate the value of transconductace (gm) By using

the formula

Gm=∆ID/∆VDS

12. Amplification factor (μ) = dynamic resistance. Tran conductance

μ = ∆VDS/∆VGS

OBSERVATIONS:

DRAIN CHARACTERISTICS:

S.NO VGS=0V VGS=0.1V VGS=0.2V

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vDS(V) ID (mA) VDS(V) ID(mA) VDS(V) ID (mA)

TRANSFER CHARACTERISTICS:

S.No. VDS =

0.5V

IP(mA) VDS=1V VDS=1.5V

vGS (V) VGS (V) ID(mA) VGS (V) ID(mA)

MODEL GRAPH:

TRANSFER CHARAC TERISTICS:-

DRAIN CHARAC TERISTICS:-

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PRECAUTIONS:-

1. The three terminals of the FET must be carefully identified

2. Practically FET contains four terminals, which are called source, drain,

Gate, substrate.

3. Source and case should be short circuited.

4. Voltages exceeding the ratings of the FET should not be applied.

RESULT:-

1. The drain and transfer characteristics of a given FET are drawn

2. The dynamic resistance (rd), amplification factor (µ) and Tran conductance

(gm) of the given FET are calculated.

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Aim of experiment

To Study the Rectifiers a] Half wave b] Full wave & draw i/p & o/p wave forms.

Apparatus: (i) A step-down transformer,

(ii) A junction diode,

(iii) 3 Load resistors,

(iv) Capacitor,

(v) Oscilloscope,

(vi) Multimeter,

(vii) Connecting wires,

(viii) Breadboard Theory/Principle:

The process of converting an alternating current into direct current is known as

rectification. The unidirectional conduction property of semiconductor diodes (junction

diodes) is used for rectification. Rectifiers are of two types: (a) Half wave rectifier and

(b) Full wave rectifier. In a half-wave rectifier circuit (Fig. 1), during the positive half-

cycle of the input, the diode is forward biased and conducts. Current flows through the

load and a voltage is developed across it. During the negative half-cycle, it is reverse

bias and does not conduct. Therefore, in the negative half cycle of the supply, no current

flows in the load resistor as no voltage appears across it. Thus the dc voltage across the

load is sinusoidal for the first half cycle only and a pure a.c. input signal is converted

into a unidirectional pulsating output signal.

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Fig.1: Half-wave rectifier circuit

Since the diode conducts only in one half-cycle (0-π), it can be verified that the d.c.

component in the output is Vmax/π, where Vmax is the peak value of the voltage. Thus,

The current flowing through the resistor, and power consumed by

the load,

Ripple factor:

As the voltage across the load resistor is only present during the positive half of the cycle, the

resultant voltage is "ON" and "OFF" during every cycle resulting in a low average dc value. This

variation on the rectified waveform is called "Ripple" and is an undesirable feature. The ripple

factor is a measure of purity of the d.c. output of a rectifier and is defined as

In case of a half-wave rectifier Vrms = Vmax/2 = 0.5Vmax

Rectification Efficiency:

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Rectification efficiency, η, is a measure of the percentage of total a.c. power input

converted to useful d.c. power output.

Here rd is the forward resistance of diode. Under the assumption of no diode loss

(rd<<), the rectification efficiency in case of a half-wave rectifier is approximately 40.5%.

Filters: The output of a rectifier gives a pulsating d.c. signal (Fig.1) because of presence of

some a.c. components whose frequency is equal to that of the a.c. supply frequency. Very

often when rectifying an alternating voltage we wish to produce a "steady" direct voltage free

from any voltage variations or ripple. Filter circuits are used to smoothen the output. Various

filter circuits are available such as shunt capacitor, series inductor, choke input LC filter and

-filter etc. Here we will use a simple shunt capacitor filter circuit (Fig. 2). Since a capacitor

is open to d.c. and offers low impedance path to a.c. current, putting a capacitor across the

output will make the d.c. component to pass through the load resulting in small ripple

voltage.

Fig.2: Half-wave rectifier circuit with capacitor filter

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The working of the capacitor can be understood in the following manner. When the

rectifier output voltage is increasing, the capacitor charges to the peak voltage Vm. Just past

the positive peak the rectifier output voltage tries to fall. As the source voltage decreases

below Vm , the capacitor will try to send the current back to diode making it reverse biased.

Thus the diode separates/disconnects the source from the load and hence the capacitor will

discharge through the load until the source voltage becomes more than the capacitor voltage.

The diode again starts conducting and the capacitor is again charged to the peak value Vm and

the process continues. Although in the output waveform the discharging of capacitor is shown

as a straight line for simplicity, the decay is actually the normal exponential decay of any

capacitor discharging through a load resistor. The extent to which the capacitor voltage drops

depends on the capacitance and the amount of current drawn by the load; these two factors

effectively form the RC time constant for voltage decay. A proper combination of large

capacitance and small load resistance can give out a steady output.

Circuit Diagram: (As shown in Fig. 1 and 2):

Procedure:

i) Configure the half-wave rectifier circuit as shown in the circuit diagram. Note down

all the values of the components being used

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ii) Connect the primary side of the transformer to the a.c. Mains and secondary to the

input of the circuit.

iii) Feed the input and output to the two channels of oscilloscope (we will use

oscilloscope here only to trace the output waveform) and save the data for each

measurement.

iv) Measure the input a.c. voltage and the output a.c. and d.c. voltages using multimeter

for at least 3 values of load resistor (Be careful to choose proper settings of

multimeter for ac and dc measurement).

v) Multiply the Vac at the input by √2? to get the peak value and calculate Vdc Using the

formula Vdc = Vmax/ π. Compare this value with the measured Vdc at the output.

vi) Calculate the ripple factor and efficiency.

vii) Connect the capacitor across the output for each load resistor and measure the output

a.c. and d.c. voltages once again and calculate the ripple factor.

Observations:

i) Code number of diode = ________

ii) Input Voltage: Vac = _________ Volt

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Table(I): Half wave rectifier w/o filter

Table(II): Half wave rectifier with filter (C = ____ µF)

Waveforms without Filter:

RL = ______ Input Output (Paste data here)

Waveforms with Capacitor Filter:

RL = ______ Input Output (Paste data here)

Discussions:

Full Wave Rectifier Circuit without & with Filter

Objectives: To construct a full-wave bridge rectifier circuit and analyze its output.

Theory: As you have seen already a half-wave rectifier circuit is unsuitable to applications

which need a "steady and smooth" dc supply voltage. One method to improve on this is to use

every half-cycle of the input voltage instead of every other half-cycle. The circuit which allows

us to do this is called a Full-wave Rectifier. Here, unidirectional current flows in the output for

both the cycles of input signal rectify it.The rectification can be either by a center tap full wave

rectifier(using two diodes) or a full wave bridge rectifier(using four diodes). In this experiment

we will study a full wave bridge rectifier.

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The Full-wave Bridge Rectifier

Another type of circuit that produces the same output as a full-wave rectifier is that of the Bridge

Rectifier (Fig. 1). This type of single phase rectifier uses 4 individual rectifying diodes connected

in a "bridged" configuration to produce the desired output but does not require a special centre

tapped transformer, thereby reducing its size and cost. The single secondary winding isconnected

to one side of the diode bridge network and the load to the other side as shown in figure.

The 4 diodes labeled D1 to D4 are arranged in "series pairs" with only two diodes conducting

current during each half cycle.

During the positive half cycle of the supply, diodes D1 and D2 conduct in series while

D3 and D4 are reverse biased and the current flows through the load as shown below (Fig. 2).

During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but diodes D1

and D2 switch off as they are now reverse biased.The current flowing through the load is the

same direction as before.

Fig. 1: Full-wave Bridge Rectifier

Fig 2: working of full wave rectifier

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As the current flowing through the load is unidirectional, so the voltage developed across the

load is also unidirectional during both the half cycle. Thus, the average dc output voltage across

the load resistor is double that of a half-wave rectifier circuit, assuming no losses.

Ripple factor: As mentioned in the previous lab the ripple factor is a measure of purity of d.c output

output of a rectifier and is defined as

In case of a full-wave rectifier Vrms = Vmax/_2 = 0.707Vmax

The ripple frequency is now twice the supply frequency.

Rectification Efficiency:

Rectification efficiency η is given by

where rd is the forward resistance of diode. Under the assumption of no diode loss

The rectification efficiency in case of a full wave rectifier is 81.1%,which is twice the value of

half wave rectifier.

Filter:

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The full-wave rectifier circuit with capacitor filter is shown in Fig. 3. The smoothing

capacitor converts the full-wave rippled output of the rectifier into a smooth dc output

voltage. The detailed description of its filtering action is already explained in half-wave

rectifier handout. Two important parameters to consider when choosing a suitable a capacitor

are its working voltage, which must be higher than the no-load output value of the rectifier

and its capacitance value, which determines the amount of ripple that will appear

superimposed on top of the dc voltage.

Apart from rectification efficiency, the main advantages of a full-wave bridge rectifier

is that it has a smaller ac ripple value for a given load and a smaller smoothing capacitor than

an equivalent half-wave rectifier. The amount of ripple voltage that is superimposed on top of

the dc supply voltage by the diodes can be virtually eliminated by adding other improved

filters such as a pi-filter.

Circuit components/Equipments:

(i) A step-down transformer, (ii) 4 junction diodes, (iii) 3 Load resistors, (iv) Capacitor,

(v) Oscilloscope, (vi) Multimeters, (vii) Connecting wires, (viii) Breadboard.

Circuit Diagram: (As shown in Fig. 1 and 3):

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Procedure:

i) Configure the full-wave rectifier circuit as shown in the circuit diagram. Note down

all the values of the components being used.

ii) Connect the primary side of the transformer to the a.c. Mains and secondary to the

input of the circuit.

iii) Measure the input a.c. voltage (Vac) and current (Iac) and the output a.c. (Vac) and d.c.

(Vdc) voltages using multimeter for at least 3 values of load resistor (Be careful to

choose proper settings of multimeter for ac and dc measurement).

iv) Feed the input and output to the oscilloscope (we will use oscilloscope here only to

trace the output waveform) and save the data for each measurement. BE CAREFUL

NOT TO MEASURE THE INPUT AND OUTPUT VOLTAGES

SIMULTANEOUSLY.

v) Multiply the Vac at the input by √2 to get the peak value and calculate Vdc Using the

formula Vdc = 2Vmax/ π . Compare this value with the measured Vdc at the output.

vi) Calculate the ripple factor and efficiency.

vii) Connect the capacitor across the output for each load resistor. Measure the output a.c.

and d.c. voltages once again and calculate the ripple factor. Trace the input and output

waveforms in oscilloscope and notice the change.

Observations:

i) Code number of diode = ________

ii) Input Voltage: Vac = _________ Volt

Table(I): Full-wave rectifier w/o filter

Table(II): Full-wave rectifier with filter (C = ____ µF)

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(III) Input and output waveforms:

Waveforms without Filter:

RL = ______ Input Output

(Paste data here)

Waveforms with Capacitor Filter:

RL = ______ Input Output

(Paste data here)

Discussions:

Precautions:

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Aim: To study Single stage common emitter amplifier & plot its frequency

response.

Apparatus: BJT,Resistors,Capacitors,DC Source,AC Source,Oscilloscope

Theory:

When in an amplifier circuit only one transistor is used for amplifying a weak

signal, the circuit is known as single stage amplifier. However, a practical

amplifier consists of a number of single stage amplifiers and hence a complex

circuit. Therefore, such a complex circuit can be conveniently split into several

single stages and can be effectively analysed

The above fig. shows a single stage transistor amplifier. When a weak a.c. signal is

applied to the base of the transistor, a small base current starts flowing in the input

circuit. Due to transistor action, a much larger (β times the base current) a.c.

current flows through the the load Rc in the output circuit. Since the value of load

resistance Rc is very high, a large voltage will drop across it. Thus, a weak signal

applied in the base circuit appears in amplified form in the collector circuit. In this

way the transistor acts as an amplifier.

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Circuit Diagram:

Circuit analyses (to be completed before coming to the Lab):

Draw the DC and AC equivalent circuit for Fig 1.1. Work out the equation for the

load line and the operating point. Draw the load line and mark the operating point

on it. Analyse the ac circuit and calculate the expression for voltage gain.

Procedure:

A. To record the operating points of the transistor and confirm that it lies on the

load line.

Assemble the circuit as shown in Fig 2. and record the base current IB, collector

current Ic and collector to emitter voltage VCE or VC and base to emitter voltage

VBE. Verify that operating point lies around the center of the load line. operating

point lies around the center of the load line.

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B. To measure the gain of the amplifier as a function of frequency

1. Assemble the circuit of Fig 1.1

2. Set the function generator for sinusoidal out put. With the help of a Tee

connector and BNC cable display the function generator output (input to the

amplifier circuit) on one of the channel of the scope. Set the function generator for

~ 1kHz and 20mV peak to peak signal

3. Display the out of the amplifier on second channel of scope.

4. Make sure that both the input and output sine waves are not clipped or distorted

in any way. If it is so then reduce the input voltage from the function generator.

until nice clean looking sine waves are displayed at both input and output

terminals. Also note the polarity of output sine wave relative to the applied input

signal.

5. Record the output signal as a function of frequency maintaining the input fixed

at 20mV.

6. Trace the input and output signals for atleast two frequencies (the scale should

be well labeled)

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7. Calculate the gain of this amplifier by taking the ratio of the output and the input

amplitude.

8. Plot the gain as a function of frequency and determine the band width of the

amplifier. Mark the cut off frequencies on the graph.

9. Slowly increase the amplitude of the input sine wave until the output sinusoidal

wave begins to clip.

Note the voltage at which clipping is observed for the positive as well negative

polarity. Trace the distorted signals for both the polarity.

The frequency response curve will be

Conclusion:

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Aim:1.To study Transistor shunt voltage regulator, when input voltage Vin is

fixed while Load resistance RL is variable.

Apparatus: 1. Analog board of Transistor Shunt Voltage Regulator

2. DC power supply +12V external source

3. Digital Multimeter (2 numbers).

4. 2 mm patch cords.

Theory:

Circuits that maintain power supply voltages or current output within specified

limits, or tolerances are called Regulators. They are designated as dc voltage or dc

current regulators, depending on their specific application. Voltage regulator

circuits are additions to basic power supply circuits, which are made up of rectifier

and filter sections (Fig. 1). The purpose of the voltage regulator is to provide an

output voltage with little or no variation. Regulator circuits sense changes in output

voltages and compensate for the changes.

There are two types of voltage regulators. Basic voltage regulators are classified as

either Series or Shunt, depending on the location or position of the regulating

element (s) in relation to the circuit load resistance. Fig 2 illustrates these two basic

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types of voltage regulators. Broken lines have been used in the figure to highlight

the difference between the series and shunt regulator

The schematic drawing in view A is that of a shunt regulator. It is called shunt

regulator because the regulating device is connected in shunt or in parallel with the

load resistance. Fig 2 illustrates the principle of shunt voltage regulation. From the

figure it is clear that the regulator is in shunt with the load resistance (RL). In a

shunt voltage regulator, as shown in Fig 2, Output voltage regulation is determined

by parallel resistance of the regulating device, the Load resistance (RL), and the

series resistor Rs. If the load resistance RL increases/decreases, the regulating device

decreases/increases its resistance to compensate for the change.

The schematic for a typical series voltage regulator is shown in Figure 3. It

employs the NPN transistor in shunt configuration in place of the variable resistor

found in Fig 2.

Since AB is in parallel across VL, we have

VL –VZ –VBE = 0 or VBE = VL –VZ (kirchoff's Voltage Law)

Also VL= VZ + VBE

i.e. The output voltage is close to the sum of the voltage across Zener and the

voltage at the base-emitter junction of transistor.

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Circuit Operation :

Case 1: when input voltage is constant while Load varies.

Since VZ is fixed, any decrease or increase in VL will have a corresponding effect

on VBE. Suppose, VL decreases then as seen from the above relation VBE also

decreases. As a result, IB decreases, hence IC ( = βIB) decreases, thereby

decreasing I and hence VB (=IR). Consequently, VL increases because at all times.

Vin = VR+ VL or VL = Vin – VR

From the above description it is concluded that when by any reason VL decreases

VR also decreases thereby keeping VL constant.

Similarly, when by any reason VL increases VR also increases thereby keeping VL

constant.

Case 2 : when input voltage varies while Load remains constant.

When the input voltage increases, output voltage across increases momentarily.

This momentarily deviation or variation, from the required regulated output

voltage of 6.3 volts is a result of a rise in the input voltage. This increases forward

bias of transistor. Recall that the voltage drop across Zener remains constant at

5.6V. Since the output voltage is composed of the Zener voltage and the base-

emitter voltage, the output voltage momentarily increases. At this time, the

increase in the forward bias of transistor lowers the resistance of the transistor

allowing more current to flow through it

Since this current must also pass through R, there is also an increase in the voltage

drop across this resistor. Due to increase in this voltage drop across R, voltage

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across VL remains close to the required regulated value of output voltage.

Similarly, when input voltage decreases, forward bias of transistor also decreases.

This decrease in bias voltage increases the resistance of transistor allowing less

current to flow through it. Since this current must pass through resistor R, there is

also an decrease in the voltage drop across this resistor. This drop in voltage across

R maintain output voltage close to the required regulated value.

Circuit diagram :

Procedure :

1. Connect + 12V dc power supply at its indicated position from external

source.

2. Connect one voltmeter between test point 1 and ground to measure input

voltage Vin.

3. Connect ohmmeter between test point 5 and ground and set the value of

load resistance RL at some fixed value [full load (1.1K), 1K, 500Ω...]

4. Connect a 2mm patch cord between test point 3 and 4.

5. Connect voltmeter between test point 5 and ground to measure output

voltage Vout.

6. Switch ON the power supply.

7. Vary the potentiometer P1 to set fixed value of input voltage Vin = 9V and

measure the corresponding values of a. Output voltage Vout between test

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points 5 and ground. b. Zener voltage Vz between test points 2 and 6. c.

Forward bias voltage VBE of transistor between test point 6 and ground.

8. Disconnect the 2mm patch cord between test point 3 and 4.

9. Repeat the procedure from step 3 for different sets of load resistance R and

note the results in an observation Table 1.

Observation Table 1 :

Sl.no Load Resistance

RL

Voltage

Across

Zener VZ

Forward bias

voltage VBE

Output voltage

Vout at constant

Input voltage Vin

= 9 volt

1 Full Load (1.1K)

2 1KΩ

3 800 Ω

4 600 Ω

5 400 Ω

6 200 Ω

7 No Load

Note : To measure Voltage at No Load disconnect 2mm patch cord between test

point 3 and 4, measure voltage between test point 3 and ground.

Calculations:

Percentage regulation is given by formula

% Regulation = [(VNL – VFL)/VFL] * 100

VNL = No-load or open-circuit terminal voltage.

VFL = Full-load terminal voltage.

Result :

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1. The result of Experiment 1 reveal that for the network of Fig 3 with a fixed

input voltage Vin the output voltage will remain close to 6.3 V for a range of

load resistance that extends from extends from _____________ to

_____________

2. Percentage regulation = __________%

Aim 2.To study Transistor shunt voltage regulator, when input voltage Vin

is variable while Load resistance RL is fixed.



3. Digital Multimeter (2 numbers)


Circuit diagram :

Procedure :

• Connect +12V dc power supplies at their indicated position from external source

or ST2612 Analog Lab.

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1. Connect one voltmeter between test point 1 and ground to measure input voltage

Vin

2. Connect ohmmeter between test point 5 and ground and set the value of load

resistance RL at maximum value.


4. Connect voltmeter between test point 5 and ground to measure output voltage

Vout.


6. Vary the potentiometer P1 to set fixed value of input voltage Vin = 7V, 8V, 9V

and measure the corresponding values of

a. Output voltage Vout between test points 5 and ground.

b. Zener voltage VZ between test points 2 and 6.

c. Forward bias voltage VBE of transistor between test point 6 and ground.

7. Repeat the procedure from step 7 for different sets of input voltage Vin and note

the results in an observation Table 2.


Sl No. Input

Voltage(Vin)

Voltage across

Zener(VZ)

Forward bias

voltage(VBE)

Output voltage

Vout at fixed

load resistance

RL=Max

1 7 V

2 8V

3 9V

4 10 V

5 11 V

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Results :

1. The result of Experiment 2 reveal that for the network of Fig 5 with a fixed Load

resistance, the output voltage will remain close to 6.3V for a range of input voltage

Vin that extends from _____ to_____ .

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Aim:To study the input and output characteristics curves of the given NPN transistor in

common emitter mode.

Apparatus: An NPN Transistor, regulated power supply, voltmeters and ammeters

Theory/Principle: A transistor is a three terminal semiconductor device The three terminals are the

emitter, the base and the collector. A transistor transfers a signal from low resistance to high resistance;

hence it is named as transistor.

In a transistor, the N and P type semiconductor sections are alternated. Though it is a continuous body of

a single crystal, the doping is done in such a way that either a P section appears in between two N sections

( the NPN transistor), or an N section appears in between two P sections ( the PNP transistor). The middle

appears to be sandwiched between the other two, namely the emitter and the collector. In comparison to

the doping level of the collector, the emitter is always heavily doped in order to provide a large supply of

charge carriers, and the base is lightly doped to minimize the recombination that occurs in it between the

electrons and the holes. The collector will be having a large area in order to efficiently gather the charge

carriers. In order to identify the terminals you should carefully observe the transistor. You can locate a

small projecting flap at its bottom edge. The nearest lead to this tab is always the emitter lead, and the

farthest one is the collector lead. The lead between the emitter and the collector is the base lead.

The emitter base junction of a transistor is forward biased whereas collector base junction is reverse

biased. If for a moment we ignore the presence of emitter base junction, then practically no current would

flow in the collector circuit because of the reverse bias. However, if the emitter base junction is also

present then forward bias on it causes the emitter current to flow. It is seen that this emitter current almost

entirely flows in the collector circuit; therefore the current in the collector circuit depends on the emitter

current. If the emitter current is zero, then collector current is nearly zero. However, if the emitter current

is 1mA then collector current is also about 1mA.

Working of NPN transistor: The forward bias causes the electrons in the N type emitter to flow towards

the P type base. This constitutes the emitter current Ie. As these electrons flow through the P type base,

then tend to combine with holes. As the base is lightly doped and very thin, therefore, only a few

electrons combine with holes to constitute base current Ib. The remainder cross over into the collector

region to constitute collector current Ic. In this way, almost the entire emitter current flows in the

collector circuit. It is clear that emitter current is the sum of collector and base currents.

i.e., Ie = Ib + Ic ……(1)

33

Current amplification factor is the ratio of output current to the input current.

Current amplification factor in common emitter configuration is the ratio of change in collector current to

the change in base current, denoted by β.

Circuit Diagram:.

Expected Graphs:

34

Experimental Procedure:

The circuit connections are made and the transistor is connected in CE configurations.

Input Characteristics: The collector emitter voltage (Vce) is kept constant. Now the base emitter

voltage (Vbe) is varied and the corresponding base current Ib readings are noted. Similarly, one

more set of readings are taken by keeping Vce constant. A graph of Vbe against Ib is plotted.

Output Characteristics: The base current Ib is kept constant and collector emitter voltage Vce is

varied and corresponding current Ic are noted. Similarly, one more set of readings are taken for

35

Table.2:- Output Characteristics

IB = in μA IB = in μA

VCE in Volts Ic in mA VCE in Volts Ic in mA

0.1 0.1

0.2 0.2

0.3 0.3

0.4 0.4

0.5 0.5

0.6 0.6

0.7 0.7

0.8 0.8

0.9 0.9

1.0 1.0

Table.3:- Transfer Characteristics

VCE = in V

IB in μA Ic in mA

10

20

30

40

50

60

70

80

90

100

36

Result: The input and output characteristics of transistor is studied and the values of β and α are found.

β =……….(from output characteristics graph)

β =……….(from transfer characteristics graph)

α = ………..

37




AIM: To Plot Input & output characteristics of transistor in CB mode.

APPARATUS: Transistor, variable DC power supply, micro-ammeter, milli-

ammeter, voltmeter.

THEORY:

Transistor characteristics are the curves, which represent relationship between

different DC currents and voltages of a transistor. These are helpful in studying the

operation of a transistor when connected in a circuit. The three important

characteristics of a transistor are:

1. Input characteristic.

2. Output characteristic.

3. Constant current transfer characteristic.

Input Characteristic :

In common base configuration, it is the curve plotted between the input current

(IE) versus input voltage (VBE) for various constant values of output voltage

(VCB). The approximated plot for input characteristic is shown in figure 1. This

characteristic reveal that for fixed value of output voltage VCB, as the base to

emitter voltage increases, the emitter

current increases in a manner that

closely resembles the diode

characteristics.

38

Output Characteristic :

This is the curve plotted between the output current IC versus output voltage VCB

for various constant values of input current IE.

The output characteristic has three basic region of interest as indicated in figure 2

the active region, cutoff region and saturation region.

In active region the collector base junction is reverse biased while the base emitter

junction if forward biased. This region is normally employed for linear

(undistorted) amplifier.

In cutoff region the collector base junction and base emitter junction of the

transistor both are reverse biased. In this region transistor acts as an ‘Off’ switch.

In saturation region the collector base junction and base emitter junction of the

transistor both are forward biased. In this region transistor acts as an ‘On’ switch.

Constant current transfer Characteristic :

This is the curve plotted between output collector current IC versus input emitter

current IE for constant value of output voltage VCB.

39

The approximated plot for this characteristic is shown in figure 3.

Circuit Diagram:

Procedure :

· Connect -5V and +12V DC power supplies at their indicated position from

external source .

40

· To plot input characteristics proceed as follows :

1. Rotate both the potentiometer P1 and P2 fully in CCW (counter clockwise

direction).

2. Connect Ammeter between test point 2 and 3 to measure input emitter current IE

(mA).

3. Short or connect a 2mm patch cord between test point 4 and 5


V BE other voltmeter between test point 6 and ground to measure output voltage

VCB.

5. Switch ‘On’ the power supply.

6. Vary potentiometer P2 and set a value of output voltage VCB at some constant

value (1V, 2V…)

7. Vary the potentiometer P1 so as to increase the value of input voltage VBE from

zero to 0.9V in step and measure the corresponding values of input current IE for

different constant value of output voltage VCB in an observation Table 1.

9. Repeat the procedure from step 6 for different sets of output voltage VCB.

10. Plot a curve between input voltage VBE and input current IE as shown in figure

1 using suitable scale with the help of observation Table l. This curve is the

required input characteristic.


SL No.

Input Voltage VBE Input current IE(mA) at constant value of

output voltage

VCB=1 V VCB=3 V VCB=5 V

1 0.0V

2 0.1 V

3 0.2 V

4 0.3V

5 0.4 V

6 0.5 V

41

7 0.6V

8 0.7V

9 0.8 V

10 0.9 V

· To plot output characteristics proceed as follows:

1. Switch ‘Off’ the power supply.

2. Rotate both the potentiometer P1 and P2 fully in CCW (counterclockwise

direction).


VCB.

4. Connect one Ammeter between test point 2 and 3 to measure input current

IE(mA) and other Ammeter between test point 4 and 5 to measure output current

IC(mA).


6. Vary potentiometer P1 and set a value of input current IE at some constant value

(0mA, 1mA...)

7. Vary the potentiometer P2 so as to increase the value of output voltage VCB

from zero to maximum value in step and measure the corresponding values of

output current IC for different constant value of input current IE in an observation

Table 2.

8. Rotate potentiometer P2 fully in CCW direction.

9. Repeat the procedure from step 6 for different sets of input current IE.

10. Plot a curve between output voltage V CB and output current IC as shown in

figure 2 using suitable scale with the help of observation Table 2. This curve is

the required output characteristic.

Observation Table 2 : SL No. Input

Voltage

VCB(V)

Input current IC(mA) at constant value of output voltage

IE=0 mA IE=1 mA IE=2 mA IE=3 mA IE=4 mA

42

1 -1.0

2 0.0

3 0.5

4 1.0

5 2.0

6 3.0

7 4.0

8 5.0

9 6.0

10 7.0

· To plot constant current transfer characteristics proceed as follows:

1. Switch ‘Off’ the power supply.

2. Rotate both the potentiometer P1 and P2 fully in CCW (counter clockwise

direction).


VCB.

4. Connect one Ammeter between test point 2 and 3 to measure input current IE

(mA) and other Ammeter between test point 4 and 5 to measure output current

IC (mA).


6. Vary potentiometer P2 and set a value of output voltage VCB at maximum

value.

7. Vary the potentiometer P1 so as to increase the value of input current IE from

zero to 10 mA in step and measure the corresponding values of output current IC

in an observation Table 3.

8. Plot a curve between output current IC and input current IE as shown in figure 3

using suitable scale with the help of observation Table 3. This curve is the

required Transfer characteristic.


43

SL No. Input

current

IE (mA)

Output current IC(mA) at

constant output voltage VCB = 10 V

1 0.0

2 1.0

3 2.0

4 3.0

5 4.0

6 5.0

7 6.0

8 7.0

9 8.0

10 9.0

Calculations :

1. Input resistance : It is the ratio of change in the input voltage VBE to change in

the input current IE at constant value of output voltage VCB or it is the reciprocal

of the slope obtained from the input characteristic.

Mathematically

Rin= 1

Slope from input Characteristic=1/𝐼𝐸/IBE=

𝑉𝐵𝐸

𝐼𝐸 at constant VCB

To calculate input resistance determine the slope from the input characteristic

curve obtained from observation Table 1. Reciprocal of this slope will give the

required input resistance.

2. Output resistance : It is the ratio of change in the output voltage VCB to change

in the output current IC at constant value of input current IE or it is the reciprocal of

the slope obtained from the output characteristic.

Mathematically

Rout = 1/ Slope from output Characteristic=1/I C/VCB=VCB/IE at constant IE

To calculate output resistance determine the slope from the output characteristic

curve obtained from observation Table 2. Reciprocal of this slope will give the

required output resistance.

44

3. Current Gain : It is the ratio of change in the output current I C to change in the

input current IE at constant value of output voltage VCB or it is the slope

obtained from the constant current transfer characteristic. It is denoted by αac.

Mathematically

αac = Slope of constant current transfer characteristic = IC/IE

To calculate current gain, determine the slope from the constant current transfer

characteristic curve obtained from observation Table 3. This slope is the required

current gain.

Results :

Input resistance Rin = ___________

Output resistance Rout =________

Current Gain αac = ___________

45




Aim: To study Two stage RC coupled amplifier & plot its Frequency response.

Apparatus:

1. BJT (BC107BP) 2No.

2.Resistors (47k, 2.2k, 10k, 10k, 1k) 10No.

3.Capacitors (10UF, 10UF, 10UF, 100UF,100UF) 5No.

4. DC source 1No.

5. AC source 1No.

6. Oscilloscope 1No.

Theory:

Due to its low cost and excellent audio fidelity over a wide range of frequencies,

an RC Coupled Amplifier is the most popular type of coupling used in a multi

stage amplifier.

It is usually used for voltage amplification.

The figure below shows two stages of an RC coupled amplifier.

46

A coupling capacitor CC is used to connect the output of first stage to the base i.e.

input of the second stage and this continues when more stages are connected

Since here the coupling from one stage to next is achieved by a coupling capacitor

followed by a connection to a shunt resistor, therefore, such amplifiers are known

as resistance-capacitance coupled amplifier or simply RC coupled amplifier.

The resistances R1, R2 and RE form the biasing and stabilisation network.

The emitter bypass capacitor offers a low resistance path to the signal. Without this

capacitor the voltage gain of each stage would be lost.

The coupling capacitor CC transmits a.c. signal but blocks d.c. This prevents d.c.

interference between various stages and the shifting of operating point.

Working of RC Coupled Amplifier:

When a.c. signal is applied to the base of the first transistor, it is amplified and

appears across its collector load RC. The second stage again amplifies this signal

and the more amplified signal appears across the second stage collector resistance.

In this way the cascaded stages amplify the signal and the overall gain is

considerably increased. However, the total gain is less than the product of the gains

of individual stages. It is because, when a second stage follows the first stage, the

effective load resistance of first stage is reduced due to the shunting effect of the

47

input resistance of second stage. This reduces the gain of the stage which is loaded

by the next stage.

Frequency Response of RC Coupled Amplifier:

The figure below shows the frequency response of a typical RC coupled amplifier.

We can notice from the above fig. that the voltage gain drops off at low (< 50 Hz)

and high (> 20 KHz) frequencies. However, it is uniform over the mid-frequency

range i.e. 50 Hz to 20 KHz.

Circuit Diagram:

48

Circuit analyses (to be completed before coming to the lab)

Draw the ac equivalent circuit and work out the expression for voltage gain for this

amplifier.

Procedure:

1. Assemble the circuit shown .

2. Apply sinusoidal signal having peak to peak voltage~ 10mV to the input of

the amplifier.

3. Display input as well as out put signal on to the CRO on two different channels

simultaneously.

4. Measure the output voltage as a function of frequency keeping the input

voltage fixed ~10mv.

5. Plot the gain as a function of frequency.

6. Determine the lower and higher frequency cut off and hence the bandwidth of

the amplifier.

7. Trace input and out put signals at two-three frequencies. (The traces should be

well labeled).

49

8. Compare the frequency response of single stage amplifier of experiment 1 and

the RC coupled amplifier.

9. Record the band width for triangular as well as square waves also.

10. Compare the frequency response of sinusoidal, triangular and square pulse

Application:

used as voltage amplifiers for example in the initial stages of public address

system.

If other type of coupling such as transformer coupling is used in the initial stages,

this results in frequency distortion which may be amplified in the next stage.

But, due to its poor impedance matching, it is rarely used in the final stages.

Conclusion:

50




AIM: To study the Characteristics of UJT

APPARATUS:- Regulated Power Supply (2Nos) (0-30V, 1A) , UJT

2N2646,Resistors 10kΩ, 47Ω, 330Ω, Multimeters ,Breadboard

and Connecting Wires

CIRCUIT DIAGRAM:

THEORY:

A Unijunction Transistor (UJT) is an electronic semiconductor device

that has only one junction. It has three terminals an emitter (E) and two

bases (B1 and B2). The base is formed by lightly doped n-type bar of

silicon. Two ohmic contacts B1 and B2 are attached at its ends. The

emitter is of p-type and it is heavily doped. The resistance between B1

and B2, when the emitter is opencircuit is called interbase resistance.

51

The original UJT, is a simple device that is essentially a bar of N type

semiconductor material into which P type material has been diffused

somewhere along its length.

The UJT is biased with a positive voltage between the two bases. This

causes a potential drop along the length of the device. When the emitter

voltage is driven approximately one diode voltage above the voltage at

the point where the P diffusion (emitter) is, current will begin to flow

from the emitter into the base region. Because the base region is very

lightly doped, the additional current (actually charges in the base region)

causes (conductivity modulation) which reduces the resistance of the

portion of the base between the emitter junction and the B2 terminal.

This reduction in resistance means that the emitter junction is more

forward biased, and so even more current is injected. Overall, the effect

is a negative resistance at the emitter terminal. This is what makes the

UJT useful, especially in simple oscillator circuits. When the emitter

52

voltage reaches Vp, the current starts to increase and the emitter voltage

starts to decrease. This is represented by negative slope of the

characteristics which is referred to as the negative resistance region,

beyond the valley point , VEB proportional to IE.

PROCEDURE:

1. Connection is made as per circuit diagram.

2. Output voltage is fixed at a constant level and by varying input

voltage corresponding emitter current values are noted down.

3. This procedure is repeated for different values of output voltages.

4. All the readings are tabulated and Intrinsic Stand-Off ratio is

calculated using η = (Vp-VD) / VB2B1 5. A graph is plotted

between VE and IE for different values of VB2B1.

OBSEVATIONS:

VB2B1=2V VB2B1=3V

53

VEB(V) IE(mA) VEB(V) IE(mA)

CALCULATIONS:

VP = ηVB2B1 + VD

η = (VP-VD) / VB2B1

η = ( η1 + η2 ) / 2

APPLICATION: UJT can be used as trigger device for SCR’s. traics

and other applications including sawtooth generator, phase control and

timing circuits.

CONCLUSION:

54




Aim:1To study Transistor shunt voltage regulator, when input voltage Vin is

fixed while Load resistance RL is variable.



3. Digital Multimeter (2 numbers).


Theory:

Circuits that maintain power supply voltages or current output within specified

limits, or tolerances are called Regulators. They are designated as dc voltage or dc

current regulators, depending on their specific application. Voltage regulator

circuits are additions to basic power supply circuits, which are made up of rectifier

and filter sections (Fig. 1). The purpose of the voltage regulator is to provide an

output voltage with little or no variation. Regulator circuits sense changes in output

voltages and compensate for the changes.

There are two types of voltage regulators. Basic voltage regulators are classified as

either Series or Shunt, depending on the location or position of the regulating

element (s) in relation to the circuit load resistance. Fig 2 illustrates these two basic

55

types of voltage regulators. Broken lines have been used in the figure to highlight

the difference between the series and shunt regulator

The schematic drawing in view A is that of a shunt regulator. It is called shunt

regulator because the regulating device is connected in shunt or in parallel with the

load resistance. Fig 2 illustrates the principle of shunt voltage regulation. From the

figure it is clear that the regulator is in shunt with the load resistance (RL). In a

shunt voltage regulator, as shown in Fig 2, Output voltage regulation is determined

by parallel resistance of the regulating device, the Load resistance (RL), and the

series resistor Rs. If the load resistance RL increases/decreases, the regulating

device decreases/increases its resistance to compensate for the change.

The schematic for a typical series voltage regulator is shown in Figure 3. It

employs the NPN transistor in shunt configuration in place of the variable resistor

found in Fig 2.

Since AB is in parallel across VL, we have

VL –VZ –VBE = 0 or VBE = VL –VZ (kirchoff's Voltage Law)

Also VL= VZ + VBE

i.e. The output voltage is close to the sum of the voltage across Zener and the

voltage at the base-emitter junction of transistor.

56

Circuit Operation :

Case 1: when input voltage is constant while Load varies.

Since VZ is fixed, any decrease or increase in VL will have a corresponding effect

on VBE. Suppose, VL decreases then as seen from the above relation VBE also

decreases. As a result, IB decreases, hence IC ( = βIB) decreases, thereby

decreasing I and hence VB (=IR). Consequently, VL increases because at all times.

Vin = VR+ VL or VL = Vin – VR

From the above description it is concluded that when by any reason VL decreases

VR also decreases thereby keeping VL constant.

Similarly, when by any reason VL increases VR also increases thereby keeping VL

constant.

Case 2 : when input voltage varies while Load remains constant.

When the input voltage increases, output voltage across increases momentarily.

This momentarily deviation or variation, from the required regulated output

voltage of 6.3 volts is a result of a rise in the input voltage. This increases forward

bias of transistor. Recall that the voltage drop across Zener remains constant at

5.6V. Since the output voltage is composed of the Zener voltage and the base-

emitter voltage, the output voltage momentarily increases. At this time, the

increase in the forward bias of transistor lowers the resistance of the transistor

allowing more current to flow through it

Since this current must also pass through R, there is also an increase in the voltage

drop across this resistor. Due to increase in this voltage drop across R, voltage

57

across VL remains close to the required regulated value of output voltage.

Similarly, when input voltage decreases, forward bias of transistor also decreases.

This decrease in bias voltage increases the resistance of transistor allowing less

current to flow through it. Since this current must pass through resistor R, there is

also an decrease in the voltage drop across this resistor. This drop in voltage across

R maintain output voltage close to the required regulated value.

Circuit diagram :

Procedure :

10. Connect + 12V dc power supply at its indicated position from external

source.

11. Connect one voltmeter between test point 1 and ground to measure input

voltage Vin.

12. Connect ohmmeter between test point 5 and ground and set the value of

load resistance RL at some fixed value [full load (1.1K), 1K, 500Ω...]


14. Connect voltmeter between test point 5 and ground to measure output

voltage Vout.


16. Vary the potentiometer P1 to set fixed value of input voltage Vin = 9V and

measure the corresponding values of a. Output voltage Vout between test

58

points 5 and ground. b. Zener voltage Vz between test points 2 and 6. c.

Forward bias voltage VBE of transistor between test point 6 and ground.

17. Disconnect the 2mm patch cord between test point 3 and 4.

18. Repeat the procedure from step 3 for different sets of load resistance R and

note the results in an observation Table 1.


Sl.no Load Resistance

RL

Voltage

Across

Zener VZ

Forward bias

voltage VBE

Output voltage

Vout at constant

Input voltage Vin

= 9 volt

1 Full Load (1.1K)

2 1KΩ

3 800 Ω

4 600 Ω

5 400 Ω

6 200 Ω

7 No Load

Note : To measure Voltage at No Load disconnect 2mm patch cord between test

point 3 and 4, measure voltage between test point 3 and ground.

Calculations:

Percentage regulation is given by formula

% Regulation = [(VNL – VFL)/VFL] * 100

VNL = No-load or open-circuit terminal voltage.

VFL = Full-load terminal voltage.

Result :

59

3. The result of Experiment 1 reveal that for the network of Fig 3 with a fixed

input voltage Vin the output voltage will remain close to 6.3 V for a range of

load resistance that extends from extends from _____________ to

_____________

4. Percentage regulation = __________%

Aim 2.To study Transistor shunt voltage regulator, when input voltage Vin is

variable while Load resistance RL is fixed.



3. Digital Multimeter (2 numbers)


Circuit diagram :

Procedure :

• Connect +12V dc power supplies at their indicated position from external source

or ST2612 Analog Lab.

60


Vin

2. Connect ohmmeter between test point 5 and ground and set the value of load

resistance RL at maximum value.



Vout.


6. Vary the potentiometer P1 to set fixed value of input voltage Vin = 7V, 8V, 9V

and measure the corresponding values of

a. Output voltage Vout between test points 5 and ground.

b. Zener voltage VZ between test points 2 and 6.

c. Forward bias voltage VBE of transistor between test point 6 and ground.

7. Repeat the procedure from step 7 for different sets of input voltage Vin and note

the results in an observation Table 2.


Sl No. Input

Voltage(Vin)

Voltage across

Zener(VZ)

Forward bias

voltage(VBE)

Output voltage

Vout at fixed

load resistance

RL=Max

1 7 V

2 8V

3 9V

4 10 V

5 11 V

61

Results :

1. The result of Experiment 2 reveal that for the network of Fig 5 with a fixed Load

resistance, the output voltage will remain close to 6.3V for a range of input voltage

Vin that extends from _____ to_____ .

62




AIM: To study the Zener Diode as Regulator& calculate load regulation

APPARATUS: Zener diode, Resistors, Power supply, Multi meter

CIRCUIT DIAGRAM:

THEORY:

Zener diode is a P-N junction diode specially designed to operate in the reverse

biased mode. It is acting as normal diode while forward biasing. It has a particular

voltage known as break down voltage, at which the diode break downs while

reverse biased. In the case of normal diodes the diode damages at the break down

voltage. But Zener diode is specially designed to operate in the reverse breakdown

region.

The basic principle of Zener diode is the Zener breakdown. When a diode is

heavily doped, it’s depletion region will be narrow. When a high reverse voltage is

applied across the junction, there will be very strong electric field at the junction.

And the electron hole pair generation takes place. Thus heavy current flows. This

is known as Zener break down. So a Zener diode, in a forward biased condition

acts as a normal diode. In reverse biased mode, after the break down of junction

63

current through diode increases sharply. But the voltage across it remains constant.

This principle is used in voltage regulator using Zener diodes.

The figure shows the zener voltage regulator, it consists of a current limiting

resistor Rs

connected in series with the input voltage Vs and zener diode is connected in

parallel with the load RL in reverse biased condition. The output voltage is always

selected with a breakdown voltage Vz of the diode.

The input source current, Is = IZ+ IL………….. (1)

The drop across the series resistance, Rs = Vin – Vz …….. (2)

And current flowing through it, Is = (Vin – VZ) / RS ………….. (3)

From equation (1) and (2), we get, (Vin - Vz )/Rs = Iz +IL ………… (4)

Regulation with a varying input voltage (line regulation):

It is defined as the change in regulated voltage with respect to variation in line

voltage. It is denoted by ‘LR’.

In this, input voltage varies but load resistance remains constant hence, the load

current remains constant. As the input voltage increases, form equation (3) Is also

varies accordingly. Therefore, zener current Iz will increase. The extra voltage is

dropped across the Rs. Since, increased Iz will still have a constant Vz and Vz is

equal to Vout.

The output voltage will remain constant. If there is decrease in Vin, Iz decreases as

load current remains constant and voltage drop across Rs is reduced. But even

though Iz may change, Vz remains constant hence, output voltage remains

constant.

Regulation with the varying load (load regulation): It is defined as change in

load voltage with respect to variations in load current. To calculate this regulation,

input voltage is constant and output voltage varies due to change in the load

resistance value.

Consider output voltage is increased due to increasing in the load current. The left

side of the equation (4) is constant as input voltage Vin, IS and Rs is constant.

Then as load current changes, the zener current Iz will also change but in opposite

64

way such that the sum of Iz and IL will remain constant. Thus, the load current

increases, the zener current decreases and sum remain constant. Form reverse bias

characteristics even Iz changes, Vz remains same hence, and output voltage

remains fairly constant.

PROCEDURE:-

A) Line Regulation:

1. Make the connections as shown in figure below.

2. Keep load resistance fixed value; vary DC input voltage from 5V to 15V.

3. Note down output voltage as a load voltage with high line voltage ‘VHL’ and

as a load voltage with low line voltage ‘VLL’.

4. Using formula, % Line Regulation = (VHL-VLL)/ VNOM x100, where VNOM =

the nominal load voltage under the typical operating conditions. For ex.

VNOM= 9.5 ± 4.5 V

B) Load Regulation:

1. For finding load regulation, make connections as shown in figure below

2. Keep input voltage constant say 10V, vary load resistance value.

3. Note down no load voltage VNL’ for maximum load resistance value and full

load voltage ‘VFL’ for minimum load resistance value.

4. Calculate load regulation using, % load regulation = (VNl-VFL)/ VFL x100

OBSERVATION TABLE:-

Calculations:

% Line Regulation = (VHL-VLL) / vNOM x100 = ------------ %

% voltage regulation = (VNL-VFL)/VFLx100 =----------%

CONCLUSION:-

65




1. Aim of experiment

To plot Forward & Reverse biased characteristics of Zener diode

2. Apparatus: Zener diode, Power Supply, dc milliammeter and dc voltmeter

3. Theory/Principle:

A PN junction is known as a semiconductor or crystal diode, so called because

PN junction is grown out of a crystal.

Biasing a diode means applying appropriate DC voltage to a diode. A forward

biased diode conducts easily whereas reverse biased diode practically conducts no

current. It means that forward resistance of a diode is quite small as compared with

its reverse resistance

Forward current is the current flowing through a forward biased diode. Every

diode has a maximum value of forward current, which it can safely carry. If this

value is exceeded, the diode may be destroyed due to excessive heat. Reverse

current or leakage current is the current that flows through the reverse biased

diode. This current is due to the minority carriers. Under normal operating voltages

the reverse current is quite small. Its value is extremely small for silicon diodes but

it is appreciable for germanium diodes. Peak inverse voltage is the maximum

reverse voltage that a diode can withstand without destroying the junction. If the

reverse voltage across a diode exceeds this value, the reverse current increases

sharply and breaks down the junction due to excess heat. Peak inverse voltage is

important when diode is used as a rectifier.

Zener diode is a heavily doped PN junction diode, meant to operate in reverse bias

condition. The breakdown or zener voltage depends upon the amount of doping. If

the diode is heavily doped, depletion layer will be thin and consequently the

breakdown of the junction will occur at a lower reverse voltage. On the other hand,

66

a lightly doped diode has a higher breakdown voltage. When zener diode is

forward biased, its characteristics are just same as ordinary diode.

Circuit Diagram:

67

Expected Graph:

Experimental Procedure:

The circuit connections are made as shown in the Fig.1. Initially the voltage

is set at 0V. The voltage is then varied in the steps of 0.1V and the

corresponding reading of the ammeter is noted. This set of reading

corresponds to the forward bias.

The circuit is then made for reverse bias as shown in Fig.2. The current is

measured by varying voltages in steps of 1V (in steps of 0.1 after the

current starts increasing).

A graph is plotted by taking voltage along X-axis and current along Y-axis.

The knee voltage is found by drawing a tangent in the forward bias at a

point where the current starts increasing at a faster rate. The breakdown

voltage is found by extending the curve to the X-axis in the reverse bias.

68

Result:

The Knee voltage = ……………………..V

The Breakdown voltage = ……………….V

Government Polytechnic, Muzaffarpurgpmuz.bih.nic.in/docs/BEE III.pdf · Basic Electronics Lab...

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