EXPERIMENT-1 FULL WAVE RECTIFIER AND BRIDGE RECTIFIER …mitmysore.in/downloads/ece/AEC_Manual_2018-2019.pdf · To construct full wave rectifier and bridge rectifier with and without

MIT MYSORE DEPT.OF ELECTRONICS AND COMMUNICATION ENGG

1 Analog Electronics Lab 17ECL37

EXPERIMENT-1

FULL WAVE RECTIFIER AND BRIDGE RECTIFIER

AIM:

1. To construct full wave rectifier and bridge rectifier with and without capacitor

filter in order to have rectified, filtered output dc voltage.

2. To calculate efficiency, percentage regulation and ripple factor.

COMPONENTS REQUIRED:

Sl. No Component/ IC/Device

/Equipment

Specification /Value Number/Quantity

1 Transformer 230V,50Hz/ 15 V, 50 HZ 01

2 Decade Resistance Box -- 01

3 Diodes BY127/IN4001 04

4 Capacitors 470 μF 01

5 Ammeter 0-500mA 01

6 Voltmeter 0- 30V 01

Center Tapped Full Wave Rectifier:

Theory:

The circuit of a center tapped full wave rectifier uses two diodes D1&D2. During

positive half cycle of secondary voltage, the diode D1 is forward biased and D2 is

reverse biased. The diode D1 conducts and current flows through load resistor RL.

During negative half cycle, diode D2 becomes forward biased and D1 reverse biased.

Now D2 conducts and current flows through the load resistor RL, during both the half

cycles and will get unidirectional current as shown in the model graph. The difference

between full wave and half wave rectification is that a full wave rectifier allows

unidirectional current to the load during the entire 360 degrees of the input signal and

half wave rectifier allows this only during one half cycle (180 degrees).

Procedure:

Without C filter:

1. Make the Connections as shown in the circuit diagram.

2. Switch on the Supply.

3. Observe the input and output waveforms of the rectifier and note down the output DC and

AC voltages (Vdc and Vac).

4. Observe the Voltage across the secondary of the Transformer (i.e Vrms). Calculate

the ripple factor, efficiency and percentage regulation at the output.



Full wave rectifier with

center tapped transformer

without filter

Waveform:

Vm

Full wave rectifier with center tapped transformer with filter

Waveform:



Calculations: From C.R.O. observations:

Vrms = Vm/2 =

Vdc =2Vm/π=

Vac = (Vrms2-Vdc2) =

r = Vac/Vdc= Vdc(NL)=

Percentage Regulation= (Vdc(NL) - Vdc(FL))×100/ Vdc(FL)=

Efficiency = η =( Pdc / Prms) ×100 = (Vdc/Vrms)2×100=

With C filter:

Connect a capacitor filter in parallel with load resistor and repeat the same procedure listed

above.

Shunt capacitor filter in FWR - r = 1/(43fRLC)

Vdc

(v)

VRMS

(v)

Idc

(A)

IRMS

(A)

Resistance

(Ω)

Efficiency

(η)

Ripple factor

(r)

%Regulation

Vdc

(v)

VRMS

(v)

Idc

(A)

IRMS

(A)

Resistance

(Ω)

Efficiency

(η)

Ripple factor

(r)

%Regulation



Bridge rectifier without C filter

Tabular column without C filter

Bridge rectifier circuit with C filter

Tabular column with C filter

Vdc

(v)

VRMS

(v)

Idc

(A)

IRMS

(A)

Resistance

(Ω)

Efficiency

(η)

Ripple factor

(r)

%Regulation

Vdc

(v)

VRMS

(v)

Idc

(A)

IRMS

(A)

Resistance

(Ω)

Ripple factor

(r)

%Regulation



Result:





EXPERIMENT-02

DIODE CLIPPING AND CLAMPING CIRCUITS

AIM: To design the different types of Clipping and Clamping Circuits and also to obtain

the transfer characteristics for the same.

Equipment and components:

SL.NO NAME OF EQUIPMENTS/

COMPONENTS

SPECIFICATIONS QTY

1. Dual Power Supply Variable 0 – 30 V / 2 Amp D.C. 01

2. Function Generator 0 – 3 MHz, 0 – 20 V(p-p) 01

3. CRO Analog, 30 MHz, Dual Channel 01

4. Connecting Board 01

5. Resistance 10 KΩ (½ Watt) 01

6. CRO Probe -- 03

7. Diode 1N4007 02

8. Patch cords, Connecting Wires, etc.

THEORY: Refer Text Book Electronic Devices And Circuit Theory By Robert L.

Boylestad / Louis Nashelsky “chapter 2 “

Procedure:

1. Circuit is wired up as shown in Fig.(1) and a sinusoidal signal of 1 KHz and

amplitude of 10 V(p-p) (Peak amplitude should be greater than clipping

level) is applied at input Vi.

2. Observe output signal on the CRO and verify it with the given waveforms.

3. Apply Vi and Vo to the X and Y channel of CRO and transfer characteristics is

obtained using X – Y mode in CRO.

Inference: ____________________



1. Diode Shunt Clipping above Vr (reference voltage) or Positive Peak Clipping

Circuit.

Fig. (1) Circuit Diagram of Diode Shunt Input output Waveforms Transfer

Characteristic Clipping above Vr

Rf =10 Ω (Forward Resistance of Diode)

Rr =10 KΩ (Reverse Resistance of Diode)

Design:

The output to be clipped above 2 V.

So Vo (max) = +2 V

From the Fig. (1)

Vo = Vo (max) – Vr + Vref

Where Vr is Diode drop which is nearly equal to 0.6 V.

So Vref = Vo (max) – Vr

= 2 – 0.6 = 1.4 V

Select the input amplitude more than 3Volts.

R = 10 KΩ



2. Diode Shunt Clipping below Vr (reference voltage) or Negative Peak Clipping

Circuit.

Fig. (2) Circuit Diagram of Diode Input output Waveforms Transfer Characteristic

Shunt Clipping below Vr

Design: Output voltage be clipped at + 2 Volt.

Vo (max) = Vref = 2 V

R = 10KΩ

Procedure:

1. Circuit is wired up as shown in Fig. (2) and a sinusoidal signal of 1 KHz and

amplitude of 10V (p-p) (Peak amplitude should be greater than clipping level) is

applied at input.




Inference: ____________________



3. Clipping two independent level or slicer

Fig. (5) Circuit Diagram Input output Waveforms Transfer Characteristic

Design: To clipping the signal below 2 Volt and above 4 Volt levels

Let VR1 > VR2

1] Vo max = 4 V, i.e. Vo max = VR1 + Vr

VR1 = Vo max – Vr

= 4 – 0.6

so

2] Vo min = 2 V i.e. Vo min = VR2 – Vr

VR2 = Vo min + Vr

= 2 + 0.6

Procedure:

1. Circuit is wired up as shown in Fig. (5) and a sinusoidal signal of 1 KHz and a

suitable amplitude (Peak amplitude should be greater than clipping level) is

applied at input Vi.




VR1 = 3.4 V

VR2 = 2.6 V R = 10KΩ



Inference: _______________

4. Double ended clipper to generate a symmetrical square wave or squarer

Fig. (6) Circuit Diagram of Double ended Input output Waveform Transfer Characteristic

Clipper or squarer

Design: To generate a symmetrical square wave

Vref =± 4 Volts

i.e. VR = VR1 = VR2 = Vref = 4 Volts

Vo max = VR + Vr

VR = Vo max - Vr = 4 – 0.6

Procedure:

1. Circuit is wired up as shown in Fig.(6) and a sinusoidal signal of 1 KHz and a

suitable amplitude (Peak amplitude should be greater than clipping level) is

applied at input Vi.




Inference: _______________

VR = 3.4 V R = 10KΩ





Clamping Circuits:

1] Positive peak or negative clamping:

Fig. (1) Circuit Diagram of Positive Peak Clamping Circuit

Design:

Clamping circuit to clamp positive peak at +3V.The input waveform has a frequency of 1

KHz sine wave or square wave with suitable amplitude.

Vo max = Vref + Vr

Vref = Vo max – Vr = 3 – 0.6

Given frequency 1 KHz

So 3101

1

T

= 1mSec

RC = 10 T

RC = (10) 1mSec = 10 mSec

C = 10 mSec / 10 K = 1µF

Vref = 2.4 V

Choose RC » T

R = 10 KΩ



Fig. (2) Input and Output waveforms of a positive Peak Clamping Circuit

Note: Set Vref = 0 and observe the output for both sine and square wave input.

Procedure:

1] Positive Clamping Circuit;

a) Connect the circuit as shown in Fig. (1)

b) Apply input sinusoidal signal of amplitude 12 V p-p and frequency of 1 KHz

[Amplitude of input signal must be greater than clamping level]

C) Connect the output to CRO in DC Mode and compare the output with the given

waveforms.

d) For the same circuit, give a square wave input and observe the output and

compare output with given waveforms

e) Make Vref = 0 and observe the output.

f) Observe the output in DC mode of CRO

3V

0



2] Negative peak or positive clamping:

Fig. (3) Circuit Diagram of Negative Peak Clamping Circuit

Design:

Clamping Circuit to clamp Negative peak of the output voltage at -3V

Vo min = Vref - Vr

Vref = Vo min + Vr = -3 + 0.6

Fig.(4) Input and Output waveforms of a Negative Peak Clamping Circuit.

Note: Set Vref = 0 and observe the output for both sine and square wave input.

Vref = -2.4 V

0 0



Procedure:

2] Negative Clamping Circuit;

a) Connect the circuit as shown in Fig. (3)

b) Apply input sinusoidal signal of amplitude 12 V(p-p) and frequency of 1 KHz

[Peak amplitude of input signal must be greater than clamping level]

C) Connect the output to CRO in DC Mode and compare the output with the given

waveforms.

d) For the same circuit, give a square input and observe the output and compare

output with given waveforms.

e) Make Vref = 0 and observe the output.

Inference: ________________

Probable viva questions:

1) What are clipping and clamping circuits

2) What is the difference between clipping and clamping circuits

3) What are the components of clipping circuit

4) What are the types of clipping circuit

5) Sketch the transfer characteristics of double ended clipper circuit

6) What are the applications of clipper and clamper circuits

7) Write the different equivalent circuits of a diode

10) What is diffusion and transition capacitance?

11) Define reverse recovery time of a diode

12) Explain the diode current equation



EXPERIMENT-03

SERIES VOLTAGE REGULATOR

AIM: To study and verify the functionality of Zener diode in forward bias and reverse

bias and to

1. Plot Volt-Ampere Characteristics of Zener Diode.

2. Find Zener Breakdown Voltage in Reverse Biased conditions.

3. Calculate static and dynamic resistance of the Zener diode in both forward and

reverse biased conditions (before, after break down voltages).

Components Required:

S.No. Name Quantity

1 Zener Diodes (1N4735A) 1(One) No.

2 Resistors (1K , 3.3 ) 1(One) No. Each

3 Bread board 1(One) No.

Equipment:

S.No. Name Quantity

1 Dual DC Regulated power supply (0 - 30V) 1(One) No.

2 Digital Ammeter (0 - 200 mA) 2(Two) No.

3 Digital Voltmeter (0 - 20V) 1(One) No.

4 Decade Resistance Box (DRB) 1(One) No.

5 Connecting wires (Single Strand)

Specifications of 1N 4735A Zener diode:

Breakdown Voltage = 5.1V

Power dissipation = 0.75W

Max Forward Current = 1A



Operation:

Zener diodes are a special kind of diode which permits current to flow in the

forward direction. What makes them different from other diodes is that Zener diodes

will also allow current to flow in the reverse direction when the voltage is above a

certain value. This breakdown voltage is known as the Zener voltage. In a standard

diode, the Zener voltage is high, and the diode is permanently damaged if a reverse

current above that value is allowed to pass through it. Zener diodes are designed in a

way where the Zener voltage is a much lower value. There is a controlled breakdown

which does not damage the diode when a reverse current above the Zener voltage

passes through a Zener diode.

The most common values for nominal working voltage are 5.1 V, 5.6 V, 6.2 V, 12

V and 15 V. We also carry Zener diodes with nominal working voltage up to 1 kV.

Forward (drive) current can have a range from 200 uA to 200 A, with the most common

forward (drive) current being 10 mA or 200 mA. In the forward bias direction, the

Zener diode behaves like an ordinary silicon diode.

In the reverse bias direction, there is practically no reverse current flow until the

breakdown voltage is reached. When this occurs there is a sharp increase in reverse

current. Varying amount of reverse current can pass through the diode without

damaging it. The breakdown voltage or Zener voltage (VZ) across the diode remains

relatively constant. The maximum reverse current is limited, however, by the wattage

rating of the diode.

Avalanche Break down:

When the diode is in the reverse bias condition, the width of the depletion region

is more. If both p-side and n-side of the diode are lightly doped, depletion region at the

junction widens. In reverse bias, the minority charge carrier current flows through

junction. As the applied reverse voltage increases the minority carriers acquire

sufficient energy to collide with the carriers in the covalent bonds inside the depletion

region. As a result, the bond breaks and electron hole pairs are generated. The process

becomes cumulative and leads to the generation of a large number of charge carriers

resulting in Avalanche Breakdown.



Zener Break down:

If both p-side and n-side of the diode are heavily doped, depletion region at the

junction reduces compared to the width in normal doping. Applying a reverse bias

causes a strong electric field get applied across the device. As the reverse bias is

increased, the Electric field becomes strong enough to rupture covalent bonds and

generate large number of charge carriers. Such sudden increase in the number of

charge carriers due to rupture of covalent bonds under the influence of strong electric

field is termed as Zener breakdown.

Zener Diode as Voltage Regulator:

The function of a regulator is to provide a constant output voltage to a load

connected in parallel with it in spite of the ripples in the supply voltage or the variation

in the load current and the Zener diode will continue to regulate the voltage until the

diodes current falls below the minimum IZ(min) value in the reverse breakdown region.

It permits current to flow in the forward direction as normal, but will also allow it to

flow in the reverse direction when the voltage is above a certain value - the breakdown

voltage known as the Zener voltage. The Zener diode specially made to have a reverse

voltage breakdown at a specific voltage. Its characteristics are otherwise very similar to

common diodes. In breakdown the voltage across the Zener diode is close to constant

over a wide range of currents thus making it useful as a shunt voltage regulator.

The purpose of a voltage regulator is to maintain a constant voltage across a load

regardless of variations in the applied input voltage and variations in the load current. A

typical Zener diode shunt regulator is shown in Figure 3. The resistor is selected so that

when the input voltage is at VIN(min) and the load current is at IL(max) that the current

through the Zener diode is at least Iz(min). Then for all other combinations of input

voltage and load current the Zener diode conducts the excess current thus maintaining

a constant voltage across the load. The Zener conducts the least current when the load

current is the highest and it conducts the most current when the load current is the

lowest.

If there is no load resistance, shunt regulators can be used to dissipate total

power through the series resistance and the Zener diode. Shunt regulators have an



inherent current limiting advantage under load fault conditions because the series

resistor limits excess current.

A Zener diode of break down voltage Vz is reverse connected to an input voltage

source Vi across a load resistance RL and a series resistor RS. The voltage across the

Zener will remain steady at its break down voltage VZ for all the values of Zener current

IZ as long as the current remains in the break down region. Hence a regulated DC output

voltage V0 = VZ is obtained across RL, whenever the input voltage remains within a

minimum and maximum voltage.

Basically there are two type of regulations such as:

a. Line Regulation: In this type of regulation, series resistance and load resistance

are fixed, only input voltage is changing. Output voltage remains the same as

long as the input voltage is maintained above a minimum value.

b. Load Regulation: In this type of regulation, input voltage is fixed and the load

resistance is varying. Output volt remains same, as long as the load resistance is

maintained above a minimum value.

Circuit Diagram:

Fig (1) - Forward Bias Condition:



Fig (2) - Reverse Bias Condition:

Fig (3) - Circuit Diagram of Zener Diode as Line Regulator:

Fig (4) - Circuit Diagram of Zener Diode as Load Regulator:

Procedure:

a) Forward Bias Condition:

1. Connect the circuit as shown in figure (1).

2. Initially vary Vs in steps of 0.1V. Once the current starts increasing vary Vs in

steps of 1V up to 12V. Note down the corresponding readings of Vzf and Izf.



b) Reverse Bias Condition:

1. Connect the circuit as shown in figure (2).

2. Vary Vs gradually in steps of 1V up to 12V and note down the corresponding

readings of Vzr and Izr.

3. Tabulate different reverse currents obtained for different reverse voltages.

c) Zener Diode as Line Regulator (for variations in supply voltage):

Connect the circuit for Line regulation as shown in figure (3).

Vary supply voltage (Vs) in in steps of 1volt from 0 - 15 volts and note the

corresponding Zener Current (IZ), Load Current (IL) and Output Voltage (VO).

Plot the graph between VS and VO taking VS on X-axis and VO on Y-axis.

d) Zener Diode as Load Regulator (for variations in load connected):

1. Connect the circuit for Load regulation as shown in figure (4).

2. Now fix the power supply voltage, Vs at 10V.

3. Without connecting the load RL, note down the No-Load Voltage (VNL).

4. Now connect the load (RL) using Decade Resistance Box (DRB) and vary the

resistance in steps 1K from 1K to10K / in steps of 10 K from10K to

100K and note the corresponding Zener Current (IZ), Load Current (IL) and

Output Voltage (VO) for 10 readings and calculate the percentage regulation.

5. Plot the graph between RL and VO taking RL on X-axis and VO on Y-axis.

Zener Diode Characteristics in X-Y mode:

1. Adjust CRO TIME/DIV knob in X-Y mode.

2. Give the input as triangular voltage waveform from Function Generator (both

positive and negative peaks).

3. Connect the CRO CH1 across the input and CH2 across resistor.

4. Zener diode characteristics can be observed.



Observations:

Table: 1 Forward Bias Condition:

RPS

Voltage Vs(volts)

Forward Voltage across the

diode Vzf (volts)

Forward Current through

the diode Izf (mA)

Table: 2 Reverse Bias Condition:

RPS

Voltage Vs(volts)

Reverse Voltage across the

diode Vzr (volts)

Reverse Current through

the diode Izr (mA)

Table: 3 Line Regulation:

Load Resistance RL = ____________ (K )

Unregulated Power

Supply Vs (V)

Zener Current

Iz (mA)

Load Current

IL (mA)

Regulated Output

Voltage Vo (V)



Table: 4 Load Regulation:

Input Supply Voltage VS = _________ Volts

No-load DC Voltage, VNL = _________ Volts

Load

Resistance RL

(K )

Zener

Current

Iz (mA)

Load

Current

IL (mA)

Regulated Output

Voltage Vo (V)

% Voltage

Regulation

Expected Graph:

Fig: V- I Characteristics of Zener Diode under Forward & Reverse Bias Conditions



Line Regulation

Load Regulation:

Calculations from Graph:

Static forward Resistance Rdc = Vf / If

Dynamic Forward Resistance rac = Vf / If

Static Reverse Resistance Rdc = Vr / Ir

Dynamic Reverse Resistance rac = Vr / Ir

For Load regulation,% Voltage Regulation = %

Precautions:

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

lead to damaging 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.

Result: The characteristics and Voltage Regulation of Zener diode are studied.

a) Forward Bias Zener Diode:

1. The Knee voltage or Cut-in Voltage (Vy) is __________________ Volts.

2. The Dynamic Forward resistance is __________________ .

3. The Static Forward resistance is __________________ .

b) Reverse Bias of Zener Diode:

1. Zener Breakdown Voltage (VZ) is ____________________ Volts.

2. The Dynamic Reverse resistance is __________________ .

3. The Static Reverse resistance is __________________ .

Outcomes: Students are able to

1. Analyze the forward and reverse bias characteristics of Zener diode.

2. Calculate static and dynamic resistance in both forward and reverse bias

condition.

3. Analyze the working of Zener diode as a voltage regulator for line regulation and

load regulation.

Viva Questions:

1. What is the difference between p-n Junction diode and Zener diode?

2. What is break down voltage?

3. What are the applications of Zener diode?

4. What is cut-in-voltage?



5. What is voltage regulator?

6. What is the doping concentration in Zener diodes?

7. Can we use Zener diode as a switch?

8. What is PIV of Zener?

9. What will happen if P-N regions are heavily doped in Zener diode?

10. List the other Zener diodes with different breakdown voltages.

11. Is the breakdown region in Zener really destructible?

12. What is a Zener diode?

13. How the name of the Zener came?

14. What is cause of reverse breakdown?

15. What is Zener voltage?

16. Write the Symbol for the Zener diode.

17. What are the differences between the PN junction diode and Zener diode

fabrication?

18. What are the applications of Zener diode?

19. What are the different types of breakdowns in semiconductor junctions?

20. Compare Zener Breakdown and Avalanche breakdown?

21. What is the max value of voltage of Zener breakdown devices?

22. Zener diode is generally used in ________ biased condition. When Zener diode is

forward biased, it acts as a _________.





EXPERIMENT-04

BJT DARLINGTON EMITTER FOLLOWER

AIM: To determine experimentally and

a) To plot the frequency response of Darlington Emitter Follower and

b) To measure Zi , Zo and find the Current gain Ai.

Equipment and components required:

SL.NO NAME OF EQUIPMENTS/ COMPONENTS SPECIFICATIONS QTY

1. Power Supply 0 – 30 V / 2 Amp D.C. 01

2. A.C.milli Voltmeter(or Digital

Multimeter)

01

3. Function Generator 0 – 3 MHz, 0 – 20 V(p-p) 01

4. CRO Analog, 30 MHz, Dual

Channel

01

5. Terminal Board -- 01

6. Capacitors 0.47 µF (Ceramic) 02

7. DRB -- 02

8. Resistors Designed values

9. Transistor SL 100 02



Boylestad / Louis Nashelsky “ chapter 5“

Procedure:

A] To find Q point:

1) Connect the circuit as shown in the Fig. No. (2)

2) Switch on the power supply and set +12V D.C. as VCC.

3) Measure the DC Voltage using CRO or DC Voltmeter at the VB2, Collector VC2 and

emitter VE2 with respect to Ground.



Then VCE2 = VC2 – VE2

IC2 = IE2 = VE2 / RE

So Q point = (VCE2, IC2)

B] To find Frequency response:

1) Connect the circuit as shown in Fig. No. (1), set VCC = 12 V d.c.

2) Apply a sine wave of 1 V peak to peak amplitude (Vi = 1V p-p) from the Function

Generator.

3) Vary the input sine wave frequency from 10 Hz to 1 MHz in suitable steps and

measure the output Vo of Darlington Emitter Follower circuit at each step using CRO or

AC mili voltmeter (The input Vi must remain constant through the Frequency range).

4) Note down the reading in table given and plot the graph of frequency v/s. Gain in

dB.

BJT Darlington Emitter Follower

Fig. No (1) Circuit Diagram of BJT Darlington Emitter Follower



Fig. No (2) Biasing Circuit BJT Darlington Emitter Follower

Tabular column:

Vin = 1V (p-p)

SL

NO

FREQUENCY

in Hz

Vo(p-p)

in Volts

Av=Vo/Vi POWER GAIN

in dB= 20log10 Av

Fig. (3) Circuit Diagram of BJT Darlington Emitter Follower with Bootstrap



Design:

Let VCC = 12 V D.C. IC2 = 4 mA, β = 100 (for SL 100)

Chose VE2 = VCC / 2 = 12/2 = 6V

RE = 6 / 4mA = 1500 Ω

VB1 = VBE1 + VBE2 + VC2

= 0.7+0.7+6

= 7.4 V

We know 21

21

RR

RVV CCB

21

2

12

4.7

RR

R

21

2616.0RR

R

R2 = 0.616R1 + 0.616R2

0.383R2 = 0.616R1

R2 = 1.61R1

Let R2 = 100 KΩ

R1 = 62.11

Choose R1 = 68 KΩ

(Nearest standard Resistance value)

Choose CC1 = CC2 = 0.47 μF

Current gain [ai]: Vo / Zo Vo Zi

[Ai] Io/Ii = ------------- = ---------. ---------

Vi / Zin Vi Zo

RE = 1.5KΩ



Since Vo = Vin , Ai = Zin / Zo

Procedure:

C] To measure Zi:

1) Connect the circuit as shown in Fig. No (5).

2) Set the following

DRB to minimum (0 Ω)

I/P sine wave amplitude to 1V (p-p)

I/P sine wave frequency to 10 KHz

3) Measure Vo (p-p). Let Vo = Va (say)

4) Increase DRB till Vo = Va/2. The corresponding DRB value gives Zi.

D] To measure Zo:

1) Connect the circuit as shown in Fig. No (6).

2) Set the following

DRB to its maximum resistance value

I/P sine wave amplitude to 1V (p-p)

I/P sine wave frequency to 10 KHz

3) Measure Vo (p-p), let Vo = Vb (say)

4) Decrease DRB till Vo = Vb/2. The corresponding DRB value gives Zo.

E] To measure Zi with bootstrap:

1) Connect the bootstrap circuit as shown in Fig No. (3).

2) Measure Zi as given in [C].



To measure zi:

Fig. No (5)

To measure zo:

Fig. No (6)

Result:

1] Q Point : ________________________

2] Bandwidth (Bw) : ________________________

3] CURRENT GAIN [Ai] : ________________________

4] Input Impedance (Zi without Bootstrap) : ________________________

5] Output Impedance (Z0): ________________________

6] Input Impedance (Zi with Bootstrap): _________________________


1) What are the important characteristics of Darlington emitter follower

2) What are the applications of the Darlington Emitter follower

3) Write the expressions for Av, AI, Vo for the Darlington Emitter follower

4) Explain Bootstrapping technique.



5) Write small signal ac model for the Darlington emitter follower circuit.





EXPERIMENT-05

COMMON EMITTER AMPLIFIER (VOLTAGE DIVIDER BIAS)

AIM: To study the frequency response of Common Emitter Amplifier and calculate its

Bandwidth.


S.No. Name Quantity

1 Transistor BC107 1(One) No.

2 Resistors (100K%u2126, 10K%u2126,

1K%u2126)

1(One) No.

Each

3 Resistors (2.2 K%u2126) 2(Two) No.

4 Capacitors (10µF) 2(Two) No.

5 Capacitors (100µF) 1(One) No.


Equipment:

S.No. Name Quantity

1 Dual DC Regulated Power supply (0 – 30 V) 1(One) No.

2 Cathode Ray Oscilloscope (CRO) (0-20MHz) 1(One) No.

3 Function Generator (0-1MHz) 1(One) No.


Theory:

The common emitter configuration is widely used as a basic amplifier as it has

both voltage and current amplification. Resistors R1 and R2 form a voltage divider

across the base of the transistor. The function of this network is to provide necessary

bias condition and ensure that emitter-base junction is operating in the proper region.

In order to operate transistor as an amplifier, biasing is done in such a way that

the operating point is in the active region. For an amplifier the Q-point is placed so that

the load line is bisected. Therefore, in practical design VCE is always set to VCC/2. This

will confirm that the Q-point always swings within the active region. This limitation can



be explained by maximum signal handling capacity. For the maximum input signal,

output is produced without any distortion and clipping.

The Bypass Capacitor:

The emitter resistor RE is required to obtain the DC quiescent point stability.

However the inclusion of RE in the circuit causes a decrease in amplification at higher

frequencies. In order to avoid such a condition, it is bypassed by a capacitor so that it

acts as a short circuit for AC and contributes stability for DC quiescent condition. Hence

capacitor is connected in parallel with emitter resistance.

The Input/ Output Coupling (or Blocking) Capacitor: An amplifier amplifies

the given AC signal. In order to have noiseless transmission of a signal (without DC), it

is necessary to block DC i.e. the direct current should not enter the amplifier or load.

This is usually accomplished by inserting a coupling capacitor between two stages.

XCC << ( Ri hie )

CC - Output Coupling Capacitor

CB - Input Coupling Capacitor

Frequency response of Common Emitter Amplifier:

Emitter bypass capacitors are used to short circuit the emitter resistor and thus

increases the gain at high frequency. The coupling and bypass capacitors cause the fall

of the signal in the low frequency response of the amplifier because their impedance

becomes large at low frequencies. The stray capacitances are effectively open circuits.



In the mid frequency range large capacitors are effectively short circuits and the stray

capacitors are open circuits, so that no capacitance appears in the mid frequency range.

Hence the mid band frequency gain is maximum.

At the high frequencies, the bypass and coupling capacitors are replaced by short

circuits. The stray capacitors and the transistor determine the response.

Characteristics of CE Amplifier:

1. Large current gain.

2. Large voltage gain.

3. Large power gain.

4. Current and voltage phase shift of 1800.

5. Moderated output resistance.

Circuit Diagram:



Procedure:

1. Connect the circuit as shown in the circuit diagram.

2. Set source voltage VS = 50mV (say) at 1 KHz frequency using the function

generator. Observe the phase difference between input and output by giving

these two signals to the dual channels of CRO.

3. Keeping input voltage constant, vary the frequency from 50 Hz to 1 MHz in

regular steps and note down the corresponding output voltage. Calculate gain in

dB as shown in the tabular column.

4. Plot the graph: gain (dB) verses Frequency on a semi log graph sheet.

5. Calculate the 3-dB bandwidth from the frequency response.

Expected waveform:

(a) The Input & Output Waveforms at 1 KHz

(b) Frequency Response Curve

In the usual application, mid band frequency range are defined as those frequencies at

which the response has fallen to 3dB below the maximum gain (|A| max). These are

shown as fL and fH and are called as the 3dB frequencies (Lower and Upper Cut-Off

Frequencies respectively). The difference between higher cut-off and lower cut-off

frequency is referred to as bandwidth (fH - fL).



Fig: Frequency Response Curve

Calculations from the graph

Bandwidth = fH – fL (in Hz)

Observation tables:

VS = 50mV

Frequency Vo(Volts) Gain = Vo/Vs Gain(dB) = 20 log(Vo/Vs)

Result:

Common Emitter Amplifier is studied and its Bandwidth is calculated.

1. Maximum Gain ( Amax ) = ___________ dB

2. 3dB Gain = ___________ dB

3. 3dB Lower cut-off frequency, fL = ___________ Hz

4. 3dB Upper cut-off frequency, fH = ___________ Hz



5. 3dB Bandwidth ( fH - fL ) = ___________ Hz


1. Calculate the Bandwidth of BJT Common Emitter amplifier.

Viva Questions:

1. What is the equation for voltage gain?

Ans:

2. What is cut off frequency? What is lower 3dB and upper 3dB cut off frequency?

3. What are the applications of CE amplifier?

4. What is active region?

5. What is Bandwidth of an amplifier?

6. What is the importance of gain bandwidth product?

7. Draw h parameter equivalent circuit of CE amplifier.

8. What is the importance of coupling capacitors in CE amplifier?

9. What is the importance of emitter by pass capacitor?

10. What type of feedback is used in CE amplifier?

11. What are the various types of biasing a Transistor?

12. What is Q point of operation of the transistor? What is the region of operation of the

transistor when it is working as an amplifier?

13. Why frequency response of the amplifier is drawn on semi-log scale graph?

14. If Q point is not properly selected, then what will be the effect on the output

waveform?

15. What are the typical values of the input impedance and output impendence of CE

amplifier?



EXPERIMENT-06

CHARACTERISTICS OF JFET

AIM: To study Drain Characteristics and Transfer Characteristics of a Field Effect

Transistor (FET).

Components:

S.No. Name Quantity

1 JFET (BFW11/ BFW10) 1(One) No.

2 Resistor (1K ,100K ) 1(One) No. Each


Equipment:

S.No. Name Quantity

1 Dual DC Regulated Power supply (0 - 30 V) 1(One) No.

2 Digital Ammeters ( 0 - 200 mA) 1(One) No.

3 Digital Voltmeter (0 - 20V) 2(Two) No.


Specifications:

For FET BFW11:

Gate Source Voltage VGS = -30V

Forward Gain Current IGF = 10mA

Maximum Power Dissipation PD = 300mW



Circuit Diagram:

Figure 1.

Pin assignment of FET:

Top View

Bottom View



Operation:

The circuit diagram for studying drain and transfer characteristics is shown in the

figure1.

1. Drain characteristics are obtained between the drain to source voltage (VDS) and

drain current (ID) taking gate to source voltage (VGS) as the constant parameter.

2. Transfer characteristics are obtained between the gate to source voltage (VGS)

and drain current (ID) taking drain to source voltage (VDS) as the constant

parameter.

Procedure:

Drain Characteristics:

1. Connect the circuit as shown in the figure1.

2. Keep VGS = 0V by varying VGG.

3. Varying VDD gradually in steps of 1V up to 10V note down drain current ID and

drain to source voltage (VDS).

4. Repeat above procedure for VGS = -1V.

Transfer Characteristics:

1. Connect the circuit as shown in the figure1.

2. Set voltage VDS = 2V/5V (BFW10/ BFW11).

3. Varying VDD in steps of 0.5V until the current ID reduces to minimum value.

4. Varying VGG gradually, note down both drain current ID and gate-source voltage

(VGS).

5. Repeat above procedure (step 3) for VDS = 4V/ 8V (BFW10/ BFW11).



Observations:

Drain Characteristics

VDD (Volts) VGS = 0V VGS = -1V

VDS(Volts) ID(mA) VDS(Volts) ID(mA)

Transfer Characteristics

VGG (Volts) VDS = 2V/5V VDS = 4V/8V

VGS(Volts) ID(mA) VGS(Volts) ID(mA)

Graph:

1. Plot the drain characteristics by taking VDS on X-axis and ID on Y-axis at a

constant VGS.

2. Plot the transfer characteristics by taking VGS on X-axis and taking ID on Y-axis at

constant VDS.



Calculations from Graph:

1. Drain Resistance (rd): It is given by the relation of small change in drain to

source voltage ( VDS) to the corresponding change in Drain Current ( ID) for a

constant gate to source voltage ( VGS), when the JFET is operating in pinch-off

region.

at a constant VGS (from drain characteristics)

2. Trans Conductance (gm): Ratio of small change in drain current ( ID) to the

corresponding change in gate to source voltage ( VGS) for a constant VDS.

gm at constant VDS (from transfer characteristics).

The value of gm is expressed in mho’s ( ) or Siemens (s).

3. Amplification factor (µ): It is given by the ratio of small change in drain to

source voltage ( VDS) to the corresponding change in gate to source voltage (

VGS) for a constant drain current (ID).

Inference:

1. As the gate to source voltage (VGS) is increased above zero, pinch off voltage is

increased at a smaller value of drain current as compared to that when VGS = 0V.

2. The value of drain to source voltage (VDS) is decreased as compared to that when

VGS = 0V.

Precautions:

1. While performing the experiment do not exceed the ratings of the FET. This may

lead to damage of FET.

2. Connect voltmeter and ammeter with correct polarities as shown in the circuit

diagram.



3. Do not switch ON the power supply unless the circuit connections are checked as

per the circuit diagram.

4. Properly identify the Source, Drain and Gate terminals of the transistor.

Result:


1. Analyze the Drain and transfer characteristics of FET in Common Source

configuration.

2. Calculate the parameters Trans conductance (gm), drain resistance (rd) and

amplification factor (µ).

Viva Questions:

1. Why FET is called a unipolar device?

2. What are the advantages of FET?

3. What is trans conductance?

4. What are the disadvantages of FET?

5. Relation between µ, gm and rd?

6. Why an input characteristic of FET is not drawn?

7. What is the importance of high input impedance?

8. Why wedge shaped depletion region is formed in FET under reverse bias gate

condition?

9. Why FET is less noisy?

10. What is the difference between n- channel FET and p-channel FET?







EXPERIMENT-07

FREQUENCY RESPONSE OF COMMON SOURCE JFET AMPLIFIER

AIM: To study the JFET common source amplifier and find it’s cut off frequencies and

Bandwidth.

Components:

Name Quantity

JFET BFW 11 1

Resistor 6.8K , 8.2K , 1M 2, 1, 1

Capacitor 10µF, 100µF 2,1

Equipment:

Name Range Quantity

Bread Board 1

Dual power supply 0-30V 1

CRO (0-20)MHz 1

Function Generator (0-1)MHz 1

Connecting Wires

Theory:

Of the possible three configurations of JFET amplifiers, common source (CS)

configuration is mostly used. The advantage of using CS configuration is that it has very

high input impedance. Circuit diagram shows the FET amplifier of common source

configuration. The biasing input and couplings are shown in the figure. The mid-range

voltage gain of the amplifier is given by A = gm(rd || RL) At the mid-frequency range,

there is n effect of input and output coupling capacitors. Therefore, the voltage gain and

phase angle are constant in this frequency range. The amplifier shown in the circuit

diagram has only two RC networks that influence its low-frequency response. One

network is formed by the output coupling capacitors and the output impedance looking

at the drain. Just as in the case of BJT amplifier, the reactance of the input coupling



capacitor, reactance increases as the frequency decreases. The phase angle also changes

with change in frequency.

As the frequency is increased beyond mid-frequency range the internal

transistor capacitance effect is predominant. For JFET’s is the internal capacitance

between gate and source. This is also called input capacitance,. The other internal

capacitance, which effects the performance it acts as a feedback circuit, which couples

both, input and output. The effect of both these capacitances is that it reduced the gain

appreciably as in the case of BJT.

Circuit Diagram:

Procedure:

1. Connect the circuit as shown in the circuit diagram.

2. Set source voltage VS = 50mV (say) at 1 KHz frequency using the function

generator.

3. Keeping input voltage constant vary the frequency from 50 Hz to 1 MHz in

regular steps and note down the corresponding output voltage.

4. Plot the graph: gain (dB) verses Frequency on a semi log graph sheet.



5. Calculate the bandwidth from the graph.

6. Calculate all the parameters at mid band frequencies (i.e. at 1 KHz).

7. To calculate voltage gain

Gain AVS = Output Voltage (VO) / Source Voltage (VS)

Expected waveform:

In the usual application, mid band frequency range are defined as those

frequencies at which the response has fallen to 3dB below the maximum gain (|A| max).

These are shown as fL and fH and are called as the 3dB frequencies are simply the lower

and higher cut off frequencies respectively. The difference between higher cut off and

lower cut off frequency is referred to as bandwidth (fH - fL).

Observation tables:

VS = 50mV

Frequency Vo(volts) Gain= Vo/Vs Gain(dB)=20 log(Vo/Vs)



Result:

Viva Questions:

1. Why FET is called as unipolar device?

2. Why the common-source (CS) amplifier may be viewed as a transconductance

amplifier or as a voltage amplifier?

3. What are the characteristics of JFET source amplifier?

4. What is the impedance of FET?







EXPERIMENT-08

INPUT AND OUTPUT CHARACTERISTICS OF N-CHANNEL MOSFET

AIM: To plot the transfer and drain characteristics of n-channel MOSFET and to find the

following parameters

1. Drain resistance

2. Mutual conductance

3. Amplification factor


SL. no EQUIPMENT RANGE

1. Power supply (0-30)V

2. Ammeter (0-100)mA

3. Voltmeter (0-30)V

4. Bread board & wires -

5. Transistor BFW 96

THEORY

MOSFETs are three terminal devices having a source, gate and drain. MOSFET is

the abbreviation of metal oxide semiconductor field effect transistor. It uses a thin layer

of silicon dioxide as an insulator between gate and channel. It is also known as

insulated gate field effect transistor (IGFET). There are two kinds of MOSFET,

depletion and enhancement type.

An N-channel depletion type MOSFET is shown in figure. A heavily doped two N-

type wells are doped on P-type substrate to form source and drain. Between these two

N-type wells a lightly doped N-type material forms a channel. A thin layer of Sio2 which

is an insulating material is fabricated on the surface above the channel and gate



terminal is attached to it. Source and drain terminals are attached to heavily doped N-

type material with metal contacts.

A positive voltage VDS is applied at the drain with respect to source to establish

drain current. When a negative voltage VGS is applied at the gate with respect to the

source, electrons under gate get repelled causing the channel effectively thinner. This

reduces the current through the channel. If the magnitude of VGS is increased, the drain

current decreases. If it is increased further drain current stops.

If a positive voltage is applied at the gate, electrons get attracted in to the

channel and drain current increases. A depletion type MOSFET can operate in depletion

and enhancement modes.

Application

MOSFETs are widely used in VLSI circuits.

Symbol and Pin Diagram



Circuit Diagram:

Drain Resistance

It is defined as the ratio of change in drain to source voltage to the change in drain

current at an operating point, when gate to source voltage remains constant.

Rd = ΔVDS / ΔID with VGS constant

Mutual Conductance

It is defined as the ratio of change in drain current to the change in gate to source

voltage at an operating point, when drain to source voltage remains constant.

Gm = ΔID / ΔVGS with VDS constant.

Amplification Factor

It is defined as the product of drain resistance and mutual conductance.

μ = Rd * Gm



Procedure:

Drain Characteristics

1. Rig up the circuit as per circuit diagram.

2. Set VGS = 0V and vary VDS insteps of 0.5 V and note down the corresponding ID.

Repeat the above procedure for VGS = -1V,-2V,-3V , for the depletion mode

3. Repeat the above procedure for VGS = 1V, 2V, 3V, for the enhancement mode.

4. Plot the graph: VDS vs ID for constant values of VGS.

5. Find the drain resistance Rd = ΔVDS / ΔID with VGS constant

Transfer Characteristics

1. Rig up the circuit as per circuit diagram.

2. Set VDS and vary VGS insteps of 0.5 V and note down the corresponding ID. Repeat

the above procedure for various values of VDS.

3. Plot the graph: VGS vs ID for constant values of VDS.

4. Find the Trans conductance. Gm = ΔID / ΔVGS with VDS constant.



Tabulation

Drain characteristics – Depletion mode

SL. No VGS = VGS = VGS =

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

1.

2.

3.

4.

5.

6.

7.

8.

Drain characteristics – Depletion mode

SL. No VGS = VGS = VGS =

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

1.

2.

3.

4.

5.

6.

7.



Transfer characteristics

VDS = _________

SL. No. VGS (V) ID (mA)

1.

2.

3.

4.

5.

Calculation

Drain resistance Rd = ΔVDS / ΔID

=

Mutual conductance Gm = ΔID / ΔVGS

=

Amplification factor μ = Rd * Gm

=



Model graph

Drain characteristics Transfer Characteristics:

RESULT:

Drain resistance Rd

Mutual conductance

Gm

Amplification factor μ

VIVA-VOCE QUESTIONS

1. What is a unipolar device?

2. What are the types of FET?

3. How a MOSFET is formed?

4. Mention the modes of MOSFET.

5. Explain enhancement mode.



EXPERIMENT NO-09

TESTING OF A TRANSFORMERLESS CLASS B PUSH PULL POWER AMPLIFIER

AND DETERMINATION OF ITS CONVERSION EFFICIENCY

AIM: To determine conversion efficiency of a transformer less class B push pull

Amplifier.


SL.NO NAME OF EQUIPMENTS/

COMPONENTS

SPECIFICATIONS QTY

1. Power supply 0-30V/2AMP D.C. 01

2. Function Generator 0-3MHz,0-20VP-P 01

3. CRO 01

4. Transistors SL100,SK100 01

5. Resistors Designed value 01

6. Terminal board 01

7. C R O 01


Circuit Diagram:

Fig: Transformer less class B push pull power amplifier

E

RL 470Ω

B

E

C

B

C

Vi (0-20VPP)

C 0.33µF

T1

SL1000

T2

SK100

VCC1

10V

VCC2

10V



Procedure:

1. Rig up the circuit as shown in the figure.

2. Adjust VCC to 10V each.

3. Vary the input signal voltage from 0V to 20VPP and observe the output voltage

(peak) using CRO across load resistor (RL) in AC mode.

4. Calculate the efficiency.

Design:

Assume RL=470Ω such that the collector current is less than the maximum collector

current of transistor. Then

ICmax=10/470 = 21.2mA (≤ICmax of transistor)

PDmax (across the transistor) = Pi (dc) max – PO (ac)

=135mW-0

=135mW (≤ PDmax of transistor)

Let output load voltage= V0 (P)

Then IC (P) =V0 (P)/RL

Idc=2IC (P)/∏

Therefore Pi (dc) =Vcc x Idc

P0 (ac) = V0 (P) 2/2RL

Pi (dc) max = VCC X Idc (max)

= 10.2(10)/ (∏) (470Ω)

Pi (dc) max=135mW



Tabular column

Sl

no

Vin(pp)

(V)

V0(P)

(V)

Idc

(mA)

Pi(dc)

(mW)

P0(ac)

(mW)

Efficiency

η%

NOTE: Maximum efficiency is obtained when V0 (P) =VCC

Result:

Output waveform

T

VO

0



Viva questions:

1) What do you understand by Operating point?

2) Why do we choose the Q point at the center of the load line?

3) Name the two techniques used in the stability of the q point .explain.?

4) Define stability factor &Give the expression for stability factor?

5) List out the different types of biasing.

6) What do you meant by thermal runway?

7) Why transistor is called as a current controlled device?

8) Define current amplification factor?

9) What are the requirements for biasing circuits?

10) When does a transistor act as a switch?

11) What is biasing?

12) What is an operating point?

13) What is d.c load line?

14) Explain about the various regions in a transistor?

15) Explain about the characteristics of a transistor?

16) Why the operating point is selected at the Centre of the active region?

17) What is an amplifier?

18) What is small signal amplifier?

19) What is a Darlington pair?







EXPERIMENT NO-10

RC PHASE SHIFT OSCILLATOR

AIM: To design and test a RC Phase Shift Oscillator for a given frequency


SL

NO

NAME OF EQUIPMENTS/ COMPONENTS SPECIFICATIONS QTY.


2. CRO Analog, 30MHz, Dual Channel 01


4. Capacitors 0.01 µF

0.47 µF

03

01

5. Potentiometer 10 KΩ 01


7. Transistor BC107 01

8. Patch cords, Connecting Wires,etc. 01


Boylestad / Louis Nashelsky “chapter 14” page No 746

Circuit Diagram:

Fig. (1) Circuit Diagram of R C Phase Shift Oscillator



Fig. (2) Circuit Diagram of R C Phase Shift Oscillator using FET

DESIGN:

Let VCC = 12 volts IC = 3mA, β = 100(for SL 100)

Choose VE = VCC / 10 = 12/10 = 1.2 V

VE = IERE = 1.2 V

RE = 1.2/Ic = 1.2/3mA = 0.400 KΩ (IE ≈ IC)

RC: Choose VCE = VCC /2 12/2 = 6V

Apply KVL in CE loop:

VCC – ICRC – VCE – VRE = 0

12 – 3mRc – 6 – 1.2 = 0

RC = 1.6 KΩ

Select

R1 and R2:

VB = VBE + VE = 0.7 +470* 3mA = 2.11 V

I2 ≈ IC/10 = .3mA

Then R2=Vb/.3mA =7.03KΩ

RE = 470 Ω

RC = 1KΩ



Let us assume R2 = 6.8KΩ

R1=vcc-vb/I1 (I1=I2) = ( 12-2.11)/.3mA

R1 = 32.9KΩ

Choose R1 = 33KΩ

Procedure:

1] Connect the circuit as shown in Fig. (1).

2] Switch on the D.C. power supply.

3] Observe the output Vo on CRO. The 10 K pot is adjusted to get a stable output on the

CRO.

4] Measure the frequency of the output wave.

5] Compare the measured frequency with theoretical value.

6] With respect to output at point P, observe the waveforms at point Q, R and S on the

CRO. We can see that phase shift at each point being 60°, 120° and 180° respectively.

7] repeat the design for different value of frequency (in Audio range only). At each

case

Compare the generated frequency with theoretical value.

Note:

a) the last Resistor in the phase shifting network is chosen to be a 10 K pot. This is done

to get a overall phase shift of 180° at frequency of Oscillations.

b) The minimum hfe required for the transistor to oscillate is

Hfemin = 23 + 29(R / RC) + 4(RC / R)

Where RC = 1 KΩ and R = 2.2 KΩ (Phase Shifting Network)

Hfe min = 23 + 29(2.2K / 1K) + 4 = 1K / 2.2K

≈ 89

The transistor should be chosen to have a value of hfe greater than 89.

By pass capacitor CE: Let 10

ECE

RX =470/10 =4

At f = 100 Hz; 102

1 ER

fce

Ce=33µF



Choose CE = 47 µF (electrolytic)

Cc1 and CC2: Assume CC1= CC2=0.47 µF (ceramic)

To design:

10

)||(1

RbhieXcc

10

)||2

RLRcXcc

12

11

CcXcc

22

12

CcXcc

CC1 =? CC2 =?

Design of phase shifting network:

The frequency of oscillations is determined by phase shifting network.

The oscillating frequency for the above circuit is given by

R

RcRC

f

462

1

The ratio Rc / R is usually < 1

Let f = 1 KHz (Audio frequency in the range 20 Hz to 20 KHz) and R = 3.9 KΩ

Since Rc = 1KΩ we get

K

KC

9.3

146109.3142.32101

1

33

202.7109.3142.321

16

C

65.2109.3142.321

16

C

6

61001.0

1090.64

1

C

C = 0.01 µF

RESULT: ______________




1) What is Barkhausen’s criterion?

2) How oscillations are generated in RC Phase Shift Oscillator?

3) Give the condition which determines the frequency of oscillation

4) Explain the phase-shift principle in Lc oscillators?

5) Compare and contrast RC & LC oscillators

6) Where do you use IC oscillators?





EXPERIMENT NO-11

HARTLEY OSCILLATOR AND COLPITT’S OSCILLATOR

AIM: To design and test a Hartley and Colpitt’s Oscillator for a given frequency




2. A.C.Milli Voltmeter(or Digital Multimeter) 01

3. CRO Analog, 30MHz, Dual

Channel

01


5. Capacitors 0.47 µF (Ceramic)

47 µF (Electrolytic)

1000pF (or DCB)

02

01


7. Inductance 1mH (or DIB)

2.2mH (or DIB)

01

01



THEORY :Refer Text Book Electronic Devices And Circuit Theory By Robert L.

Boylestad / Louis Nashelsky Chapter“14 “ page no 751.

Procedure:

1] Connect the circuit as shown in Fig. (1)


3] Observe the output on CRO screen.

4] Measure the frequency of the output wave.


6] Repeat the design for different value of frequency. At each case compare the

generated frequency with theoretical value.



Hartley oscillator

Fig 1: Hartley Oscillator

Design:


Choose VE = VCC / 10 = 12/10 = 1.2 V

VE = IERE = 1.2 V

RE = 1.2/Ic = 1.2/3mA = 0.400 KΩ (IE ≈ IC)

RC: Choose VCE = VCC /2 =12/2 = 6V



RE = 470 Ω



12 – 3mRc – 6 – 1.2 = 0

RC = 1.6 KΩ

Select

R1 and R2:

VB = VBE + VE = 0.7 +470* 3mA = 2.11 V

I2 ≈ IC/10 = .3mA

Then R2=Vb/.3mA

=7.03KΩ

Let us assume

R1=vcc-vb/I1 (I1=I2)

= ( 12-2.11)/.3mA

R1 = 32.9KΩ

Choose R1 = 33KΩ

By pass capacitor CE:

Let 10

ECE

RX =470/10 =47

At f = 100 Hz; 102

1 ER

fce

Ce=33µF



RC = 1KΩ

R2 = 6.8KΩ



To design:

10

)||(1

RbhieXcc

10

)||2

RLRcXcc

12

11

CcXcc

22

12

CcXcc

CC1 =? CC2 =?

Tank Circuit Design:

Oscillator Frequency f = 100 KHz

f = ..2

1

CLeq = 100 x 103 =

..2

1

CLeq

Where Leq. = L1 + L2

In designing Split Inductors, the ratio

21

2 L

L or L2 = 2L1

Let L1 = 1 mH L2 = 2.2 mH Leq = 3.2 mH

D) 22 .4

1

fLC

eq = 791.57 pF

Choose C=1000 pF

RESULT: ______________



Colpitt’s Oscillator:





Channel

01


5. Capacitors 0.47 µF (Ceramic)

47 µF (Electrolytic)

2200pF (or DCB)

1000pF (or DCB)

02

01

01

01

6. Resistors

7. Inductance 3.6mH (or DIB) 01


9. Patch cords, Connecting Wires,etc.


Boylestad / Louis Nashelsky Chapter“14“ page no 750.

Procedure:



3] Observe the output waveform on CRO screen.

4] Measure the frequency of the output waveform.


6] Repeat the design for different values of frequency. At each case compare the

generated frequency with theoretical value.



Circuit Diagram:

Fig. (1) Circuit Diagram of Colpitt’s Oscillator

Design:


Choose VE = VCC / 10 = 12/10 = 1.2 V

VE = IERE = 1.2 V

RE = 1.2/Ic = 1.2/3mA = 0.400 KΩ (IE ≈ IC)




RE = 470 Ω



12 – 3mRc – 6 – 1.2 = 0

RC = 1.6 KΩ

Select

R1 and R2:

VB = VBE + VE = 0.7 +470* 3mA = 2.11 V

I2 ≈ IC/10 = .3mA

Then R2=Vb/.3mA

=7.03KΩ


R1=vcc-vb/I1 (I1=I2)

= ( 12-2.11)/.3mA

R1 = 32.9KΩ

Choose R1 = 33KΩ

By pass capacitor CE:

Let 10

ECE

RX =470/10 =47

At f = 100 Hz; 102

1 ER

fce

Ce=33µF



To design:

10

)||(1

RbhieXcc

10

)||2

RLRcXcc

RC = 1KΩ



12

11

CcXcc

22

12

CcXcc

CC1 =? CC2 =?

Tank Circuit Design:

Oscillation frequency f = 100 KHz

eqLC

f2

1

Where 21

21

CC

CCC

Assume C1= 1000 pF and C2 = 2200 pF

12

24

10)22001000(

1022001000

C

pFC 5.687)22001000(

1022001000 12

Cf

L)(4

122

L = 3.6mH

RESULT: _______________




1) what is a tank circuit

2) what are the important components of tank circuit

3) What are the advantages of Hartley oscillator compared to RC-phase shift

oscillator.

4) What is the frequency range of Hartley Oscillator

5) Explain the conditions required to produce sustained oscillations

6) What are the basic properties of an inductor and a capacitor?

7) Explain Barkhausen criteria for sustained oscillations

8) What is the difference between Hartley and colpitts oscillator

9) Explain the condition for sustained oscillations for colpitts oscillator

10) Write the equation for frequency for oscillations and conditions for sustained

oscillations

11) Write the frequency range of colpitts oscillator





EXPERIMENT NO-12

CRYSTAL OSCILLATOR

AIM: To design and test the crystal Oscillator






Channel 01


5. Capacitors

0.1µF (Ceramic)

.01 µF (Ceramic)

330pF (or DCB)

02

01

01

6. Resistors


9 Crystal 3.547MHz 01



Boylestad / Louis Nashelsky Chapter “14“, page No 753.

Procedure:



3] With the crystal disconnected measure the D C potentials using the Multimeter at

the Base, emitter and collector of the transistor.

4] The 1KΩ pot is adjusted to get a stable sinusoidal output on the CRO screen.

5] Measure the frequency of the output wave and compare with the value indicated

on the crystal.



Circuit Diagram:

Fig 1: Circuit

Diagram of Crystal

Oscillator

Design:

Let VCC = 12 volts IC

= 3mA, β = 100(for

SL 100)

Choose VE = VCC / 10 = 12/10 = 1.2 V

VE = IERE = 1.2 V

RE = 1.2/Ic = 1.2/3mA = 0.400 KΩ (IE ≈ IC)




12 – 3mRc – 6 – 1.2 = 0

RC = 1.6 KΩ

Select

R1 and R2:

VB = VBE + VE = 0.7 +470* 3mA = 2.11 V

RE = 470 Ω

RC = 1KΩ



I2 ≈ IC/10 = .3mA

Then R2=Vb/.3mA

=7.03KΩ


Use bypass and coupling capacitor of very small value say 0.1µF

RESULT:

PROBABLE VIVA QUESTIONS:

1) Explain the piezo electric phenomena

2) What are the characteristics of a quartz crystal

3) Name some crystals which exhibit piezo electric effect

4) Explain the characteristics of quartz crystal

5) What are the merits and demerits of crystal oscillator

6) What are the applications of crystal oscillator







LABORATORY VIVA QUESTIONS & ANSWERS

1. What is a rectifier? Differentiate between Half Wave and Full Wave rectifier

Rectifier is a circuit which converts AC signal to DC signal (with ripples)

HWR has one diode and conducts during positive half cycle and FWR has 2

diodes and each diode conducts for one half cycle

2. Explain the working of a half wave rectifier

3. Explain the working of full wave rectifier and full wave bridge rectifier.

4. Differentiate between full wave and full wave bridge rectifier

FWR needs two diodes and PIV is 2Vm and center tapped transformer is

required and hence difficult to find the center tapped

FWBR needs 4 diodes and PIV of each diode is Vm and does not center tapped

transformer

5. Explain the following 1. Ripple factor 2. Efficiency 3. regulation

Ripple factor is defined as the ratio of ac component to dc component.

It gives the measure of amount of ac content in the dc output

Efficiency is defined as the ratio of dc output power to ac input power

Efficiency= Pdc/Pac

Regulation is defined as the change in dc output voltage with change in dc load

current

(Or other words change in load).

6. What are filters? Mention the different types of filters

Filters are those circuits which are used to reduce the ripples or ac component in

the dc output or rectified output

Types: ‘C’, Choke (L), Pi filters

7. Explain the working of ‘C’ filter



8. Explain the advantages of filter.

Filters are used to reduce the ac component so that the output is pure dc and Pi

filters are widely preferred as it filters ac component completely

9. Why filters are widely preferred over other types of filters.

10. Define load regulation and line regulation

Load regulation is nothing but it maintains constant output voltage with changes

in Load current (load resistance.)

Line regulation is a measure of how constant the output voltage with changes in

the input voltage.

11. Explain TUF

12. What is thermal runaway

Self-destruction of transistors or diodes, with increase in temperature is due to

regenerative effect. (As temperature increases leakage current and in turn

collector current increases which increases junction temperature, and this

processes repeates.)

13. What is a bipolar and unipolar devices. Give examples

Flow of current is due to both minority and majority charge carriers is called

bipolar device. Ex: transistor.

In Unipolar device flow of current is due to majority charge carriers only. Exam:

UJT.

14. Differentiate between voltage and current controlled devices. Give examples

Output current changes w.r t to changes in the input current is called current

controlled device. Ex: transistor.

Output varies w.r t changes in the input voltage is called voltage controlled

devices.

Ex: FET.



15. Differentiate between Uni directional and bidirectional devices. Give examples.

Diode conducts in only one direction i.e. when forward bias and hence are called

unidirectional devices.

Triac and diac conducts in both the directions and hence are called bidirectional

devices.

16. Define resolution

It is the performance parameter of DAC.

Resolution is defined as the change in output voltage resulting from a change in

one LSB at the digital inputs. It is also defined as reciprocal of step size.

17. Explain the block diagram of a CRO

18. What are Lissajous Patterns?

Lissajous patterns or figures are used to measure the phase difference between

the two signals and to measure the frequency of unknown signal if the frequency

of the other signal is known.

19. How do you measure voltage, current and frequency of a signal using CRO?

20. Explain 1. Phase measurement 2. Frequency measurement of an unknown

signal using

Lissajous Patterns

21. What is the need for diodes in Astable Multivibrator?

22. Give the ideal characteristics of a diode.

23. Give the application of voltage or emitter follower and also give its Advantages

The gain of voltage/emitter follower is one and hence called buffer amplifier.

The input impedance of emitter follower is very large and output impedance is

very small. It basically finds application in impedance matching.

24. Explain totem pole logic in TTL gates.



25. What is an oscillator?

Oscillator is a circuit which provides oscillations i.e. it generates waveforms

without any input signal.

26. Give the classification of oscillators based on

i. Frequency of the signal generated

ii. on the tank circuit or oscillatory circuit

27. Explain the working principle of a LC tank circuit

28. Explain the working of following oscillators. Also indicate there frequency

range of

operation and applications

1. Crystal oscillator 2. RC phase shift oscillator 3. Colpitts and Hartley

oscillators

29. Give The Ratings Of A Resistor.

Ohmic ratings, Tolerance rating and Wattage rating

30. Name the two types of capacitor

a) Non-Electrolytic Capacitor

b) Electrolytic Capacitor

31. Give the rating of a Capacitor

a) Voltage rating

b) Farads rating

32. Give the rating of a rectifier diode

a) Cut in Voltage rating

b) Forward Current rating

c) Peak Inverse Voltage rating (PIV rating)



33. Give the ratings of a Zener diode

a) Power ratings

b) Voltage ratings

34. Name the different types of transistor

(a)BJT- Bipolar Junction transistor

(b)FET- Field affected transistor

(c)UJT- Uni-junction transistor

(d)MOSFET- Metal Oxide Semiconductor field effected transistor

35. List the different Types of diodes

a) Rectifier diode

b) Zener diode

c) Detector Diode (Point Contact diode)

d) Light Emitting diode (LED)

e) Photo diode

f) Tunnel diode

g) Gun diode

h) Aviator diode

36. How the transistors are classified on the basises frequency handing Capability

a) Audio frequency Transistors (ex: BC148, SL100, Ac128 Ac127 etc, )

b) Radio frequency Transistors (Bf194, BF195 etc,)

c) VHF/UHF Transistors (BD115 etc,)

37. Give the Classification of Transformers

a) Isolation Transformer (1:1)



b) Power supply Transformer

(i) Step-up Transformer

(ii) Step-down Transformer

(iii) Secondary Centre tap Transformer

(iv) Secondary mulitapping Transformer

c) IFT’S (Intermediate frequency tunings transformer)

d) Audio Driver and Audio output transformers

e) Auto transformer.

38. Name different signal Sources

a) DC Source

b) A.C. Source (i) Sinusoidal (ii) Square (iii) Triangular

c) Pulse and Ramp

39. Write the Symbol for the following

a) Resistance

b) Variable resistor

c) Diode (Rectifier)

d) BJT (NPN-PNP)

e) FET

40. Name the three Transistor Configurations

a) Common Base Configuration

b) Common Emitter Configuration

c) Common Collector Configuration



41. Define , ,& with respect to transistor

a) is Current amplification factories common base configuration

b) is Current amplification factor in common emitter

configuration

c) is Current amplification factor in common collector

configuration

43. Name the three terminals of a bipolar junction transistor

Three terminals of BJT are Base, Emitter, and Collector

44. Explain the meaning of Bipolar Junction Transistor?

Both holes and electrons are responsible for the performance of transistor

(i.e, current conduction in bipolar transistor is due to both holes & electrons)

45. When transistor is used as an amplifier, how transistor should be biased?

Input circuit must be forward biased and output circuit must be reverse biased

46. Why CE configuration is widely used in amplifier circuit

i) The CE configuration is the only configuration which provides both voltage

gain as

well as current gain greater than unity

ii) In a Common emitter circuit, the ratio of output resistance to input

resistance is

small. This makes Configuration an ideal for coupling between various

transistors

stages.

47. Name the three operating regions of transistors?

i) Cut off ii) Active iii) Saturation.



48. Why the biasing is required for a transistor?

To operate the transistor in desired region.

49. What is operating point means?

Operating point means zero signals Ic and Vce.

50. Where the transistor operating point must be chosen?

The Q-point (operating point) is chosen along the Load line.

51. What is stabilization with reference to transistor amplifier circuit?

Stabilization means making transistor circuit independent of temperature

Variation or

independent of transistor parameters.

52. What is thermal runaway in transistor?

The increase in the collector current increases in power dissipated at the

Collector

junction. This, in turn further increases the temperature of the junction and

hence

increases the collector current. The process is cumulative. The excess heat

produced at the collector base junction may even burn and destroy the transistor.

This situation is

called “Thermal runway” of the transistor.

53. What should be the stability factor value?

Ideally, stability factor should be perfectly one to keep operating point stable.

54. Name the three biasing techniques?

i)Fixed Bias (Base resistor biasing)

ii) Voltage divider biasing (potential divider biasing)

iii) Feedback resistor biasing (collector to Base bias)



55. What is ac load line?

If we draw a new load line for an transistor amplifier circuit to describe the

Circuit performance when signal is applied. Such load line is called “ac load line”

56. Summarize the h-parameters for the entire three configurations?

Parameters CB CE CC

Input resistance hib hie hic

Reverse voltage gain hrb hre hre

Forward current gain hfb hfe hfc

Output admittance ho

b

hoe hoc

Important Points

1) The decrease in voltage gain with frequency is called roll –off.

2) The band width of an amplifier is determined by cut-off frequencies.

3) The voltage gain of a amplifier is maximum at “midrange frequencies”.

4) The audio frequency range is 20Hz to 20 KHz.

5) The voltage gain is given by 20 logs Av.

6) If Av is greater than one, the DB gain is positive.

Documents

EXPERIMENT-1 FULL WAVE RECTIFIER AND BRIDGE RECTIFIER …mitmysore.in/downloads/ece/AEC_Manual_2018-2019.pdf · To construct full wave rectifier and bridge rectifier with and without