EXPERIMENT NO:- - Government Engineering …gecbharuch.com/tutorial/New EDC Experiments 1 to 4.docx · Web viewTrainer Kit Theory: A p-n junction diode conducts only in one direction

EXPERIMENT NO:- 1

Verify and Compare characteristics of PN Junction Silicon and Germanium Diode

EXPERIMENT NO:- 1

AIM : Verify characteristics of PN Junction Diode

Apparatus: P-N Diode 1N4007, Regulated Power supply (0-15V) , Resistor 1KΩ , Ammeters (0-200mA, 0-200µA) , Voltmeter (0-20V) , Breadboard

Connecting wires OR Trainer Kit

Theory :

A p-n junction diode conducts only in one direction. The V-I characteristics of the diode are

curve between voltage across the diode and current through the diode. When external voltage is

zero, circuit is open and the potential barrier does not allow the current to flow. Therefore, the

circuit current is zero. When P-type (Anode is connected to +ve terminal and n- type (cathode) is

connected to –ve terminal of the supply voltage, is known as forward bias. The potential barrier

is reduced when diode is in the forward biased condition. At some forward voltage, the potential

barrier altogether eliminated and current starts flowing through the diode and also in the circuit.

The diode is said to be in ON state. The current increases with increasing forward voltage.

When N-type (cathode) is connected to +ve terminal and P-type (Anode) is connected –ve

terminal of the supply voltage is known as reverse bias and the potential barrier across the

junction increases. Therefore, the junction resistance becomes very high and a very small current

(reverse saturation current) flows in the circuit. The diode is said to be in OFF state. The reverse

bias current is due to the minority charge carriers.

Procedure:

(1) Forward Biased Characteristics

1. Connect the circuit as per Figure-1. Keep current range on Digital Current meter at 20mA range.

2. By increasing the voltage across the diode in steps, note down corresponding current in the table shown Below.

3. Now connect series resistor 1K instead of 100 Ohms and observe the forward Current. Similarly use variable resistor 470E. The forward current will change as per series resistor value

Observation Table

S.No.Input Power Supply

Voltage (V)Silicon Diode Germanium Diode

Current (IF) in mA

Voltage (VF) in V

Current (IF) in mA

Voltage (VF) in V

1. 0.12. 0.23. 0.34. 0.45. 0.56. 0.67. 0.78. 1.09. 2.010 3.011. 6.012. 8.013 10.0

(2) Reverse Biased Characteristics:

1. Connect the circuit as per Figure-2. Keep current range on Digital Current meter at 20uA range.

2. By increasing the voltage across the diode in steps, note down corresponding current in the table shown Below

Observation Table

S.No.Input Power Supply

Voltage (V)

Silicon Diode Germanium DiodeCurrent (IR)

in mAVoltage

(VR) in VCurrent (IR)

in mAVoltage

(VR) in V1. 12. 23. 34. 45. 66. 87. 108. 129. 1410 1611. 1812. 20

Note : reverse bias current in Silicon diode is negligible. Add observation if it requires

Note : Plot the graph on separate graph piece of paper according to observation reading on laboratory session

Diode Trainer board

CONCLUSIONS:-

Tutorial

1. Define depletion region of a diode. 2. What is meant by transition & space charge capacitance of a diode? 3. Is the V-I relationship of a diode Linear or Exponential? 4. Define cut-in voltage of a diode and specify the values for Si and Ge diodes. 5. What are the applications of a p-n diode? 6. Draw the ideal characteristics of P-N junction diode. 7. What is the diode equation? 8. What is PIV? 9. What is the break down voltage? 10. What is the effect of temperature on PN junction diodes?

Signature of Faculty

EXPERIMENT NO:- 2 P-N Junction diode as half

wave rectifier.

EXPERIMENT NO:- 2

AIM: To examine the input and output waveforms of half wave Rectifier and alsocalculate its load regulation and ripple factor.1. with Filter2. without Filter

APPARATUS:

Digital multimetersMultimeter - 1No.Transformer (6V-0-6V) - 1No.Diode, 1N4007 - 1No.Capacitor 100μf/470 μf - 1No.Decade Resistance Box - 1No.BreadboardCRO and CRO probesConnecting wires or Trainer board

THEORY:

In Half Wave Rectification, When AC supply is applied at the input, only Positive HalfCycle appears across the load whereas, the negative Half Cycle is suppressed. How this can be explained as follows:

During positive half-cycle of the input voltage, the diode D1 is in forward bias and conducts through the load resistor RL. Hence the current produces an output voltage across the load resistor RL, which has the same shape as the +ve half cycle of the input voltage.

During the negative half-cycle of the input voltage, the diode is reverse biased and there is no current through the circuit. i.e., the voltage across RL is zero. The net result is that only the +ve half cycle of the input voltage appears across the load. The average value of the half wave rectified o/p voltage is the value measured on dc voltmeter.

For practical circuits, transformer coupling is usually provided for two reasons.1. The voltage can be stepped-up or stepped-down, as needed.2. The ac source is electrically isolated from the rectifier. Thus preventing shock hazards in the secondary circuit. The efficiency of the Half Wave Rectifier is 40.6%

Theoretical calculations for Ripple factor:Without Filter:

Vrms=Vm/2Vm=2Vrms

Vdc=Vm/π

Ripple factor r = √( V rms

V dc)

2

−1=1.21

With Filter: Ripple factor, r=

12√3 f ∙C ∙R

CIRCUIT DIAGRAM:

A) Half wave Rectifier without filter:

B) Half wave Rectifier with filter:

EXPECTED WAVEFORMS:A) INPUT WAVEFORM

B) OUTPUT WAVFORM WITHOUT FILTER

C) O UTPUT WAVEFORM WITH FILTER:

PROCEDURE:Without filter:

01. Connect the circuit as per the circuit d i a g r a m 02. Connect CRO across the load 03. Note down the peak value VM of the signal observed on the CRO 04. Switch the CRO into DC mode and observe the waveform. Note down the DC shift 05. Calculate Vrms and Vdc values by using the formulae

Calculate Vrms & Vdc by using the formulas

V rms=V m

√2 , I rms=

I m

√2 , V dc∨Avg.=

V m

π , I dc∨ Avg .=

I m

π

06. Calculate the ripple factor by using the formulae

Ripple factor=VacVdc

=√( V rms

V dc)

2

−1

With filter:

01. Connect the capacitor filler across the load in the above circuit diagram 02. Proceed with the same procedure mentioned above to measure Vac value

from the CRO and also dc shift from CRO

03. Calculate Vac & Vdc by using the formulas


= 12√3 f ∙C ∙ R

Where Vac is the peak to peak amplitude of filter output 04. Calculate ripple factor

REGULATION CHARACTERSTICS: 1. Connections are made as per the circuit diagram. 2. By increasing the value of the rheostat, the voltage across the load and current flowing

through the load are measured. 3. The reading is tabulated. 4. Draw a graph between load voltage (VL ) and load current ( IL ) taking VL on X-axis and

IL on y-axis 5. From the value of no-load voltages, the % regulation is calculated using the formula,

% Regulation=(V NL−V FL)

V FL× 100

PRECAUTIONS:1. The primary and secondary side of the transformer should be carefully identified2. The polarities of all the diodes should be carefully identified.3. While determining the % regulation, first Full load should be applied and then it

should be decremented in steps.

OBSERVATIONS: WITHOUT FILTER:

RL

(K)Vac

(Volts)Vdc

(Volts) Ripple factor=V ac

V dc

% Regulation

¿(V NL−V FL)

V FL× 100

WITH FILTER:

RL

(K)Vac

(Volts)Vdc

(Volts)Ripple factor=¿

V ac

V dc= 1

2√3 f ∙ C ∙ R

% Regulation

¿(V NL−V FL)

V FL× 100

MODEL GRAPHS: HALFWAVE RECTIFIER (WITH & WITHOUT FILTER):

Note : Plot the graph on separate graph piece of paper according to observation reading on laboratory sessionConclusion:

Tutorial

1. What is the PIV of Half wave rectifier? 2. What is the efficiency of half wave rectifier? 3. What is the rectifier? 4. What is the difference between the half wave rectifier and full wave Rectifier? 5. What is the o/p frequency of Bridge Rectifier? 6. What are the ripples? 7. What is the function of the filters? 8. What is TUF? 9. What is the average value of o/p voltage for HWR? 10. What is the peak factor?

EXPERIMENT NO:-3 P-N Junction diode as centre

tapped full wave rectifier

EXPERIMENT NO:- 3

AIM: To examine the input and output waveforms of Full Wave Rectifier and also

calculate its load regulation and ripple factor.1. with Filter2. without Filter

APPARATUS:Digital multimetersMultimeter - 1No.Transformer (6V-0-6V) - 1No.Diode, 1N4007 – 2 No.Capacitor 100μf/470 μf - 1No.Decade Resistance Box - 1No.BreadboardCRO and CRO probesConnecting wires or Trainer board

THEORY:

The circuit of a center-tapped full wave rectifier uses two diodes D1&D2. During positive half cycle of secondary voltage (input voltage), the diode D1 is forward biased and D2is reverse biased. So 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 in the same direction. There is a continuous current flow through the load resistor RL, during both the half cycles and will get unidirectional current as show in the model graph. The difference between full wave and half wave rectification is that a full wave rectifier allows unidirectional (one way) 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 degree).

THEORITICAL CALCULATIONS:

Without Filter:Vrms=Vm/2Vm=2Vrms

Vdc=2Vm/π


V dc)

2

−1=1.21

With Filter:

Ripple factor, r=1

4√3 f ∙C ∙ R

CIRCUIT DIAGRAM:A) FULL WAVE RECTIFIER WITHOUT FILTER:

B) FULL WAVE RECTIFIER WITH FILTER:


B) OUTPUT WAVEFORM WITHOUT FILTER:

C)OUTPUT WAVEFORM WITHOUT FILTER:

PROCEDURE:Without filter:1. Connect the circuit as per the circuit diagram 2. Connect CRO across the load 3. Note down the peak value VM of the signal observed on the CRO 4. Switch the CRO into DC mode and observe the waveform. Note down the DC shift 5. Calculate Vrms and Vdc values by using the formulae


V rms=V m

√2 , I rms=

I m

√2 , V dc∨Avg.=

V m

π , I dc∨ Avg .=

I m

π



=√( V rms

V dc)

2

−1

With filter: 01. Connect the capacitor filler across the load in the above circuit diagram 02. Proceed with the same procedure mentioned above to measure Vac value from the

CRO and also dc shift from CRO 03. Calculate Vac & Vdc by using the formulas


= 14√3 f ∙C ∙ R






V FL× 100

PRECAUTIONS:

1. The primary and secondary side of the transformer should be carefully identified 2. The polarities of all the diodes should be carefully identified. 3. While determining the % regulation, first Full load should be applied and then it



RL

(K)Vac

(Volts)Vdc


V dc

% Regulation

¿(V NL−V FL)

V FL× 100

WITH FILTER:

RL

(K)Vac

(Volts)Vdc


V ac

V dc= 1

4 √3 f ∙ C ∙ R

% Regulation

¿(V NL−V FL)

V FL× 100

MODEL GRAPHS: FULLWAVE RECTIFIER (WITH & WITHOUT FILTER):


Tutorial

01. Define regulation of the full wave rectifier? 02. Define peak inverse voltage (PIV)? And write its value for Full-wave rectifier? 03. If one of the diode is changed in its polarities what wave form would you get? 04. Does the process of rectification alter the frequency of the waveform? 05. What is ripple factor of the Full-wave rectifier? 06. What is the necessity of the transformer in the rectifier circuit?

07. What are the applications of a rectifier? 08. What is meant by ripple and define Ripple factor? 09. Explain how capacitor helps to improve the ripple factor? 10. Can a rectifier made in INDIA (V=230v, f=50Hz) be used in USA (V=110v, f=60Hz)?

EXPERIMENT NO:- 4 P-N Junction diode as full

wave bridge rectifier.

EXPERIMENT NO:- 4

AIM: To examine the input and output waveforms of Bridge Rectifier and also calculate its

load regulation and ripple factor.1. with Filter2. without Filter

APPARATUS:Digital Multimeter - 1No.Transformer (6V-0-6V) - 1No.Diode, 1N4007 – 4 No.Capacitor 100μf/470 μf - 1No.Decade Resistance Box - 1No.BreadboardCRO and CRO probesConnecting wires or Trainer board

THEORY:

The bridge is seen to consist of four diodes connected with their arrowhead symbols all

pointing toward the positive output terminal of the circuit.

During the positive half cycle of input voltage, the load current flows from the positive

input terminal through D1 to RL and then through RL and D4 back to the negative input

terminal. During this time, the positive input terminal is applied to the cathode of D2 so it

is reversed biased and similarly D3 is also reverse biased. These two diodes are forward

biased during negative half cycle; D1 & D4 are reverse biased during this cycle. And finally

both half cycles are rectified.

THEORITICAL CALCULATIONS:

Without Filter:Vrms=Vm/2Vm=2Vrms

Vdc=2Vm/π


V dc)

2

−1=0.48

With Filter: Ripple factor, r=

14√3 f ∙C ∙ R

CIRCUIT DIAGRAM:A) FULL WAVE RECTIFIER WITHOUT FILTER:

B) FULL WAVE RECTIFIER WITH FILTER:


B) OUTPUT WAVEFORM WITHOUT FILTER:

C)OUTPUT WAVEFORM WITHOUT FILTER:

PROCEDURE:Without filter:

1. Connect the circuit as per the circuit diagram 2. Connect CRO across the load 3. Note down the peak value VM of the signal observed on the CRO 4. Switch the CRO into DC mode and observe the waveform. Note down the DC shift 5. Calculate Vrms and Vdc values by using the formulae


V rms=V m

√2 , I rms=

I m

√2 , V dc∨Avg.=

V m

π , I dc∨ Avg .=

I m

π



=√( V rms

V dc)

2

−1

With filter: 01. Connect the capacitor filler across the load in the above circuit diagram 02. Proceed with the same procedure mentioned above to measure Vac value from the

CRO and also dc shift from CRO 03. Calculate Vac & Vdc by using the formulas


= 14√3 f ∙C ∙ R






V FL× 100

PRECAUTIONS:

4. The primary and secondary side of the transformer should be carefully identified 5. The polarities of all the diodes should be carefully identified. 6. While determining the % regulation, first Full load should be applied and then it



RL

(K)Vac

(Volts)Vdc


V dc

% Regulation

¿(V NL−V FL)

V FL× 100

WITH FILTER:

RL

(K)Vac

(Volts)Vdc


V ac

V dc= 1

4 √3 f ∙ C ∙ R

% Regulation

¿(V NL−V FL)

V FL× 100

MODEL GRAPHS: FULLWAVE RECTIFIER (WITH & WITHOUT FILTER):


Signature of Staff.

Tutorial

01. Define peak inverse voltage (PIV)? And write its value for Bridge rectifier? 02. What are the applications of a bridge rectifier? 03. Why is a bridge type full wave rectifier preferred over centre tap rectifier?04. Show that ripple factor of bridge rectifier is 0.48 and efficiency = 81%.

Documents

EXPERIMENT NO:- - Government Engineering …gecbharuch.com/tutorial/New EDC Experiments 1 to 4.docx · Web viewTrainer Kit Theory: A p-n junction diode conducts only in one direction