Manual - SSC I

Ex No:

Date:

POWER SUPPLY-HALF WAVE RECTIFIER

Aim:To design a power supply using half wave rectifier with simple capacitor filter.

Components Required:

S.No Components Required Range Quantity1 Regulated Power Supply 2 Transformer3 Diode4 Resistor5 Capacitor6 CRO7 Bread Board8 Connecting Wires

Theory:Half Wave Rectifier converts an ac voltage into a pulsating d.c. voltage using only

one half of the applied ac voltage. The rectifying diode conducts during one half of the ac cycle only.

During the positive half cycle of the input signal, the anode of the diode becomes more positive with respect to the cathode and hence, diode conducts. For an ideal diode, the forward voltage drop is zero. So the whole input voltage will appear across the resistor.

During the negative half cycle of the input signal, the anode of the diode becomes negative with respect to the cathode and hence, diode does not conduct. For an ideal diode, the impedance offered by the diode is infinity. So the whole input voltage will appear across the diode.

Procedure:1. Connections are made as per the circuit diagram with correct polarity provision given

to the circuit from the transformer.2. The peak voltage (Vm) is noted using CRO.3. Corresponding efficiency (η) and ripple factor (Γ) are calculated and compared with

the theoretical value.

Circuit Diagram:

Without Capacitor:

With Capacitor:

Model Graph:

Theoretical Calculation

Ripple Factor

Ripple factor is defined as the ratio of rms value of ac component to the dc component in the output.

Ripple Factor (Γ) =RMS value of t h e accomponentDC value of t hecomponent = V r ,rms

V dc

V r , rms = √V rms2 −V dc

2

Where, V rms=V m

√2

Γ = √(V rms

V dc)

2

−1

Vav the average or the dc content of the voltage across the load is given by

V av=V dc=1

2π [∫0π

V m sinωtd (ωt )+∫π

2π

0.d (ωt )] =

V m

2π[−cosωt ]0

π =V m

π

I dc=V dc

RL=V m

π RL

RMS voltage at the load resistance can be calculated as

V rms=[ 12π∫0

π

V m2 sin2ωt d (ωt )]

1 /2

=V m ¿¿= V m

2

Γ = √(V m

2V m

π)

2

−1 = 1.21

Efficiency

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

η = DC output powerAC input power = Pdc

Pac

=

V dc2

RL

V rms2

RL

=[V m

π ][V m

2 ]2

2

= 4π2 =0.406=40.6 %

Tabulation:

Amplitude(V) Time Period(ms)

Input Signal

Without Filter

With Filter

Practical Calculation:

Tabulation

Ripple Factor (Γ)Theoretical 1.21

Practical

Efficiency (η)%Theoretical 40.6

Practical

Ex No:

Date:

POWER SUPPLY-FULL WAVE RECTIFIER

Aim:To design a power supply using full wave rectifier with simple capacitor filter.


S.No Components Required Range Quantity1 Regulated Power Supply Dual (0-30)V 12 Transformer (9-0-9)V 13 Diode4 Resistor5 Capacitor6 CRO (0-20)MHz 17 Bread Board - 18 Connecting Wires few

Theory:Full Wave Rectifier converts an ac voltage into a pulsating d.c. voltage using both

half cycles of the applied ac voltage. It uses two diodes of which one conducts during one half-cycle which the other diode conducts during the other half-cycle of the applied voltage.

During the positive half cycle of the input signal, anode of the diode D1 becomes positive and at the same time the anode of the diode D2 becomes negative. Hence, diode D1 conducts and the diode D2 does not conduct. The load current flow through the diode D1 and the voltage drop across the load will be equal to the input voltage.

During the negative half cycle of the input signal, anode of the diode D1 becomes negative and at the same time the anode of the diode D2 becomes positive. Hence, diode D2 conducts and the diode D1 does not conduct. The load current flow through the diode D2 and the voltage drop across the load will be equal to the input voltage.

Procedure:1. Connections are made as per the circuit diagram with correct polarity provision given

to the circuit from the transformer.2. The peak voltage (Vm) is noted using CRO.3. Corresponding efficiency (η) and ripple factor (Γ) are calculated and compared with


Circuit Diagram:

Without Capacitor:

With Capacitor:

Model Graph:

Ripple Factor



V dc


2

Where, V rms=V m

√2

Γ = √(V rms

V dc)

2

−1


V av=V dc=1π [∫0

π

V m sinωt d (ωt )] =

V m

π[−cosωt ]0

π =2V m

π

I dc=V dc

RL=

2V m

π RL


V rms=[ 1π∫0

π

V m2 sin2ωt d(ωt)]

1 /2

=V m ¿¿= V m

√2

Γ = √(V m

√22V m

π)

2

−1 = 0.482

Efficiency



Pac

V dc2

RL

V rms2

RL

=[ 2V m

π ][V m

√2 ]2

2

= 8π2 =0.812=81.2 %

Tabulation:


Input Signal

With Filter

Without Filter


Tabulation:


Practical


Practical

Ex No:

Date:

POWER SUPPLY-BRIDGE RECTIFIER

Aim:To design a power supply using bridge wave rectifier with simple capacitor filter.


S.No Components Required Range Quantity1 Regulated Power Supply Dual (0-30)V 12 Transformer 13 Diode4 Resistor5 Capacitor6 CRO (0-20)MHz 17 Bread Board - 18 Connecting Wires few

Theory:The bridge rectifier has four diodes connected to form bridge. The ac input voltage is

applied to diagonally opposite ends of the bridge. The load resistance is connected between the other two ends of the bridge.

For the positive half cycle of the input ac voltage, diodes D1 and D3 conduct, whereas diodes D2 and D4 do not conduct. The conducting diodes will be in series through the load resistance R1. So the load current flow through R1.

During the negative half cycle of the ac input voltage, the diodes D2 and D4 conduct, whereas diodes D1 and D3 do not conduct. The conducting diodes will be in series through the load resistance R1. So the load current flow through R1 in the same direction as the previous half cycle. Thus the bidirectional wave is converted to unidirectional one.

Procedure:

1. Connections are made as per the circuit diagram with correct polarity provision given to the circuit from the transformer.

2. The peak voltage (Vm) is noted using CRO.3. Corresponding efficiency (η) and ripple factor (Γ) are calculated and compared with


Circuit Diagram:

Without Capacitor:

With Capacitor:

Model Graph:

Theoretical Calculation

Ripple Factor



V dc


2

Where, V rms=V m

√2

Γ = √(V rms

V dc)

2

−1


V av=V dc=1π [∫0

π

V m sinωt d (ωt )] =

V m

π[−cosωt ]0

π =2V m

π

I dc=V dc

RL=

2V m

π RL


V rms=[ 1π∫0

π

V m2 sin2ωt d(ωt)]

1 /2

=V m ¿¿= V m

√2

Γ = √(V m

√22V m

π)

2

−1 = 0.482

Efficiency



Pac

V dc2

RL

V rms2

RL

=[ 2V m

π ][V m

√2 ]2

2

= 8π2 =0.812=81.2 %

Tabulation:


Input Signal

With Filter

Without Filter


Tabulation


Practical


Practical

Ex No:

Date:

CHARACTERISTICS OF SCR

Aim:

To obtain the V-I characteristics of Silicon Controlled Rectifier (SCR).


S.No Components Required Range Quantity1 Regulated Power Supply2 SCR 2P4M3 Resistors4 DC Ammeter5 DC Voltmeter6 Bread board7 Connecting Wires

Theory:

Silicon Controlled Rectifier is a four layer PNPN device with three terminals namely Anode (A), Cathode (K) & Gate (G). Keeping the gate open, if the forward voltage is applied across the SCR. It will remain in off state, but if the applied voltage exceeds break over voltage the current flows through it. The break over voltage can be reduced if a small voltage is applied at the gate, as the gate current increases the break over voltage decreases. Once SCR is fired, the gate losses the control over the current through the device, in this even if the gate is disconnected the anode current cannot be brought back to zero. To turn off an SCR anode current should be made less than the holding current.

Procedure:

1. Connections are made as per the circuit diagram.2. By keeping the gate current constant, the anode voltage is varied and corresponding

anode current is noted.3. When the SCR is fired, then the gate current is made zero and the anode current is

reduced and at which the SCR is turned off is noted (called holding current).4. A graph is plotted with VAK on X-axis and IA in Y-axis.

Pin Diagram:

Circuit Diagram:

Model Graph:

Tabulation:

S.No

Gate Current IG= Gate Current IG= Gate Current IG=

Anode-

Cathode

Voltage VAK

(V)

Anode

Current

IA (mA)

Anode-

Cathode

Voltage VAK

(V)

Anode

Current

IA (mA)

Anode-

Cathode

Voltage VAK

(V)

Anode

Current

IA (mA)

Ex No:

Date:

CHARACTERISTICS OF UNI JUNCTION TRANSISTOR

Aim:

To obtain the characteristics of an Uni Junction Transistor.

Components required:

S.No Component Specification Quantity1 Uni Junction Transistor 2N2646 12 Voltmeter (0-10) V, (0-30) V 13 Ammeter (0-50) mA 14 Resistor 1KΩ 15 DC Regulated power supply (0-30) v 26 Bread board and patch card - As req

Theory:

UJT is a three terminal semiconductor switching device. As it has only one PN junction and three leads, it is commonly called as Uni junction transistor. It consists of a lightly doped N-type silicon bar with a heavily doped P-type material alloyed to its one side closer to B2 for producing single PN junction. The emitter leg is drawn at an angle to the vertical and the arrow indicates the direction of the conventional current.

Three other important parameters for the UJT are IP, VV and IV and are defined below:

Peak-Point Emitter Current Ip. It is the emitter current at the peak point. It represents the minimum current that is required to trigger the device (UJT). It is inversely proportional to the interbase voltage VBB.

Valley Point Voltage VV The valley point voltage is the emitter voltage at the valley point. The valley voltage increases with the increase in interbase voltage VBB.

Valley Point Current IV The valley point current is the emitter current at the valley point. It increases with the increase in inter-base voltage VBB.

Special Features of UJT:

1. A stable triggering voltage (VP)— a fixed fraction of applied inter base voltage VBB.2. A very low value of triggering current.3. A high pulse current capability.4. A negative resistance characteristic.5. Low cost.

The static emitter characteristic curve of UJT shows that for emitter potentials to the left of peak point, emitter current IE never exceeds IEo. The current IEo corresponds very closely to the reverse leakage current ICo of the conventional BJT. This region is called the cut-off region. Once conduction is established at VE = VP the emitter potential VE starts decreasing with the increase in emitter current IE. This Corresponds exactly with the decrease in resistance RB for increasing current IE. This device, therefore, has a negative resistance region which is stable enough to be used with a great deal of reliability in the areas of applications listed earlier. Eventually, the valley point reaches, and any further increase in emitter current IE places the device in the saturation region.

Procedure:

1. The circuit connections are made as per the circuit diagram.2. By adjusting the RPS, V B 1B2 is kept at 4V.3. By adjusting the RPS, V BE is varied and IE is noted.4. The above steps are repeated for different values of V B 1B2and the graph is plotted.

Device Symbol:

Pin diagram:

Circuit Diagram:

Model Graph:

Tabulation:

S.No V B 1B2=¿ V B 1B2

=¿ V B 1B2=¿

V BE(V) IE(mA) V BE(V) IE(mA) V BE(V) IE(mA)

Manual - SSC I

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