LAB NO 1

INTRODUCTION OF ORCAD 10.5,USE,CIRCUIT DESIGN & SIMULATION

Step 1: Software opens by clicking an option “CAPTURE CIS” in the start menu.

Step 2: clicking file menu in the menu bar displays the option for creating “new project”.

Step 3: Assign the project name & select the option “Analog or mixed A/D” and give the location (by

clicking browse option) to save the project in the software directory.

Step 4: Draw the circuit for simulation in the window shown below by using the options in the right side

command bar. Assign the values of parameters by double clicking them as cleared from the following

figures.

Draw the circuit.

Create NEW SIMULATION PROFILE

Choose the analysis type (here we want Time domain analysis, set frequency and suitable range time

period)

SIMULATION

Step 5: click “RUN” option and the simulation results are:

Step 6: Here an interesting option “Trace” is present by clicking it we can trace other quantities

required, e.g. here we add power trace.

LAB #2 Find the Ripple Factor of Half Wave Rectifier

Components

1) - Diode with code number D1N4001

2)-Resistor of 1k

3)-Voltage source (Vsin)

4)-Connecting wires.

Procedure:

Open the software and draw a neat and clean diagram of half wave rectifier, using

the components described above. Now apply the voltage markers from the tool

bar across the resistor, now the circuit becomes ready for simulation. Make new

simulation profile and save it. Now simulate the circuit and take readings.

Circuit In the Orcad:

Simulation Results:

OBSERVATIONS & CALCULATIONS:

Vmax 14.127 V

Vrms2

Vac2

Vdc2

Ripple FactorVac

Vdc

Vrms2

Vdc2

Vdc

Vrms Vmax 0.5 14.127 0.5 7.06V

VdcVmax 14.127

3.14164.5V

Ripple factor7.06

24.5

2

4.51.21

Therefore required ripple factor of half wave rectifier is 1.21

LAB#3

a)-Design and simulation of Full wave rectifier by Center Tapped Transformer

b)-Calculation of ripple factor:

Theory:

A popular full-wave rectifier appears in Fig. with only two diodes but requiring a center-tapped (CT)

transformer to establish the input signal across each section of the secondary of the transformer. During

the positive portion of Vi applied to the primary of the transformer, the network will appear as shown in

Fig. D1 assumes the short-circuit equivalent and D2 the open-circuit equivalent, as determined by the

secondary voltages and the resulting current directions. The output voltage appears as shown in Fig.

Center Tapped Full Wave Rectifier

For Positive Half Cycle

For Negative Half Cycle

Components:

1) - 2 Diodes with code number D1N4001

2) - Center tapped transformer of code number XFRM_LIN/CT-SEC

3) - AC Voltage source with code number Vsin

4) - Load, A resistor R = 1K is being assumed.

Procedure:

Open the orcad software and draw neat and clean diagram in the window, Assign the

parameter values and after making the simulation file run the simulation and note down the

maximum value of the voltage is 5V.At the end find out the ripple factor by the following

procedure.

Ripple FactorVac

Vdc ………………………………………. (Q)

also, Vdc Vmax 0.6366 Vmax 5V Vdc 5 0.6366 3.183V

since VrmsVmax

2

5

23.535V

Vac Vrms2

Vdc2

1.537

Therefore required Ripplefactor isVac

Vdc

1.537

3.1830.483

Circuit in ORCAD Software:

Simulation Results:

Vmax= 5V by the toggle cursor.

Conclusion:

Therefore the ripple factor of the center tapped Rectifier is 0.483.

LAB NO# 04 Design & Simulation of Full Wave Bridge Rectifier

THEORY:

This type of single phase rectifier uses four individual rectifying diodes connected in a closed loop

"bridge" configuration to produce the desired output. The main advantage of this bridge circuit is that it

does not require a special centre tapped transformer, thereby reducing its size and cost. The single

secondary winding is connected to one side of the diode bridge network and the load to the other side

as shown below.

Positive Half Cycle Negative Half Cycle

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

unidirectional the same as for the previous two diode full-wave rectifier, therefore the average DC

voltage across the load is 0.637Vmax and the ripple frequency is now twice the supply frequency (e.g.

100Hz for a 50Hz supply).

COMPONENTS:

1) - Connecting Wires

2) - 4 diodes (D1N4001)

3) - AC Voltage Source (V sin)

4) - Load (A resistor of value 1k)

Procedure:

Draw the circuit diagram clearly and apply the differential voltage markers over the load (i-e R) make the

simulation file and run the file to obtain the simulation results as shown. Then with the help of toggle

cursor find Vmax=15V and thus find the Ripple factor.

RipplefactorVac

Vdc ………………………………………………………………. (A)

VrmsVmax

2

15

210.605

where Vdc Vmax 0.6366 &Vmax 15V 9.549 V

Vac Vrms2

Vdc2

………………………………………………………………… (B)

Vac 10.6052

9.5492

V 4.61 V

Ripplefactor4.61

9.5490.482

CIRCUIT DIAGRAM IN ORCAD:

SIMULATION RESULTS:

Vmax= 15 V (by Toggle cursor)

Conclusions: There fore the required ripple factor is 0.482 of full wave bridge rectifier.

LAB#05 Design & Simulation of half Wave Controlled Rectifier

Theory:

When rectification is used to provide a direct voltage power supply from an alternating source, the amount of ripple can be further reduced by using larger value capacitors but there are limits both on cost and size. For a given capacitor value, a greater load current (smaller load resistor) will discharge the capacitor more quickly (RC Time Constant) and so increases the ripple obtained. Then for single phase, half-wave rectifier circuits it is not very practical to try and reduce the ripple voltage by capacitor smoothing alone, it is more practical to use "Full-wave Rectification" instead.

Disadvantage

In practice, the half-wave rectifier is used most often in low-power applications because of their major disadvantages being. The output amplitude is less than the input amplitude, there is no output during the negative half cycle so half the power is wasted and the output is pulsed DC resulting in excessive ripple.

Apparatus

1- Voltage Source (Vsin) 2- Capacitor (of any 5 different values) 3- A load Resistor (R) 4- Diode for rectification(D1N4001) 5- Connecting Wires

Procedure:

Open the Orcad software and draw neat and clean diagram in the window, Assign the parameter

values and after making the simulation file run the simulation and note down the maximum value

of the voltage is 5V.At the end find out the ripple factor by the following procedure.

Mathematical Calculations:

Vrpp Vcmax Vcmin …………………………………………………… (1)

As, Vmax Vcmax

VrppVmax

fRC

Vdc 11

2frcVmax

RipplefactorVrpp

Vdc

Circuit Designing In Orcad

Simulation Results:

Observations & Calculations:

Ser#N0. R C Vmax Vr(p-p) Vdc RF Vmin

01. 1k 50uF 14.391V 4.091 12.345 0.18 10.300

02. 1k 100uF 14.301 2.24 13.181 0.16 12.061

03. 1k 200uF 14.352 1.191 13.756 0.09 13.161

04. 1k 300uF 14.349 0.775 13.961 0.07 13.574

05. 1k 500uF 14.341 0.615 14.034 0.05 13.726

Result:

As the value of the capacitor increases the Ripple Factor Decreases.

Lab No#06 Design and Simulation of full wave Controlled Rectifier by

1) - Bridge method

2) - Centre tapped Transformer method

Components:

1) - 4 diodes with code number D1N4001

2) – Load Resistance (100 ohm)

3) - Filter capacitor

4) - Ac voltage source (Vsin)

5)-Connecting wires

Procedure:

Draw neat and clean diagram of the circuit under test and put the voltage marker

across the load resistance. Now connect a capacitor parallel to the load resistor,

make the simulation file and after saving it, simulate the circuit and find the ripple

factor.Simiarly repeat the experiment for centre tapped transformer.

Full wave Bridge Controlled Rectifier:

Simulation Results:

Observations & Calculations:

OBSERVATIONS

CAPACITANCE( F )

100 200 300 400 500

Vmax 3.6102 3.5839 3.5781 3.538 3.5239

VrppVmax

2fRC

0.361 0.1792 0.1193 0.088 0.0705

Vdc 11

4fRC Vmax

0.8195 0.9104 0.9404 0.956 0.9648

RipplefactorVrpp

Vdc 0.4405 0.1968 0.1268 0.093 0.0731

Controlled Full Wave Rectifier by Centre Tapped Transformer Method:

Components:

1) - 2 diodes with code number D1N4001

2) - A filter capacitor(C)

3) - Load resistor (R)

4) - Voltage source (Vsin)

5) - Centre tapped transformer with code number XFRM_LIN/CT-SEC

Circuit for Simulation:

Simulation Results:

Observations & Calculations:

OBSERVATIONS CAPACITANCE( F)

100 200 300 400 500

Vmax 1.8327 1.8098 1.8055 1.792 1.7868

VrppVmax

2fRC 0.1833 0.0905 0.0602 0.045 0.0357

Vdc 11

4fRC Vmax 0.9084 0.9548 0.9699 0.978 0.9821

RipplefactorVrpp

Vdc 0.2018 0.0948 0.0621 0.046 0.0364

Lab#07 Design and simulation of half wave controlled rectifier

Theory:

Being a unidirectional (one-way) device, at most we can only deliver half-wave power to the

load, in the half-cycle of AC where the supply voltage polarity is positive on the top and negative

on the bottom. However, for demonstrating the basic concept of time-proportional control, this

simple circuit is better than one controlling full-wave power (which would require two SCRs).

With no triggering to the gate, and the AC source voltage well below the SCR's breakover

voltage rating, the SCR will never turn on. Connecting the SCR gate to the anode through a

standard rectifying diode (to prevent reverse current through the gate in the event of the SCR

containing a built-in gate-cathode resistor), will allow the SCR to be triggered almost

immediately at the beginning of every positive half-cycle: (Figure below)

We can delay the triggering of the SCR, however, by inserting some resistance into the gate

circuit, thus increasing the amount of voltage drop required before enough gate current triggers

the SCR. In other words, if we make it harder for electrons to flow through the gate by adding a

resistance, the AC voltage will have to reach a higher point in its cycle before there will be

enough gate current to turn the SCR on. The result is in (Figure below)

Resistance inserted in gate circuit; less than half-wave current through load.

With the half-sine wave chopped up to a greater degree by delayed triggering of the SCR, the

load receives less average power (power is delivered for less time throughout a cycle). By

making the series gate resistor variable, we can make adjustments to the time-proportioned

power: (Figure below)

Increasing the resistance raises the threshold level, causing less power to be delivered to the

load. Decreasing the resistance lowers the threshold level, causing more power to be delivered

to the load.

Unfortunately, this control scheme has a significant limitation. In using the AC source waveform

for our SCR triggering signal, we limit control to the first half of the waveform's half-cycle. In

other words, it is not possible for us to wait until after the wave's peak to trigger the SCR. This

means we can turn down the power only to the point where the SCR turns on at the very peak of

the wave: (Figure below)

Components:

1)-Ac voltage source (Vsin)

2)-Two resistors R-load and Rvar

3)-A Thyristor (2N1595)

4)-A diode (D1N4001)

5)-Connecting Wires and ground

Procedure:

Open the software and draw neat and clean figure for the required circuit (whose simulation is required) as explained in the theory. Now put differential voltage markers across the load resistor and change the values of the variable resistor and note down the values of the firing angle time then by using the formula ‘ t ‘find out the firing angle .Repeat the simulation 5 times. At the end we will find out that as the value of Rvar goes higher the firing angle of the thyristor increases.

Circuit for Simulation:

Simulation Results

Observations & Calculations:

S.No# RL(load) Rvar(variable) Firing Time(t) Firing angle

t 2 f, f 50Hz

1 10 6k 3.26ms 58.68

2 10 5k 2.61ms 46.98

3 10 4k 2.0ms 36.0

4 10 3k 1.56ms 28.08

5 10 1k 774.2us 13.93

Result:

It has been observed that as the value of the variable resistor R var goes down the

firing angle decreases significantly or in other words the thyristor on earlier.

LAB #08 Design & Simulation of Full wave Bridge Rectifier with

Thyristors

Theory:

When multiple SCRs are used to control power, their cathodes are often not electrically

common, making it difficult to connect a single triggering circuit to all SCRs equally. An example

of this is the controlled bridge rectifier shown here:

In any bridge rectifier circuit, the rectifying diodes (or in this case, the rectifying SCRs) must

conduct in opposite pairs. SCR1 and SCR3 must be fired simultaneously, and likewise SCR2 and

SCR4 must be fired together as a pair. As you will notice, though, these pairs of SCRs do not

share the same cathode connections, meaning that it would not work to simply parallel their

respective gate connections and connect a single voltage source to trigger both.

Apparatus:

1) - Voltage source (Vsin)

2) - 4 thyristors with code no. 2N1595 for BR and 2 for CT rectifier

3) - Load resistor (R)

4) - Center tapped transformer (XFRM_LIN/CT-SEC)

5) - Pulse signal (V pulse)

Procedure

Open the software and draw a neat and clean diagram of half wave rectifier, using

the components described above. Before any thing else define the width of the

pulse and assign specific values to the pulse signal. Now apply the voltage

markers from the tool bar across the resistor, now the circuit becomes ready for

simulation. Make new simulation profile and save it. Now simulate the circuit and

take readings.

a) - Perform first with center tapped transformer

b) - Perform with full wave bridge rectifier

FULLWAVE CONTROLLED RECTIFICATION WITH CENTER TAPPED TRANSFORMER:

Circuit Diagram In Orcad:

Simulation Results:

FULL WAVE CONTROLLED RECTIFICATION BY BRIDGE METHOD:

Circuit Diagram In Orcad:

Simulation Results:

Result:

There fore we can control the dc output in each half cycle by using the thyristors.

LAB#09 Design and Simulation of the Three Phase Rectifier with Filter

Capacitor

Apparatus:

1- 3 voltage sources (Vsin) with phase difference of 120 respectively

2- 6 diodes with code number D1N4001

3- Connecting Wires

4- 3 limiting Resistors (R)

5- 3 inductors (one for each phase)

6- A filter capacitor (C)

Procedure:

1)-Open the software and after creating a blank project draw neat and clean

figure for the circuit under simulation.

2)-Then save the project and create a new simulation profile.

3)-Run the simulation profile and observe the output waveform.

4)-Change the capacitor values and find how it removes the ripples.

Circuit Diagram:

Circuit in Orcad:

Simulation Results