EEE DEPARTMENT

ECE103 / T191

EXPERIMENT NO. 3

Submitted by: Etorma, Ariston A.

Course/Year: BSCpE/2nd

Date of Experiment: December 4, 2015

Date of Submission: December 11, 2015

ENGR. MELANNIE MENDOZAProfessor

II. Objective

To become familiar with half-wave and full-wave rectification

III. Introduction

The Half wave rectifier is a circuit, which converts an AC voltage to DC voltage.

It can be used to obtain the desired level of dc voltage (using step up or step down

transformers). It provides isolation from the power line. During the positive half cycle of

the input voltage the polarity of the voltage across the secondary forward biases the

diode. The forward biased diode offers a very low resistance and hence the voltage drop

across it is very small. During the negative half cycle of the input voltage the polarity of

the secondary voltage gets reversed. As a result, the diode is reverse biased. Practically

no current flows through the circuit and almost no voltage is developed across the

resistor. All input voltage appears across the diode. In this lab we address about the

characteristics of half wave diode rectifier circuits. In half wave rectification of a single-

phase supply, either the positive or negative half of the AC wave is passed, while the

other half is blocked. Because only one half of the input waveform reaches the output,

mean voltage is lower. Half-wave rectifiers produce far more ripple than full-wave

rectifiers. In view of this advantage, the half wave rectifier circuit can often be seen

within large items of electronics equipment. The operation of a half wave rectifier is

pretty simple. From the theory part, you should know that a pn junction diode conducts

current only in 1 direction. In other words, a pn junction diode conducts current only

when it is forward biased. The same principle is made use of in a half wave rectifier to

convert AC to DC. The input we give here is an alternating current. This input voltage is

stepped down using a transformer. The reduced voltage is fed to the diode ‘D’ and load

resistance RL. During the positive half cycles of the input wave, the diode ‘D’ will be

forward biased and during the negative half cycles of input wave, the diode ‘D’ will be

reverse biased. We take the output across load resistor RL. Since the diode passes current

only during one half cycle of the input wave, we get an output as shown in diagram. The

output is positive and significant during the positive half cycles of input wave. At the

same time output is zero or insignificant during negative half cycles of input wave. This

is called half wave rectification.

IV. Discussion

V. Procedure

PART 1. Threshold Voltage

Choose one of the four silicon diodes and determine the threshold voltage, VT, using

the diode-checking capability of the DMM.

VT =

PART 2. Half-Wave Rectification

a) Construct the circuit of Fig.2.2 using the chosen diode of Part 1. Record the measured

value of the resistance R. set the function generator to 1000 Hz, 8 Vp-p sinusoidal

voltage using the oscilloscope.

Figure 2.2 half –wave rectifier

b) The sinusoidal input of fig2.2 has been plotted on the screen of fig.2.3.Figure 2.3

Vertical Sensitivity=

Horizontal Sensitivity=

c) Using the threshold voltage VT of Part 1, determine the theoretical output voltage vO

for the circuit of fig 2.2 and sketch the waveform on fig 2.3 for one full cycle using the

same sensitivities employed in Part 2(b). Indicate maximum and minimum values.

d) Using the oscilloscope with the DC position, obtain the voltage vO and sketch the

waveform on fig.2.4. Before viewing vO be sure to set the vO = 0 V line using the GND

position of the coupling switch.

e) Calculate the DC level of the half-wave rectifier signal of Part 2(d) using eq.2.1

VDC (calculated) =

f) Measure the DC level of vo using the DC scale of the DMM and find the percent

difference between the measured value and the calculated value of part 2(e) using the

following equation:

VDC (measured) =

(% Difference) =

g) Reverse the diode of fig.2.2 and sketch the output waveform obtained using the

oscilloscope on fig.4.5 be sure the coupling is in the DC position.

h) Calculate and measure the DC level of resulting waveform of fig.4.5. Insert the proper

sign for the polarity of VDC as defined by fig.4.3 using eq.2.1

VDC (calculated) =

VDC (measured) =

PART 3. Half-Wave Rectification (continued)

a) Construct the network of fig.2.6. Record the measured value of resistor R.

b) Using the threshold voltage of Part 1, determine the theoretical output voltage vo

for fig.2.6 and sketch the waveform on fig.2.7 for one full cycle. Indicate the

maximum and minimum values on the output waveform.

c) Using the oscilloscope with the coupling switch in the DC position, obtain the

voltage vo and sketch the waveform on fig.2.8.

d) Calculate the DC level of the waveform of fig.2.8 using the following equation:

VDC (calculated) =

e) Measure the output DC voltage with the DC scale of the DMM and calculate the

percent difference using the same equation appearing in Part 2(f).

VDC (measured) =

(% Difference) =

VI. Sample Computation

VII. Problems

VIII. Interpretation of Data

IX. Conclusion

X. References

http://akademik.bahcesehir.edu.tr/eee2116/documents/expWeek3NEW.pdf

http://www.circuitstoday.com/half-wave-rectifiers

http://www.evalidate.in/lab1/pages/diode/DiodeHalfwave/DiodeHalfwave_I.html

XI. Data Sheet

XII. Index