EEE DEPARTMENT

ECE103 / T191

EXPERIMENT NO. 4

Submitted by: Etorma, Ariston A.

Course/Year: BSCpE/2nd

Date of Experiment: December 11, 2015

Date of Submission: December 18, 2015

ENGR. MELANNIE MENDOZAProfessor

II. Objective

To become familiar with full-wave rectification

III. Introduction

The conversion of AC into DC is called Rectification. Electronic devices can

convert AC power into DC power with high efficiency. A drawing of a full-wave bridge

rectifier is given below. The bridge is composed of four diodes in a diamond shape.

During the positive half-cycle of input voltage vin the terminal ‘A’ is at positive potential

with respect to the terminal ‘B’ and because if this diodes D1 and D2 are forward biased

whereas diodes D3 and D4 are reverse biased. The current therefore flows through diodes

D1, D2 and load resistor RL. During the negative half-cycle of input voltage waveform,

on the other hand, the diodes D3 and D4 are forward biased whereas the diodes D1 and

D2 are reverse biased. Like the half wave circuit, a Full Wave Rectifier Circuit produces

an output voltage or current which is purely DC or has some specified DC component.

Full wave rectifiers have some fundamental advantages over their half wave rectifier

counterparts. The average (DC) output voltage is higher than for half wave, the output of

the full wave rectifier has much less ripple than that of the half wave rectifier producing a

smoother output waveform. In a full wave rectifier circuit we use two diodes, one for

each half of the wave. A multiple winding transformer is used whose secondary winding

is split equally into two halves with a common center tapped connection. Configuration

results in each diode conducting in turn when its anode terminal is positive with respect

to the transformer center point C produces an output during both half-cycles. Full rectifier

advantages are flexible compared to that of half wave rectifier. The full wave rectifier

circuit consists of two power diodes connected to a single load resistance (RL) with each

diode taking it in turn to supply current to the load resistor. When point A of the

transformer is positive with respect to point A, diode D1 conducts in the forward

direction as indicated by the arrows. When point B is positive in the negative half of the

cycle with respect to C point, the diode D2 conducts in the forward direction and the

current flowing through resistor R is in the same direction for both half-cycles of the

wave.

IV. Discussion

On the part 1 of the experiment we determine the threshold voltage using

The diode-checking capability of the DMM. After checking the threshold voltage we

construct the full wave bridge rectifier and we’re stuck on part 2 for 30 minutes because

even if we did construct the circuit correctly it still having short circuit when we put the

transformer on the diodes. We kept on changing the layout of the circuit and it still has

the same result. Until on the last 30 minutes somehow we make it work and it has too

much noise on the oscilloscope. So on our data we just imagine what would be the figure

that will be flashed on the oscilloscope. After getting the values that is needed on the

experiment we computed for the calculated values and we got a 6-9% percentage error

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 = 0.7v

PART 2. Full-Wave Rectification

a) Construct the full wave bridge rectifier of fig 4.14. Be sure that the diodes are inserted

correctly and that the grounding is as shown. If unsure ask your instructor to check your

setup. Record the measured value of the resistor R

In addition, measure the rms voltage at the transformer secondary using the DMM set to

AC. Record that rms value below. Does it differ from the rated 12.6V?

Vrms (Measured) =12.45v

b) Calculate the peak value of the secondary voltage using the measured value

(Vpeak=1.414Vrms)

Vpeak (Calculated)=17.60v

c) Using the VT of Part 1 for each diode, sketch the expected output waveform vo on fig.4.15. Choose a vertical and horizontal sensitivity based on the amplitude of the secondary voltage. Consult your oscilloscope to obtain a list of possibilities. Record your choice for each below.

Vertical Sensitivity=5vHorizontal Sensitivity=7ms

d) Using the oscilloscope with the DC position, obtain the waveform for Vo and record on

fig.4.16. Use the same sensitivities employed in part 5(c) and be sure to preset the Vo

=0V line using the GND position of the coupling switch label the maximum and

minimum values of the waveform using the chosen vertical sensitivity.

How do the waveforms of part 5(c) and 5(d) compare?

e) Determine DC level of the full-wave rectified waveform of fig.4.16

VDC (calculated) =10.74v

f) Measure the DC level of the output waveform using the DMM and calculate the

percent difference between the measured and calculated values.

VDC (measured) =10.06V

(% Difference) =6.54%

g) Replace diodes D3 and D4 by 2.2 kOhm resistors and forecast the appearance of the

output voltage. Including the effects of VT for each diode. Sketch the waveform on

fig.4.17 and label the magnitude of the maximum and minimum values. Record your

choice of sensitivities below

Vertical Sensitivity=5v/divHorizontal Sensitivity=2ms/div

h) Using the oscilloscope obtain the waveform for V0 and reproduce on Fig 4.18

indicating the maximum and minimum values. Use the same sensitivities and determine

in Part 5(g)

How do waveforms of figs. 4.17 and 4.18 compare?

i.)Calculate the DC level of the waveform of fig. 4.18

VDC (calculated)=

j.) Measure the DC level of the output voltage using the DMM and calculate the percent

difference

VDC (measured)=

(% Difference)=

k) What was the major effect of replacing the two diodes with resistors?

Part 3. Full wave center tapped Configuration

a. Construct the network of fig.4.19. Record the measured value of the resistor R

Measure the two secondary voltages of the transformer with the DMM set on AC. Record

below. Do they differ from the 6.3V rating?

Vrms (measured)=

Vrms (measured)=

Using an average of the two rms readings, calculate the peak value of the overall

secondary voltage.

Vpeak (calculated)=

b. Using the VT of part 1 for each diode, Sketch the expected output waveform

Vo on fig. 4.20. Choose a vertical and horizontal sensitivity based on the

amplitude of the secondary voltage. Consult your oscilloscope to obtain a list

of possible settings. Record your choice for each below.

Figure 4.20

Vertical Sensitivity=

Horizontal Sensitivity=

c. Using the oscilloscope with the coupling switch in the DC position obtain the

waveform for Vo and record on fig. 4.21. Use the same sensitivities employed

in part 6(b) and be sure to preset the Vo= 0V line using the GND position of

the coupling switch. Label the maximum and minimum values of the

waveform using the chosen vertical sensitivity

Figure 4.21

How do the waveforms of figs 4.20 and 4.21 compare?

d. Determine and compare the calculated and measured valies of the DC level

associated with Vo

(calculated)=

(measured)=

VI. Sample Computation

VII. Problems

VIII. Interpretation of Data

IX. Conclusion

Rectification is the process of converting an alternating voltage or alternating current into

direct voltage or direct current. The device used for rectification is called rectifier.

Rectifiers are mainly two types, half wave rectifier and full wave rectifier. Full wave

rectifier is a circuit which rectifies both half cycles of the a.c. when P of 1st diode is

positive, the 1st diode is forward biased and will conduct. Now the 2nd diode will not

conduct as it is reverse biased. In all the half cycles either of the two diodes will be

conducting. It is clear to see that full wave rectifier offers better efficiency since the full

ac supply cycle is used to supply power to the load although this is at the cost of an

additional three diodes. However, if a relatively smooth dc level is required the full wave

rectifier offers a dc output with much less ace ripple superimposed upon it, this means

that smaller and therefor cheaper smoothing capacitors could be used to make the

waveform much closer to a dc value

X. References

http://dei.vlab.co.in/?sub=22&brch=60&sim=1113&cnt=2150

http://www.elprocus.com/full-wave-rectifier-circuit-working-theory/

http://www.electronics-tutorials.ws/diode/diode_6.html

http://academic.cankaya.edu.tr/~mgokce/files/ece_246_dosyas%C4%B1/

ece246_lab3.pdf

XI. Data Sheet

XII. Index