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INTRODUCTION

Most electronic equipment requires the use of dc voltage to operate properly. Since most equipment is connected to the 240 ac power line, this ac voltage must somehow be converted to the required dc voltage. This can be achieved through the use of a rectifier circuit. There are two types of rectifier circuits, namely the half-wave rectifier circuit and the full-wave rectifier circuit.

The unfiltered output from a rectifier circuit is a pulsating dc voltage. For most applications, this dc voltage must be filtered or smoothed if it is to be useful. One way to smooth out the pulsations in this dc voltage is to connect a capasitor at the output of the rectifier circuit. The filtered output is a ripple dc voltage whose degree of “flatness” depends on both the capasitance of the capasitor and the resistance of the load.

In this experiment, we examine the half-wave rectifier circuit and the bridge rectifier circuit ( a full-wave rectifier circuit ). For each, we investigate the effect on the output of including capasitor in the rectifier circuit. We also inspect the effect on the output of varying the capasitance of the capasitor and resistance of the load.

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THEORY OF RECTIFIER CIRCUIT AND FILTERING

A) PRELIMINARIESBefore we proceed to the theory of rectifier circuit and filtering, we review some fundamental facts about diode and capacitor.

DIODE1. A diode is a near-unidirectional electronic device that allows current to flow through it easily in one direction and impedes current to flow through it in the opposite direction.2. Fig. 1 shows the schematic symbol of a diode. Its two terminals are known as anode and cathode respectively. The diode allows conventional current to flow through it easily from anode to cathode but impedes current to flow through it from cathode to anode.

Figure 1 Schematic symbol of a diode.

3. When the anode is maintained at a positive potential with respect to the cathode, the diode is said to be in forward bias and current flow is easy. When the cathode is maintained at a positive potential with respect to the anode, the diode is said to be in reverse bias and current flow is difficult.4. Fig. 2(b) shows the voltage-current characteristic of a diode where V and I are defined as in fig. 2(a).

(a) (b) Figure 2 (a) Definition of Vand I. (b) Voltage-current characteristic of a diode.5. When analyzing diode circuit, we often model a diode as an ideal diode

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defined by the voltage-current characteristic shown in fig. 3 and summarized as below: For all V £ 0, I = 0 For all I ³ 0, V = 0 ( Equation 1 )

(a) (b) Figure 3 (a) Ideal diode. (b) Voltage-current characteristic of an ideal diode.

6. To facilitate our analysis and yet keep it quite accurate, we will model a diode as shown in fig. 4 and summarized as below: For all V £ VB, I = 0 For all I ³ 0, V = VB ( Equation 2) VB is known as the potential barrier of the diode.

Figure 4 (a) Second model of a diode. (b) Voltage-current characteristic of the second model of a diode.

7. It is an immediate result that our model is equivalent to an ideal diode in series with an ideal voltage source VB as shown in fig. 5.

Fig. 5 Equivalent of the second model of a diode.

CAPASITOR1. A capasitor is formed by a pair of conductors seperated by an insulator (

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also known as dielectric ). It can charge or discharge in a circuit when the voltage across it varies.

2. Fig. 6 shows the schematic symbol of a capasitor.

Figure 6 Schematic symbol of a capasitor. 3. The capasitance C of a capasitor, refering to fig. 7 is defined as C = Q / VAB ( Equation 3 ) where Q ³ 0 is the net charge on one of the conductors.

Figure 7 Definition of capasitance. Q ³ 0.

4. A number n of capasitors in parallel, each with capasitance C1, C2, ... , Cn, are equivalent to a single capasitor with capasitance n

Ceq = S Ci ( Equation 4 ) i=1

as shown in fig. 8. Derivation:

Suppose Ci has charge Qi and voltage Vi. Since all the capasitors are parallel, we have for all i, Vi = V i.e. Qi / Ci = V. Hence from the property of fractions that if a / b = c / d = f, then ( a + c ) / ( b + d ) = f, we have S Qi / S Ci = V Q / S Ci = V where Q = S Qi

S Ci = Q / V From the last relation, we see that Ceq = S Ci.

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Fig 8 Equivalent capasitor for capasitors in parallel.

5. In fig. 9(a), the capasitor has a voltage V0 initially. At time t = 0, the switch is closed and the capasitor starts to discharge through the resistor as in fig. 9(b). After time t, the voltage of the capasitor, V, is given by V = V0exp[-t / (RC)] ( Equation 5 ) Derivation: After a time t, we have Q = CV and V = IR. Differentiating Q = CV with respect to t and substituting dQ / dt = - I = - V / R, we have - V / R = C dV / dt i.e. dV / V = - dt / ( CR ) which upon integration from t = 0 to t = t and after some manipulation gives V = V0exp[-t / (RC)]

Figure 9 (a) The capasitor-resistor circuit before the switch is closed. (b) The capasitor-resistor circuit after the switch is closed.

6. We call t = RC the time constant or the relaxation time.7. It is an immediate result from eq. 5 that the time constant or the relaxation time is the time from t = 0 taken for the capasitor to discharge to a voltage of 1/e of its original voltage V0.

(B) RECTIFIER CIRCUIT

HALF-WAVE RECTIFIER CIRCUIT

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1. The circuit shown in fig. 10 is called a half-wave rectifier circuit.

Figure 10 Half-wave rectifier circuit.

2. T is a step down transformer with a turns ratio of Np : Nswhich transforms the primary sinusoidal voltage of rms value Vp and frequency fp to the secondary sinusoidal voltage of rms value Vs and frequency fs

according to the relations Vs / Vp = Ns / Np ( Equation 6 )

and fs = fp ( Equation 7 ) Vs is the value that would be measured if a ac voltmeter were placed across the secondary.3. The peak secondary voltage Vs(pk) is given by Vs(pk) = Ö2Vs ( Equation 8 )4. Fig. 11 shows the waveform of the secondary voltage.

Figure 11 Waveform of the secondary voltage.

5. During the positive alternation of the secondary voltage, the diode D is forward biased and a current flows in the load R. Fig. 12 shows the equivalent circuit of the circuit in fig. 10 at the instant the secondary voltage is at its positive peak value Vs(pk).

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Figure 12 Equivalent circuit of the half-wave rectifier circuit at the instant the secondary voltage is at its positive peak value. Using Kirchhoff’s Voltage Law, Vo(t)(pk) = Vs(pk) - VB ( Equation 9 )6. During the negative alternation of the secondary voltage, the diode D is reverse biased and no current flows in the load R. The diode acts like an open switch. Hence Vo(t) = 0.7. Therefore, the output voltage Vo(t) is a series of positive pulses as shown in fig. 13.

Figure 13 Waveform of the output voltage.

Note that the frequency of Vo(t), fo = fs = fp. ( Equation 10)8. The average dc voltage across the load R is given by VL(dc) = Vo(t)(pk) / p ( Equation 11) VL(dc) is the value that would be measured if a dc voltmeter were placed across the load R.9. The average dc current through the load R is given by IL(dc) = VL(dc) / R ( Equation 12) IL(dc) is the value that would be measured if a dc ammeter were placed in series with the load R. 10. The average dc current through the diode D is given by ID(dc) = IL(dc) ( Equation 13)11. Fig. 14 shows the equivalent circuit of the circuit in fig. 10 at the instant the secondary voltage is at its negative peak value -Vs(pk).

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Figure 14 Equivalent circuit of the half-wave rectifier circuit at the instant the secondary voltage is at its negative peak value.

Using Kirchhoff’s Voltage Law, the peak inverse voltage PIV = -Vs(pk) ( Equation 14 ) In order that the diode is not damaged, we must have | PIV | < peak inverse voltage rating.

FULL-WAVE RECTIFIER CIRCUIT1. The circuit shown in fig. 15 is called a bridge rectifier circuit.

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Figure 15 Bridge rectifier circuit.

2. T is a step down transformer with a turns ratio of Np : Nswhich transforms the primary sinusoidal voltage of rms value Vp and frequency fp to the secondary sinusoidal voltage of rms value Vs and frequency fs

according to the relations Vs / Vp = Ns / Np ( Equation 15)

and fs = fp ( Equation 16) Vs is the value that would be measured if a ac voltmeter were placed across the secondary.3. The peak secondary voltage Vs(pk) is given by Vs(pk) = Ö2Vs ( Equation 17)4. Fig. 16 shows the waveform of the secondary voltage.

Figure 16 Waveform of the secondary voltage.

5. During the positive alternation of the secondary voltage, the diodes D2 and D3 are forward biased and a current flows in the load R. Note that the diodes D1 and D4 are reverse biased and act like open switches. Fig. 17 shows the equivalent circuit of the circuit in fig. 15 at the instant the secondary voltage is at its positive peak value Vs(pk).

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Figure 17 Equivalent circuit of the full-wave rectifier circuit at the instant the secondary voltage is at its positive peak value. Using Kirchhoff’s Voltage Law, Vo(t)(pk) = Vs(pk) - 2VB ( Equation 18)

5. During the negative alternation of the secondary voltage, the diodes D1 and D4 are forward biased and a current flows in the load R in the same direction as during the positive alternation of the secondary voltage. Note that the diodes D2 and D3 are reverse biased and act like open switches. Fig. 18 shows the equivalent circuit of the circuit in fig. 15 at the instant the secondary voltage is at its negative peak value -Vs(pk).

Figure 18 Equivalent circuit of the full-wave rectifier circuit at the instant the secondary voltage is at its negative peak value.

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Using Kirchhoff’s Voltage Law, Vo(t)(pk) = Vs(pk) - 2VB ( Equation 19)

7. Therefore, the output voltage Vo(t) is a series of positive pulses as shown in fig. 19.

Figure 19 Waveform of the output voltage.

Note that the frequency of Vo(t), fo = 2fs = 2fp. ( Equation 20)8. The average dc voltage across the load R is given by VL(dc) = 2Vo(t)(pk) / p ( Equation 21) VL(dc) is the value that would be measured if a dc voltmeter were placed across the load R.

9. The average dc current through the load R is given by IL(dc) = VL(dc) / R ( Equation 22) IL(dc) is the value that would be measured if a dc ammeter were placed in series with the load R. 10. The average dc current through each of the diodes D1, D2, D3 and D4 is

given by ID(dc) = IL(dc) / 2 ( Equation 23)11. Fig. 20 shows the equivalent circuit of the circuit in fig. 15 at the instant the secondary voltage is at its negative peak value -Vs(pk).

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Figure 20 Equivalent circuit of the full-wave rectifier circuit at the instant the secondary voltage is at its negative peak value.

Using Kirchhoff’s Voltage Law, the peak inverse voltage PIV1 = - ( Vs(pk) - VB ) ( Equation 24) PIV4 = - ( Vs(pk) - VB ) ( Equation 25) Likewise, for the negative alternation of the secondary voltage,the peak inverse voltage PIV1 = - ( Vs(pk) - VB ) ( Equation 26) PIV4 = - ( Vs(pk) - VB ) ( Equation 27) In order that the diodes are not damaged, we must have | PIV | < peak inverse voltage rating.

(C) FILTERING

HALF-WAVE RECTIFIER CIRCUIT FILTERING1. Fig. 21 shows a half-wave rectifier circuit with a capasitor filter

connected at the output.

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Figure 21 Half-wave rectifier circuit with capasitor filter.

2. Fig. 22 shows the waveforms of the secondary voltage and the corresponding output voltage beginning from the instant the secondary voltage is applied. The waveform of the output voltage is known as ripple voltage.

Figure 22 (a) Waveform of the secondary voltage since the instant it is applied. (b) Waveform of the corresponding voltage since the instant the secondary voltage is applied.

3. Initially,during the first half of the first positive alternation of the secondary voltage, i.e. from t = 0 to t = 1 / (4fs), the diode D is forward biased and a current flows in the load R. At the same time, the capasitor C charges to a maximum voltage of

Vo(t)(pk) = Vs(pk) - VB ( Equation 28)4. Next, during the second half of the first positive alternation of the secondary voltage, i.e. from t = 1 / (4fs) to t = 1 / (2fs), the diode D is reversed biased due to the voltage of the capasitor C. It continues to be in reverse bias during the first negative alternation of the secondary voltage. It remains so for a portion of the first half of the second positive alternation of the secondary voltage due to the voltage of the capasitor C. In short, the diode D stays in reverse bias from t =1 / (4fs) to t = t0 as

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depicted in fig. 22(b). During this time, the capacitor C discharges through the load R causing a current flow in the load R. The output voltage Vo(t) for this time interval is, by eq. 5,given by Vo(t) = Vo(t)(pk)exp{-[t- 1/(4fs)] / (RC)} ( Equation 29)5. At time t = t0, the positive-going secondary voltage is equal to the voltage of the capasitor C. The diode is again forward biased and a current flows in the load R. At the same time, the diode charges again to a maximum voltage of Vo(t)(pk) = Vs(pk) - VB at t = 5 / (4fs). 6. Then the cycle begins again as in point 4 above. 7. Note that the output voltage has a frequency fo = fs.8. The peak-to-peak ripple voltage, Vr(pp) is given by Vr(pp) = Vo(t)(pk){1- exp[-T / (RC)]} ( Equation 30) where T = 1 / fo = 1 / fs is the period of the ripple voltage. Derivation : From fig. 22(b), it is clear that Vr(pp) = Vo(t)(pk) - Vo(t)(pk)exp{-[t0 - 1/(4fs)] / (RC)} Since t0 » 5/(4fs), the result follows by noting that T = 1/ fs.9. If the time constant t = RC is large which is usually the case, eq. 30 simplifies to Vr(pp) = Vo(t)(pk)T / (RC) ( Equation 31) Derivation : If RC is large, exp[-T/(RC)] = 1 - T/(RC) + T2/(2!R2C2)- T3/(3!R3C3) + ….. » 1 - T/(RC) Substituting into eq. 30 yields the result.10.The rms ripple voltage, Vr(rms) is given by Vr(rms) = Vo(t)(pk)T / (2Ö3RC) ( Equation 32) Derivation: The rms value of a triangular pulse is A/(2Ö3) where A = amplitude of the pulse. Since the ripple voltage is quite close to a triangular pulse, the result follows by noting that Vr(pp) is twice the amplitude.11.The average dc voltage across the load R is given by VL(dc) = Vo(t)(pk) - Vr(pp)/2 ( Equation 33) and when RC is large, by VL(dc) = Vo(t)(pk)[1- T/(2RC)] ( Equation 34) Derivation: The first result is immediate from fig. 22(b). The second result is a consequence of eq. 30. 12.From eq. 30 and eq. 31, we see that if R increases or C increases, then

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Vr(pp) decreases.13.From eq. 32, we see that if R increases or C increases, then Vr(rms)

decreases.14.From eq. 33 and eq. 34, we see that if R increases or C increases, then VL(dc) increases.

FULL-WAVE RECTIFIER CIRCUIT FILTERING1. Fig. 23 shows a bridge rectifier circuit with a capasitor filter connected at

the output.

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Figure 23 Bridge rectifier circuit with capasitor filter.2. Fig. 24 shows the waveforms of the secondary voltage and the

corresponding output voltage beginning from the instant the secondary voltage is applied. The waveform of the output voltage is known as ripple voltage.

Figure 24 (a) Waveform of the secondary voltage since the instant it is applied. (b) Waveform of the corresponding voltage since the instant the secondary voltage is applied.

3. Initially,during the first half of the first positive alternation of the secondary voltage, i.e. from t = 0 to t = 1 / (4fs), the diodes D2 and D3 are forward biased and a current flows in the load R. Note that diodes D1 and D4 are reverse biased. At the same time, the capasitor C charges to a maximum voltage of

Vo(t)(pk) = Vs(pk) - 2VB ( Equation 35)4. Next, during the second half of the first positive alternation of the secondary voltage, i.e. from t = 1 / (4fs) to t = 1 / (2fs), the diode D2 and D3 are reversed biased due to the voltage of the capasitor C. Note that diodes D1 and D4 are still reverse biased.5. During a portion of the first half of the first negative alternation of the secondary voltage, diodes D1 and D4 are reverse biased up to t = t0 due to the voltage of the capacitor C. Note that diodes D2 and D3 are reverse biased now.

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6. In short, the all the diodes stays in reverse bias from t =1 / (4fs) to t = t0

as depicted in fig. 24(b). During this time, the capacitor C discharges through the load R causing a current flow in the load R. The output voltage Vo(t) for this time interval is, by eq. 5,given by Vo(t) = Vo(t)(pk)exp{-[t- 1/(4fs)] / (RC)} ( Equation 36)7. At time t = t0, the negative-going secondary voltage is equal to the voltage of the capasitor C. The diodes D1 and D4 are forward biased and a current flows in the load R.Note that the diodes D2 and D3 are still reverse biased. At the same time, the diode charges again to a maximum voltage of Vo(t)(pk) = Vs(pk) - 2VB at t = 3 / (4fs). 8. During the second half of the first negative alternation of the secondary voltage, D1 and D4 are reverse biased due to the voltage of the capacitor C. Note that diodes D2 and D3 are still reverse biased.9. During a portion of the first half of the second positive alternation of the secondary voltage, diodes D2 and D3 are reverse biased up to t = t0 + 1 / (2fs) due to the voltage of the capacitor C. Note that diodes D1 and D4 are reverse biased now. 10.In short, all the diodes stays in reverse bias from t =3 / (4fs) to t = t0 + 1 / (2fs) as depicted in fig. 24(b). During this time, the capacitor C discharges through the load R causing a current flow in the load R. The output voltage Vo(t) for this time interval is, by eq. 5,given by Vo(t) = Vo(t)(pk)exp{-[t- 3/(4fs)] / (RC)} ( Equation 37)

11.At time t = t0 + 1 / (2fs), the positive-going secondary voltage is equal to the voltage of the capasitor C. The diodes D2 and D3 are forward biased again and a current flows in the load R. Note that the diodes D1 and D4 are still reverse biased. At the same time, the diode charges again to a maximum voltage of Vo(t)(pk) = Vs(pk) - 2VB at t = 3 / (4fs). 12.Then the cycle begins again as in point 4 above.13.Note that the output voltage has a frequency fo = 2fs.14.The peak-to-peak ripple voltage, Vr(pp) is given by Vr(pp) = Vo(t)(pk){1- exp[-T / (RC)]} ( Equation 38) where T = 1 / 2fo = 1 / 2fs is the period of the ripple voltage. Derivation : As above.15.If the time constant t = RC is large which is usually the case, eq. 29 simplifies to Vr(pp) = Vo(t)(pk)T / (RC) ( Equation 39) Derivation :

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As above.16.The rms ripple voltage, Vr(rms) is given by Vr(rms) = Vo(t)(pk)T / (2Ö3RC) ( Equation 40) Derivation : As above.17.The average dc voltage across the load R is given by VL(dc) = Vo(t)(pk) - Vr(pp)/2 ( Equation 41) and when RC is large, by VL(dc) = Vo(t)(pk)[1- T/(2RC)] ( Equation 42) Derivation: As above.18.From eq. 38 and eq. 39, we see that if R increases or C increases, then Vr(pp) decreases.19.From eq. 40, we see that if R increases or C increases, then Vr(rms)

decreases.20.From eq. 41 and eq. 42, we see that if R increases or C increases, then VL(dc) increases.21.Note that for the same secondary voltage, the discharge time for the capasitor C of a half-wave rectifier circuit is nearly twice that for the capasitor C of a bridge rectifier circuit. Note also that for the same secondary voltage, Vr(pp) and Vr(rms) of half-wave rectifier circuit are respectively greater than Vr(pp) and Vr(rms) of a bridge rectifier circuit whereas VL(dc) of half-wave rectifier circuit is less than VL(dc) of a bridge rectifier circuit.

EXPERIMENT : RECTIFIER CIRCUIT

OBJECTIVESIn this experiment, we investigate 1. the half-wave rectifier circuit

2. the bridge rectifier circuit3. the similarities and differences between the half-wave rectifier

circuit and the bridge rectifier circuit4. the effect of filtering on the output of the rectifier circuit 5. the effect of varying the resistance and capasitance on the

output of the rectifier circuit6. the similarities and differences between the responses of the

half-wave rectifier circuit and the bridge rectifier circuit to the actions in 4 and 5

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EQUIPMENT1. Electronic universal trainer2. Set of leads3. Set of bridging plugs4. Multimeter5. Oscilloscope

(A) HALF-WAVE RECTIFIER CIRCUIT

PROCEDURES

1. The center and the lower part of the printed circuit board of module SO 4201-8D are used. The circuit is as below

Circuit for the half-wave rectifier circuit.2. An ac voltage 13 V is applied to the input of the half-wave rectifier

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circuit.3. The values required to complete table 1 are determined using multimeter and oscilloscope. 4. The output voltages given above the graticules are displayed on the oscilloscope and sketched in the graticules. The experiment record beside the graticules are completed.5. The accompanying exercises are completed.

RESULTS

Table 1 Ue = 13 V , 50 Hz (36.8 Vpp)RL (W) UaAV (V)

MultimeterUaS (V)Scope

UBrpp (V)Scope

IL (mA)Multimeter

No-load 0 8.40 18.0 18.0 0.00Without CL 150 6.86 17.0 17.0 0.52Without CL 330 6.91 17.0 17.0 0.53Without CL 1500 6.96 17.0 17.0 0.54CL = 47 mF 150 4.83 17.0 14.5 2.45CL = 47 mF 330 3.33 17.0 10.0 2.46CL = 47 mF 1500 1.11 17.0 3.0 2.46CL = 100 mF 150 3.43 17.0 10.5 2.45CL = 100 mF 330 2.04 17.0 6.0 2.45CL = 100 mF 1500 0.57 17.0 1.5 2.45CL = 147 mF 150 2.69 17.0 8.5 2.46

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CL = 147 mF 330 1.50 17.0 4.5 2.46CL = 147 mF 1500 0.39 17.0 1.0 2.46

1. Explain the operation of the half wave rectifier circuit with charge capasitor and load resistance in the “on” state with reference to the measurement circuit. Answer: The half-wave rectifier circuit with charge capasitor and load resistance can be represented by the cicuit below.

Half-wave rectifier circuit with charge capasitor and load resistance. The input and output voltages are as below.

The input and output voltages. Initially, during the first positive alternation of the input voltage Ue, the diode D is forward biased and a current flows in the load R. The capasitor C charges to a maximum voltage of UaS. Then the diode D is reverse biased up to t = t0 due to at first the voltage of the capasitor C and then the negative alternation of the of the input voltage Ue and then again the voltage of the of the capasitor C. During this interval, the capasitor discharges through the load R and the output voltage drops exponentially up to t = t0. UBrpp is the difference between UaS and the value of the output

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voltage at t = t0. At t = t0, the positive-going input voltage is equal to the voltage of the capasitor C, the diode D is again forward biased and a current flows in the load R.The capasitor C again charges to a maximum voltage of UaS. Then the cycle repeats as above. UaAV is the average dc voltage across the load L whereas IL is the average dc current through the load L.

2. Graticule 1 : Ua without CL, with RL = 1.5 kW

3. Graticule 2 : Ua with CL = 47 mF, with RL = 1.5 kW

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4. Graticule 3 : Ua with CL = 100 mF, with RL = 1.5 kW

5. Graticule 2 : UBrpp with CL = 147 mF, with RL = 1.5 kW

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6. Interpret the measurement results from your completed table and the voltages observed using the oscilloscope.

Answer :From the table and the voltages observed using the oscilloscope,we see that (1) Irrespective of the capasitance and resistance, the peak output voltage UaS is always the same and is slightly less than that of the input voltage.(2) For a fixed RL, when CL increases, the peak-to-peak ripple voltage UBrpp

decreases.(3) For a fixed CL, when RL increases, the peak-to-peak ripple voltage UBrpp

decreases.

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BRIDGE RECTIFIER CIRCUIT1. The upper part and the lower part of the printed circuit board of module SO 4201-8D are used. The circuit is as below

Circuit for the bridge rectifier circuit.

2. An ac voltage 13 V is applied to the input of the half-wave rectifier circuit.3. The values required to complete table 1 and table 2 are determined using

multimeter and oscilloscope. 4. The output voltages given above the graticules are displayed on the oscilloscope and sketched in the graticules. The experiment record beside

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the graticules are completed.5. The accompanying exercises are completed.

RESULTS

Table 1 Ue=13.90 V, 50 Hz (39.3 Vpp) without filter element (short-circuit R1 with B6)

RL (W)U1 (V)Multi-meter

U1

(Vpp)Scope

U2 (V)Multi-meter

U2

(Vpp)Scope

UBrpp

(V)Scope

IL (mA)

0 2.00 21.0 1.50 21.0 21.0 3.00without 150 5.51 17.0 5.51 16.5 16.5 2.76CL 330 5.56 17.0 5.57 17.0 17.0 2.80

1500 5.60 17.0 5.60 17.0 17.0 2.750 0.00 17.5 0.00 17.5 0.0 2.72

CL: 150 5.02 16.5 5.02 16.5 14.5 2.5010 mF 330 4.18 17.0 4.18 17.0 12.0 2.62

1500 1.79 17.0 1.79 17.0 10.5 2.530 0.00 17.5 0.00 16.5 0.0 2.48

CL: 150 3.02 17.0 3.02 17.0 9.0 2.4747 mF 330 1.86 17.0 1.86 17.0 6.0 2.47

1500 0.54 17.0 0.54 17.0 2.0 2.47

Table 2 Ue=13.90 V, 50 Hz (39.3 Vpp) without filter element (short-circuit R1 with B6)

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RL (W)U1 (V)Multi-meter

U1

(Vpp)Scope

U2 (V)Multi-meter

U2

(Vpp)Scope

UBrpp

(V)Scope

IL (mA)

0 0.00 18.0 0.00 18.0 0.0 2.51CL: 150 1.51 17.0 0.99 15.0 4.0 2.51100 mF 330 0.83 17.0 0.56 16.0 2.0 2.52

1500 0.22 17.0 0.15 17.0 1.0 2.520 0 18.0 0.00 18.0 0.0 2.52

CL: 150 1.38 17.0 0.61 14.0 2.0 2.51200 mF 330 0.76 17.0 0.34 15.5 1.0 2.51

1500 0.21 17.0 0.10 17.0 0.5 2.51

3. Graticule 1 : UBrpp ;without CL ,without filter element, RL = 150W

4. Graticule 2 : UBrpp ;with CL = 57 mF, with filter element, Cs =200 mF, RL=150W

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5. Explain the operation of the bridge rectifier circuit with charge capasitor, filter element and load resistance in the “on” state with reference to the measurement circuit.

ANSWER : The bridge rectifier circuit with charge capasitor ,fiiter element and load resistance is represented by the circuit below

Bridge rectifier circuit with charge capasitor ,fiiter element and load resistance

The input and output voltages are as follows

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The input and output voltages

During the positive half of the input voltage, the diodes D2 and D3 is forward biases and a current flows in the filter element F and the load R. The capasitor C is charged to a maximum voltage which is the peak value for U1 displayed by the oscilloscope. Just beyond the peak of the input voltage , the diodes D2 and D3 are reverse biased due to the voltage across capasitor C. That means all the diodes are now shut off and the capasitor C discharges through the filter element and the load exponentially. A similar process occurs during the negative half of the input voltage. U2 is the output across the load only and is equal to RV/(R+F) where V is the output voltage across both the filter and the load. UBrpp is the difference between the maximum value and minimum value of the output voltage across the load.

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DISCUSSIONS

The result of our experiment is in agreement with our theory of rectifier circuit and filtering. For instance, the waveform of the output voltages for both the half-wave rectifier circuit and bridge rectifier circuit are as predicted by the theory. Moreover the differences of these output as stated in the theory are verified by the display on the oscilloscope. We also find that as the load resistance and capasitance increase, the ripple voltage become less significant. This is consistent with equations 30, 31, 38 and 39 presented in the theory.

CONCLUSIONS

In this experiment, we have looked at the half-wave rectifier circuit and the bridge rectifier circuit. The importance of diode as a rectifier has been understood. So is the role of capasitor in filtering the output voltage to produce a smoother output. The differences between the two types of

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rectifications and the influence of the filtering capasitor and load resistance on them have also been explored.

REFERENCES

The following references have been used

Mitchel E. Schultz : “Electronic Devices”, Macmillan / McGraw-Hill, 1994

Savant, Roden, Carpenter : “Electronics Design”, Addison-Wesley Publisher Donald A. Neaman : “Electronic Circuit Analysis and Design”, Times Mirror International Publishers

Floyd : “Electronic Fundamentals : Circuits, Devices and Applications”, Prentice Hall Floyd : “Electronic Devices ”, 5th edition, Prentice Hall, 1996

Malvino : “Semiconductor Circuit Approximations”, Tata McGraw-Hill Publishing Company, 1993

B. L. Theraja, A. K. Theraja : “A Text-Book Of Electrical Technology”,Nirja Construction & Development Co. (P) Ltd, 1994 W. H. Hayt, Jr. : “Engineering Electromagnetics” 5th Edition, McGraw-Hill

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