# CHAPTER THREE DIODE RECTIFIERS

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1 CHATE THEE DODE ECTFES 3. Single-hase Half Wave ectifier: Single phase half-wave rectifier is the simplest circuit, this circuit is not used in precise practical applications due to high voltage ripples, and low efficiency. Therefore discussing this circuit aims to compare further electrical circuits with this circuit. erformance arameters : 3... Half Wave ectifier: The electrical circuit is shown in fig.3.-a, where a resistive load is energized throughout this rectifier, and the waveforms obtained from this circuit is illustrated on fig.3.-b fig.3.-a: Electrical circuit

2 Fig.3.-b: Circuit waveforms( source voltage, output voltage, and load current) As well shown from the figure the diode will conduct only for the positive half wave of the supply voltage, during the negative half wave the diode is in reverse biasing and there is no output voltage. During the negative half wave the source voltage is applied across the diode, therefore the diode must carry the peak value of the source voltage. There are different types of rectifier circuits and the performances of a rectifier are normally evaluated in terms of the following parameters:. The average value of the output ( load) voltage dc, and current dc: T sin( ω t) dω t. The rms value of the output ( load) voltage rms, and current rms: rms T ( sin ωt) dωt rms rms 3. The load average and rms power: These are the power energized the load in form of (average) and AC (effective) values. Usually the effective value is greater than the average value due to output waveform shape.

3 AC MS.. MS 4. The ectifier Efficiency : This parameter characterized the ratio between the average and effective power, and depends on the rectifier type and configuration: η 5. The Transformer Utilization: This parameter characterized the ratio between the average power and transformer secondary ( source) volt-ampere rating ( A)rating, and characterized the ratio between average output power and the appearance power energized the system ( transformer, rectifier, and load): TUF Where (A) rating s.s, s the secondary ( source) rms voltage, sms : the secondary rms current. (A 6. The Form Factor: This parameter characterized the ratio between the rms and average voltage (the physical mean of this parameter is the difference between the root mean square of the signal shape and the average value of this shape, therefore if the shape has pure dc value, there is no difference and FF. FF 7. The ipple Factor: This parameter characterized the difference between the ac component of the output voltage and dc component of the same voltage : MS ) MS rating F AC MS FF 8. The Harmonic Factor: This is a measure of the distortion of a waveform, which characterized the difference between the total rms ac current ( secondary current s) and fundamental component of ac source current, which can be defined by decomposing the secondary current into Fourier series ( Harmonics Specter): HF S S S n the case of pure sinusoidal source current ss, therefore HF. S S

4 9. The Displacement angle: This parameter characterized the angle(φ) between the fundamental current s and the source voltage : DF cos φ. The ower Factor: This parameter defines the input power factor: F S.S S.cos φ cos φ S.S S As well mentioned in the F equation the input power factor depends on the load character and on the source current shape.. The Crest factor: This parameter defines the measure of the peak input current (S) peak as compared with its rms value S: (S) CF S peak Example 3.: Single phase rectifier has a purely resistive load of W, energized by voltage source of throughout two windings transformer with ratio :. Determine: - the average and rms voltage and current - the efficiency, TUF, 3- FF, F, and the peak voltage across the diode (). 4- the CF, and the input F. Solution:. The average and MS voltage and current: ; A rms rms rms A. The efficiency and TUF:. η AC MS.MS TUF ( A ) rating where s MS s.s A % 4. 5 %

5 3. The FF, F, and MS FF F FF s. The CF, and input F %. ( S) peak CF S ac F ( A ) rating / S S. S lag S S / S.cos φ. S.5 /.5.77 / Summary: Taking into account the obtained rectifier parameters we conclude that this type rectifier characterized with bad parameters presented by :. Low ( poor) transform utilization 8.6%, which means that the transformer must be / times larger that when it is used to deliver power from a pure ac voltage.. Low ( poor) rectification efficiency 4.5% 3. resence of current dc component in the secondary current causing additional losses ( winding and core heating). 4. High ripples % greater than that when the source is pure dc 5. High ripple factor, which means that a filter with large capacitance is required for smoothing the output voltage, therefore this yield high capacitor starting current problem. Therefore this type rectifier is rarely used due to the weakness in quality of it's power and signal parameters The effect of freewheeling diode on the output voltage: When a rectifier energized L load, the conduction period of the diode D will extend beyond 8 o until the current becomes zero at wtp+f. Figure 3. illustrates the electrical circuit consist of L load ( W, LmH) ( fig.3.a), and the obtained simulation performance of the waveforms ( s, o, s) where it's shown the diode will conduct for the time of p+f.

6 (a) The diode conduction for ms+φ Fig.3.: Electrical circuit(a) and the circuit waveforms. As well shown from the previous figures it's noted that the diode will conduct in the negative half cycle for the time of F, therefore the average output voltage decreases due to load inductance. The average voltage taking into account F can be expressed as follows: φ tan T + φ ω L. sin( ω t) d ω t ( + cos φ ), As q increases, the output average voltage decreases, which the main drawback of existing inductance in the rectifier circuit. Avoiding this drawback requires connecting a freewheeling diode Dm across the load as well shown on fig.3.3(a,b) where the negative portion of the output voltage is eliminated, and the result is keeping the average voltage at the rated value despite of the existing of L load.

7 Fig.3.3-a: Electrical circuit with L load. The diode conduction for ms+φ The diode conduction for ms only. Fig.3.3-b: Circuit waveform The effect of back voltage ( charger) the circuit performance: When the rectifier energized Charging circuit with back voltage E, the conduction period of the diode D will be less than half period (d <p) depending on the back voltage value. n this case the diode will conduct when s > E. Figure 3.4 illustrates the electrical circuit energized battery charger with resistance play the role of current limiter( fig.3.4a), and the obtained simulation performance of the waveforms ( s, o, s) where it's shown the diode will conduct for the time of d. The front angle a and back angle b depends on the peak value of the secondary voltage and back voltage. The diode D will turned off when s< E. Fig.3.4-a: Electrical charging circuit ( principle).

8 The diode conduction angle δ Back angle β Front angle α Fig.3.4-b: Circuit waveforms The values of a, b and d are defined as follows: E α sin ; β α ; δ β α. The charging current can be expressed : s E.sinωt E io for α<ωt <β. Example 3.: The battery voltage of fig.3.4a is E4 and its capacity is Wh. The average charging current be dca. The primary input voltage is p4, 5Hz, and the transformer has a turn ratio of n:. Calculate: (a) the conducting angle d of the diode, (b) the current-limiting resistance, (c) the power rating of, (d) the charging time ho in hours, (e) the rectifier efficiency h, and (f) the of the diode. Solution: E, p, sp/n4/, and s 69.7, (a) the conducting angle d of the diode: α sin E sin ;

9 β α δ (b) the average charging current : β. sin ω t E d ω t, α ( cos α + E α E ). ( cos α + E α E ).. ( 69.7 COS (8.3 ) Ω ) (c) the power rating of,. MS MS β.sin ωt E α + E dωt α ( ) + sin α 4.E.cos α A + 4 (.4) sin cos W (d) the charging time h O in hours: E. 44W h W; ho h / / hrs. (e) the rectifier efficiency h: (f) the of the diode: η % +E

10 Summary: Taking into account the obtained rectifier parameters we conclude that :. The conducting time of the diode is less than p. Low ( poor) rectification efficiency 33.34% due to small conducting time 3. Great heating losses due to current limiting resistance 4. The diode must carry total voltage > in reverse biasing Single-hase Full Wave ectifier: Single phase Full-wave rectifier is the popular circuit, applied in most industrial application due to good rectifier parameters. There are two types rectifiers - Full- wave center tap rectifier ( Midpoint) - Full- Wave bridge rectifier (Gretz) circuit. Both circuit characterized with identified rectifier parameters, except the secondary current, efficiency and diode voltage. erformance arameters : Full-Wave Center tap rectifier The electrical circuit is shown on fig.3.5-a, where a resistive load is energized throughout this rectifier, and the waveforms obtained from this circuit are illustrated on fig.3.5-b. Fig.3.5-a: Electrical circuit Each diode will conduct for half period ( D- positive cycle, D negative cycle). The output voltage of full wave rectified with less ripples. As well shown from the illustrated waveforms on fig.3.5-b, the output rectified voltage has a unidirectional form among the full period, also the voltage applied across the diode is twice the source value.. The average value of the output ( load) voltage dc, and current dc:. T sin( ω t) dω t.6366

11 Fig.3.5-b: Electrical circuit Fig.3.5-b: Electrical circuit. The rms value of the output ( load) voltage rms, and current rms: rms rms T rms ( sin ωt) dωt S Example 3.3: Single phase center tape rectifier with the Ω resistive load,, : transformer ratio. Determine: - the average and rms voltage and current - the efficiency, TUF, 3- FF, F, and the peak voltage across the diode (). 4- the CF, and the input F.

12 Solution:. The average and MS voltage and current: rms rms rms A. A.3 ;. The efficiency and TUF: η AC TUF (A ) where s MS MS.. rating / s.s MS A. 8 % 57.3% 3. The FF, F, and FF F MS FF %. s The diode average and rms current and DA D ( m sin ω t) s. m sin ωt dωt dωt MS 5. The CF, and input F (S) CF F peak S ac (A ) rating / MS /.77 lag. / /. S.S S.cos φ. S.S S / /

13 Summary: Taking into account the obtained rectifier parameters we conclude that the rectifier parameters has been improved comparing with previous case, but not all as follows:. Low transform utilization 57.3%, which means that the transformer must be / times larger that when it is used to deliver power from a pure ac voltage.. good rectification efficiency 8% 3. No dc component in the secondary current, therefore no additional losses in the transformer core. 4. Acceptable ripples % greater than that when the source is pure dc 5. Acceptable ripple factor 48%. 6. The diode must carry twice voltage in the backward biasing, which the main drawback of this type rectifier. 7. Overcoming of this drawback and improving the transformer utilization, force us to use Full-wave bridge rectifier, as well going to be described hereinafter Full-Wave Bridge rectifier The electrical circuit is shown in fig.3.6-a, where a resistive load is energized throughout this rectifier, and the waveforms obtained from this circuit are illustrated on fig.3.6-b fig.3.6-a: Electrical circuit

14 fig.3.6-b: The main waveforms ( s,o, s). fig.3.6-c: The diode waveforms ( d, d), and load current o. With purpose to evaluate the harmonic spectrum of the output voltage, and how can the higher order harmonics can be predicted or fully eliminated, an example is going to be described as follows: Example 3.4: Finding the Fourier Series of the Output oltage for a Full-Wave ectifier energizing -L load. Let s, Ω, LmH. Determine: - determine the mathematical expression of the output voltage - the percentage value of the highest order harmonics. Solution: The rectifier output voltage may be described by a Fourier Series as follows: - The Fourier series: o (t) + n,4,.. (A n.cos nω t + B n sin nω t),

15 where: o (t ) d ( ω t).sin ω t d ( ω t) The Fourier coefficients can be found as follows: A n o. cos n ω t d ( ω t ). cos n ω t d ( ω t ) 4 for n, 4,6,.. n, 4 ( n )( n + ) for n,3,5,... B n o. sin n ωt d ( ω t ). sin n ω t d ( ω t ). Substituting the values of An and Bn, the expression for the output voltage is: o (t) cos ω t cos 4ω t cos 6ω t The percentage values.. There are even harmonic numbers only nd order harmonic is : o 3 3 o o % This means that this harmonic is characterized with great negative effect and leads to additional large losses and waveform deformation. The harmonic speed ( angular frequency) ω ( f) rad/s Which means that the harmonic frequency is twice the rated i.e. f fs Hz. 4 nd order harmonic is : o o o 4 % % The negative effect of this harmonic is negligible comparing with previous one. The harmonic speed ( angular frequency) ω 4( f) rad/s, which means that the harmonic frequency is 4 times the rated i.e. f44 fs Hz. Now the issue is to eliminate ( reduce) the effect of the strongest harmonic which is in our case the fourth one. This could be realized by connecting filtering circuit with C filter as shown on fig.3.7a: % Fig.3.7a: Single phase full wave rectifier with C- filter

16 Fig.3.7b: The obtained waveforms illustration the capacitor effect on the output voltage and load current. As well shown from fig3.7c, the magnitude of nd harmonic has been reduced to 4, which equals to 5.4% from the fundamental one. Fig.3.7c: The Fourier analysis of the output voltage Full-Wave Bridge rectifier with L & EMF: Bridge rectifier energized L load with back induced voltage E is illustrated on fig.3.8a, where the load current flows when the supply voltage being greater than the back voltage E. The circuit waveform are illustrated on fig.3.8b. The load current has discontinuous character due to the value of back emf E.

17 Fig.3.8-a: Bridge rectifier energized -L-E load Fig.3.8-b: Circuit waveforms ( s, o, o). By increasing the back induced voltage, the conducting time of the diode decreases with reduced magnitude as well shown on fig.3.8-c.

18 Fig.3.8-c: Circuit waveforms ( s, o, o) for L load with back emf. At the same time keeping E at constant value, increasing the load inductance leads to further deformation in the output voltage and current as well shown on fig3.8d. Fig.3.8-d: Circuit waveforms ( s, o, o) with high inductive load and back emf..

19 Summary: n general Full wave rectifier is characterized with enhanced parameters as follows :. High transform utilization 8%, which means that the transformer must be /.8.3 times larger that when it is used to deliver power from a pure ac voltage.. good rectification efficiency 8% 3. No dc component in the secondary current, therefore no additional losses in the transformer core. 4. Acceptable ripples % greater than that when the source is pure dc 5. Acceptable ripple factor 48%. 6. The diode must carry only the applied voltage in the backward biasing, 7. n the case of L load and back emf, the current mode ( continuous or discontinuous), therefore the diode conducting time depends on the load impedance and phase shift. 8. Single phase rectifiers are used for load up to 5 kw. For larger power output, three-phase and multiphase rectifiers are used. Despite the good performances of the rectifier, but there is a need to improve the system performances, reducing the transformer weight and capabilities, and energized loads with larger power. This could be achieved by applying three phase rectifiers as well described in the coming sections:

20 3.4. Three hase ectifiers Three-phase rectifiers are classified into Half-wave, and Full-wave energized loads with various impedances and back emf. Applying three-phase rectifiers aims to realize smooth rectified voltage, increasing efficiency, utilization, and minimizing the parameters of the filter. 3.4.: Three-hase Half-Wave ectifier Figure 3.9-a illustrates three-phase half wave rectifier energized load, where three diodes ( D, D, and D3) operates in series sequence, each one for a time of. The operation sequence is determined by the criteria : " the diode with maximum positive voltage applied across it's terminals will conduct ", therefore each phase will pass the current for, where for the rest of time this phase will be off and doesn't participate in the rectification process. Fig.3.9-a: Three-phase half-wave rectifier- Wye connected

21 Fig.3.9-b: Three-phase voltage, and output rectified voltage. Fig.3.9-c: hase current, and instantaneous rectified current.

22 Fig.3.9-d: Diode voltage Fig.3.9-e: hase current, and instantaneous rectified current at L load.. The average, rms voltage and current: / q. cos ω t / q q MS 3 in the case 3 3. / / q.8468 MS MS.. The efficiency and TUF: of (.cosωt) Three phase q 3 MS q three. 869 d ( ω t ) d( ωt) phase q. + q q sin sin q q ;

23 η AC AC ; η TUF ( A ) S ( A ) TUF rating MS AC. ( A ) 3 3. rating ( A ) (.869 ) (.8468 ). MS / q rating S. m ; S cos ω t d( ω t) rating m q.95 + sin % q % 66 % %.4854 m 3. The FF, F, F and MS.846 FF.869 F FF 8.4% %.65% AC.846 F.6844 (A ) rating The average and rms diode current and / 3 DA m. cos ω t d ( ω t) m sin m D / 3 ( m. cos ω t ) d ( ω t ) MS / s Summary: Taking into account the obtained rectifier parameters we conclude:

24 . The output average voltage is 8% of the phase magnitude.. Satisfied transformer utilization 66%, which means that the transformer must be / times larger that when it is used to deliver power from a pure ac voltage. 3. good rectification efficiency 96.66%. 4. There is a dc component in the secondary current, therefore additional losses in the transformer core. This reason explain the small value of TUF. 5. Good form factor %, and Acceptable ripples factor 8.4% greater than that when the source is pure dc. 6. The diode must 33% of the total average dc current, 57% of the total rms current, and must carry.73 in the reverse biasing. With purpose to enhanced the rectifier parameters ( reducing the ripples, increasing TUF, increasing the rectifier capability, and elimination the dc component in the secondary current) a three-phase full-wave rectifier is applied as follows:

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