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VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) Rectifiers, Filters and, Regulators 1. Explain the principle of operation of Half-Wave Rectifier (HWR), Centre tap and Bridge FullWave Rectifiers (FWRs) with the help of circuits and waveforms. Half-wave Rectifier This is the simplest rectifier arrangement, as shown in Fig. 1 (a). The secondary voltage of the transformer (being sinusoidal) can be expressed as vs = Vm Sin ωt = Vm sin α where α = ωt This is shown in Fig.1 (b), where Vm is the peak value of voltage. During the positive half cycle, when the voltage of the point A is positive w.r.t. point B, the diode D1 is forward biased and will conduct. This current, which flows through the load resistance RL, the diode (having forward resistance Rf) and secondary winding of the transformer (having resistance Rs can be expressed as i = Im sin α for 0 ≤ α ≤ π where Im = Vm Rf +Rs +RL (This assumes that the cut-in voltage of the diode is negligible compared to Vm ) During the negative half cycle, when the voltage of the point A is negative w.r.t. point B, the diode D1, is reverse-biased and will not conduct. Therefore, the current i = 0 for π ≤ α ≤ 2π The waveforms of current i during both the half cycles are shown in Fig.1(ç). The current, though not a good d.c. is unidirectional and has an average value which is non-zero (actually positive in this case). Fig.1. Half-wave rectifier input voltage and load current waveforms. Centre-tap Full-Wave Rectifier Fig.2 (a) shows a full-wave rectifier circuit. The transformer secondary has a centre-tap and each half give a peak voltage of Vm . In each half (here is one diode i.e. D1 and D2 The load resistance RL is common to both halves. This can be seen to comprise of two half-wave circuits. On the positive half cycle, when the point A is positive w.r.t. B, the diode D1 conducts and current i1 flows through RL. During this half-cycle, the point C is negative 1 VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) w.r.t point B and hence the diode D2 does not conduct. Therefore i2 = 0. On the negative half-cycle, the point C is positive w.r.t. point B. Hence the diode D2 conducts and current i2 flows through RL. During this half-cycle, the point A is negative w.r.t. point B and hence the dode D1 does not conduct. Therefore i1 = 0. Fig.2 (b) and (c) shows the waveforms of currents i1 and i2 since both i1 and i2 flow through the load RL. The total current i through RL is i = i1 +i2, which is obtained by adding the two waveform and is shown fig.2 (d). Fig. 2 Full-Wave Rectifier and the current waveforms Bridge Full-Wave Rectifier Fig. 3 shows the bridge rectifier circuit, there are four diodes D1, D2, D3 and D4 which form the four arms of the “bridge”. The a.c. from the transformer secondary, in fed to the two corners and the load resistance RL is connected to the other two corners. Please note the directions of all the diodes and also the polarity of d.c. output. Fig. 3 Bridge rectifier circuit During positive half-cycle, the point A is more positive than the point B. The current path is point A, through D1, RL, D3 and back to point B. During this time, the diodes D2 and D4 are reverse-biased. In the negative half-cycle, the point B is more positive than point A. The current path is point B, through D2, RL, D4 and back to point A. During this time, the diodes D1 and D2 are reverse biased. It is interesting to note that in both the halfcycles, the direction of current flow through RL is the same. Thus we obtain full-wave rectification, the waveform being same as shown Fig. 2(d). 2. Derive the following HWR and FWR a) Average value b) RMS value c) PIV d) Efficiency f) TUF Half-wave Rectifier a) Average value: If we put a d.c. ammeter in series with RL, it cannot show a reading the readings as shown in the Fig.1 (c), because a meter cannot (i) Average Current (Idc) 2 VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) respond to such fast changes in wave form. (However an oscilloscope can show the exact waveform). Instead, the d.c. ammeter will show an average reading. For finding out the average value of such an waveform, we have to determine the area under the curve and divide by 2π. Since the value of the current i=0 for π ≤ α ≤ 2π , therefore the second term above is equal to zero. Idc = Im π ∫ sin α 2π 0 2 1 2π 1 2π Idc = 2π ∫0 i dα = 2π ∫0 Im sin α π 2π Im (∫ sin α dα + ∫π sin α dα) 2π 0 Im [cos α]π0 2π I = πm dα = = Idc = Im 2π dα = (ii) D.C. Output Voltage (Vdc) The d.c. (i.e. average) output voltage appearing across RL is Vdc = Idc R L Vdc = Im R π L V = π(R +Rm +R f s L) RL Normally Rf and Rs are quite small in value. If Substiting the value of Idc from equation 2 in the above equation, Rf +Rs RL ≪1 Vdc = Then Vm π b) RMS Value Because i = Im sin α Root-means-square (RMS) current can be obtained by carrying out the following operations in the sequence s-m-r i.e. (i) Square Since current is given by i , therefore its square is i2; Irms = 1 2 1 2 Im 2 π 1 Im 2 π 1 Irms = [ ∫ dα − ∫ cos2αdα ] 2π 0 2 2π 0 2 Im Irms = 2 1 Irms Im 2 π [ 2π ∫0 sin2 α dα] (as i =0 for π ≤ α ≤ 2π 1 2 As sin α = 2 (1 − cos2α) (ii) Mean By calculating the area under the curve (by integrating) and then taking the mean over the period of 2π; (iii) Root: take the square-root. 2 1 2π = [ ∫ i2 dα] 2π 0 1 2 Im 2 2π 2 Irms = [ ∫ sin α dα] 2π 0 1 2 1 2π Irms = [ ∫ Im 2 sin2 α dα] 2π 0 C) PIV In the reverse bias, each diode has maximum voltage across it as it is open. Peak Inverse Voltage is defined as the maximum voltage to which a diode is subjected in the reverse bias. Referring to Fig.1 (a) in the negative halfcycle, the point B is positive Vm volts w.r.t. the point A. Therefore, in half-wave circuit, PIV of the diode Vm volts. D) Efficiency 3 VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) 4 1 × 100% 2 π Rf + Rs + 1 RL If we assume (Rf + Rs) ≪ RL then 4 η = 2 × 100% = 40.6% π This means, even under ideal conditions (i.e. RL and Rs equal to zero), only 40.6% of the a.c. input is converted into d.c. power. In general it can be understood that higher the rectification efficiency, lower would be the ripple content. Rectification efficiency is defined as η= d. c. power delivered to the load a. c. input power Pdc η= × 100% Pac 2 Pdc = Idc RL Pac is the rms power delivered into the circuit, comprising of RL, RS, and Rf 2 Pac = Irms (R L + R s + R f ) 2 Idc RL η= 2 × 100% Irms (R L + R s + R f ) I 2 ( πm ) R L η= × 100% Im 2 ( 2 ) (R L + R s + R f ) e) Ripple factor η= r. m. s. value of the alternating component of the wave r= average value of the wave The purpose of a rectifier is to convert a.c. into d.c. But the simple circuit as used in halfwave rectifier does not fully achieve this purpose. r= We redraw the Fig.1 (c), but this time Idc is drawn in firm line and the rectified output in dotted line. From this, we can consider that an alternating component (of course nonsinusoidal) has been superimposed on Idc. r= 2 2 √Irms − Idc Idc = ′ ′ Irms Vrms = Idc Vdc 2 − V2 √Vrms dc Vdc 2 − V2 Vrms dc = √ 2 Vdc Im r = √( 2 ) − 1 = 1.21 Im π Idc = Average value of the waveform i’ = The value of alternating component of the current wave form I’rms = r.m,s value of the alternating component, i.e. of i’. If the above expression is expressed as a percentage, it indicates that the amount of the alternating component present in the output is 121% of the d.c. value. In other words, it means the undesired ac. component is in fact more the desired half wave rectifier is not a good rectifier. Ripple factor, r. is defined as f) TUF (Transformer Utilization Factor) Most of rectifier circuits make use of a transformer, whose secondary feeds the ac. power. It is necessary to determine the rating of transformer while designing a power supply. TUF is useful for this purpose. TUF is defined as 4 VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) TUF d. c. power delivered to the load a. c. power rating of transformer secondary Pdc = Pac (rated) = I 2 ( πm ) R L TUF = Im (R f + R s + R L ) Pac (rated) = Vac(rms) Irms Im √22 If (R f + R s ) ≪ R L , then I 2 ( πm ) R L TUF = Vm Im √2 2 TUF = √22 = 0.287 π2 Full-wave Rectifier 1 a) Average value: (i) Average Current (Idc) 2π Idc = 2π ∫0 i dα If we put a d.c. ammeter in series with RL, it cannot show a reading the readings as shown in the Fig.2, because a meter cannot respond to such fast changes in wave form. (However an oscilloscope can show the exact waveform). Instead, the d.c. ammeter will show an average reading. 1 2π ∫ I sin α dα 2π 0 m π 2π Im = (∫ sin α dα + ∫ sin α dα) 2π 0 π = Since the value of the current i for π ≤ α ≤ 2π is the same as the period 0 ≤ α ≤ π, therefore the curve will be twice that of the half wave case. For finding out the average value of such an waveform, we have to determine the area under the curve and divide by 2π. Idc = 2 Im π (ii) D.C. Output Voltage (Vdc) The d.c. (i.e. average) output voltage appearing across RL is 2 Im RL π Vdc = 2V Vdc = π(R +Rm+R Vdc = Idc R L f s L) RL Normally Rf and Rs are quite small in value. Rf +Rs If ≪1 R substiting the value of Idc from equation 2 in the above equation, L Then 2 Vm π Vdc = b) RMS Value Root-means-square (RMS) current can be obtained by carrying out the following operations in the sequence s-m-r i.e. (i) Square Since current is given by i , therefore its square is i2; (ii) Mean By calculating the area under the curve (by integrating) and then taking the mean over the period of 2π; (iii) Root: take the square-root. 5 VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) 1 (1 − cos2α) 2 2 2π Im 1 1 Irms As sin2 α = 2 1 2π = [ ∫ i2 dα] 2π 0 Irms = [ 2π 1 1 2 Im 2 2π 2 Irms = [ ∫ sin α dα] 2π 0 Irms = 1 2 dα] α ≤ 2π 0 2 dα 1 2 Im 2 2π 1 − ∫ cos2αdα ] 2π 0 2 Im Irms = √2 2 1 2π Irms = [ ∫ Im 2 sin2 α dα] 2π 0 Because i = Im sin α Im 2 2π [ 2π ∫0 sin2 α ∫ (as i =0 for π ≤ C) PIV In the reverse bias, each diode has maximum voltage across it as it is open. Peak Inverse Voltage is defined as the maximum voltage to which a diode is subjected in the reverse bias. D) Efficiency FWR the non conducting period, the inverse voltage across the non conducting dode is 2Vm 2I 2 ( m) R L π Rectification efficiency is defined as η= d. c. power delivered to the load η= a. c. input power η= Pdc × 100% Pac η= 2 Pdc = Idc RL 2 (R + R + 2R ) Irms L s f 8 1 × 100% 2 π Rf + Rs + 1 RL η= 2 Pac = Irms (R L + R s + 2R f ) η= × 100% If we assume (2Rf + Rs) ≪ RL then Pac is the rms power delivered into the circuit, comprising of RL, RS, and Rf 2 Idc RL I 2 ( m ) (R L + R s + 2R f ) √2 8 × 100% = 81.2% π2 This means, even under ideal conditions (i.e. RL and Rs equal to zero), only 81.2% of the a.c. input is converted into d.c. power. In general it can be understood that higher the rectification efficiency, lower would be the ripple content. × 100% e) Ripple factor i’ = The value of alternating component of the current wave form I’rms = r.m,s value of the alternating component, i.e. of i’. The purpose of a rectifier is to convert a.c. into d.c.. Idc = Average value of the waveform 6 VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) Im r = √( √2 ) − 1 = 0.483 2 Im π Ripple factor, r. is defined as r. m. s. value of the alternating component of the wave r= average value of the wave If the above expression is expressed as a percentage, it indicates that the amount of the alternating component present in the output is 121% of the d.c. value. In other words, it means the undesired ac. component is in fact more the desired half wave rectifier is not a good rectifier. ′ ′ Irms Vrms r= = Idc Vdc r= 2 2 √Irms − Idc Idc = 2 − V2 √Vrms dc Vdc 2 − V2 Vrms dc = √ 2 Vdc f) TUF (Transformer Utilization Factor) In FWR, the average TUF is obtained by considering yhe primary and secondary windings separately. In each half of the secondary winding, the current flows through 1800, whre in the primary winding the current flows through complete 3600 (ii) TUF (primary) Since the primary winding carries a full wave current, its rated capacity is utilized to the same extent as the rectification. TUF = 0.812 (i) TUF (secondary) (iii) Average TUF Pac(rated secondary) = Pac(rated) one half + Pac(rated) other half = Vm √2 = X Im 2 Vm Im √2 + Vm √2 X Average TUF = Im 2 = = 0.574 TUF (secondary)+TUF (primary) 2 0.574+0.812 2 = 0.693 3. Explain why a bridge rectifier is preferred over a centre tap rectifier (i) The current flow through both the primary and the secondary windings are sinusoidal (i.e. full cycle). Due to this, the TUF of the both the primary and the secondary are 0.812 and the overall TUF is 0.812. This is higher than the TUF of full-wave rectifier (0.693) and hence for the same d.c. output power, a smaller transformer can be used in the bridge circuit. (iii) The PIV of each diode = Vm. Let us consider the positive half cycle when diodes D1 and D3 are conducting and therefore the voltage drops across each of them is negligible. Therefore the voltage appearing across D2 is equal to Vm. Also the voltage appearing across D4 is equal to Vm In a similar way considering the negative half-cycle, it can be understood that PIV of D1 and D3 are also equal to Vm. Thus, for a given voltage rating of the diodes, these can be used for highvoltage applications. (ii) A centre-tap is not required in the transformer secondary. 7 VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) (d) Bridge circuit requires four diodes. At any given time two of them conduct in series, causing voltage drop due to two diodes. This additional voltage drop cannot be ignored in low voltage d.c. power supplies. 4. Derive the expression for ripple factor for FWR with L-section filter. Explain the necessity a bleeder resistor. 2 Vm This is also known as the choke-input filter or L Vdc = − 𝐼𝑑𝑐 (R c + R s + R f ) C filter. Figure shows the filter circuit, in which π an inductor L is used at the input of the filter and a capacitor in parallel with the load where R f , R c and, R s are the resistances of resistance RL. Also shown in the dotted line is diode, choke and secondary winding. Both the Rb the bleeder resistance. above equations are similar to those of only inductor filter. However the ripple is much lower in L-section filter than inductor filter. Ripple Factor: The choke L is series with the parallel combination of the capacitor C and the load resistance RL. In order the choke can suppress the ripple components, the reactance of the choke must be very large compared to the parallel impedance of the capacitor and the resistance. This parallel impedance is made small by making the reactance of C much Figure : L-section filter smaller than resistance RL. So we can The inductor L offers series impedance to the reasonably assume that the entire a.c. ripple ripple currents and the capacitor C offers low passes through the capacitor and none impedance in shunt with the load. A corn through the resistance. hin4tion of these two effects result in a load current which smoothed much more effectively Due to the above consideration, the total impedance between point A and B is than either L or C alone. practically that of the choke alone and is If L is small, then the capacitor C will get approximately, charged to peak voltage Vm and diodes will 𝑋𝐿 = 2𝜔𝐿, since ripple frequency = 2𝜔 conduct only for short periods in each half cycle. If the L is large, the diodes will conduct for longer period due to smoothening effect of The ripple voltage has the magnitude 4𝑣𝑚 and 3𝜋 the inductor if the value of L is more then therefore the ripple current, certain minimum value (called critical inductance), each diode will conduct for 4𝑣𝑚 1 complete half-cycle. The output d.c. voltage is 𝛤= 3𝜋 𝑥𝐿 given by Vdc = 2 Vm π The r.m.s. value of ripple current is 𝛤𝑟𝑚𝑠 = The output d.c. voltage on load is 8 4𝑣𝑚 1 √23𝜋 𝑥𝐿 𝛤 √2 so, VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR ELECTRONIC DEVICES & CIRCUITS (UNIT-3) The a.c. ripple voltage the load RL is the same as that across the capacitor and this is equal to 𝐿𝑐 ≥ ′ ′ 𝑉𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑋𝑐 = A practical value (taking into account the approximations made in the derivation) would be more than the above and a figure of √2 𝑉𝑑𝑐 𝑋 3 𝑋𝐿 𝑐 𝐿𝑐 ≥ Where Xc = reactance of the capacitor at the ripple frequency i.e. 2𝜔 By defining ripple factor √2 𝑉𝑑𝑐 3 𝑋𝐿 𝑉𝑑𝑐 𝑋𝑐 = √2 𝑋𝑐 3 𝑋𝐿 = r= ′ 𝑉𝑟𝑚𝑠 𝑉𝑑𝑐 𝑅𝐿 3𝜔 𝑅𝐿 800 Bleeder Resistance: The output voltage of L-section filter at load is Vm whereas when some load current flows the 2𝑣 value is 𝜋𝑚 . This means in the region of very low current, the output voltage drops sharply, as shown in figure = √2 12 𝜔2 𝐿𝐶 We note the ripple factor is dependent on the values of L and C, but is independent of load Critical Inductance : All the foregoing calculations are based on the assumption that value of the inductance is more than the critical value so that the current i does not fill to zero at any time. For this to happen, the value of Idc must be more than the peak value of the a.c. (ripple) component. Figure: (a) D.C. Output Voltage Vs. Load Current. ′ 𝐼𝑑𝑐 > √2 𝐼𝑟𝑚𝑠 In order to avoid this sharp change in output voltage, a resistance Rb (called bleeder resistor) is connected in parallel to RL so that a current of Imin, bleeds (i.e. flows) through Rb. Making use of equation Lc, the value of the bleeder resistance is 𝑉𝑑𝑐 2𝑉𝑑𝑐 ≥ 𝑅𝐿 3 Let the critical value of inductance be Lc. therefore 2𝜔𝐿𝑐 ≥ 𝑅𝑏 ≤ 800 𝐿𝑐 2 𝑅𝐿 3 9