POWER ELECTRONICS MODULE 2 Explain the significance of freewheeling operation in a semiconverter (KTU JUNE 2022) Due to inductive load, there will be a flow of load current even after phase reversal i-e., after compl is because, due to the large inductive nature of the load, the g each half cycle. load current has a tendency to flow continuously by finding an alternative path through the diode of the same arm of conducting SCR even after phase reversal. Thus there will be no control over the thyristor once itis triggered as seen in continuous and discontinuous modes, This problem can be overcome by connecting a diode in parallel with tiYoad{this, dibde is known as a fly-back diode or freewheeling diode. When there is a s diode acts as the short-circuit path for inductive load current rather than though the thyristors Hence the thyristor regains its blocking state once there is a a in the supply. The free-wheeling diode improves the load curredfwaveform, and increases the input power factor. It also eliminates the damage of the hehe x the formation of voltage spikes during the tum-OFF period sa A.230V, SOHz, 1-pulse SCR eyes isttriggered ata firing angle of 400 and the load current extinguishes at an angle o ind ecu in oe time, te. (KTU JUNE 2022) te = (2n-B)/w te= (2m-120y(2n"50)\._) \ SN te = (360-210)/(100"180) te = 150/18000_» WAN tc = 0.008333 s WN tc = 8.33 ms ;plain the advantage of freewheeling diode in halfwave converter, when the load is inductive in nature. (KTU JUNE 2023) (Same question repeat (Qn No.1 )) (i) Output DC voltage is improved (i) peak to peak output voltage ripple is reduced (ii) load current waveform is improved (iv) SCR is protected as FD reduces reverse power flow/reverse voltage.6. A single phase full bridge converter, connected to 230 V, SOHz. source is feeding a load having resistance R=10Q with a large inductance that makes the load current ripple free and continuous, Calculate the average value of load current for a firing angle of 60°. (KTU JUNE 2023) Output voltage, VO=2V me cosa = 103.95V Average load current 10=V0/R =10.39 A Derive the expression for average output voltage of single phase half-wave-controlj@dectifier feeding RL load, without freewheeling diode. (KTU MAY 2023(5)) © ») © fr. sinar dion \ ad = (cosa - cos) an x . A thyristor is,cOnnetted in series with a resistance of 100 and an ac supply of 230 V at 50 Hz. CAmpiie,the average current delivered if the thyristor is triggered at 45° in every positive half- (RTU MAY 2023(S)) ~ Average voltage, V, 2A +0081 H=(1 +c08a) = BOL As ont5r =88.35V L =VdIR = 88.35/100 = 0.8835 A! «Sia UNIT - I q PHASECONTROLLEDCONVERTERS — 1 phase half wave controlled converter with R load —» 1 phase half and fully controlled converter with R,RL RLE load —> 3 phase half wave controlled converter with R load — 3 phase half controlled converter with RLE load —> 3 phase fully controlled converter with RLE loadES cn cer er ee oo PART-B MODULE ~ II PHASE CONTROLLED CONVERTERS ee —> Phase Controlled Rectifiersare devices which converts fixed AC to controlled DC power by controlling the phasé « angle of SCR. ‘ — Here SCR is used as switching device. Which can b& « commutated by the supply phase AC voltage by providing reverse bias and hence current fallen below holdings —— < c — This type of commutation is called natural or line current. commutation. Hence it does not requires extemal commutation circuits. i ! — Therefore phase controlled rectifiers are simple, les’ ‘ expensive and widely used in industries. —> Some applications of phase converters are given below. _ ( > Steel rolling mills, paper mills, textile mills. > Electrochemical and electrometallurgical process.» Magnet power supplies. » Portable hand tool drives. -—> HVDC transmission. Classification of Controlled Converter [3 Hews} 1. Input Supply » Single phase converter > Three phase converter 2. Quadrant of operation » Single quadrant converter (Semi-controlled) » Two quadrant converter-(Fully controlled) » Four quadrant converter (Dual Converter) 3. Number of Pulses per cycle » Two pulse converter (1 ® SCC & FCC) » Three pulse converter (3 ® SCC) » Six pulse converter (3 ® SCC)SINGLE PHASE SEMI-CONTROLLED CONVERi#:K eo (h" (HALF CONTROLLED CONVERTER) _ — A semi controlled converter uses half number of switche as SCR which are controllable and the remaining as diode which are uncontrollable. — As only half number of switching devices is controllable, 1 is called half controlled or semi controlled converter. — Itis a two pulse, single quadrant converter. R=LOAD @. Kexplacs edoout 1. pase semremliolled convester cotty R lead-O Construction [= va] — Figure shows the circui diagram of semi converter. ce — It consists of two SCRs, twi diodes and a resistive load. — The load voltage and curren are marked as V, and I,.we ee Operation Mode A (a At ot =1+ a, gate pulse is given to T2. — Hence T2 moves to forward conduction mode. — Therefore load is connected to source. — The path for load current‘aveform AK WwSteady state analysis — Average output voltage V, is given as 1 p20 . Vo= So Vm sinot dot =" Vm sinot dot rea = = [-cosat]% = = [- cos x —(— cos a)] V.= = [1 + cosa] — > RMS output voltage-Vims is given as Vams = = fe Va sinZat dot =Vn { ma sinday 2n 4nSS 1 Q: Keaqpléen | about I Phase Senxcovtsolled, Commesten cattle vollage 1 te yond] } ew Load. Derive autpul: Construction RL—LOAD + D1 are forward bias. « « ( « —— 2 — Figure shows the circuit diagram of semi converter. ¢ ( — It consists of two SCRs, two diodes and a RL load. ( — The load inductance is assumed to be large hence load current is continuous. ; ‘ — In positive cycle T, an > At ot = a, gate pulse i given to T). | — Hence T; moves t forward conduction mode. — Therefore load is nov connected to source. 7 ¥To / Sry Kr, i sr | Ove Vo er ee 3h Zpp Ady — Operation Mode A (a At ot =12+ a, gate pulse is given to T>. =. — Hence T2, moves to Vo forward conduction 3 mode. — Therefore load is connected to source. =~ —> The path for load currentWaveformSteady state anal — Average output voltage V, is given as _1 20 * Vo= Ig Vmsinot dot =f" vmsinot dot na = = [-cosat]y Vi = = [- cos m—(— cos a)] Vm [1 + cosa] — RMS output voltage Vims is given as V7 be V2 sin2ot dat ma , sin2a fala tio 4n }M@. allustiale the core Construction : “wud RUE Node ad darts output vollage eqyat ico 40 disCoulinneny mode 9 “SiGe Pass CA Cael alps RLE— LOAD [le moans -tery aune 2035) UST Been Cate Ade & |1. — Figure shows the circuil diagram of semi converter. Sr — It consists of two SCRs, twe 3 diodes and anRLE load. , L — The load inductance ii assumed to be large henc« load current is continuous. — Operation Mode A (a <@t < 2) — In positive cycle T, anc Ty D2 D1 are forward bias. — At ot = a, gate pulse i R given to T;. — Hence T; moves te Yo forward conduction mode. —> Therefore load is noy a connected to source. — The path for load current . PT1—Load—D1— NMode B (x < wt < a+) —» T, remains in conduction state Ta D2 T To D1 as firing pulse is not given to ~_T>. Hee x — But D, is reverse bias and D, is forward bias. —+ Hence T, and Dy starts conducting. 3u _L, — As two devices from the same leg starts conducting, supply is shorted. — Therefore output voltage will be zero. Mode C (x + a <@t <2%)-—- — In this mode T, and D, + are forward bias. tT To’ — At ot=2+ a4, gate pulse is given to T. — Hence T, moves to ' forward conduction 3h mode. — Therefore load is connected to source. — The path for load current N-T2—Load—D2—P |WavetormMs en (A yo) Steady state analysis — Average output voltage V, is given as 1 p20 i: —— ¥, = 5 Vm sint dot a =1 ("vm sinot dot na = = [-cosat]i -= [- cos x —(— cos a)] Vi= = [1 + cosa] SINGLE PHASE FULLY CONTROLLED CONVERTER Baeplaio about Scrgle —qloase pally Cavtnlled, comvedlen curt R-Load,atse devine avollage eqzuatico; SB Noaaks Grnw - way a4 % wrens} — Figure shows the circuit Construction an T3 = diagram of full converter. | — It consists of four SCRs and Ve oe a resistive load. — The load voltage and current are marked as V, and I,.nna : ¢ cut Operation Mode A (a In positive cycle T; and Tx are forward bias. « i | 1 Ty T3 = | vs To R NTA | T% 72 A | Mode B (x + a <@t< 27) — At wot = a, gate pulse is given to T,and T2. — Hence T,and T, moves to forward conduction mode. | ‘eo _» Therefore load is now connected to source. , — The path for load current c P—T,—-Load—T,—-N — In this mode T; and Ty ar forward bias. Ty 13 ! dy *_f CON Ipy SR Nit > Ta ta) —> Atot=2+ a, gate puls . is given to T; and Ty. — Hence T; and Ty moves t forward conductio Vo mode. — Therefore load connected to source. —> The path for load current = N>T;—Load—>T,—PWaveformi Steady state analysis eet — Average output voltage V, is given as —1 p20 7 Vo ade Vm sinot dot =1s" Vm sinat dot mea = = [-cosat]= =" [-cos x—(— cos a)] Vi= = [1 + cosa] Q- wit a céacwik diagraa ercplaeo Abe, ecoattérg. - Ply cangdled base competes ong ee SE LOAD ve ad Feo : ; 10 marks —KTO anne 20a Construction soa4— q wan) 1, —> Figure shows the circu diagram of full converter. x > It consists of four SCRs an a RL load. 3, —> The load inductance j assumed to be large henc load current is continuous.Operation Mode A (a In positive cycle T; and T; are forward bias. _ — At ot = a, gate-pulse is given to Tjand T>. — Hence T,and T, moves to Vo forward conduction mode. — Therefore load is now connected to source. — The path for load current Mode B (x + a 7 PT\|—Load—-T,--N — In this mode T; and T, are T K z Ir, To) 12 forward bias. — At ot =71+ a, gate pulse is given to T; and Ty. — Hence T; and Ty, moves to Vo forward conduction mode. — Therefore load is connected to source. — The path for load current N—T3—> Load > Ty PSteady state analysis Se YSIS — Average output voltage V, is given as Vote “vm sinot dot = eS Vm sinat dat = m [-cosat]zt+ = " [-cos(x+a) — (— cos a)] V; = al cos a + cos a)] 2Vm V.= = cosa. — RMS output voltage Vins is given as Vims = : fr" Va sin2at dot =V, Yn v2Q Explan a Seng RUE a Construction 45 Ronee RLE-LOAD — G raaes -K 10 beey 2003] a wave Ccantadled bridge aealipes nS , fp vllage £. cuareob), 2 ~—» Figure shows the circuit ¥To z Fl diagram of semi converter. Rk —» It consists of two SCRs, two diodes and anRLE load. 3 — The load inductance is +e assumed to be large hence load current is continuous. ~_ Operation Mode A (a T,—N—> In this mode T; and Ty a fa * forward bias. — At ot = 1+ a, gate pul: 3 7 is given to T; and T4. i L — Hence T; and T, moves | forward conductic mode. ae — Therefore load connected to source. I — The path for load current N-T3—Load—T,—P Ty @ | Io¥ Vo NT 3u [Pa T Steady state analysis--- aitady state anatysis- —> Average output voltage V, is given as —1 20 i Vo=s Jo) Vmsinat dot 1 : = s™*" vm sinat dot Tea eens eotytte =—[-cosat]y Vi = a [- cos (+a) — (— cos a)] vi = =I cos a + cos a)] 2Vm | V.= — cosa T! Waveform (Rectification Mode 0<90")TS Waveform (Inversion Mode u>90")@-Draco dhe Clacitk dtagram and woe poses 4 Q B- phax ser Conticlled Converter canta RUE lodh 7 fo mand] Cero mayag THREE PHASE SEMI-CONTROLLED CONVERTER rats Construction — It consists of three_thyristors T,, Ts, Ts which forms +ve group and three diodes D,, D2, i. D; which forms -ve group. — Positive group thyristors will turned on when supply voltage is +ve and gate pulse is given 7s to it. — Negative group diodes will turned on automatically when supply voltage is negative. TE i —> The load inductance is assumed to be large hence load current is continuous. Operation — The circuit operation may explain in six modes for a=0. Mode A (30° < @t < 90") — At wt = 30°, voltage of phase R and B are equally positive i.e line voltage Vp-p or Vp.p is zero.— After this instant, phase R becomes more positive wher compared to other phases and hence SCR T; moves tc forward blocking mode. oe > Therefore T, start conducting when a gate pulse is given tc it.During this interval, phase Y is more negative and hence the return path is provided by diodeD2. — The load current path is given as R-T, > Load > D, > Y Mode B (90° < at < 150°) — During this interval phase R remains more positive anc hence T; remains in conduction. -— Now phase B is more negative and hence diode D; wil start conducting and it provides the return path. — Path for the load current is given by R-—T, — Load > D; >BMode C (150" < wt < 210") — During this interval phase Y becomes more positive and hence T; becomes forward biased and when a gate pulse is given, it starts conducting. — As phase B remains more negative, the return path is provided by D3. — Path for the load current is given as Y — T, — Load > D3; > B Mode D (210° < wt < 270") — During this interval phase Y remains more positive and hence T; remains in conduction. —> Now phase R is more negative and hence diode D, will start conducting and it provides the return path. — Path for the load current is given as Y —>T)— Load— Di > RMode E ( 270" < @t < 330") —— OS) — During this interval phase B becomes more positive an hence T3 becomes forward biased and when a gate pulse given, it starts conducting. provided by D,. — Path for the load current is given as — During this interval phase B remains. more positive an hence T; remains in conduction. — Now phase Y is more negative and hence diode D) wi start conducting and it provides the return path. — Path for the load current is given as B — T;3 — Load > D, — Y i | | ! — As phase R remains more negative,’ the return path |Porn aerenenes[Sis We cial ageing Spoaw poly Coble coats | Git RLE load Derive beget ation 9 Geae xo une 205) — CONTROLLED CONV ERLE! Construction — It consists of six thyristor, whe "Ty, Ts, Ts forms +ve group and 1 T4, Ts forms -ve group. — Positive group thyristor are turn¢ on when supply voltage is +ve ar negative group are turned on whe supply is —ve. — Each thyristor in the positive and negative group conduct ! for a period of 120 degree. — It is mainly used for large power drives, wher i | regeneration of power is required. | — The load inductance is assumed to be large hence loa | gi qj current is continuous. Operation — The circuit operation may explain in six modes.Mode A ( 30" < @t < 90") — At wt = 30°, voltage of phase R and B are equally positive i.e line voltage Vapor Vpris zero. > After this instant, phase R becomes more positive when compared to other phases and hence SCR T, moves to forward blocking mode. — Therefore T, start conducting when a gate pulse is given to it.During this interval, phase Y is more negative and hence the return path is provided by SCR Tg. —> The load current path is given as R-—T, — Load > T5 — Y Mode B (90° < at < 150°) — During this interval phase R remains more positive and hence T, remains in conduction. — Now phase B is more negative and hence when a gate pulse is given to T;, it provides the return path. — Path for the load current is given byR-T, > Load > T, > B Mode C ( 150" < wt < 210") |=» During this interval phase Y becomes more positive an - hence T; becomes forward biased and when a gate pulse given, it starts conducting. — As phase B remains more negative, the return path 1 provided by T, and it continues to conduct. — Path for the load current is given as Y —T; — Load > T,; —> B Mode D ( 210° < wt < 270") — During this interval phase Y remains more positive an hence T; remains in conduction. — Now phase R is more negative and hence when a gat pulse is given to T,, it provides the return path. — Path for the load current is given as Y —>T,:— Load > Ti > RMode E ( 270" < wt $330") — During this interval phase B becomes more positive and hence T; becorfies forward biased and when a gate pulse is given, it starts conducting. — As phase R remains more negative, the return path is provided by T, and it continues to conduct. — Path for the load current is given as Mode F (330° < @t < 390") —» During this interval phase B remains more_positive and hence T; remains in conduction. : — Now phase Y is more negative and hence when a gate pulse is given to Tg, it provides the return path. ‘> Path for the load current is given as B-T; — Load > T; > Y[to ae ree