ed SINGLE PHASE AND THREE PHASE CONVERTERS Bee su cok cn POULT” SHORT QUESTIONS WITH SOLUTIONS. iin, gi. Write a short notes on, (a) Sensing Unit (b) Control Unit, Ans: . Model Paper-ll, Q1(a) @) Sensing Unit ‘Sensing unit senses the output of the load and converts it into the signal that can be understood by the control unit. Generally two types of sensors are used. They are, () Speed Sensors Example : Analog tachometer, Digital tachometer. (i) Current Sensors . Example : Hall effect sénsors, sensors employing non-inductive resistance shunt in conjunction with an isolation ‘amplifier. (0) Control Unit ‘The function of the control unit is to control the power modiilator so as to‘obtain the desired speed-torque characteristics. The input command is fed to the control unit in order to achieve the desire¢ speed-torque characteristics. It continuously monitors the output of the load with the help of sensing unit and calculates the deviations in the ‘output from the input and ‘modifies the control signals accordingly. ~ ‘Thenature ofthe control unit depends on the type of power modulator: empoyed For example ia semiconductor converter isused asa power modulator, then a control unt is employed which generetes firing pulses for triggering the devices. @, Give some applications of thyristor controlled drive. ., a > Ans: Following are some of the applications of thyristor controlled drive, |. Itis employed in traction 2 tis used in stee! mills, rolling mills, textile mills, sugar and paper mills 3. tis also used in coal mines as importing conveyors. 4 Italso employed in transport agriculture. 5. Itfinds its applications in flight and landing control systems of aircraft. °%. ‘Discuss the advantages of thyristor controlled drives. t Anes Model Paper ata) Thytistor controlled drives are advantageous because of the following, Simple and reliable operation ~~ High efficiency (ie., upto 95%) size, , a Low initial cost 7 te ‘SPECTRUM ALL|N-QNE JOURNAL FOR ENGINEERING STUDENTS. 5 9) J —CC NMA_ Sooo 12 Less weight of the drive 6. Flexible packing 7. Easy to carry anywhere 8. Can be adjusted in small area 9. Faster time response. Q4.Discuss the di; of thyristor Controlled drives. Ans: 1. Due to high ripple content motor will be heating quickly. 2. Commutation problem in output 3 4 5. 6. In thyristor converters, regeneration control circuit is complex. May-46, (R13), Ha) Rectification mode is the mode in which the power is delivered from source to load and the motor terminal voltage and current are positive. This mode is also known as motoring mode or first quadrant mode of operation. Q6. Whats the principle of phase control? May-16. (R13). QAO) Phase control technique is based on the principle of ON- OFF switching. It utilizes the thyristors as switches to connect the load circuit to the input A.C supply for a friction of each input cycle. Itis a method used to control output RMS voltage across the load (a). Thyristors are tumed ON by controlling the phase angle of input A.C signal and turned OFF by using A.C line commutation (or) natural commutation. In this type of technique, commutation circuitary is not required. Q7. Draw the circuit symbol of thyristor and mark its terminals. Ans: The circuit symbol of SCR is shown in figure. an c Figure: Symbol of SCR ee Q8. Drawa neat sketch of full converter feeding D.C series motor with free wheeling diode. Ans: D.C. series motor fed from a full converter with free wheeling diode as shown in figure. ‘May-17, (R13), Qt(a) POWER SEMICONDUCTOR DRIVES (JNTU-HYDERABa, Figure: Full Converter Feeding D.C. Series Motor with Freewheeling Diode Q9. Write conduction drives. Ans: “The advantages of continuous conduction ofa drive, as follows, 1. Good speed regulation of the drive 2 Harmonic content is less 3. Speed control range is more. 4. The commutation capability of the motor is good. 5. No-load speed is less. | G10. Sketch the speed-torque characteristics off; half controlled bridge D.C shunt motor. Model Papa, Oi | The speed-torque characteristics of a D.C. shunt mor | is shown below. . Ans: Speed (7) 160. vo} ara a 120 —_ dl foge) 1203 45 6 78 Figure: Speed-Torque Characteristes Inccan be observed from the above fgets of D.C. shunt motor decreases rapidly with ner This can be explained mathematically 3, speed, w= Haze WARMING: aoPoucyi o sbokiaCRMNAL et Anos ogy ABL wo =HUNIT-1 Canto of D.C Motors by Single Phatland Three Phase Converters 1.3 ‘cited D.C motor for fully controlled bridge Ls ‘Two quadrant operation can be achieved by single phase fully controlled bridge converte. ‘When the firing angle (a) is less than 90° the terminal voltage e, and current i, both are positive. Hence the motor draws wer from supply which is delivered from source to load. Now, the converter is said to be operated in first quadrant shown in figure During the period m T. adhe. drive operates in continuous conduction mode. In _hismode, th speed-torque curves are parallel straight ines, Hence sped gepuaton is good. The slope of these speed-torque \ > curves depends on armature resistance, — Discontinuous Conduction Mode gee” ‘When the operating torque is less than the rated torque(7 7 =U tcoseyEn Ret x (Keen (ab) 7 Where, Armature curreng 1,5 we Speed of motor at no-load, _ (+0084) Em Katyn /* (Where, T= 0] Also, from speed-torque characteristics shown, it is observed that, continuous armafire current is necessary to obtain good speed fegulation., N, {teas Q26. Derive the speed and torque expressions for a single phase semi-controlled converter connected to a separately excited D.C motor. Ans: (June-14, (R09), Q4 | Model Papers, 02(a)) Let, V- Terminal voltage of the motor E,~ Back e.m.f of the motor [,— Armature current of the motor R,- Armature resistance 4 - Flux/pole co- Angular-speed k= Motor constant a. Firing angle of converter V,,~ Peak value of phase voltage. We know that for a D.C separately excited motor. Back emf, E, % 0, = £7K,90, => 7K, % Torque, Tx $1, = T=Kohy => 7-K,I, Terminal voltage, V = £,+ 1 Re 1g equation (1) in equation (3), o- (I) (2) -» 3) we get, ‘Substitutin, v=K,0,+ JR, 2 Ko,-0-1%, V —1eRo 0,° Km Speed, , = vipe rad/sec SPECTR 1.11 But, ‘The average input voltage of a separately excited motor fed by a 1-6 semi-converter is given by, v= % (1 +c0sa) Now, From equations (3), we have, ‘Armature current, = Ge = L_ Kaen [- From equation (1)] _ fe + cosa) ~ Kniom 4 eo a) Substituting equation (4) in equation (2), we get Torque, 7=K, 1, Yn (1 + cosa) — Kniom = Ka| = 1 +0084) - K20y a 27. Draw and explain the speed-torque characteris tics at different firing angles for a full converter feeding a separately excited D.C motor. ed Ans: The relation between speed and torque is given by, ya 2émgosa. __ReT wee) 7K (Kz)? A baf Con The no-load speed is given by, 2eq C08 Te, n= eege [v9] when all the “This no-load speed can be achieved o thyristors fail to conduct. This can be done when E, > Vi.e., E,>¢, fora< F E,>e, sinot for a> 5 ‘Therefore, no-load speed is given by, N,= F fordsas F = (sina) for Fsasn “The spéed-torque characteristics drawn based on above data is as shown in figure, UM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS1.12 POWER SEMICONDUCTOR DRIVES [JNTU-HYDERABAD] mt ae ‘ | ea Region (1) Reglon (2) Dscontinuous: ‘Continuous Ee anes aie cous 0 Figure: Sj torque Characteristics of i Full Converter fed D.C Separately Excited Motor . 2 ‘The maximum value ofthe averages “4? The no-load speed at this voltage is Dep : The speed-torque characteristics graph consists of two regions ie., region (1) and region (2). Region (1) is called the discontinuous conduction mode, whereas region (2) is called continuous conduction mode, For remaining answer refer Unit-1, Q12. Q28. Explain the different modes of operation of 1 fully controlled rectifier-fed separately excited motor for motoring mode of operation. ‘AprilMay-12, Sat-4, 2(0) OR Explain how the speed control of a D.C motor is achieved illustrating the triggering circuits of the thyristors. Ans: ‘A-Single phase fully controlled converter operates in 1 quadrant and as well as in IV quadrant, Hence, a single phase fully controlled converter is also known as two quadrant con- verter, In I quadrant, both the voltage and current are positive and in TV quadrant the voltage is negative and the current is, positive, A single phase fully controlied converter consists of four thyristors 7,, T,, 7, and T,, Among the four thyristors, the thyristors 7, and 7, forms positive group and conducts during the positive half cycle and the remaining thyristors T, and T, form negative group and conducts during negative half cycle. A single phase fully controlled converter fed from a D.C separately excited D.C motor can be in two modes, one is continuous mode and another one is discontinuous mode. In continuous current mode, the load current flows continuously without being zero, which is because of large inductance, ie., the inductor is large enough to supply load current tll the next thyristor pair is triggered. A single phase fully controlled converter fed separately excited D.C motor’is shown in figure (a). k, k, 4, Phase Fully Controlled Convertor Fd Separately Excited D.C Motor ‘Continuous Conduction Mode During positive eee ae T, and 7, ae forward biased. When these thgristors are triggered at ot = a, the supply voltage appears across the output and the SCRs will be in conduction till the period of a < of <(n +a), At, f=, the thyristors 7, and 7, are reverse biased, the curent continues to flow from supply to load. At the same instant the thyristors 7, and 7, are forward biased. When these thyristors are triggered at (x + @) the motor is connected to supply through the thyristors 7, and T, and at this instant the current is transferred to the thyristors 7, and 7, from 7, and 7,, Hence, the supply voltage appears at the output: The thyristors 7, and T, will be in conduction till the period of (n + a) Substituting V, value from equation (1) in above equation, we ge’, Wa, ye Leos = Hae Ko Now, in separately excited D.C motors the torque is given by, (4) T=Kgl, > I, we (5) ‘Substituting equation (5), in equation (4), we get, ‘Vey — Ne Boe TR Ke Ke Ke 2a cose = R oR Ke _ 2Wqcosa RT N= ee bE 6) Thus, from equation (6) it is clear that, the voltage is directly proportional to speed and the voltage can be controlled by varying the firing angle. Hence, the speed can also be | controlled. The speed-torque characteristics are shown in figure. a. increasing Characteristics for Single Phase Fully Figure: Speed-torque Controlled Fed Separately Excited D.C Motor NE JOURNAL FOR ENGINEERING STUDENTS oss,POWER SEMICONDUCTOR DRIVES [JNTU-HYDERABAD, from dlscontinugg 0 ion for critical speed which separates continuous conduct Conduction for a 1-4 full converter fed separated excited D.C motor- Ana May$6 (19), cay Expression for Critical Speed ive operates in the ie none complete eyele of motor terminal voltage the single phase fully controlled converter ace sete two modes. Mode-l Duty interval for a < of < ri.e., When the motor is connected to source (V, = V,) Here, V= Reig tly te+8 =V,sin ot ~(I) ‘0 at re Mode-2 Free wheeling interval for x < ot < when the motor is rot connected tothe source (7, = 0 , Here, xX Rit l,te+B=0 o»() dt Mode-3 Zero current interval for BS at Sm +0. ie., when the armature current j, = 0 and ¥, = Back em£(E,) V,=Eand =) V, Equation (1) has two components in it, one due to source ("ps0r-0)] and the other due to counter e.m.f (-E/R). Each of these two components has a trénsient component.-Let these components be represented by a single component Ke" Fesitar-9)— Feet 5 all) Where, ‘ Ao Aye” Z= {R?+(wl,)* en oe $= tan (LR) K, can be obtained from the initial conditions, i.e., substituting f= and i, = 0 in equation (3), we get, Vn Fig oe O= Pala Q)—FP+ Ke oe wat 2 lings 5 Kent = sna) +R" Vi gr £5 oso [fsme- Al WARNING: XeroulPhotoeopyin of his bok i CRIMINAL et. Anyone fund guity it UABLE to tnéa LEGAL erecendvat aituting the value of K, in equation (4), we get, van E, i, [ rm] [SacBee me oles E,|_[¥, = [ 7 sneer mm Vn = FP sinter -9)-sin( ag) e(0-ene4y 2 —¢er-aoy (5) Substituting «wf = m in equation (5), we get y : "tsi 4)—siga geo Bags gto-orey (6) Now, the solution for equation (2) CF and P1is, E, keen A =) K, can be obtained from the initial conditions i.e, substituting of = x in equation (7), we get, Van ei & E, F sin (x — 4) ~sin(a.— ge 9%] FEL “Ie = Keron Fe (From equation (6)) ¥, => Flsind—sin(a- get] n- et NOH] = Keto V,, = bing -sinfa peony Bee = eeo 2 k= Fler sing 2° sina) eM ‘Substituting the value of K,, in equation (7), we get, . 5, ow fet" sing —e.sin(a—gj]eraent + Beate eet feel Pah sing A sina ))~ REL H “O. ‘Now substituting or = Band i, = 0 E, 0= Bape O-netsing OO sia -o)- 3e- O-oxah) ” ‘The value of B can be obtained by iterative method, We know that, o ¥,=E,+1R, 3, But for discontinuous conduction waveforms, we haves aya ue i J v,sinond(o) * J E,alon) (xta-BEy 7 y (MI) = [cosa — cos] * epecrai aot iin on ENOEENE STUDEATS a1.16 POWER SEMICONDUCTOR DRIVES [JNTU-HYDERag, OD we get, ¥, mn +a- F le0sa ~cosn]+F4=B) p You q [eosa-(-1)]-1,R, = Ey faa +oosa)-1,R, = a[ke] . “ We know that, Back em.fis also given by, E,=Ko, a Also torque, ~(U) ‘Substituting equations (13) and (14) in equation (12), we get, b, T, _, [Bea PU + cosa) - FR, = xo [Ee Vn (1 +0080) R, 0. EGS Bea} The boundary between continuous and discontinuous conduction is reached when B= n+ a. On substituting B= n+ aia equation (9) we get the critical speed which separates continuous ¢onduction from a discontinuous conduction, From equation (9), we get, O= J [ewe sing — et 2sin(a — 4)] — z [1 -etreexon) o= F - E, = J [em sing — e*sin(ae— )] — Rem] > 5, Yo rot KU e= F [ee sing “eminga 9) Vale **sing ~€* sina 4)} = Relate“ singe" sin(a - @)} = A Zl RiVale** sing - &°* sin(a— 4)) = Seale singe sina) , > Ko, Ze] (From equation (13) o, = Be Yale sing — e*4 sin a - gy] = Z eK ‘The rteal speed (@,) which separates continuous conduetion fom € discontinuoy “Ry, fee sing —e-*°4sin(a— ) Fp Enle sing =e *sin(a 4) = 2 [1-e""]K us conduction is, WARNING: Xerox/Photocopying ofthis book is @ CRIMINAL act, Anyone found guity i ABLE to fac LEGAL procadngs. d4 Control of D.C ingle PI Three Pt or col Motors by Single Phase and Three Phase Ci 2 inverters UN re fect OF armature Wyre Pnase Converters Explain the effect of armature inductan ice on the performance of a D.C drive is ‘The various effects of armature ind uct areas follows, ance, Lona D.C anst May-16, (R13), 03(0) ev it improves the speed regulation of a D.C dri operating over a torque without limitation, men 4, Domain of discontinuous motor current gets reduced Develops the performance of a motor and a semi converter (or) full converter system, me Reduces the peak current of a D.C drive and at 63% of rated speed, the ratio of peak-to-average motor current becomes maximum. Decreases the RMS to average current ratio and armature heating without the use of additional designed motors. 6, Power factor is improved constantly. The ripple current is undesirable for the motor but makes the supply current permanently sinusoidal. 32. Two independent single-phase semi-converters are supplying the armature and field circuits of the separately excited D.C. motor for controlling its speed. The firing angle of the converter, supplying the field, adjusted such that maximum field current flows. The machine parameters are: armature resistance of 0.25 0, field circuit fesistance of 147 0, motor voltage constant, K, = 0.7032 ViA-radisec. The load toraue is 1.245 Nem at 1000 r.p.m. The converters are fed from a 208 V, 50 Hz A.C. supply. The friction and windage losses are neglected. The inductance of the field and armature circuit are sufficient enouigh to make the armature and field currents continuous and ripple free. Determine, (a) The field current (b) The delay angle o| (c) Input power factor of the armal converter. 1 fthe armature converter ture circuit Ans: Given that, 1-4 semi-converter feeding separately ex Amature resistance, R= 025% Field circuit resistance, Ry Motor voltage constant, K, = Load torque, 7= 45 N-™ Speed, N = 1000 r.p.m Input voltage, Vins 208. V Frequency, f= 50 H2 sPECTRU ited D.C motor iM ALLAN-ONE J ]OURNAL FOR ENGINEERING STUDENTS To fin (a) The field current, J,=? =? (b) The delay angle of the armature converter & (©) _ Input power factor of the armature circuit conver- tor, p.f=? (a) The field current (1,) Field voltage is given by, (ld Von Fe (cosa +1) ) of the converter at field But, given that firing angle (ot flows in field (and also side is such-that maximum current voltage). It is possible only if = 0° Equation (1) becomes, y,= ie [-c080° = 1] = 2XAX208 fs yy VPVend = 187.26 V .. Field current, -% Law = 187,26 ‘147 = 127A (b) The Delay Angle of the Armature Converter (0) Armature current, L-F By - 45 2 = proaaxtaT 0384 Armature voltage, 1 =V41R, Where, = Back emf V,=K, ol, “ 2nN _ 2n 1000 - 0 = FEM = EE = 108.7 V, = 0.7032 * 104.7 1.27 ¥,=93.50V ‘Armature voltage, V, = 93,50 + (50.38 * 0.25) = 106.09 V ‘1.18 POWER SEMICONDUCTO! Also armature volt we, Vou (1 + cosa) 2 x 208 x (It eosa) ( 106.09 = 93.63 (1+ cos a) y 106.09 VV, u Leos a = 106.02 licos a = 1.13 l4cos a = 1.13 -1 3 cos a= cos 0.13 a= 82.53° 82.53° Delay angle (a rmature Circuit Converter (©) __ Input Power Factor of At Given that, the armature and field currents are: ‘continuous and ripple free. Therefore, the output power is given by, =VL 06.09 x 50.38 = 5344.81 Watts ‘Also, given that, friction and windage losses are neglected. => Input power = Output power . P.=P, P, = 534481 W ‘The rms value of input current ofthe armature converter is given by, ; Ing [: Jf Bacon 33 I’ we L, = 50.38 [ Ing, = 37.0TA Input V.A (voltage-amipere) rating is given by, 1 =Vi = 208 x 37.07 = 7710.56 Therefore, Input power factor, Pr, pi= Vf ) R DRIVES [JNTU-HYDERAByy “ rter Is ! ¢ voltage at the terminals of {nyristor is triggered continugys” nal. The resistance of armats" by auit Ig 10 € and because of fixed m nd Sxeltation and high Inertia, the motor g fe considered constant s0 that the back e., is 60 V. Find the average value of the armaty,, te prent neglecting armature Inductance, ~ (Mayilune-13, (R09), 1(0)| Mode! Paper, any motor. The by a D.C sig Ans: Given that, D.C separately excited motor is fed from 1-)hay controlled converter. = 120 V (rm.s) Input voltage, ¥; ‘Armature resistance, R, = 102 Back emf, V,=60V Average armature current, J Given the thyristor is triggered continuously Firing angle, «= 0° Peak voltage, V,, = V2 Vems = ¥2120=169.7V Vn( + cosa) a Motor terminal voltage, V, (1+ 0080") = 169.7(2) oe V, = 108.03 V For semi-converter feeding D.C separately excited m> tor, armature voltage equation is, a= Rale+ La YE + V _ Where, L, is armature iniductance. Neglecting armature inductance (L, = 0) => V=RL+Y, 108.03 = (10) I, + 60 WARNING: XeroxiPhotocopying of this book is CRIMINAL Average armature current, /, = 4.80 A ‘act, Anyone found guilty is LIABLE to face LEGAL proceedings.1 Control of D.C Motors by si ¥ Single Phase and, Three Phase Converters mature circuit half-controlled re¢ S60 cpm, TAB Roo eT Prase Convers inductance 7220 V, 960 BAse| : om of Pie, excited D.C motor hi resistance and inductance ctifier with an A.C source (2) Motor torque for = 60: and speed = 600 (bp) Motor speed for 0 = 60° and T = 20 N; p.m. Im. AprivMay-12, ‘Set-2, QB st A iven that, Rated voltage, V= 220 V Speed, N, = 960 rp.m Current, = 12.8 ‘Armature resistance, R, = 2. Inductance, L = 150 mH A.C source voltage, V, = 230 V Frequency, f= 50 Hz qodetermine @ o ® Motor torque For « =60° and speed N, = 600 rpm Motor speed For a = 60° and T= 20-N-m Peak voltage, V,= V2, = ¥2x230 = 325.27 Volts Motor Torque . Itis necessary to know, in which mode the motor is operating: For this, the critical speed must be determined. :. Critical speed is given by, acot Asin moot oes") Genk NG | ao) We know that, :. Inductive Reactance, X= 2nfL = (2n)(50)(150 * 10°) = 47.1242 Impedance, Z= (mex ore = 47.162 ‘We know that, oar [| 124 wn | 74] = 87.57° SPECTRU RNAL FOR ENGINEERING STUDENTS JM ALLAN-ONE JOUR DRIVES [JNTU-HYDER Ag), 10 POWER ‘SEMICONDUCTO! cot § = cot (87.57) = 0.0424 Induced e.rn.f at rated operation E-V-IR, = 220~-(12.8)(2) = 194.4 E _ 1944 om ins eh = x00 = 0.8752 cee = g O*fgr008 = 0.9565 Substituting the values in equation (1), we get, (2)G25.27)_[ sin(87.57)(0.9565) - sin (60-8 11.933)(47.16)| 8752 7.136){10.9] 7.78 rad/sec or = 742.74 rpm Critical speed is greater than the motor speed (600 rp.m). Thus, the motor is For continuous conduction mode, the average output Voltage is given by, is operating in continuous conduction meds. tn (+0080) * = 275.21 4 40860) 7 = 155.3 Volts We know that, 7 4.% Eom Induced voltage, 0 600 194.4 — = ——x194.4 E, XE, = 969 = 121.5 Volts Motor torque, 7= KI, + = 32,66 Nm 5 Ry jit isa is WP) WARNING: Xerox/Photocopying ofthis book is CRIMINAL act. Anyone found uity is LIABLE to face LEGAL proceedings.Control of D.C Motors by R yT- Y Single Pha; e Motor Speed S® and Three Phase Converters 1.21 induced E.MLF for critical speed, Maximum voltage, V,, = 72 *¥, ty 2 x 230 E.= Sexp x ' =325.27V TA2T4 142.14 At rated operation back e.m.f is given by, 960 TT 969% 194-4 E,=V-IR = 150.4 Volts =220-13 x2 . Vv. = 220-26 Critical torque, T= K [| E,=194V = sap S83-1s0.21 2 =4.73N-m Critical torque 7. is less than motor tox ical torque 7, rque (20 N-m), ‘she drive is operating under continuous conduction u=g ae Toay 710346 Induced voltage at 20 Nem o LR, = 155.3 - (103462) = 134,608 Volts Required speed, E = 960 Neg 134.608 56 194.4 = 664.73 035. A 220 Volts, 960 r.p.m, 13 Amps separately excited D.C motor has armature resistance of 2 ohms. It is fed from a:single-phase half controlled rectifier with an A.C source of 230 Volts, 50 Hz. Assuming continuous conduction, calculate motor torque for a = 60° and speed 600 r.p.m. Ans: Given that, Rated voltage, V = 220 V Speed, N, = 960 rp.m Current, J= 13 A ~~ Armature resistance, R, = 29, ACC source, V,= 230 V Frequency, f= 50 Hz ~ Delay angle, a = 60° Speed, N, = 600 rp.m ‘The average voltage for continuous operation is given by, He-(1 + cosa) 32521 (1 +c0s 60") = 325.26 4 5) = 487.89 V, = 155.3 We know that, B® EO ™ N, 600 = xg, =, $00 Bo PB = S6y % 194 Backemf, E,=121.25V | ‘We know that, in separatcly excited D.C motor the torque is directly proportional to J, (1) = 194% 60 2m X 960 = 1.93 Torque, 93 {15 - 34.05 193 x 4g = 1.93 * 17.025 T= 32.85 N-m SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS " X1.22 A200, 875) p.m, 150 A separately excited D.C Ra has an armature resistance of 0.06 ohm. It is fed from a single phase fully controlled rectifier with an A.C source voltage of 220 V, 50 Hz. Assuming continuous conduction mode. Calcul 0) Firing angle for rated motor torque and 750 rpm, (li) Firing angle for rated motor torque and 500 p.m, (ili) Motor speed for a firing angle of 160° and rated torque. Ans: ‘May-46, (R13), Q2(b) Given that, Motor input voltage, Y= 200 V Speed, N= 875 r.p.m ‘Armature current, f,= 150A Armature resistance, R, = 0.06 @ Rectifier input voltage, V, = 220 V Frequency, f= 50 Hz To calculate, (i) Firing angle (a) for rated motor torque and 750 rpm (ii). Firing angle (a) for rated motor torque and 500 rpm (iii) Motor speed (N) for a firing angle of 160° and rated torque. At rated operation, EMF induced, E= V—I,R, = 200-150 « 0.06 =191V At speed N, = 750 rp.m We have, @ =163.71V Motor terminal voltage, V=E,+LR, 163.71 + 150 x 0.06 =172.71 by, © WARNING: Xerox/Photocopying of this book is @ CRIMINAL act, POWER SEMICONDUCTOR DRIVES [J INTU-HYDERA conduction, But for continuous 2h, = He cosa aa 2x220V2 oe cose (“! 172.71 ze = cosot = 0.872 = c0s(0.872) = (ii) At speed N, = 500 rp.m = 220x191 = 109.14 875 1,8, Vi=E,+1,R, = 109.14 + 150 * 0.06 = 118.14 But for continuous conduction, 2, —cosa F 119.14 = 2222092 cose in . = cosds = 0.596 => — a=cos (0,596) Firing angle, «= 53,41° ” At Firing Angle, o = 160° For continuous conduction, —cosa eos 2x22 = ADO? seg . — 186.12 ‘We know that, motor terminal voltage is alo gi VA +18, ~ 186,12 =E, +150 0.06 => &£ 195.12 V Anyone found guy is UABLE to face LEGAL proceedings:=~ 893.87 rp. _, speed, N,=— 893.87 p.m. ow. Ans: ‘A 220 P excited motor is controlled by a 1 -phase ful controlled rectifier with an A.C source Sclee of 230 V, 50 Hz. Enough filter Inductance is gdded to ensure continuous conduction of any torque greater than 25 percent of rated torque, R222. . (a) What should be the value of the firing angle to get the rated torque ‘at 1000 rp.m? Calculate the firing angle for the rated braking torque and —1500 r.p.m. Calculate the motor speed at the rated torque and o = 160° for the regenerative braking in the second quadrant. May-47, (R13), Q2 (b) (2) Given that, Rated motor voltage, V,= 220 V Rated motor speed, Njugq= 1500 rpm Rated motor current, /, = 11.6 ‘A.C source voltage, V, = 230 V Source frequency, f= 50 Hz ‘Armature resistance, R,= 2 Atrated speed, . ‘The back e.m.f of the motor is given by, Est I, Re =220- (11.6 *2) = 157.08 rad/sec w= 157.08 rad/sec wy spECTRUM ALLAN @) ONE JOURNAL FOR ENGINEERING STUDENTS 1.23 Exod = Bug = KOO E, = Kee 196.8 = KO" 75708 = K,o= 1.252 V-seolrad Motor voltage constaft, K,g= 1.252 Vsecirad. c1=? so that Rated Torque is Obtained Firing angle at N, = 1000 rp.m We have, E, By , Firing angle, cy = 41.84°oN ICONDUCTOR DRIVES LINTUHYDERABAy, a POWER SEMI (b) Firing angle, a= ? at rated Braking Torque and _ NE X230 0461 60° —(11.6 «2) N,=- 1500 rpm = Nn We have, Ey = 194.58 23.2 Fos 1778 | By, braking operation in the seo, rated * N _ 1968x(-1500) - 1500 =~ 196.8 V. N, => Ae = , =-196.8V 22, © = *cos a, By tik, (eB oho, “ERD 22v, 22M cosa, * 2 X230 cog © ~ 1968+ (11.6 « 2) ~ 173.6 = 207.07 cos a, 173.6 207.07 cosa, =~ 0.838 0, = cos" (~ 0.838) ar, = 143.92° cosa = = angle, a, = 14% (©) Speed, N,=? at rated torque and : = 160° for regenerative braking in second quadrant At rated torque, 1, = 116A We have, quadrant i. the forward regeneration can E obtained eth, by the field reversal oF the armature reversal ‘Thus, under this condition, K, 9 == 1.252 Veseclrad We have, Ey, = 90, E,,=K,90, -217.78 T1282, =o, 173.945 rad 2nN + = 173.945 [: - "60 251.659 x 60 eM On N, = 1661.05-p.n ‘A200 V, 1000 r.p.m, 10 A separately, excited D.C motor Is fed from a single-phase full converter with A.C source voltage of 230 V, 50 Hz. Armature circuit resistance is 1 ©. Armature current is continuous. Calculate firing angle for, (a) Rated motor torque at 500 r.p.m. (b) "Half the rated motor torque at (~500) p.m. ‘AprivMay-12, Sta! Qs Ans: Given that, ‘Separately excited D.C motor Rated voltage, V= 200 V Rated Speed, N= 1000 rpm Rated armature current, J, = 10 A AC source voltage, V, = 230 V Frequency, f= 50 Hz Armature circuit resistance, R, = 1 0 To determine, ° Firing angle, =? (PP WARNING: Xerox Photocopying of this bok is @ CRIMINAL act Anyone found f i: . ‘Quilty is UABLE to sroce face LEGAL proceedings.‘yen fo) Rated motor trae at 500 r.p.m trated operation, packemf, £,=V—1,R, =200-(10« 1) = 190 volts ated Motor Torque at S00 rp.m 0 We know that, = 95 volts Motor terminal voltage, V.=E+1,R, =95-+(10* 1) =95+10 - = 105 We also have, y,= 2m cosa % xt = cosa = 2xVm 105x% 2 NOS __ 2x (42x20) =0.53 = cos" (0.53) = 57.99 ©) Half the Rated Motor Torque at (500) rpm We know that, AM = Ms N, Bah F,= 2x19 (000 =-95 Volts SPECTRUM ALLAN Half the rated motor torque at (~$00) rpm, 1-4 ut contro! of D. C Motors by Single Phase and Three Phase Converters ‘Motor terminal voltage, B+ IR, 10 --95+(—*! 95+ ( 2 ) [v° Half the rated torque] =~90 Volts We also have, cs = 937, 90x at v= v2¥¥,) 2x(¥2x220) a 0.45 cos (- 0.45) = 116.74 4.3 SINGLE-PHASE SEMI AND FULLY CONTROLLED CONVERTERS CONNECTED © TO D.C SERIES MOTORS — CONTINUOUS CURRENT OPERATION — OUTPUT VOLTAGE AND CURRENT WAVEFORMS - SPEED AND TORQUE EXPRESSIONS - SPEED TORQUE ~ CHARACTERISTICS — PROBLEMS ON CONVERTER FED D.C MOTORS d explain speed-torque characteristics feeding a D.C series motor. Q39. Drawan of semi-converter Ans: Mode! Paperstl, 2(b) ‘The speed of a D.C series motor can be controlled by using set of thyristors involved in a converter. The converter ‘used can either be a semi-converter or full-converter. ‘The circuit arrangement ofa 1-p semi-converter feeding a D.C series motor is as shown below. Ry Supply u + D.C. series BY motor ries Motor Fed from 1-} Semi- Figure (al: Speed Control of a D.C convorter ONE JOURNAL FOR ENGINEERING STUDENTS1.26 Jn figure (a), the field is placed in series with the motor armature. The speed of the D.C series motor can be controlled with appropriate triggering of the thyristors. Basically, the speed of a D.C drive is controlled by varying its terminal voltage. Here in this case, the terminal voltage of D.C motor is varied with the help of thyristors. Depending upon the circuit parameters and operating conditions of the motor, the D.C serits motor operates under, (Continuous armature current (fi) Discontinuous armature current. Discontinuous armature current mode is considered in separately excited D.C motor due to the presence of large back emf but, in D.C series motor, the back emf (E,) decreases ‘with decreasing armature current. Hence, discontinuous current mode of operation of D.C series motor can be neglected as iscontimuous conduction takes place for very short period of time. The voltage and current waveforms of the continuous armature current of a D.C series motor fed from 1- semi- converter is shown in figure (b). During the postive half cycle ofthe supply the thyristor Tis wiggered ata firing angle a. T, and D, conducts and current ‘flows through field and armature of the motor upto the period (xa), T, is triggered in the very next cycle at an angle +a. Now. T, and D, conducts and current flows through 7, and D, into armature. Curent freewheels due to the presence of freewheeling diode (D,) between the period x < ot< (n+a), The motor terminal voltage, for continuous current operation is given by the equation, Val * cosa) Vo =RL+KILN+KN N= Speed K, = Motor constant K,,=e:m{ due to residual magnetism (V-Sirad) From equation (1), speed of motor, Torque developed is proportional to the square of the motor rn.s current. To lins T= Kiltas By assuming continuous and ripple free current the tonque-speed characteristics ofa D.C series motors shown in POWER SEMICONDUCTOR DRIVES [JI NTU-HYDERABAp, Figure (b: Current and Voltage Waveform for Continuous Conduction Figure (cl: Speed-torque Characteristics By reversing either the field winding or the armatat terminals, the speed of D.C series motor can be reversed. Due to change in load current there is an immediat change in speed, that is why the speed control is complex DC series motor. Hence, D.C separately excited motor is prefer™ over D.C series motor for controlling. ‘of speed. (240. Explainin detail the operation ofaful-conve™ feeding a D.C series motor with referen? voltage and current wave-forms, Assuming! the motor current is a continuous one. oes pars > fects? shown in figure (1). it consists of figure (c).; ono 4 / int il J Aa ot r Yeo, bt the thyistors 7, and 7. must be fired wine bal ele and sinilaly 7nd 7 inthe Sato 7 ROR ts WW 00 L Figure (1) ree ae basically two modes of operation based on coat curent BOW. They are, {) Continuous conduction ji) Discontinuous conduction. 4) Continuous Conduction Caxzimous conduction mode occurs when the value of cases large enough to supply the load. In this mode the {elements continuous and never becomes zero. The current | clterasfered from outgoing SCRs to the incoming SCRs. The voltage and current waveforms for continuous sabia are shown in figure (2). Figure (2) 1 g, Ding the positive half cycle at ot = 0, thyristors” og vad biased But as no triggering is applied initially, Sein OFF sae, At ot = a, the thyristors are triggered = So, the motor terminal voltage E, follows the supply suring the period a tox. At OF he biased whereas the thyristors 7, 7, Bets f The load voltage becomes 2er0 but the load current “tes due wo the effect of inductor and hence, the load “flows the negative part of the supply SPECTRUM ALLIN-ONE JO ).C Motors by Single Phase and Three Phase Co ng the negative half of the supply. For 16 Converters 1.27 Now, at or= (n+ a) the negative group of thyristors qr, and 7, ae simultaneously fired and hence they start conduction Due to the natural commutation process the positive group © thyristors 7, and T, gets turned OFF and stops ‘conduction. The current gts transferred from the outgoing SCRs, 7, 7, tothe incoming SCRs T,, 7, The load current now follows the path b-T,-R-L-M-T.,a. During the negative half cycle, thyristors T, and T, operate and the motor terminal voltage follows the supply: voltage from ( + a) to 2n. Derivation Let, R, ~ Armature resistance of D.C series motor J, ~ Armature current N ~ Average speed of D.C motor T — Torque developed in a D.C motor E, ~ Average motor terminal voltage E, ~ Supply voltage a Firing angle E, ~ Back em fof D.C motor 6, - Residual flux 4, — Flux produced due to armature current KK, K, Kay K,are Constants. The average motor terminal voltage is given by, E, sin(at) dot) = a (-coset)s** = 2 Leos(n +a) + cosa] E, = 72 cosa (1) The average armature voltage is given by, , E, “LR,+E, ~) Back e.m4, E, is given as, E, KNI,+tK,N=V-LR, > VrKNI+KNtLR, > V=I(KNtR)+K,N > 1, x(K.NtR)=V-KN V—K,N > 17 KN (6) Substituting equation (4) in equation (6), we get, Va +2080) KN RK ‘Substituting the above value of J, in equation (5), we ge, Yul + cosa) yy Roth aa Explain the use of free wheeling diode in the converter fed D.C drives. Take an example of 1-phase fully controlled converter for explanation. How it is going to affect the machine performance? Ans: Model Papers, cai) When a D.C drives fed from fully controlled convert, there is a need of connecting a freewheeling diode to avoid regenerative breaking, improving power factor and reductin in reactive power component and voltage ripple. . Consider a D.C series motor fed from a full convett with free whesling diode as shown in figure (a) Supply ko, D.C series motor Figure (a): Full Converter Feeding D.C Series Motor with Fret Wheeling Diode _ Let us assume that the motor is operating it? continuous current mode and there is no free wheeling di (D,) present in the circuit. When the thyristors 7, and 7," triggered simultaneously at a fring angle a, they conduct! the period n + (0 to m is positive half cyele), At an ins WARNING: Xerox/Photocopying of this book is a CRIMINAL act. Anyone found guilty is LIABLE to face LEGAL. proceedings.| y control of D.C Motors by Single Phase and Three Phase Convert ters , is T, and T, a i | sonen thyristors 7, and 7, are triggered, they st 4 "This makes T, and T, OFF and the current flows cttigg Tt the motor, 8 | ang positive half cycle ie, from 0 to x the power fom supply tothe motor and this power is positive. But ors oT od x to M+ ck (negative half cycle) some of the ai Hin inductor present near motor i fo ecko the at) results in regenerative braking of motor and also sal “he voltage ripple and reactive power component ie ot desired. To avoid these problems, a freewhecling wih coasted atthe terminals of «converter By the use seeing diode the voltage which is appearing inthe ride of the waveform can be ceased, The current se as ed back to the supply from motor now freewheels meg ike freewheeling diode. Hence, the continuous ‘operation of the motor is achieved. With the help of eeting diode, power factor is improved, voltage ripple Fpnased ad reactive power is reduced, The voltage waveform of full converter feeding D.C ceesmotor with and without freewheeling diode is shown in figure). v ing Diode \ : ot With FreewheelingDi 7 Fg i Waveorms____——_— | AD.C series motor has R, = 3.0, R, = 3 0 an M, = 0.15 H. The motor speed is varied by a phase-controlled bridge. The firing angle is Tidand the average speed of the motor Is 14 tpm. The applied A.C. voltage to the bridge © 330 sinut. Assuming continuous motor CUTTER find the steady state average mote” current and torque, Sketch the waveforms for outp Voltage, current and gating signals. jes motor is fed from 1-o Anas Letus assume that the D-C seri ‘enter. Given that, Armature resistance, R, = 312 SPECTRUM ‘ALLAN-ONE JO! = 1.29 Field resistance, R, My Firing angle, «= /4 = 45° Average speed, N= 1450 p.m Input voltage (A.C), V = 330sinor Comparing given input voltage with V~ V,, sin of ‘Assume continuous conduction mode. Required to determine, Steady state average motor current torque, 7? We get, V, = 330V Armaturevoliag, 7 = “24! * 2080 v,= 17931 Angular speed, InN. _, (Valt)(1 +0084). o* 0 Magla* Mres Where, " e RL=R,+R, =343=60 wA = 2x 0% 1450 awe = 9301n) (1+ cos45")— 44) ~ 0.15 Xa [Neglecting M,_] 315184222318). 5X Te 22.771,= 1793161, 1,(22.77+6) = 179.31 = 17938, = “9g97 = 623A ‘Average armature current, J, = 6.23 A _ Motor torque, 7 = M, 1? 15 « (6.23 82 N-m 0 ot Figure: Output Current Voltage Waveform of 1-4 Semi- ‘converter fed D.C Series Motor URNAL FOR ENGINEERING STUDENTSPOWER SEMICONDUCTOR DRIVES LINTUHYDERAg, RS CONNECTED vou 'D - TORQUE CHARACT; converters over single phase converters? erter. Converter ‘Three Phase Converter Single Phase “The average output Voltage is igh 1, | The average output voltage is low. Applicable for large power loads like in high 2, | Cannot be used for large power loads } power variable speed drives +. | Converter operation leads to the production Converter operation leads fo the production of ; of lower order ripples. higher order ripples {harmonics). Hor smoothing out the lower order ripples and 1p make load current and load voltage ripple free a’ large smoothening reactor is employed. Filtering of higher order ripples so as to smoothen | 4, ‘out the load current and load voltage is simple. Cost towards filtering the A.C ripples is lower, | 5. } Cost towards filtering A.C ripples is higher because of smoothing reactor. 45. With a neat circuit diagram explain the operation of a three-phase seml-controlled ‘converter div hy ‘sketch and explain the following waveforms, (a) Output voltage and output current at «= 90° (>) Output voltage and output current at a = 120°. OR Explain the: ft i " dies ree phase semi converter connected to a series excited D.C motor. Drawty Ans: . Three-phase Semicontrolled Converter Drive For answer refer Unit-1, Q15. Operation The 3-6 semi-converter fed to a D.C motor and terminals, 1 « the =| in i shown in figure (1). freewheeling diode (D) is in panlkliowe io fsa att he shorn inf across the output termi thyristors Positive groupand the diodes D, D andD, for the negative grup and ey ase beryl the com M 4 te v, koh 1, 7 5 vi] R I Dee — K May/.Junee13, (RO), Figure (1): 3-4 Semi Controlled Converter Fd to D.C Series Mot - tor WARNING: Xerox!Photocopying of this books & CRIMINAL act, Anyone found guilty is UaBLE vg ' MABLE ta face LEGAL procoading1 the sequence of numbers. If is triggered either one SCR diod ime, ve group and diode fre or one diode is triggered at a time, m negative group should be triggered simultaneous! angles c= 90° and «= 120° is shown in figure (2) in . form of voltage a d also shown. As shown, the diode D, conducts from 4, to eet if the thyristors 7,, 7, and T, acts as diodes and the output voltage conduct from ¢, to t,t, to t,and f,t01,. Therefore 1,1, and f, acts as crossing For a = 90° @) Figure 2(a) shows the conduction for @ = 99° pene x + = 120° i Pee for the period @7 = 120° to ay Joy yee tor, is triggered at F+a = +90" = 120° and conducts with the 3 . : 0 + 90 = 210° ice,, till 1, At of, , the output voltage V, is zero and from this time Vt be negative. As the thyristor, 7, and D, are connected to the phases A and C’so the voltage V,. appears across the load. Each periodic cycle is of 120° in which the output voltage is equal to line voltage for only 90° and for rest 30° the freewheeling diode (D) gets forward biased and conducts and at the instant the output voltage V, = 0 and the current flows through the diode until the next SCR in sequence is triggered. Ifthe freewheeling diode would not have been used, then when the output or load voltage becomes zero, at that time a negative group diode D, would start conducting through 7, from cf = 90° to 120° at wf = 90° where V,= Veg This means that in the absence of freewheeling diode T, conducts for 120° from wt = 30° to 150° and D, for 90° from wr = 30° to 120? for this cycle of 120°. It is to be noted that in 3- semiconductor, the thyristors are fired at an interval of 120° in aproper sequence. The fundamental component of source current i, is also shown in figure. With the increase in firing angle 0, the displacement angle 4, also increases. 1, 3 1 q tee a w ros +» | —_} vA . VA Vy Ve Va 4 4 4 4 eo 0 . > ot we t, t, & u 4 be ! i i ;* Oe > Pop Ps ‘ “KAY Vac. Paves | Pe PNY > ot ot (a) For a = 90° ua) ENR ENGINEERING STUDENTS ==| | | POWER SEMICONDUCTOR DRIVES QNTU-HYp,, 1.32 ret at inuous i, Cont 1 120° For a (© For a= 120°, i, Discontinuous Controlled D.C. Motor Figure (2k: Output Waveform for 3- Semiconverter For.a = 120° ) © and at 120° the motor current may be continuous or discontinuous. At The conduction of thyristors for a so oot fo output voltage V, is divided in three parts for each periodic ¢ i & 3 3 3 8 no com E. At this instant, back emf Eic., V, Seee8, baad}: see Rez eggeei Pigiis a gsse gabeee Soe88s gfege ageeie qazese Se3ees- eifhaa 3 ges 3 afi £ 8 K ‘The motor average terminal voltage is given by, (1 + cosa) 3 “7 % Ifthe motor current is continuous, The speed is given by, R, T 3h, = Fai (i +e0sa)~ WARNING: Xerox/Photocopying ofthis book is @ CRIMINAL act An J + Anyone found gui 'Y iS LIABLE to fare GAL procest Let, y - Terminal voltage of the motor ,-Back em. ofthe motor 1, -Armature current of the motor Armature resistance § Flux established o ~Angular speed K,Ky-Constants, fora D.C separately excited motor, Buckemf, E,=K,go =Ko [* $s constant] E,=Ko ofl) Torque, T=K, 61, =Kl, 4 is constant] T=K,1, «= Q) Teminal voltage, V= E, + 1,R, = G) From equation (2), we have, 1,=TIK, (4) Substituting equations (1) and (4) in equation (3), we es 2K; / The average input voltage of a series motor fed by 3h “Sonverter is given by, aac) v= sta +cosa) cc) av, Speed, = Be +0080) RT raajsec 7K, 7 From equation (3), we have, V-E, 6 SPECTRUM ALLIN-ONE JOURN |AL FOR ENGINEERING STUDENTS av art (1 + cosa)-K eo atl eos), ai L, Now, Substituting equation (7) in equation (2), We get, », mn (1 + cosc)—K °| av, K,ait(1 + cosa)-K20 Torque, T = Ava *eosa)-o Nem mo Q47. With a neat circuit diagram, explain the opera- tion of a three-phase fully controlled converter drive. Also.sketch and explain the following waveforms, (a) Output voltage and output current at «= 60° (b) Output voltage and output current at c = 90° (©) Output vottage and output current at a= 120°. Ans: Three-phase Fully Controlled Converter Drive ‘The 3-} full converter drives (or) 6-pulse converter drives gives satisfactory response when compared to single phase drives and 3- scmiconverter drives, as in 3-p full converter drives the output ripple is small, the ripple frequency is large and the filtering process required is reduced to larger extent and the armature current obtained is mostly continuous. So, 3-6 full converter drives are usually preffered for huge Power requirements and to reduce ripple content. It is a two- quadrant drive. Operation The 3- full converter drive fed to separately excited de motor and three-phase input supply given through terminals 4, Band Cis shown in figure (1), There are two groups of SCR’s, thyristors. 7,,7,,T, are positive group SCR’s and thyristors T,, T, and T, are negative group SCR’s. X, Bn Oy Figure (1): 3-9 Full Controlled Converter Drive Fed to Separately Excited D.C Motor VY1.34 7 ‘The motor ripple is 6 pulses per cycle. SCR’s are triggered according tothe sequence of gating of thyFistors, soma) Scheme is adopted. If only one SCR is triggered at time no current flows, sotwo SCR’s, one from positive BFOUp and ote negative group must be fired simultaneously, For 6-pulse converter drive, each SCR should fire twice in ts conduction oy and the firing interval of the thyristor is 60° and each SCR conducts for 120°. The conduction sequence of various thyting shown in figure (2). At a= 0°, the thyristor 7, is fired at wf = 30°, T, at of = 30 + 60 = 90°, 7, at wf = 90 + 60 = 150°, ang .5'"| Figure 2(b) shows the conduction for a = 60°, here 7, is triggered at a/ = 30 + 60 = 90°. 7, at of = 90 + 60> 150° ands) ‘When thyristor 7, is fired, the thyristor 7, is tuned OFF and 7, is tuned ON. ‘The thyristor 7, which was already conducting conducts with the thyristor 7, As 7, is connected to phase A and 7, to phase B. 0 the voltage 1, = ap APPERS 208 te ‘Now when 7, is triggered, then the thyristor 7, is tumed OFF and T, is turned ON, the thyristor 7, Sra ae ready Condy ‘with T, now conducts with 7, hence 7, 7, conduets. As 7, is connected to phase A and 7, t0 phase Les ca ~ Vet aeross the load. Now when 7 is triggered, 7, is tuned OFF and 7, is tured ON: the thyristor 7, whieh i ay cond now conducts with 7, hence 7,7, conducts and voltage 1, = V ppea across Ne joe Toa ie eachSCR eT negative group arestriggered at an interval of 120°, but from both the groups SCR's are fired at it, ee conduc jg | 120° and commutation occurs for every 60° and the SCR’s conducts in pairs, one from positive and one from negative i.e, Tr) T, T,and soon. ‘The converter operates in two modes, ( _ Rectifying (or) Motoring mode (ii) Inversion (or) Braking mode. | ‘When the power is delivered from source to load the motor terminal voltage and current is positive and this is cay rectifying (or) motoring mode. ‘When the polarity of motor terminal voltage or current reverses, then the power is delivered from load to source ani is called inversion (or) braking mode, ot ot (©) Fora = 60° "WARNING: Xerox Photocopying ofthis book isa CRIMINAL act. Anyone fou uty is LIABLE to proceedings. IMINAL act, found guilty is ity is Ui face LEGAL irrT-1 Control of D.C Motors by Single Phase and Three Phase Converters 1 y Single Pha PI ve vA NN 1.38 PARRA AAA TYP} INA l TrTryry > ot h ()Fora=90° ~ 0 . c A i INARI AON OIA ONO OY INO ; . at A ‘ i >ot (d) For a = 120° Figure (2): Output Waveforms for 3-4 Flly Controle Converter for D.C Metor Figure (2) shows the output waveform for output voltage ‘E,” and-output current ‘i,’ at a. = 60°, 90° and 120°, ® Fora=60° Figure 2(b) shows the conduction for a= 60°, 7 is triggered at of = 30+ 60 = 907, T, at dr = 90 + 60= 150°, T,ator triggered, 7 s reversed based and ur OFF. 7 is tune ONand T, which was already ferred to-T, and T, T, conducts. Since phase A =150+60=210° and so on. When 7, is ducing with 7, ow conducts Tian 7, and hence the curent from 7, i tans | sounected to T, and phase B is connected to T, So voltage V4 appears across the load. Now 7, is triggered and 7, turns OFF cr ono bese fn ree rn etm ines wih 7, an henee 7 77, conducts ore phase A and Care connected tT, 7, S0 voltage Y, appears across the load. Way the SCR’s turn-ON and turn OFF in sequence both from positive and negative groups and conducts in pairs ie. 7, TT, 7,7 and so on represent source CuTEn! mee eis positive when 7, conducts and negative when 7, conducts, The m current i, is almost continuous: ®) Fora =s0° Figure (c) shows the waveform for a = 90°, here the load voltage V, is symmetrical about its axis, i. itconducts equally Positive and negative grouP, and hence the average value-is zero. —_— /AL FOR ENGINEERING STUDENTS ee SPECTRUM ALLIN-ONE JOURN.Ad POWER SEMICONDUCTOR DRIVES (INTU-HYDER,, ©) Fora=120° Aa 20°, T, is triggered at ot = 1/3 + 120°, 7, ator = 180" + n/3 = 240° and so on. Inthe OUPUL Waveform gy, figure 2(€, the motor terminal voltage is negative i, its polarity is reversed. This i called as inversion mode ofthe ogy) ‘tmeans the power is delivered from load to source or from motor to source by reversing the field current. The converte gpg" as 344 full converter drive for a= 0° to 90° and from a= 90° to 180° it operates as an inverter and the current direction foe ‘converter and inverter does not change, only the output voltage polarity is reversed. ‘ ‘The output average voltage of motor is given by, Vo==t Yacosal ‘The average speed is given by, Since torque, T= Qa8. Draw the circuit diagram and explain the operation of a three-phase full converter drive: Also ska, and explain the output voltage and output current waveforms at firing angles of 30° and 140°, Ska the speed-torque characteristics of the drive, Ans: For answer refer Unit-l, Q47, Topics: Three-phase Fully Controlled Converter Drive, Operation, Excluding figue 2, The output waveform ie. output voltage and output current of 3- fully controlled converter drive fed to separately ext D.C. motor is shown in figure (1) for a = 30° and a = 140°, E 1, 1 T, Tt 1, Tt “T, (a) Fora =30° 77 WARNING: XeroPhotocopying ofthis book s a CRIMINAL ect. Anyone found guty is IABLE to face LEGAL prceeds-—— ot uh am) at (b) For a = 140° Figure (1}: Output Waveform of 3-4 Fully Controlled Converter for Separately Excited D.C Motor Fera=30° Figure I(a) shows the conduction of thyristor for a= 30°. When the thyristor 7, is triggered, T, immediately reverse biases adtums OFF. 7, is tumed ON and the transfer of current from outgoing SCR T,, to incoming SCR 7, occurs ifthe line voltage tussuch polarity that it should not only reverse bias outgoing SCR 7, but also it should forward bias the incoming SCR T,. T,, ‘ich was already conducting with T,, now conducts with’T, and hence 7, T, conducts. As 7, and 7, are connected to the phases ‘and B respectively so load voltage E, = V,, appears across load: The firing angle ‘a for 7, is measured from the instant Vy = (tillit reaches maximum or increases. When the thytistor 7, is triggered, 7, turns OFF immediately and 7, is turned ON. The thrstor 7, which was already conducting with T, now conducts with 7, and hence 7, T, conducts. The thyristors 7, and 7, are cunnected to phases A and C respectively so voltage E, = V,- appears across the load. As from waveform, itcan beseen that SCR ‘onducts for 120° and commutation occurs for every 60° and SCRs conduct in pairs, one from positive and one from negative it,1,1, 1, Ty T, T, and $0 on. The current waveform is represented by ‘i,’ Fora = 149° ‘eform a = 140°, The voltage becomes negative as the power is delivered from D.C merge roa isrnce the converter operates in its inversion or) braking mode, This causes the motor voltage hence the motor slows down, Thus the triggering angle ‘ais to be adjusted. he Chara drive. As it is a two quadrant dive, so it operates i characteristics of 3+ full converter drive. As itis a two q ve, mee (3) shows the speedtora coe sor discontinuous. The boundary between continuous and discontinuous is | and IV. Conduction may be an onduetion occurs ifthe torque is below the rated value. The speed-torque ented by a dated line, The discontinue oar on, quadrant the current and voltage polarities are same ic pig tiie are parallel straight lines for conte converter here opeatesin motoring mode ic. a8 3+ fully controled Ve $0 is positive and fring nel sure oad (to) Butin quant 1V th polarity ofthe yolags ene oq Motor and the power is deliv fi ed fore > 90°, and bence the machine here operates as generator and prods “vs lrg gay Ted but current J, remains Oy or rivered from D.C motor1o A.C supply and, is negative kant thus known as braking mode: The PONE. nye. The dsconiuous made her is not wen in account tions as inv’ ‘gets inversion operation of| cate won ‘Therefore te speed is given 8 a very narrow region during = Te oy t| Where T= Ki, ECTRUM ALL IW-ONE JOURNAL FOR ENGINEERING STUDENTS sP' l- baa POWER SEMICONDUCTOR DRIVES LINTUHYD ER, Boundary between continuous and discontinuous conduction @ increasing Speed ‘a (radisec) Torque (Nm) —> Discontinuous conduction Figure (2: Spoed-torque Characteristics of 3-4 Fully Controlled Converter Fed D.C Separately Excited Motor Q49. Describe how the speed of a separately excited D.C motor is controlled through the use of two 3-phas. full converters. Discuss how two quadrant drive can be obtained from the scheme. Derive expressiow. for r.m.s values of source and thyristor currents. 3 Ans: ‘Model Papers, ‘Three-phase Fully Controlled Converter Fed to Separately Excited D.C Motor ‘The speed of a separately excited D.C motor is controlled through the use of two 3-phase full converters, one to coer the terminal voltage and the other to control the field flux by varying the field terminal voltage. For remaining answer refer Unit-1, Q47, Topic: Operation. Bridge 1 Bridge 2 . i, Fi + eee) | eS wy fe | | \s Z Veo —T ——T.— + af ep oa | Ke An Kum | * Sue ue a dn \ a | I= We know that, Thus, co ~O ‘The r:m.s value ofthe thyristor current is given by, Wa cos Me 08a R, The ram.s value of source current is given by, . Explain the speed-torque characteristics of a Separately excited D.C motor connected to a ‘semi-controlled converter. RR Fei Figure (1): Three-phase Semi-controlled Converter Fed Separately Excited D.C Motor1.40 POWER SEMICONDUCTOR DRIVES !JNTU-HYDERAt n (3), ‘A 3 semi-controlled converter drive is a single quadrant drive which operates in J" quadrant, In I* quadrant, the voltage and current are positive. With these type of drives, the regeneration is not possible and these drives are employed fF the motors in the range of 15-150 HP. Figure (1), represents a three-phase semi-controlled converter fed separately excited D.C motor, which can be operated in two modes. One is continuous mode and another is discontinuous ‘mode. Incoptinuous mode, the load currents continuous, which is because of inductor in the circuit ic, the inductor is large enough to supply load current til the next thyristor is ied. ‘three-phase semi-controlled converter fed separately ‘excited D.C motor consists of three thyristors 7,, 7, and 7, ‘which forms positive group and three diodes D,, D, and D, ‘which forms negative group anda freewheeling diode Each SCR will conduct fora period of 120° Dring I* postive ha eel, the thyristor 7, andthe diode D. are forward biased andthe remaining pars are reverse thane When the thyistor 7, istrggered, the SCR 7, and diode [Fee 55 and hence, the voltage Eye APPEAS AES the load, At 5E ine diode D, ges reverse bias and at same time , will be in conduction state for a period of NK + NK, = Ua tesa) |p (1+ > NK L+I,R, Sat Fcose) —NK, (+ S LuvK,+R) = Hal #00s0) _ yy, Yall *c080) _ yy, rm s 1,= NK Re i The torque current relationship for a series motor is &iven by, Torque, T= K, 12 ”, tx, Stbsttuting the above value of J, in equation (6), we 3Vq(1 + cosa) - NK Torgue,7= x ee on(8) u Charge Si8 Equation (8), the nature of the speed-oraue tics of three phase semi-controlled converter fo cs y i X-axis is “ries motor with speed on Y-axis and torque on “rnin figure SPECTRUM ALLIN-ONE JOURNAL FOR ENGINEERING STUDENTS ’ hree Phase Converters N (ep.m) ; —> T(t) Figure: Speed Torque Characteristics Q52.. Explain the speed-torque characteristics of a D.C series motor connected to a three-phase fully controlled converter. ‘Ans: = » ___ Athree-phase fully controlled converter is also known as 6-pulse converter, The output consists of 6-pulses per cycle. The fully controlled drives operate in fwo-quadrants i.e, in T* quadrant and IV* quadrant. In general’ when 3-6 drives are compared with single phasé drives they does not require any filtering components. By using 3-9 drives in motor good performance can be achieved, which is because of continuous current in the armature. A @+R) vy : u T, x cn 5 . oR “Figure (1): 3+) Fully Controlled Converter Fed D.C Series Motor A344 fully controlled converter fed D.C series motor is shown in figure (1), which consists of six ‘thyristors, two on each limb. The thyristors 7,.7,, 7, form a positive group and the thyristors 7,, 7,, 7, form a negative group. At every instant, two thyristors, one from positive group and another from negative group will onduct. Each thyristor conducts for 120° and the thyristors from the alternate ‘group is triggered at an interval of 60°. Now, the thyristor 7, is triggered1.42 POWER SEMICONDUCTOR DRIVES (INTU-HYDERAB A, the voltage V,, is as triggered, the thyristor 7, is already in conduction stat Pome (eatalicneae at the output terminals, during the interval (n/6 + a) to (n/2 + a), When the thyristor 7, is trig rn at an angle of 30°, Befor £ B + a). the thyristors 7, and T, are in Conduct thyristor 7, is reverse biased and turns off. During the interval | $-+ a. |to ‘ the thyristor 7, is red at (# et a] even, mode. The voltage appeared at the output will be V,. When the thyristor 7, is trigee! events tn pili, ¢ appeared biased and tum’s off. Now, the thyristors 7,, 7, will be in para and the voltags iased and tum’s off. Now, "y : ip SE, id current wave for 9. LE respectively. The voltage an 7 {s continued by triggering remaining thyristors a 2, shown in figure (2). t 60 120 180 240 300-360 Figure (2h: Voltage and Curent Waveforms fra 3 Full Control Converter Fed. Series Motor Speed-Torque Characteristics . Let, $,— Flux due to armature current 4,.,~ Flux due to residual magnetism J, Armature current in amps E,- Back e.m.f in volts, Then the total flux is given by, $6.44, WARNING: Xerox/Photocopying of this book is a CRIMINAL act. Anyone found Auilty is LIABLE to face LEGAL proceedings:ynit-1 Conte’ OES Motors by Single Phase and Thre But, in series motor the armature flux is direct e Phase Converters 1.43 lirect ie “ . 2K, 'Y Proportional to armature current, 54,01, ‘Therefore, equation (1) becomes, Ook 144, Now, the back eam.Fof the motor is given by »Q) E,= Kn Substituting 6 value from equation (2) in (3), we get, ae ELK (Kd, +4,,) N BE, KKIN+K,6,,N B= (KG*K, 6) Also, a The back exm.fis given by, E,=V-1,R, By Comparing equation (4) and equation (5), we have, . V-I,R,=(Kyl,+K, oN 2 N= 6 Now, the terminal voltage is given by, 00 = ved ff rasinada woe ~ = He (cosa) SSE = r= =H [c0s(150 + a) - c0s(30" +a) > = (605 150° cos sin 150 sina) ~ 6830 cos a- sin 30 sna = y= =3¥fe05(90-+ 60) cosa sn(90+ 60) sin a~ 608 30" cos + sin 30" sina) ™ = 3a [sin 60" cosa ~c05 60 sin a cos 30°08 2+ sin 30sin a] ™ an 3 cosat bei * he [ Saenapabe-Foes be] : yor a yee R =3%m | -9. 8 cos. cos [22 a2 3 ya8le «fF cosa 5A Bu, ,=/2 ¥, a Vm O° Therefore, . oT fi : = SYS x 1B BROS = 35x) A) y= 316% xcosa DENTS JOURNAL FOR ENGINEERING STU SPECTRUM ALL:N-ONE J ez1.44 Substitute equation (7) in equation (6), we get, v=o Kyla Ree > > = a (9) But, ina series motor the torque is directly proportional to the square of the to current, i. redl, But, ¢«/, = Tel,.1, > Tel? = Te. (10) ‘Substituting /, equation (9) in (10), we get, | on (Il) Kop N¥Re From equation (11), itis clear that, when K,,, << K, the torque and speed in a D.C series motor are inversely proportional to cach other. The speed-torque characteristics of D.C series motor is shown in figure (3). N aso a= 60" a= 90° 0 T Figure (3): Speed-Torque Characteristics of 3. “Fully Ce Converter Fed D.C Series Motor td or POWER SEMICONDUCTOR DRIVES [JNTU:HYDERABAp, 1.5 PROBLEMS ge comprising thre from a 277 £.m-S, line to neutr, an provides an adjustable pc Hy Pine terminals of a separately oxciteg OOo: ‘The motor has R,= .02, L, =0.004 : oc i tand fulltoad |, = 500 A. Find the tin Korig'so that the motor Operates 2 fullogy Gurrent and at rated speed of 200r.p.8. Assum, curimuous conduction and neglect thyrst, forward voltage drop. if wave brid Q53. A 3-$ hall thyristors 60 Hz supply and Ans: Given that, RMS input voltage, Vq.~277V Frequency, f= 60 Hz *anmature resistance, R,= 0.02 2 ‘Armature inductance, L, = 0.001 H Constant, K,= 1.2 Full-load current, J, = 500 A Rated speed, N = 200 rps Under continuous conduction and neglected thyristor {forward voltage drop, the firing angle at full-load current ad atrated speed, a=? _ Maximum voltage, Ve V2 Ving =¥2 *277 =391.74V Back e.m.f, E,=K,N =12*200 =240V The terminal voltage of a D.C it wiry .C separately excited motor VaE+IR, = 240 +500 x 0.02 =240+10 =250V Phase half wave rectifier circuit, avers ven by, For a three- Output voltage is gi Vv = 3h, : = F(t cosa) Substituting the values of V., V.= Vin the abo quation, we get, don WARNING: XerouPhotocopying ofthis book ea CRIMINAL act A 250= 3391.7 srt (1+ cosa) ‘nyone found guilty is }yapy cynit-1 contol of D.C Motors by Single Phase and Three Phase Converters 1.45 = 2nX250 > LH c0se = C307 a y= 47833-2999 > IF cosat = 1.3366 448.34 2 cos o = 1.3366 ~1 eee \ N= 298.89 radi/set = a = cos"'(0.3366) 298.89 x 60 2 a= 70.33° Ne’ In pm ‘he fring angle, c De 55, A80 KW, 440 V, 800 r.p.m D.C motor is operating ——, speed of a separately excited D.C motor The speed of a Separately excited D.C motor is controlled by means of a 3-phase semi converter from a 3-phase 415 V, 50 Hz supply. The motor constants are inductance 10 mH, resistance 0.9 ohm and armature constant 1.5 Viradis. Calculate speed of the motor ata torque of 50 Nem when the converter is fired at 45°. Neglect losses in the converter. Frequency, f= 50 Hz Inductance, L, = 10 mH Resistance, R, = 0.9 ‘Armature constant, K, = 1.5 Virad/s Torque, T= 50.N-m Converter delay angle, a = 45° Maximum value, V,= V2 *V = /2 «415 - = 586.89 V From the torque and current relation, we have, T=K,I, T PLE 50 115 1,=33.33A Now, the speed of semi-control converter fed by asepa- ‘ately excited D.C motor is given by, A ‘SPECTRUM ALL-IN-ONE Jol URNAL FOR ENGINEERING STUDENTS “ere at 600 r.p.m and developing 75% rated torque is controlled by three-phase six pulse thyris- tor converter. If the back e.m-f, at rated speed is 410 V, determine the triggering angle of the converter. The input to the converter is three- phase, 415 V, 50 Hz A.C supply. Given that, Power, P= 80 kW * Speed, N= 800 rp.m N,= 600 rp.m Torque = 75% rated torque = 75% J, R, Back exm.f Bp, at rated speed N, = 410 V To determine, ‘Triggering angle ‘a’ for converter input, AC supply, = 415 V Frequency, f= 50 Hz. From the relation, Fy LM Boi M Back é:mif, Be, = ERP > = 410x $0 E_=3075V We know that, Back em, E,=V-1,R, => 410=440-1,8, = 1R,= 440-410 LR,=30V But, P=V1, 2 > 1, = 181824 1R,=30VPOWER SEMICONDUCTOR DRIVES [JNTU-HYDERABar 30 SORE TLR The terminal voltage of D.C motor at 800 p.m and 75% rated torque E,= Ey, + (I,R,) 75% E,= 3075+ ad 181,82 x 0.165 307.5 + (0.75 x 181,82 ¥ 0.165) = 3075 +2250 Peak voltage, Wan Where, V,, = RMS value of A.C voltage reais ¥,= 338.84 E> 38 (338.84) cos a => 330= 1.654% 338.84 cos a = 330 =560.44 cos a = cosa =0.589 = a=cos'(0.589) sn era Q56. A200 V, 1500 r.p.m, 50A separately excited mo- torwith armature resistance of 0.59, is fed from three-phase fully controlled rectifier. Available ‘ACC source has a line voltage of 440 V, 50 Hz. Astar-delta connected transformer is used to feed the armature so that motor terminal volt- ‘age equals rated voltage when converter firing angle Is zero. f : (a) Calculate transformer tums ratio. {b) Determine the value of firing angle when, ()) Motor Is running at 1200 r.p.m and rated torque (i) Motor Is running at - 800 r.p.m and twice the rated torque. Given that, ‘Average terminal voltage, V,= 200 V Speed, N= 1500 rp.m ‘Asmature current, J,= 50A WARNING: XwPteoprig iso's CRIMIUAL ae Ayo oud pity is ABLE o ac LEGAL proces @ @) “Armature resistance, Line voltage, Frequency, f Converter firing angle, = 0° To calculate, (a) Transformer turns rato? (b) Firing angle, (At 1200 opm and rated torque (i) At-B00 pam and swice the rated torque Back emf, £=¥,- 1,8, = 200-50 0.5 =175V se fully controlled rectifier, For three-pha: v,=2V cosa Vou RMS voltage, Fins f° 209.43 7 =148V Vas" 2 * ‘Transformer Turns Ratio Phase voltage, 440 Vie = BB = 254.03 V For star-delta transformer connection, Yn Vem ‘Tums ratio= Firing Angle ‘a? () At 1200 rp.m and Rated ‘Torque For rated motors, E=e-h = 200-50 0.5 E=115¥- At 1200 rpm, Back em.f, B= 2g «175 E=140Vcos & = 1.047 x 0.788 cos a = 0.8250 = cos-"(0.8250) @=3441] (i At-800 p.m and Twice the Rated Torque ‘At-800 rpm. E = 7800 x 175 =~0.533 x 175 E=-9333V Terminal voltage, V, =E +1, R, (il) No-load speed, ifthe firing angles are same as In part (i) and the armature current at No-load is 10% of the rated value. (ili) Speed regulation. Ans: Given that, Power, P= 25 H.P=25 x 746 W (co LHP = 746 W) = 18,650 W Voltage, Y= 380 V Speed, N= 1800 rp.m AC. input voltage, V, Supply frequency, f° Armature resistance, R, =210V 0 Hz =0.150, Field resistance, R,= 250 0 Motor voltage constant, K, To determine, .2.ViA-rad/seo () \ Firing angle ‘a at rated power and rated speed (i) No-load speed ‘0, at 10% rated value of J, and ‘a’ as V, =—93.33 + 2(50) «0.5 (Twice the | obtained in (i) rated torque s0'2 (1) R} | (il) Speed regulation. \ % V, =~ 93.33 +100 x05 Phase voltage, Vp, = ~93.33 +50 . Maximum voltage,Y, = V,,x /2 ve =17145V @ Firing Angle ‘a’ at Rated Power and Rated Speed Rated armature current, * aa Fed E > cosa =~ 0.2167 pe > a = cos*(- 0.2167) lag 20 = 102.51" The field controlled voltage equation is, ENS O87. The speed of a 25 H.P, 380 V, 1800 r.p.m, p= Wile oo Separately excited D.C motor is controlled ; ye cone, by a 3-phase fully-controlled converter. The Since, field current is maximum, a, = 0° field circuit is also controlled by 3-phase fully aficinuas controlled converter.and the field current is ARITA, Set to the maximum possible value. The es z inputis a 3-phase star-connected 210 V, 50 Hz ‘Supply. The armature and field parameters are Fiidccmen, i= R= 0150, R, = 250 Q and K, = 1.2 Minete: I L, Assume that the armature and field curren are continuous and ripple free. By neglecting the system losses determine the following, 7 it ()) Firing angle of the armature converter, Armature currer the motor supplies the rated power at the Back ems, =, 0 rated speed. . SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS PRAYane ona H 1800 60 =2xnx = 188.4 E,=12% 1.134% 1884 =256.4V Also. E,=E,-1,R, Terminal voltage, £,= E, +, R, = 256.4 + (49,07 x 0.15) =256.4 + 7.36 E, = 263.76V But, E= Y3Me cos & = 25376 = SEIT. cosa > 263.76 = 283.58 cosa — 263.76 = 0S = 283.58 > cos a 93 a =cos"'(0.93) (ii) No-load Speed ‘a, Given, [,= 10% of rated value - io, 4.= Jog * 49-07 1,=4.907A Back em£, E,=E,-I,R, = 263.76 — (4.907 x 0.15) = 263.76 — 0.736 63.02 V The no-load speed is given as, E, 0,= 7 Kr 263.02 SF %* T2x1 134 26302 - > = T56 => — 0,= 193.39 radls = 0,=19339x-2 ‘ => 0,= 193.39 «9.54 . = 1844.94 = 1,845.00 [too = 1845 r-p.m _ WARNING: Xerox/Photocopying of this book is a’ CRIMINAL act, PO} WER SEI INTU-HYDI Whew MICONDUCTOR DRIVES [J! ER ABA, q ii) Speed Regulation Speed regulation , No load speedFull load speed ° Full load speed = @o-N = oa 1845-1800, 1800 = 0.025 io speed regul 5%) G58. The speed of a25H.P, 320, 960 r.p.m separ excited D.C motoris controlled by a three pha, fully controlled converter. The field current , also controlled by a three-phase fully control, * converter and is set to the maximum possity, value. The A.C. input is a three-phase sta, connected 210 V, 50 Hz supply. The armatun, resistance is R, = 0.2 9, the field resistance, = 130.0, and the motor voltage constant isk ='1.2ViAradis. The armature and field curren, are continuous and ripple tree. Determine, (i) The firing angle of the armature converte if the free converter is operated at thy maximum field current and the develope torque is On-mat at 960 r.p.m. (ii) The speed of the motor if the field circut converter is set for the maximum fie current, the developed torque is 111 N-m and the firing angle of the armatue converter is 0°. (ii) The firing angle of the field converter iftte speed has to increased to 1750 r.p.m, fr the same load requirement as in part (i Neglect the system losses. Ans: Given that, Power, P=25 HP ‘Voltage, = 320 V Speed, N= 960 rpm ‘A.C input voltage,V,, = 210 V Supply frequency, f= 50 Hz Armature resistance,R, = 0.2. Q Field resistance, R/= 1300 Motor constant, K, = 1,2 V/A rad/s To determine, (Firing angle. ‘a’ wien the converter operates! © maximum field current and developed torav® M0N-m (i) Speed +N’ when the converter operates maximum, field current, the developed tors" MON-m and firing angle a = 0°. ‘Anyone found guity is LIABLE to face LEGAL proceedings.0 peak voltage, V2% Vp, = 2 121.24 =171.46V v= Firing Angle Forfield controlled converter, 3 5,= 28h cosy, As given that field current is set to maximum, a, y= © £,= 3X13 2171.46 cos 0° 9 3a =283.74V ‘We have, E, Field curent, J, = 39 = 283.74 | 130 L=218A The relation between torque and current i, T=1K I, Back em E,=K, 1,0 Where = 2n0N, o= "00 = 100.53 E,= 1.2 * 2.18 x 100.53 E,= 262.99 V. Motor terminal voltage, E,=E,+1,R, E, = 262.99 + (42.04 0.2) E,=214v Also, 2n 6H * 960 wy, di) 3V3 x 171.46 ose 0 271.4 = 283.74 cosa 271.4 283.74 cosa = 0,9565 cosa = cos"'(0,9565) [-. Firing angie, a= 16.96" Speed (N) Field current is maximum, i.e., B = 0° Converter firing angle, «= 0° Torque, T= 110 N-m . For a 3-6 fully controlled converter, B= BX cose E= 3Y3x10146 co: 80 E,=283.2V Back é.m.f, , BMEALR, \ = 283:72 (42.04 « 0.2) hy E,=21531'V ' Speed, = 105.24 rad/see * = 0 = 105.24 x $0 . Speed, N= 1004.9rp.m| Firing Angle (c,) for Speed = 1750 p.m 2n @= &X 1750 © = 183.259 rad/s Backem§ £,= Ko = 27531512« J, x 183.259 (> from part (ii)} > 27531=21991 1, = 27531 > 4 2991 > L=125A SPECTRUM ALL-IN-ONE JOURNAL FOR ENGINEERING STUDENTS |) He1.50 Q59. Aseparately exci POWER SEMICONDUCTOR DRIVES [. We have, E r= t Ry E,= 1.25% 130 E, = 162.5V 33 xX hy Also 5,-35khe cy => 162.5 = 283.59 cosa, 3¥3-x 171.46 SEA cos a, > 1625= — 1625 > 005 Oy = 283.59 => cos a, = 0.573 => ' a,=00s"(0.573) ited D.C. motor rated at 10 kW, is supplied with power from a fully controlled, three-phase bridge rectifier The ideal three-phase power supply Is rated at 220 V, 50 Hz. This motor has an armature resistance R, = 0.2 0 and sufficient added in- ductance to maintain continuous: conduction. The motor constant is 1.38 volt seconds/radian, and it delivers rated power at a )°. If the SCR firing-angle is retorted to a = 30°, calculate the speed, power factor and efficiency of operation if the load torque is constant. 300 V, 1000 F.p.m Ans: Given that, Power, P= 10 kW= 10x 10 W Voltage, V= 300 V Speed, N= 1000 rp.m 3-4 A.C supply voltage = 220 V Supply frequency, f= 50 Hz “Anmatute resistance, R, = 0.22 Motor constant, K, = 1.38 volts-sec/radian Rated power is delivered at a= 0° Load torque, 7, = constant To determine, (i) Speed Wat a= 30° (ii) Power factor PF ata Efficiency ‘1 Phase voltage, Vp, = ” =127 Maximum voltage, ¥,= Ym * ¥2=179.6 oO the co) : JNTUHYDERAR, : Speed ‘N’ at a= 30° For a 3+ fully controlled converter, = 33 V,, 008 & Ata =30, y ~ 3¥3. « 179.6 c08 30° ame v,=2573V At a =0° 34 «17961 Es x ¥, =297.1V load torque is constant, so with constant Since the I ortional to average armature voli, no-load speed is prop’ ie, N&!, «2 Speed at a= 30° is, Nese Taa=30" a= Vaa= yy > Newser = Va,a=0" 2573 = do71 * 1000 = 866.03 Power Factor at ‘Assuming zero speed regulation The power developed by motor with constant load ton =m Load torque, T = S26 601010? ‘2m 100 = 95,5 N-m P.,, = 95.5 * 866 x 2E = 8660.63 P.,, = 8661 W TN=E,I, => 8661=E > B= Ass. alll Wehave, V, =F, +12, VIR, ( => £,=2573-021 of > &, ain, voc IPhatocopying of this book is a ‘CRIMINAL act. Anyone found guilty is LIABLE e0dings. to face [EGAL proce aequations (I) and (2) we ge ‘py oom 3661 = 257.3 - ca =257.31,-0.2 1 2 8661 257.31, 0 20, 4 02 2 - 257.31, + 8661 =0 asolving above equation, we get, 2 1252 A or, 346 A (Neglecting higher value) pking, 346A B= Me £,=250.3,V assuming that armature current, J, inductance Z, is unknown) c ‘Armature copper loss = 13 Ra = 4.67 «0.2 =239.4W ‘The input power is, Pi=Pygt Ra = 8661 + 239.4 P= 8900.4 W The supply current is, - VF Ascurrent ripple is neglected, so J, =I, x1, x 34.6 [= 28.2A) Input power, P, = 3 V,,1,c08 > cos $= a7 a 8900.4 = JX179.6 X346 $0.48 Power factor, cos $= 0.48. © SPECTRUM ALLIN-ONE JOURNAL FOR ENGINEERING STUDENTS ™ py three-phas full converter. The ications of the con- 0 V, 300 A. The input to the con- verter ia athe "415V, 50 Hz A.C supply: () Firing angle of the converter and power factor at rated speed (ll) Firing angle and power factor at 10% rated speed (il Active and reactive power drawn from the system at rated speed (lv) Active and reactive power drawn at 10% rated speed (v)_ Ratio of reactive power drawn at 10% and rated speed. Neglect the system losses and effect of commutation angle. Ans: @ uuu UYU Given that, Power, P='100 kW = 100 x 10° W Speed, N= 1000 rp.m ‘Armature terminal voltage, E, = 460 V Armature current, J, = 300 A 3-6 A.C source voltage, V, = 415 V (Line value) To determine, (Firing angle and power factor at rated speed (ii) Firing angle and power factor10% rated speed (ii) Active and reactive power at rated speed (iv) Active and reactive power at 10% rated speed (v) Ratio of reactive power at 10% and rated speed. Firing Angle and Power Factor Phase voltage, V,, = ‘Maximum voltage, V, V2 xV,,= V2 x239.6 =338.84V For 3-, fully controlled converter, the voltage £,= 393 «338840080 460 = 560.44 cos. a cosa = — a =cos'(0.082) Power factor, pf = cos 0 = cos (85.3°) = 0.082) ‘Active and Reactive Power at Rated Speed. Active power = #3 V,I/cos a Now, 1=J, 1816, Wehave, = > t 1,=21739A 1=0.816 * 217.39 1=17739A At rated speed cos a = 0.82 ‘+. Active power = 73 VJcosa. = YB x 415 x 177.39 « 0.82 = 104556,6 104.5 kW} ower | POWER SEMICONDUCTOR DRIVES ~™" 0) P Reactive powe! sina. = 1-082)? =0572 Reactive power = {3 * AIS * 177.39 «0.579 ve powel ‘Active and Reactive Po" wer at 10% Rated Speey wy) ‘active power= v3 *F,* Feosa At 10% rated speed cos a = 0.082 ‘Active power= 13 * 415% 177.39 x 0.08) = 10,455.6 W Reactive power = V3 * 7, x/sina sina = Vi-costa = Vi-(0.082)? = 0.9966 Reactive power = 73. 415 x 177.39 x 0.9966 = 127,074.5969 VAR ‘Reactive power = 127.07 KVR) (v) Ratio of Reactive Power at 10% and Rated Speed Reactive power at 10% rated speed _ 127.07 "Reactive power at rated speed 72.9 =14 EE Q61. The speed of a 50 KW, 500 V, 120A, 1500 rp | a separately excited D.C. motor is controlledby a three phase fully controlled converter andis fed from a 400 V, 50 Hz supply. Motor armatut resistance is 0.1 0. Findthe range of firingandé required to obtain speeds between 1000rp.naié (- 1000) r.p.m at rated torque. MaylJune-t3, (R09, 24 Given that, 3-4 fully controlled converter fed sepam®™ excited D.C motor, Speed, N, = 1500 rp.m Power rating, P=SOkW * Terminal voltage, 7 = 500 V Armatuie current, J, = 120A Armature resistance, R, = 0.1 2 ‘AC supply voltage to converter, V,= goove® Frequency, f= 50 Hz GD warmine: rrr ik RINAL per fn ‘ Guilty fs LUABLE'to fata LEGAL proceed proceedings:control of D.C Motors by Sin, gle Phase an rT ww Fofind, Three Phase Converters 1.53 range of firing angles for speeds aN, ‘=~ 1000 rp.m Peds between W, = 1999 ie when speed, N, = 1500 rpm packesm.Fof the motor is given by, Ey = VLR, = 500~ (120 « 0.1) = 488 V . By = 488V packeamfat speed, N, = 1000 rpm is given by, Rxa, EM Ei in Np 1000.x 488 1500 =3253V 1. B= 3253V Now, The average D.C output voltage of a 3-6 full converter, isgiven by, y= 313 Ya ‘cosa. ww (1) But, Fy=V—-H,R, > 353 =¥,-(120*0.1) [J,= 120, since motor runs at rated torque] > V,=325.3+12 > V,=3373V V,=3373V Let, ‘a,’ be the firing angle when speed is 1000 r.p.m Then, Equation (1) becomes, 3¥3 Vm Na > 3373= | v2 x 400 Jose cos 2 Bae , cpeBity to OX | 3373 x n= 3¥2 x 400 cos a, y 337.3%" £0805 499 x3 x 2 > SPECTRUM Al LLAN-ONE JOURNAL => a, = eos (0.624) = 51.39% (=a, = 51.39} Now, When speed, N, =~ 1000 r.p.m Back emf, B= Mey = =1000 x 488 1500 =-3253V Hence, armature voltage at ~ 1000 r.p.m is given as, y= V,-1,R, => Y= Bt, V, =~ 3253 + (120 0.1) = V,=-3133V u V,=-3133V Average D.C voltage of converter at speed, N, is given 313 Vm v, = SPeosas ~ 3y3| ¥2 x 400 = -3133= APY AATES |x cosas 3133 = = 3133 Xn ©0805 3 x 2 x 400 = cosa, =-0.58 > a= cos" (- 0.58) = 125.45° 03 = 125.45) Hence, range of firing angles = 0, = 40 YB = 254.03 V ‘The peak voltage for a 3-phase fully controlled rectifier @ ai) When N= 800 rp.m and at rated torque, For rated motor speed, N= 1500 rpm Back em4, £,=V-1R, =220-(50« 0.5) = 195 Volts ‘At 800 p.m, = 129 Volts For a 3-6 fully controlled converter, > =058 a= cos" (0.58) =54.55° ‘When N= 800 p.m and twice the rated torus, At ~ 800 rpm, mu) E=—xiss T0095 =0.53 x 195 ~ 104 Volts ASV, =E+IR, == 104 + [2(50) x 0.5] [Twice the rated om