Question

O bructe CONFIDENTIAL 2 limitatio of Flyb Q2 (a) With the aid of respective topology, point out three (3) basic de-de converters ( 3 marks) (b) Discuss the energy transfer in on each circuit by refering the circuits drawn in Q2(a) (3 marks) (c) Design a boost converter for continuous inductor current and an output nipple voltage of less than 1% that will have an output of 30 V from a 12 V source, Assume the load resistance of 50 2. and ideal components for this design Discuss the applications and limitations of flyback converter (8 marks) (d) Design a buck converter to produce a constant output voltage of 10 V when the input de voltage source varies from 20 V to 25 V. The output power is fixed to 50 W. Design the range of required duty cycle of the PWM switch for each range, the average inductor current, the maximum current needs for the output, and the minimum inductor current where the inductor ripple should respond to continuous condition mode. At the meantime, it is required, the capacitance must limit the output ripple less than 0.2%. The switching frequency is 20 KHz and use design criterion of Ain = 40% of Il. (11 marks) Q3 (a) Inverter can be operated in two (2) mode conditions which are the force and phase control mechanism Currently, the phase controlled is more widely used due to less harmonic output produced by the inverter (1) Explain with the helps of diagram the different functions of the D- STATCOM and the DVR as an inverter based regulator in the power sytem operation (4 marks) buck D = les Il=vde A phase controlled inverter requires a firing angle in order to change the output voltage magnitude of the inverter Demonstrate this condition in the form of Pulse Width Modulation (PWM) technique (2 marks) boost D= 1- VE vo IL = VS (1-0)=RL Train= O(1-D²RL zfs all foot CONFIDENTIAL R L.(min) = A- ORL All= (us-bo)DT

Answer 1

DC - DC converter - converts DC Power (sined) to DC Power (variable BUCK converter Boost converter Buck- Boust converter BUCK Converter L CH Load -Vs NC CH IS ON (oradi) Ne =(vs-vo) CA OPR V2 = Vo Apply volt sec balance elcyols inductor Vs-V) DT +f-vo) (T-NT) 20 (Vg - Vo) D-Vo(1-0) 20 Vo = DV D-Duty cycle =TON T Boost conventer L mm V When Chtis on VS CH VL=US When ch 18 OFF Volt - see balance across L VL Vs-vo VS CDT) + (Us -vo) (1-0)T zo Apply Amp-bec in capacitor Noa Vs Step up chopper 1-D -Bo DT & (IL-ID)(1-1/T20² 12,= 2 | -

.CH árin Yec 0 C when Ampere-sec belunce guals caput BUCK-boust converter - VS VL CH is on V₂ =Vg Ve=0 Ic=f(L+[d) Volt -sec balance V COT) + Vo (1-0)=0 Vo - - DV T-D Ice-TO When at og OFF :10 IO CDT) [ 24+Pd) C1-01750 I ID 1 BOOSH converter V = (a V Avo=11. Vo =300 Re="son Voz uso 30 10-A 1-0 =12 » D=1-12 L-20-6 Il Do (1-D) = 1.5A (Avoiman = 15. و - 4 without f = 9 HEL we cant go forthes ool=

> buck convertey 200 vina 20to g5v. Vozlov Po - sow XRD RO-D (21) No=Dug 10- -0(20) D- Ž D= 10 20.4 25 0.5 20 D range from o. Io = Vo (6) Po= No to To so 25AIL f = 20 kHz, DI, = 401. PL -49 xt=&A 100 Imone - Do tsh Iminz Do 4 ore = 5-1 Ripples 302). Ave - Vin D(ID) 5+1 6A = 4A Eco 84600 fz 01) AVC=T4Q Ve=(1-R 0.2of to— TX401.12 8 cc 0.2 X10 = 1 X 4 X 20X103 8 X Cc 0.02 = STONE 8Xce '8οχα - Le » 2018 Ke= 6.25 mF