Power Electronics Assignment - Math Problem Example

School of Engineering & Mathematics Power Electronics Assignment Submitted by: Name Introduction The purpose of converters is to convert a direct current to AC which is the power that is required for most applications, most converters help in converting low voltage direct current. They have Diodes a component that allows component to flow much more easily in one direction than in the other. An idea diode has zero resistance for current in one direction, so that the current flows without any voltage drop across the diode and infinite resistance for current in the other directions so that no current flows. The circuit symbol for the diode has an arrow headed to indicate the direction of allowed current. Detailed analysis and design of the converters (ZETA and fly back) The design of the power stage of the self-oscillating fly back converter follows the well established design rules for the fly back converter operating at the boundary of continuous conduction mode and discontinuous conduction mode (Irving and Jovanović, 2002). Vin = 5 V … 30 V , Vout = 15 V Output voltage ripple 2% peak to peak, Switching frequency 20 kHz Load 20 W … 150 W T= = = 0.05ms This is the time that it will take foe one circle. It is assumed that the switch will have 0 voltages drop in on the situation. By using the equivalent turns of the indicators to their voltage value, the duty ratio is always 50%. Therefore we determine when Vin = 5. = 0.025ms To find the value of c when the diode is off, we calculate as follows Ic = 20W/15 = 1.33A Ic = C dv/dt C = (1.33 x 0.025x 10-3)/0.3 = 22.2F Therefore the capacitor for the fly back converter should be greater than 22.2 . if we consider 10% ripple in current output we get = 0.532A In designing L2 becomes very important. Therefore L2 output becomes V = L Thus L=( 15 x 0.025 x 10-3) / 0.532 = 57.8mH For L1 by using the relation between the Vin and Vout of 1000:399 which was 1.33:0.532.so, the inductance is related in square base relation instead of the inductance turns common used relation. Therefore the L1 can be calculated as 2^2=4 L1=4*57.8*10^-3=231.2mH R=/w= 11.25Ω So, Now we start calculating the other VIN which is the VIN=30v T=0.05ms ∆t=0.025ms Ic= 150/15=10A Ic = C dv/dt C=(10*0.025*10^-3)/0.3 =833.3µF ∆i=1.33A L2=15*25*10^-3 / 1.33= 2.09375 mH L1=4*2.09375*10^-3 = 8.375*10^-3 R=12.5 Ω Calculations of the efficiency of the converters To show how the efficiency of the linear supply is affected by changes in input voltage. Calculate the efficiency of the supply using equation below when the output current and the output voltage is 15V .The results for each input voltage are shown in Table below where it is clear that the linear supply's efficiency drops very quickly as the input voltage increases. Efficiency of converters = x100(Glover, Sarma and Overbye, 2007). Vin Iin Vout Iout x100 5 1.33 20 15 1.33 75% 10 2.67 40 15 2.67 66.7% 15 5.33 80 15 5.33 58% 24 8.00 120 15 8.00 47% 30 10 150 25 10 43% When switch, turns on and off so that the correct amount of energy is transferred to the output. The input signal is chopped into pulses which are averaged to give a rectified DClevel. The longer the on time of the pulse with respect to its off time the higher is the regulated dc voltage. Suppose a circuit connected to the input terminals supplies a mixture of a dc potential difference plus ac voltages at a range of frequencies. The reactance of the capacitor is large at low frequencies , so most of the voltage drop for low frequencies occurs across the capacitor ,most of the high frequency voltage occurs across the resistor and thus across the output terminals(Glover, Sarma and Overbye, 2007). Combinations of capacitors and inductors are also used as filters. For both RC and LC filters, there is a gradual transition between frequencies that are blocked and frequencies that pass through .The frequency range where the transition occurs can be selected by choosing the values of R and C (or Land C). Detailed analysis and design of the ZETA converter GIVENS: Vin=(5v)….(30v) Vout= 15v Output voltage ripple 2% Switching frequency= 20KHZ Load= (10w)….. (150w) Time is D2T T=1/f= 1/150KHZ = 50 µs We start with calculating the VIN=5v 15=D1 (5) / (1-D1) (15-15D1)=5D1 20 D1=15 D1=15/20=0.75 D2=1-0.75=0.25 So, Now we start calculating the other Vin which the Vin=30v 15=D1 (30)/(1-D1) (15-15D1)=30D1 45D1=15 D1=15/45=0.33 D2=1-0.33=0.67 Average output current (10)/ (15) =1.33A To (150)/ (15) =10A Iout= 1.33A to 10A D2 range from 0.25 to 1.33/0.67 Is could be higher as 10/0.25=50 Or could be 10w as 1.33/0.67=1.99 The input current depends on the duty lets allow 10% inductor current ripple. VL=15=L Di/Dt=LAi/At We want (15v) (D2T)/L=Δi Δi< (0.1) (1.33) D2 L< 0.33*(0.67)^2*(50*10^-6) L Read More