Question

Please answer 2 and 1.

Assignment No.6 10mH, C-20μF, an 1. A boost converter as the following parameters: V,-20V, L Find the current spec of the inductor for continuous current mode (Max an Min current) -20 2. Switching frequency is 50 KHz and conduction duty cycle is 0.6 a. b. Calculate the output voltage and the ripple voltage An array has been formed with three modules in a string and two strings in parallel. The overall characteristic of the array is shown below. V 2. of each module is 44.2 V and Isc iS 6 A. The (a) Find Pmax of the array (use graph) (b) What is the matched load of the array? (c) If the VMI ofthe array is approximated to be 80% ofVc (general assumption). array is driving a 10 2 heating element in a hot water heater, select a suitable y is driving a 10 Ω heatin converter so that max power is delive average current from the source, and the output current. red to the element. Specify its duty cycle, (d) If one module is fully shaded, what would be the max power? Redesign the duty eycle if necessary to deliver max power to the heating element. The modules are provided with bypass diodes. 35. 3 VOLTAGE 3. A buck-boost converter has the following component values: DC input 12V, L- 10uH C 20uF, R (Load) 102, Switching Frequency 200 KHz, and Duty Cycle- 0.6 a. Find output voltage, ripple voltage across the load. b. Find the maximum, minimum, and average inductor current. c. Determine if the converter is operating in CCM or DCM. The open circuit voltage of apanel is 20V. The VMPPs 80% of 18W. Design a converter to deliver max power to the load if load resistance varies between 5 ohm and 25 ohm. Assume input voltage is VMPP. Switching frequenc KHz. 4 Voc. The max power is s. Deseribe a controlot temlevel t aecomplish 0.4

Answer 1

F_s = 50 KHz D = 0.6 0 < t < DT switch is ON (closed) V_L = V_i I_c = -I_0 DT < t < T switch is open V_L = V_i - V_0 V_L = V_i - V_0 I_c = I_L - I_0 Ampere-sec balance equation -I_0 DT + (I_L - I_0) (T - DT) = 0 -I_0 DT + I_L T - I_L DT - I_0T + I_0 DT = 0 I_L = I_0/1 - D Volt-sec balance V_i DT + (V_i - V_0) (T - DT) = 0 V_i DT + V_i T - V_i DT - V_0T + V_0 DT = 0 V_0 = V_i/1 - D \r\n (2) V_0 = V_i/1 - D = 20/1 - D = 20/1 - 0.6 = 20/0.4 = 200/4 = 50 V (b) Output voltage V_0 = 50 V I_0 = V_0/R implies I_0 = 50/20 = 2.5 (a) when switch is closed 0 < t < DT V_L = V_i L di/dt = V_i implies L delta I_L/DT = V_i implies delta I_L