Select all the things you should do before running your buck converterQuestion 2Answera.Connect the power supply to the boxb.Set up your code for closed-loop controlc.Set the power supply to 60Vd.Check the output voltage of the commutation cell is correcte.Check the PWMs are correct

Where should you connect the voltage sense input to in your buck converter circuit?Question 2Answera.groundb.The input to the resistor (between the inductor and the resistor)c.DC+d.The input to the inductor (between the commutation cell and the inductor)

To solve this problem, we need to determine the value of the filter inductance \( L \) for a Buck converter that will ensure continuous conduction mode (CCM) at one-third of the maximum output power. Given: - Input voltage, \( V_{in} = 20 \, \text{V} \) - Output voltage, \( V_o = 12 \, \text{V} \) - Maximum output power, \( P_o = 72 \, \text{W} \) - Switching frequency, \( f_s = 400 \, \text{kHz} \) First, calculate the maximum output current \( I_o \): \[ I_o = \frac{P_o}{V_o} = \frac{72 \, \text{W}}{12 \, \text{V}} = 6 \, \text{A} \] At one-third of the maximum output power, the output current \( I_{o, \text{min}} \) is: \[ I_{o, \text{min}} = \frac{I_o}{3} = \frac{6 \, \text{A}}{3} = 2 \, \text{A} \] The duty cycle \( D \) for the Buck converter is given by: \[ D = \frac{V_o}{V_{in}} = \frac{12 \, \text{V}}{20 \, \text{V}} = 0.6 \] The inductor current ripple \( \Delta I_L \) is given by: \[ \Delta I_L = \frac{V_{in} - V_o}{L} \cdot D \cdot \frac{1}{f_s} \] To ensure CCM, the peak-to-peak inductor current ripple \( \Delta I_L \) should be less than twice the minimum output current \( I_{o, \text{min}} \): \[ \Delta I_L \leq 2 \cdot I_{o, \text{min}} \] \[ \Delta I_L \leq 2 \cdot 2 \, \text{A} = 4 \, \text{A} \] Rearranging the inductor current ripple equation to solve for \( L \): \[ L = \frac{(V_{in} - V_o) \cdot D}{\Delta I_L \cdot f_s} \] Substitute the values: \[ L = \frac{(20 \, \text{V} - 12 \, \text{V}) \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}} \] \[ L = \frac{8 \, \text{V} \cdot 0.6}{4 \, \text{A} \cdot 400 \times 10^3 \, \text{Hz}} \] \[ L = \frac{4.8 \, \text{V}}{1.6 \times 10^6 \, \text{A} \cdot \text{Hz}} \] \[ L = 3 \times 10^{-6} \, \text{H} \] \[ L = 3 \, \mu\text{H} \] Therefore, the value of the filter inductance \( L \) should be \( 3 \, \mu\text{H} \) to ensure the converter remains in CCM at one-third of the maximum output power.

You suspect that an uninterruptible power supply (UPS) is malfunctioning. You need to test the output from the UPS using a multimeter. Which multimeter setting should you use?Question 41Answera.Ohmsb.AC Voltagec.Continuityd.DC Voltage

Describe how the dc voltage at the buck chopper output varies when the duty cycle of the switching controlsignal is increased

If operating a buck converter in closed loop, what signal should we feed into the controller?Question 4Select one:a.(Reference voltage - Measured voltage)b.Reference voltagec.Measured voltaged.(Measured voltage - Reference voltage)