I wanted to dive in Power Electronics and create a 24V DC-DC Boost Converter that would run from 2.5 to 5 V power supply. 80% efficiency was assumed, and 30 kHz frequency chosen for the PWM.

Initial Parameters






Duty Cycle

Assuming the forward voltage drop of a diode was 0.5, duty cycle was calculated in the following manner.


Inductor & Capacitor

I only had 220μH inductor & 100μF capacitor around, so those two have been chosen for the circuit.



Here is the DC-DC Boost Converter circuit with the inrush current protection.


PWM signal is generated by PIC10F200, and PWR is 2.5 - 5 volts.


I have made the circuit on a perfboard, and it is working perfectly. However, the output voltage is much higher than expected, around 75 volts, and has fried my regulator as I have omitted the Zener regulation.

Were my calculations wrong, should the PWM be tweaked?


2 Answers 2


Your calculations are correct when the converter is in continuous conduction mode (CCM), ie when there is always some current flowing in the inductor.

However, if there is no load, the boost goes in discontinuous conduction mode (DCM), and your calculations are not applicable. Theoretically, in DCM with no load, the output of a boost goes towards infinite voltage. Actually, because of several kind of losses (stray resistance of the inductor...), the voltage is limited. Still, it can reach high values.

A solution is to feed back the output voltage to the MCU (through a voltage divider) and implement some kind of regulation loop in the code.


Lowering the duty cycle and adding a load resistor has resolved the issue.

  • \$\begingroup\$ Always "soft start" with a low PWM duty cycle and always have a "look" on output voltage ... \$\endgroup\$
    – Antonio51
    Commented Nov 6, 2021 at 15:23

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