I'have to build a buck converter with the following requirements:

$$\text{1) Input voltage (Uin): 12 Volt}$$ $$\text{2) Output voltage (Uout): 5 Volt}$$ $$\text{3) Voltage ripple (on the output) 0,1 Volt}$$ $$\text{4) Output current of 0,5 Ampere (when you short the output resistor)}$$

The Dutch word functiegenerator is a functiongenerator, that produces the input puls.

The MOSFET is a P-Channel MOSFET IRF9540n. And the voltage drop on the diode is 0.6 volts

The question I've are the following:

The formula for getting the right duty cycle (that you've to set on your fuctiongenerator) is that given by:


And how can I calculate the minimum value of my capicator and inductor?

enter image description here

Thanks in advance

  • \$\begingroup\$ Is this homework or are you making a real product? Depending on this you might need a textbook answer or an app-note answer. \$\endgroup\$ Commented Oct 30, 2015 at 15:35
  • \$\begingroup\$ If it was homework I didn't asked it over here,so it is a real project and I can't find out \$\endgroup\$ Commented Oct 30, 2015 at 15:37
  • \$\begingroup\$ Ok, what's your "functiongenerator" in the real product? \$\endgroup\$ Commented Oct 30, 2015 at 15:39
  • \$\begingroup\$ I dont know yet, I've to use a chip that produces the puls, but fist of all I've to know what the puls has to be (duty cycle etc.) \$\endgroup\$ Commented Oct 30, 2015 at 15:40
  • \$\begingroup\$ You can just download focus.ti.com/download/aap/utilities/buck.xls and plug the values in. \$\endgroup\$ Commented Oct 30, 2015 at 15:41

2 Answers 2


The ESR value of your capacitor and inductor affect the ripple output voltage. The output ripple voltage can be estimated based on the inductor ripple current (delta IL) and the ESR of your capacitor.
The inductor shall be used to reduce the ripple current (delta IL).
delta IL = (1/F*L)Vout(1-Vout/Vin). Where F is the switching frequency. In the same time the capacitor is selected with the following calculation: C=Ip/(F*Vripple); where Ip is the peak output current and F your switching frequency. The Ripple voltage is based on requirements and performance expectation.

  • 3
    \$\begingroup\$ "L is the switching frequency"???? \$\endgroup\$
    – Andy aka
    Commented Oct 30, 2015 at 15:59
  • \$\begingroup\$ Sorry a typo error \$\endgroup\$
    – R Djorane
    Commented Oct 30, 2015 at 16:04

The inductor is as explained by codo. Now for the output capacitor, the most important parameters are the ripple current allowed by the inductor and the ESR of the capacitor.

Vpp = Ipp*ESR. Vpp- peak to peak ripple on the output dc rail. Now, you have to calculate the ESR based on the ripple voltage you can tolerate. For low ripple voltages you have to go for very low ESR.

Now, with this Ripple volateg(CAP) = IP-P/ (8 x f x c). Where Ip-p is ripple current (as a thumb rule take it as 20-30% of max load current in your system).

f - switching frequency of the buck you have designed.

From this you can find the output capacitor value.

High ESR has a down fall of not just increasing ripple voltage but also dropping a voltage and due to the ripple dissipating heat.


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