# Minimum Inductance for Buck converter to maintain CCM

In a buck converter, consider all components to be ideal; the output voltage is kept constant at 5 V by controlling the duty cycle. The input voltage varies between 10 V to 40 V, the output power is greater than 5 W, and the switching frequency is 50 kHz. Calculate the minimum inductance value required to keep the converter operation in the continuous conduction mode.

Been provided this question and I immediately think of v=Ldi/dt. The answers explained in figure below have an equation that I've never seen before. Does anyone know where it has been derived from? I also know that the inductor value must be chosen such that the inductor current doesn't reach zero during the converter's operation. • Where does that image comes from? Please edit your question and provide a link or reference - Don't post the reference in the comments. Apr 22 at 23:00

In a buck converter operated in continuous conduction mode or CCM, the inductor current ramps up to the peak value $$\I_p\$$ at the end of the on-time and goes down to the valley current $$\I_v\$$ when the switching cycle is complete. The difference between the peak and the valley is called the inductor ripple current and noted $$\\Delta I_L\$$. The inductor average current is the output current $$\I_o\$$ of the buck converter and, in CCM, it equals the sum of peak and valley currents divided by 2.

When the output current decreases, the inductor peak and valley currents are shifting down and, if the load gets lighter, the valley current reaches 0 A. At this exact point, the converter is at the border between continuous and discontinuous conduction mode (DCM). This operating point is called borderline or boundary conduction mode (BCM) and often represents a wanted operating mode such as in boost power factor correction stages (PFC) for instance. But here, we want to determine the value of the inductance which guarantees CCM for the given operating point. In other words, we must determine the inductance value which brings the valley current to 0 A. By keeping our inductance above the minimum we compute, CCM is always ensured down to the specified output current.

I have gathered the derivation below and it is not very complicated: draw the inductor current and identify the peak and valley currents: When the load current decreases, you can see in the right side of the picture how the whole inductor current is shifted down until the valley current cancels. Write the corresponding equations and solve for $$\L\$$ which ensures this result. Rearrange the expressions linking $$\V_{out}\$$ and $$\V_{in}\$$ and you are done.

You have: u=L*(di/dt)

=> Convert the formula to L: L=(u*dt)/di

• If you have 40V at the input, you get the fastest rise of the current and the difference between 40 and the output (5V) is 35V.
• The time is the calculated on-time during the current can rise.
• Id you want to have 1A at the output, then your dIL in the coil will be 2 Ampere.

This solution is only right if you can forget the schottky diode (or voltage drop at the MOSFET which works as a diode in synchronous rectifier), because of the diode you have to generate a 0.55V higher voltage and the on-time must be higher and because of this you will need around 49µH as a inductor.