Buck converter output ripple voltage calculation

Papabravo

Joined Feb 24, 2006
17,548
As the inductor is being charged during the 'ON' position of the switch, the current reaches the average value of (Imax - Imin)/2 at time DTs/2. As the inductor supplies current to the output capacitor during the 'OFF' position of the switch, it reaches the average value of (Imax - Imin)/2 in a time of D'Ts/2. If we add these two times together we get:

$$\cfrac{DT_{s}}{2}\;+\;\cfrac{D'T_{s}}{2}\;=\;\cfrac{(D\;+\;D')T_{s}}{2}\;=\cfrac{T_{s}}{2}$$

because

$$D\;+\;D'\;=\; 1$$

It is also worth noting that the shaded area above the average current is equal to the area below the average current, so the long term average charge on the inductor is zero.

• hoyyoth

hoyyoth

Joined Mar 21, 2020
192
Hi Pap,

Thank you.

How that becomes exactly half the on time and half the off time.
Is the on time and off time are equal.

Regards
HARI

Last edited:

Papabravo

Joined Feb 24, 2006
17,548
Hi Pap,

Thank you.

How that becomes exactly half the on time and half the off time.
Is the on time and off time are equal.

Regards
HARI
It is invariant for a fixed frequency. If the duty cycle is not 0.5 (or 50%) then the slopes of the current rise and the current fall are unequal, but it is still one half of the 'ON' time for the current to rise from Imin to the steady state average, and during the OFF time it takes one half of the OFF time for the current to fall from Imax to the steady state average current.