Buck converter output ripple voltage calculation

Thread Starter

hoyyoth

Joined Mar 21, 2020
170
Dear Team,

I have a question regarding buck converter output ripple .In the below figure,could you please tell me how the area shown in the red circle
have a time duration of Ts/2

1625851773836.png
Regards
HARI
 

Papabravo

Joined Feb 24, 2006
17,242
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.
 

Thread Starter

hoyyoth

Joined Mar 21, 2020
170
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
17,242
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.
 
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