Electronic – Output capacitor in buck converter

buckdc/dc converterlow passpassive-filter

I am working with the buck converter AOZ1284 Datasheet

I want to get 5V with 12V in input so I chose R1=52.3k and R2=10k (Vout = 0.8 x (1+R1/R2) =0.8x(1+52,3k/10k)=4.984V).
Also I fixed L=4.7uH.

How do I calculate the value of Co and Cin (input and output capacitors?)

I found a section in the datasheet about the output capacitor but I don't know

  • the acceptable value for DLETA_Vo
  • the cut off frequency for the LC filter in the output

enter image description here

Best Answer

For future use, use Q = C * V (charge = capacitance times voltage).

If you take the derivative with respect to time, you get

  • dQ/dT = C * dV/dT + V * dC/dT

If we define dQ/dT = current (I), and define dC/dT = ZERO, we now have

  • dQ/dT = I = C * dV/dT

Now let us use that for ripple of 0.01 volt, for 1microSecond delay time, with 1 amp load:

  • I = C * dV/dT ===>> C = I * dT/dV

and we substitute

  • C = 1 amp * 1uSecond / 0.01 volt = 1uF/0.01 = 100 microfarad

Notice this does not bring up the Resonance of the Inductor with this Capacitor which affects the SwitchReg regulation_loop (negative feedback) stability and phase margin.

=======================================

Read the comment by "BobT" for appropriate caution on Capacitor ESR.

Related Topic