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

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  • \$\begingroup\$ If the device has an eval board, try those values. It is pretty odd that the datasheet does not even give an example with specs and component values. But surely you know what you are powering and how much or litte ripple it tolerates, or at least can you figure it out? \$\endgroup\$ – Justme Sep 2 at 8:26
  • \$\begingroup\$ @Justme Because it is cheep. I am using for USB to UART converter,do you think 1% output voltage ripple and 30% ripple current in the inductor will be enough? \$\endgroup\$ – walid24 Sep 2 at 8:28
  • \$\begingroup\$ The typical design point for most convertors is 30% current ripple, if you want less ripple by using a larger inductor, then there is more loss in the copper of the inductor, or you might need the next size up inductor, which will be bigger and cost more. The 1% ripple will be fine with a serial convertor , it would probably still work with 10% ripple. as TTL and USB thresholds are quite low , around 1.5v so will usually work down to 3v on the 5v line. \$\endgroup\$ – BobT Sep 2 at 8:57
  • \$\begingroup\$ @BobT I appreciate your taking the time to write this \$\endgroup\$ – walid24 Sep 2 at 9:06

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.

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  • \$\begingroup\$ thank you ,I appreciate your help \$\endgroup\$ – walid24 Sep 2 at 9:09
  • \$\begingroup\$ The voltage ripple specified by the OP was 1% which is 0.01 x 5v = 0.05v , On that basis C = 20uF , that assumes a perfect capacitor, , ESR in a real capacitor could be expected to contribute, very roughly, the same amount of ripple so double the capacitance to get a reasonable estimate of Cout. . \$\endgroup\$ – BobT Sep 2 at 14:37

You are expected to specify the value of \$\Delta V_O\$ yourself.

How much ripple do you want to allow on your output?

  • Running a microprocessor from it? You might allow 50mV of ripple.
  • Running an audio amplifier? You might want ripple lower than 1mV.

The \$f\$ referred to in the equations for \$\Delta V_O\$ is the switching frequency of the regulator.

You might attack it backwards. Take a common capacitor value, then calculate the ripple and ask yourself if it is acceptable. No, pick a bigger capacitor and try again.

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  • \$\begingroup\$ thank you ,I appreciate your help \$\endgroup\$ – walid24 Sep 2 at 9:09

The capacitor values depend on whether you are trying to make a physically small converter , or if the input is from a transformer + rectifier, or you want a low cost design.

If the input is already smooth DC , then for Cin you just need a small low ESR capacitor. one or two 22uF 35v ceramic capacitors , or a handful of 10uF 35V ceramic capacitors, X5R would be cheapest option. If you have a transformer / rectifier then you will need some bulk energy storage, like 1000uF electrolytic.

For Cout you will need another couple of 22uF ceramic capacitors, e.g 2 x 22uf or 4 x 10uF if running at 400kHz , half that if running at 800kHz . Running at a higher frequency allows smaller capacitors, but reduces efficiency (i.e. runs hotter). The TI website has some useful tools , including web-bench. Here is a snip from the LMR36520 datasheet , this is commonly used on the Adafruit and pololu DC-DC modules. LMR36520 data sheet snip

As with most power supply electronics, don't try running on the hairy edge, if you need a 1A supply , then use a 2A rated device.

The Pololu 12v-5v 1A module is here https://www.pololu.com/product/2831 (you can see the schematic if you scroll down) it uses 4 x 10uF ceramics on the output, and only one on the input. You can get 10uF ceramics in an 0805 package at a better cost point than 22uF in a 1206, for roughly the same space. Designers and manufacturers try to minimise the variety of parts in their designs, for example I use 10uF/35v in 1206 , 10uF/10v in 0805 and 1uF/50v in 0805. A full reel (5000) of 10uF/35v cost less than a half reel of 22u/35v.

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  • \$\begingroup\$ Thank you I found a section in the datasheet LMR36520 talks about the selection of the components \$\endgroup\$ – walid24 Sep 2 at 8:50

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