Buck converter questions (24V to 5V)

[apolloneo]

Hello,

I want to use MP1584 buck converter to convert 24V DC power supply to 5V regulated one. I try to follow the MP1584 datasheet and the online design tool from monolithicpower.com as well. Here are my questions:

From the datasheet typical application recommendation, the circuit looks like this,

but from the online tool, they look like this
for Vin from 8V to 28V,

for Vin from 22V to 26V,

my power supply used in project is 24V (range from 22V to 26V). So I wonder if I should follow the online design tool or application from datasheet. I understand probably, all of those 3 are okay, just wonder if there is one slightly better and the reason behind it.

[Wawa]

@apolloneo

I want to use MP2338 buck converter.

Not as easy as you might think. Circuit board layout and parts choice is critical.
Much easier to use a buck converter module.
Pololu has an excellent range.
Leo..

[jremington]

+1 for Pololu, and getting on with the actual project.

[belmont1591]

MP2338 or MP1584?
Text says MP2338 but images says MP1584

[apolloneo]

good catch. I was thinking MP1584. fixed in my original post.
MP2338 is another option, but so far, MP hasn't put it on their online design tool yet. So, it wouldn't bother me to use which schematics as the MP1584 did.

[Wawa]

Don't forget that parts choice is critical too.
Leo..

[belmont1591]

First of all, I wouldn't choose the MP1584 because...

However, let us ignore this warning.

The MP1584 Datasheet (page 15) shows a reference design using a 15uH inductor and from the online tool, the inductor's value is 9.1uH. Why?

If we go to page 10, we will find this equation:

But, what is Delta_IL (inductor ripple current)? Please refer to this Analog Devices article (read this ENTIRE article).

So, from the above equation, we know the switching frequency (910 kHz), we know the output voltage (5V), we know the input voltage (24V)

And how I do calculate Delta_IL? Please refer to this Texas Instruments article
specifically, here (page 3)

Basically

Delta_IL = k * Iout (max)

From the online tool, the inductor current ripple is 16% (0.16) Therefore Delta_IL will be:

Delta_IL = 0.16*3A = 0.48 A 

What happen if we replace all these values in the inductor equation?

Now, what will be the value of the ripple current in the datasheet typical application so that the inductor's value is 15uH?

Let's first determine what is the switching frequency. From the datasheet page 9, we get this another equation

If we isolate the Fs(kHz) variable we get:

From the example, we get a 100kOhm resistor value, then

Now, let's isolate Delta_IL value from inductor's equation:

replacing values we get:

this value in percentage will be:

k = Delta_IL / Iout(max) = 0.28998779/3 = 9.66%

If we put this value in the online tool we will get:

Therefore, the application shown in the datasheet is assuming a ripple current of 9.66% while you set a ripple current of 16% in your design

[apolloneo]

Nice explanation. I appreciate it.

[oldcurmudgeon]

The explanation by @belmont1591 is why I buy switching regulators rather than build them. It is usually cheaper and a lot less work.