Selection of Inductor value

[FreshmanNewbie]

I have a LED driver IC.

My requirements is Input Voltage (typ) 12V. Output Voltage is 22V (typ) and LED forward current typ is 160mA.

For an efficiency of 0.8 of the driver and considering the 0.5V schottky diode drop, considering the formulas mentioned from page 15 of the datasheet, my calculated inductor value comes around 27uH, 2.55A (considered ripple current of 40%, as mentioned on page 15).

Would the selected inductor value be sufficient for my application?

 

[KlausST]

Hi,

"sufficient" regarding what?

I mean: You said you did the math, so why are you worried?

Klaus

 

[FreshmanNewbie]

Hi,

"sufficient" regarding what?

I mean: You said you did the math, so why are you worried?

Klaus

Just want to confirm whether my math is correct for the inductor value? Is my inductor value correct?

 

[KlausST]

Hi,

I didn´t do the math.
My opinion: Inductance seems reasonable, current not.

What about using eXcel?

***
But let´s imagine I did the math ... and my values are different.
Then what´s your conclusion?
My values could be wrong, your values could be wrong. Still you don´t know which one is correct.

To find out one needs to do it step by step and compare the results of each step.
If one sees a difference, then maybe because of different input values or because of wrong math.

So the best way is to show your math, step by step.
So anybody can validate the input values as well as the math.
The error can be located and discussed.

***

My way:
Excel.

Here I can use the given formulae. And get step by step output.
I can just use the values from the example schematics (usually in the datasheet) and verify my excel result.

But going deeper one can compare the values with design_notes, application_notes, or online simulators (all of them often provided by the IC manufacturer).

***

Reliability of informations (my opinion): best on top, worst on bottom
* manufacturer datasheet
* additional documents provided by the manufacturer
* double checked documents provided by universities
* sources from reputable electronics designers
* documents provided by students
* internet sources that provide full documentation (professional designers, good hobbyists)
* internet sources without information how they derive their results (hobbyists)

In my opinion 80% of random internet sources of hobbyists include a lot of mistakes and mis-conceptions.

Klaus

 

[FreshmanNewbie]

Sure, do you think the current rating of the inductor is higher or lower?

 

[KlausST]

Hi,

it´s you who worte 160mA plus 40% of ripple. (Which shoud also be 160mA +/- 20% ripple)
How does this match with 2.55A?

Again: we don´t see your math!

But regarding efficiency (which you mentioned in post#1) the current rating does not matter. --> It´s the ohmic resistance that matters, the RMS current and the core loss.

Klaus

 

[D.A.(Tony)Stewart]

The selection of L described here is for continuous conduction mode.Eq. (6)

paper napkin estimate.
Pout = 22V*0.16A = 3.52 W
Iin = Pout/(Vin*eff) = 3.52 / (12*0.8) = 0.367

Current should be limited by the peak ripple and stay in CCM

Your estimate was Ipk= 2.55A !

--- Updated Nov 18, 2023 ---

Although the Inductor start-up current will be much higher charging 2*4.7 uF even if the FET is off.
Let's look at startup. Current is sensed on the load and FET but not sensed on the inductor thru diode to cap.

The total series resistance will be different than this model but that will limit the current and the inductor will be saturated and act as a resistor at some point.

The LC values chosen result in the energy of the inductor being transferred to the capacitor and quickly charge to the expected voltage. The IC soft-start cuts out after 1.5 ms so its duty cycle starts low and rises in some exponential RC time constant.

The ideal inductor power is lossless going in 1 cycle here while the DCR dissipates the heat.
The diode dissipates the most power yet the time duration is low that it should not be a problem with a steady input.