Calculate the ripple current in my inductor.

Thread Starter

Zouglou LeMagicien

Joined Feb 12, 2019
33
Hello,

For a project, I need to design a power circuit like this one :
1575234211324.png
With the input pusle, I can say that I want this value of current in my LED. But this current won't be stable, it will have some ripple.
By the way, that's a simple schematic, the pulse isn't the real input source, it's only to show you how it works. I know that at the start it will not be a ripple but I'm speaking here when the current is in his stable phase.

I want to calculate this ripple to after, calculate a good frequency for my switch (nmos). But I don't know how can I calculate this.
I have calculated the 2 formula for the inductor which are:
1575234628543.png
This is an exemple of what it does like:
1575234947850.png
Does someone know how can I calculate this ? :)
 

ronsimpson

Joined Oct 7, 2019
1,125
I think D1 has no current. Can be removed.
D3 will have the same current as the LED but only when the MOSFET is off.
To reduce ripple current in D2 you can put a 100uF cap across D2+R1.
If you increase the frequency the ripple current will be smaller.
If you increase L1 inductance the ripple current will be smaller.
Ripple is what you see because, when the MOSFET is on the current increases, when the MOSFET is off the current decreases.
Your ripple is actually small and just fine for a LED. I have made this circuit with much higher ripple.
 

ci139

Joined Jul 11, 2016
1,696
Your ripple is actually small and just fine for a LED. I have made this circuit with much higher ripple.
↑↑↑ it's better to have a small ripple
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
↓↓↓ than to build additional control to stop/redirect the inductor current incase of the switch(MOS-Fet) - passing too high current - and then fails closed(not conducting) = your cap may get over charged . . . and such also sets delay in control loop - thus - complicating the control loop . . . a lot !
To reduce ripple current in D2 you can put a 100uF cap across D2+R1.
________

if the LTC4441-1 has not built in "Soft Start" option - then you already are in trouble (even without the 100µF ripple filter)
 

ronsimpson

Joined Oct 7, 2019
1,125
For a project, I need to design a power circuit like this one :
We can not see the big picture. Why the LTC4441?
There are ICs that include the MOSFET and diode. They don't just blindly output a duty cycle but actually watch the LED current.
Most of the LED constant current PWM run much faster so the inductor can be much smaller. 500khz to 2mhz
The LED drivers do not have start up problems.
 

Thread Starter

Zouglou LeMagicien

Joined Feb 12, 2019
33
Thanks for all your responses, I might have the solution right now.
To answer to your question for the inductance and the LTC4441, it's an obligation from my teacher, I don't know why :/
 
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