LMR54410: Maximum Output Current

Hello

What is your frequency and inductor value ??

Thanks

Hello

Frequency: 1.1MHz because of LMR54410F

Inductor value: 6.8uH

Best regards,

Shotaro

Hello

You can use the frequency and and inductor value to calculate the peak and valley of the inductor current.

Then compare with the worst case current limits and determine if the current limit will be tripped at your load of 1.1A.

Thanks

Hello

I run Power Stage Designer. Your opinion is that "low side current limit" will be tripped, right?

Low side

High side

Best regards,

Shotaro

Hello

Based on your inputs, the inductor ripple current is about 0.344A

Your peak current will be 1.1+(0.344/2) = 1.272A

Your valley current will be 1.1-(0.344/2) = 9.928 A

So, the valley current limit will clear, while you are very close to the peak current limit.

Thanks

Hello

I also calculated the maximum current based on the below note. And the result is here. Can you review my calculation?

https://www.ti.com/lit/ta/ssztay5/ssztay5.pdf

<Equation>

<Result>

• Iout(max) = 1.43A (with Ipeak)
• Iout(max) = 1.27A (with Ivalley)

--> 1.27A is maximum current which is allowed by current limit values.

*Ipeak=1.6A, Ivalley=1.1A

Best regards,

Shotaro

Hello

Your equations and calculation are correct for the current limit values you selected.

However, in my opinion it is best to use the minimum value of peak current limit as a worst case, since 

that value is guaranteed.

Also, keep in mind that if the valley current limit is tripped, the frequency will reduce, but the output voltage

will stay in regulation.  If the peak current limit is tripped, then the output voltage will fall out of regulation.

Thanks

Hello

OK, I understood. I will inform the maximum current as below to my customer.

• Iout(max) = 1.08A (with Ipeak)
• Iout(max) = 1.07A (with Ivalley)

--> 1.07A is maximum current which is allowed by current limit values.

*Ipeak=1.25A, Ivalley=0.9A

Best regards,

Shotaro