LM5155: Question about inductor value for LM5155.

Hi Kazuya,

Thanks for using the E2E forum.

Regarding the selection of the inductor

1. The value calculated using the equation No. 5 from the App note is correct. The value calculated by power stage calculator is also 2.5uH; thus, you can use 2.5uH. 

2. The QuickStart calculator is using different strategy for calculating the inductor value, if the duty cycle is greater than 33%.

The recommended inductor is 2.5uH.

Best Regards,

Hassan  

Hi Hassan,

Thank you for your reply.

Could I ask you a few additional questions as the follows?

Q1.
I understood LM5155/56 Boost Controller Quick Start Calculator uses a different strategy
when the on duty is greater than 33%.
Could you please tell me the strategy roughly?

Q2.
Why does the calculator use the different strategy?

Q3.
Do you not recommend us to use the calculator if the on duty is greater than 33%?   

Thank you again and best regards,
Kazuya.  

Hi Kazuya,

Thanks for your interest in the topic.

I am sorry for my misunderstanding regarding the QuickStart calculator. 

As mentioned earlier there are two strategies to select the inductor.

Strategy 1. Says that the peak inductor current ripple ratio occur at minimum Vin (usually this is used by power stage designer or SLVA372D document).

Strategy 2. Says that in CCM operation the maximum ripple ratio typically occurs at 33% duty cycle. When the duty cycle at the maximum input voltage is greater than 33% the maximum ripple ratio occurs at Vin,max. When the duty cycle at the minimum input voltage is less than 33% the maximum ripple ratio occurs at Vin,min.

Thus, the max ripple ratio is not always at minimum Vin, it changes with the duty cycle, which is taken account in strategy 2. The inductor value is important for the copper losses and the RHPZ frequency of the control loop. For better understanding of the strategy 2, kindly check the app note (page 3 to 5).

https://www.ti.com/lit/an/snva991/snva991.pdf

In your case, the duty cycle is always greater than 33%; hence, strategy 2 will be better for the selection of inductor (it is recommended to use the QuickStart calculator recommended inductor value). 

Best Regards,

Hassan 

Hi Hassan,

Thank you very much for your supports.

Can I ask you an additional question about LM5155 on this thread?
If I should ask it on a new thread, please let me know.

The customer is designing their LM5155 power supply circuit using SNVC224 calculator.
Also they refer to SNVA824 document for the design too.

At this moment, they have the following questions.
Could you please give me your reply?

Q.
They want to add Rfil and Cfilt because the following sentence are descried in SNVA824.
"For all designs is it recommended to add the low pass filter to the current sense signal."

When Rsl is much smaller than Rfilt, such as Rsl=0ohm,
does Rfilt not affect the Recommended Minimum Isat value?

They selected 0ohm for Rsl because the device datasheet shows Rsl = 0ohm on Table 10-2.

There is no place to input Rfilt on the SNVC224, but they worry that it might have an effect.
Could you please give me your reply?

Thank you again and best regards,
Kazuya.

Hi Kazuya,

Thanks for reaching out again.

Selecting the Rsl = 0 will not effect the current limit. It is okay to use Rsl = 0, if the Duty cycle is not very large.  

A small external RC filter (RF, CF) at the CS pin can be added to overcome the large leading edge spike of the current sense signal. Select an RF value in the range of 10 Ω to 200 Ω and a CF value in the range of 100 pF to 2 nF. Because of the effect of this RC filter, the peak current limit is not valid when the on-time is less than 2 × RF × CF. To fully discharge the CF during the off-time, the RC time constant should satisfy the following inequality.

3 * Rf * Cf < (1-D) / Fsw

Best Regards,

Hassan 

Hi Hassan,

Thank you for your reply and I7m sorry to be late response.

Could I ask you an additional question?

Q.
The customer usage conditions are the follow.
Vin=2.6V~3.6V, Vo=12V, Io=5A.
So the deal duty cycle at the minimum Vin is 78%.

Is this 78% not very large duty cycle?
Can Rsl be set to 0ohm at this 78% duty cycle?

Thank you again and best regards,
Kazuya.

Hi Kazuya,

Thanks for reaching out again.

If the duty cycle is greater than 50%, it is recommended to use the slop compensation. 

In your case, you need to use the slope compensation. Check the page 19 and 20 for the slope compensation calculations. 

https://www.ti.com/lit/ds/symlink/lm5155.pdf?ts=1743756862706&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FLM5155

Best Regards,

Hassan

Hi Hassan,

Thank you very much for you reply.

Could I ask you an additional question?

Q.
The duty cycle of the sample circuit which shown in section 10-2-1 Design Requirements of LM5155 datasheet is more than 50%.
But RSL is 0ohm in Table 10 on page 25. Why is the RSL 0 ohms at the sample circuit?

Thank you again and best regards,
Kazuya.

Hi Kazuya,

Thanks for your query regarding the slope compensation.

For the reference design in datasheet, the internal slope compensation is enough because of the larger load currents.

But if you have lower load current then external slope compensation will be needed. 

You can check the Quick start calculator regarding slope compensation resistor selection based on different load currents.

https://www.ti.com/tool/download/SNVC224

Best Regards,

Hassan