LM5071: Application clarification

Gerald

Part Number: LM5071
Other Parts Discussed in Thread: TPS23755

Hi Experts,

We are looking at the LM5071 application circuit diagrams on the datasheet, specifically the non-isolated buck configuration
https://www.ti.com/lit/ds/symlink/lm5071.pdf

There's a few things we wanted to confirm:

1. What are the recommended ratings for the parts used?

2. How to connect the circuit if no temperature compensation is required?

3. Just wanted to double check, is R18 meant to be 1.24 ohms, or is it another value?

Our customer is basically just trying to implement a standard class 0 5V buck configuration using the LM5071, as we'd imagine that would affect some of the part specs.
We hope that you can help us. We will wait for you advise/confirmation. Thank you.

Kind regards,

Gerald

over 3 years ago

+1 Diang Xing over 3 years ago

TI__Mastermind 37121 points

Hi Gerald,

Nice to talk to you again!

Another 13-W 5-V Vout Buck converter example (using TPS23755) you could find is:

/cfs-file/__key/communityserver-discussions-components-files/196/TPS23755-Buck-PSIL048E1_2800_001_29005F00_Sch-_2800_3_2900_.PDF

1. For the part ratings, you could find the part# from the design above.

2. You can direct connect the base of Q2 to the 5-V RTN.

3. According to the blocking diagram of LM5071, without the temperature compensation, to have 5-V V_o and assuming V_EB of the PNP is 0.7 V, you should have the relationship of R1 and R2 as shown below.

Best regards,

Diang

0 Gerald over 3 years ago in reply to Diang Xing

TI__Genius 9570 points

Hi Diang,

Thank you for the response. Apologies, but the customer decided to use the temp compensation after all, so would R2 be 1.24R and will the equation be the same in this case?

Another question actually. Should the LM5071 COM be isolated from the rest of the circuit? Thank you.

Kind regards,

Gerald

+1 Diang Xing over 3 years ago in reply to Gerald

TI__Mastermind 37121 points

Hi Gerald,

If you connect the temp compensation PNP back, your circuit will become the case below:

V_EB and V_BE cancel each other. Let's use Vo = 5V, R1 = 5k, then we should have R2 = 1.25k. So I guess the R18 in the schematic should be 1.24k rather than 1.24.

The COM and 5V RTN should be different ground. Your load circuit should refer to 5V RTN and be isolated with COM.

Best regards,

Diang

0 Diang Xing over 3 years ago in reply to Diang Xing

TI__Mastermind 37121 points

Hi Gerald,

Since your question has been answered, I will close this question for now. If you have any further questions, please open a new thread and we will be glad help you!

Best regards,

Diang

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LM5071: Application clarification