LM74202-Q1: LM74202 Overload R(ILimit) understanding

Sanket Patadiya

Part Number: LM74202-Q1

Hi,

I am Sanket Patadiya from Volansys Technologies.

We are working on one design including the LM74202 part for protection circuitry.

Have a look at the circuit below.

I would like to understand the functionality of the LM74202 when we set R155[R(ILimit)] = 3.9k (datasheet recommends >5.36k).

What will be the ILimit if we have R(ILimit) = 3.9k?

Will it be 0.055A considering R(ILimit)=Open or 0.095A considering R(ILimit)=Short?

Please let us know the functionality for R(ILimit).

Thanks & Regards,

0 Rakesh over 2 years ago

Hi Sanket,

Thanks for reaching out

3.9k sets 0.095A current limit

Best regards

Rakesh

0 Sanket Patadiya over 2 years ago in reply to Rakesh

Hi,

Thanks for the quick response.

Can you provide us an understanding of the R(ILimit) using the functional block diagram?

Like How the LM74202 sets the ILimit to 0.095A when R(ILimit) is 3.9k?

Thanks & Regards

0 Sanket Patadiya over 2 years ago in reply to Sanket Patadiya

Hi Rakesh,

Please refer to the image of a functional block diagram of LM74202.

I want to understand the functionality of R(ILimit) using the block diagram. How the LM74202 sets the ILimit to 0.095A when R(ILimit) is 3.9k?

Please let us help with that.

Let me know if you have any questions.

Thanks & Regards

0 Rakesh over 2 years ago in reply to Sanket Patadiya

Hi Sanket,

When Rilim value is <5.36kOhm, the "short detect" block of the controller detects it as short and sets the current limit reference as 95mA.

Best Regards,

Rakesh

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