Why does the L5973D burn up if I turn R1 into 10K in this circuit?

[Jessicale]

Hi, I want to ask for your help on my L5973D.

I used to use L5973D to produce voltage of 3.3V/4.2V according to the typical circuit in the datasheet. The figures are same while the feedback resistance are different. The input is 24V DC. They worked well without anything wrong.



(Test application circuit in the datasheet)

However, I met some problems when I want to produce 5V voltage according to the circuit above. I tried to turn the R1 into 10K. Theoretically,it will produce 5V, but when I connected it with 24V power,L5973D burned up. In the same circuit, the voltage is 3.3V when R1 is 5.6K. It worked well. I am wondering why...So I’m here to ask for your help.Sorry for my poor English,but can anyone give me some suggestions?

Here are some details of L5973D:

Series:L5973

Operating-Temperature:    -40°C ~ 150°C (TJ)
Mounting-Type:    Surface Mount
Output-Type:    Adjustable
Function:    Step-Down
Supplier-Device-Package:    8-HSOP
Number-of-Outputs:    1
Topology:    Buck
Frequency-Switching:    250kHz
Current-Output:    2.5A
Output-Configuration:    Positive
Voltage-Output-Min-Fixed:    1.235V
Voltage-Output-Max:    35V
Synchronous-Rectifier:    No
Voltage-Input-Min:    4V
Voltage-Input-Max:    36V
Output-Voltage:    35 V
Input-Voltage-MAX:    36 V
Output-Current:    2.5 A
Switching-Frequency:    212 kHz to 280 kHz

 

[Elex518]

Based from datasheet of L5973D, the pin5 should get 1v2, replacing R1 to 10K will get pin5 to 0v8 only and can possible stress the regulator and blown. Recalculate the voltage to pin5 that it can get 1v2. If you want use 10K for R1 and regulate 5V, your R2 should be 31K (based from my calculation). 

Check here for more repair tips --> https://elex518.blogspot.com/