LP38852: Heat Loss of LP38852MRX-ADJ/NOPB

Hello Ryusuke,

Can you please share the calculation that resulted in a 40°C increment?

What is the regulator's package? {KTW (DDPAK/TO-263), NDZ (TO-220)  or DDA (SO PowerPAD)}

Regards,

Daniel

Hi Daniel,

Thank you for your reply. I will confirm the calculation to the customer.

Their package is DDA. PN is LP38852MRX-ADJ/NOPB.

Best Regards,

Ryusuke

Hello Ryusuke,

Thank you for sharing the package type.

Does the customer have the ability to capture a thermal image of the device mounted in the PCB? 

Was the 1.5C increase observed at the top of the device?

Keep in mind that the junction to case (top) thermal impedance is 54.6°C/W, which is greater than the junction to case (bottom) at 4.3 °C/W, so it is expected to have more heat dissipated thru the thermal pad at the bottom of the device into the PCB. This would result in lower temperature observed at the top of the device.

Regards,

Daniel

Hi Daniel,

I sent the measurement method and calculation formula below.

I would appreciate any comments.

[Test Method]
A point type thermometer is used for the "(contact type) temperature rise comparison test". It is attached to the IC surface with tape.
Device: LP38852MRX-ADJ/NOPB
⓪ Current consumption: 0.9 A
① Ambient temperature outside the enclosure
Di:25.4°C
Di removal: 25.0°C
② Ambient temperature of LP38852 on the substrate
Di:31.3°C
Di after removal: 31.9°C
③ Surface temperature of LP38852
Di:40.0°C
Di after removal: 42.1°C
④ Temperature including Di (3) - (2): 8.7°C
⑤ Temperature when Di is removed (3)-(2): 10.2°C
⑤ The difference between -(4) is 1.5°C

Calculation formula:
θja: Calculation is difficult because copper foil has solid GND.
Calculated using 10 mm*0.25 mm of the surface as the minimum range. (0.4*0.1inch) =0.04
FAN with minimum air flow of 0.69m3/min is attached at both ends, and considerable air flows. Based on the data sheet, it is considered that θja =98°C/W (saturation value).

Relationship between total consumption current and θja:
Vdo:500mV
Pd(pass)=(2.5-0.5-1.5)*0.9=0.45 W
Pd(bias)=(5.3*0.003)= 0.0159 W
Pd(in)=(1.5*0.007)= 0.0105 W
Pd=0.45+0.0159+0.0105= 0.4764 W

ΔTj=125-(45+15) =65°C * Assuming ambient temperature on the substrate is +15°C
Θja<ΔTj/Pd=65/0.4764=136.4°C/W

Evaluation result of consumption current: 0.9 A
θja adopts 98°C/W: 0.4764 W*98°C/W=46.6872°C temperature rise
Tjmax-45°C - 46.6872°C=+33.3128°C

Determine the environmental temperature tolerance X from θja <ΔTj/Pd
98°C/W< (125°C - 45°C - X)/0.4764 W
98<(125-45-33.3128)/0.4764
98<98, and the environmental temperature tolerance is +33.3128°C.

The temperature rise is considered to be about 46°C.
The difference of Vdo is 100 mV⇒500 mV in the actual machine evaluation result.
The difference of IC temperature rise is +1.5°C.

Best Regards,

Ryusuke

Hello Ryusuke,

Thank you for providing the detailed analysis. Please allow an additional business day while I perform the analysis on the provided data.

Regards,

Daniel