LDL212

[amansinghaljpr]

LDL212-3.3v DFN (2x2) Running Superhot 

 With LDL212

1. Heat Dissipation (Power Dissipation)

The power dissipation PP can be calculated using the formula:

P=(VIN−VOUT)×IOUTP = (V_{IN} - V_{OUT}) \times I_{OUT}

Where:

• VINV_{IN} = 6.65V (input voltage)
• VOUTV_{OUT} = 3.3V (output voltage)
• IOUTI_{OUT} = 0.22A (output current)

P=(6.65V−3.3V)×0.22A=0.726WP = (6.65V - 3.3V) \times 0.22A = 0.726W

1. Junction Temperature (TJT_{J})

The junction temperature can be calculated using the formula:

TJ=TA+(P×θJA)T_{J} = T_{A} + (P \times \theta_{JA})

Where:

• TAT_{A} = Ambient temperature (let's assume 25°C)
• PP = Power dissipation (0.726W)
• θJA\theta_{JA} = Thermal resistance, junction-to-ambient (for LDL212 package, typically around 65°C/W)

TJ=25°C+(0.726W×65°C/W)=25°C+47.19°C=72.19°CT_{J} = 25°C + (0.726W \times 65°C/W) = 25°C + 47.19°C = 72.19°C

1. Case Temperature (TCT_{C})

TC=TA+(P×θJC)T_{C} = T_{A} + (P \times \theta_{JC})

Where:

• θJC\theta_{JC} = 15°C/W

TC=25°C+(0.726W×15°C/W)=25°C+10.89°C=35.89°C

All these numbers are theoritical 

But in reality the LDL runs in 44 degree zone and easily goes beyond 55 degrees with just 220mA of current , where the exposed pad is connected to GND Zone with an aproximate area of 150^2 mm.

Attaching the landing pattern as well 

[Peter BENSCH]

Welcome @amansinghaljpr, to the community!

Well, you are burning almost 750mW in a DFN 2x2. Have you determined the thermal resistance of the 150mm2 GND cooling area, which has to be added to Rthjc of the device?

Have you considered adding further cooling through thermal vias near the LDL212?

Regards
/Peter

[amansinghaljpr]

Hey @Peter BENSCH  , thanks for reply. 

I just wanted to verify a couple of things . 

1) Are my calculations correct ? 

2) I dont know the thermal resistance of that area , its just a big chunk of copper area connected to the exposed pad intended to increase the surface area for heat dissipation. I thought i had overdesigned , but after looking at the temp i guess some major issues going on 

3) Please enlighten me about any numbers i am missing out on 

Also is 750mW too much heat ? I thought a DFN 2x2 package should be capable enough to handle it 

I have added the schematic i am using as well just for your reference 

[Peter BENSCH]

1. Well, the complete consideration of the effective temperature, especially in your calculation 3 (you called it Case Temperature (TCT_{C})), also includes the thermal resistance of the cooling area, which still needs to be added. Using a thermal camera, you can then look at the heat distribution on the board and probably recognise a hotspot directly at the LDL212, while the temperature quickly decreases as the distance increases.
2. As already mentioned, it is advisable to calculate or simulate the thermal resistance of the cooling area. Regardless of this, you should use additional cooling area, e.g. on the other side of the board, and as many thermal vias as possible, especially in the close vicinity of the LDL212 to distribute the heat as widely as possible. You can also consider using a thicker copper layer, which makes the board more expensive but also improves the heat transfer to the cooling area (heat build-up around the LDL212).
3. With 150mm² you seem to have a lot of cooling surface, but as you realise, this is obviously not enough. The amount of heat generated and its removal is often underestimated. To get an impression of the heating of a device, it is sufficient to load e.g. an ohmic resistor with e.g. 750mW and measure the resulting temperature (or hold it in your fingers). You will find that the temperature is inversely proportional to the component size, which is simply due to the thermal resistance of the device (and also the thermal resistance of the environment, e.g. the fingers). Yes, the DFN 2x2 package can handle this power dissipation - if you ensure appropriate cooling.

4. Speaking of heat distribution: if you know the current consumption of your 3.3V load very well, you can possibly change R23 from 0ohms to a larger value. At 8.2ohms, for example, a voltage of 1.804V drops at 0.22A, i.e. a power loss of 397mW. The power dissipation that the LDL212 has to burn is then reduced by this value. Of course, you then have to use a resistor with a corresponding power capacity - and ensure that it is cooled. So the problem remains the same, it is just distributed.

Hope that helps?

Regards
/Peter