LMV331: Thermal calculation method

Hello Tomoaki,

Which package are you using? This is very important. Is this the LMV331 or LMV331-N (ex National version?)

Any power dissipated by the die will raise the die temperature above the ambient.

You need to find the total power dissapated in watts. P=I*V. This includes the comparator quiescent current and the load current.

So if the quiescent current of the LMV331 is 150uA, and you have a 5V supply:

 150uA * 5V = 750uW

So that is 750uW for just the comparator supply current alone.

Load current will also cause dissipation. Load current dissipation is calculated from the voltage across the output and the load current.

So if you have a 10mA load current, the voltage drop will be about 280mV (from Fig 3 of the LM331-N Datasheet):

 10mA * 280mV = 2.8mW

Total power dissipation is 750uW + 2.8mW = 3.55mW

To find the temperature rise, you multiply the power dissipation by the package thermal resistance.

So assuming the worst-case SC-70 (DCK) thermal resistance of 478°C/W:

 3.55mW * 478 = 1.7°C

So the total power dissipation the die to increase in temperature just 1.7°C over the ambient temperature.

The assumption with low power devices like comparators and op-amps is Ta ~= Tj. So if the ambient temp is 35°C, the die temp would be 36.7°C.

If the load was 50mA, this would result in a 19.1°C rise over ambient (or 54.2°C). So the load makes the biggest difference.

The above is also assuming continuous dissipation. If you had a 10% sinking duty cycle, then the output power would only be 10% (but you still have to stay within abs-max current).

If this were a multi-channel device (dual/quad), then the supply current for all the comparators and all the loads must be combined.

For more info about package thermals, please see the Packaging Appnotes section:

 http://www.ti.com/support-packaging/packaging-resources/SMT-and-application-notes.html#image