LTC2870IFE Thermal Data

hi Naveen Afzal

Thanks for your brief explanation regarding Power dissipation !
But it seems like you have taken ICC value from LTC2871 Characteristics graph. We are using LTC2870IFE for our design. Is it right way to take ICC value from the graph you have mentioned? If this graph cannot be used to determine supply current of LTC2870IFE, kindly suggest the other way.

Hi Rama,

You're welcome. You can take the ICC value from the graph in the row above it.  (50mA worse-case)

Power dissipated in the device (PDISS) = Input Power (PIN) – Output Power (POUT)

Input power, PIN = VCC * ICC

Output power, POUT, is what is dissipated in the load resistor. POUT =  VOD^2/RL

Reworking the numbers out for the VCC =  3.3V  (max VCC = 3.6V)

PDISS = PIN – POUT

PDISS = (VCC * ICC) – (VOD^2 /RL)

I spotted another mistake here:

VOD is 2.4V in this case not 2.2 at RL=100 ohms

PDISS = (3.6V * 50mA) – (2.4^2/100)

PDISS = 180mW – 57mW

PDISS = 123mW

-Naveen

Hi Rama,

You're welcome. You can take the ICC value from the graph in the row above it.  (50mA worse-case)

Power dissipated in the device (PDISS) = Input Power (PIN) – Output Power (POUT)

Input power, PIN = VCC * ICC

Output power, POUT, is what is dissipated in the load resistor. POUT =  VOD^2/RL

Reworking the numbers out for the VCC =  3.3V  (max VCC = 3.6V)

PDISS = PIN – POUT

PDISS = (VCC * ICC) – (VOD^2 /RL)

I spotted another mistake here:

VOD is 2.4V in this case not 2.2 at RL=100 ohms

PDISS = (3.6V * 50mA) – (2.4^2/100)

PDISS = 180mW – 57mW

PDISS = 123mW

-Naveen