Output load current less - DCDC Converter [closed]

Question

I'm using this Boost converter. 3.6V to 12V with a load current 300mA. 10uH is the inductor used.

But I am only able to get a maximum load current of 120mA. What might be the reason?

Answer 1

I've drawn an orange circle on the efficiency graph (input supply voltage of 3.3 volts): -

• That orange circle is my estimation of where the 12 volt output graph might be somewhere near its current limit.
• I've chosen the 3.3 volt input supply graph because it's likely your power source will not be able to sustain 3.6 volts under the heavy current needed to power your expected load (circa 1 amp from the source).
• 12 volts on the output (the orange circle) corresponds with about 200 mA load current and this is an output power of 2.4 watts
• The efficiency is only about 70% hence the power dissipated in the chip might be as high as 1.03 watts
• The efficiency will get worse at higher load currents.
• If 300 mA could be delivered, the output power would be 3.6 watts and the chip dissipation would probably be more like 2.4 watts.

Given that the thermal characteristics of the device will cause it to warm by 90°C / watt it will rapidly overheat and shut down.

In other words, the chip is unlikely to be suitable for your application without some form of heatsinking (if that is indeed possible).

Answer 2

Pout = 12v * 0.3A = 3.6W therefore Pin = 3.6V * 1A minimum + losses=20% thus 1.2A in

Report all V,I's accurately during switching with case temperature and cooling method for 20% of 4.2W = 820 mW and needs cooling heatsink , with Cu plane or equiv. See datasheet for details.

Comparing Voltage gain and impedance gain, expecting 300 mA demands very low DCR inductor rated for >3A but 75% efficiency is expected rather than 80% .