Hi,

ADuM3070 datasheet contains following efficiency graph:

I plan to use 5V input voltage and 3.7V output voltage with 5mA maximum (1.5mA typical) load current. I would like to create effective isolated power supply for light loads (1-5 mA). Could you please recommend me right transformer or isolated solution for achieving maximum efficiency at small loads (3.7 V, 1-5 mA)?

Regards,

Ivan

Ivan:

I have looked at the application for high efficiency at low output current and it is very challenging for the ADuM3070 because it was designed for high efficiency at 2W output power, not at 3.7V and 5mA which is 18.5mW output power. The problem is the quiescent current as shown in the datasheet table is typically 3.5mA at 5V in to 3.3V out (close to your 5V to 3.7V application) which uses 17.5mW in quiescent power.

If the efficiency of the dc to dc converter were perfect with no losses in the components, the total input power at 5mA load would be the output power plus the quiescent power or 18.5mW + 17.5mW = 36mW. The overall efficiency would be 18.5/36*100 = 51%, which may not meet your requirement for high efficiency.

If we assume 70% efficiency as stated in the datasheet, the input power to deliver 18.5mW output power would be 26.4mW, and if we add the quiescent power to this the total input power would be 43.9mW. The overall efficiency would then be 18.5/43.9*100 = 42%. This is actually a little better than what you see in the datasheet curve figure 6.

From these two cases you can see that changing the transformer or another component like the schottky diode would have a small effect on increasing the efficiency at this light load condition, because 17.5mW quiescent power loss will dominate.

Regards, Brian