Hello group, I am not a step down converter specialist and I would like your opinion on the true consumption of TPS62203 http://www.ti.com/lit/ds/symlink/tps62207.pdf
I need to do 3.3 VDC from 6V (4 AA batteries) to power an autonomous system. When this system is in sleep mode it consumes 1 µAmp (MSP430 in LPM3 mode). The data sheet curve stops at 50 µA. So according to you how much the regulator would consume when it has to supply 1µ Amp.

 

The efficiency at 55μA (182μW) output is about 65% according to the chart.
This means the input power is approximately 182μW/.65 = 279μW.
The power used by the converter is then 279μW - 182μW = 97μW.
Assuming the converter power doesn't change significantly for a 1μA (3.3μW) output, the input power would be 98μW + 3.3μW = 102μW.
This would give an input current from the 6V supply of 102μW/6 ≈ 17μA.

 

The efficiency at 55μA (182μW) output is about 65% according to the chart.
This means the input power is approximately 182μW/.65 = 279μW.
The power used by the converter is then 279μW - 182μW = 97μW.
Assuming the converter power doesn't change significantly for a 1μA (3.3μW) output, the input power would be 98μW + 3.3μW = 102μW.
This would give an input current from the 6V supply of 102μW/6 ≈ 17μA.

Crutschow thanks for your answer .
One question more : Do you think I have to add the Typical Quiescent Current (15µA) ?