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I'm planning to use a MAX640 (Maxim 9 V -> 3.3 V Step-Down DC-DC Converter) to supply a couple of ICs with 3.3 V.

In active mode the current consumption of the ICs are around 20 mA and according to the datasheet the efficiency of MAX640 is around 90 %. Hence the total current consumption will be around (20 mA/ 90 %) = 22-23 mA, right?

But most of the time I want to shutdown ICs (and MCU) to get a consumption of maybe 10 uA.. looking at the datasheet (EFFICIENCY vs OUTPUT CURRENT) it is not even plotted... what will happen? Can I not use the MAX640 if I drop the current consumption that low?

A final question, anyone have a tip of a Step-Down DC-DC converter (or with equal functionality) that also can measure current consumption (analog, spi, i2c)

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In active mode the current consumption of the ICs are around 20 mA and according to the datasheet the efficiency of MAX640 is around 90 %. Hence the total current consumption will be around (20 mA/ 90 %) = 22-23 mA, right?

No, that doesn't make any sense.

Look at the graphs in the data sheet typically this one: -

enter image description here

With a load current of 20 mA (not the chip current consumption) the efficiency is around 90% for the MAX640 so, if the output voltage is 3.3 volts, then the output power at 20 mA is 66 mW and the input power to the device and load is 66 mW/0.9 = 73 mW. So, if the input voltage is 9 volts and a power power of 73 mW is taken, the current is 8.11 mA.

But most of the time I want to shutdown ICs (and MCU) to get a consumption of maybe 10 uA.. looking at the datasheet (EFFICIENCY vs OUTPUT CURRENT) it is not even plotted... what will happen? Can I not use the MAX640 if I drop the current consumption that low?

No, it won't be plotted because there is no need - the device is in standby mode and the load is not powered. In standby, the device takes 10 uA typically and the load is not powered.


However, if your load took only 50 μA, there would still be a significant power consumption just to keep the MAX640 operating in non-standby mode. At 50 μA load current (see the graph above) the efficiency is 50% hence input power will be 2 x 3.3 volts x 50 μA = 330 μW or, from 9 volts, the overall load and device current would be 37 μA.

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  • \$\begingroup\$ But if the device (MAX640) is in standby .. doesn't that mean that all circuits "after" the MAX640 will be completely shutdown? \$\endgroup\$ – RedSmolf Apr 30 at 13:35
  • \$\begingroup\$ @RedSmolf that's what I said in my answer. If your load took only 50 uA there would still be a significant power consumption just to keep the MAX640 operating in non-standby mode. At 50 uA load current (see the graph above) the efficiency is 50% hence input power will be 2 x 3.3 volts x 50 uA = 330 uW or, from 9 volts, the overall load and device current would be 37 uA. \$\endgroup\$ – Andy aka Apr 30 at 13:36
  • \$\begingroup\$ You two are talking about different things. OP is talking about load sleeping. Andy is talking about SMPS sleeping \$\endgroup\$ – DKNguyen Apr 30 at 13:43
  • \$\begingroup\$ @DKNguyen I covered that in my comment directly above. \$\endgroup\$ – Andy aka Apr 30 at 13:45
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    \$\begingroup\$ @DKNguyen - I decided to add that to my answer also. \$\endgroup\$ – Andy aka Apr 30 at 14:35
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The curves don't go down that far because it takes about 50uA just to keep the part running. This is due to the required switching (gate capacitance) as well as the internal processing. Your efficiency would thus be about 17% at 10uA. If this won't work for you, you might be better off switching to an LDO or choosing a part with a discontinuous mode.

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