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I need to generate 3.3 V from a 3.6 V battery to power a sensor. I choose the Rohm buck-boost converter BD8306MUV for that, but I discovered that it consumes nearly 5 mA of current in idle mode without a load. The circuit is as recommended by the manufacturer. A novice question; Is this normal, or do I have an error somewhere?

Schematic:

enter image description here

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    \$\begingroup\$ Where in the datasheet do you read 5 mA? \$\endgroup\$
    – MiNiMe
    Commented Nov 23, 2023 at 8:12
  • \$\begingroup\$ @MiNiMe No .. datasheet says 500 uA, but in real life I get 5 mA. I measured it without any load in the output .. that is someway strange. \$\endgroup\$
    – Tiith
    Commented Nov 23, 2023 at 8:16
  • \$\begingroup\$ In the schematic, the thermal pad is connected to RT, is it a mistake? Datasheet does not say anything at all about the pad, so I doubt it should be connected. \$\endgroup\$
    – MiNiMe
    Commented Nov 23, 2023 at 8:29
  • \$\begingroup\$ @MiNiMe no -- thermal pad is connected to the ground. In the reference schematic and PCB layout is thermal pad also connected to the ground. \$\endgroup\$
    – Tiith
    Commented Nov 23, 2023 at 8:35
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    \$\begingroup\$ Ok .. I try to replace with more similar part with the same size as my part right now is. \$\endgroup\$
    – Tiith
    Commented Nov 23, 2023 at 9:23

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Is this normal, or do I have an error somewhere?

Looks normal to me.

Apparently, you copied the first application circuit given on the p.11 of the datasheet. There are some performance graphs on p.13 and onwards. The efficiency graph shows an efficiency of 13~14% for 1 mA output current when the input is 3.7 VDC. This is close to your case for comparison. Now let's work out the input current at this "minimum" load:

$$ \eta=0.13 = \frac{3.3 \cdot 0.001}{3.7 \cdot I_{in}} \Rightarrow I_{in}=6.9 \ \text{mA} $$

1 mA and idle (zero load) are different things but should be enough to give an idea:

For 3.3 mW of output power (3.3V - 1 mA) the circuit consumes roughly 25 mW, according to the performance graphs. Now let's think straightforward: If we assume the 3.3 mW is transferred to the output with no extra loss then the input power at exactly zero load (idle) would be 25 - 3.3 = 21.5 mW, making the input current roughly 6 mA.

So it looks normal to see 5 mA input current at zero load.

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  • \$\begingroup\$ I think that is the case. I choosed just wrong chip with too large standby current for battery operation. Thanks for looking at the problem. \$\endgroup\$
    – Tiith
    Commented Nov 24, 2023 at 7:21

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