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I'm struggling to understand where 5.7mA may be leaking in this very simple circuit based on ST1S10 step-down regulator and LP2985IM5-3.3 LDO: enter image description here

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Long story short: ST1S10 drops down the voltage to 3.8V to supply the modem and the LDO that in turn drops down the voltage to 3.3V to supply the microcontroller. There is nothing connected to the circuit, the ST1S10 is loaded only with LP2985IM5-3.3 that is turned on but idle (not loaded by anything).

When connected 6V to K1 (left side) I observe the current of 7.4mA. When I connect the voltage in place of the PWR_FLAG near C7 I observe the current of 1.7mA consumed. That would indicate that the LP2985IM5-3.3 consumes this 1.7mA (another question is if it's possible that it consumes that much current doing nothing - I couldn't identify this info in IC's application note).

That is under assumption that no power is consumed by ST1S10 when it has revere voltage applied to SW (pin 7). May not be true.

But anyway, from the simple math comes the conclusion that 5.7mA (7.4 - 1.7) is lost somewhere between PWR input and C7. Is that possible?

Of course the next step is desoldering the circuit to look for the leakage, but maybe there is no point and it just seems correct.

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2 Answers 2

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That looks pretty good to me. The datasheet indicates that the chip draws 1.5-2.5 mA when not inhibited and not switching, and if it only draws another 4 mA when switching, I'm actually rather impressed.

The additional current is mostly used to charge and discharge the gates of the two big (5A peak) switching transistors inside the regulator chip. This current ultimately gets dumped to ground; it can't contribute to the output current.

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  • \$\begingroup\$ Ok, it seems I will have to live with it and replace switching step-down with a linear LDO. Thanks for the explanation! Daily (taking into consideration the duty cycle of my system) the sole idle power module of my device consumes more (0.192A) than the microcontroller altogether with sensors and GSM modem (0.16A), that's sad :( \$\endgroup\$
    – eltomek
    Jan 26, 2014 at 17:21
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Check the LP2985 datasheet carefully.

As the input voltage on an LDO regulator is brought closer to the desired output voltage, its current consumption increases dramatically (because it has to start saturating the regulator transistor) With only 0.5V between input and output, that might be happening here.

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  • \$\begingroup\$ Would you help me to understand which chart in the datasheet (ti.com/lit/ds/symlink/lp2985-n.pdf) shows the current depending on the input to output voltage? In my case the voltage difference is 500mV while this chip's minimum value is 300mV, I'd agree that's quite close. \$\endgroup\$
    – eltomek
    Dec 4, 2014 at 20:11
  • \$\begingroup\$ The datasheet doesn't show it exactly, but take a look at graphs 18 and 19. Graph 19 shows Ignd (which is mainly base current for the power transistor) increasing as output current increases - to about 1.1ma for 150mA out. But notice Vin = Vout+1V here. Graph 18 shows a jump (0.3ma) in Iin as Vin reduces, as it tries to saturate the output transistor. But note, that's at a 1ma output current, you can expect worse at higher currents. But these only hint : no graph shows Ignd for 150mA out as you reduce Vin. You may have to measure that yourself. \$\endgroup\$
    – user16324
    Dec 4, 2014 at 21:33

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