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I am trying to understand my LDO current consumption when the system is sleep. So we are using the ld6806 to output 3v, with input 2-AAA batteries in series .

(should we use the 2.9v LDO for 2-AAA batteries? i know that after a while they are under 3v? should we use LDO at all ?)

Most of the time the mcu and rest of the system is sleep , and i couldn't understand from the data sheet , what would be the current consumption when nothing is working, only batteries connected to regulator ?

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The actual current consumed by the LDO is given in the datasheet as the quiescent current.

In table 10 (page 7) we see:

Quiescent current for LD6806

So at typical loads (some current but Iout < 200mA) you're looking at a negligible current of 250uA maximum to power the LDO itself.

From TI's helpful paper "Understanding the Terms and Definitions of LDO Voltage Regulators" the definition of quiescent current is:

Quiescent current consists of bias current (such as band-gap reference, sampling resistor, and error amplifier currents) and the gate drive current of the series pass element, which do not contribute to output power. The value of quiescent current is mostly determined by the series pass element, topologies, ambient temperature, etc.

Note I am assuming that you are not using the shutdown / enable feature of the LDO. If you are you should look for the standby current in the datasheet instead.

Do you need an LDO? It depends. You don't mention what MCU you are using, but a typical low power 3.3V microcontroller can run directly from batteries without regulation. You will need to take the following into account:

  • The maximum voltage from all of your batteries when fully charged, compared to the maximum allowable voltage for all parts in the system.
  • The minimum voltage when your batteries are reaching the end of their life, compared to the minimum voltage all parts in the system require.

There are other drawbacks to an unregulated supply, for example if you intend on using an internal analogue to digital converter which needs a fixed reference. However these limitations can often be worked around without the need for an LDO.

If you do use an LDO you need to ensure that the minimum input voltage never drops below the output voltage plus the dropout voltage. In this case the LD6806 has a typical figure of 60mV, which is very small. So you need to ensure your batteries never drop below 3000mV + 60mV (3.06V).

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  • \$\begingroup\$ Wow what a great and helpful answer. Thanks a lot. Mcu used is Atmega328p, it also has to read battery voltage. I feel for some reason that not using regulator is not safe or proffesional, but from fully charged aaa batteries you get 3.3V and after some time they DO go below 3v, so i wonder whats the point of a regulator here.. \$\endgroup\$
    – Curnelious
    Oct 31, 2015 at 15:40
  • \$\begingroup\$ The LDO current consumption in case of application in sleep mode is: quiescent current + (current draw of you application * voltage difference). If the input voltage gets below the regulation voltage the LDO will still create a voltage drop, the output voltage will drop linear with input voltage minus voltage drop and you will still have a power dissipation of voltage drop multiplied by the drawn current. \$\endgroup\$
    – optronik
    Oct 31, 2015 at 16:14
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    \$\begingroup\$ Operating voltage of Atmega328p is 1.8 to 5.5V if you do not have other components in your application that have a tighter supply voltage range I would not use a regulator at all. If there is a danger of having the batteries reverse connected I would simply add a series diode in the positive supply path or a 5V Zener diode in parallel to your battery input. \$\endgroup\$
    – optronik
    Oct 31, 2015 at 16:16

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