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I have a system that is powered with 3 AA Alcaline batteries Varta Industrial Pro (1,5V@2950mAh).

The batteries are connected to a this 3V3 LDO that supplies power to a MCU. I have taken a minimum dropout of 150mV (so LDO would work down to 3V45). The MCU system will activate (using a FET) a electronic lock of this type, which consumes a maximum of 1000mA during 1 second and admits a voltage from 3V to 5V. As my LDO only supports up to 250mA I would like to know.

I would like to know if there would any problem connecting the lock directly to the three batteries power supply, as they are a not fixed voltage power supply.

In case there's no problem, would these three batteries in series be capable of maintaining the required voltage level when the lock is activated?

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  • \$\begingroup\$ The lock admits from 3v to 5V \$\endgroup\$ – LazyTurtle May 27 '20 at 16:39
  • \$\begingroup\$ @jsotola My question is not about the lock maximum ratings, is about connecting the lock directly to the batteries and not using a voltage fixed power supply \$\endgroup\$ – LazyTurtle May 27 '20 at 16:43
  • \$\begingroup\$ @jsotola You're right. Thanks for that. I've changed the link. For some reason I put the wrong one. \$\endgroup\$ – LazyTurtle May 27 '20 at 16:47
  • \$\begingroup\$ it is unclear what problem you are having ... if the lock operates on a voltage between 3 V and 5 V, why would you think that 4.5 V is somehow excluded from that range? \$\endgroup\$ – jsotola May 27 '20 at 16:48
  • \$\begingroup\$ The problem is I have never used this type of mechanisms and I don´t know if they need a fixed voltage source or what implications are there if you connect it directly to the batteries. On the other hand, if there's a voltage drop due to the load demanded by the lock, maybe the LDO will not have enough input voltage to give the 3V3 to the MCU \$\endgroup\$ – LazyTurtle May 27 '20 at 16:51
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It is a much better design to connect the lock directly to the batteries, this way the lock operation will not inject current spikes or other disturbances downstream of the LDO where the MCU is connected.

When the batteries are new they will maintain the appropriate voltage level (3-5V there is ample margin). At some point when they become depleted the lock will not operate anymore and the MCU may reset if the voltage goes too low during the lock operation.

On this site, you can compare the voltage vs current of AA batteries https://lygte-info.dk/review/batteries2012/CommonAAcomparator.php

set the current to 1A instead of the default 2A

They list some Varta models but not your specific model, but Varta should be able to supply the curves

When the battery is new enough, the voltage per cell drops with 1A drawn but is above what you need as there is a need of at least 1.15V per cell to keep the LDO working (3 x 1.15 = 3.45).

At some point during the discharge, the 1A curve crosses down the 1.15V level, the batteries will have to be replaced to maintain proper operation of the LDO + MCU.

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  • \$\begingroup\$ Thank you very much for your answer. And can the batteries support that current spike during a maximum of 1s without voltage drop? \$\endgroup\$ – LazyTurtle Jun 3 '20 at 17:08
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enter image description here If you were using a switching power supplier buck/boost you can use 3 AA Batteries maximum

enter image description here

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    \$\begingroup\$ I have no idea how this relates to the question \$\endgroup\$ – pipe Jun 6 '20 at 15:41
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    \$\begingroup\$ Adding slightly more explanation will help people see the relevant points you are tring to make. \$\endgroup\$ – Russell McMahon Jun 7 '20 at 8:50
  • \$\begingroup\$ You are using LDO that reduce significantly lowers battery life \$\endgroup\$ – Koren Reuben Jun 7 '20 at 15:42

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