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LDO heating up

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As I suggested a few posts ago, the regulator doesn't do much for you. Here is a sim of a "white" 3.2V LED discharging a battery. The Battery voltage is the X-axis. Notice the current I(D1). The battery discharge current goes to ~zero naturally when the battery reaches ~2.4V. The LED will be visibly bright while the LED power is between 50mW and ~20mW.

100's of millions of solar powered garden lights cannot be wrong... ;)
 

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At least I will have a stable voltage supply to the leds
No, you won't. Read Mike's explanation in post #19 carefully.
And yes, you can place the via as shown.
 
How can I apply such a voltage driven shutdown? Quote from "http://dics.voicecontrol.ro/process_mails/arata_discutia/134698/LDO_with_low-voltage_cutoff.html":

Some regulators have shutdown pins. Shutting down the regulator when
Vout OR Vin got too low would work.

This may be able to be just 2 x r plus a bipolar.
Almost certainly precise enough for this task.
Check at desired temperature extremes.
3 small components.

Or a power supply monitor IC driving shutdown.
Possibly 1 small (3 pin) device.

Or 2 x R and a TLV431 or TL431. Or (extra points) a bipolar and 2 r's
connected to shut down when Vdropout got too low. (Noting that
Vdropout MAY increase in dropout with some ICs).

For a regulator with no disable a preceding bipolar or FEt with
appropriate drive. Boy count getting up BUT one bipolar and 2 x R may
suffice.
 
You missed my point. You don't need an automatic cutoff. It happens automatically as the battery discharges below a voltage where it stops drawing current...
 
What if the battery discharges too often below the current drawing level, could it be damaged? Bear in mind, that I have three batteries in series, so one battery may fall down to that level, but the other two must not - so the battery circuit still will be closed and current will flow through the discharged battery. I wanted to prevent such an occurrence.
 
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