Switching off an LDO to avoid damages?

Question

Say I'm using an LDO to power an MCU such as the ATmega328p with 3.3V from 3 NiMH AA batteries.

What it isn't clear to me how to deal with brown-out situations.

I can think of 2 possible risks (i.e. when using an LDO such as the MCP1700):

1. deep discharge of the batteries when their voltage drops to MCU brown-out levels (say 2.7 V) and the MCU continues to draw some current in reset mode
2. damages to the LDO itself when V_in drops below V_out

Is it a good idea to look for LDO with on/off pin (such as the S-1132) and connect it to a voltage detector such as the MCP112 or TC54?

In that way one would work-around the brown-out detection of the MCU, and basically implement an external brown-out detection that just switches the MCU off (instead of keeping it in reset).

Or are there LDOs available that already include a voltage detector and just turn off V_out if V_in drops below a threshold?

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Regarding LDO damages, I stumbled over a paragraph in the S-1132 datasheet:

In the S-1132 Series, a low on-resistance P-channel MOS FET is used as the output transistor. Be sure that VOUT does not exceed VIN + 0.3 V to prevent the voltage regulator from being damaged due to reverse current flowing from the VOUT pin through a parasitic diode to the VIN pin, when the potential of VOUT became higher than VIN.

(Section Operation, page 13)

How would V_out exceed V_in? I mean, I thought if V_in drops to V_out + V_dropout or less, then the LDO works in its dropout zone where it isn't regulating anymore, i.e. V_out has some linear relation to V_in. See for example the graph for the S-1132VB50 (V_out=5V) variant (page 16), where V_Out/V_in pretty much equals 1 for V_in=2.5..5:

So is the paragraph warning against (accidentally) connecting another power-source to V_out that exceeds V_In+0.3V?

Or is it possible to damage an LDO somehow merely by letting V_in drop under a minimum threshold for an extended time?

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Another thing I don't understand in this datasheet are the output voltage/output current graphs on page 16, e.g. for S-1132B30:

I mean, e.g. for I_out=400 the curves have two V_out values?

Answer 1

The LDO won't be damaged if VIN falls -- VOUT will follow VIN down. There is an extreme case where if you had a huge (say 1000 uF) capacitor on VOUT, and VIN was suddenly shorted to ground, then the 1000 uF would discharge through the LDO and potentially damage it. With the normal values of VOUT capacitor, this won't be an issue.

If you are concerned about over discharging the NiMH cells, then you will need a lot voltage detector/protector.

Answer 2

Vout can exceed Vin in circuits with complication power supply schemes, such as multiple inputs. You may, for example, use USB power as well as a battery, and create a situation where your MCU is powered by the battery, but the 3.3v regulator accepting USB power has its input unpowered. This is unhealthy for your LDO.

This situation can also happen transiently when your power supply input is switched off, but your circuit has a large tank capacitor with a very low discharge rate somewhere on the 3.3V rail. This could be large enough to keep the voltage present on 3.3V long after 5V is gone.

In general, for non-critical circuits, these problems have fairly simple workarounds.

A reverse diode across the LDO can send your 3.3V to the input, and make sure that the the reverse voltage seen by the LDO is only 0.7V or so. Many LDOs these days include the diode inside them.

Wherever you have bulk capacitors, it can be generally safer to have a resistor in parallel to accelerate the discharge of the capacitor when the power disappears. Note that this resistor will always discharge, though, so it can be terrible for efficiency / battery life.

If you care about battery life / efficiency, or you have safety or other considerations which prevent you from simply pushing your 3.3V power to the external DC input side, you may need to add active measures instead. These measures can be relatively more expensive. You may need to choose an LDO with an enable pin and/or add a mosfet to discharge bulk capacitors. Many LDOs and switching controller ICs include input voltage monitors and output voltage monitoring, and provide some of this information on pins like Power Good (PG) and similar. You can use this digital outputs from the regulator to control your logic to turn on / off the appropriate switches.

Usually, a brownout isn't going to cause damage to most ICs themselves. It might cause memory corruption or make your MCU's firmware behave unusually, and this unusual behavior could, in certain specific situations, cause damage in some other part of the circuit. Turning off power to the MCU isn't necessary if you can detect the brown out and ensure your firmware knows it should not rely on any of the inputs it is seeing and should avoid trying to write to memory. Holding the MCU in reset is a good way to do this, and the brown out detectors are included for that very reason.