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I am regulating a lithium ion battery to 3.3V and I want the regulator output to shut off when the regulator can no longer supply 3.3V. I am having trouble finding an IC that would do this, and this charger with a low battery disconnect is all I could find: http://cds.linear.com/docs/en/datasheet/4071fc.pdf

It says there is low battery disconnect from VCC at 3.2V. It says under "Applications" that it is for energy harvesting, so I don't understand if it's really suitable for what I want to do. I need a charger in my circuit too, so that would be useful. I want to make sure I'm looking at this correctly...is there a better way to do this?

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  • \$\begingroup\$ Do a search on LDO regulator. TI Maxim and Linear Tecnologies all have multiple low dropout regulators. \$\endgroup\$
    – steverino
    Commented Jun 7, 2015 at 21:31
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    \$\begingroup\$ I want "undervoltage lockout" when the regulator output is below its specified 3.3V, and I don't see that. \$\endgroup\$
    – cheeto
    Commented Jun 7, 2015 at 21:37
  • \$\begingroup\$ I'm confused. If you have a lithium iron battery connected to the input of a regulator with a 3.3 volt output and the output of the regulator is connected to some load, wouldn't it make more sense to disconnect the battery from the regulator when the regulator's output voltage fell below 3.3 volts, then connect it to the charger and charge it until it got topped off and then disconnect it from the charger and connect it back up to the regulator? \$\endgroup\$
    – EM Fields
    Commented Jun 7, 2015 at 21:44
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    \$\begingroup\$ Lots of LDO parts have an enable input, that may be used to turn the regulator off. Set up a reference feed it to a comparator and latch the part off. \$\endgroup\$
    – steverino
    Commented Jun 7, 2015 at 21:52
  • \$\begingroup\$ It doesn't matter to me whether the battery is disconnected from the regulator or the regulator output is off because I want the device I'm making to be portable, so the input voltage to the charger wouldn't be connected necessarily at the time it's needed. The device would be off and the user would know that it's time to charge up. The reference concept is confusing to me because the battery would be input to the reference, and if the battery voltage is dropping then the reference would be affected, right? \$\endgroup\$
    – cheeto
    Commented Jun 7, 2015 at 21:58

2 Answers 2

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Why not you are going with a Switching regulator. Instead of LDO if you will go with the switching regulator, the backup for the device will be more since the efficiency of switching regulator is better than that of LDO's.

For example you can use TPS62240DDCRG4. This regulator has the provision of under-voltage lockout. If the voltage at enable pin is less than 1.85 Volt the device will be shutdown. In your case the regulator should shut down at 3.2V, so you can use a voltage divider at regulator Enable pin in such a way that when your battery voltage is 3.2V, the enable pin should sense a voltage of around 1.8V.

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schematic

simulate this circuit – Schematic created using CircuitLab

There are a lot of situations where a reference voltage is created inside a circuit to insure it behaves correctly under a wide variety of conditions of input voltage and temperature. As an example, you want your multimeter to measure voltage correctly from the time you put in a new battery, until you have to replace the battery. Or a temperature sensor should give the same output at the same temperature regardless of the battery pack. There are simple circuits to do this. Google zener voltage reference. And there are more sophisticated versions for more demanding jobs. see bandgap reference. These are heavily enough used that these circuits are made into 3 terminal parts. A commonly used one is the tl431v. A quick search turned up the circuit shown in eevblog for low battery cutout. Adjust r2 and r3 to get the setpoint to kill vout. http://www.eevblog.com/forum/projects/li-ion-battery-low-voltage-cut-off-circuit-needed-for-project/ Note, I used an SCR since there is no 3 terminal voltage reference.

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  • \$\begingroup\$ Thanks for the information. After searching for a while I found a low cost Microchip IC (MCP1825) that is a 3.3V regulator with a "power good output" that "is an open-drain output used to indicate when the LDO output voltage is within 92% (typically) of its nominal regulation value". This is the datasheet: ww1.microchip.com/downloads/en/DeviceDoc/22056b.pdf That could be a substitute for using a comparator shut off, right? I want to use as few parts as possible, so that seems more convenient. \$\endgroup\$
    – cheeto
    Commented Jun 8, 2015 at 0:45
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    \$\begingroup\$ This is not really a micro-power circuit since it loads the battery with 47k + 18k all the time, even after it cuts off the power from the rest of the board. Depending on what the OP's intention is, it might work for him or her, but it does not protect the battery from over-discharge, and it will result in terrible standby life for the device (unless the battery is huge). \$\endgroup\$
    – user57037
    Commented Jun 8, 2015 at 1:56
  • \$\begingroup\$ To @cheeto, the powergood signal could possibly be useful to turn off a FET and prevent over-discharge of the battery. But what will cause the FET to turn back on? You need to think it through to make sure that the regulator will be re-enabled or the FET will be turned back on after the battery is charged back up. \$\endgroup\$
    – user57037
    Commented Jun 8, 2015 at 2:51
  • \$\begingroup\$ There is a shutdown pin on the MCP1825 that disables the LDO, so I can input the power good signal to that pin to disable the LDO until the battery voltage is high enough again. \$\endgroup\$
    – cheeto
    Commented Jun 8, 2015 at 3:29

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