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I have an application where Vin comes from a battery pack (3V - 7V) but on occasion and under heavy load, Vin drops to just above 2V for ~ 1 seconds before recovering.

There is a 3.3V LDO that provides power to the microcontroller. The LDO uses a pass transistor to regulate the voltage, and due to the body diode, current can travel from output to input when Vout > Vin + 0.2

To prevent the microcontroller from resetting, my intent was to add a capacitor large enough to keep the microcontroller alive until the battery voltage recovered.

However the body diode would conduct at a certain point, and the energy that would have gone to the microcontroller, would now be going to the LDO as well. I am not sure how much energy would be lost here.

This both an energy and cost sensitive project.

What can possible by done other than throwing a diode at the output of the LDO (Would like to minimize that energy loss) ?

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  • \$\begingroup\$ There is. A regular PMOS for reverse polarity won't work in this case since it doesn't guard against the load being able to push current in reverse towards the source. There is a circuit on here with a PMOS that modifies it so that it does. Lemme find it. You will know when you have found it because it uses two matched PNP BJTs. \$\endgroup\$ – DKNguyen Jul 23 '19 at 23:15
  • \$\begingroup\$ Ugh. I can't find it. You can google images for it. I could have sworn it was on this site somewhere. \$\endgroup\$ – DKNguyen Jul 23 '19 at 23:25
  • \$\begingroup\$ I'll see if I can search for something what your describing \$\endgroup\$ – efox29 Jul 23 '19 at 23:40
  • \$\begingroup\$ How does the LDO work when the battery voltage is on the low side of the range you listed, eg 3.2V (range is 3V - 7V)? Just wondering, given that you need higher input than output for an LDO. \$\endgroup\$ – Big6 Jul 23 '19 at 23:52
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    \$\begingroup\$ @DKNguyen electronics.stackexchange.com/q/223935/9409 \$\endgroup\$ – efox29 Jul 24 '19 at 0:33
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There is. A regular lone PMOS for reverse polarity won't work in this case since it doesn't guard against the load being able to push current in reverse towards the source. There is a circuit on here that modifies it so that it does:

Understanding an 'ideal' diode made from a p-channel MOSFET and PNP transistors

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  • \$\begingroup\$ Simulation of this works well enough and the Iq is small enough to not pose much of an issue. \$\endgroup\$ – efox29 Jul 24 '19 at 15:14
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The capacitor needs to be on the input of the LDO so that the LDO keeps operating when the input voltage drops out. You will also need a diode between the battery pack and the capacitor; as @DKNguyen says you can construct an "ideal" diode using a PMOS transistor.

As @Big6 mentioned, you almost certainly need the voltage on the input to the LDO to stay above 3.6V or so if you want the LDO to function properly. It's not obvious how you will do this if your battery voltage really falls to 3.0V.

Even worse, you will need to maintain a significantly higher voltage on the capacitor if you expect to operate through a loss of input voltage for 1s or more. You may need a relatively large capacitor.

At the end of the day, you may be better off replacing the LDO with a buck/boost switching regulator that can provide 3.3V as the input ranges from 2V to 7V.

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  • \$\begingroup\$ Agree with replacing the LDO with a suitable buck-boost. After all, you find them very easily these days, at a good price. Just need to add a couple of resistors, an inductor, and about two caps. \$\endgroup\$ – Big6 Jul 24 '19 at 1:20

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