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Suppose I have a circuit running at 3.30V, supplied via an LDO regulator (e.g., ADP124), powered by a Li-ion battery. In other words:

Layout

Now, I would like to implement an undervoltage cutoff to "turn off the power", because of these two coexistent reasons:

  • This prevents the load (i.e., microcontroller, ADC, etc.) running at anything less than a regulated 3.30V
  • Equivalently, this also prevents the battery from discharging to anything lower than 3.50V (i.e., equal to 3.30V + 0.20V dropout) thus ensuring that it doesn't enter deep discharge.

Question 1: Which of the following options is preferable for the undervoltage monitor's sense-LOCATION (i.e., point where it measures the voltage)?

  • Option 1) Option A1
  • Option 2) Option A2
  • Option 3) Other?

Question 2: Which of the following options is preferable for the undervoltage monitor's switch-ACTION (i.e., the way it implements the cutoff)?

  • Option 1) Option B1
  • Option 2) Option B2
  • Option 3) Something else?

To me, Option 1 appears best/most-direct for both Question 1 and 2 above, but perhaps I'm missing some considerations.

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  • \$\begingroup\$ What's the part number of the regulator? (Or is this question about the generic case?) \$\endgroup\$ – Rocketmagnet Sep 12 '12 at 9:34
  • \$\begingroup\$ @Rocketmagnet: A generic case LDO regulator indeed; but I've updated the question with the particular example of a very-low-dropout one (ADP124) that I have in mind. \$\endgroup\$ – boardbite Sep 12 '12 at 9:57
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Question 1: Option 1. This is a more reliable indicator of the actual battery voltage. The drop out of the regulator will be affected by current.

Question 2: Use the regulator's enable input if it has one. This saves you having to add an extra transistor. This will reduce the amount of current drawn by the regulator.

There are two ways you might implement this:

1) Use a potential divider to create the Enable voltage. When the output voltage of the divider goes below the enable threshold voltage, then the regulator switches off. The problem with this is that the voltage threshold of the enable input is pretty flexible:

Enable Input Voltages

The threshold is somewhere between 0.4v and 1.2v! This isn't going to be accurate enough for your purposes.

Instead, I'd suggest using a comparator:

Voltage Comparator

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  • \$\begingroup\$ The comparator tip for this type of problem is new to me and much appreciated! (Although I do have to note that since the undervoltage monitor has a CMOS-compatible output signal that stays LOW until the voltage rises above the threshold again, I could just use that instead!) \$\endgroup\$ – boardbite Sep 13 '12 at 16:04
  • \$\begingroup\$ Sorry, I didn't realise that you had an undervoltage monitor already. \$\endgroup\$ – Rocketmagnet Sep 13 '12 at 16:27
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Q1. Option 1 - because the regulator output would need to be monitored for say 3.25V to allow some variation, and worst case temperature variation needs dealing with, and load transients may be an issue (or not) whereas monitoring at the regulator input achieves a "Minority Report" (Movie :-) ) effect of "detecting the problem before it causes a problem" and acting on a smoothly falling variable ( = Vin).

Note that 3.5V is well above safe Vmin for the LiIOn or LiPo battery and only the regulator dropout needs to be considered.

Q2: Various issues. Needs more discussion.
Option 1 OK.
Option 2 probably also OK. Has power down current considerations.

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Question1: Option 1 is better for the following reasons: You can precisely set the cutoff voltage and you don't depend on the manufacturing tolerance of the regulator. Second: you can easily include a small security zone because unvervolting LiIon is quite bad in terms of capacity loss (afair!).

Question 2: I'd go for the cutoff on the regulator as any switch will have a small current flowing throught it. Regulators have (almost every time) a low quiescent current and don't add additional resistance in the path between the power source and your load wasting energy

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For question 1, I would go with the second option. 3.5V cutoff is not near a deep discharge (this would be at 3.0V). So your first reason for switching off the circuit is not needed, since you could discharge the battery safely to 3.3V. Measuring after the the LDO ensures that your second reason is considered, since you then aren't depended on the dropout voltage of the regulator. If it is larger than 0.2V, you switch of too late, and the MCU supply voltage drops below 3.3V. If the dropout voltage is smaller than 0.2V, you switch of too soon, and could have run longer on the battery.

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