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I have put together the push button circuit explained in TI's SCEA048B application report with an under voltage protection using TI's LMS33460 chip. I ended up with the circuit attached. The idea is to have the LMS33460 under voltage detector monitoring the +PACK line from the battery so it pulls down the regulator enable pin to power down the device in case the battery voltage falls bellow 3v, but I have several doubts:

  1. After doing the schematic I realized that I should use an under voltage detector that triggers at 3.4v instead of 3v. I think this because the 3.3v LDO regulator has a maximum dropout voltage of 250mV and the TM4C123GH6PM has a minimum operating voltage of 3.15v, so 3.15v + 0.25v = 3.4v.. under this battery voltage the microcontroller will not have enough voltage supply. But, since I am using a 3.7v lithium battery, this would only give 3.7-3.4 = 300mV of margin for the battery to discharge.. I don't know much about lithium batteries but that doesn't seem to be a lot... is that a problem?

  2. I haven't found the input resistance of the TLV70033's Enable pin in the datasheet, so how can I know the maximum pull down resistor (R33) to minimize consumption and still pull down the voltage to turn off the regulator. Anyway, this pull down resistor is not needed anymore after connecting the enable line to the flip-flop, since the flip-flop's output is not an open drain and it will command the line high or low without ambiguity, right?.

  3. Do you think this would work? : )

  4. Is there a better way for protecting the battery?

Thanks

ON/OFF push button circuit with battery under voltage protection

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  • \$\begingroup\$ I have also a boost converter for getting 4v from the battery, since there is one module that requires a higher voltage supply. I am thinking about connecting the 3.3v LDO regulator to those 4v instead of to the battery's 3.7v. I lose some efficiency because I am rising the voltage from 3.7 to 4 with the boost converter and then decreasing it from 4 to 3.3v in de LDO, but then I will still have a valid voltage supply although the battery voltage falls bellow 3.4 volts and the system will work for a longer time \$\endgroup\$
    – jap jap
    Commented May 12, 2018 at 17:18

1 Answer 1

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That is what "D" input on SN74LVC1G175 is for. In application note it always transfers Vcc ("1") to the output. If you feed it the state of voltage detector you will essentially transfer battery state to the output (although you need inverted detection signal from the one of LMS33460)

As a side note, I don't see a reason for not using something like STM6601 which does everything you want and more in one tiny package.

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  • \$\begingroup\$ Thanks @Maple, I didn't know that IC, very interesting. Have you worked with it before? Although in page 5 it says that it also disables the power with one button (PB) long push, in the rest of the document it always says that the two buttons have to be pressed in order to turn off the power. I just have one button so I have to be sure about that.. I could connect both inputs to the same single button but it is not clear what side effects that could have.. to bad I didn't know this two days ago when I bought some components from DK! It would have been interesting buy one of this for testing! \$\endgroup\$
    – jap jap
    Commented May 15, 2018 at 7:46
  • \$\begingroup\$ No, I did not use this particular chip before. There are quite a few power management chips, this one (or STM6600) just looked like best fit for your application. The SR input and combinations with it are used to assert RST. It usually connected to a micro that you can press with a needle trough a tiny hole on the back of your device to reboot it. The documentation only seems contradictory if you ignore "depending on the option used" at the end of every other sentence. See options on p. 46,48 and choose the behavior that you want. \$\endgroup\$
    – Maple
    Commented May 15, 2018 at 19:48

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