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I'm designing a low-power, battery-powered sensor node using an ATtiny1626 microcontroller (1.8V~5.5V supply with low clock speed), powered directly from 2x AAA batteries (various chemistry, let's say 2V up to 3V over usable life). For providing supply for other parts of circuit, a hi-efficiency boost converter XC9141B/42B is used. The chosen version with B suffix, bypasses the input voltage to the VOUT pin when the CE pin is low and the VOUT would be more or less equal to VBATT in that case. When the CE goes up, VOUT also goes up from VBATT to 3.3V.

In order to get rid of level translations, I thought about supplying the MCU from the VOUT pin itself; so the MCU (still powered from pass-through capability) holds the supply voltage at VBATT and when needed, turns the boost converter on, and the supply of MCU and the other part of circuit jumps to 3.3V.

  • The other parts have their own sleep modes, so it's not a problem if they still get powered by VBATT or when the voltage jumps between values.
  • The idea is to boost voltage when the battery level is low and unable to power other parts sufficiently, and maybe for better TX power of LoRA (SX1278 module.

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What kind of problems would arise from this sudden switch of voltage for the MCU itself?

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  • \$\begingroup\$ Depends, what input voltage range can your MCU accept? Can it boot from 2 V - diode drop of the boost converter before it has started? \$\endgroup\$
    – winny
    Nov 27, 2023 at 16:23
  • \$\begingroup\$ @winny 1.8V with lower than 5MHz system clock. I can't find any related info about that drop on the regulator, however, it mentions a 180 ohms bypass switch resistance. \$\endgroup\$ Nov 27, 2023 at 16:30
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    \$\begingroup\$ The idea is to boost voltage when the battery level is low and unable to power other parts sufficiently - if the batteries are dying, boosting the voltage won't help much for any sufficient time. \$\endgroup\$
    – Eugene Sh.
    Nov 27, 2023 at 16:40
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    \$\begingroup\$ Then you need 1.8 + 0.7 = 2.5 V to boot it. Scrap the idea and fix the root problem instead? Perhaps 3xAA instead? \$\endgroup\$
    – winny
    Nov 27, 2023 at 17:40
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    \$\begingroup\$ @EugeneSh. Not necessarily. The lower you can take your batteries, the more power you can get out of them. This is especially true of alkalines. For example, what if the senors needed 2.8V+, and he's using 2 AA's. Running only to 2.8V would leave something like 80% of the power in the cells unused. \$\endgroup\$
    – Drew
    Nov 27, 2023 at 22:09

1 Answer 1

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I've done that with an ATtiny84 with no problems.

When in sleep mode parts of the circuit (such as the analog section) are powered down and the supply voltage is dropped from 5V to 1.8V with only the MCU running.

When the processor periodically comes out of sleep from a timer interrupt it sets the supply to 5V, enables power to the analog sections then waits an appropriate time for the supply to stabilize before using the analog section.

This is in a device powered from solar photovoltaic cells that charge a single NiMH cell to power the circuitry through a boost converter controllable to output either 1.8V or 5V. A MCU GPIO controls the output voltage. Energy is very limited although instantaneous power is not so much as the battery can provide that power.

You need to watch out for sneak paths in the sleep state. For example the I/O pins often have ESD diodes to VCC. You also may need to put the MCU GPIOs into tri-state mode, high or low as needed.

If the voltage is too low for the peripherals in the sleep state they may need to be power-switched to avoid operating out of guaranteed operating range. If you power down a peripheral the MCU GPIOs to that peripheral will probably need to be put to low.

Buses like I2C can be tricky or there may be leakage into the peripheral or you may need to power-switch the pull-up resistors.

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