1
\$\begingroup\$

The ADXL354 with Analog Outputs include three power pins: Analog Power Source V1P8ANA, Digital Power Source V1P8DIG and Logic High Level VDDIO, besides the VSUPPLY Main Supply Pin.

According the Power Sequencing datasheet feature (p.22), when VSUPPLY is tied to ground, the Analog V1P8ANA and Digital V1P8DIG Power Sources can be supplied externally.

Which is the purpose of this feature in the ADXL354?

For which reason should I use that complex Power Sequencing instead of simply plugging in the VSUPPLY to 3.3V?

If the maximum level of these pins is 1.98V, which is finally the purpose of them? It cannot be changed into 3.3V either.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ possibly uses the 1.8 V external power supply for low power application \$\endgroup\$
    – jsotola
    Sep 25, 2019 at 4:34

1 Answer 1

1
\$\begingroup\$

The V1P8DIG and V1P8ANA provide 1.8V to digital and analog sections respectively. They are internal LDO's that drop the main VCC to around 1.8V, which the IC needs to function. If you already have 1.8V then the LDO's don't work, so you'll bypass the LDO's. My guess is the LDO's are provided because most people use 3.3V for Vcc, you would be less likely to use this accelerator in a product if you had to provide your own 1.8V LDO which adds cost and space (especially since most acellerometers go into products that have tight weight/space requirements like cellphones and drones).

The LDO regulators are operational when VSUPPLY is between 2.25 V and 3.6 V. V1P8ANA and V1P8DIG are the regulator outputs in this mode. Alternatively, when tying VSUPPLY to VSS, V1P8ANA and V1P8DIG are supply voltage inputs with a 1.62 V to 1.98 V range.

Source: https://www.analog.com/media/en/technical-documentation/data-sheets/adxl354_355.pdf

\$\endgroup\$
1
  • \$\begingroup\$ Yes i agree, but.. after the shown paragraph, some explanation about timings in the sequence of powering over the device, if one is willing to use an external source for them. My question is why should i need that? \$\endgroup\$
    – Brethlosze
    Sep 28, 2019 at 4:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.