NRF52820 as 3LED high-side driver using internal LDO?

[magicsmoke]

hello nordic, we are looking at nrf52820 to satisfy some requirements. 

1: We plan to power off single cell Lithium application which can be 2.7V - 5V trhu VBUS / VDDH and use the internal regulator. 
2: Plan to use Internal Linear regulator @ 2.7Vout to drive the LEDs with very low value LED as this is a sweet spot for the led IV characteristic, so we could drive it with zero or low value resistor and burn the power in the regulator. 


I have a few questions. 

1: From my reading, driving the LEDs with up to 15mA is well within the LDO's capability?

2: We could use high current drive to drive the LEDs with ~5mA without issue?

3: Any additional consideration such as increased Decap etc? 

4:  Could the NRF52805 operate down to VDDH=2.7V or lower even though we set the linear regulator to 2.7V? Would it passthrough the voltage and then VDD(reg)=VDDH? 


check out this artwork 

[Kazi Afroza Sultana]

Hello,

1. ''driving the LEDs with up to 15mA is well within the LDO's capability?''

Internal LDO regulator is designed to supply the nRF52840 and it's peripherals. But the maximum current that can be sunk by all GPIOs is 15mA. You should not exceed this limit. 

''2: We could use high current drive to drive the LEDs with ~5mA without issue?''

Yes, you can use the high drive setting on the GPIOs to drive LEDs at around 5 mA without issue, as long as the total current sunk by all GPIOs does not exceed 15 mA. The internal pad resistance should be considered when calculating the series resistor for the LED, as the voltage drop across the pin increases with current.

''3: Any additional consideration such as increased Decap etc? ''

For adding decap, you need to follow our reference design. If you want to switch multiple output at a time and use the high drive mode it is important to have sufficient decap close to the VDD pins. 

4:  Could the NRF52805 operate down to VDDH=2.7V or lower even though we set the linear regulator to 2.7V? Would it passthrough the voltage and then VDD(reg)=VDDH? 

You may look at this case (+) VDD and VDDH out-of-spec consequences - Nordic Q&A - Nordic DevZone - Nordic DevZone.