Current Consumption RTC & RTC + 64 kB RAM Retention

RE: Current Consumption RTC & RTC + 64 kB RAM Retention

FrankBi — Wed, 14 Jul 2021 08:19:30 GMT

Dears,

How to measure 64K RAM‘s power consumption， that marked 0.48 uA， how to get it ？

Thanks.

RE: Current Consumption RTC & RTC + 64 kB RAM Retention

Stian Røed Hafskjold — Thu, 12 Jul 2018 08:34:51 GMT

Sorry for the confusion. The RTC implies that the LF clock is running, and in the current consumption scenarios they've used numbers for the internal LF (LFRC). Sorry, this should have been specified.

So then you have:

SysON + RTC + LFRC = 1.2 + 0.1 + 0.6 = 1.9 uA
SysON + RTC + LFRC + 64k RAM = 1.2 + 0.1 + 0.6 + 0.48 = 2.4 µA
The radio itself consumes 5.3mA. When adding all the other peripherals needed to run the radio you get to 7.1mA
TX 2Mbps is the same as TX 1Mbps.

RE: Current Consumption RTC & RTC + 64 kB RAM Retention

Mr_BLE — Thu, 12 Jul 2018 06:53:43 GMT

Hi Stian,

thanks for your efforts. But this is confusing: I found in the datasheet V1.4 of the nRF52832 different values. Additionally there are different values on the first page of the datasheet compared to the values in the middle of the sheet.

To 1.: The datasheet sais on page 77: ION_RAMOFF_RTC System ON, No RAM retention, Wake on RTC 1.9 μA. Why is this different to your value? Do I also have to add LFXO with 0.25 µA -> 1.9 + 0.25 = 2.15 µA?

To 2.: ION_RAMOFF_RTC System ON, No RAM retention, Wake on RTC 1.9 μA + 64 k RAM-Ret. 0.48 µA = 2.38 µA? Also have to add LFXO? -> 2.38µA + 0.25µA = 2.63 µA?

3.: First page sais: TX current 5.3 mA @ 0 dBm; Page 76 sais: TX current 7.1 mA @ 0 dBm. So is it correct to asume that the 7.1 mA is the real value?

4.: The datasheet has no values of TX and RX power consumption @ 0dBm and @ 2 Mb/s. Could you tell me the values?

Thank you!

RE: Current Consumption RTC & RTC + 64 kB RAM Retention

Stian Røed Hafskjold — Wed, 11 Jul 2018 12:16:03 GMT

Hello,

System ON base current: 1.2 µA
RAM retention per 4 kB block: 30nA (64k RAM = 0.48 µA)
RTC running: 0.1 µA
LFXO running: 0.25 µA

Then you can just add the numbers:

SysON + RTC + LFXO = 1.2 + 0.1 + 0.25 = 1.55 uA
SysON + RTC + LFXO + 64k RAM = 1.2 + 0.1 + 0.25 + 0.48 = 2 µA