LDO current consumption, low voltage mode.

[cocoa]

Hi Nordic guys,

How much is the current consumption of the LDO only (quiescent current), without DC / DC converter, in uA? How much is dropout voltage and PSRR (ripple rejection) ?

When powering nRF51822 in low power mode, LDO has a quiescent current or is true disconnected?

What are the advantages and disadvantages when powering nRF51822 in low voltage mode ?

How much is leakage current of a gpio pin when enabled as output and as input ?

Thanks.

-c

[Ole Morten]

I don't have a number for the current consumption of the LDO itself, since you will never see just this. The lowest current consumption you'll see for the chip in total is system off current with no RAM retention, which is specified to be 0.6 µA.

I'm not quite sure what you mean by low-power mode, but if you mean low-voltage mode (as described in section 3.4.1 in the PS), this will incur an additional 1 µA current consumption in system on and system off. In addition, the ripple tolerance is lower.

In practice, I'd therefore recommend using the regular mode, the internal LDO, on current chip revisions. This gives the lowest current consumptions in sleep, and the best ripple tolerance. As you probably know, the DC/DC is currently not recommended for use unless your chip continuously draws enough current to justify having it on. See section 11.1.1 of the RM for details.

The leakage current of any GPIOs is negligible, in the femtoampere range or lower.

[cocoa]

Thank you Ole, I've questions again.

In the event that the supply voltage was 1.8V, LDO is disconnected or continues to consume the current Iq (quiescent).

I do not want to evaluate the power consumption in the case of System Off chip but functional.

When LDO is working, (i.e. 3.3 V as Vin) how is its consumption to reduce the voltage down to 1.8V, assuming the fact nRF51822 internal operating voltage is 1.8V and this is why there is an LDO inside.

I can not understand how it can consume less using internal LDO respect to an external power supply of 1.8V, which should exclude LDO.

Can I exclude forcibly the internal LDO using a register as in the case of the DC / DC converter?

Can you help me understand better?

Thank you.

-cocoa

[Håkon Alseth]

Hi Cocoa,

There's some logic inside that "disconnects" the LDO if VDD == DEC2, and this cannot be set in firmware. It has to be connected on the hardware, as per our reference schematic.

I think you may have misunderstood parts of Ole's answer. All peripheral/power modes, except SystemOff and SystemOn, will have the same current numbers in LDO and low-voltage mode.

Do you have some troubles with 1.8V low power mode? Or are you asking because you're evaluating both LDO and 1.8V mode?

Best regards Håkon

[cocoa]

Hi Håkon, I'm asking 'cause I'm evaluating both LDO and 1.8V mode.

My question is, how can I evaluate an external LDO compared with internal one ?

Does my LDO have better performances compared to nRF51822 LDO ? Impossible to say without electrical characteristic.

Another point of view, LDO ripple reduction, which is better? External or internal one? How much is the maximum admitted VDD ripple for nRF51822?

Thank you.

-cocoa

[Ole Morten]

I'm still not quite sure I understand your point. You have to compare the current consumption of the complete system, not just your external LDO against the internal one, i.e. either external LDO + nRF51822 in low voltage mode or just nRF51822 in internal LDO mode.

For the latter, you must use the values from the PS as is. For the former you must use the current consumption of your external LDO, plus the values from the PS, making sure to add the additional 1 µA consumed in system on and system off.

The general recommendation on ripple voltage when using low-voltage mode is to make sure it's less than 10 mV, which is pretty low, and may be problematic to achieve.

In summary, using an external LDO and low-voltage mode is almost never beneficial in real applications, since you'll have some current consumption from the external LDO in addition to the extra 1 µA, which comes directly on top of the numbers you'll have if you just used the internal one.