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Ultra low-power Vin regulation

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Stratofour

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Dear members,

I would like to power a Microchip PIC from a 4.2V li-ion cell. I have a few constraints on my PIC selection which direct me in the group of PIC's with a maximum supply voltage of 3.6V. (The availability of a internal clock for USB comm. is one of them, no ext. osc. is needed)

Of course the voltage exeeds the PIC's maximum ratings (Vdd.abs.max:4.0V) when the li-ion cell is fully charged.

The controller is most of the time sleeping with a 32KHz RTC osc. which consumes a rough 10uA. About 6 times a day the current consumption will rise to roughly 10mA for 4 seconds. As you can see the controller is mostly sleeping.

I would like to keep the current consumption to a absolute minimum.

The LDO's I have found expect a Iq of 50uA, which is almost 5 times the standby current the PIC consumes. A DC/DC converter isn't very efficient at these current consumptions.

Does someone know a descent Vin regulation which keeps the powerloss to a absolute minimum in comparison to the controllers consumption? I would also like to use the complete Voltage range of the li-ion (4.2v->3v)Thanks!
 

If you want to use the complete voltage range of the Li-ion battery, you need to use a buck-boost converter. You have no other choice.

Another option is to place a supercapacitor across the battery, to supply your temporary high current needs. I calculated that if you draw 10mA for 4s and accept a voltage droop of 0.1V, you need a 0.4F capacitor.
 

Unfortunately you have created conflicting requirements. You have picked a micro with the lowest power consumption but in order to use it you need to consume extra power to convert the voltage. Using a PIC which can work at 5V with slightly higher power consumption may be a better solution than trying to use the lower power PIC and wasting the power somewhere else (in a regulator).

I guess you might be able to construct some sort of crude voltage limiter with a couple of MOSFETs but you would still waste power because it will need some resistors which themselves will consume power.

Keith.
 

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