Powering 1.8V-3.6V chip from 3xAA batteries

[Slowmo]

I need to power a chip with 1.8V-3.6V operating voltage from 3xAA batteries. And I need a solution to be the most power efficient as it can be. I know I can use for example an LDO regulator for this, but that will be very inefficient considering it must regulate down to around 2.6V to take a full advantage of available battery capacity. And what's more important - the closer the voltage is to 3.6V, the better.

One thing that came into my mind is to power it directly from MCU, which has a much greater voltage range. I/O pins have enough source current for this purpose. So I could use two pins for that. One with series diodes to drop the voltage if it is above 3.6V and the other in parallel without any diodes. Battery voltage could then be monitored and chip could be powered either from one pin or another depending on voltage. Will this work and what are other options I have here? DC-DC converter will add unnecessary complexity and noise to the circuit.

 

[ronv]

Why not use 2 of the 3 batteries to power it?

 

[Slowmo]

Because batteries will not be discharged evenly this way and I need a 3 battery setup to have higher voltage and overall capacity for other things on the circuit.

 

[Ian Rogers]

3x rechargeable AA batteries (1.2v) = 3.6 volts

 

[Slowmo]

Yes, I know, but I want to keep an option to use alkaline batteries as well.

 

[ronv]

If the current is low enough you can power it from a micro output I wouldn't worry about 20ma. more from 2 of the 3 batteries. It's only .1% of the capacity of the battery. Or is the current much higher and the micro outputs much stronger?

 

[Slowmo]

The problem is that although current will be below 20ma, device will not be constantly on and in most cases that chip will be the only thing draining batteries.

 

[languer]

Take a look at NCP1402 or NCP1403; using one battery to create the desired voltage.

 

[Slowmo]

Capacity of one cell will not be enough. The whole thing must be powered from 3xAA batteries and there is not much I can do about that.

 

[languer]

3xAA in series (to get +4.5V) are still limited in current, correcT?

That being said; with the little information you've provided, your solution sounds great. If you care to provide more information, then perhaps others can entertain better solutions.

 

[Mr RB]

Just use a 3.3v low Q current LDO. With good batteries you are running from 4.5v, and the regulator won't drop out until 3.3v (1.1v per cell) which is pretty "flat" anyway. And even then at 0.9v per cell the LDO will still put out the full 0.9v (ie 2.7v).

 

[nickelflippr]

What is the mystery chip, a data sheet? How long does it need to be on, or how often must it be sampled? Is a micro already involved? There must be some sort of communication to the outside world, and how is that effected?

 

[ronv]

Maybe I'm missing something but I think if you run it from 3 batteries and a diode drop for 3.8 volts it will still draw 20 ma. (maybe a little more since the voltage is higher) Same 20 ma. from 2 batteries = same time to a dead battery(s). The only difference is you wasted the power in the diode.

 

[Slowmo]

Yes, I know that, the benefit of that is that at one point I can switch directly to battery voltage without wasting anything further on. Unfortunately that point is quite low and batteries have lost most of their capacity already.

I'll explain what I have here.
That chip is an RF transceiver from Silicon Laboratories. Chip draws about 15mA at the power I will be transmitting. There will be very short messages once every 1 to 5 minutes and a small Rx time each period. Increased voltage and additional capacity is needed for a mall DC motor which will occasionally need to be switched on for a couple of seconds. There will be an MCU which can operate from more than 5V so I can power it from the batteries directly. The whole solution must last at least a year on one set of batteries and that is why I want to be super efficient here.
RF chip must be kept in sleep mode to retain internal register values. The problem with LDOs is that they have quite high quiescent current while turned on. Sure it is in in uA range, but that is still enough to discharge the batteries in a long run. The same for DC-DC converters, whose quiescent current will be at least 10 times greater that that of LDO.
I could shut down the chip completely by cutting down the power, but then there will be a lengthy initialization process and I have to do calculations on what is better - losing power in standby or doing the same in initialization process.

 

[nickelflippr]

@Slowmo,
Much better explanation, thank you.

I like the original idea of using a port pin to power the transceiver. I would just use a resistor divider off one pin at say 0.8 x Vdd, that should cover the range required. Some micros have a DAC built in that could do just that. Considering the actual on time over a year, the waste of the divider is minimal. Also, the time spent re-initializing and powering up (a millisecond or two?) should be negligible compared to the 1-5 minute wait time between transmissions, so why not shut everything down?

 

[languer]

Yes much better explanation. I would both motor and transceiver from pnp switches. The MCU should be used to turn them on or off. That way you matain isolation betwenn the MCU and the loads (key for the motor, and possibly still needed for the transceiver since it may want it's supply well filtered). Use a diode (or two) to drop the voltage where you want it. If you very little leakage current look at some of the logic level MOSFETs (as opposed to BJTs). For the sake of completeness, what is the voltage required for the motor, and what is the voltage required for the transceiver.

 

[languer]

BTW, this is one of those; don't leave home without it kind of website: **broken link removed**

 

[dougy83]

RF chip must be kept in sleep mode to retain internal register values. The problem with LDOs is that they have quite high quiescent current while turned on. Sure it is in in uA range, but that is still enough to discharge the batteries in a long run. The same for DC-DC converters, whose quiescent current will be at least 10 times greater that that of LDO.

You can get a 3v3 LDO regulator that has <3uA quiescent current. Assuming your AA batteries are alkaline, with capacity > 1500mA.Hrs, you have a power budget of at least 171uA average, drawn continuously over the whole year. The <3uA is not really worth worrying about.

 

[Mr RB]

Good point!

It also might be possible to find a 3.3v LDO regulator IC with an "enable" pin.

 

[ronv]

Can someone fix me up? I just can't see any advantage of using a linear regulator over just 2 of the 3 batteries. Here is how I get there.
3 batteries = 4.5 volts 9900 mw / hours
2 batteries =3 volts 6600 mw / hours
2200 ma/hr batteries

2 batteries = 3 volts at 20 ma. = .06 watt hours. 3 batteries at 20ma = .09 watt hours - .03 to heat in the regulator and .06 to the load. The only better way I can see to get there would be a buck convereter at say 80% would save about .015 watt hours. Lot of stuff for 1+%.
What am I missing?