Simplest way to boost voltage from AAAA battery

MisterBill2

Joined Jan 23, 2018
8,455
If you consider four AAAA cells to be excessively bulky, consider also that any voltage step-up circuit will be bulky as well. And the performance of the system using the noise-free low impedance battery supply will be very consistent and repeatable.
 

MrAl

Joined Jun 17, 2014
8,149
If you consider four AAAA cells to be excessively bulky, consider also that any voltage step-up circuit will be bulky as well. And the performance of the system using the noise-free low impedance battery supply will be very consistent and repeatable.
All SMD parts?
 

BobTPH

Joined Jun 5, 2013
3,106
I believe you would be better off with two lithium coin cells in series to get 6V. The popular CR2032 would have more capacity than an AAAA. If one AAAA is enough, two CR2032 is at least twice as much. If you want smaller, there are smaller sizes that could natch the capacity of the AAAA, and no boost converter is needed.

Bob
 

Audioguru again

Joined Oct 21, 2019
2,988
Why not boost the battery cell voltage with the little 4-pins IC used in a solar garden light? The inductor is the size of a 1/4W resistor. I add a Schottky diode and 0.1uF ceramic filter capacitor so it can provide 10mA or more at a few volts to a colors-changing LED. It still works when the AAA Ni-MH battery voltage drops to 0.8V.

Instead of using a rare AAAA cell and booster circuit I would use a dual cell Li-PO battery that is lightweight and is rechargeable.
 

Thread Starter

Dritech

Joined Sep 21, 2011
873
Actually I like the ideal of using two CR2032 cells as suggested by @BobTPH

I found the CR2032 MFR.IB , which is quite cheep at $0.34 per cell. My only concern is the capacity which is stated as 225 mAh.

In the datasheet, it is mentioned that the Max. Cont. Discharge Current is 3mA. Does this mean that this battery should not be used for circuits which require more than 3mA continuous current?

Also, assuming the circuit will consume approx 2mA and that two batteries will be used in series followed by a LDO 5V regulator, how can I calculate the average time before the voltage per cell drops to 2.55V ? (which is the minimum required at the input of the regulator, that is 2.55V+2.55V = 5.1V)

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Audioguru again

Joined Oct 21, 2019
2,988
Yes, the CR2032 little coin cell drops on its face with a current near 3mA. With any load current more than 0.2mA its voltage srops even when it is new so it will quickly drop to your 2.55v minimum.
A CR2450 is thicker and might last 3x longer.
 

MisterBill2

Joined Jan 23, 2018
8,455
One question is just exactly how critical is the supply voltage? It might work with just one lithium cell. And that could be rechargeable which could be handy indeed. Just think of all of those cell-phone batteries available. Lots of different shapes.
 
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