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replacing 2 D cells with 18650 batteries

I have a couple of pro radiation meters (Ludlum 12) that run on a pair of D cells outputting 3V. Since I have a bunch of unused 18650 batteries lying around and I like their performance in other items I have, I have been thinking about adapting those to the geiger counter. Problem is the voltage. 3.7V per cell and these are 3000 mA/hr. The meter seems to draw up to 100 mA current if scintillation probes are attached and forced into near saturation. So, I am considering using two of the 18650's either in series or parallel. What would be a cheap but effective way to drop the resulting 3.7 V (or 7.4 volts in series) down to the 3V needed for the unit. I first thought maybe several silicon 3A diodes in series to get the necessary voltage drop, but not sure if this would be a good way. There isn't much room in the battery compartment and it is sealed so don't really want to drill holes and such. I could drop the 18650's in there, but very little room for anything else unless I used just one. Thanks in advance. Just trying to keep things simple with what I already have here.

Reply to
Witherspoon
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"Mini360" buck module: 50c.

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  Jasen.
Reply to
Jasen Betts

Buy a little switcher brick, 7 volts to 3.3 volts.

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John Larkin      Highland Technology, Inc 

The best designs are necessarily accidental.
Reply to
jlarkin

On a sunny day (Mon, 8 Feb 2021 23:07:27 -0500) it happened Witherspoon wrote in :

It all depends, maybe your meters are perfectly happy with 3.7 or even 4.2 V? For example in all my remotes that take 2 x AAA I now use one AAA lifepo4 (

Reply to
Jan Panteltje

Correction:

It all depends, maybe your meters are perfectly happy with 3.7 or even 4.2 V? For example in all my remotes that take 2 x AAA I now use one AAA lifepo4 (

Reply to
Jan Panteltje

Two D cells gives 3V when fresh, 2V when depleted, so it should suffice to insert a 2.5V linear regulator, three-terminal type, after the power switch. If it really only takes 100 mA, a TO-220 package regulator will not require a heatsink.

Check, though; maybe those radiation meters already have a regulator, and will just run fine on a single 18650.

Reply to
whit3rd

I have a few things that required 2 AA cells running instead on one

14500 LiFePO4 cell and one "blank" cell (a short circuit cell, available from the same suppliers). It works really well.

Not sure if anyone makes LiFePO4 cells in 32600 size to match a D cell.

Clifford Heath.

Reply to
Clifford Heath

On a sunny day (Tue, 9 Feb 2021 22:05:46 +1100) it happened Clifford Heath wrote in :

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;-)

Reply to
Jan Panteltje

John Doe wrote in news:rvtth0$2su$ snipped-for-privacy@dont-email.me:

You are using a buck regulator mindset. These are high efficiency DC to DC converters and there is no "wasted current" you dopey dipshit.

Reply to
DecadentLinuxUserNumeroUno

Thanks to all so far. Let me start off by asking what was wrong with using the diode(s) in series to drop the voltage to what is needs to be (from 3.7 to 3V)? That would have been the simplest way and took up the

least amount of space, plus no tapping into switching like I would need with any regulator.

Several folks have linked to tiny regulators. Problem is the reviews with some having runaway problems, not regulating, etc. I might go with

one of those, or even a linear regulator, but I wouldn't be comfortable unless I also added some sort of protection circuit. Wouldn't I be able

to use a zener across the output, say of 3.1V, in case the input voltage

started going awry? For current, a fast blow fuse maybe. Just some ideas.

Finally, someone in another group suggested using AA with D reducer cells of this type:

tinyurl.com/19wlgreg

Not ideal as it brings more batteries into the equation since I already have a bunch of 18650's, but if its going to cost more to order the regulators and make protection circuits, might be the way to go.

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Reply to
Witherspoon

Witherspoon wrote in news:rvui6v$e24$ snipped-for-privacy@dont-email.me:

Those little "watchdog" regulators ahve been around for a while and are all but perfected, and cheap and voila worries all gone!

Dropping it with diode junction potential drops woud likely work too though. But then YOU have to monitor charge state and HALT input upon near full charge. Those batteries do not last long woth constant or often overcharge events.

Reply to
DecadentLinuxUserNumeroUno

Another thought I had is maybe just a zener diode regulator. Again, no extra switching or power. Let me give some specs and maybe someone can help come up with a suitable circuit. I would be using either 3.7V (or

7.4 volts if the batteries are in series). The circuit runs at 3 V with

up to 100 mA draw. What suitable zener circuits could I use and especially what zener diode type that would handle the current? This could easily be fitted close to the batteries if it could work.

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Reply to
Witherspoon

Hmmm, not sure I follow here. Not quite set it and forget it then, maybe not the best idea even though the batteries would always be removed and charged up remotely if that makes any difference. What about a zener diode regulator I mentioned below?

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Reply to
Witherspoon

Part of the problem is that you asked for 3V, so people proposed solutions that give 3V.

Diodes are not precise voltage drops, neither are LiFePO4 batteries precise sources, but then neither are dry cells.

Dropping 0.6 to 0.7V in an ordinary silicon diode combined wuith the

3.7 to 2.7 range of the lithium cell would get 3.1 to 2V, which is not really 3V, but it is basicaly the same voltage range that the two dry cells would make.

So yeah, do that.

--
  Jasen.
Reply to
Jasen Betts

Maybe a 3.0V LDO and a single **protected** 18650 cell (including discharge protection).

Eg. S-13A1A30-A6T1U3

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That will regulate down to about 3.07V, which is about where you should be shutting off the battery anyway.

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Best regards,  
Spehro Pefhany
Reply to
Spehro Pefhany

Yes, I didn't mention that, because the capacity is so much less. One

14500 LiFePO4 cell has roughly equivalent energy to two alkaline AA cells.
Reply to
Clifford Heath

The LDO will want a small (10uF?) output capacitor, typically, but yes, that's a good solution. If you only expect 100 mA load, just choose an LDO with 200-500 mA current limit; the battery won't blow up. Remove or disconnect the battery when not in use, or for charging.

MCP1825S-3002 comes up on a Digikey search.

Reply to
whit3rd

I like the idea of using a battery with discharge protection because forgeting it 'on' just once can damage the battery.

The protected cells use a little coin-shaped PCB under the shrink wrap, usually with some MOSFETs and a DW01 or similar.

You could also put a part like the TI bq29700DSE on a tiny PCB with the MOSFETs.

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Best regards,  
Spehro Pefhany
Reply to
Spehro Pefhany

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The best designs are no accident - sw
Reply to
Steve Wilson

On 2/11/2021 2:45 AM, Steve Wilson wrote: [snip]

Free shipping with Prime. Maybe not to Canada?

Reply to
Bob Engelhardt

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