12v dc to 3v dc to replace 2 d batteries ..#2

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Builddesign

Joined Mar 26, 2025
1
I'm upgrading to a tankless water heater in my home and the one I got requires 2 d batteries to function. I have 120v and 12v available at the location so I was thinking that it would make sense to put in a converter or transformer to step down the 12v to 3v and hardwire that in so I'm not replacing batteries all the time. I estimated that about 3 amps of 3v DC would be similar to 2 d batteries. Are there any unforeseen consequences of doing this and would 3a be reasonable or would dropping it lower make sense.
Is this the Marey unit? I just installed one and was wondering how to do this as well. I’ve been using mine for almost 2 months on the initial D cells, but this morning it wouldn’t light so I’d love to hear or see what you came up with as a hardwired solution? Thanks!

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https://forum.allaboutcircuits.com/...ictable-behavior-in-robot.147227/post-1254372
 
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crutschow

Joined Mar 14, 2008
38,384
A simple 3V wall-wort with a super-capacitor at the output to provide the peak starting current should work.
You may have to experiment to determine how much capacitance you need.

If you want to use the available 12Vdc then you could use a simple LM317 regulator with the capacitor output to provide the surge current.
No need here for the efficiency (and lower reliability) of a switching buck regulator.
But that will only work if the ground points for the 12Vdc and 3V battery are the same.
 
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MisterBill2

Joined Jan 23, 2018
27,258
Given that we are not told just what the 3 volts from two "D" batteries does, it is difficult to decide what sort of approach would be best to power the device from 12 volts.
My choice, if the 3 volts is only needed for a short time periodically, would be a series resistor followed by a voltage shunt regulator. That is based on a trade-off between efficiency and effectiveness.
So is there anyone who actually knows just what the three volts supply does in the tankless water heater?? And does that person know what portion of the time the current is actually drawn??
My guess is that it serves to power an igniter for the gas heater function, but that is only a guess. IF that is the case then my concept will work.
 
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