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I have a motion clock that runs off of two C cell batteries. One battery runs the movement at 1.5 volts. That other battery then continues in series to run the motion part delivering 3 volts to it. I would like to use a 3 volt dc plug in power supply to run the entire clock, but do not know how to separate 1.5 volts from it. I would not be opposed to using a 4.5 volt power supply if that is what I would need to do - 3V + 1.5V. The movement is quartz, so I'm sure the amperage draw is small. The motion part is an electromagnet that pulses and pushes a permanent magnet away from it to move the pendulum. If possible I would like to put this on a board that would press fit into where one of the batteries go, so it could be returned to battery operation in the future if desired. I'm also planning to use a jack plug that accepts the power supply/dc transformer, so it could be easily replaced in the future if it burns out. Thanx in advance for any help.

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    \$\begingroup\$ I'd buy a 3v 'plug top' power supply. Make a mechanical arrangement that fitted in place of the batteries, with a socket to accept the 3v supply, and either an LDO linear regulator to get the 1.5v, or a couple of silicon diodes and a resistor as a simple regulator. If the current demand is low enough (which it should be) this is one of those very few places where a resistive voltage divider would be suitable as a low voltage power supply. \$\endgroup\$
    – Neil_UK
    Sep 22, 2017 at 5:49
  • \$\begingroup\$ There are OTS solutions available you can tailor to suit or get ideas. The powersupplies do seem rather over priced but the various adapters are a good starting place for your non-invasive requirement. You may find that with just the clock running on one C-cell you may have many years of run time and only need to mains power the movement. - batteryeliminatorstore.com/… \$\endgroup\$
    – KalleMP
    Sep 22, 2017 at 7:04
  • \$\begingroup\$ I do believe you're right about the C cell lasting for years. I have another clock that uses 1 C cell without motion. We've been in this house for almost 5 years and I can only remember changing it once even though it chimes every 15 minutes. \$\endgroup\$ Sep 23, 2017 at 9:39
  • \$\begingroup\$ If I knew the values or part numbers of the diodes & resistor and how to wire them up that would be helpful. I've tried 2 1N4004's in series using 2 C cells, but they do not drop the voltage enough; it was over 2 volts. I have the 1N4004's that I use to reduce arcing across the 12 volt winding contacts on vintage car clocks that I repair and have those, but no others. \$\endgroup\$ Sep 23, 2017 at 10:04

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I have run battery powered quartz electric clocks on mains derived low voltage DC for about 2 decades .They work fine .I made my own low idle current linear regs from garden variety BJTs like BC337 ,BC327 .The battery clock does not need a highly accurate DC voltage because it is designed to keep time as the 1.5 volt cell ages .Now days you could use a LDO P channel reg .I did not use LM317 because the idle current wastes much more power than the clock .I used a combination of supercaps and electrolytic caps to allow operation during short power outages which are common in my area .Some years later I hooked up a smoke alarm to the system .

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There are very cheap and easy to use voltage regulator modules available on Ebay for like $ 1 each, I suggest you get two of those.

This is the (LM317 based) module I'm talking about: enter image description here

Then connect them like this:

schematic

simulate this circuit – Schematic created using CircuitLab

You should use a DC adapter of at least 5 V so you could use a USB phone charger. I would not go higher than a 12 V input voltage.

Before you connect the clock, use the small screw on the blue variable resistor to set the output voltage to 1.5 V and 3 V.

Indeed as Neil suggests in his comment, you could also use a resistive divider or (what I'd do) a couple of diodes in forward mode (and a resistor to 3 V) to make 1.5 V from 3 V. You'd still need a 3 V adapter though and these might be hard to find.

Also the cost of the individual components to make that will not be less than just buying two modules. We EEs often have these components in our part drawer so for us they're "free" but since my guess is that you would have to buy everything, it might be easier to just get two cheap LM317 based regulator modules.

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  • \$\begingroup\$ This seems like a very economical solution (LMC317) that would end up as a clean installation. I appreciate the help with this! :) \$\endgroup\$ Sep 23, 2017 at 9:27
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2 Silicon diodes in series step down 1.5 Volts (.7V * 2). Use diodes rated at twice the current or more than you believe the clock movement needs.

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  • \$\begingroup\$ This is more of a comment, keep in mind the post is over 2 years old \$\endgroup\$
    – Voltage Spike
    Dec 20, 2019 at 3:55
  • \$\begingroup\$ No, it is not a comment. 2 diodes in series lowers the voltage exactly 1.5 volts which is the exact voltage drop he needs. \$\endgroup\$ Jan 31, 2020 at 14:26
  • \$\begingroup\$ Exactly 1.5V? the voltage will be dependent on the current. Diodes make up for poor regulators, check out art of electronics figure 9.2 and chapter 9 \$\endgroup\$
    – Voltage Spike
    Jan 31, 2020 at 15:57
  • \$\begingroup\$ The current matters little. Anything over ~50 mA will forward bias most Silicon diodes .7 Volt. In such low power applications as a clock motor, I've already used Silicon diodes for this purpose. They work JUST FINE! I also use a diode to reduce voltage from 5 USB to 3.5. \$\endgroup\$ Feb 1, 2020 at 22:29

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