Converting Christmas Village from DC to AC

Question

So this was the year we bought a AC-powered O-gauge Lionel train and a battery-operated village to add to our Christmas tree decor. The problem, the village eats batteries like a kid eats cookies.

Here is the setup:

• 3 village houses running on (3) AA batteries each [1.5V * 3 = 4.5V each house]—batteries last the longest here, two houses have 1 LEDs, the church has 2 LEDs.
• 1 village skating rink that also plays music and goes through batteries faster (3) AA batteries [4.5V]. There are 2 bright LEDs, 2 lamp LEDs, and 16-17 LEDs that blink to the music.
• 2 street lamp posts running on (1) 3.5V CR2450 button lithium battery [3.5V each lamp). The box calls it 3V—these batteries go fast too.

The bulbs are not designed to be replaced and there is no such indication of the watts or mA consumption.

The options as I see it are to buy a AC-to-DC power adapter for each of the two circuits above—3.5V circuit/adapter and 4.5V circuit/adapter. I have found for example, an AC adapter with a DC output of 3.5VDC@400mA. I have no idea if that would work for the lamp circuit and if done in parallel to keep volts the same at 3.5, does that 400mA become 800mA and what that means to the bulb that is currently running on 3.5V@650mAh. I guess what I need to learn is what rating that bulb could be on a unit where a bulb is not designed to be replaced or its mA use documented.

The other option of course is to create a custom circuit board and get my hands dirty in the details—I have 10.5 months so I'm open to it. I would imagine a single AC input split into two circuits that are individually transformed. I would love to have a setup on a board where I had screw terminals, one for each circuit to add each unit (house or lamp) to the board. Are there kits for this?

I greatly appreciate any thoughts, directions, and/or lessons.

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UPDATE #1 The current in mA for each device is as follows:

• 2 Street Lamps draw 10mA@3V each (each one has 3 small LEDs)
• 2 Houses draw [email protected] each (each with 1 LED)
• 1 Church draws [email protected] (2 LEDs)
• 1 Skating Rink draws between [email protected] (16-17 LEDs, 2 constant, the rest flashing, plus plays music)

Total per circuit:

• 20mA@3V
• [email protected]

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UPDATE #2 With confidence gained from suggestions here, I've cracked open each unit.

• The lamp posts with 3 tiny LEDs in each, have a 51Ω±5% inline resistor.
• The houses have no resistors (1 LED in each house).
• The church (with 2 LEDs) has 2 lines to the positive battery post, one without a resister, the other with a 220Ω±5% inline resistor—my guess is this is going to an LED in the tower and dropping the power/brightness a bit for the small space to equalize the appearance of the 2 lights.
• The skating rink has no in-line resistors, but a printed circuit board and speaker.

Answer 1

The simple answer for the houses that run from 3* 1.5V cells is to simply use an off-the-shelf cell phone charger. These are nominal 5 Vdc output. Note that most alkaline cells start off at about 1.6 Vdc but decay quickly down to the nominal 1.5V value.

If you are concerned that 5V is a tad bit too high, simply add a standard silicon diode in series with either lead, making sure to observe polarity. This will drop the voltage down to near 4.3 Vdc or so.

The street lamps are a tiny bit more work - just use more silicon diodes in series with the power supply leads that feed those lamps. Two diodes will drop your 5V power supply down to near 3.3 Vdc.

I think highly of Genuine Samsung cell phone chargers: nominal 5 Vdc at 2 Amps. You even may have some hanging around from old phones. If not, they are readily available from eBay and other sources. Just be sure to purchase Genuine chargers - some of the Asian-made knockoffs are complete trash.

Answer 2

3 village houses running on (3) AA batteries each [4.5V * 3 = 13.5V total] ...

This is most likely incorrect unless the battery packs really are in series. Why? Because there is no need for 13.5 V to drive the LEDs.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A very simple approach.

A transformer will supply AC which isn't suitable for the LEDs. You would need a DC power supply. An economical solution, and readily available, is a USB phone charger all of which give out 5 V at, typically, at least 500 mA which is plenty for your application. You now just need to drop the voltage or limit the current to protect the LEDs.

Silicon diodes have a "forward voltage drop" of about 0.6 to 0.7 V. Adding one in series with each of your 4.5 V loads should work and add three in series with the 3 V street lamps to create a 2 V drop. Any common diode such as 1N4148 or 1N4001 should do the trick.

Just watch on initial power-up. If the LEDs seem much brighter than when battery powered we may need to add in some resistance.

^{simulate this circuit}

Figure 2. Using resistors.

The original battery supplies may be relying on the internal resistance of the batteries to limit the current. The diode solution may not limit it enough so a resistor solution may be safer. The 100 Ω resistor will limit the current to a safe value for the LEDs. Add resistors in parallel to increase the current where the models have more than one LED. Use two in parallel for two LEDs, etc.

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For wiring, get an old USB lead, cut the 'B' plug off and use the red (+) and black (-) wires. To avoid screw terminals you could consider wiring each model individually to one or more USB hubs.

Answer 3

The first thing you need to do here are to check each unit's wiring. Do just two wires leave the battery compartment to the test of the unit or are there possibly three. Two will be the best thing for an upgrade solution. Three wires means that the batteries are being used in a way that multiple voltages are being tapped so more complicated.

The second thing needed is to get a good grip on how much current is supplied from fresh batteries in each of the units. For this you are going to need a digital multi-meter (DMM) [Do not get scared here. Decent enough meters can be had on Amazon or eBay for around 30USD.] By inserting some paper or tape to isolate one battery contact from its corresponding battery terminal you can then use the DMM in current measuring mode. Use the meter leads to probe on both sides of the temporary barrier to complete the circuit from the battery terminal to the battery socket contact through the meter. Obviously do this when the unit is switched on. These readings will give you a starting point as to how much current you will require for all the units plus say a 20% margin. Then this will allow you to size the source current requirement.

Your simplest solution will be to find a wall wart or AC powered brick that supplies 5VDC. A good bet would be a 5V supply from an old USB hub. These often have a rating of 2.5A which is a good guess to be more than adequate for your battery replacement application (still do measure your loads to make sure though). Take the output of that 5VDC supply and run the + side through a forward biased power Schottky diode. Use one for each of your loads to spread the current across multiple diodes. (A 1N5819 diode is suitable and can handle up to 1A and can be purchased from a place like mouser.com). These diodes will drop the 5V from the brick down into the range of 4.5 to 4.3V if you not trying to run too much current through them.

For the one unit that uses 3.5V you can then run its power through two more series 1N5819s to further drop the voltage.

Should be a fun simple project.

Answer 4

LM317 can provide 4.5V from >6.5V input and 3V add R=4.5V-20mA*R=75 Ohms with one per house either added to house or supply side.

Given load

• 20mA@3V
• [email protected]

^{simulate this circuit – Schematic created using CircuitLab}

This means if your supply is fixed then use R3=50 ohm, but if variable then R must be reduced to some value, possibly 5~10 Ohms if 8Vmin or directly to U1 input. But then worst case heat is 400mA * (18V-4.5)= 5.5W which needs a finned heatsink > 2" So if you'd rather not have any heatsink or just a little 2W heatsink, then you need a lower input DC supply like 7~9Vdc 400mA