electric signs

Sure, you just need to replace the battery with a similar voltage wall supply.  I would have thought they'd have a power plug, many of the buildings with Miller signs do.

They don't have a jack for the power...just a battery holder. You have to cut off the battery tray and wire the remaining length of red (+) and black (-) wires to 4.5 volts DC. Miller Engineering also sells a 4.5 VDC module which you can power with AC voltage 5 to 17 VAC, and also a couple of wall warts that run off of 120 VAC and plug into the wall.

Buy the Miller #4803 AC/DC adapter. It will run up to 10 signs. Believe me when i say you will eventually want several of their signs; this way you will have enough power to meet those future requirements. A #4805 Power Distribution Board will make installing those signs much easier. If you really think you will NEVER have more than 3 signs then buy the #4802 AC/DC adapter; it will handle up to 3 signs.

jackson

I'm using the 4804 and feeding my accessory voltage source to it...about 14 vac. It'll do up to 6 signs.

You can use a 7805 voltage regulator circuit, then on DC out install a diode in series in proper polarity. That will give you a 4.4 volt DC regulated output. You can also use an LM317 regulator circuit made for 4.4 volts.

Dale H

I made this circuit using an LM317 regulator:

Input is 10 VAC passed thru a bridge rectifier. R1 is 270 ohms, R2 is 680 ohms.  This calculates to an output voltage of 4.4V.  But I am measuring an output voltage of 7.5 volts with no load.  I have not yet powered a sign with it.  Did I make a mistake in the circuit, or do I need to measure it with a load?  How?

Bob

That looks right. What size and where are the capacitors physically? I think you are measuring with a multimeter and cannot see that the circuit is oscillating. Try to put the .1uF ceramic right at the input of the chip.

Put a light bulb load on it and measure the output voltage.