Drop D C voltages

Maybe 4 diodes in series.

@David Johnston posted:

Maybe 4 diodes in series.

. . . mmm, or a few more, since a diode will drop the voltage about 0.7 volts each. So, to drop about 5 volts, as the OP desires, that'd be about seven diodes in series.

Or, you could just drop in a small value resistor, or even easier, just drop in a buck converter to produce a regulated 12 volts. Even better, set the buck converter to drop the voltage to 9 volts, for smoother operation.

@Steve Tyler posted:

... even easier, just drop in a buck converter to produce a regulated 12 volts. Even better, set the buck converter to drop the voltage to 9 volts, for smoother operation.

Considering how inexpensive buck converters are - some are less than $2 each in bulk - and you can dial in the specific voltage, why do anything else?

Here is a convertor I use.  I like the digital readout.  Rated output 2.5 amps

@trainman129 posted:

I have a rack mounting box full of power supplies which run my layout. The one for my turnouts (tortoise) is making about 17+ volts both pos. and neg. This seems maby a bit too much. Does anyone have a means of dropping the outputs down to about 12 volts? Seems like the turtles will like that better.\

Thanks

Jay

Since you mention DC, I'm assuming that positive is one direction and negative the other, if that is so, diodes would not work in one direction or you would have to have 2 series strings paralleled with anodes going one way and cathodes the other, double the parts count.  Is the power supply supposed to make 17VDC or has something failed in it?  A switch motor like that would be a constant load in operation, with perhaps a current draw of 250ma,  a simple 20 ohm 2 watt resistor would drop 17 to 12, but if a bunch of turnouts run from that supply and run at the same time, you would need a resistor in the line to each turnout.  If the current draw is higher or lower, you have to adjust size and wattage and higher is more of a problem than lower, if it drew 1 amp, that would take 5 ohms at 5 watts minimum, in that case the converter would be a better bet but adds a complicated failure element to the mix.

@Steve Tyler posted:

..

Or, you could just drop in a small value resistor...

The resistor would limit the current only, resulting in a different voltage drop depending on the load. The drop would vary as the current demands went up & down.

I use diode pairs all the time to drop AC voltage, for stuff like taming motors they work great.  Resistors are not the answer in most cases as the voltage varies with the current, usually not the desired result.  Diodes work as well for DC, if it's bi-polar then the same diode pairs, if it's one polarity, then just a single string of diodes.