Resistor to bump down voltage

I would use an adjustable voltage regulator instead.

Resistor value will depend on impedance of the accessory involved and may change as it operates.

Also you would need a rather large wattage rating resistor and it is going to get hot.

To calculate the proper resistance the current is needed.  Using the voltage drop in diodes  might be a better way to do this.  I do not think you will get near the heat you will get with resistors.

Is this AC or DC?  Adjustable voltage regulators are for DC. For AC diodes would work.

The diodes drop 0.6 volts across each diode in a series chain.  Therefore to reduce 4 volts would require approximately 7 diodes (7 x 0.6v = 4.2 volts).  Depending on current flow, the power is: current times voltage = watts.  This is required to assure the proper selection of the diodes with respect to power handling capability. An example would be current is 0.25amps @ 0.6 volts = 0.15 watts or 150 milli watts. Six 1/4 watt (250 mw), diodes would be sufficient in this example.

Thanks guys,

I needed to know this because I was buying a scratch built station that had 12V tiny  bulbs in it. I was thinking either bumping down the voltage or....replacing the lamps with LEDs. Turns out it's not going to be necessary. The guy I was going to buy it from didn't like my questions.....I asked him if he had larger voltage bulbs (he couldn't understand why I didn't want to go 12DC and have to use a separate power supply for this). I'd rather not give my money to someone who is not accomodating.   But at least now I know how to get the voltage down next time I need it. Thanks!

Roger

With 16 VAC in a single diode would yield pulsed DC with about 8 volts RMS. Bulbs do not care if current is AC or DC. 16 VAC can be reduced to 12 VDC  with a bridge rectifier, then 5 or 6 diodes in series in proper polarity. Dropping voltage with diodes is explained here

www.jcstudiosinc.com/BlogShowT...=413&categoryId=

If bulbs are hard to change I would use LEDs

Several posts here on LEDs

www.jcstudiosinc.com/BlogCategoryMain?catId=426

Dale H

Hi Dale,

Changing out the bulbs with LEDs (which I do alot) would have been my first choice if I couldn't find 14V bulbs. But getting out the bulb socket and fitting in the LEDs would have been quite a project in the small "street lamps" that would have come with it. The rectifier and diodes would have been the ticket. Glad I didn't end up doing business with that seller. 

Actually, the diodes will dissipate the same amount of power as the resistor(s).  If you have 1 amp at 16 volts coming in, and you drop 4 volts and have 1 amp at 12 volts on the output, any passive component will have four watts of power dissipation.

You guys are way over - thinking and over - complicating this. If you put one diode in series with the bulbs, running them at half wave, you will reduce the effective voltage by about 30%.

16 volts AC times .7 = 11.2 volts. The 12 volt bulbs will be dimmer, but they will last a long time and stay cool. 

Just answering the question asked Roy.   Besides, if the accessories have more than just lamps, he may not want to run them on half-wave DC or an effective voltage of 8 volts.

Your calculation is also incorrect, the heating value of that diode modified waveform will be half the RMS value of the 16 volts, not .7 times it.

A diode string dropping voltage from 16 to 12 will use the same wattage as a resistors,however it will always maintain about 12 volts regardless of load so it acts in effect like a variable resistor. For a DC single string dropper the the wattage used would be the number of diodes times .6 times the amperage drawn.

A single diode dropping the voltage in effect by half (a little less due to the forward drop of the diode) will use much less wattage than the resistor. The wattage used would be only the forward drop of the diode,about .6 times the amperage drawn. The reverse current is completely blocked with no current flow in the opposite direction. A small insignificant amount of heat might be generated blocking it in reverse.  A resistor of half the load value would use as much amperage as the load itself. So with this half wave dimming method we kind of get a free lunch.

Dale H

I agree Dale, the diodes are the solution if you want to drop the voltage.