Resistors for panel lights

[Brian Degulis]

I'm putting in some new instrumentation and want to wire the ilumination into the existing sytem which is 32V. The gauges have 12 bulbs anyone what resistor to use?

Brian

[SKYCHENEY]

Can you replace the bulb with a 32v that has the proper base?

Another question: Are you sure you have 32v insturment lights? My boat is a 1985 as well and all of the instrument lights are 12v. They run off of a converter under the each helm. Put a tester on them if you don't know for sure.

[TedZ]

Need to know the resistance of the bulb and the wattage. If it is were a 12 watt bulb it would be 1 amp, i.e. Watts divided by volts. The resistance is low but it is likely .16 ohms or something similar. As soon as you take a meter and know that, i'll do it or send you the link i've posted before which has plug ins and does the math for you.

Don't get nervous if the numbers get very small...just do the math.

Stupid me...i'll put it in here now.

http://ourworld.compuserve.com/homepages/Bill_Bowden/r2.htm

And ther are many more. Search for voltage dividers.

Ted

PS the resistor should be double the wattage of the bulb or greater.

[Brian Degulis]

Can you replace the bulb with a 32v that has the proper base?

Another question: Are you sure you have 32v insturment lights? My boat is a 1985 as well and all of the instrument lights are 12v. They run off of a converter under the each helm. Put a tester on them if you don't know for sure.

Yes Sky they are 32 V and so far I can't find anyone who has the bulbs unless I want to buy 100 Pcs. A resistor would be the easiest way.

Brian

[Boatsb]

Why not just step the whole panel down to 12Volts so there is no need for individual resistors?

[yachtsmanbill]

Why not just step the whole panel down to 12Volts so there is no need for individual resistors?


Thats what I did and so far so good... all those gage lites are only about 2 amps total. When I was playing under the helm last winter, I pinched the
22 ga. wire for the binnacle light and and instantly had a wheelhouse full of smoke and didnt blow the breaker either!
Not to rehash the old thred, but why on earth did bubba pick 32 vdc for everything? Even 36 for starting with a 12 volt tap for everything else wouldve made much more scent (?) ws

[Fanfare]

I think those were the engineers at Rockwell. It really would have been a better idea if anyone else had followed their lead. Instead it was a dead end. Still, I keep seeing references to higher voltage DC systems coming out of Detroit for automobiles. 48 volts? Well, we'll see.

I am pretty happy with my 12v systems, as Jack Hargrave and Hatteras originally laid out in 1965.

[yachtsmanbill]

For years that was the problem with the "JEEPS for $49" thing. The military stuff all had 24 volt lighting which is good for vision but bad for a crash. An expensive adventure to de-militarize it.
Ever see a military convoy coming from 2 miles away; thats 24 volt lighting.
48 volts outa Detroit seems like a jap snuck in there ;-(
Lets see; a 4 cam, 16 valve, 12000 rpm buzz box with magnesium paddle shifters (? whatever those are) in a 4 cylinder doosh wagon, and youve still got a 4 banger doosh wagon.
Sorry everyone! I mustve overdosed on the cold medicine... I'm feelin' kinda "anxious" but still lousy. ws

[Brian Degulis]

Thats what I did and so far so good... all those gage lites are only about 2 amps total. When I was playing under the helm last winter, I pinched the
22 ga. wire for the binnacle light and and instantly had a wheelhouse full of smoke and didnt blow the breaker either!
Not to rehash the old thred, but why on earth did bubba pick 32 vdc for everything? Even 36 for starting with a 12 volt tap for everything else wouldve made much more scent (?) ws

The Tapping 12V from a 32V bank doesn't work very well cause when the 12V is being drawn only some of the cells in the 32V bank are being discharged. That causes the voltage to drop on the bank and the 32V charger to kick in. So what ends up happening is the cells that where not discharged by the 12 V draw get over charged and the cells that were discharged by the 12V draw don't get charged enough.

My guages are all 12 V powered by a 32V to 12V power supply so I could change the lights to that also. But it would be easier to just find a resistor to bring down the voltage to the 4 new guages I'm adding.

Brian

[chris piazza]

Brian,

Try this site, it will tell you at least what resistor you need . I just did this on my boat on the bridge for the five 24v lights for the pumps. Simple just soldered them in-line and dropped voltage from 32/24v.
http://www.kpsec.freeuk.com/components/resist.htm
Just be advised that Radio Shack don't carry crap any more, I had to go to an electronic parts specialist to get the wattage I needed.

Chris
Superior Nights 53C

[yachtsmanbill]

The PO did a really half-assed job of tapping off the 12vdc generator battery for electronics.
Aside from the electronics, both Atlantes heads and a potable water pump are now 12 vdc. I am still in the process of ripping out some of Bubbas handiwork, but most of the routing is done with new wire. Almost all helm stuff has been converted as the terminus blockus are right there. This all started because of the deck courtesy lights being 32 vdc "festoon" bulbs are like $20 each GMAFB !
I am going to leave the nav lites 32 vdc for higher visibility. As for now, the 12vdc battery is run on a charger only, but I do have a Delcotron alternator thats getting mounted to the starboard engine for charging while underway. ws

[REBrueckner]

Brian,
Your question involves several subtle issues.

First is that each 12 volt bulb with a different wattage requires a different size resistor inserted. Second, you can't measure the resistance of a cold bulb; it's much less than the rated (hot) resistance. Third, when you calculate the size of the resistor to insert, you need also determine its power consumption as you'll likely have to specify a POWER resistor or it may burn up.

The reference by Ted does the calculations...be sure you record also the power requirements calculated there. Likely a 10 watt (power) resistor will be big enough, but don't guess. I can do the calculations if you have the data posted here: All I really need is the rated wattage of each of your 12 volt illumination bulbs.

[SKYCHENEY]

You could just buy 100 of the 32v bulbs and post the other 99 here for sale. Just a thought.

[Brian Degulis]

Need to know the resistance of the bulb and the wattage. If it is were a 12 watt bulb it would be 1 amp, i.e. Watts divided by volts. The resistance is low but it is likely .16 ohms or something similar. As soon as you take a meter and know that, i'll do it or send you the link i've posted before which has plug ins and does the math for you.

Don't get nervous if the numbers get very small...just do the math.

Stupid me...i'll put it in here now.

http://ourworld.compuserve.com/homepages/Bill_Bowden/r2.htm

And ther are many more. Search for voltage dividers.

Ted
PS the resistor should be double the wattage of the bulb or greater.

I went to the sight and it apears to be showing 2 resistors one on the + side and 1 on the -. Why would this require 2 resistors or am I reading it wrong?

Brian

[TedZ]

The one on the plus side is the bulb...the other the resistor. If you want to supply 12 vdc to all of the gauges, wire them in series - you will do that anyway - and measure the resistance as a group and add up the wattage.

this is not cancer research. Buy a 100 resistor assortment from radio shack and come close to what the numbers show. If a resistor gets hot, parallel them with resistors 2x the rating of the one, as two resistors in parallel end up being 1/2 the total. The hot bulb phenomenon is correct but play and see if it gets dim...likely won't.

[Brian Degulis]

I'm only needing to power 4 12V bulbs with 32V I don't want to wire in series cause then one goes out and none work. The Bulbs are G3 or # 5 and there will be 4 being powered. What resistor to buy?

Brian

[yachtsmanbill]

How about one of those old Chrysler ignition resistors with the ceramic block.
General motors used a resistance wire for the same thing... X number of inches in length of SS wire dropped the volts to the coil from 12 to 9 for steady voltage when cranking. ws

[MikeP]

My helm/FB lights/gauges (1980 53 MY) is like Sky's - 12v provided by several 32v to 12v converters that are the oem Hatt setup in the helm/FB. Seems odd that your boat doesn't have them in a 32v configuration. Wonder if a PO removed them in a misguided moment?

[Brian Degulis]

It has them for the guages themselves but the lights are 32V looks OEM to me but can't be sure.

Brian

[krush]

X number of inches in length of SS wire dropped the volts to the coil from 12 to 9 for steady voltage when cranking. ws

Got it backwards :) 12v during cranking (notice the wire that comes off the starter solenoid to the positive side to the coil), 7-9volts during running to keep the points from burning. Don't let a young punk of the computer age show you up on points!

[TedZ]

I'll do it for you...got to get to high speed...later today.

Ted

[fixin-to-cruise]

The simple solution is to just put groups of 3 bulbs in series connection. They will each get 32/3=10.7 volts. When the batteries are being chargesd they will see abut 37/3=12.3 volts. That should make them last longer too, though they won't be quite as bright, especially when your charger and engines are off.

[TedZ]

OK

32 vdc to a 25 ohm resistor to the lights positive. On the ground side put a 21 ohm resistor. Just get close to those values. Check it with a meter. Those are low resistances but the load is only .56 amps. Watch for heat.

The four lights in parallel are about 10 ohms given that they are about 42 ohms each.

Don't forget the feeds are from the resistor and ALL of the grounds are to the other resistor and then that is grounded.

Wiring them in series does not divide the voltage by three or four or whatever. Try it if you haqve plenty of bulbs!!

Ted

[REBrueckner]

Four bulbs in series should work...and with less than 12v across each, they should last a long,long time....

If you put in a resistor with an individual bulb, makes no difference if the bulb or resistor comes first in the series hook up....the current through the pair will be identical....

[TedZ]

I don't know if anyone cares but a resistor in series with anything does nothing to the voltage. A resistor is a current limiting device. Lowering the current in circuits can lower the voltage, i.e. a voltage divider according to V = IR. This is just basic stuff...first day E101.

Hope this helps.

Ted

[Brian Degulis]

OK

32 vdc to a 25 ohm resistor to the lights positive. On the ground side put a 21 ohm resistor. Just get close to those values. Check it with a meter. Those are low resistances but the load is only .56 amps. Watch for heat.

The four lights in parallel are about 10 ohms given that they are about 42 ohms each.

Don't forget the feeds are from the resistor and ALL of the grounds are to the other resistor and then that is grounded.

Wiring them in series does not divide the voltage by three or four or whatever. Try it if you haqve plenty of bulbs!!

Ted

In a series circuit the voltage is most certainly divided in other words wire 3 12V bulbs in series power them with 36V and each bulb will be getting 12V

Brian

[fixin-to-cruise]

It is important to note that the resistance of a light bulb will change when it is on. You can measure the resistance with an ohm meter when the bulb is not powered but you would need a voltmeter and ammeter to measure the resistance when the power is on. Then resistance is calculated by Ohm's Law. V=IR which can be stated R=V/I or Resistance=Voltage/Amps.

I think resistance goes up when the filament is heated. That makes it a little more complicated to choose the correct resistor. Also, if you have a 5 Watt bulb of 12 volts and are dropping 24 volts across the resistor, you will be dissapating 10 watts outof that resistor. That's a lot of heat and will require one of those large ceramic resistors.

Multiply 10 bulbs x 10 watts and you have the equivalent of a 100W light bulb in heat generation behind the panel. Not a good thing. The series connection of groups of 3 bulbs will cut the heat by 2/3. It's a simple solution.

[TedZ]

In a series circuit the voltage is most certainly divided in other words wire 3 12V bulbs in series power them with 36V and each bulb will be getting 12V

Brian


I'm not sure why you think this...maybe because the current sort of follows your hypothesis.

Maybe this will help.
http://www.rkm.com.au/ANIMATIONS/animation-light-bulbs-series.html

Do the math and see what the voltage does. I have no problem if when you ask a question you say, "Ted need not respond." You don't leave any wiggle room when you say "most certainly." It is hard to be professional when you say stuff that any EE knows is wrong, in general. Now could there be a combination of bulbs that when hot happened to divide the supplied voltage by the number of bulbs? Unlikely, but of course there could be but that is not a rule you want to follow.

All you have to do is come close to the resistors suggested and see what happens when the bulbs get hot. Then you can vary the resistance to make them burn appropriately. I believe i used the hot resistance in my calculations but i could be wrong. Try your system, try various resistors and make it work. This is what you want...do it. Report back.

Ted

[Brian Degulis]

Just to be clear are you saying that if I take 3 identical 12V bulbs and wire them in series + to - + to - + to - then put 36Vs to the un taped + and - each bulb will not be running on 12 volts?

Brian

[REBrueckner]

Brian,
You are correct in saying that three bulbs in series in a 36 volt circuit will be running on 12 volts each. I don't understand Ted's comments.

Anytime you put three resistors of equal value in series, each shares an equal portion of the overall total voltage drop...and each has the identical current thru it...

[TedZ]

I was wrong. Sorry. Hell to get old.

If the resistance is equal and the lights are in series, the voltage will be divided equally.

What i was trying to say and the values i gave are fine is if you supply the three bulbs from a resistor with three connections (wires) to each bulb, i.e. a series - parallel hookup, then if a bulb burns out the rest will not go out but they will get brighter, maybe too bright. Then the other resistor in the ground side and you will have a typical voltage divider.

It is one thing to be wrong and another to be wrong and a jerk about it! That's be me!

Ted

[Brian Degulis]

2 27 ohm 2 watt resitors wired in parellel on each 12V bulb brought the voltage down to 12V. I ended up using 2 cause it's what I had on hand.

Brian