I have a number of bulbs running off a 36V transformer. 36V bulbs are quite expensive and so I wanted to use 12V automotive bulbs in the circuit. Has anyone here ever done this?

If I place 3 12V bulbs in series I should be able to make them work, the problem is that when one blows, the other 2 will follow.

ADDITIONAL INFORMATION: I have dismantled one of the 4 lamp units and I'd like to provide additional information, mainly for future readers that want to address a similar problem. I made a silly mistake! I thought they were 36V because I was reading from an oxidised base lamp, but it turns out the bulbs are actually 150W 32V Now it seems this is a "designer transfomer/lamp set for pool illumination". I have only found one company that sells this ODD voltage lamp/transformer. The fact that the bulbs blow out very frequently (only a few hundred hours of operation) makes me think that maybe they are poorly engineered so they can sell more lamps.

The Transformer is 32V 4.7 A 150VA 240-250V 50Hz. I assume it is AC. The cost of the bulbs are around $30, hence my desire to find an alternative solution.

I thank all the contributors for the useful tips and for reminding me that when they are in series, one will blow and the others will be turned off, which is probably not so bad.

  • The voltage drop across a 12VDC bulb will not be exactly 12V. Therefore the voltage supplied to the bulbs in series will be >12V. One will blow and break the circuit for the rest. – Chenmunka Mar 6 '16 at 10:39
  • @DanielGriscom, Resistive bulbs (such as incandescent) don't care if they receive AC or DC. LEDs, on the other hand can depend on whether they have a rectifier circuit built in. Most good LEDs have a rectifier circuit so they can work on reverse DC voltage or AC. Without a rectifier circuit, an LED will only work if the power is connected in the correct polarity. – Maxfield Solar Mar 6 '16 at 14:26
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    You're trying to fill the void of one bulb with three? Are the 36V bulbs really that much more expensive? -- Also, series blowing of one doesn't mean they will all go. They will all go out, but only one will be bad. You'll just have to find and replace only it at that point. – TFK Mar 6 '16 at 15:05
  • @TFK no, she's trying to reconfigure an array of lights into groups of 3 in series. – Harper - Reinstate Monica Mar 8 '16 at 22:10
  • @DanielGriscom, a previous comment had said "3 in series makes 36V per light" and I replied "You mean parallel" and then he deleted his message... misaiming my pronouns. I should have used the @ feature. I deleted my message too since it was misaimed and irrelevant. – Harper - Reinstate Monica Mar 8 '16 at 22:11

Yes. Three lamps in series will work, provided they are reasonably well matched in terms of current flow and resistance. Automotive bulbs are actually meant to run on 13.8 volts, which is the voltage of a car's electrical system when the generator is running. That gives you about a 20% margin between bulbs on 32v. Can't see a problem. If one of the bulbs blows, all 3 series bulbs will go out. The other two are not blown. If you can't get 3 that match, replace all 3 with another kind.

Another option is to look at LEDs intended for automotive or RV use, but rated for a range of voltage, e.g. 12-30VDC. This multi-voltage rating means they have an electronic switching power supply inside, which adapts the incoming voltage to the constant current the LED needs. Same concept as those 100-240v laptop power supplies, or 100-277v fluorescent ballasts. They commonly specify 24 or 28 VDC as their max voltage, but there's a fair chance any particular brand will work fine on 32V.

Yet another option is to get a 12V DC power supply to replace the 32V power supply. 12V DC switching power supplies are common thanks to the popularity of custom LED lighting. You would not need to rewire; simply leave the lamps in parallel. However, since voltage dropped by a factor of 3, current will increase by a factor of 3 (assuming you stay with incandescent bulbs of the same brightness) - so make sure your wires are thick enough. I would actually recommend also converting to 12V LED at that point, which draws much less power and solves the wire-thickness problem.

Historically, electric streetcars/trams used five 120V bulbs in series powered by the 600V DC trolley wire. 600VDC arcs like crazy, so the bulbs are special "transit grade" types designed to snuff a 600v arc when they burn out.

Resistors are a bad idea. Railways do that too, commonly for headlights, and it's a pain - the resistor must be matched to the bulb. Changing bulb types requires recalibrating and often redesigning the resistor. Also, resistors run at spec have a 700 degree F surface temperature - dust accumulation or a bit of paper could start a fire. Even if you wildly oversize to bring the temperature into safe range, you still have to deal with the waste heat.

By the way, last I was in an auto parts store, they wanted $6 for a 2-pack of bulbs. Don't pay that. Online, they are much cheaper.

32v is a largely obsolete voltage once used in railways. Bulbs are getting very hard to find. If it has anything to do with a pool, I would switch to 12v.

For incandescent lights, heaters and other resistive loads, you don't need to worry about AC vs DC. AC voltage is labeled based on the DC voltage it behaves like: to a resistive load: "32VAC" is that voltage which will make the bulb the same brightness as 32VDC.

  • I have just added additional information to the question so I am adding a solution based on the advice of you experts. – Clarissa Mar 7 '16 at 5:06
  • Sorry but i am not practical with the forum format. So I am adding the comment again with more information: based on the advice I read here, I would exclude the resistor option as it would waste energy without the light. Now that the transformer delivers 150VA at 32V I should be able to add 3 50w 12v bulbs in series. My logic says this should yield less light but make the bulbs last much longer. Just to give you an idea, the cost of a 12v 50w halogen bulb is less that $1 – Clarissa Mar 7 '16 at 5:16
  • @Clarissa I have updated. – Harper - Reinstate Monica Mar 8 '16 at 22:18

3 12v lamps in series will work 3 x 12 = 36. There can be additional sets in parallel if your supply is large enough. The wattage for each set of 3 will be the wattage of 1 lamp. Example a 20 watt 12v set of 3 will draw 20 watts at 36v. A second set of 3 in series with each other would draw a total of 40w and a 3rd set the total would be 60w. If 1 of the lamps in a 3 bulb set burns out the other 2 will go out until the bad lamp is replaced in that series circuit. Additional parallel sets would not be affected.

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    Ed. Great answer except for the wattage of each set will actually add up. A 20 watt 12 volt lamp would draw 1.66 amps and have 7.2 ohms of resistance. 3 in series will add up to 21.6 ohms and still draw 1.66 amps on a 36 volt supply. Since they are getting 12 volts and 1.66 amps they would consume 20 watts each. The key here is for the lamps to be the exact same rating if they are in series and then they will split the voltage and burn at their designed wattage. So the first set will draw 60 watts and each additional set would draw another 60 watts until the supply is maxed out. – ArchonOSX Mar 6 '16 at 12:26
  • Not 60w at 36 volts each lamp at 12v draws 20w so the total wattage is dvides by 3 with 36 volts. In a series circuit the current flow is the same with every thing in that leg the voltage drop on each equal the total – Ed Beal Mar 7 '16 at 4:43

Yes, you are correct. There are a few of ways to run your 12V lights off the 36V supply.

  1. Wire the 12V bulbs in series in sets of three (you need to make sure that the bulbs are the same wattage). This way each bulb will work just as normal unless one blows. If one blows, it breaks the circuit and the other two will go out until the bad one is replaced, just like the old christmas lights. Each bulb will still use the same amount of power (watts) but the current (amps) through the whole string will be the same.
    Eg. If you use 12w 12V bulbs, the current (amps) would be 1 A. (12V * 1A = 12W). So if you use 3 12W 12V bulbs in series on a 36V system, your current (amps) will still be 1 A but your total power will be 36W (12W *3).
  2. If you don't want your bulbs in sets of three, you can us a resister in the place of one or two of the bulbs. You want to make sure that the resistance is equal to that of the bulb(s) that you are replacing. So if you want to use a 12W 12v bulb, the resistance of this bulb would be 12 Ohms (12V ÷ 1A = 12 Ohms). You could also check this with a multimeter. So if you want to just have one 12V 12W bulb, you can just wire it in series with a 24 Ohm resister (the resistance of two 12V 12W bulbs would be 24 Ohms[12 Ohms + 12 Ohms = 24 Ohms])
    Notice, this is not an efficient way to do this because your string with a 12V 12W bulb and the resister will still consume 36W but only 12W is being used by the bulb and the rest is being lost in the resister.
  3. You could use a 3:1 transformer to supply 12V. This is probably the most expensive method but it makes the wiring of the 12V bulbs simpler.
  • I believe #2 is the way to go here. It's a shame wasting that power heating the resistor, but with it in series with the first bulb, you can then place the rest of bulbs in parallel with the first and never wonder which one goes bad. This will also pull the constant 12V along each and every one of them. – TFK Mar 6 '16 at 15:01
  • @TFK I would think if you have the wherewithall to source and wire up one or more 25W resistors (and deal with the heat they dissipate), you should have no trouble ascertaining which of three light bulbs is bad. Seems like a large price for a three-fold reduction in efficiency. – Joel Keene Mar 6 '16 at 17:00
  • @JoelKeene Bulbs are largely variable on resistance based on their heat through startup. It might add a little heat, but the resistor seems more exact of a route. It also seems wiser to maintain a constant voltage along the lights - putting them in parallel. – TFK Mar 6 '16 at 17:26
  • @TFK I would think light bulbs would have to be somewhat accurate in their resistance, at temperature at least, since otherwise their light output would be unacceptably different. Resistors, inn the other hand are anywhere from +-1% to +-10%, with larger, heavier resistors such as these being on the wider end. That's just speculation though. Supposing you're right and resistors are the more exact route, I would still argue that it would probably be a better long term solution to change the power supply or just get the more expensive bulbs, rather than go the resistor route. – Joel Keene Mar 6 '16 at 17:40
  • I guess the point is just that, if you're installing resistors in series with the bulbs, you're creating your own power supply, albeit a very inefficient one, and you have to deal with all the issues coming along with that: housing, wiring, heat dissipation. If the objective is to have fun and see if you can do it, then great, but if the objective is to save money over buying 36V bulbs, then it seems easier to buy a premade, efficient power supply. – Joel Keene Mar 6 '16 at 17:55

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