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Fri Apr 21, 2023 10:20 am

Does anybody know what in-line resistance (ohms) I need to run 6 volt trafficators on 12 volts please? I bought a 56 Oval a couple of years ago, which has been converted to run on 12 volt 20years ago, and has the original trafficators in place, but I don't think they have worked for 51 years, which is the length of time the previous owner owned the car.

I have some spares if they are totally shot, and as we're meeting up with this previous owner at the Lavenham VW show in June, it would be good to have working trafficators again - that would surprise him!

https://www.vintage-volkswagens-lavenham.co.uk/

Fri Apr 21, 2023 12:01 pm

My understanding of ohm's law dictates that you need double the resistance to double the voltage on a DC circuit. Whatever the ohmic value of your (existing) device is, it needs to double if you're going to double the voltage.

Fri Apr 21, 2023 1:01 pm

Answered in your post in the split forum?

Fri Apr 21, 2023 2:10 pm

you want to reduce the voltage reaching the solenoids/semaphores from 12 to 6. this involves adding a resistor in series with the semaphore solenoid and the semaphore light bulb. to calculate the correct resistor value you'll need to know the current/amps in these wires, which you can measure with your meter. voltage = current x resistance. do the math and i think you should be alright... also make sure that the wattage of the resistor is correct as well, watts = amps x volts.

Fri Apr 21, 2023 4:07 pm

Hi Martin,

Nice to hear them called Trafficators, reminds me of my Dad.

One thing I will say is the 12V aftermarket trafficators are rubbish in every respect and I use the term respect pretty loosely.

They flop about, don't close properly due to too much side play and mine even open and float half mast when driving - total garbage.

I think I will refit my 6V units and do what you are doing.

Hope you have a good result.

Regards,

Mark (RHD is the go)

Fri Apr 21, 2023 4:45 pm

If it's just temporary to show them off I'd just tap the 12V to them and run them only with the headlights on (lol?). If you're looking for a *solution*, I'd find a digital / solid state voltage-drop or convertor.

Using one of the common 12-6V reducers that turns the voltage into heat hardly ever seems to give any acceptable function, but they are cheap and easy.

Sat Apr 22, 2023 5:36 am

Thanks for the various comments. Just come across this article regarding heat-sinks, and how to calculate what is required. Ideally, it would be good to have a submersible heat-sink, that could be situated in the washer bottle, so as to have warmer water for the screen cleaning fluid!

https://electronics.stackexchange.com/questions/35...w-resistor

Sat Apr 22, 2023 6:05 am

adaysnight wrote: you want to reduce the voltage reaching the solenoids/semaphores from 12 to 6. this involves adding a resistor in series with the semaphore solenoid and the semaphore light bulb. to calculate the correct resistor value you'll need to know the current/amps in these wires, which you can measure with your meter. voltage = current x resistance. do the math and i think you should be alright... also make sure that the wattage of the resistor is correct as well, watts = amps x volts.

An easier way it to use this tool:

https://www.calculator.net/ohms-law-calculator.html

Sat Apr 22, 2023 3:45 pm

My 55 is converted to 12V and has the original trafficators which operate just fine. I just don't leave them on for more than about 30 seconds because I suspect they would get too hot and might be damaged. I did replace the bulbs with 12V bulbs.

Sat Apr 22, 2023 5:18 pm

Martin Southwell wrote: adaysnight wrote: you want to reduce the voltage reaching the solenoids/semaphores from 12 to 6. this involves adding a resistor in series with the semaphore solenoid and the semaphore light bulb. to calculate the correct resistor value you'll need to know the current/amps in these wires, which you can measure with your meter. voltage = current x resistance. do the math and i think you should be alright... also make sure that the wattage of the resistor is correct as well, watts = amps x volts.

An easier way it to use this tool:

https://www.calculator.net/ohms-law-calculator.html

Below is edited from earlier comment in order to avoid future confusion:

I measured the current across my trafficator circuit with a 12V battery and it is 8A. That's probably much more power (96 watts!!) than I should be running through the wires which explains why that circuit is using a 16A fuse instead of 8A.

Sun Apr 23, 2023 10:04 am

ecallaway wrote: Martin Southwell wrote: adaysnight wrote: you want to reduce the voltage reaching the solenoids/semaphores from 12 to 6. this involves adding a resistor in series with the semaphore solenoid and the semaphore light bulb. to calculate the correct resistor value you'll need to know the current/amps in these wires, which you can measure with your meter. voltage = current x resistance. do the math and i think you should be alright... also make sure that the wattage of the resistor is correct as well, watts = amps x volts.

An easier way it to use this tool:

https://www.calculator.net/ohms-law-calculator.html

So just for reference here are the actual numbers. I measured the current across my trafficator circuit with a 12V battery and it is 8A. That's probably much more power (96 watts!!) than I should be running through the wires which explains why that circuit is using a 16A fuse instead of 8A.

From this I can calculate the resistance in the circuit, which is 12V/8A=1.5 Ohms.

So if I were to run the correct voltage through the circuit the current would be 6V/1.5=4Amps.

So the desired power to the trafficator is 6V times 4A = 24 Watts.

To get 24 Watts out from the 12V battery the current should be only 24W/12V=2Amps.

To draw 2Amps from the 12V battery the resistance needs to be 12V/2A=6 Ohms.

So I need to add 4.5 Ohms to the existing 1.5 Ohm circuit.

This 50 Watt 4 Ohm resistor should work.

https://www.jbugs.com/product/9384.html

This is irrelevant math as you are using a current measure on a 6 volt device being powered by 12 volts..

The real math is what most have said, use a 1 ohm resistor. I have been restoring semaphores for nearly 20 years and test every one of them on a current limited supply. A 6 volt semaphore, if undamaged will pull 5.8-6 amps at 6 volts. Do the math (as has been done in this thread) and a 1 ohm resistor will do the trick. It will drop 6 of the 12 volts in the circuit.

Running 6 volt sems on 12 volts, even if for a few seconds at a time is a very bad idea. As heat is generated in the coil, the enamel on the copper windings will start to break down and eventually adjacent windings will short together creating MORE current, MORE heat and LESS magnetic pull for the solenoid and there will be a snow ball effect as the coil degrades.

Sun Apr 23, 2023 12:17 pm

X2 on John's comments.

Anyone who thinks running 6V semaphores directly on 12V is a good idea is only fooling themselves....it is a bad idea. The 6V solenoid coils can get plenty heated on 6 volts alone, touch one and see!

Since a good 6V semaphore solenoid ohms out at roughly an ohm, it makes perfect sense to add this same value in series to maintain the desired current when doubling the voltage to 12V.

A heat-sinked 1 ohm resistor of at least 50W rating, (preferably an even higher wattage rating) should be used.

Sun Apr 23, 2023 1:07 pm

johnshenry wrote:

This is irrelevant math as you are using a current measure on a 6 volt device being powered by 12 volts..

The real math is what most have said, use a 1 ohm resistor. I have been restoring semaphores for nearly 20 years and test every one of them on a current limited supply. A 6 volt semaphore, if undamaged will pull 5.8-6 amps at 6 volts. Do the math (as has been done in this thread) and a 1 ohm resistor will do the trick. It will drop 6 of the 12 volts in the circuit.

Running 6 volt seems on 12 volts, even if for a few seconds at a time is a very bad idea. As heat is generated in the coil, the enamel on the copper windings will start to break down and eventually adjacent windings will short together creating MORE current, MORE heat and LESS magnetic pull for the solenoid and there will be a snow ball effect as the coil degrades.

Thanks for the correction. Nothing wrong with the math, just the assumption. I assumed that the device in question needed a particular amount of power, which works for things light bulbs. But for a solenoid the amount of force generated depends on the amount of current regardless of voltage. So we need 6A of current regardless of voltage.

Sun Apr 23, 2023 1:47 pm

I had a set of NOS semaphores in my 1950 Split, running on 12 volts..They worked fine, just don't leave them out too long. PS- I almost got shot by a Policeman early one morning on my way to work, even wrote a "kaferdave" editorial about it in VW Trends, called "Assault With A Deadly Semaphore" It really happened; I can't make this stuff up :shock:

Sun Apr 23, 2023 3:03 pm

Dave wrote: I had a set of NOS semaphores in my 1950 Split, running on 12 volts..They worked fine, just don't leave them out too long. PS- I almost got shot by a Policeman early one morning on my way to work, even wrote a "kaferdave" editorial about it in VW Trends, called "Assault With A Deadly Semaphore" It really happened; I can't make this stuff up :shock:

I agree that it's generally not a good idea to run the semaphores on 12V and it is an easy fix to add the resistor so I will be doing that. But they do seem to be pretty robust. I have left mine on at 12V for 20-30 seconds at a time a few times a week for a couple years and the measured current is higher than expected if wires had melted together and reduced resistance. I am measuring 8A at 12V = 1.5 ohms. Reported value above of 6A for 6V gives just 1 ohm. If there has been any damage it has increased resistance rather than decreasing.

Sun Apr 23, 2023 6:20 pm

First, when you’re running your car on 12V you would never put in a 24 volt battery!

I have replaced quite a few cooked coils on semaphores because they were run on 12 V.

Just don’t do it!!

Matt

Mon Apr 24, 2023 8:50 am

ecallaway wrote: Dave wrote: I had a set of NOS semaphores in my 1950 Split, running on 12 volts..They worked fine, just don't leave them out too long. PS- I almost got shot by a Policeman early one morning on my way to work, even wrote a "kaferdave" editorial about it in VW Trends, called "Assault With A Deadly Semaphore" It really happened; I can't make this stuff up :shock:

I agree that it's generally not a good idea to run the semaphores on 12V and it is an easy fix to add the resistor so I will be doing that. But they do seem to be pretty robust. I have left mine on at 12V for 20-30 seconds at a time a few times a week for a couple years and the measured current is higher than expected if wires had melted together and reduced resistance. I am measuring 8A at 12V = 1.5 ohms. Reported value above of 6A for 6V gives just 1 ohm. If there has been any damage it has increased resistance rather than decreasing.

What does your coil resistance read cold, with an ohm meter? The resistance rises with heating, so in this case, deducing resistance from the measured current is an invalid approach. The coil may still be around 1 ohm or so if there has not yet been any breakdown in the insulation. But when the coil does cook you can be sure that the resulting short circuits will cause resistance to drop and current to soar.

Stock: 6V across 1 ohm coil = 6A
12V Conversion: 12V across 2 ohms* = 6A

(* 1 ohm coil + 1ohm resistor)

Bottom line: 8A is too high and has to be reduced with a series resistance.

Mon Apr 24, 2023 9:37 am

Zwitterkafer wrote: ecallaway wrote: Dave wrote: I had a set of NOS semaphores in my 1950 Split, running on 12 volts..They worked fine, just don't leave them out too long. PS- I almost got shot by a Policeman early one morning on my way to work, even wrote a "kaferdave" editorial about it in VW Trends, called "Assault With A Deadly Semaphore" It really happened; I can't make this stuff up :shock:

I agree that it's generally not a good idea to run the semaphores on 12V and it is an easy fix to add the resistor so I will be doing that. But they do seem to be pretty robust. I have left mine on at 12V for 20-30 seconds at a time a few times a week for a couple years and the measured current is higher than expected if wires had melted together and reduced resistance. I am measuring 8A at 12V = 1.5 ohms. Reported value above of 6A for 6V gives just 1 ohm. If there has been any damage it has increased resistance rather than decreasing.

What does your coil resistance read cold, with an ohm meter? The resistance rises with heating, so in this case, deducing resistance from the measured current is an invalid approach. The coil may still be around 1 ohm or so if there has not yet been any breakdown in the insulation. But when the coil does cook you can be sure that the resulting short circuits will cause resistance to drop and current to soar.

Stock: 6V across 1 ohm coil = 6A
12V Conversion: 12V across 2 ohms* = 6A

(* 1 ohm coil + 1ohm resistor)

Bottom line: 8A is too high and has to be reduced with a series resistance.
That's what I said! :idea:
slayer61 wrote: My understanding of ohm's law dictates that you need double the resistance to double the voltage on a DC circuit. Whatever the ohmic value of your (existing) device is, it needs to double if you're going to double the voltage.

Mon Apr 24, 2023 11:57 am

Indeed, Slayer61 and others gave the correct solution.

But in addition, ecallaway's statement was being addressed:

"If there has been any damage it has increased resistance rather than decreasing."

It is a common pitfall to forget about the temperature-resistance relationship, no biggie, nothing to be embarrassed about. All good, carry on!

Mon Apr 24, 2023 2:24 pm

Who can forget the heating effect of a current, i.e. I²/R from their school days? This web-site explains: https://uk.farnell.com/i2r-loss-definition

I²R Loss

In an ideal circuit, all the power applied to the input terminals would reach the critical load with no energy wasted or dissipated in the wiring or components along the power path. In real circuits, however, these components always have some resistance, however small. This occurs with both AC and DC supplies, causing electrical losses which are dissipated as heat. These losses can be calculated as below:

Ohm’s Law: V = IR where V = voltage (Volts) across a component, R is the component’s resistance (in Ohms) and I is the current in Amps through it.

Power Law: W = VI where V and I are as above, and W = power dissipated in Watts.

From combining these, we can see that the loss W = (IR)I or I²R. This is known as copper loss. (Or heating effect)

Copper losses can be significant in AC circuits involving wound components like transformers. Such losses occur in their windings, so they are sometimes called winding losses. However, further losses will arise from induced currents flowing through resistance in the components’ iron core; these are called core losses.

Transformer or motor copper losses can be reduced by increasing the conductor’s cross-sectional area, improving the winding technique, and using materials with higher electrical conductivities.

Now you know - Please pay attention at the back of the class!!