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Reduce 12vdc to 6 v dc


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You need to be more specific, what do you want to power?

If it's very low-power say <10mA then a zener diode would do.
Higher power levels < 1.5A would need a regulator IC, eg. 7806, for even more power an external pass transistor would need to be added.

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You could run 2 - 6 Volt batteries in series, and tap the batteries to provide 6VDC for the glow plugs.
...or...
If the proper connections are present on the glow plugs (two connection terminals on the glow plug itself, instead of an internally grounded terminal and a single external terminal), then run two in series. I'd imagine that you don't have an odd number of cylinders on that thing.

Are you sure that tractor started out with a 12 Volt system? Most I've seen (of course it's been decades) were 6 volt system machines.

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My mechanical knowledge is very poor - I don't know much about engines so may I ask a stupid question, what are glow plugs?

If they're anything like spark plugs you need a high voltage in the order of about 10kV, not a low voltage.

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Alun:

Glow plugs are simply "heaters"...they are used in diesel engines to provide the initial heat source (in each cylinder) to start combustion. Once the diesel engine is running, the glow plugs turn off as the heat of compression is enough to sustain combustion. (Diesels are relatively high compression engines).

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Thanks for the info jrcfg,

I agree with EnigmaOne, connect them in series if you can.

Do they all have a common ground i.e. the tractor body or do they have separate +/- terminals, if they have a common ground a voltage regulator might be the only option.

How much power do they use? - to me being heaters it sounds like they're pretty high powered so at 6V a linear regulator would have to dissipate half the power, so a switcher would be a good choice.

Come to think of it if they're heaters then they would behave electrically as resistors, then you might be able to getaway with just pulsing them with 12V at 25% duty cycle to cause them to dissipate the same amount of power as if they were connected to 6V.

Edit: 50% changed to 25% , I remembered Power = V^2/R!

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Hi,

Why not use a resistor on each glow plug to drop the voltage, then use a timer circuit to supply power to the glow plugs via relays, you only need around 5 to 10 seconds depending on your climate. The resistor will only need to withstand the current being drawn through them for that short period of time. Someone else will have to provide the answer for the resistors values and wattage needed,

(1) pushbutton to start timer, LED comes on.

(2) LED turns off start engine.

(3) time to get back to work. ;D ;D

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Hi Ante,

Yes that's right, this is for warming up your hands on those cold winter mornings ;D ;D ;D.

The right wattage resistor should handle the heat for this short period of time?

Maybe two 10 watt resistors for each glow plug, then mount them to a basic heat sink, a small sheet of aluminium plate.

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Hi Dazza,

Yes it would work, the easy way would be to use one extra glowplug in series with each as a resistors. A glowplug have a dynamic resistance (it changes with temperature) and is not easy to make a good voltage divider with a conventional resistor. The extra plugs can be fitted under the seat for cold winter days. ;D

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