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I have an old John Deere tractor that is 24v. It uses 24v to start. The lights are 12v. The charging output indicator lamp is 12v as is all other lights. The lamp has stopped working. I tested the bulb and it works fine on my 12v power supply. Ground is good. Lamp socket has been cleaned. This particular lamp has a marble sized round resistor inline just a couple inches from the lamp. The John Deere wiring diagram shows this resistor but it is obsolete. I was told on a forum that it is 50ohm. I know this is not correct. I need to replace it. What size resistor(s) would I need to replace this resistor in order to drop the voltage from measured 24.5v to 12v not knowing the value of the original?

Added - John Deere wiring diagram:

John Deere wiring diagram

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    \$\begingroup\$ Does the John Deere wiring diagram show a part number or value for the resistor? If so, could edit the question to include either a picture of the relevant part of the wiring diagram as that could help to identify it. \$\endgroup\$ Commented Nov 16, 2023 at 18:04
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    \$\begingroup\$ Are u sure the problem is the resistor? Have u tested for voltage before the resistor? is it 24v? \$\endgroup\$
    – Rodo
    Commented Nov 16, 2023 at 18:06
  • \$\begingroup\$ The wiring diagram does not show a value. The JD part no. is R39550. I found a dealer out of state that had one and he wants $76.00!! I measured the voltage in and out of the resistor at 24.5v. \$\endgroup\$
    – russT
    Commented Nov 16, 2023 at 18:38
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    \$\begingroup\$ @russT Did you test the voltage with the lamp in the socket? The resistor should have 24.5 V on the input side of it and 12 V on the lamp side when the lamp is in the circuit. If it's showing 24.5 V on each side then it would either be shorted (which should burn out the lamp) or it has no load on it (lamp not in socket or bad connection to lamp if it is in the socket). If the lamp is in and the voltage on the output of the resistor is very low or zero then the resistor is bad. \$\endgroup\$
    – GodJihyo
    Commented Nov 16, 2023 at 20:33
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    \$\begingroup\$ use another 12 V lamp in series \$\endgroup\$
    – jsotola
    Commented Nov 16, 2023 at 20:58

2 Answers 2

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Since there have been a few good suggestions with no "official" answer, here's a summary of those "inline" ideas.

Use a resistor to dispense with 12V

(As suggested by The Photon)

Calculate the resistance of the lamp, by measuring current through it while operating at 12V:

schematic

simulate this circuit – Schematic created using CircuitLab

Assuming measured current is \$I=240mA\$, resistance \$R\$ of the lamp will be:

$$ R = \frac{V}{I} = \frac{12V}{240mA} = 50\Omega $$

This same resistance will used in series with the lamp when powered from 24V. Having the same resistance as the lamp, it will therefore also develop 12V, the remaining 12V being across the lamp. Power \$P\$ dissipated by the resistor will be:

$$ P = \frac{V^2}{R} = \frac{(12V)^2}{50\Omega} = 2.9W $$

schematic

simulate this circuit

Use a zener diode, or an active equivalent, to dispense with 12V:

(As suggested by hacktastical)

As calculated above, a 12V zener diode would dissipate 3W:

schematic

simulate this circuit

Use active 240mA current limit

(Also suggested by hacktastical)

Current limiting has the benefit of working even when battery voltage is very different from 24V:

schematic

simulate this circuit

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  • \$\begingroup\$ Please note that tossing in "naked" zeners or BJT etc into tractor electronics isn't really recommended. The first component in the beginning of each of these circuits should be a TVS. Also the raw 24V battery/alternator voltage may or may not be awful, mostly depending on how many tools the tractor got and in case they are running simultaneously or not. Stuff like front loaders that draw a lot of current likely cause surges - the starter will for sure. It's not uncommon that the voltage is actually some 26-28V and oscillating quite a bit up and down. \$\endgroup\$
    – Lundin
    Commented Feb 23 at 15:19
  • \$\begingroup\$ That being said, the LM317 current source solution is likely to work quite nicely even in such conditions and probably the best one IMO. Just add a TVS on the input. \$\endgroup\$
    – Lundin
    Commented Feb 23 at 15:22
  • \$\begingroup\$ My concern is only that the lamp will vary in light strength to the point where it becomes annoying. I've used something very similar to the LM317 solution for driving LEDs on a concrete truck, where you could also count on lots of surges caused by big currents. We first used a voltage regulated circuit similar to the plain series resistor solution, but it became a pain due to voltage drop over long cables, as well as fluctuations in the supply. In general for these environments: current regulation = good, voltage regulation = bad. \$\endgroup\$
    – Lundin
    Commented Feb 23 at 15:27
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What size resistor(s) would I need to replace this resistor in order to drop the voltage from measured 24.5v to 12v not knowing the value of the original?

This isn't possible to know with the given information. You need to know the resistance of the lamp (when operating) or the current it draws in order to choose the resistor.

Because 12 is half of 24, you want the resistor to have the same resistance as the lamp does. Because incandescent lamps increase in resistance as they heat themselves up, you need to measure the lamp resistance while it is operating.

You could measure the lamp current when powering it from your 12 V supply. If this current is I, then the resistor you need has resistance 12 / I. You'll also need a fairly beefy resistor, with a power rating of more than 12 I in watts (I'd choose at least 2x this value to get good reliability).

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    \$\begingroup\$ Ok, so I measured the lamp, which by the way is a tiny little dash indicator lamp, at 250 mA. Not being electronically inclined, I am to divide 12 by .250 which is 48. So I found some resistors in the back of a drawer that are 10W 5% 10ohm and 20ohm. I series two 20 ohm and one 10ohm which resulted in 51ohm. Am I going in the right direction? \$\endgroup\$
    – russT
    Commented Nov 16, 2023 at 19:09
  • \$\begingroup\$ That'll work. If the light looks exceptionally bright then disconnect quickly and add in some more resistance. \$\endgroup\$
    – Transistor
    Commented Nov 16, 2023 at 19:43
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    \$\begingroup\$ Maybe a 12V zener diode in place of the resistor? Or, a 12V regulator? \$\endgroup\$ Commented Nov 16, 2023 at 19:51
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    \$\begingroup\$ @hacktastical Technically yes. In practice a Zener would add a third connection and an additional wire. + you would need a very large one to support dissipation. A DC-DC buck regulator would be the perfect solution as it would save current and regulate much better than resistors. But you would have to redo the entire circuitry. Feasible for someone with experience in automotive electricity and who has time to do it. Even better would be replacing every lights with LED;s. Then you can use much smaller resistors and save a lot of current. \$\endgroup\$
    – Fredled
    Commented Nov 16, 2023 at 20:39
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    \$\begingroup\$ If the Zener is installed in-line with the indicator lamp it will have an internal reverse-breakdown drop of about 12V. No third terminal needed. \$\endgroup\$ Commented Nov 16, 2023 at 20:56

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