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My 1009 left me stranded

GSP

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**** thing wouldn't start, i had to bum a ride. So i need some help. My trucks glow plugs are on a push button. Tonight it wouldn't start, and something was smoking. If your looking at the air cleaner, the thing (highly technical term) behind it, behind the heat shield was smoking. Pretty sure thats not good. Whats back there and could it be my problem? Any help is most appreciated.
 

welpro222

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I haven't messed with mine before, but see if you can get a DC tester and check the voltage on the drivers wire exiting from the resistor pack. Im guessing it should be 12-13v, if its 24v its burned out. Might be a good time to change over to 12v, if you want to buy a 12v starter?
 

GSP

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Thanks everyone! Ill go to it tonight and look at bypassing the resistor. I changed it over to AC Delco 60G about a month ago, they should be still be ok right?
 

cucvrus

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Maybe . Maybe not. over voltage is a real killer of glow plugs. See if you can get the stock circuit card / glow plug system set up and working. It is cheaper then over heating the glow plugs every couple months. I know shut up cucvrus. You are annoying. But I tell the truth. Thanks for looking. I am trying to help no matter what you think.
 

rsh4364

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What I would like to do is pull GP power from both batterys instead of 12v from just front batt.Is there an upgrade for the ballast resistor which is no longer produced? Or can we install 24v glows and still use existing GP solenoid system? And pull power from 24v.Mine is currently manual only gp.No card.After doing a search,I retract my ? Sorry to bring up an old topic.
 
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welpro222

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The best thing to do is to study and learn about how your vehicle works. I'm still learning, but since I changed to 12volt I don't have to worry about that too much. I have owned a 6.2 diesel truck for over 5 years now and Im still learning. Knowing about your truck and having a few commons tools/parts on board comes in handy if you break down somewhere. The 12 volt system is nice because I could jump my glow plug system if the solenoid fails or change out a lift pump on the side of the road if needed.
 

rsh4364

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Maybe . Maybe not. over voltage is a real killer of glow plugs. See if you can get the stock circuit card / glow plug system set up and working. It is cheaper then over heating the glow plugs every couple months. I know shut up cucvrus. You are annoying. But I tell the truth. Thanks for looking. I am trying to help no matter what you think.
cucvrus is right when my resistor bank went bad I lost 7 new glows in a matter of 5 days,its called cascade effect if you care to search.I did the resistor bypass and have had good luck ever since. On my second 1009 it was already bypassed by the fire department that used it before me.Other than pulling GP power from the front batt. the only problem with the bypass mod is pulling the 12v power from the 12v power source on firewall,I ran separate power source from batt.to gp solenoid and 12v.power port.
 
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Gunfighter1

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cucvrus is right when my resistor bank went bad I lost 7 new glows in a matter of 5 days,its called cascade effect if you care to search.I did the resistor bypass and have had good luck ever since. On my second 1009 it was already bypassed by the fire department that used it before me.Other than pulling GP power from the front batt. the only problem with the bypass mod is pulling the 12v power from the 12v power source on firewall,I ran separate power source from batt.to gp solenoid and 12v.power port.
Got a picture of how you wired it up. I was contemplating bypassing my resistor when i put in new glow plugs.
 

tim292stro

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If you source power from 24V, and use 12V glow plugs they will die very quickly.

Brace yourselves, this is a long post with TONS of math...


The point of the ballast resistor bank behind the air cleaner it to literally resist half of the voltage powering the glow plugs - the voltage overcoming that resistance is work that is directly converted to heat.

Each glow plug is about 200Watts (assuming AC60G). This means the total (x8 ) glow plug power is 1600Watts at 12V (200 * 8 = 1600). If the glow plug are only using half of the power, the resistor is using an equal amount of power to allow a double voltage supply (power "Watt" = Volts x Amps).

To explain the circuit better...

Ohm's Law is what you want to understand in these failure modes:

Volts
-------------------------------
Amps x Resistance (Ohms)

That's Volts over Amps times Ohms. You can solve for any one if you have two.

So for a glow plug at rating: 200Watts / 12V = 16.667Amps, further 12Volts / 16.667Amps = 0.72Ohms (720milliohms)

To figure out the total resistance of a parallel set of resistors we use the equation:

Code:
            1
-------------------------
    1              1
----------  +  ----------
Resistance     Resistance


So why is all of this important at all? 8 functioning glow plugs provides a total resistance at 12V of:


Code:
                                 1
-------------------------------------------------------------------  =  0.09Ohms
  1        1        1        1        1        1        1        1
----  +  ----  +  ----  +  ----  +  ----  +  ----  +  ----  +  ----
0.72     0.72     0.72     0.72     0.72     0.72     0.72     0.72
So why is the resistance lower? There are more ways from the battery to the ground... If you had a jug of water and poked one hole in it, it would drain much faster if you poked seven more holes. More holes in parallel allow the flow of water (electrons as current/Amps) less resistance against the pull of gravity (voltage).

So doing the math again to solve for the current: 12Volts / 0.09Ohms = 133.333Amps, and for power 12Volts * 133.333Amps = 1600Watts.

The ballast resistor has to have the same total resistance as the 8 functioning glow plugs to "use" half of the voltage up. Once that resistance is selected and installed it doesn't change (barring a failure of some sort). For a series chain of resistances, we just add them up:

R[SUB]1[/SUB] + R[SUB]2[/SUB] = R[SUB]total[/SUB]

So:

R[SUB]ballast[/SUB](0.09Ohms) + R[SUB]glowplugs[/SUB](0.09Ohms) = R[SUB]total[/SUB](0.180Ohms)

We can do the math again: 24Volts / 0.180Ohms = 133.333Amps (see it's the same!!), and further 24Volts * 133.333Amps = 3200Watts. The system voltage has doubled, but the total circuit current stayed the same - the ballast resistor in this type of electronic circuit is what we call a "current limiting resistor". A very important detail is that the current limiting resistor is designed for a specific load and supply voltage, change either of those and the system get into trouble quickly.

Important note: The ballast resistors in this setup use as much power as the glow plugs do, so the total power has doubled - and remember they will put out as much heat as those glow plugs do!!! Keep them clean and free of debris!!!

Where the cascade failure comes into play is the loss of a glow plug due to burn out. Let's take one out of the parallel resistance calculation and see what happens:

Code:
                            1
----------------------------------------------------------  =  0.103Ohms
  1        1        1        1        1        1        1
----  +  ----  +  ----  +  ----  +  ----  +  ----  +  ----
0.72     0.72     0.72     0.72     0.72     0.72     0.72
Notice that the total resistance for the glow plugs has gone up - your ballast resistor will not change, so while there's resistance to the supply voltage, the limiting is less than the load's demand current as the circuit has become unbalanced:

R[SUB]ballast[/SUB](0.09Ohms) + R[SUB]glowplugs[/SUB](0.103Ohms) = R[SUB]total[/SUB](0.193Ohms)

This has an effect on the total current in the circuit: 24Volts / 0.193Ohms = 124.352Amps, and 24Volts * 124.352Amps = 2984.456Watts. This doesn't seem dangerous, we are using 300Watts less when we lose one 200Watt glow plug - until you see where that power is being used:

The ballast resistor doesn't change, but:

0.09Ohms * 124.352Amps = 11.192Volts... Uh oh... that means the ballast resistor is no longer using half of the system voltage, that voltage now gets used by the glow plugs which are only rated for 12V but they are getting:

24 - 11.192 = 12.808Volts

Now with one glow plug out of commission, your poor glow plug is seeing almost 13Volts across its 0.72Ohm resistance:

12.808Volts / 0.72Ohms = 17.789Amps, and 12.808V * 17.789A = 227.840Watts, or 114% of its design rating.

Lose a second glow plug (down to 6) and each glow plug is burning at 13.714Volts/261.22Watts or 130.6% of its rating.

When a resistor (glow plug) is running over 125% of its rating for any sustained length of time (like a few glow cycles) it'll fail pretty quickly.

So here is the "cascade" ("...a large number of things that happen quickly in a series...") in the cascade effect:
With 5 plugs left, your remaining plugs are at: 14.769V/302.954Watts (150% of rating)
With 4: 16V/355.556Watts (178% of rating)
With 3: 17.455V/423.152Watts (212% of rating)
With 2: 19.2V/512Watts (256% of rating)
With 1: 21.333V/632.089Watts (316% of rating)


The higher the power dissipated in the glow plugs the faster they will fail, so if you don't catch this early, you'd be replacing more plugs.

You should also see from the equation of one glow plug remaining, where the ballast resistor is actually only resisting 2.667Volts out of that 24Volt system, why just taking out the ballast resistor and leaving the 24Volt supply is really bad idea - your plugs will die a horrible death if you put it in a hostile electrical circuit like that (I imagine similar to something like putting a poor goldfish in a tank of boiling bleach).


So, take-aways in sumary:

1) Maintain your glow plug system and it'll work fine, leave it to failure and you'll curse at it like it's a crappy system (where it's actually an operator error).

2) If you remove your ballast resistor, you MUST convert the glow plug supply to 12V, or get 24V plugs.

3) If your ballast resistor shorts out, it will have the same effect as having one glow plug left in the system, or a piece of wire rather than a resistor - they will all fail extremely rapidly.


I hope you found this post informative [thumbzup]

Cheers! :beer:
 
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antennaclimber

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I wonder why my 8 GP's only draw 100 amps?
They are AC60G plugs.

The first picture is 2 plugs, the second picture is 4 plugs. Full scale is 50 amps. These readings were taken from a truck that is not using the resistor. These readings were taken with the GP system is being powered from the front battery and without the truck running.

Thanks for the great information on why the ballast resistor is needed and a fantastic explanation of basic DC theory.
Karl
 

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tim292stro

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S.F. Bay Area/California
I'll admit to a little over simplification for the sake of clarity - I presume a fixed DC resistance. AC60G plugs are dual coil with a PTC coil that regulates the current based on temperature - lower resistance (more current) when cool, higher resistance (lower current) when hot.

The reason I simplified for the above post is that all 8 plugs have temperature variable resistances - if you (okay, not likely you, I'm familiar with your efforts in glow plug cards [thumbzup]) look at the parallel resistance calculation above, one can see that one cold plug (or one mismatched plug) can affect the voltage and power of all 8 plugs. It's for this reason I like individual voltage regulators of the 24V/12V systems rather than how GM did it, if you're using self regulating plugs. The "dumb" single coil plugs are probably better suited for a ballast resistor system IMHO.
 
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