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  #1  
Old 25-09-08, 02:18
greg anderson (RIP) greg anderson (RIP) is offline
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Default after market resistor

When I wired my carrier I bought an after market resistor which was to have been the correct replacement and it did not work well -I put the old one back in and my carrier starts well (touch wood) -nuff said -Greg
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  #2  
Old 25-09-08, 04:17
Big D Big D is offline
Darryl
 
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Default Carrier ignition

Thanks for the replies.

I checked things again with a second voltmeter just to be sure. This showed 2.8 volts at the coil when the ignition is switched on and 1.6 volts when the engine actually starts.

I've found now though that the engine starts instantly first time even when stone cold.

Mike - voltage at the battery when running at a fast idle is 7.2-7.4 volts so this seems right eh?

Ajays in Katikati have NOS 6 volt resistors on the shelf so I am getting one of those just to try and see how this affects things.

Cheers

Darryl
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Old 25-09-08, 05:04
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Mike K Mike K is offline
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Quote:
Originally Posted by Big D View Post
Thanks for the replies.



Mike - voltage at the battery when running at a fast idle is 7.2-7.4 volts so this seems right eh?


Cheers

Darryl
Yes, that sounds Ok.. But why isn't the 7.2 Volts getting to your resistor It should be there on the switch side of the resistor. Sounds like dirty-oxidised connections somewhere . You need to run it for a while to get an idea of what's going on - a short run from cold won't tell you much . This would be a good option

Mike
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Old 25-09-08, 10:49
Lynn Eades Lynn Eades is offline
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Default Daryll

Some of the wire used to rewire your carrier is too light for a six volt system. (blue at resistor goes to a skinny white) It's might be fine for a twelve volt system.
The cross sectional area of the wire looks too small.
If you apply six volts to that resistor , you should get 4.2 out the other end.
If you have a high resistance (voltage drop) before the resistor ( in a switch, or corroded terminal etc)under load it will drop your input voltage.
Hook your resistor up direct from a battery, through a test light or bulb, and then apply your volt meter, to check your output voltage. Then you know its not your resistor.
Note. The voltage in your primary ignition circuit varies depending on points being open(0 volts) or closedResistance of coil), changes when cranking, and again when running.
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Last edited by Lynn Eades; 25-09-08 at 10:59.
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  #5  
Old 25-09-08, 12:16
Big D Big D is offline
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Default Carrier ignition

Hi Lynn,

Yes, you are right. The wire to the coil is very fine in comparison. What I will do is run a jumper using decent sized cable, in parallel with the existing wire from the resistor to the coil, and measure it again. That will be a good test to see if that is where the voltage drop off is.

Cheers

Darryl
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Old 25-09-08, 14:12
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Quote:
Originally Posted by Big D View Post
Hi Lynn,

Yes, you are right. The wire to the coil is very fine in comparison. What I will do is run a jumper using decent sized cable, in parallel with the existing wire from the resistor to the coil, and measure it again. That will be a good test to see if that is where the voltage drop off is.

Cheers

Darryl
The coil is drawing about 4 amps from memory.. not really enough to be worried about thin wires .. The main wires to & from the battery - reg. and ammeter and lights need to be heavy though . I would be checking the ignition switch and the connections on it... put your voltmeter on either side of the IG switch .
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  #7  
Old 01-10-08, 16:22
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Bob Moseley (RIP) Bob Moseley (RIP) is offline
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Default Carrier Panel

Hi Darryl - this probably won't help you much, but by way of comparison here are two images of my latest aquisition. Whowever removed this fro the Carrier was very carefull by retaining all the original wiring.

Bob
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File Type: jpg 000_0142 sm.jpg (60.3 KB, 52 views)
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