How alternator light works, a more detailed description |
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How alternator light works, a more detailed description |
Tom |
May 3 2014, 01:30 PM
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#1
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Advanced Member Group: Members Posts: 2,139 Joined: 21-August 05 From: Port Orchard, WA 98367 Member No.: 4,626 Region Association: None |
After reading many accounts of how this circuit works, I felt compelled to investigate further as I did not understand how two positives would cause a light to operate. They won't. One must be somewhat negative to complete the circuit. Internet searches turned up the same basic explanation, still was not buying it. I think it was being oversimplified.
This is how I think the alt light works: When the key is on and engine not running, there is 12 volts + at the alt light power side coming from the fused side of fuse #9. The other side goes to a junction on the relay board with D+. With the key to off and a meter connected between D+ and ground at the relay board, the reading is 12 ohms. As soon as the key is turned to on, the reading jumps to 12.5 meg ohms and the light comes on. If the wire for D+ to the alt is removed, the reading stays the same and the light stays on. Removing the VR caused the reading to jump to infinity and the light goes out. For the light to work, there has to be power to one side of the light and some resistance reading to ground for the other. Looking at the wiring diagram, one can follow the blue wire to the junction at the relay board at D+, then up thru the VR to a set of relay contacts, then down thru a ( resistor ?, not sure) and then down to the DF connection and on to the rotor where the current will produce a magnetic field. After the rotor, it goes to ground. When the alt spins enough RPM's, a voltage is produced and fed back to the VR, causing the relay to open and removes the ground path for the alt light. I could be entirely wrong here, but this is what I see and my readings more or less confirm it. If you see an error in my thinking, please post and let me know. Thanks, Tom Attached thumbnail(s) |
Tom |
May 4 2014, 12:44 PM
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#2
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Advanced Member Group: Members Posts: 2,139 Joined: 21-August 05 From: Port Orchard, WA 98367 Member No.: 4,626 Region Association: None |
stugray,
OMG, you are so right! I don't know what I was thinking yesterday. I do know that you can't take resistance readings with power applied. Must have forgot yesterday. Dave_Darling, The light must have a path to ground, even if there is a resistance between the light and ground. I measured 12 ohms initially. I was talking about the two rectangular "boxes" in the voltage regulator. They don't look like diodes. One is all blacked in, the other is left uncolored or clear. I am thinking they are resistors. toolguy, The light doesn't get connected to a potential less than 12.5 volts thru the diodes, it's connected thru the voltage regulator and then thru the field to ground. I still think the relay in the voltage regulator has the major part in keeping the light off when there is voltage being produced by the alternator. With the normally closed contacts open, there is no path for the light negative side to get to ground. I guess it is like the chicken or the egg. If no voltage is present, light lights, if there is voltage present, the relay operates opening the path to ground, but the light would not light anyway as the two voltages would be the same. Although, looking at the regulation side of the circuit, if the relay did not close and remove the ground, it would be almost like the initial "bootstrapping" was in place and the alternator would not vary it 's output to keep up with demands, so the relay does have an important part in this circuit. I guess I am just being too technical, but when I see something I don't fully understand, I reach out to the guys who may have the answer for me. Often times I have found out information that has helped me in other projects. Thanks for the reply, Tom |
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