How alternator light works, a more detailed description Options

[Tom]

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

[mikesmith]

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.

The functioning of the regulator is a bit sneaky and not necessarily obvious (though honestly you can say this of a lot of pieces of heavily cost-optimised technology).



Things to know:

- The alternator will only produce an output when current is already flowing through the field coil (marked G). This is the coil that spins when the alternator input shaft is turned by the engine.
- There are a couple of capacitors (close vertical lines) in the regulator drawing; I've covered some of their functions here but almost certainly not all of them.
- I only marked two of the resistors in the drawing, there's another that's blacked in - being blacked in almost certainly means something, but it's not clear what.
- Current takes time to start up / slow down, especially when it's flowing through coils like the field or stator coils in the alternator.
- It takes more current flowing through the coil to turn a relay on than it does to keep it on (so the 'turn off' current is lower than the 'turn on' current).
- Like most folks I'll use positive-to-negative when describing current flow; don't be upset that electrons actually move the the other way. 8)

When the engine is off, and immediately after it's started, current flows from the battery, through fuse #9, through the alternator warning lamp (K2), through the D+ terminal on the regulator, through the relay in the regulator and out the DF terminal, to the DF terminal on the alternator, then out the alternator ground back to the battery (current always flows in circles). This provides the necessary field coil current to bootstrap the alternator.

Once the engine has started the voltage across the stator coils begins to rise. Current will start to flow out the D+ terminal once the voltage rises above a certain point (tough to calculate due to the way the field coil behaves, but probably a few volts), and then out the B+ terminal once it rises above the current battery voltage.

While current is flowing out both terminals the voltage on each terminal will be nearly the same (since in both cases the current's coming from the stator coils). B+ will never fall below battery voltage; D+ will never go (much) higher than battery voltage / B+, but may be quite a bit lower at times.

As B+ is connected to the battery, and thus to the other side of the warning lamp (K2) this means there will be no current flow through the lamp and it will go out once D+ reaches battery voltage (no voltage difference -> no current flow).

The voltage on B+ and D+ will continue to rise (assuming the alternator is spinning fast enough) as the battery charges.

As the voltage on the D+ terminal continues to rise, the current flowing through the regulator relay coil and resistor R1 (circled in red) also rises.
Eventually this current is sufficient to energize the relay and it switches, disconnecting the field coil (and in fact shorting it to D- to ensure that no leakage current continues to energize it). The upper of the two capacitors in the diagram helps prevent damage to the relay contacts during this switching operation.

Removal of the field coil current causes the stator coil to stop generating (this isn't instantaneous; the field coil current falls, the magnetic field starts to collapse, the stator induced current falls...).

As the stator voltage falls the current out the B+ terminal falls, reducing charge current to the battery / rest of the vehicle.

At the same time the voltage on the D+ terminal and thus the current through the relay coil / R1 also falls; some current flows via K2 but this is insufficient to keep the relay energized. This causes the relay to switch back, returning the alternator / regulator to a state similar to the bootstrap state. At this point, there are several sources of current for the field coil; the bootstrap path through K2, the residual charge in the lower of the two capacitors, and the the stator via D+ and R2; one or more of these (probably all three contribute) cause the field coil current to rise and the cycle repeats.

The on/off process repeats very quickly; it can be several hundred times a second. During the off phase some current may flow through K2, but the average current will be low enough that the lamp won't light. The duty cycle (ratio of on time to off time) will vary based on the load on the system; more load will cause the B+ / D+ voltage to rise more slowly, delaying the time until the relay switches. With less load, it will rise more quickly. The time the charge circuit spends off will (tend to) be more constant, as it's related to the value of the lower capacitor and the characteristics of the relay coil, which don't vary with load.

HTH, and apologies for any errors or confusion this might cause.