M
Member 10617
Guest
Guest
Offline
KVH,
Seems to me it should be entirely off when you are running.
Seems to me it should be entirely off when you are running.
Hi Guest!
smilie in place of the real @
Pretty Please - add it to our Events forum(s) and add to the calendar! >> Here's How << 
There are different connections required for different tests. For checking the regulated voltage, the voltmeter needs to be connected from the D terminal to ground (which can be either the E terminal or any good solid ground point).Let me ask once more. When hooking up the voltmeter to the control box, how is this done? What is the setup for the wire connections and what is the process?
No. The voltmeter draws essentially no current and so using even very small wire will not affect the reading. It's best not to go too small though, just because the wires become mechanically fragile. Bigger wires are stiffer though, which becomes a hassle if you go too big. So I'd say anything from, oh, 24 AWG up to maybe 16 AWG will work fine.Also, in connecting the terminals, is there a specific gauge of wire you need to use?
There are two relays in the control box. One of them (the one with the screw on the upper contact) regulates voltage and current, by virtue of having two windings on it. It has "normally closed" contacts, so when the relay is not energized, the contacts bypass the resistor in the field circuit and apply full generator output voltage to the field coils inside the generator. When either voltage or current rises high enough to move the armature, the contacts open and the resistor limits the current through the field coils to a small value.a) what are the two cut-outs in the Voltage Regulator, and what function is each performing?
Yeah, I guess you could look at it that way.b) how is it that bringing voltage to one side of the ignition light from the Voltage Regulator and to the other side from the Battery, via the solenoid and ignition switch, somehow turns the light off? I'm imagining some balancing of power from two opposite sides, like a reverse tug of war,
Normally, the light comes on when you turn the key on, because the ignition is supplying 12v to one side of the lamp, and the other side is grounded through the generator armature. When the engine starts, the generator starts producing voltage and the voltage on the 'D' terminal (which is connected to the armature through the brushes) comes up. When it gets close to 12v, the light goes out because it now has 12v on both sides.but what happened to the positive and negative that turned the light on in the first place?
. . . . so when the relay is not energized, the contacts bypass the resistor in the field circuit and apply full generator output voltage to the field coils inside the generator. When either voltage or current rises high enough to move the armature, the contacts open and the resistor limits the current through the field coils to a small value.
. . . The generator will actually double as an electric motor and try to turn the engine. . . .but then when current starts to flow backwards into the generator, the current winding fights the voltage winding and the relay is released. When it opens, it interrupts the current, and then stays open until the voltage comes back high enough to close it again. . . .
Yes, through the voltage regulator contacts in the control box. It seems like an odd way to do it, but there is actually a method to the madness. By making the generator self-exciting (rather than drawing power from the battery), it won't draw from the battery when the engine isn't running, which might overheat the coils (and run the battery down).The generator applies its own output voltage to the field coils in the generator itself?
Sorry, that was my bad. The part of the relay that moves is also known as an armature, but it was a poor choice of words. My point was that when either voltage or current rises too high, the regulator relay contacts open and reduce the generator field current, which reduces the output of the generator. Normally, this open-close-open cycle happens too fast to see, but under some circumstances it can be slow enough to make the ammeter jump around.And when you say the output will rise high enough to move the armature, that confuses me. I thought only the fan belt moves the generator's armature.
Yes, exactly. A big diode would do the same thing (more or less), but solid state diodes hadn't been invented yet when this circuit was designed. Alternators have diodes, so they don't need a cutout relay.So it's analogous to a back flow valve or anti-siphon valve? Charge can't go back to the generator?
Sorry, Ed, I forgot that part since I don't do it that way. The temporary connection only needs to carry a few amps (just to run the ignition coil) and some voltage drop here won't affect what you are measuring. So the wire gauge is not critical and one of those Radio Shack "Test Jumpers" would do fine. Or if you have a somewhat larger alligator clip, you can just stuff both wires into the same clip and not worry about the wire at all (except to be sure it can't short to anything).Join the A and A1 leads to provide an ignition feed." -- This is the connection that I'm not sure about. Use a wire and two alligater clips? But what gauge wire joined by the clips?
Any questions about that part? BTW, the screw setting is very sensitive to tiny changes. Move it only a tiny fraction of a turn before rechecking the voltage. If you have the earlier style box where there is a lock nut on the adjusting screw, I recommend also tightening the lock nut before making the check, as tightening it will make some small difference.The instructions continue: "Start the engine and run up to charging speed; the voltage reading will increase until the setting point of the regulator is reached and there should then be no further increase. If the voltage does not conform to specification for the particular model, turn screw (Fig. 7-1) inwards to increase the voltage, or outwards to lower it, then re-check reading."
Sorry, I guess I don't, really. I've mentioned before that I do not agree with Moss on these points, so I'm not interested in trying to explain their point of view. 12.7 is simply not enough to fully charge the battery, IMO. You'd have to ask them why they would recommend such a thing. Unless you are using an Optima battery, the battery chemistry (and hence voltages) are just the same as they were in 1950.And, to further complicate matters, the setting given by Lucas is not the same as the setting given by Moss. The explanation from Moss is that times have changed....
According to an article by Moss (Oct. 2010) "The point at which the contacts open (around 8.5 to 11 volts) is known as the drop-off point.... The point when the contacts close is usually adjusted so that the internal voltage of the regulator is about 12.7 to 13 volts."
But the latest instructions from Moss indicate that "The cut-out relay should be adjusted so that the points open at 10 to 12 volts." And ... "The regulator relay should be adjusted so that the points open at 13 to 14 volts."
According to Haynes, the cut-in voltage is 12.7 to 13.3 v. And the drop-off voltage is 8.5 to 10.0 v. (Same as the Service Manual).
See my problem?