Deactivating Alternator Field wire

[Noel]

I am going to install a switch to turn off power to the alternator field wire.

Coming out of the alternator are 2 wires, Black w/ White Stripe, and Yellow with Red Stripe. I don't want to cut them at the alternator, this is already a very hard place to reach, I don't want to add a failure point that far down in the engine bay.

I THINK I have followed this up to the harnes on the top of the engine bay, passanger side.

Do they keep the wire colors consistant? I don't want to cut into the wrong one.


...Plan:

Switch inside to simply turn off the alternator.
Deep Cycle in the back feeding inverter to a battery charger to charge main battery while under way. This way, the vitals always have full voltage.

If the deepcycle gets past 50% SOC, I can turn off the rear inverter, and turn on the alternator, that will just keep the starting battery topped off, not wasting ineficiencies on charging the rear deep cycle.

Once at shore power, charge rear deep cycle. (using 2 chargers to avoid having to swap cables)

[rgathright]

Wouldn't it be easier to just disconnect the alternator at the battery? I can easily travel 30 miles on a traditional lead-acid battery and 60 miles on a Group 24 Gel-Cel battery. Using a good battery charger, you can top off in four hours.

Here is the unit I use... the efficiencies are as high as 88% which means less kilowatt usage. Expect to use .65KW with each battery fill up:
Black&Decker 10 Amp Switching Power Supply Based Battery Charger


On the subject of the inverter and deep cycle battery.

You do not need an inverter and a battery charger. Get one of these battery switches, from your local marine supply store or online at WestMarine. With it you can connect your two batteries together. In the event of 50% charge turn the switch on and you will get the advantage of two batteries.

[Noel]

No.


I will not send low voltage to items made to have 13.6volts... run a fuel pump that is rated to get full voltage, and run it at say 12.2v, amperage goes up, along with heat and early failure. I want the computer and air bags to always have full voltage as well.

Paralleling the batteries only prolongs my low voltage.

I will have a selenoid to jump the batteries together, so if I WANT to, I can either jump start myself, or charge the deepcycle from the alternator if I so desired.

I wish I had pictures of my old minivan from years ago..... looked like a cockpit.

[rgathright]

Yet, in my current "no alternator" setup I am lucky to achieve 12.8 Volts for longer than 15 minutes. I understand your concerns but find them unnecessary.

Have you considered LED lighting for your brakes and turn signals? I found this upgrade drastically reduced my battery consumption. I used to fear stop and go traffic!!

[Noel]

My commutes in the winter are in the dark both ways, and sometimes top 3 hours when I get stuck in weather with idiots.

I already have the equipment needed to do this. I'll still be using shore power, just not as efficiently, but trading that for higher voltage and unlimted trip time.

Originally Posted by rgathright

Yet, in my current "no alternator" setup I am lucky to achieve 12.8 Volts for longer than 15 minutes. I understand your concerns but find them unnecessary.

Have you considered LED lighting for your brakes and turn signals? I found this upgrade drastically reduced my battery consumption. I used to fear stop and go traffic!!

[johnmyster]

I don't think a fuel pump draws more current at lower voltage.

If your concern is running your system at less than 13.8 volts...

Car audio guys used to install 7 cell battery systems (a 3 cell golf cart 6 volt, and a special 4 cell 8 volt battery in series) in order to have 14 nominal volts. They'd charge this one of two ways to around 15.5 volts. Either they'd use an inverter to step up from their 12 volt system, or they'd have a secondary alternator set up to run the higher voltage. Extra voltage into the amplifiers meant they could push extra power out of them.

Nowadays, car audio classes aren't based on amplifier rated power. Thus, the stimulus to do such things no longer exists. If you could find the needed batteries to run a 14 volt setup, you could run battery only and not have your "low voltage" concerns.

Or if you're doing two batteries anyhow (no weight concerns) then get a semi truck battery. They'd handle your commute with no voltage sag at all.

[bodyshop]

Originally Posted by johnmyster

I don't think a fuel pump draws more current at lower voltage.

I'm afraid you need to check ohm's law all electric motors will draw more current when the voltage drops. Also when the current is forced higher that means more amps are flowing through a wire that was designed for a lower current which means more HEAT. That heat leads to failure! If both voltage and current drop then your power drops lowering fuel pressure till the point the engine is not running efficiently or quits

[johnmyster]

Originally Posted by bodyshop

I'm afraid you need to check ohm's law all electric motors will draw more current when the voltage drops. Also when the current is forced higher that means more amps are flowing through a wire that was designed for a lower current which means more HEAT. That heat leads to failure! If both voltage and current drop then your power drops lowering fuel pressure till the point the engine is not running efficiently or quits

Um, yeah, ohm's law agrees with me.

You're saying that heat is directly proportional to current, which means you're thinking of a motor as a resistive device. Cool. Ohm's law works great for resistive devices. Actually, it works perfect.

Well, ohm's law says that for your resistive motor, when I decrease voltage, current decreases. V=I*R. So current is directly proportional to voltage. Voltage drops. Current drops. Ohm's law, right? Yup. I checked it alright. There you go.

You're thinking of a motor as a constant power device. Thus, whatever it can't get in terms of voltage, it makes up in current. However, your very assumption of a resistive motor contradicts this. Resistive and constant power are two different things. But then again you contradict yourself, because you say that power will fall when current and voltage decrease. So does current drop, or does it increase? Which is it? Pick a side here.

Motors are somewhat resistive and somewhat inductive by nature. However, for DC motors, it's safe to say the inductive contribution falls out.

Wowzers, and I wonder why people think HHO is actually a viable concept.

[bodyshop]

Originally Posted by johnmyster

Um, yeah, ohm's law agrees with me.

You're saying that heat is directly proportional to current, which means you're thinking of a motor as a resistive device. Cool. Ohm's law works great for resistive devices. Actually, it works perfect.

Well, ohm's law says that for your resistive motor, when I decrease voltage, current decreases. V=I*R. So current is directly proportional to voltage. Voltage drops. Current drops. Ohm's law, right? Yup. I checked it alright. There you go.

You're thinking of a motor as a constant power device. Thus, whatever it can't get in terms of voltage, it makes up in current. However, your very assumption of a resistive motor contradicts this. Resistive and constant power are two different things. But then again you contradict yourself, because you say that power will fall when current and voltage decrease. So does current drop, or does it increase? Which is it? Pick a side here.

Motors are somewhat resistive and somewhat inductive by nature. However, for DC motors, it's safe to say the inductive contribution falls out.

R drops with voltage and current increases DC motors can NOT be looked at as pure resistive device. Inductive resistance (back EMF) plays a big part as it opposes the current.
Newton's laws say a force applied to (a DC motor in this case) should continue to accelerate. When we put current through a DC motor a torque is induced on the motor shaft. The only thing opposing this torque is some small amount of bearing friction. Yet, the shaft does NOT continually accelerate (as Newton suggests would happen if an unbalanced force is applied to a system). The motor quickly reaches some steady speed at which it will continually operate. The phenomenon that keeps the motor from continually accelerating is the second electromagnetic principle mentioned above. Because the coil is moving in an external magnetic field, there is an induced voltage in the coil. This voltage polarity is such that it opposes the battery and, hence, reduces the current through the coil. In reality, the current going through the motor will only be enough to generate enough torque to overcome friction and the inertial load on the shaft...that is if the motor is strong enough to turn the shaft at all. If we apply such a heavy load to the shaft that the motor cannot turn it, we say the motor is "stalled." Imposing such conditions on a motor for any length of time is a sure way to ruin a motor. A DC motor will "try" to stay at it's constant power, conditions may or may not permit it.

So you have "R" which is made up of the (back EMF) plus the pure resistive part that equal the total resistance. If conditions cause "V" to drop then back EMF also drops which is a part of "R", So if "V" and "R" have dropped now ohms law tells us "I" has increased (proportionately.)

With increased current let look at the pure resistive part of "R". P=R*I^2 Now there is more power going to the resistive part of "R" which is where the heat comes in.

Now the part I said your loosing voltage and current I wasn't talk about the motor but the source (battery) not being able to supply both and it's effect on the motor. (note: if you take a lead acid bat that low you can damage it)

It's to bad that magnetic barrier is there like the speed of light barrier, other wise Newton's law might let a 12 volt battery speed you past the sound barrier

[dcb]

AFAIK, you can (sort of) think of a DC motor as a constant current device.
1. The resistive component will tend to decrease the current in the motor if you decrease the voltage (ohms law).
2. The decreased voltage also leads to slower turning speed and less counter electromotive force (thus bringing the current back up).

Nonetheless, the power consumption goes down (slower = less work being done).



I've never heard of a fuel pump quitting due to an alternator deactivation myself, but that isn't saying much. I know I've driven home without an alternator belt on a few occasions. EOC with bump starts and a few cooling off breaks plus pulling the RAD fan relay was the only thing that got me all the way home. No complaints from the rest of the systems, knock on wood.