Does any body know if there should be a .9 volt drop between the back of the alternator and the battery for alternators with external regulators?

IT is hightly possible, if the car has the factory wiring, and it is in less than perfect shape, I could see it happening. The set up in this cars was unique and odd. Look for any abraisions in any of the heavy guage wire on the long route between the alt and the battery.

Wow. Almost a full volt is quite a bit! You'd be better off switching to an internal regulator and then upgrading the alternator. Just my .02 Running your accessories at a higher voltage (ideally 14V) means less current (amperage) draw and that helps extend alternator life. -Jabin

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Gmachinz Sales and Performance
"...updating the level of performance..."

Thanks for the info. I just replaced the 10 gage wire that runs from the alternator to the voltage regulator and then back again to the battery. Both the alternator, voltage regulator and battery are less than 6 months old. Tha battery charges to 12.5 volts. I just don't see why I have the .9 voltage drop.

I just don't see why I have the .9 voltage drop.

Way back in the 1800's Georg Simon Ohm discovered that the voltage drop across a device was directly proportional to the current and the resistance, as in E = IR. You have a 0.9V drop from your alternator to your battery because the product of the current and resistance of the wire equals 0.9.

Which is a good way of saying that either your contacts between the wire and the device present resistance or that the wire itself presents resistance. Try busting out an ohmmeter on the wire, and/or use a good electrically conductive grease at the terminals.

Due to my experience as an engineer, I am fully aware of ohms law. I have a pretty thorough understanding of electronics, yet after analyzing the circuit I cannot come up with a good explanation for the voltage drop. I'm curious if the voltage regulator functions in a way that I am not familiar with. Anyhow, I suppose I will just have to learn to accept it, seeing how the battery is maintaining a healthy charge.

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http://community.webshots.com/user/cmeyer124

I don't have tons of theoretical experience with this stuff but I have traced down electrical gremils from time to time. I know 10ga sounds big- it seems like it would be plenty big to me too, but remember that on many GM cars with HEI ignitions they use a wire that big JUST TO POWER THE DISTRIBUTOR!

As a test.... might break out your big a$$ jumper cables and run a parallel connection (temporarily) with that 10ga wire. Then check the voltage drop again. If something as big as a jumper cable can't get rid of all/most of the voltage drop you're seeing then there's got to be something else going on besides wiring resistance.

Nothing beats actually doing some simple experiments to see if the results agree with one's theory of what SHOULD happen. Often you find the theory is dead-wrong (although I'm not going to argue against Ohm's law!) or there's some ALTERNATE phenomenon going on at the same time that you never thought about and is skewing or counteracting things.


[This message has been edited by Damon (edited June 06, 2005).]

One really nice thing about electricity, it follows the laws of physics, and Ohm wrote one of those laws. The only way to get a voltage drop is to have resistance and current flow, it doesn't work any other way.
10 gauge wire has a resistance of about 1 ohm per 1000', so a 10' section of it will have a resistance of about .01 ohms. If your alternator is delivering 90 amps, you will have 0.9 volts drop across the wire. If you add in the resistance of the crimp connections on each end you probably triple the resistance. Now you're dropping 0.9v with just 30 amps from the alternator.