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Alternator Stator coil resistance

I just had a Marchal alternator rebuilt. Upon examining it, I find that the resistance between any two of the six wires coming from the stator coil is very low, less than about .2 ohms, which is the accuracy of my meter. Is that right? It seems awfully low. Is there some electrical test that doesn't require taking the alternator appart to determine whether the stator is open or shorted?

-Juan
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Old 01-04-2005, 05:18 PM
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"I find that the resistance between any two of the six wires coming from the stator coil is very low, less than about .2 ohms, which is the accuracy of my meter. Is that right?"

It's very much less than .2 ohms, .2 X 55 amps = 11 volts. NOT GOOD for a voltage loss!

Try .01 to .02 ohms. You need to run a constant (& known) current - 1 amp.
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Old 01-04-2005, 06:55 PM
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Hi Loren,

That makes sense -- the resistance has to be low, lest there be a large voltage drop in the stator. The larger the voltage drop, the hotter and more inefficent the alternator would get. Thanks for the insight!

-Juan
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Old 01-04-2005, 09:10 PM
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Juan,

A bit of disinformation from Loren ... his 'calculation' is meaningless! A measured stator winding resistance of 0.01 - 0.02 Ohms would indicate a SHORTED winding!!!

The stator coil specs from Vol. II of the factory service manual are 0.26 Ohms +/- 10 % for the Bosch alternator. No spec in the supplement section for the Marchal alternator ... implying no significant change in the data!

The stator coils are the Voltage source, so the peak [maximum] Voltage values are developed accross those windings!!!

Result ... only significant Voltage drops are in the positive and negative rectifier diodes for each of the three windings ~ 1.2 V - 1.4 V for each pair.

15 V - 1.2 V = 13.8 Volt output ... 15.0 V / 55.0 A = 0.27 Ohms ...

These are DC calculations that approximate the R.M.S. values of the actual AC circuit in operation.
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Last edited by Early_S_Man; 01-05-2005 at 10:31 AM..
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"The stator coil specs from Vol. II of the factory service manual are 0.26 Ohms +/- 10 % for the Bosch alternator."

WRONG!

As usual, you need to do a measurement before making inaccurate comments.
The current MUST flow thru what resistance is in the winding TOO.

"15 V - 1.2 V = 13.8 Volt output ... 15.0 V / 55.0 A = 0.27 Ohm"

That's a dynamic/effective resistance calculation and NOT the static resistance
measured with an ohmmeter. It's the effective output impedence of the alternator.

Don't guess and use a meter. Also, take a basic electronics class.
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Last edited by Lorenfb; 01-05-2005 at 06:12 PM..
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Warren, Loren,

Thanks for your insight. This is my take on it: the resistance actually affects the efficiency of the alternator, not it's output voltage. The regulator adjusts the output of the alternator to compensate for any voltage drop due to the resistance of the stator.

The stator can still produce a 15V output voltage at 55A while still having a 15V voltage drop due to a .27 ohm resistance. What would happen in this case is that the EMF generated in the stator would be 30V, but then 15V of it is lost to resistance, leaving only a 15V output. Thus the stator would be only 50% efficient, with the 15V voltage drop simply generating heat instead of output voltage. (Same caveat about RMS vs. DC calculations.)

Common sense tells you that the resistance can't be much greater than .27 ohms, lest the efficiency of the alternator drop so much that the alternator would generate more heat than output. Thus it is reasonable to expect the resistance of the stator to be only a few tenths of an ohm. It all make sense!

Warren, Thanks for pointing out that the service manual section about the Bosch alternator specifies a stator resistance. I didn't think to look there when I didn't find it specified for the Marchal.

-Juan
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"Common sense tells you that the resistance can't be much greater than .27 ohms, lest the efficiency of the alternator drop so much that the alternator would generate more heat than output. Thus it is reasonable to expect the resistance of the stator to be only a few tenths of an ohm. It all make sense!"

You got it. You're correct about what the regulator would do, i.e. force the alt. to max.
But if the internal resistance is too high, the output voltage would be less than 12 volts.
Thus it would never charge the battery.

The diodes are baiscally a .75 to 1.0 (X2) or 1.5 to 2.0 volts drop at the 55 amps.
The key output voltage is determined by the stator series resistance, the RPM,
the number of windings, and the field current.
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Old 01-05-2005, 06:44 PM
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Hmmm I don't know. it looks like 16 ga wire. 0.0042 ohms per foot.

How many feet are there in each coil? 2 feet sounds to short(0.01 / 0.0042) and 60 (0.27/0.0042) sounds like too much.

I don't have a number for the two solder joints.
Old 01-05-2005, 09:15 PM
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Measure it! It's easy. Just use a known current.

Resistance = Voltage measured divided by the current.

I use this method to find shorted windings (bad stators).
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Out of curiosity (not the original poster) I measured the coil resistance with the method suggested by Loren. I connected a resistor to a big power supply I have and simultaneously (two voltmeters) measured the voltage across the coil and the current through it. The raw data and calculations are below.

The alternator is out of a 1988 Carrera Cab. I removed the terminals from the diode arrays.
I labeled the terminal numbers as follows; With the regulator at the bottom terminal 1 is above and to the right. Terminals 2 7 are sequential around the circle counter-clockwise.

Terminals__Amps___mVolts____ohms
1 - 2______2.371___187.5___0.079081
1 - 3______2.294___350.4___0.152746
1 - 4______2.295___350.3___0.152636
1 - 5______2.295___364.5___0.158824
1 - 6______2.292___363.3___0.158508
1 - 7______2.294___362.5___0.158021

2 - 3______2.378___189.2___0.079563
2 - 4______2.375___188.7___0.079453
2 - 5______2.371___203.3___0.085744
2 - 6______2.372___202.4___0.085329
2 - 7______2.372___201.3___0.084865

3 - 4______2.298___352.2___0.153264
3 - 5______2.290___365.1___0.159432
3 - 6______2.290___364.4___0.159127
3 - 7______2.287___362.8___0.158636

4 - 5______2.287___364.6___0.159423
4 - 6______2.288___364.0___0.159091
4 - 7______2.290___363.1___0.158559

5 - 6______2.293___349.1___0.152246
5 - 7______2.292___348.4___0.152007

6 - 7______2.292___347.9___0.151789

It is apparent terminal 2 is the center of all the windings. With the two diodes conducting it looks like there would be 165 milli ohms in the windings.

I wonder if there are so many windings because they are physical shifted 30 degrees so instead of getting 6 current pulses per cycle you get 12 for smoother DC.

With the alternator putting out 55 amps that thing is disapating close to 500 watts. (55^2 * 0.165) I guess that is why they put big fans on them.


oops make that 0.155 ohms

Last edited by rick-l; 01-06-2005 at 08:10 PM..
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Rick,

That's really nice work! Presumably the lower .15 ohms on the Carrera alternator vs. the .27 of the Bosch is necessary in the design of the Carrera alternator because of its higher current capacity.

One thing I don't get though is the seven terminals. In the Marchal, the Stator has three terminals. The stator windings form a triangle in the schematic. Each terminal goes to two diodes, one going to D+/B+, and the other going to ground. Actually, the stator appears to have six wires coming out of it. But presumably each terminal on the stator goes to two wires.

So how many terminals does the Carrera alternator have, seven? Why is the restance from terminal 2 different than all the other?

-Juan
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Old 01-06-2005, 08:37 PM
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Quote:
Originally posted by logician
So how many terminals does the Carrera alternator have, seven? Why is the restance from terminal 2 different than all the other?
I'm guessing it is two three phase Y connected windings physically shifted so the phases of one Y is between the phases of the other.

The tie point of the 2 Ys (which should carry no current ??) is brought out. That would be the windings with 1/2 the resistance.
That would be 7 wires, 3 from each Y and the center or neutral of both connected together. (2x3+1)

If I could figure a way to spin the thing I would check the phase relationships and see if that is right.

I was just thinking this thing isn't very efficient. output power is 14 volts * 55 amps = 770 watts. It loses 500 watts just in the winding resistance alone. Thats up to 1270 watts. Thats less that 63% efficient! Does about 2 horseposer to drive the alternator sound about right?
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"With the alternator putting out 55 amps that thing is disapating close to 500 watts. (55^2 * 0.165) I guess that is why they put big fans on them."

Nonsense!

"I was just thinking this thing isn't very efficient. output power is 14 volts * 55 amps = 770 watts. It loses 500 watts just in the winding resistance alone."

You're right. That's why it's nonsense!
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Old 01-07-2005, 07:43 AM
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Quote:
Originally posted by Lorenfb
You're right. That's why it's nonsense!
Which part of the above reasoning is nonsense?

Does the guess about the two 3 phase Y connections sound reasonable?
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Yes, you're very correct about the 55 amp alternator being a 3 phase "Y".
The resistance from the center of the "Y" to each tap is .07 ohms.
The later Marchel 70 amp alt. is wound as a "delta" which has a lower
effective series resistance but also lower peak voltage per tap.

Because the alternator is a three phase unit, you need to determine the efficiency
(losses) dynamically. To do this, you load the alt. at an initial current and measure
it output voltage. Then you load it to another current and measure the voltage.
The change in voltage divided by the change current determines the internal
impedence. Using this, you can determine its' losses. Obviously, this is done open
loop without a regulator. Another method is to measure its' no load voltage
minus the loaded voltage divided by the load current.

You can still use the series resistance of .07 to determine a minmum loss value, i.e. the
losses will be greater than this value, e.g. 55 x 55 x .07 = 212 watts. Using its'
rating of 55 amps @ 14 volts, results in 770 watts or a best case efficiency of;
770 divided by 770 + 212 = 78%. Actually, the losses of the diodes (3 in series
on the 55 amp unit) are big factor too, i.e. 3 volts x 55 amps = 165 watts,
further reducing the efficiency.

Obviously, the best determination of efficiency is:

Power out (Watts) divided by Power in (HP)

1 HP = 746 watts
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Last edited by Lorenfb; 01-07-2005 at 09:39 AM..
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Actually, the Carrera alternator that that Rick measured puts out 92A, not 55A, so that's the current that you have to use when computing its efficiency at peak output. The Bosch alternator puts out 55A, but it's resistance is .27 ohms, not the .07 ohms that Loren used. So this is what I get when I apply the right resistance and current figures for each type of alternator.

Power dissipated by Bosch stator @55A = 55*55*.27=817W. Net output power = 55A*14V = 770W. Efficiency = 770/(770+817) = 48%.

Power dissipated by Carrera stator @92A = 92*92*.082=694W. Net output power = 92A*14V = 1288W. Efficiency = 1288/(1288+694) = 64%.

Actually these estimates are optimistic because they do not include the power dissipated by the diodes.

One other point. This assumes that the Bosch stator uses a Y type circuit like the Carrera. If the Bosch uses a triangle type circuit like the Marchal, then the stator efficiency is even lower because the resistance of each stator segment is actually 1.5x the resistance measured when all the segments are connected together in the triangle configuration.

-Juan
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"Actually, the Carrera alternator that that Rick measured puts out 92A, not 55A, so that's the current that you have to use when computing its efficiency at peak output. The Bosch alternator puts out 55A, but it's resistance is .27 ohms. So this is what I get when I apply the right resistance and current figures for each type of alternator."

Do what????????????
Where do you get .27 ohms? Stop guessing and "playing" with the numbers.

Now, you need to get the multimeter out.
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Loren,

I'm just going by the specs. The Carrera alternator produced more power than the original Bosch alternator used in the early cars. The Carrera alternators produced 92A vs. the 55A produced by the Bosch. Yes?

-Juan
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Old 01-07-2005, 08:22 PM
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"The Carrera alternator produced more power than the original Bosch alternator used in the early cars."

This thread is about the Marchel/Motorola 55 amp alt. Now we're changing the topic
to the 3.2 alternator? The thread was started to determine how to check for shorted
winding.

By the way, the spec is wrong. Don't believe everything you read.
Measure it for yourself!
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Old 01-07-2005, 08:25 PM
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The stator that Rick measured was a Carrera stator, not Bosch. For the Bosch, the .27 ohm resistance is specified in the factory manual.

To do the efficiency computations, you have to use the Carrera current with the Carrera resistance, and the Bosch current with the Bosch resistance. The principles are the same either way, just you have to plug in the right numbers.

The interesting observation is that these alternators really aren't that efficient -- approaching 50%.

-Juan
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