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Just purchased 12V alternator to put on my '26 model T ford. Instructions said without wiring through switch the alternator will slowly discharge the battery when not running. Never encountered this before. Alternator is only rated at 35A. If I put a 40V,10A diode in series with the output will this keep the battery from discharging and will the diode handle that output. (I've never seen one of these charge more than 17-20 Amps and then only briefly.)

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  • Is this an Alternator or a Generator? For a 1926 Model T Ford... Is the smaller terminal labelled "F" ? A picture would help. – Solar Mike May 3 '18 at 17:55
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An alternator already contains diodes before the output terminal that will prevent reverse current flow, in most automobiles the alternator output is wired directly to the battery.

The risk of slow discharge is the alternator input. This line often includes the "check alternator" lamp in series, is is this connection that will slowly discharge your battery if left on with the motor stopped. Only a small switch is required, as the current in this line is no more than is required to light the warning lamp.

If your alternator kit includes an external voltage regulator, it is the input to the voltage regulator that should be switched. But most alternators have the voltage regulator built in.

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A diode will stop reverse current flow.
A well heatsunk 10A diode MAY survive. I'd recommend a 20A diode or 2 x 10A in parallel OR a MOSFET wired to act as diode - ask if interested.

A diode will drop about 1V so at say 15A it dissipates 15W.
That's equivalent to a resistor of R = V/I = 1V/15A =~ 6- milliOhms, so a MOSFET with around say 10 milliOhms om resistance will dissipate far less energy. (FET at 10 milliOhms. Power = I^2.R = 15^2/0.010 = 2.25W.

It seems unlikely than an alternator would do what you say. A standard 12V alternator has diodes in it as of right as it is an AC machines with conversion to DC using diodes. An olde school generator with a commutator has no explicit reverse current blocking path so MAY discharge a battery. Usually the good olde days mechanical regulator handled this with a contact that was open when not charging. Supply model and brand for a better comment on your new fangled gear.


  • Using MOSFETS as blocking diodes.

A MOSFET may be connected in a manner that allows it to act as a diode.
It is connected so that when votage is connected in the desired "forward" direction, current flows through the drain-source "body diode" PLUS the gate source junction is biased on so the FET conducts. When polarity is reverse the body diode blocks and the FET turns off.QED.

Here is a webpage covering Using MOSFETs as blocking diodes. They provide practical advice as well as the theoretical basis. Very importantly, the FET gate-source junction must NEVER be subjected to > Vgsmax - often 20 to 24 V but sometimes as low as 10V.

From that page:

The alternator is in the position shown as a battery in these diagrams. Note that in each case the polarity of the drain-source connection is the opposite of what is used in practice. (I've used several hundred thousand examples of this circuit - it works :-)).

If a PFET is used it is placed in the alternator positive lead.

enter image description here

If an NFET is used it is placed in the alternator negative lead.

enter image description here

As noted above - the gate voltage must ALWAYS be "in spec" - this should need as little as a resistor or two and a zener diode and maybe a small capacitor.

  • 2 x 10A in parallel...you sure about that? Didn't think the diodes would share very well but have never tried it. You can get diodes larger than 20 A. Like VS-100BGQ100 - schottky rectifier. 100 A – Robert Endl May 3 '18 at 11:02
  • @RobertEndl More horsepower ! :-) - a single 20A diode is better. If using two diodes then to assist current sharing a small resistance can be included in series with each diode - and by simply having a short length of non shared wire per diode you'll probably get enough to balance. Devices within batches are usually "reasonably well" matched. At 1A you only need 0.01 Ohm to get 0.1V drop and that is probably enough to balance approximately matched diodes.. – Russell McMahon May 6 '18 at 12:35

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