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baymee

Tecumseh voltage regulator problem

baymee
12 years ago

The voltage regulator of this Tecumseh 7HP engine is not working. (VH70 135008) The voltage at the battery with the engine running is 12.2 VDC. The battery is in good condition and fully charged.

In the picture below you'll see the red wires coming from the stator on the left side. The voltage from each wire to the left side of the diode is 15.6VAC.

On the right side, the battery voltage of 12.2VDC is measured at each end of the fuse and on the right side of each diode.

{{gwi:357077}}

I don't have any manuals to troubleshoot this condition and would appreciate any advice of what to do next.

Comments (16)

  • exmar zone 7, SE Ohio
    12 years ago

    I'd pull the fuse with the engine running and see what voltage you have after the diode. The lower diode seems connected correctly as + would go into/through the diode (into the arrowhead) to the battery. The top diode "arrow" arrangement should be the same. The preceeding was just to determine if the diodes are working, e.g. not "open." Alternately, you could just take them out and check with an ohmeter, reading one way you'll get a relative short, reverse the leads and you'll read infinity that proves the diodes are working. The diodes aren't shorted or you'd have the same voltge at both points.

    Never seen an arrangement like that-pop out diodes. Looks like they were meant to be easily replaced?

    Typically you'll not see a 3V drop across a diode, looks like you're just reading battery voltage so I'd be very suspicious of them. To restate, the arrow on the diode will pass + the way it's pointing, alternately, in the reverse direction it will pass -.

    I'm sure someone will come along who's more familiar with this arrangement. Just trying to help since it's a long weekend.

    Ev

    Here is a link that might be useful: checking a diode

  • tomplum
    12 years ago

    The bridge rectifiers of the 70's are covered in the manual I have. Google tecumseh 4 cycle engine manual 692509 pdf and there will be a few places to find it. Or e me and I'll send a copy.

  • mownie
    12 years ago

    15.6A at what RPM? At high RPM (wide open)there should be something like 28 VAC or a bit higher.
    That is not actually a "regulator" per se. It is actually a "two diode rectifier".
    Apparently the stator has 2 coils. Each diode rectifies its own coil's output and feeds it to the collector buss that feeds to the fuse.
    It sounds to me like you have a couple of open diodes unless you (or I myself) are missing something in the description of symptoms.

  • baymee
    Original Author
    12 years ago

    I found some information online and the AC voltage coming off the stator is 27 volts at about 3400 RPM. I removed the fuse on the right side and measured 27 volts going into the left side of the diode and absolutely zero coming out of the right side of either diode. So, I guess the diodes are bad.

    I found a posting that showed a drawer full of these diodes and I took the numbers (IN 1084) off the photo and googled them. They are $52 each and I'd hate to spend that kind of money and find out that my diodes are OK.

    I don't know if I did the right thing but I applied 12 volts DC to one side of the diode and 10 VDC came out the other side. I reversed the ends and also had 10 VDC come out. Ohms resistance through the diode is 5M.

  • baymee
    Original Author
    12 years ago

    Another question: Since the power to the points comes off the board separate from the charging system, will the magneto supply power to the points? Thus, the diode end of the system only charges the battery?

  • ericwi
    12 years ago

    To my knowledge, magneto ignition is totally independent of any battery charging system. You could remove the battery and all of the related parts & wiring, and the engine would still run OK. A functional test of the diode in question would be to set up a series circuit, with the diode and a 12 volt lamp, powered by a 12 volt dc battery. The lamp should light when the diode has correct polarity, and fail to light, or even glow, if the polarity of the diode is reversed. Anything you can do to keep the diode/voltage regulator assembly cool, when the engine is running, should extend diode life.

  • mownie
    12 years ago

    I agree, $102 for a pair of diodes is robbery at gun point.
    Of course the Pentagon might call that a fair price.
    Might be time to research diodes and learn some terminology/facts because you may be headed toward adapting some standard (think Radio Shack) diodes to this application.
    Having said that.............I suggest diodes rated at about 6 amp with a Peak Inverse Voltage (PIV) of 36 volts or higher would do the trick. A PIV of 36 or higher (because you state you can read 27 VAC at 3,400 RPM) would assure no AC in the rectified output feeding into the fuse.

  • baymee
    Original Author
    12 years ago

    I learned some stuff through a post on another forum and they suggested a radio shack diode. However, I would have to do some soldering and eliminate the pop-out feature. I believe they are military spec diodes.

    I have a pinpoint wire test lamp, 12VDC, which has a ground clip on the other end. I placed one end of the diode on the 12VDC + terminal of the battery and the light did not light up. I reversed the diode and the light did not light up.

  • ericwi
    12 years ago

    The type of test lamp you are using matters. An incandescent type lamp, with a filament, will glow, as opposed to light up, if the voltage is somewhat lower than 12V. Since a normal diode in good condition will drop at least one volt, your lamp will see at most 11 volts. It should glow, and that would indicate the diode is good, if polarity is forward.

  • mownie
    12 years ago

    You would not necessarily be required to eliminate the pop out configuration.
    You could use 1/4 inch copper tubing cut in 3/8 inch lengths (need 4) and put regular crimp type eyelet wire terminals on the ends of the replacement diode wire pigtails.
    Then you use a machine thread screw and nut to secure the eyelet terminal to the short length of copper tube by running a screw through the tube lengthwise. Then secure the eyelet connection to the tube with a nut. Might need a flat washer under the nut. Pop the tube into the existing diode clips and you are set to go.
    Of course you must observe correct polarity.

  • mownie
    12 years ago

    Below is an edited (fast & rough) version of your original to give a bit of graphic explanation to my suggestion on copper tubing to adapt replacement diodes to the existing diode clips.

    {{gwi:357079}}

  • baymee
    Original Author
    12 years ago

    Thanks mownie. I couldn't make any sense at all without the picture. And, yes, the engine magneto runs fine without the battery.

  • baymee
    Original Author
    12 years ago

    And for ericwi, I was using a Snap on test light with a 12VDC filament bulb....no light in either direction.

  • tomplum
    12 years ago

    Gotta love those Mownie diagrams! I remember when these things were about new, those diodes were robbery at like 12$ a piece. 52 yikes!. The ones we had were 3 amp rectifiers on our machines. They weren't over designed by any means if a person used lights etc. How about trying a basic regulator in it's place? Like if you have an extra Kohler one laying around. Providedyou have a full size L&G battery in it. Or if you wanted to use the red star diodes, it may be a neater install just to delete the board and add a fuse holder.

  • baymee
    Original Author
    12 years ago

    I'm already creating mownie's trick set-up. The clearance is close on the right side, so I'm using a flat head bolt. Also, the clips on the right side of the diode prevent you from putting the diode in backwards. I had to do a little Dremel grinding to allow some lateral movement.

  • baymee
    Original Author
    12 years ago

    Installed the diodes today and got 13.5 V at the battery. Thanks for the help, men.

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