Voltage regulator help

Discussion in 'The Projects Forum' started by neo8222, Mar 29, 2011.

  1. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
    Hi all, I'm in need of a little bit of help with a voltage regulator project. My friends motorcycle has a alternator with a internal voltage regulator, a previous owners bad wiring has resulted in the regulator burning out. the factory wants $300 to replace it!! I would prefer to just build one myself. I have a basic idea of electronics and I can build a basic regulator but my problem is that I'm unsure of what regulator chip would be best suited for the task and once made would the battery get over charged? The desired output voltage is between 14.1 and 14.5 volts
     
    Last edited: Mar 29, 2011
  2. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
    ok well i think i have a design that will work but if anyone else wants to chime in on whether it will function like i want, they are more than welcome to

    [​IMG]

    [​IMG][​IMG]

    i added diodes for protection against short circuit, and adjusted the R2 resistance. would this function to charge a motorcycle battery and not over charge once the battery is at full power? would it also be able to handle a alternator spinning up to 11,000rpm in short bursts?
     
    Last edited: Mar 30, 2011
  3. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The LM317 regulator will go into shutdown. It is not capable of supplying the current needed to charge the battery and run the lighting.

    I have no clue how the generator on your particular motorcycle works. It might be faulty, or the electrical system could be faulty, which may be why the regulator failed to begin with.

    Automotive alternators control the output voltage by controlling the current through the field winding (which is actually the rotor in most alternators). Some motorcycle generators use rotating permanent magnets inside or outside a field winding, and an SCR circuit to more-or-less regulate the output. I don't know enough about your situation to give any advice - except that your LM317 circuit will be fried to a crisp.

    The charge voltage can be 14.1-14.2v for starters, but should drop down to ~13.8 after a few minutes, or the battery will be overcharged.

    A headlamp may be rated for anywhere between 30 to 60 Watts. Then you have an assortment of other lights; parking, turn-signal, brakes, etc. Then you have the battery to charge. Then you also have to supply the ignition with power, and whatever other accessories (like instrument panel lighting, etc) possibly fuel injection, etc. ad nauseum.

    It's going to be mighty tough for us to suggest anything without a LOT more information from you.
     
    Last edited: Mar 30, 2011
  4. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
    ok some more info. the alternator is NOT a permanent magnet alternator and i have checked to make sure all the other parts of the alternator function (including both windings, rectifier, and brushes. and yes like you said the power is adjusted by controlling the field winding voltage to regulate power output. the original regulator was a TO-3 package regulator (since that is the original regulator i can't see the field winding circuit being a high current draw) and i have no idea what happened to it.

    here is a semi-schematic out of the service manual to show how it's hooked up on the inside. the diodes in the regulator are providing somewhere around 650 ohms one direction
    [​IMG]

    the previous owner replaced that with what seems to be some random one that just didn't work (wrong unit). looking at that schematic from the service manual it seems as though the load is not carried by the alternator but instead by the battery and the alternator just charges the battery. correct me if i'm wrong but it also seems that the large current output of the alternator is not going through the regulator. that being the case it would seem as though a low current regulator could still handle it. ( i could be wrong here) what other kind of information do you need? also im having a brain fart, E is ground/adjust F is output and IG is input correct?
     
    Last edited: Mar 30, 2011
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    I think that you meant that the diodes in the rectifier assembly when using the "diode test" function give somewhere around 650mV in one direction".

    You need to ensure that each of them indeed conduct one way, but not the other.

    E = chassis ground; the frame of the bike, common with the battery negative terminal.
    I = the current source for the regulator, and where the regulator reads the voltage of the system.
    F = where the current is sunk by the regulator from the field, constantly adjusted to keep the system at about 13.8v; but the current sunk from the field must also be limited, or the regulator could easily burn up if the ignition were on and the engine were not turning.

    Ahhh, do you mean the regulator or the rectifier?

    The alternator has to provide enough power to charge the battery, PLUS operate all of the accessories (lights, ignition, horn, etc. ad nauseum). Once the engine starts, the battery basically just acts like a big capacitor (not exactly, but that's the easiest way to explain it).

    Correct. A relatively small current in the field creates an "adjustable electromagnet"; the rotation and torque of the engine induces a much higher current in all three of the stator windings.

    It's actually 3-phase AC at the ends of the "Y" in the middle of the rectifier bridge, but the full-wave 3-phase bridge converts it to rippled DC.

    If any of the rectifiers in the bridge are open or shorted, that could result in high current in the field winding.

    I cannot tell you offhand how much current is required in the field winding to produce usable output current @ 13.8v. It will vary significantly over the RPM range of the engine.
     
    Last edited: Mar 30, 2011
  6. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
    no i don't have a diode tester at my disposal so i used a resistance test and all the diodes are providing the same resistance in one direction but a way higher in the opposite so they should be good.

    the previous owner replaced the regulator with some random unit i guess thinking all of them worked the same...

    as far as the case of the field winding being relatively low current in comparison to the rest of the system, do you think i should just use that circuit i made and see what happens? worse thing that could happen is the regulator blowing right?

    so my question is if u were in my situation with the limited info i can get would you build the regulator and try it out? any suggestions on next course of action?
     
    Last edited: Mar 30, 2011
  7. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    This is the circuit of a rev engineered Lucas Alt reg which is same wiring as your Reg. Will look into using an of the shelf BOSCH RE55 regulator but it will require altering the Field wiring.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Thanks Debe, you caused me to go back and correct my entry for "F =" a couple posts ago. Being distracted, I'd put that the regulator sources current to the field, which it does not; it sinks current.

    Neo, your regulator would not be suitable for a field regulator; it's not close to what might work.
     
  9. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    This is by far the cheepest & easiest, Use a Bosch RE55 Alt REG & alter the Rotor/Field wiring by earthing one side of the rotor per diagram so its compatable with the verry common Bosch reg, The Lucas style is not very common.
     
  10. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    This circuit is a Bosch type Reg if you realy want to build one.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Very interesting, Daryl.

    I'm wondering why they haven't experienced any problems with that last circuit you posted - if they had turned on the ignition but the engine was not running, it seems like maximum current would be flowing through the field; since the transistor is an emitter follower, quite a bit of heat would be dissipated in the 2N3055 transistor - unless there was a centrifugal switch on the rotor, perhaps? I don't know; I'm just speculating.
     
  12. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    I have built one & used it, but you are right if the engine isnt running but switched on the transistor does get fairly warm & needs a heat sink. Most regulators have some feed back in the circuit & oscillate to reduce loading. I dont have such a circuit. But the RE55 Bosch reg at least here in Australia is cheep & readily avalable. Perhaps some one can design a suitable circuit using feed back based on that circuit. Im not into design!! Very hard to reverse engineer the Bosch reg as its set in tough ressin.
     
  13. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
    well since there is doubt to whether the idea i had will work i will look to Daryl for advise since he has the schematics. that Bosch type as you called it (last post on first page) i looked for a Bosch RE55 and i can't find it so i guess i will have to build one following that schematic, the other problem is now with finding the parts to build it.

    i would prefer to get all the parts from the same vendor to help with shipping time. that being said the zener diode (BZY88C8V2) is a little hard to find. anyone know of a alternate i can use? same deal with the BC108. the other stuff (BD139,2N3055,1N4004, and the resisters) i can get easily from digikey

    i also have a few questions about some of the parts on the schematic. the resistor that says "set voltage 500Ω" is that a potentiometer? and the 51Ω 10W resistor, can i use a wire wound resistor for that?
     
  14. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    Set voltage is a pot. 5w & 10w resistors are wire wound. It may be worth waiting as im about to try & reverse engineer this Reg which is an adjustable RE55 type reg. Probably easier to get parts to. Will take a little while.
     
  15. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
  16. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    Use this circuit its mutch better all resistors are half watt Zenner is also half watt. The TIP125 transistor needs a small heat sink & must be insulated from hearsink.
     
  17. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
    which one did you mean? the bosch type or the lucas?
     
  18. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    OOPS. forgot circuit.
     
  19. neo8222

    Thread Starter New Member

    Mar 29, 2011
    12
    0
    hmmm i'm liking this circuit, especially since it uses easy to find parts. i guess this is the one i will attempt to use thanks to everyone that helped. i get a list compiled and get it to my friend so he can get me the cash and i'll buy up some parts and assemble it. i'll definitely let you guys know the outcome. oh yeah, the heatsink what kind of thermal disipation are we looking for? would a 3W@80C* work? http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=HS106-ND and then use a thermal adhesive like they use to attach heatsinks to RAM
     
    Last edited: Mar 31, 2011
  20. debe

    Well-Known Member

    Sep 21, 2010
    946
    184
    6 posts up is a pic of the regulator it has small aluminum plate rivited to the shell of the reg for a heat sink. just remember if the heat sink is grounded, the transistor will need to be insulated from it. If you use this Reg modify the rotor wiring in the modified circuit of yours that i posted.
     
Loading...