Bicycle Dynamo with Battery backup problems.

Discussion in 'General Electronics Chat' started by bogyman57, Jun 28, 2013.

  1. bogyman57

    Thread Starter New Member

    Sep 1, 2011
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    Hi!

    I'm trying to wire a 6v DPDT relay to switch between 2 power sources.

    My application is going to use a bicycle dynamo (rectified/regulated to 6vdc) to power the lights - and powered by a 6v battery when the bike is not moving.

    Can the same 6v coming from the dynamo be used as BOTH the switching voltage and the "other" power to the lights?

    I want to power my lights from a 6v battery until the dynamo kicks in.

    I have tried using 1N4001 diodes but I must be doing something wrong - I am having major power sucking issues - the lights come on very brightly at first, and then go dim very quickly (regardless of the power source).

    Could someone please explain (hopefully with a circuit/picture) the correct way this should be wired?

    Thanks!
     
  2. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,969
    744
    Yep sure post your circuit so we can assist .
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,716
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    Unfortunately my crystal ball is in the shop, so I can't divine how you have anything hooked up.

    Now, if you were to, say, post a schematic (just a sketch will do) of how you presently have things wired we might be able to figure out what your problem is.
     
  4. bogyman57

    Thread Starter New Member

    Sep 1, 2011
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    I drew it in eaglecad.
    Sloppy, but I hope you get the gist...
     
  5. blueroomelectronics

    AAC Fanatic!

    Jul 22, 2007
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    Omit the relay and just use the diodes. As long as the dynamo output is higher than the battery output it'll power the circuit.
     
  6. bogyman57

    Thread Starter New Member

    Sep 1, 2011
    7
    0
    wow!
    I never thought of that!
    are 1n4001's ok for the job or should I be using 1n4734's (for example) instead?

    Thanks!!!!
     
  7. Ramussons

    Active Member

    May 3, 2013
    557
    92
    As far as I know, one end of a bicycle dynamo is internally connected to the body of the bicycle: so there cannot be a bridge rectifier connected at its output the way it's shown.

    If a bridge rectification needs to be done, then the Entire Circuit beyond the Bridge needs to be electrically floating (Capacitors, Regulator, Relay, Battery and Lamp with its Holder)


    If the dynamo output is floating, the present wiring shows a discountinity on the return path - the "ground" of the 3 pin regulator needs to be grounded.

    As blueroomelectronics says, "Omit the relay and just use the diodes. As long as the dynamo output is higher than the battery output it'll power the circuit."

    You don't need the regulator - by pass D2 to float the battery across the dynamo output. It regulates (Purists - please ignore the minor variations) the voltage and charges the battery too.

    Ramesh
     
  8. crutschow

    Expert

    Mar 14, 2008
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    You might want to use large Schottky diodes to minimize the diode voltage drop.
     
  9. ian field

    Distinguished Member

    Oct 27, 2012
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    The relay needs more energy to pull in than to hold, so you may have a problem at the crossover point - a charge pump voltage doubler for the relay coil drive might be worth a try.
     
  10. ian field

    Distinguished Member

    Oct 27, 2012
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    The rubwheel (bottle) dynamos are usually grounded - hub dynamos are usually floating.
     
  11. bogyman57

    Thread Starter New Member

    Sep 1, 2011
    7
    0
    Thanks for all the replies!

    It's a hub dynamo that I'm using. (floating ground)

    My original intent for the relay was to conserve battery life.

    Without the relay and only pair of diodes in a "Y" configuration, the battery is always draining regardless of whether the dynamo is powering the circuit or not.

    Am I wrong in this assumption?
     
  12. crutschow

    Expert

    Mar 14, 2008
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    Probably. If the dynamo's output voltage is higher than the battery's, than the battery diode will be reversed-biased and no current will flow from it.
     
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