can anyone redo this for me???

Discussion in 'The Projects Forum' started by junior_tm, Apr 9, 2012.

  1. junior_tm

    Thread Starter New Member

    Mar 27, 2012
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    im trying out a project and i have a diagram of but have no idea how to do it because it does not show color or how is it. i never built anything alectronic or circuit board.

    here is what im suppose to follow.


    can anyone redraw this but with color as positive and negative?? and add anything that will make this project easier



    thank you alot
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    The 9V battery symbol uses a longer line for the positive side and a shorter line for the negative side. Other than that, all of the items shouldn't care about polarity, just wire them as the diagram shows.
     
  3. junior_tm

    Thread Starter New Member

    Mar 27, 2012
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    The potentiometer has 3 little "legs" how do I wire it? And how do I wire the switch which has also 3 "legs" just like the "u1"
     
  4. mlog

    Member

    Feb 11, 2012
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  5. panic mode

    Senior Member

    Oct 10, 2011
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    switch with three legs: usually you would use middle and one of the sides
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    If you have an ohmmeter or continuity tester (or just a battery and a small lamp, as from a flashlight), then you can wire them up and see which two legs of the switch work as you want them to in both the "off" and the "on" position. If it is a SPDT (single pole, double throw), then when you use a particular two of the terminals, you won't get any continuity (i.e., the lamp won't light) no matter which position you use. Switching either of the leads to the unused pin will yield a workable solution. If you find two pins that light the lamp no matter which position you use, then you have a switch that is probably a SPST (single pole, single throw) but that has two pins connected to one side of the switch. These aren't too common, but they exist.
     
  7. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    [​IMG]
     
    chimera and junior_tm like this.
  8. wayneh

    Expert

    Sep 9, 2010
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    Well la dee dah !! Mr. Big Shot artist!

    Seriously, that's sweet. No diode around the motor, to protect the regulator?
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    I think that qualifies as "anything that will make this project easier"!

    Even the resistor and the caps are the correct value.

    Now, is the pot the correct value? I couldn't quite read it. Where's CSI when you need them!
     
  10. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    I don't think a diode is necessary, it's not a switching regulator.

    Then, it doesn't seem to make sense to use a 9V battery to drive a 3V motor, the rest is dissipated in the regulator.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    The motor itself will generate some EMF, due to the rotor switching polarity. A diode may not be needed, but it couldn't hurt.

    Basically this is a linear power supply.
     
  12. junior_tm

    Thread Starter New Member

    Mar 27, 2012
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    thank you very much for taking you time and doing this for me. i will be doing this tomorrow when i get back from work.

    again thank you so much
     
  13. junior_tm

    Thread Starter New Member

    Mar 27, 2012
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    i followed the new drawing which was alot better than the previous one i posted. i did just what is here and there is no power going to the light. if i touch the positive wire to the wire going to the potentiometer the lights turn on.
     
  14. chimera

    Member

    Oct 21, 2010
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    wow man.. im impressed with the determination! Keep it up. A lot of people here dont really bother any more with replies and stuff. BTW.. the diode is absoluately necessary. The EMF from the motor, though small, may still cause havoc on the regulator. 1N4001 should do the trick.
     
  15. junior_tm

    Thread Starter New Member

    Mar 27, 2012
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    if i have the r2 the other way around will that be a problem?


    where would i put the diode at?
     
  16. wayneh

    Expert

    Sep 9, 2010
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    Do you mean by jumpering pins 1 (green wire to pot) and 3 (red 9V supply) of the regulator, or by moving the red wire (so it is no longer connected to pin3) to touch the green?

    Also, if you can reconnect things properly and then report back here voltages at various points in your circuit, leaving the negative probe for your meter on the negative pole of the battery as ground. Battery voltage, regulator out voltage, voltage on adjust pin, voltage between the fixed and variable resistors, etc, etc. Very helpful for diagnosis. You may have toasted your regulator.
     
  17. junior_tm

    Thread Starter New Member

    Mar 27, 2012
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    wayneh thank you for asking me to check the voltage on my circuit, i was tryinh to make some led to light up but they wherent so i tryed on the motor im using and it seems to be working fine. i will post the voltage in a min
     
  18. junior_tm

    Thread Starter New Member

    Mar 27, 2012
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    here is the voltage on my circuit. the numbers on red are the voltage. seems like the max out put voltage is 14.9 at the motor. i checked to see and it goes down as i turn the knob. thank you all you guys for replying.
     
  19. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    Put the negative probe of your meter to the minuspole of the battery and measure again with the positive probe. The minus pole will be your reference.

    Measuring directly on the plus pole of the battery your meter should read about 9V.
     
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