Circuit troubleshooting

Discussion in 'The Projects Forum' started by CitrPug, Apr 4, 2011.

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  1. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Hope someone can help - I have a headache!

    I've been trying to prototype this circuit on breadboard when suddenly a friend brings to my attention a flaw. The circuit itself is to fold wing mirrors in on a car when the car is locked, and to unlock them when it is unlocked.

    It is possible for both relays to be energised at the same time due to the capacitors keeping the circuit live for a few seconds. This is undesirable.

    Neither of us can work out how to prevent this from occuring.

    Circuit is below - note the two wires on the left are the lock motor wires. On locking one is +, one is - When unlocking polarity is reversed.

    [​IMG]

    Any help is greatly appreciated.
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Why do you have those large capacitors in the first place? If they are used to absorb back-emf voltage surges, could you use zener diodes instead?.
     
  3. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Capacitors are there to keep the relay coils energised for approximately 3 seconds as the locking signal is very brief.
     
  4. BillB3857

    Senior Member

    Feb 28, 2009
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    What if you turned your relays over and instead of using the common to drive the mirrors, you used the NO contacts to go to the mirrors. You could then use the NC of each relay to LOCK out the opposing relay.
     
  5. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    My apologies for asking but could you do perhaps a quick sketch of such a set up. I can't work out how to lock out the opposing relay with the NC contacts

    Thanks
     
  6. mbxs3

    Active Member

    Oct 14, 2009
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    Couldnt you just change the location of the diode that is furthest to the left? Instead of having it above the relay coil, move it below the relay coil. If you know what I mean. I am on my iPhone so I can't make a diagram...

    How are those capacitors going to fully charge during the "locking" phase or "unlocking" phase?
     
    Last edited: Apr 5, 2011
  7. mbxs3

    Active Member

    Oct 14, 2009
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    The way you describe the "locking" and "unlocking" circuit that is going on on the left side of your diagram is confusing to me. What I think you are saying is this: (Using the "locking" scenario) When the lock button is pressed, the wire on the far left is of positive potential and the wire to the right of it is of negative potential.

    If this is the case, I don't understand how the desired capacitor would fully charge or how you would keep a ground signal for the capacitor to discharge through. It sounds like the + and - signals you described are momentary, long enough to lock or unlock the doors.
     
  8. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Moving the diode will have no effect that I can see. The problem arises when the relay for locking is energised, the capacitor keeps it energised for approx. 3 seconds. During this time its possible to unlock and so both relays would be energised simultaneously. Same problem arises vice versa.

    From the limited testing I've done on the breadboard, it seems the capacitors will charge from that momentary signal and discharge correctly through the relay coils.
     
  9. mbxs3

    Active Member

    Oct 14, 2009
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    Wouldn't moving that diode as I described before keep the left relay from being energized when the wire that is 2nd from the left has the positive signal? The capacitor would also be blocked.
     
  10. mbxs3

    Active Member

    Oct 14, 2009
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    Nevermind. I am dumb. Need sleep. I see that it makes no difference.
     
  11. mbxs3

    Active Member

    Oct 14, 2009
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    What if you incorporated a limit switch into the design. If the mirror is not fully extended to the normal position then the switch is open. You could wire it into which ever leg of the circuit was necessary to keep capacitor/relay from working.
     
  12. BillB3857

    Senior Member

    Feb 28, 2009
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    Here is something I worked up. After noticing that your relay coils did not use a ground, but a reversible bus pair, I had to rethink the situation. Borrowing from Sgt. Wookie's great motor reversing circuit using only SPDT relays, I incorporated the extra set of contacts in your relays to provide interlocking. If either relay is still picked up, its NC contacts will be open, thereby preventing the opposite relay from being able to pick up.
     
  13. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Thanks for that, I can see how that works now.

    I do have one concern however. In your suggested circuit, if the + was on the right hand wire of the bus pair, would it not blow left most capacitor?

    If it would, would a simple diode fix this?
     
  14. BillB3857

    Senior Member

    Feb 28, 2009
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    The leftmost diode protects the leftmost cap from reverse charge and the rightmost diode protects the right most cap from reverse charging. The diodes will allow only one relay to energize upon LOCK and the other to energize on UNLOCK. Chase the polarity that the appropriate coil will see in each situation. The coils are not polarity sensitive, but the caps are.
     
  15. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Of course it does. I'm being an idiot.

    I traced the circuit and when I got to the capacitor I thought 'Oh!' I didn't go further to see the diode would not let this happen.

    My apologies - I'm just a bit cautious as I don't want to blow a 4700uF cap. To convince ourselves yesterday that the stripe on the caps is the negative side, we sacrificed a 1000uF cap. That went up well enough, let alone a 4700uF! :)

    Thanks for your time. It is much appreciated.
     
  16. BillB3857

    Senior Member

    Feb 28, 2009
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    No problem. Hope things work out with your project. BTW, what software did you use to draw your original schematic?
     
  17. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Thanks. I'll let you know the results.

    The schematic in the first post of this thread was done in MSPaint. I have used a demo version of TINA in the past but find Paint easier. For component symbols I simply image search on Google and copy and paste a suitable symbol.
     
  18. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    Have a look at Bill_Marsden's PaintCad.
    It's basically just a bunch of symbols and templates for use with MS Paint or other graphics program. The symbols are sized to be legible, yet not take up more space than they need to.

    His blog about PaintCad and the downloads are here: http://forum.allaboutcircuits.com/blog.php?b=49
     
  19. CitrPug

    Thread Starter Member

    Feb 20, 2011
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    Excellent tool. Thanks for that.

    Another quick question if I may.

    I've discovered that 4700uF is not enough to keep the relays powered for the time needed. As I've only got 2 4700uF caps I experimented and stuck the other one in parallel to give me 9400uF. This, of course gave pretty much double the time of 4700uF, which is enough.

    Now 4700uF caps are large enough and space is constrained so I can't just parallel loads of caps together until I get the value I need. It's looking likely I'm going to have to buy bigger caps. 8200uF and 10,000uF seem common sizes. If I need to bump the 8200uF up with an extra 1000uF I can. 10,000uF seems excessive. Even 8200uF caps are going to be large. My concern with even larger caps is charge time.

    My question...is there a better way of doing this or are caps pretty much my only option?
     
  20. CitrPug

    Thread Starter Member

    Feb 20, 2011
    83
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    I've now tested the new circuit and all is well except the capacitors.

    Now, I know these capacitors have a 20% tolerance and temperature will affect them. This means I have to make sure I have a delay that is longer than the time I need the relays to be held energised, say 4 seconds.

    In this circuit, one capacitor provides me with about 1.5 seconds and the other nearly double, just under 3 seconds.

    Can anyone make an educated guess if 4700uF really is enough and that there are some resistances in the breadboard and test wiring that are affecting the result or am I going to need bigger capacitors.

    Secondly, why the difference in delay? I don't understand it.
     
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