Turn off power when voltage reached?

Discussion in 'General Electronics Chat' started by doller, Sep 8, 2014.

  1. doller

    Thread Starter New Member

    Sep 28, 2012
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    Hi, I am charging a 5.7V supercapacitor with a solar panel. I want to cut the power when the cap's voltage reaches 2.5V.

    I detect the voltage with a comparator and a 2.5V reference, this turns on a transistor which then switches off a relay.

    My problem is that the relay won't stay off. The millisecond the cap's voltage drops below 2.5V, the comparator goes low and the relay gets back on. So I end up with a buzzer :rolleyes:

    I know this is expected behaviour, what I need is a way to extend the circuit to keep the relay switched off.

    My schematic is attached.

    Thanks
     
  2. #12

    Expert

    Nov 30, 2010
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    You need hysteresis. The method is called positive feedback. Here is one way to do it. The actual output voltage of your op-amp will affect the feedback resistor. I set this one for 1/10th of a volt difference. You can also add a diode to the feedback loop so it only affects the set point voltage in one direction.
     
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  3. wayneh

    Expert

    Sep 9, 2010
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    You need to build in some hysteresis, so that the charging turns off at, say, 2.55V but won't turn back on until it falls to, say, 2.4V.
    Just look for comparator circuits with hysteresis - it just requires a high-ohms resistor from the output to feedback to the input.

    Gaaaa, scooped again by #12 !
     
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  4. ScottWang

    Moderator

    Aug 23, 2012
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    a 2.7V capacitor can't adding a 30V power, it could blow up.
    You need to redesign the charge circuit.
     
    Last edited: Sep 8, 2014
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  5. #12

    Expert

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    6 minutes late and no schematic? :p
    I do wish more helpers on AAC would learn how to do schematics. They save so much time and verbiage when trying to communicate!
     
  6. ian field

    Distinguished Member

    Oct 27, 2012
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    Most solar chargers I've seen used a shunt regulator - its advisable to use an anti-backflow diode in case the shunt element fails S/C.
     
  7. #12

    Expert

    Nov 30, 2010
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    Probably a valid point. I was too focused on the switching to think about fail-safes.
     
  8. wayneh

    Expert

    Sep 9, 2010
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    Haha, well in my case it's not so much about learning how, as about outright laziness.

    But on the other hand and in my defense, teaching a noob how to search around and find things for themselves is worthwhile - sometimes all they need is the right keywords. Sometimes we learn a lot more that way than if we jump right to the "answer".
     
  9. doller

    Thread Starter New Member

    Sep 28, 2012
    14
    3
    omg omg

    It works...

    IT'S ALIVE!

    Thanks for the help


    You can charge a capacitor with voltages above its rating. These ratings give the maximum alowed voltage across the device, or its voltage drop. Until it is reached, you can use 2.7V or 2700V. It won't make a difference.
     
  10. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    A higher voltage across a capacitor than its rated voltage WILL damage it.
    It will explode will all kind of stuff (electolites and metal parts) spitting around.



    Bertus
     
  11. ScottWang

    Moderator

    Aug 23, 2012
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    You are playing the dangerous game.
    When you adding the voltage higher than the rating voltage, it may not explode temporarily, but it doesn't mean that it won't explode, it could be explode anytime, and it just like a bomb when it exploding.

    So a 2.7V ultra capacitor just works less than 2.5V, then it will keep it safe.

    The situations just like these:
    If you keeping awake for 24 hours, you may not need to lie down, but if you keep doing that and day after day, how long will you lie down?

    You will lie down someday that's for sure.
     
    Last edited: Sep 10, 2014
  12. #12

    Expert

    Nov 30, 2010
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    Are you guys missing the circuit I did that sets the charge level on the capacitor to 2.5 volts +/- 0.1 volt?
     
  13. doller

    Thread Starter New Member

    Sep 28, 2012
    14
    3
    I see what you mean but capacitors just don't work that way. When a discharged capacitor is connected to a power source, the voltage across its leads is 0V at first, we essentially have a short. Then, as opposing charges begin to accumulate on both sides of the cap's "dielectric", the voltage starts to rise. This increase is not instantaneous. It is slower if the power supply is 50V and faster if it is 500V. It is slower if the capacitor is 5 F and faster if it is 5µF.

    So at first the dielectric sees 0V potential difference, then 0.01V, then 1V, then 1.5V etc. If at some point in time this potential difference exceeds what the dielectric can handle, the capacitor fails, sometimes explosively. But if you cut off power before the critical potential difference is reached, there is no reason for any damage to occur.

    The maximum potential difference that my supercaps can handle is 2.7V ±20% and #12 correctly allowed for a safety margin when illustrating a resistor ratio to introduce hysteresis.
     
  14. wayneh

    Expert

    Sep 9, 2010
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    I think what is missed is that the switch drawn in that circuit is not manual and is really controlled by the cutoff relay. If you only see the drawing, it does look a little like the voltage source might be connected to the capacitor indefinitely, regardless of what the comparator does. In combination with the OP's circuit the meaning is clear, but not so much in isolation.
     
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  15. ScottWang

    Moderator

    Aug 23, 2012
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    You also to see what others said about the rated voltage for the super capacitor -- Supercapacitor charging voltage?
     
  16. #12

    Expert

    Nov 30, 2010
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    Thanks, wayneh. I'm as guilty as anybody about getting lazy and not reading the whole thread or not drawing the relay coil under the contacts it controls.
     
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