Discharging a capacitor automatically

Discussion in 'General Electronics Chat' started by mguptamel, Dec 17, 2012.

  1. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    In attached circuit, I have RC circuit which turns-on transistor as soon as Cap Voltage is >=0.6. Cap voltage keeps going up until it reaches 2.8V. So far so good.

    I want to extend this circuit that capacitor discharges itself as soon as it has reached 2.8V. In essence I want to add capacitor discharge sub-circuit to attached circuit.

    Would anybody be able to help me out.
     
  2. #12

    Expert

    Nov 30, 2010
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    I think that would be called an "oscillator". My first instinct is to use an SCR with a minimum hold current more than the current through the resistor on the upper left. If it's difficult to find that, you can increase the resistance until the current is low enough to starve the SCR after it has discharged the capacitor.

    How you trigger the SCR is dependent on how accurate you need the circuit to be. It might be as simple as a 2 resistor voltage divider or as complicated as a comparator circuit.
     
    Last edited: Dec 17, 2012
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Google "555" and see how they do exactly what you want to do.
     
  4. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    ErnieM: I have checked "555", and getting understanding of it. However, internally it uses Voltage comparator which is far too complex to understand as per my current level of understanding in electronics.

    What I am trying to do it to have basic understanding, and build simple circuits.
     
  5. #12

    Expert

    Nov 30, 2010
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    I think I should apologize for telling you a slick trick. You probably aren't thinking at that level yet.
     
  6. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    #12: I think that would be called an "oscillator"
    You are correct. Can it be achieved without an SCR? Sorry I haven't got any SCR understanding.
     
  7. #12

    Expert

    Nov 30, 2010
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    As usual, there are a dozen ways to do it. They vary in accuracy, expense, and complexity. Look up, "2 transistor model of an SCR". (You can build and SCR out of 2 transistors and some resistors.) Doing that will educate you about SCR's.

    Why do I keep beating the SCR horse? Because it's the easiest way to do this.

    Edit: I just remembered the Programable Unijunction Transistor. You can make a capacitive discharge oscillator out of those. Then there is the neon relaxation oscillator. These circuits do not resemble the circuit you posted, but they oscillate.
     
    Last edited: Dec 17, 2012
  8. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    Just went through SCR's. Hmm!!! How it can be used in provided schematic? The answer could be simple; still looking for an answer.
     
  9. #12

    Expert

    Nov 30, 2010
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    Maybe like this. I have guessed at all the resistor values. You can play with it in a simulator?
     
  10. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    Thanks..I will give it a try.
     
  11. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    I can't get it to work. Would you be able to try to check if works?
     
  12. #12

    Expert

    Nov 30, 2010
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    What results are you getting? Stays stuck "on"?
     
  13. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    yeap!!! That's correct
     
  14. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    As soon as Capacitor reaches approx 0.6V (transistor Vbe), SCR turns-on and remains turned-on.
     
  15. #12

    Expert

    Nov 30, 2010
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    I am attaching a list of SCR's that require 30ma to 80ma to stay on. If you set the supply resistor value to 9 ma, the capacitor will charge up to the gate turn on voltage of the SCR and then the SCR will dump the charge in the capacitor. It will then shut off because the current supplied by the resistor is way less than its advertised hold current.

    What we have here is 9 ma (decreasing) going into 470uf. Two resistors set the leakage to ground in the SCR trigger circuit to 1 ma. When the capacitor voltage rises to about 6.2 volts, the SCR will trigger "on". That will dump the charge in the capacitor. The 15 ohm resistor limits the dump rate so the current will not melt the SCR. When the dump current drops below the hold current, the SCR will turn off and the capacitor will resume charging.
     
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  16. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    This is one of the problem that I find very difficult to comprehend because I am from computing background. In computing, we usually apply logic like IF something Then do something.

    I am trying to do same here, that is, if SCR turned-on then drop Capacitor voltage. Please note, I don't want to use AND/NAND/OR chips as yet. If I do, then that would defy my purpose of learning.

    Thanks in advance for helping me out, and putting in your inputs.
     
  17. #12

    Expert

    Nov 30, 2010
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    I'd fix the circuit I gave you first but that would require thinking.
     
  18. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    Thanks a lot #12
    Trying to understand this, and your comments. Will get back with more questions.
     
  19. mguptamel

    Thread Starter Member

    Dec 14, 2012
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    Understood :)

    Understood

    Didn't get this. Shouldn't it be around approx 2.7volts, given SCR trigger voltage is 0.7 volts? Please ignore my ignorance.

    Haven't got this either. Shouldn't current will flow from 1K resistor into SCR?

    Sorry what is dump current and hold current?
     
  20. #12

    Expert

    Nov 30, 2010
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    The trigger voltage on an SCR is 1.5 volts.
    The current from the 1k resstor does flow into the SCR, and it does that while the charge in the capacitor is flowing through the SCR.
    Dump current is what the SCR does with the charge that has accumulated in the capacitor. Dump current is also a noun. The dump current flows through the SCR.
    Hold current is the minimum current required to keep the SCR turned on.
     
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