How does capacitor works in this circuit? (MakeElectronics page61)

Discussion in 'General Electronics Chat' started by Pithikos, Jan 7, 2013.

  1. Pithikos

    Thread Starter New Member

    Jan 7, 2013
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    [​IMG]

    I understand how this works without the capacitor but with the capacitor I am totally lost. I understand that a capacitor just accumulates energy and lets it pass once enough energy is accumulated. So how does it work in parallel?


    1. Shouldn't the capacitor be between the minus(-) terminal and the on edge of the relay?
    2. The capacitor is parallel to he relay so that means that the current can flow through the relay all the time. So what's the purpose then of the capacitor and how can it still affect the circuit?
    3. Is the capacitor "stealing" energy, while charging, from the energy going to the relay or not?
     
  2. WBahn

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    Mar 31, 2012
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    Please explain what you mean by "lets it pass once enough energy is accumulated" as this makes no sense to me.
     
  3. WBahn

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    I get a "404 Error" when trying to view your picture. Please reupload.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I re upload the picture
     
    • aa2.PNG
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  5. WBahn

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    Mar 31, 2012
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    Which is the normally open and which is the normally closed contact?

    The capacitor is there to provide a path for the inductor current when the contacts are opened. Otherwise sufficient back-emf could develop to damage the relay coil and/or the LED.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    T1 is a normally closed contact.
    And this circuit act just like a astable multivibrator
     
  7. Pithikos

    Thread Starter New Member

    Jan 7, 2013
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    I mean that electrones can't pass through the capacitor until enough electrones are gathered at one edge of it. That's how capacitors work from what I understand.

    Not sure what "inductor current" and "back-emf" mean. I would appreciate if you explain in easy terms as I am just learning.

    I am providing the circuit a steady DC 12V all the time which is the voltage at which the relay works so how is it possible to get damaged?
     
  8. WBahn

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    No. That is how capacitors are destroyed. Start with an uncharged capacitor. What happens when you "pass a current through it" is that electrons enter one terminal and a different set of electrons leaves the other terminal. To the outside world it looks just as though this current passed through the capacitor. But what actually happened is that the electrons entering one side collect on one plate, creating a net negative charge, while electrons move off of the other plate and leave the other terminal of the capacitor, leaving a net positive charge. This creates an electric field, and hence a voltage, between the plates. As more charge collects, the voltage grows. At some point, well beyond the point that you should be attempting to operate the capacitor, the electric field becomes so strong that charges are ripped from on plate and pass through the dielectric that separates them and make it to the other plate. This is generally a desrtuctive and often catastrophic process that damages or destroys the capacitor.

    In normal operation you do not let the capacitor charge to this point. Instead, at some point, you reverse the flow of current and discharge the capacitor, possibly to the point of reversing the polarity and charging it up the other way.

    The voltage across an inductor is proportional to the rate at which the current is changing. If you open the contacts that are powering the coil, the inductor (a coil IS an inductor) will produce a voltage across it that is proportional to the rate at which the current is changing. Let's say you have 1A of current in the coil and you try to reduce that to zero in 1μs by opening a contact. The coil will produce something on the order of one million volts in response, known as the inductive kickback. That is enough voltage do damage or destroy most components. To prevent this, a capacitor is places across the coil (or across other components, depending on the details of the circuit and how it is being operated) which allows this current to continue flowing and to decay at a much slower rate, thus keeping the kickback voltage tolerable.

    But you are switching that voltage via the pushbutton and the relay contacts, so it is anything by "steady" or "all the time" (as seen by the coil of the relay).
     
  9. vk6zgo

    Active Member

    Jul 21, 2012
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    The switch is not a momentary action device--once pushed,it stays on.

    In the relay's unoperated position,t1 contacts are closed,t2 contacts are open,

    (a) When the switch is operated the supply voltage appears across the series circuit current of t1 contacts,the Green LED,& the 680 Ohm resistor.
    Current flows,illuminating the Green LED.
    At the same time,the supply voltage,again via t1 contacts appears across the relay coil & capacitor,which are in parallel.
    Because the capacitor is discharged,it initially looks like a short circuit,& draws most of the current,but after a quite short time,depending on the internal resistance of the supply,it charges up sufficiently,that most of the current can flow through the relay coil,& the relay operates.

    (b) When the relay operates,t1 contacts open,removing the supply from the relay coil/capacitor combination,as well as the Green LED circuit.
    At the same time,t2 contacts make,so the supply voltage appears across the series circuit current of t2 contacts,the Yellow LED,&the 680 Ohm resistor.
    Current flows,illuminating the Yellow LED.

    As t1 contacts are now open,there is no supply connected across the relay coil/capacitor combination,so the capacitor begins to discharge through the relay coil.
    This discharge current is in the same direction as the original operating current,so holds the relay in the operated condition while the capacitor discharges.
    After a time dependent on the CR value of the capacitor & the coil resistance,the discharge current become less than the "hold in" current of the relay,which returns to the initial unoperated condition.

    (c) At this point,the sequence begins again.

    Note: The capacitor will not be completely discharged,so its initial charge time will be less than when starting from scratch.
    At any time,the Green LED will be illuminated for a considerably shorter time than the Yellow LED,so the duty cycle is not equal .
     
  10. Bob T.

    Member

    Oct 22, 2012
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    I think it is for making a stronger induction for a lower power.
     
  11. ramancini8

    Member

    Jul 18, 2012
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    Be careful to put a small resistor, about 51 ohms, in series with the capacitor or the surge current can blow the capacitor.
     
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