How to have capacitor power only one part of parallel circuit?

Discussion in 'The Projects Forum' started by tjohnson, Dec 23, 2014.

  1. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    I added a light to a small battery-powered hovercraft that I made. To do so, I wired a fan and an LED with a resistor in parallel (see attached diagram). I want to have a capacitor in the circuit that only powers the part with the LED, but I can't figure out how to do that.

    Currently, the capacitor powers both the fan and the LED, which is not what I want. Because the fan uses 6 times as much current as the LED, it uses 83% of the power from the capacitor and the LED doesn't noticeably fade out at all. I really don't want the fan to be getting any power at all from the capacitor.

    How can I wire this so that the capacitor will only power the LED and not the fan? Thank you.
     
    Last edited: Dec 27, 2014
  2. #12

    Expert

    Nov 30, 2010
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    Place a diode in the top line between the fan and the capacitor.
     
    tjohnson likes this.
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    And how long do expect that a 100uF capacitor would run a LED, even if it wasn't for the fan?

    I assume we are talking about a standard 20mA 2.2Vf RED LED?

    Answer, it will only take ~0.1s for the LED to go out, and it will be at 1/2 current in less than 0.05s. Is that what you were expecting?

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  4. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    Thanks, I'll have to try that.

    Yes, it's a 20mA LED and it's blue. The capacitor doesn't power the LED for very long, but it's noticeable. A larger one would be better, though.
     
  5. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    Should the resistor be moved in front of the LED or kept where it is in the last branch of the parallel circuit?

    I tried testing both ways on a breadboard using two green LEDs (with resistors) instead of the fan and blue LED, and the second LED didn't light up at all. I must be missing something basic here, but I'm not sure what.
     
  6. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    I got the circuit working now on a breadboard with the resistor kept where it was. I'm going to have to get a pack of assorted capacitors this weekend to try out, because the 100μF one is barely noticeable.
     
  7. Brevor

    Active Member

    Apr 9, 2011
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    Using a high brightness LED and a larger resistor will help, I have seen indicators using a blue led and a 4.7 K resistor running from 9 Volts that were clearly visible. How long do you need the LED to stay on?
     
  8. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    The blue LED that I'm using has a brightness of 2600mcd. I'm not sure if that's considered high brightness or not. I think I want it to stay on (gradually fading out) for about 4.8 seconds.

    I'm not using the LED as an indicator light, but as decorative illumination for the hovercraft, so I want it to be as bright as possible. It was barely visible when I tested it with a 470k resistor, so it would probably be visible with a 4.7k one, but I really don't want to use a larger resistor with it than necessary. So I think I want to use a larger capacitor instead. Thanks for the suggestion though. I learned something from it.
     
  9. ronv

    AAC Fanatic!

    Nov 12, 2008
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    There may be easier ways to do what you want. It sounds like you want the LED to fade when you disconnect the battery. Is that it? Or is there something else that switches it?
    To get 4.5 seconds at 20 ma. will take about 8,000 ufd.
     
  10. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    Yes, having the LED fade when I disconnect the battery is exactly what I want. When I calculated the capacitance value using the equation for RC circuits, I also got a value of about 8000μF, which seems quite high.
    What are the possible easier ways to do this that you mentioned?
     
  11. ScottWang

    Moderator

    Aug 23, 2012
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    Replacing with 4 small super capacitor and in series with them.
    In series with four 1F/2.7V and you can get 0.25F/10.8V.
    1F=10^6uF.
     
  12. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    Do you mean to have 4 1-farad capacitors wired in series? I don't think that would work very well, because the capacitor needs to fit on the bottom of a small styrofoam plate (the "skirt" of the hovercraft) and weigh as little as possible. Also, the LED only really needs about 3V, so I'm not sure what the benefit would being of wiring capacitors in series to get 10.8V.
     
  13. ronv

    AAC Fanatic!

    Nov 12, 2008
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    I think your way will be the best. You need something to store the energy, so a cap or battery plus circuits.
     
  14. ScottWang

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    Aug 23, 2012
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    The diameter of capacitor about 8mm and the height about 12mm, so it's very small, unless you want to reducing the voltage otherwise from your 9V circuit that the capacitor still needs to using 4 capacitors to in series with them.
     
  15. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    Were you referring to ScottWang or to me when you said "your"?
     
  16. WBahn

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    Mar 31, 2012
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    Placing capacitors in series in order to achieve higher voltages than the individual voltage ratings is not a good idea unless a means of balancing them is incorporated.
     
  17. WBahn

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    Mar 31, 2012
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    Hard to believe it's noticeable since you have a time constant of about 50ms.

    Why not use a small coin cell with a diode to prevent it being charged and another to prevent it from driving the fan. When the main battery is connected it will power both, but when the main power is removed the coin cell will power the LED for quite some time. You can remove the cell when you are done to extend its life. If you want it to just light for a short time then you can use a 555 or even an 8-pin micro to power the light and then effectively go to sleep.
     
  18. tjohnson

    Thread Starter Active Member

    Dec 23, 2014
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    At first I thought the effect of the capacitor seemed noticeable, but after making a circuit like #12 recommended, I could see very little if any effect, which is why in a later post I said barely noticeable. That's why I've been thinking I should use a larger capacitor.

    Thanks for the suggestion to use a coin cell. It's a good one that I hadn't thought of, but I don't really want to do something that complicated for just a small hobby project.

    EDIT: When the hovercraft is turned off, it takes a while for the fan blade to stop rotating (naturally, without a capacitor), and I thought it would be neat to have the LED fade out simultaneously. (I mean that the LED wouldn't completely go out until around the time when the fan blade completely stops.)
     
  19. ScottWang

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    Aug 23, 2012
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    If you care about the voltage balance then you can in parallel with a resistor for each capacitor, or you have some other better ideas?
     
  20. ScottWang

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    Aug 23, 2012
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    Using a coin cell that you can't keep the led to turn on for more time and too expensive.
     
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