Question about capacitator and LED in serie

Discussion in 'The Projects Forum' started by lordeos, Jul 27, 2015.

  1. lordeos

    Thread Starter Member

    Jun 23, 2015
    33
    1
    Hi Guys,

    I'm new at Electronics

    Since "the working" of capacitators i started experimenting with them in the hope it would become more clear to me. In fact i get's more mysterius every second.

    In the circuit in Picture A (picture included) i connected a 390uf capacitator (elco) in series with a red Led (voltage drop +-1.8) . As expected when i turn on the voltage , the led flashes for a second while the capacitator is charging:

    Remarks :

    1) In this case the LED is not gradually dimming ... it just flashes on and of --> is this normal ?
    2) in the case i put attach the crocodile clamps (source voltage is 4V, as shown in circuit A) my capacitator only charges up to a MAX of 2.5 V and across the LED i have a voltage of 1.3V --> I tought that the capacitator charges up to the same voltage as voltage source, in this case 4 V --> Does anyone knows the reason while it only charges partially ?
    3)
    In the case i attach the crocodile clamps directly to the + and - pole of the capacitator (Shown in picture circuit B) , the capacitator does charge up to full 4 V ????

    If the capacitator is charged (2.5V or 4V) and i remove the battery clamps and replace them by a wire , should't my LED be flashing for a fraction of a second ??? --> discharging capicator ???? Thi doesn't work either ..; the only way to discharge the capacitator (according to my test) is putting a wire directly from + pole to the - pole of the capacitator.

    Thx in advance for the help

    Regards,
    Mike
     
  2. DickCappels

    Moderator

    Aug 21, 2008
    2,664
    634
    1) This is normal -but the LED lights very quickly and goes off quickly because the LED is a low impedance across the capacitor, so it discharges down to a voltage below that needed for the LED to glow visibly. The discharge has a slope but it is so quick that it looks like it goes off instantaneously.

    2) That is because the LED is dropping (eating up) some of the voltage from the source.

    3) Yes, it should charge up to the source voltage.

    No, the LED should not flash when you short out the battery terminals because you charged the capacitor up such that when the battery connection is shorted, the capacitor voltage reverse biases the LED; negative to the anode and positive to the cathode.
     
  3. Bernard

    AAC Fanatic!

    Aug 7, 2008
    4,176
    397
    Remember that D in LED stands for diode, one way current flow. In second part of your test, reverse LED & then discharge the cap.
     
  4. wayneh

    Expert

    Sep 9, 2010
    12,154
    3,060
    For that arrangement, yes. The LED stays on as current flows into the capacitor. When the voltage across the capacitor is high enough, the drop across the LED is no longer high enough to trigger it to pass current. So, it goes out quickly. If you had an oscilloscope, you'd be able to see the current and brightness fading slightly before it shuts off, but to the naked eye it just goes out.
    As Dick said, some to the circuit's total voltage drop is at the LED. If you shorted across the LED to bypass it, the capacitor would reach battery voltage.
    The total energy stored in the capacitor might not be enough to light up the LED for more than an instant. You should see the voltage of the capacitor reduced, for instance from 4V to 1.3V, where the LED quits conducting.

    Oh wait, are you saying you shorted the capacitor thru a wire from pole to pole? That burns off all the energy in the wire and none goes to the LED. If you want to try to get a flash, you have to 1) remove the battery, 2) reverse the leads on the LED, because the "+" side of the LED is initially facing the "-" of the capacitor but now you need it to face the "+" side of the capacitor, and 3) replace the battery by a segment of wire to complete the circuit. The capacitor takes the role of your battery.
     
  5. lordeos

    Thread Starter Member

    Jun 23, 2015
    33
    1
    Hi,

    Thanks for the replies

    Indeed at forgot the fact that the led is reversed biased when i connect it ... so when i put a lightbulb (6v) instead of the LED it should work ??? taking into account i charge the capacitator to more then 6V.

    So in case i attach the crocodile clamps directly to + and - pole of the CAP i leave the LED out of the circuit and prevent it from using some voltage from the source --> am i correct ?

    Question out of curiosity --> how come the LED eats up some voltage --> i was tought a capacitator blocks everything of DC.

    cheers
    Mike
     
  6. lordeos

    Thread Starter Member

    Jun 23, 2015
    33
    1
    @wayneh

    To discharge the capacitator i shorted the + and - pole of the CAP. Otherwise the voltage stayed on the capacitator. When i tried to the LED without battery i putted a wire instead of the crocodile clamps (at the + and - entrance). As noted here i forgot the fact that the LED is reversed bias and therefore it didn't work.

    Well i must say that i searched a long time for a good forum ... since i'm really interested in learning electronics and started a paid internet course to learn the basics --> but many good information is left out on purpose the simplify the course --> The left out information is the information you need to have a good grasp on how it all works --> thats pitty.

    So thanks alot for the help and looking forward to the day i also can answer questions to electronics enthousiasts --> which will not be for tomorrow :):):)
     
    wayneh likes this.
  7. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hi,

    First let me just point out real quick that the capacitor is not a vegetable <chuckle>, it is an electronic component spelled as :"capacitor".
    To be honest though i have been tempted in the past to refer to it as a "capacitator" jokingly :)
    As you probably guessed, i like to joke around a little too.

    For a more serious view of your circuit, as others have pointed out, when you first turn on the circuit the LED flashes once and then goes dim, and that is because the capacitor charges up and once it charges up enough it no longer passes any current so the LED goes dim.

    What is not as obvious is that the capacitor may not charge up to the full input voltage because the LED acts as a sort of constant voltage drop at some point where it does not light but still drops some voltage. Thus the cap may only get to a certain voltage and stop charging. If there was no leakage in the cap this would not happen, but electrolytics usually have at least a little leakage so some current still flows in the diode and that causes a small but measurable voltage drop. The final capacitor voltage depends on the leakage of the cap and the amount of voltage the LED drops at that particular leakage current. For a quick example, if the cap leakage at the final voltage is 1ua and the LED drop with 1ua is 1 volt, then the cap will only charge to 4v with a 5v power supply (1v less than 5v). If the LED drops 1.1v at 1ua, then the cap will only charge to 3.9v with a 5v power source.

    You could also try using a series resistor to make the charge time a little longer and thus keep the LED on longer before it goes out. How long depends on the value of the cap and resistor mostly, but the resistor can not be too large or the LED will not get enough current to light brightly.
    Some of the white LEDs (and probably others) will still light a little at 100ua, but with very little brightness.
     
    RamaD likes this.
  8. wayneh

    Expert

    Sep 9, 2010
    12,154
    3,060
    It depends. A lightbulb filament requires a lot of current to heat it up and make light. A typical junk-drawer capacitor won't have enough capacity to do the job.

    I can't understand. Perhaps you could sketch a diagram?

    Those are two questions. The capacitor one is easy; you've been taught correctly. At steady state, no DC will pass a capacitor. Your LED experiment shows that some current can pass (to light the LED) before steady state is reached. That's a transient or AC effect.

    The reason that diodes have a voltage threshold before they conduct is a complicated explanation beyond my abilities! Go read the Wiki on diodes.
     
  9. lordeos

    Thread Starter Member

    Jun 23, 2015
    33
    1
    @wayneh

    My picture "circuit B" shows how i connect the 1 clamp to the + pole and the other one to the - pole of the CAP

    cheers mike
     
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