Lighting an LED using Output of amplifier(common emitter)

Discussion in 'General Electronics Chat' started by BLOBY, Apr 27, 2011.

  1. BLOBY

    Thread Starter Member

    Mar 10, 2011
    37
    1
    Hi,

    I just hav this very basic amplifier circuit ( in attachment ) which gives 4 volts peak to peak output... I just need to light up an LED fed by the output of the amplifier.. The Problem is, the LED doesnt light up!! :confused: ( even if 4 volts is way above the forward voltage of LED)..

    I need help on lighting up the LED from the output of the amplifier.. I dont want to use any transformers.. What adjustments can I make so that it lights up ( it wil blink in 10 kHz actually)..

    Thanks for ur help in advance... :)
     
  2. mbohuntr

    Active Member

    Apr 6, 2009
    413
    32
    First, LEDs are DC devices, but can be ran on a DC square wave. Second, I might be wrong, but, I don't believe it will work because LEDs are current driven , not voltage, and I don't think the cap (c3) will allow it. Where did you get this circuit?
     
    Last edited: Apr 27, 2011
  3. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    You are thinking of using AC to drive the LED. LEDs can have really low PIV, a bad thing. Put a second LED facing the opposite way.

    I don't think it will work well, but it will work. You want a better way try this...

    [​IMG]

    Just tie pin 4 to pin 8, the exact schematics will allow digital control of the lights. Basically it is joule thief, and a primitive buck boost converter. It can use 1½V or 3V, but not more.
     
  4. marshallf3

    Well-Known Member

    Jul 26, 2010
    2,358
    201
    There does not appear to be nearly enough gain to up a 0.03V signal to an LED drive level.
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Good call, R2 is one of the current limiting resistors, as well as R4. You need a low impedance output as well as high gain. To get any significant lighting from that LED you will need at least 1ma, more is better. That 1 ma + is going to have to be provided by your circuit (and flow through the caps).
     
  6. bigmike2020

    New Member

    Oct 29, 2010
    2
    0
    Hmm.. are you trying to make the LED blink at 10 kHz or be triggered to blink at 10 kHz.

    If you are trying to make it blink at 10 kHz, you will not see it blink because our eyes can only detect up to about 50-60 Hz. However, if you are having it triggered to blink at 10 kHz, a circuit utilizing a 555 should do the trick.

    [​IMG]

    For more information on how it works, here is the link:

    http://wild-bohemian.com/electronics/flasher.html

    Note: Be certain that the LED has a current-limiting resistor to limit it to a maximum of 20 mA. Anything greater makes it vulnerable of burning out.
     
  7. Wendy

    Moderator

    Mar 24, 2008
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    That is the same circuit I have had for a very long time. It is published in the AAC book, as a matter of fact.

    I suspect the OP is trying to use an AC source to drive an LED. If you do it with line current the thread promptly gets closed on this site.

    Experiments don't have to make sense to be fun.
     
  8. BLOBY

    Thread Starter Member

    Mar 10, 2011
    37
    1
    Thanks for quick replies..
    I designed this circuit on my own after seeing Common Emitter mode amplifiers..
    Yes, I am trying to get the LED to turn on only when the input voltage is given.. It should be off at all other times..
    I designed the bias conditions such that around 2mA flows through collector of the transistor ( This current itself is transferred to output isnt it?)
    And Bill said that you need a low impedence output.. What does that mean? (Im new to these technical words)
     
  9. Audioguru

    New Member

    Dec 20, 2007
    9,411
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    When the collector of the transistor is cutoff (at +9V) then the output capacitor charges into the LED. When the collector of the transistor saturates at about 1V than the LED is reverse-biased and does not discharge the output capacitor.
    After a few cycles the output capacitor is fully charged and no more current flows into the LED. So the LED blinks extremely quick when the signal is applied but it might not be seen.

    If you add a second LED connected parallel to the first lED but connected in reverse then it will discharge the capacitor each cycle. Then both LEDs will dimly light when there is enough input signal.
     
  10. mbohuntr

    Active Member

    Apr 6, 2009
    413
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    Again, my foot tastes wonderful!! When will I learn.....:rolleyes: It does indeed work.....
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    Now reduce the input to 1.75V RMS (4V p-p like in your circuit) and increase the resistor to 3.6k like in your circuit to see the LEDs glow dimly.
     
  12. mbohuntr

    Active Member

    Apr 6, 2009
    413
    32
    Sorry BLOBY, not trying to hijack your post, Audioguru, Multisim doesn't like that, no lights.... 1.75vrms X 1.414 = 2.47vP 2.47vP-1.2v (Vf) = 1.27v 1.27v/ 3.6k = .354mA Right? not enough current? @ BLOBY... I noticed the Vf for red LED's was listed at 1.66V in Multisim, I changed it to 1.2 to see if there was any better result.
     
    Last edited: Apr 28, 2011
  13. Audioguru

    New Member

    Dec 20, 2007
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    Most red LEDs are 1.8V to 2.0V. Some infrared LEDs are 1.2V but you can't see them.
     
  14. BLOBY

    Thread Starter Member

    Mar 10, 2011
    37
    1
    Thanks... I really like the idea of connecting the 2nd LED in parallel but with opposite polarity.. I checked in Multisim and it worked.. Gotta see that in my actual breadboard circuit.. thanks...
     
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