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Flashing led circuit explanation

Discussion in 'Electronic Basics' started by Ian H., May 4, 2004.

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  1. Ian H.

    Ian H. Guest

    Could someone please provide an alternate explanation of how this
    flashing LED circuit works? The description on the site is doubtless
    correct but I can't get a grip on it. Thanks!

    http://www.madlab.org/kits/lights.html

    Ian
     
  2. There are 2 keys to understanding how this works. One is that the
    base to emitter acts as a diode (conducts in one direction, only).
    The other is that when voltage changes at one end of a capacitor very
    fast, the voltage at the other end must also change by the same
    amount, regardless of what voltage drop there is across the cap when
    this sudden swing started.

    Let's also assume that the power has been applied for some time and
    cycles have been going on long enough to stabilize, with TR1 on and
    TR2 just about to turn on. If the explanation can get us back to this
    point, we will have followed a full cycle.

    R2 and R3 are each capable of holding one of the transistors on
    indefinitely, so the question becomes, what turns them off. The only
    other thing connected ot the bases are the cross coupling capacitors
    from the opposite transistor. With TR1 on, the voltage on its
    collector is near zero (if we assume the bottom rail is called zero
    volts or ground). So R2 is charging the right end of C1 positive
    while the left end is grounded. Once the right end of C1 rises to
    about +.6 volts, the base emitter junction of TR2 becomes forward
    biased and detours all current from R2 away from C1 so that this
    current turns on TR2.

    As TR2 switches on, the right end of C2 swings from almost 9 volts to
    nearly zero. This sudden negative swing must also appear on the left
    end of C2, but starting from wherever the voltage was before the
    swing. Since we have been assuming that TR1 started out by being on,
    its base voltage must have been about +.6 volts, so the negative swing
    passing through C2 drives this to almost -8.4 volts. This has two
    immediate effects. The current through R3 that had been keeping TR1
    on is now detoured to C2, instead and TR1 turns off, allowing its
    collector voltage to be pulled up by current through L1 and R1. But
    the collector voltage cannot instantaneously rise all the way to +9
    because the right end of C1 cannot rise above a diode drop from zero,
    because of the conducting junction in TR2. So the current through R1
    simply adds to the base bias current passing through TR2, turning it
    on very completely.

    After several R1*C1 time constants have passed, the current through R1
    has charged the left end of C1 up to a high enough voltage that L1 is
    almost completely starved of current, and the additional base current
    to TR2 is an insignificant addition to that supplied through R2, so
    TR2 is just on with its steady state bias.

    As this settling of voltage across C1 has been taking place, a much
    slower charging process has been taking place across C2. Current
    through R3 (boosted by the large negative voltage on the left end of
    C2 has been starting a ramp up in the voltage on the base of TR1.
    But, since the time constant is 10 times longer longer for this
    charging process than the one that involved R1
    (4.7k*47uF=.22 seconds versus 470*47uf=.022 seconds)
    The base voltage on TR1 will remain below ground for some time.

    Eventually, the current through R3 will bring the voltage on the base
    of TR1 0.6 volts higher than ground and that current will suddenly
    detour to the base. At that moment, TR1 will turn on and the second
    half cycle begins with C1 driving the base of TR2 way below ground
    which turns off TR2, allowing the current from R4 to overdrive TR1 on
    via C2, etc. and C1 begins the long climb back to +.6 volts.
     
  3. Ian H.

    Ian H. Guest

    Thanks for the reply John! A few questions if you don't mind :

    You said : "Once the right end of C1 rises to about +.6 volts, the
    base emitter junction of TR2 becomes forward biased and detours all
    current from R2 away from C1 so that this current turns on TR2."

    How or why is the current "detoured" away from C1?

    Also : "As TR2 switches on, the right end of C2 swings from almost 9
    volts to nearly zero. This sudden negative swing must also appear on
    the left end of C2, but starting from wherever the voltage was before
    the swing."

    Why does this negative swing have to appear on the left end of C2? My
    understanding of capacitors is that they store and release charge.

    Lastly, could you clarify : "But the collector voltage cannot
    instantaneously rise all the way to +9 because the right end of C1
    cannot rise above a diode drop from zero, because of the conducting
    junction in TR2."

    Thanks very much!

    Ian H.
     
  4. bench

    bench Guest

    what would happen if R1 and R4 where increased to 470K,
    I know the leds will not light, but assuming we took the leds
    out and just look at the waveform, would it continue to
    oscillate ?
     
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