Connect with us

Flasher circuit

Discussion in 'Electronic Basics' started by [email protected], Sep 13, 2003.

Scroll to continue with content
  1. Guest

    Hi, I'm trying to understand the following simple flasher circuit:

    I don't really know what I'm doing here, but I think I have a vague
    idea of how it works... I assume you have to take the forward
    resistance of the LED into account, because otherwise I don't think
    the circuit would do anything. (Let's say a voltage drop of 2V and
    resistance 5 ohms.) When the circuit is first turned on, Q1 and Q2
    turn on, because there isn't enough voltage across the 2M resistor to
    keep Q1 off. C1 begins charging up (the wrong way!) through the 1K
    resistor, approaching a steady-state voltage. (Assuming the power
    supply is 4.5V, hFE for both transistors is 100, and the above LED
    values, I calculate this to be about -1.4V.) However, any capacitor
    voltage above -1.3V is enough to keep Q1 off once it is turned off,
    because that would produce a voltage over 3.8 (4.5-0.7) across the 2M
    resistor. Thus, I'm guessing that the capacitor gets charged to near
    -1.4V, Q1 turns off, the capacitor discharges to -1.3, and the cycle
    repeats. This would account for the "flasher" behavior, since during
    the on cycle, C1 is charging through the smaller 1K resistor and
    during the off cycle, it's discharging through the 2M resistor. If
    this is correct, then I have some questions:

    Why is the capacitor oriented the wrong way in the circuit?

    When the capacitor is charged between -1.3 and -1.4V, it seems
    entirely arbitrary whether the transistors are off or on and the
    capacitor is charging or discharging, so what decides when the circuit
    changes states?

    TIA, Karl
  2. You are right that you can not make the simplifying assumption that
    the LED has a constant voltage drop, independent of current. The
    changing voltage drop across the LED is the feedback voltage to the
    input of the two stage amplifier.
    Oops. This is exactly what you cannot assume. This circuit varies
    the LED current from almost zero (where the voltage drop will be much
    less than 2 volts) to severe overload current (limited only by the
    current carrying capability of the second transistor), where the wire
    bond resistance and the die resistance will come into play. I can
    only guess whether 5 ohms is a good approximation of those, unless I
    made a measurement of voltage and current, simultaneously during the
    peak, or extrapolated from the LED data sheet.
    A clearer way to think of this state is that the capacitor can be
    assumed to start with no voltage drop across it. If we guess that Q2
    will be slammed on (we'll check this later), then it is safe to also
    assume that with the battery voltage used up across the series
    combination of the LED and Q2, then the voltage across the LED must be
    quite a bit more than the .6 volts or so it takes to turn on Q1 via
    the 1K resistor. The small current through R2 is not of much
    consequence in the presence of this >2 volt LED drop feeding the base
    of Q1 through 1K. With Q1 driven on by this current, the capacitor
    starts to build up a voltage drop that gradually decreases this base
    current. But while the base current flows, Q1 is on, and dumps base
    current to Q2, so our original guess that Q2 starts by being blasted
    on looks good.
    So far, so good.
    You generally have it right. As soon as the base drive for Q1 gets
    low enough to let Q2 come out of saturation, the resistive and diode
    drop of the LED suddenly decreases, and the voltage at both ends of
    the capacitor head positive (assuming this happens fast enough that
    the voltage across the capacitor does not have time to change much).
    This produces a positive feedback situation that snaps Q1 completely
    off, and continues to a little reverse bias.
    It has so little voltage across it that it will probably work in
    either orientation. Try it the other way and see if the timing cycle
    changes. If it gets faster, it means that leakage current is
    interfering less with the slow charge time. I agree that it is
    The capacitor does its fast charge through Q1's base. It does its
    slow discharge through R2. Q1's base has to be reverse biased during
    this process to isolate the transistor. When the base of Q1 starts to
    have enough forward bias to start turning on Q2, the current from R2
    has to be gig enough to get the 'switch on' positive feedback going
    before all its current has been detoured, or the transistors will just
    stall at some slight conduction and the capacitor will discharge no
    further (because all of R2's current is going to the base, instead of
    the capacitor).

    I would add an emitter resistor to Q2 to reduce the peak base current
    from Q1 and limit the peak LED current. 10 to 47 ohms ought to work.
  3. This circuit should have a resistor betwen the C of Q1 and the B of Q2
    to limit the current. Often these flasher circuits seem to have the
    caps in backwards, and if you check the voltage across them, the
    voltage reverses.

    @@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
    ###Got a Question about ELECTRONICS? Check HERE First:###
    My email address is whitelisted. *All* email sent to it
    goes directly to the trash unless you add NOSPAM in the
    Subject: line with other stuff. alondra101 <at>
    Don't be ripped off by the big book dealers. Go to the URL
    that will give you a choice and save you money(up to half). You'll be glad you did!
    Just when you thought you had all this figured out, the gov't
    changed it:
    @@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day