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Transistor question

Discussion in 'Electronic Design' started by Efthimios, Jan 1, 2009.

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  1. Efthimios

    Efthimios Guest

    I found the following LED flashing electrical diagram in the net

    I assume that the flashing bit is done by the transistors and the
    friquency from the RC components.

    1. Can some one explain me why these two transistors (2222 2907) will
    make the LED to flash ????

    2. Also it requires about 30uA. If i want to reduce this what should i

  2. Efthimios

    Efthimios Guest

    Thanks Jan,

    But what makes the 2222 transistor switch between on-off states?

  3. MooseFET

    MooseFET Guest

    The transistors provide the amplification needed and the RC parts
    provide the timing control.

    Do you know basically what a transistor does?

    Examine the 2N2222 and just the 4.7K on its base and the 10K on its
    collector. If you wiggle the end of the 4.7K up and down, what
    happens to the voltage on the collector?

    Now look at the 0.68uF and the 680 Ohms. What happens to the AC part
    of what is at the collector of the 2N2222?

    What happens to the 2N2907 when the AC on the collector of the 2N2222,
    goes down towards zero volts?

    What does this do to the 2N2222?

    Now try to do the above with DC instead of AC.

    Now try to fit a "by how much" into each of the above answers.
    Notice that the "by how much" leads to an interesting situation.

    Can you stand a slower blink or less light?

    You can gain a little by scaling the 4.7M up by some factor like lets
    say 1.5 or 2 and the 0.68uF down by the same factor. You can't go
    very far on this before the circuit will stop working.
  4. Efthimios

    Efthimios Guest

    Can work as a switch or an amplifier.
    If you increase the resistor the voltage will drop so the amplified
    signal output from the transistor will drop.
    What? Voltage will oscillate?
    N2907 works as a switch. It switch off and stops capacitors from
    It terns 2N2222 off.
    With DC when the capacitor is charged will switch 2222 to conducting
  5. Guest

    An interesting circuit, seemingly it's not just you thats baffled. The
    lithium cell has a very high internal resitance so the current for the
    led is supplied by the capacitor across it. Once that has discharged
    the circuit resets. If you were to run this from a power supply the
    led would come on after a delay and stay on.
  6. Efthimios

    Efthimios Guest

    what is a blocking oscillator?

  7. Jamie

    Jamie Guest

    on Initial power to the circuit, the capacitor is nearly discharge due to
    the base and emitter on the PNP (2907) and through the 680 and 10KR's
    and LED network
    The initial voltage across the cap at best will be the forward break
    down of the BE of the (2907).

    You'll get an inrush of current flow through the capacitor from the
    680 R and 10KR+LED that has the (+) rail. This will not be enough to
    get the LED started.

    Because the cap was discharged to start with on the initial cycle,
    you'll also see (+) voltage at the base of the PNP (2907), this will
    reverse bias the PNP and turn it off.

    The NPN (2222) depends on the PNP(2907) to be on for forward
    biasing.. So at this point, the NPN is also off.

    When the cap finally gets charged enough and removes the (+) forward
    inrush of voltage from the charge cap down to ~ 0.6 or less at the base
    of the PNP, the 4.7 Meg R will then forward bias the PNP, this will
    then turn on the PNP which will also turn on the NPN (2222).

    Now, the initial on cycle has been complete, and this is what happens
    Because the cap has been fully charged and now the 2222 is coming
    on, the 2222 will also shine the LED of course and also the cap will
    start to discharge how ever, since there is a charge in the cap now, the
    polarity on the other end of the cap going to the PNP base will generate
    (-) voltage, not (+) as it did in the initial power on cycle.

    This (-) voltage will force the PNP on even harder, thus biasing the
    2222 more, until full saturation is accomplished on the 2222.

    At some point, the cap will deplete and start losing output to the PNP
    base, when this happens, the PNP will drive the 2222 less. when the
    2222 gets out of saturation state, it'll start to turn off. The 10R and
    LED network will start to raise on voltage and also start charging the
    cap, this charge is like the first initial charge. because the cap was
    drained, it'll now force (+) volts on the PNP base thus turning it off
    and forcing the 2222 to go off even faster..

    The trick is to not over bias the PNP with the R, that is why you see
    4.7M there. it's more or less a tickler to get it started and handle
    forward drop off voltage of the base on the PNP.

    The 680R being like it is, I suspect the on cycle to be much shorter
    than the off cycle."
  8. Efthimios

    Efthimios Guest

    Thanks Jamie for your detailed analysis.

  9. MooseFET

    MooseFET Guest

    Right so the signal on the collector is bigger than the one on the
    base and is also inverted (goes negative when the base goes positive).

    Just imagine some AC is at the collector of the 2N2222 by magic and
    then see what the 0.68uF and 680 Ohms will do with it. We'll get back
    to how the AC voltage appears at the collector later.

    No look again at the circuit. The 2N2907 has its emitter at the +3 and
    is a PNP. The 2N2907 turns on when the collector of the 2N2222 moves
    down towards zero.

    Right for the backwards case above. 2N2907 on turns the 2N2222 on.
    2N2222 on turns the 2N2907 on. As far as the AC case goes we have a
    positive feedback loop.
    No I meant for you to consider the capacitor as an open circuit to the
    DC. This means that the positive feedback situation doesn't apply to
    the DC case.

    The interesting situation is that if the 2N2907 turns on a little, the
    2N2222 turns one quite a bit which turns the 2N2907 on a lot which
    turns the 2N2222 on hugely ... etc.
  10. MooseFET

    MooseFET Guest

    The efficiency starts with a roughly 50% starting point because the
    LED has a forward drop about half that of the battery. From that
    point down, it isn't all that bad. The 4.7M passes about a microamp
    and the transistors leak. For most of the time, that is the only
    current in the thing.

    Replacing the 47 Ohms with an inductor and a diode would greatly
    improve the efficiency but the inductor size for the simple case is
    way too big. Making the 2N2222 make a burst of oscillation would be a
    better way to go than a 2ms pulse.
  11. MooseFET

    MooseFET Guest

    No it doesn't. The circuit will oscillate with a zero impedance power
    supply. The 2N2907 doesn't have enough bias current in the DC case to
    keep the 2N2222 saturated so once the 0.68uF is discharged, the LED
    goes off.
  12. Tolstoy

    Tolstoy Guest

    the circuit has negative feedback and positive feedback.

    negative feedback: when the led is off, the big cap charges and
    eventually turns the transistors and led on;
    conversely, when the led is on, it drains the cap and at some point
    everything turns off. somebody pointed out that you need a high
    impedance battery to make this work.

    the positive feedback makes the circuit tend to snap either to full on
    or full off condition. the cap and resistor from the npn collector to
    the pnp base provide this feedback. in other words, when negative
    feedback causes the npn transistor to start to turn off, this will
    pull the base of the pnp down, turning the npn off even more -- what
    you might call a "snowball effect." Conversely, if the npn transistor
    starts to turn on, it will pull down the pnp base and make the circuit
    turn on more and more, like a snap-action switch.
  13. Tolstoy

    Tolstoy Guest

    What's your problem?
  14. MooseFET

    MooseFET Guest

    This is not correct, the circuit works with a low impedance supply.
  15. Jasen Betts

    Jasen Betts Guest

    feedback with no net phase shift will not cause oscillation

    if you've got over unity gain in a phase-shift free loop you end up
    'pegged' against one rail or the other.
  16. JosephKK

    JosephKK Guest

    Care to try that analysis again? Charged through the 4.7 Meg,
    discharge through Schmidt trigger 1st base.
  17. Jamie

    Jamie Guest

    Is that what you think?"
  18. JosephKK

    JosephKK Guest

    John, it is a blocking oscillator, and about 90% efficient.
  19. Sylvia Else

    Sylvia Else Guest

    This would only be true if the circuit is phase-shift free all the way
    down to DC. Though off hand I can't imagine a simple circuit that would
    be phase shift free at higher frequencies, but not DC.

    The circuit in question is not phase shift free at any frequency. It
    contains a 360 degree phase shift, and unity gain, at its switching
    frequency. The output waveform is not sinusoidal (by any stretch of
    imagination) because of extreme non linearity.

  20. JosephKK

    JosephKK Guest

    No way in hell spud. The 4.7M is the only possible charge path. Or
    did the reverse polarity designation fool you?
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