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Capacitors in DC Circuits?

Discussion in 'Electronic Basics' started by CaVeDoG, Sep 16, 2003.

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  1. CaVeDoG

    CaVeDoG Guest

    I'm trying to understand how a simple flashing LED circuit works
    (, but don't get how the capacitors work
    with DC (the only thing I don't get is how they discharge). In all of the
    many articles I've read on capacitors, they say that the electrons build up
    on the positive plate until it's completely charged, at which point the
    current stops...and that's where they start on about discharing in an AC
    circuit. How and when does a capacitor discharge in a DC circuit?
  2. Flashers require some switching circuit to dump the charged capacitor
    through the LED, and then switch off, allowing the cap to recharge.
    In low voltage designs, the cap charges to almost the battery voltage
    and then is reconnected in series with the battery, to effectively
    double the voltage to supply the LED. Other designs use an oscillator
    and a step up transformer to generate higher voltage to a cap that
    gets dumped to an LED after it reaches some voltage. Others use an
    inductor much like an ignition coil (connecting the coil to the
    battery till a given current builds up, and then opening that circuit,
    allowing the inductance to generate a larger, reverse voltage that is
    delivered to a cap.
  3. Jim Large

    Jim Large Guest

    It's only just a few weeks since I was scratching my head
    trying to understand that same circuit.

    Let's start with Q2 turned on, and Q1 is off. The voltage
    on the collector of Q1 (and the left hand terminal of C1) is
    pretty close to Vcc, and the voltage on the base of Q2
    (right hand terminal of C1) is about one diode drop above
    zero (somewhere around 0.7 volts).

    NOTE: You haven't labelled Vcc or ground in your
    diagram, but it's easier for me to think in those
    terms. I'm calling the "+" terminal of the battery
    Vcc, and I'm calling the "-" terminal ground.

    Now, Q1 suddenly turns on. Don't worry about why, we'll get
    to that soon. The voltage on the collector of Q1 is pulled
    to ground, which means that the voltage on the left hand
    side of C1 is pulled down too.

    What does the capacitor do? Well one equation says that the
    voltage across the terminals of the capacitor is equal to
    the charge divided by the capacitance. The capacitance
    hasn't changed, and since we're only looking at just one
    instant in time (i.e., before any current gets to flow) the
    charge hasn't changed either. So that means that the
    voltage between the two terminals must be the same as it was
    a moment ago.

    Well the voltage on the left just dropped by almost the full
    supply voltage, so if the difference between the two
    terminals hasn't changed, that means that the voltage on the
    right MUST have dropped by the same amount. The voltage on
    the right is now NEGATIVE (i.e., less than ground).

    Since the voltage on the right terminal is also the voltage
    on the base of Q2, that means that Q2 will switch off.

    O.K., the voltage on the base of Q2 can't stay negative
    forever because R2 is trying to pull it up to Vcc. This is
    where we might say "the capacitor charges." When the
    voltage on the base of Q2 eventually comes back up to that
    magic 0.7 volts (a few tens of milliseconds given the
    component values in your circuit), Q2 will switch on, and
    ITS collector will be pulled low, and the base of Q1 will be
    driven negative, and Q1 will turn off.

    After a while, C2 "charges", allowing Q1 to suddenly turn
    on, and there we are! One full cycle.

    -- Jim Large

    P.S. Here's your circuit in SwitcherCAD. Cut everything
    below the dotted line, and paste into a file named FLED.ASC.
    SwitcherCAD is an awesome toy. Get it from this page if you
    don't already have it. (
    Version 4
    SHEET 1 880 680
    WIRE -112 400 -112 304
    WIRE 336 400 336 304
    WIRE -464 400 -464 128
    WIRE -464 48 -464 -288
    WIRE -112 -288 -112 -240
    WIRE 48 -288 48 -112
    WIRE 176 -288 176 -112
    WIRE 176 -32 176 64
    WIRE 176 64 224 64
    WIRE 48 -32 48 64
    WIRE 48 64 0 64
    WIRE -64 64 -112 64
    WIRE -112 64 -112 -32
    WIRE -112 208 -112 64
    WIRE 336 208 336 64
    WIRE 336 -112 336 -288
    WIRE 288 64 336 64
    WIRE 336 64 336 -32
    WIRE 176 64 176 144
    WIRE 176 144 64 256
    WIRE 64 256 -48 256
    WIRE 48 64 48 144
    WIRE 48 144 160 256
    WIRE 160 256 272 256
    WIRE -112 -176 -112 -112
    FLAG -112 400 0
    FLAG 336 400 0
    FLAG -112 -288 Vcc
    FLAG 48 -288 Vcc
    FLAG 176 -288 Vcc
    FLAG 336 -288 Vcc
    FLAG -464 -288 Vcc
    FLAG -464 400 0
    SYMBOL npn 272 208 R0
    SYMATTR InstName Q2
    SYMATTR Value 2N2222
    SYMBOL npn -48 208 M0
    SYMATTR InstName Q1
    SYMATTR Value 2N2222
    SYMBOL res -128 -128 R0
    SYMATTR InstName R1
    SYMATTR Value 330
    SYMBOL res 32 -128 R0
    SYMATTR InstName R2
    SYMATTR Value 47k
    SYMBOL res 160 -128 R0
    SYMATTR InstName R3
    SYMATTR Value 47k
    SYMBOL res 320 -128 R0
    SYMATTR InstName R4
    SYMATTR Value 560
    SYMBOL cap 288 48 R90
    WINDOW 0 0 32 VBottom 0
    WINDOW 3 32 32 VTop 0
    SYMATTR InstName C2
    SYMATTR Value 10µ
    SYMBOL cap 0 48 R90
    WINDOW 0 0 32 VBottom 0
    WINDOW 3 32 32 VTop 0
    SYMATTR InstName C1
    SYMATTR Value 2.2µ
    SYMBOL voltage -464 32 R0
    WINDOW 123 0 0 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR InstName V1
    SYMATTR Value 12
    SYMBOL LED -128 -240 R0
    SYMATTR InstName D1
    SYMATTR Value QTLP690C
    TEXT -498 506 Left 0 !.tran 1.5
  4. CaVeDoG

    CaVeDoG Guest

    Ahh, thank you very much, it makes a lot more sense now. My only question
    now is how is one transistor on when you start?
  5. ---------------
    The two transistors and their RC circuits are never equal, and one
    goes HI or LO first, and it starts itself. If they were too equal,
    they wouldn't start as easily. It's wise to use unequal components
    for symmetrical oscillators so they will start easily.

  6. Circuit which you mentioned is a Astable Multivibrator, and is
    commonly used to generate square waveform which glows that LED

    Frequency at which LED glows depends upon charging and discharging of
    capacitors through resistor R2 and R3.

    Frequency of asymmetrical Astable Multivibrator (as in your case) is
    given by
    1 / [0.69(R2*C1+R3*C2)].

    Animesh Maurya
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