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What is the Capacitor and Resistor formula to delay the turn-on, saturation, of a transistor?

Discussion in 'Electronic Basics' started by Mr. J D, Aug 18, 2006.

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  1. Mr. J D

    Mr. J D Guest

    An Electrolytic Capacitor in series with a Resistor attachced to the
    base of a transistor would provide a simple timing circuit, yes? What
    is the formula to derive the needed Farads and Ohms?
  2. kfel

    kfel Guest

    T=RC, where R=Resistance in Ohms and C=capacitance in farads and T in
    secs. Note that this is the time it would take for the capacitor to
    reach 63.2% of the supplied voltage.
  3. John Fields

    John Fields Guest

    I've arranged your post as if you had bottom posted.
    Please bottom post.

    If this is the circuit Mr. J D had in mind: (View in Courier)

    | E
    [C] |
    | |

    then T = RC won't work

    Note that the base-to-emitter junction is a diode with its anode
    connected to the base and its cathode connected to the emitter,
    which won't allow the cap to charge any higher than Vbe.
  4. R
    E >------/\/\/\/\/-------+----------> Ec
    Er |
    ------- C
    E = 5v
    t = 1m sec
    R = 1K
    C = 1uF

    RC = R * C Example: 1K * 1uF = 1m sec note: One Time
    Constant, 63.2%
    TC = e-( t/RC ) Example: e (1m sec/1m sec) = 367.88m sec
    Ec = E * ( 1 - TC ) Example: 5v * ( 1 - 367.88m sec ) = 3.16v
    Er = E * TC Example: 5v * 367.88m sec = 1.84v

    At 2m sec

    TC = e( 2m sec / 1m sec ) = 135.34m sec note: Two Time Constants
    Ec = 5v * ( 1 - 135.34m sec ) = 4.32v
    Er = 5v * 135.34m sec = 676.67mv

    .. . .
    note: Three through five Time Constants

    At 6m sec

    TC = e-( 6m sec / 1m sec ) = 2.48m sec note: Six Time Constants
    Ec = 5v * ( 1 - 2.48m sec ) = 4.99v
    Er = 5v * 2.48m sec = 10mv

    Hope this helps.


    Mr. Bill
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