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Half power point problem help please

Discussion in 'Electronic Basics' started by ChadMan, Oct 24, 2003.

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  1. ChadMan

    ChadMan Guest

    Hello All
    Trying to figure out the 1/2 power point of the circuit
    below. This is my homework, so I just need a nudge please.

    When VO is not connected to the load, and points A & B are
    connected, I figure the 1/2 power point (where the amplitude
    is reduced by 3dB ?) to be: f = 1 / 2Pi x R5 x C3.

    Now here is where I get stuck. With points A & B disconnected
    and VO being taken across RL how do I figure the f response
    @ 3dB down? The above formula no longer works because R5 and
    C3 are no longer there, f was never stated (unless I'm supposed
    to use the previous answer).



    VCC
    +
    |
    o----------o
    | |
    | .-.
    | | |
    .-. RC| |
    | | '-'
    RB1 | | |
    '-' | + #| C2
    | o------#|------o o---------o
    | | #| VO |
    | | o .-.
    |# + | |/ | RL| |
    o-----|#----o--------| NPN === | |
    |# | |> GND '-'
    VI | | |
    o | | |
    | | | + #| C3 ===
    === | A o B o---#|----o GND
    GND | | #| |
    | | |
    | | |
    .-. .-. .-.
    | | | | RE | |
    RB2 | | | | R5 | |
    '-' '-' '-'
    | | o
    | | |
    | | |
    o----------o-----------------o
    |
    ===
    GND

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
     
  2. ChadMan

    ChadMan Guest

    Do I Use f = 1 / 2Pi x RL x C2 ?
     
  3. What about RE?
    With RL connected to C2:

    Draw a voltage source feeding RC, in series with C2 and RL. Now
    calculate where the low frequency rolloff point is.

    Kevin Aylward

    http://www.anasoft.co.uk
    SuperSpice, a very affordable Mixed-Mode
    Windows Simulator with Schematic Capture,
    Waveform Display, FFT's and Filter Design.
     
  4. ChadMan

    ChadMan Guest

    The answer the were looking for was based on this:
    1 / 2Pi x C2 x (RL + RC).

    I am not sure why they used RL + RC !?!

    ChadMan
     
  5. Dummy

    Dummy Guest

    In AC analysis, RC is parallel to RL. I guess, f = 1/2*pi*(Rc//RL)*C2.
     
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