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Voltage Divider Rule using Admittance

Discussion in 'Electronic Basics' started by golson, Oct 14, 2004.

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  1. golson

    golson Guest

    I have a question about the Voltage Divider rule. I am looking at a
    example in a textbook to calculate a transfer function for this
    circuit.


    ----inductor------|------------|
    | |
    | |
    C R
    VIN a E VOUT
    p S
    | |
    | |
    ------------------|------------|



    Transfer Function is

    Vout(s) (1/sL)
    ------ = --------------------------------
    Vin(s) (1/sL) + [sC + (1/R)]



    looking at the transfer function it appears that the voltage divider
    rule for admittances is like.

    ------ g1 ----------|-------|
    | |
    Vin g g Vout
    2 3
    | |
    --------------------|-------|

    g1
    Vout = Vin * ( 1/(g1 + g2 + g3) ) * g1 ) = Vin *
    ---------------------
    g1 + g2 + g3


    Assume above are resistors where : g1 = .5 mhos, g2 = .25 mhos, and
    g3 = .1666 mhos or r1=2 ohms r2=4 ohms r3=6 ohms

    then using admittances and my theory above:

    .5
    Vout = Vin * ------------ = 0.545 Vin
    .5 + .25 + .1666


    Using Voltage Division for Resistance:

    Vout = Vin * (r2 || r3) / ( r1 + ( r2 || r3) ) = 2.4 / 4.4 = .545 *
    Vin


    Is the approach I took for using admittances correct for all cases.
    It seems to be working for the example at the end.

    I was looking at page 11 of RC Active Filter Design Handbook by F.W.
    Stephenson.

    It makes the equation much easier using admittances over impedences.

    I can not find admittance example in my textbooks.
     
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