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dc or ac - subscript conventions revisited+

Discussion in 'Electronic Basics' started by Active8, Apr 2, 2005.

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  1. Active8

    Active8 Guest

    Followup *was* set for SEB, but I don't care. This is relevant to
    both groups, really. This may also be of benefit to beginners
    confused with KCL and KVL node and mesh methods - like the poster
    with the LED/resistor problem.

    I just stumbled across another set of lecture notes and it ocurred
    to me that there was something I failed to mention in the original
    thread - the order of subscripts with more than one unique
    designator.

    If you can deal with handwritten notes and have a clue, this looks
    pretty good.

    http://ece-classweb.ucsd.edu:16080/winter05/ece102/

    Reference for this topic is:

    http://ece-classweb.ucsd.edu:16080/winter05/ece102/ECE102_S_BJT_largesignal.pdf

    I'll refer to page 15 since it's about NPN BJTs. IMO, it's easier to
    do the arithmetic mentally - not having to turn the BJT upside down.

    The only screw up I see so far is using lower case v for v_XX. It
    should be V_XX, thus my "have a clue" qualifier. He's talking about
    large signal conditions and leading up to biasing, which sets the
    operating point and determines the conditions for small signal
    analysis where v_X and v_x are correct. No offence to current
    intended by my omission ;)

    The point is that the first designator in the subscript is a value
    from which you subtract the value of the second subscript.

    Note how he uses V_BC. Look at the "inequality equation" (an
    oxymoron!) Check his math. It's correct by convention. No confusion
    possible. Try it for page 7 on PNPs where he uses both V_BC and
    V_CB. Hint: maybe read it standing on your head.

    This subscript convention is important - it keeps the signs correct.
    You may need to write simultaneous equations or even a netlister.

    R_1
    ___
    +------|___|------+ node 3 -- unnecessary, but relevant
    | | V_3 to the discussion - for this
    | | lame example - unless you
    | .-. need to know this voltage
    | | | R_3
    | | |
    | '-'
    | R_2 |
    | ___ |
    node 1 +------|___|------+ node 2
    V_1 | | V_2
    | |
    | .-.
    ^ / \ | | R_4
    I_1 | ( ) | |
    | \ / '-'
    | | |
    | |
    +-------+---------+
    |
    o node 0 - reference node

    So in a node analysis, for node 1 connected to R_1 and R_2, the
    current leaving that node and going into R_2 (charge flow flow?!)
    would be

    V_1 - V_2
    ---------
    R_2

    V_1 - V_2 is the voltage drop across R_2, V_R2 or V_1,2. V_1 could
    be written as:

    V_1,0 IOW, V_1 - V_0 = V_1 - 0 = V_1 - assuming V_0 is O V which is
    what you try to shoot for in these simple problems.

    If you do the same for R_1 at node 1,

    V_1 - V_3
    ---------
    R_1

    and simplify the node equation, you'll see that the easiest way to
    do it is to take all the resistors connected to node 1 as
    contributions to the current out of node_1 caused by V_1 and sum
    them

    1 1
    --- + --- ... they're on the bottom
    R_1 R_2

    to write the term for that node. Then take the nodes connected to
    those resistors and subtract their contributions. That gives the
    term for the other relevant nodes. Make it easier and convert to
    conductances (1/R for the uniniated).

    sum of current sources in, I_1 = (G_1 + G_2)V_1 - G_2*V_2 - G_1*V_3

    Now you have the first node equation with the node voltages in
    order. Then you'd write the other nodes under that so the voltages
    appear directly under the same voltage and plug it into your matrice
    or do your gaussian elimination without getting all fouled up.

    The same conventions apply in all math and science. length_2,1 ...
    the distance to point 2, starting at point 1 or x_2 - x_1.

    You can do the same thing with mesh (loop) analysis.

    sum of voltage sources in a loop = sum of voltage drops around that
    loop minus any voltage drops associated with that loop that are
    caused by another loop current ... (R_1 + R_2 + ... + R_n)I_1 minus
    say, R_2*I_2 and R_3*I_3 -- where the loop currents are, by
    convention assumed to be in a clockwise direction. The signs will
    work themselves out.

    HTH and special thanks to google for archiving all this blathering
    in case I loose it and need to retrieve it when I need it for a web
    page :) I'm such a user ;) I should use their free storage space at
    www.gmail.com
     
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