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Voltage divider calculation using superposition

Discussion in 'Electronic Basics' started by MRW, Jul 19, 2007.

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  1. MRW

    MRW Guest

    Hello!

    I have this simple voltage divider circuit:

    VCC
    +
    |
    |
    .-.
    | |
    | |R1
    '-'
    |
    ---- V_OUT
    |
    |
    .-.
    | |
    | |R2
    '-'
    |
    ---
    VREF

    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    First I found V_OUT using the traditional way and got the following
    relationship:

    V_OUT = VREF + (VCC - VREF) * (R2 / (R1+R2))

    Just for the heck of it, I tried superposition. I first shorted VREF
    and go the following:

    V_OUT1 = VCC * (R2 / (R1 + R2))

    Then, I shorted VCC:

    V_OUT2 = VREF * (R1 / (R1 + R2))

    So, V_OUT = V_OUT1 + V_OUT2 .... but I can't seem to simplify it to
    what I got using the other method.

    Any insights?

    Thanks!
     
  2. V_OUT1 + VOUT_2 =
    VCC * (R2 / (R1 + R2)) + VREF * (R1 / (R1 + R2))

    add and subtract VREF * (R2 / (R1 + R2))

    the subtracted term turns the first part into:

    (VCC - VREF) * (R2 / (R1 + R2))

    and the second part into:

    Vref * (R1 / (R1 + R2) + VREF * (R2 / (R1 + R2))

    which simplifies to:

    VREF * (R1 + R2) / (R1 + R2)

    which simplifies to:

    VREF

    And both parts together are:

    VREF + (VCC - VREF) * (R2 / (R1 + R2))


    The reason this is hard to arrive at is that the goal is not
    the simplest form, which is more like:

    ((VCC * R2) + (VREF * R1)) / (R1 + R2)
     
  3. Jan Nielsen

    Jan Nielsen Guest

  4. MRW

    MRW Guest


    Thanks again, John!

    --
     
  5. MRW

    MRW Guest

    Thanks, Jan! Awesome website.


    --
     
  6. You've already got a good post or two on this. My own personal view
    of 'superposition,' is more like how Spice looks at it. And as a
    mental model, I think it's a little better than the 'shorting'
    approach you used. I'll explain a little.

    Take some circuit:
    The way I look at it is to think of the voltage as both "spilling
    into" and "spilling out of" the Vx node.

    First, look at the inward spilling situation. V1 spills into Vx via
    R1. V2 spills into Vx via R2. V3 spills into Vx via R3. To set this
    up into an expression, take the resistances as conductances and we
    then have: V1*(1/R1) + V2*(1/R2) + V3*(1/R3). In other words, V1
    flows through conductance 1/R1, V2 flows through conductance 1/R2, and
    so on.

    Second, look at the outward spilling situation. In this case Vx
    spills outward via the same conductances. So we then have this
    expression: Vx*(1/R1) + Vx*(1/R2) + Vx*(1/R3).

    Inward spills are superimposed upon the outward spills and, since we
    know that electons aren't accumulating at node Vx, the net
    accumulation of charge at Vx must be zero. This means:

    V1*(1/R1) + V2*(1/R2) + V3*(1/R3) = Vx*(1/R1) + Vx*(1/R2) + Vx*(1/R3)

    or,

    V1*(1/R1) + V2*(1/R2) + V3*(1/R3) = Vx*(1/R1 + 1/R2 + 1/R3)

    This is easily solved for Vx, as:

    Vx = (V1/R1 + V2/R2 + V3/R3) / (1/R1 + 1/R2 + 1/R3)

    I chose a somewhat more complex case, just to point out that it's
    pretty easy to see what is going on in more complex situations.

    -----

    In the case of two resistors, this is:

    V1*(1/R1) + V2*(1/R2) = Vx*(1/R1 + 1/R2)

    Which becomes:

    Vx = (V1/R1 + V2/R2) / (1/R1 + 1/R2)

    Multiplying through by R1*R2/(R1*R2), you get:

    Vx = (V1*R2+V2*R1) / (R2+R1)

    Now, this isn't exactly your equation, which takes a different form.
    But it is algebraicly identical to it. You can see that with simply
    algebra.

    Let's take your expression:
    and replace some of the terms with my above briefer terminology. In
    other words, as this:

    Vx = V2 + (V1 - V2) * (R2 / (R1+R2))

    Now, let's see if I can transform what I wrote above into that one.

    Vx = ( V1*R2 + V2*R1 ) / ( R1+R2 )
    = V2 + ( V1*R2 + V2*R1 ) / ( R1+R2 ) - V2
    = V2 + ( V1*R2 + V2*R1 ) / ( R1+R2 ) - V2*( R1+R2 ) / ( R1+R2 )
    = V2 + [ ( V1*R2 + V2*R1 ) - V2*( R1+R2 ) ] / ( R1+R2 )
    = V2 + [ V1*R2 + V2*R1 - V2*( R1+R2 ) ] / ( R1+R2 )
    = V2 + ( V1*R2 + V2*R1 - V2*R1 - V2*R2 ) / ( R1+R2 )
    = V2 + ( V1*R2 - V2*R2 ) / ( R1+R2 )
    = V2 + ( V1 - V2 ) * R2 / ( R1+R2 )

    I think you can see that this is the same thing.

    Anyway, that's how I prefer "seeing" superposition.

    Jon
     
  7. me

    me Guest


    or you can arrange the first equation to get the second...


    V_OUT = VREF + (VCC - VREF) * (R2 / (R1+R2))

    = VCC * (R2 / (R1 + R2)) + VREF(1 -(R2 / (R1 + R2)))

    = VCC * (R2 / (R1 + R2)) + VREF( (R1+R2)/(R1+R2)) -(R2 / (R1+R2)))

    = VCC * (R2 / (R1 + R2)) + VREF * (R1 / (R1 + R2))
     
  8. MRW

    MRW Guest


    Thanks, Jon! Very nice. I'm also looking for alternative ways to
    perceive a method. Just with the help of this newsgroup, I think I've
    slowly adapted to some "circuit seeing."



    --
     
  9. Thanks for the kind comment. And I hope it does help a little. The
    concept is broadly applicable, and may help in 2D resistive surfaces
    and 3D resistive volumes with point voltages introduced on or within
    them.

    Jon
     
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