Voltage divider calculation using superposition

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)

MRW skrev:

If it is just because you want the result, MiscEl can do the calculations.

/Jan

Thanks again, John!

Thanks, Jan! Awesome website.

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.

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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

John Popelish wrote in news:eeGdnbNIUKSEUQLbnZ2dnUVZ snipped-for-privacy@comcast.com:

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))

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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."

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