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?
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:
,----Vx
|
R1 | R2
V1----/\/\--+--/\/\----V2
|
|
\
R3 /
\
/
|
|
V3
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:
V_OUT = VREF + (VCC - VREF) * (R2 / (R1+R2))
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
