That circuit looks exactly like what we've done before for
the non-inverting operation amplifier example.
And what we have in that case
is that V out is equal to 1 plus
R sub F over R1 times V sub S.
That's what we get if we solve this circuit with V sub S in the circuit,
VRef taken out, replaced with a short circuit.
We get the basic configuration for the non-inverting amplifier,
which has this output-input relationship.
Now if we want to add the contribution from our reference voltage,
we would do the same thing as what we did just a minute ago.
We would look at VRef and we would remove all other sources,
we'd short circuit the voltage sources.
So we'd have a short circuit across this V sub S and
we would open circuit the current source which we don't have in this problem.
So if we look at that configuration where we have VRef, R1, R sub F and
a short circuit to ground at the non-inverting input.
That is exactly what the circuit looked like when we did the example for
the inverting op-amp configuration.
And so, the output that we received from VRef,
the contribution to the output voltage is R sub F divided by R one,
times VRef.
So that's what the output of this circuit would give us.
It will give us contribution from V sub S and a contribution from VRef.
So if we compare this to what we're looking for,
it fits pretty well to that scheme.
Where we have a 10 which has to be set as 5 by 1 plus R sub s divided by R1.
And a 5 which has to be solved using
R sub F divided by R 1 times VRef.
So we had to make some decisions at this point, we know that 1 plus R sub
F over R 1 is gotta be equal to 10.
That's one equation we have and we have another equation where
Rf of the feed back side
divided by R 1 times VRef
is gotta be equal to 5.
So we have to make some decisions on what values to use for
resistors since we're not given those values.
So, if we choose R1 equal to 10K
from our first expression,
R sub F has to be equal to 90K, so
R sub F is equal to 90K.
If R sub f is equal to 90K and R sub 1 is equal to 10K,
our second equation leads us to a value for VRef.
And we've got VRef,
equal to five-ninths of a volt.
So we can plug these values back into our circuit, our initial circuit and
if we had the circuit tied to the non-inverting input of our op-amp.
It's our V sub S and it's a 0 to 1 volts signal, so
it's a 0.5 to 1 volt signal at V sub S.
Then the output be a signal which range from 0
volts to 5 volts just as we had hope with our design.
