0:23
So, as the oxygen concentration changes from 1 to 250 parts per million,
the sensor itself, puts out a signal that, which has a range of 0.5 to 1 volts.
But we need to do something with that signal and to boost it up and
to get it over the proper range for our downstream circuitry.
So want to design a sensor interface circuit that provides an output
range of 0 to 5 volts for this biosensor system.
1:20
We know that we have two conditions for
our circuit based on our problem statement.
That is if we want the output of the system to be equal to 0,
as you put the sum m plus our V sub S.
Which is at the 0 value of output as a V sub S
value of 0.5 plus this intercept.
We can also from our problem statement utilize the fact that
we want five volts out when the sensor is putting out one volt.
1:59
So we have two equations and we have two unknowns.
Our two unknowns are m and b, so if we take these two equations and
we solve them.
And we end up with a m which is equal to 10, and
a b which is equal to minus 5.
So we can rewrite our equation for
the output in terms of the input as 10 times V sub s minus 5,
so that's our input, output relationship.
2:31
This all came from the problem statement.
So what we need to do is we need to design an interface circuitry that allows us to
do this.
We can see that our output is a function of the input.
It's not inverted, so we might use a non-inverting configuration for
make this relationship between the output and the input.
We have this other factor which is a constant and it's minus 5,
where the output has an inverted relationship to some input which
gives a minus 5 contribution to the output voltage.
4:32
So, we can use superposition if we wanted to solve this problem for
the contribution to V out for V sub S and the contribution of V out from VRef.
So if we first look at the contribution for Vs,
what will we do with the reference voltage for easy superposition.
We know that when we're solving problems with superposition,
if we're looking at the contribution from one source.
That we short-circuit all other voltage sources and
we open-circuit all other current sources.
So if we do this, this circuit will look like what we have here,
except for we have VRef taken out and a short circuit to ground at this point.
5:15
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.
Dr. Bonnie H. FerriProfessor
Dr. Joyelle HarrisDirector of the Engineering for Social Innovation Center
Dr Mary Ann Weitnauer
