0:23

So, as the oxygen concentration changes from 1 to 250 parts per million,

Â the sensor itself, puts out a signal that, which has a range of 0.5 to 1 volts.

Â But we need to do something with that signal and to boost it up and

Â to get it over the proper range for our downstream circuitry.

Â So want to design a sensor interface circuit that provides an output

Â range of 0 to 5 volts for this biosensor system.

Â 1:20

We know that we have two conditions for

Â our circuit based on our problem statement.

Â That is if we want the output of the system to be equal to 0,

Â as you put the sum m plus our V sub S.

Â Which is at the 0 value of output as a V sub S

Â value of 0.5 plus this intercept.

Â We can also from our problem statement utilize the fact that

Â we want five volts out when the sensor is putting out one volt.

Â 1:59

So we have two equations and we have two unknowns.

Â Our two unknowns are m and b, so if we take these two equations and

Â we solve them.

Â And we end up with a m which is equal to 10, and

Â a b which is equal to minus 5.

Â So we can rewrite our equation for

Â the output in terms of the input as 10 times V sub s minus 5,

Â so that's our input, output relationship.

Â 2:31

This all came from the problem statement.

Â So what we need to do is we need to design an interface circuitry that allows us to

Â do this.

Â We can see that our output is a function of the input.

Â It's not inverted, so we might use a non-inverting configuration for

Â make this relationship between the output and the input.

Â We have this other factor which is a constant and it's minus 5,

Â where the output has an inverted relationship to some input which

Â gives a minus 5 contribution to the output voltage.

Â 4:32

So, we can use superposition if we wanted to solve this problem for

Â the contribution to V out for V sub S and the contribution of V out from VRef.

Â So if we first look at the contribution for Vs,

Â what will we do with the reference voltage for easy superposition.

Â We know that when we're solving problems with superposition,

Â if we're looking at the contribution from one source.

Â That we short-circuit all other voltage sources and

Â we open-circuit all other current sources.

Â So if we do this, this circuit will look like what we have here,

Â except for we have VRef taken out and a short circuit to ground at this point.

Â 5:15

That circuit looks exactly like what we've done before for

Â the non-inverting operation amplifier example.

Â And what we have in that case

Â is that V out is equal to 1 plus

Â R sub F over R1 times V sub S.

Â That's what we get if we solve this circuit with V sub S in the circuit,

Â VRef taken out, replaced with a short circuit.

Â We get the basic configuration for the non-inverting amplifier,

Â which has this output-input relationship.

Â Now if we want to add the contribution from our reference voltage,

Â we would do the same thing as what we did just a minute ago.

Â We would look at VRef and we would remove all other sources,

Â we'd short circuit the voltage sources.

Â So we'd have a short circuit across this V sub S and

Â we would open circuit the current source which we don't have in this problem.

Â So if we look at that configuration where we have VRef, R1, R sub F and

Â a short circuit to ground at the non-inverting input.

Â That is exactly what the circuit looked like when we did the example for

Â the inverting op-amp configuration.

Â And so, the output that we received from VRef,

Â the contribution to the output voltage is R sub F divided by R one,

Â times VRef.

Â So that's what the output of this circuit would give us.

Â It will give us contribution from V sub S and a contribution from VRef.

Â So if we compare this to what we're looking for,

Â it fits pretty well to that scheme.

Â Where we have a 10 which has to be set as 5 by 1 plus R sub s divided by R1.

Â And a 5 which has to be solved using

Â R sub F divided by R 1 times VRef.

Â So we had to make some decisions at this point, we know that 1 plus R sub

Â F over R 1 is gotta be equal to 10.

Â That's one equation we have and we have another equation where

Â Rf of the feed back side

Â divided by R 1 times VRef

Â is gotta be equal to 5.

Â So we have to make some decisions on what values to use for

Â resistors since we're not given those values.

Â So, if we choose R1 equal to 10K

Â from our first expression,

Â R sub F has to be equal to 90K, so

Â R sub F is equal to 90K.

Â If R sub f is equal to 90K and R sub 1 is equal to 10K,

Â our second equation leads us to a value for VRef.

Â And we've got VRef,

Â equal to five-ninths of a volt.

Â So we can plug these values back into our circuit, our initial circuit and

Â if we had the circuit tied to the non-inverting input of our op-amp.

Â It's our V sub S and it's a 0 to 1 volts signal, so

Â it's a 0.5 to 1 volt signal at V sub S.

Â Then the output be a signal which range from 0

Â volts to 5 volts just as we had hope with our design.

Â