This post describes the design and function of the circuit represented by the schematic in post #13. Supporting information is also included.
(1)
Requirements
The first thing to do in any design is to establish the requirements as definitively and completely as possible.
In industry this is normally documented in a Requirement Specification.
Your professor has obviously given you the basic requirements to design a circuit as an academic exercise to help you understand the techniques involved, but he has not given a full specification. For example, what is the output impedance of the pressure transducer and the voltage reference. What is the rate of change of pressure that is required. What is the accuracy required. What type of pressure transducer and voltage reference. What are the environmental conditions. What are the size and weight limits. What is the power consumption limit. What voltage rails are available. What is the maximum cost. You could go on for ever.
Having said this, it is impossible to define on paper the whole requirement for any item when you take into consideration all the variables. I certainly have never seen a full specification, even in the the thickest military requirement specifications. It is bit like trying to write a requirement specification for your girlfriend or wife.
Often the customer simply does not know the full requirement or states it incorrectly. He also makes mistakes of all kinds (just like me).
So you have three choices. Either say the design cant be done. Alternatively you can or go on asking endless questions. On the other hand, you can take a positive approach and make educated guesses. Also you can make the performance of the design the best that you can as a balance between cost, complexity and performance. That is what I have done in this case, hopefully.
Here is a list of the requirements you have been given.
(1.1) Pressure Transducer: Offset: 1.683V. Transfer function: 131mV per pound pressure
(1.2) Voltage reference: 1V
(1.3) Circuit output: 1V per pound of pressure
This is a classic situation where the circuit has to deal with a sensor offset and produce an amplified output of the parameter of interest, in this case a voltage proportional to pressure in pounds.
(2)
Concepts and Laws
In order to follow the circuit description a few fundamental concepts and laws need to be understood.
(2.1) Ohm's law: I=V/R
(2.2) Kirchhoff's first law: The current flowing into a circuit junction is equal to the current flowing out of the junction
(2.3) Voltage versus Current
Many electronics newbees are comfortable following a circuit using a voltage analysis, but they find a current analysis alien, which is a shame because often the latter is simpler, as is the case with the schematic of post 13.
(2.4) Operational amplifier
Because an operational amplifier (opamp) is the central component in the design of post#13 and is liable to be the central component in any other pressure transducer conditioning design, it is important to understand their primary function and parameters.
Put quite simply, an opamp does not care what you want it to do. Instead it is single minded and hates its two inputs to be at different voltages, even slightly. As a result it will do everything it can to make its two inputs the same. But the only thing it can do is to change its output voltage and it will do this very fast.
So a designers task is to place a circuit around an opamp so that when it balances its inputs it carries out the function that he wants.
What I have described is a perfect opamp, but real opamps have shortcomings that complicate the issue. But the OPA192, in practical terms, is perfect and has no shortcomings. That means it is ideal for a very wide range of applications. But more importantly in this case, it makes the circuit simpler to explain (note that all opamps have limitations, including the OPA192, it is a matter of application and degree though). If you are interested in taking a closer look at the OPA192 see this ETO article:
https://www.electro-tech-online.com/articles/game-changer-opamp-opax192.768/
(2.5) Shunt Feedback Amplifier (Virtual Earth Amplifier, Summing Amplifier)
In the schematic of post 13, the opamp is configured as a shunt negative feedback amplifier, where R1 provides the negative feedback. Because the non inverting input of the opamp is connected to 0V, the opamp will keep its inverting input at 0V too. It does this by altering its output voltage to meet this condition. Thus the inverting input is a virtual earth (0V), hence one of the names for this amplifier configuration.
(3)
Circuit Description
The schematic of post 13 uses a single operational amplifier instead of three and only three resistors. The loading of the transducer and voltage reference is as light as possible, by using high value resistors, without going too high, which may compromise the operation of a practical circuit.
After further analysis though, it turns out that all resistor values in the schematic of post 13 could be multiplied by ten to reduce the loading on the transducer and voltage reference by a further factor of ten. This is possible because of the superior performance of the OPA192: vanishingly low input current and input offset voltage. There are dangers with high resistance circuits which need to be catered for, but it would over complicate the picture to describe these here.
(3.1) Offset Cancelling
The first task of the circuit is to cancel the transducer output offset.
The offset cancelling is simplicity itself. N1 maintains the junction of R2/R3 at 0V. Thus the current flowing through R2 is, - 1V683/168K3 =-10uA) . Similarly the current flowing through R3 = 1V/100K =10uA. From this it follows from Kirchoffs first law that the currents at the junction of R2/R3 sum to zero; even without the opamp, the voltage at the junction of R2/R3 will be 0V anyway.
(3.2) Gain
The second task is to amplify the transducer signal (in the correct sense) to produce an output to meet the given requirement.
The pressure transducer has an output voltage of 131mV per pound pressure. This causes a current of -131mV/168K3= -778.37 nA to flow through R2. This in turn causes the voltage at the junction of R2/R3 to drop but the opamp cannot tolerate this and will immediately make its output voltage sufficiently positive to inject exactly 778.37 nA into the junction of R2/R3 via the feedback resistor, R1, to return the junction of R2/R3 to 0V.
For a one pound pressure increase the output of the opamp is required to increase by 1V. It is now just a matter of calculating the value of R1, the negative feedback resistor, to produce a voltage of 1V when 778.37 nA is flowing through it. Thus R1= 1V/778.37nA= 1.285M.
That is all there is to it.
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