You are designing a compressor safety circuit that shuts down a compressor when
ID: 2081009 • Letter: Y
Question
You are designing a compressor safety circuit that shuts down a compressor when the tank pressure exceeds a maximum safe level. The safety circuit is a comparator circuit is shown below. A pressure sensor is attached to the compressor tank to indicate the pressure inside the lank. The pressure sensor outputs a voltage V_ from the compressor that is a function of pressure m the tank. When the pressure sensor senses 30 PSI. its output voltaic V_ is 4.5V. When the pressure sensor senses 150 PSI. its output voltage is 10 5V. The sensor's output voltage varies, linearly between these two points, and linear interpolation cm he used between these two points. (For example, at 90 PSI. the sensor's output is 7.5 V.) The shutdown control voltage V_SHUTDOWN is an input to the compressor When the voltage V_SHUTDOWN is high (+5V), the compressor will turn off. When the V_SHUTDOWN is low (0V), the compressor will nun. Ulf is a comparator integrated circuit. Input A is either the non-investing input or the inverting input, and Input B is either the inverting input or the non-inverting input (you will decide this). Design requirement: The compressor safety circuit shall turn off the compressor when the tank pressure exceeds 130 PSI. For the compressor safety circuit to function as required, which input (A or B) is the comparator IC's inverting input, and which input (A or B) is the comparator IC's non-inverting input? For the compressor safety circuit to function as required, find the value of resistor Rv (in ohms).Explanation / Answer
30 psi – 4.5V
150psi – 10.5V
90psi – 7.5V
130psi - ? V
Since the variation is linear, we can write a y=mx + C equation for PSI and V
V= (10.5-4.5)/(150-30) * PSI + C
=0.05PSI + C => 4.5 = 0.05*30 + C ( substituting 30, 4.5 in the above equation, we can determine the constant C )
ð C= 3
Therefore V= 0.05PSI + 3 ( this equation satisfies 90, 7.5 co-ordinate as well. Hence it is the correct equation ).
Now, at 130PSI, V= 0.05*130 + 3 => V= 9.5V
Let the voltage at terminal A be Va. Va= 12V * Rv/ ( Rv+2) ( by applying voltage divider for Rv and 2K ).
The compressor should shut off or the opamp should give 5V, when the voltage at terminal B exceeds 9.5V. That means the reference voltage at terminal A should be 9.5V.
Substituting Va=9.5, we get 9.5=12* Rv/(Rv+2) => 9.5 Rv + 19 = 12Rv => 2.5Rv= 19 => Rv= 7.6 kohms
Let A be the +ve terminal. Then when Va>Vb, then output of the comparator will be 5V. But this is not what we want. Let B be the +ve terminal. Then the output of the opamp will be 5V ( to switch off the compressor ), when Vb>Va. i.e., when Vb> 9.5V. Which is matching with what is required.
Thus B is the +ve terminal.
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