Question #ec537

Dec 22, 2017

Adjusting the alarm to trip at the proper level of smoke.

Explanation:

I am not familiar with the symbol in the diagram tied to the potentiometer. I will assume that is the smoke sensor.

Because gate B has one input held low by the switch, its output is high. So with one of gate A's inputs high, that gate inverts the level of the other input.

What is the voltage at which gate A will switch from recognizing that input (coming from the junction of the pot and the sensor) changing from being low to being high or changing from high to low? (It will be a voltage between what is considered a definite high, a logic 1, and a definite low, a logic 0.) And what is the resistance of the sensor when it is smelling the amount of smoke such that you would want it to report "I smell smoke!"? (You would need to look at the data sheet of the sensor. There would probably be a chart.)

It would be annoying for the circuit to alarm every time someone with cigarette smoke in their clothes walked by. At the other extreme, you do not want it to require a raging fire before alarming.

You will not be so lucky that you do not need to intervene to make the alarm come on at the desired level of smoke. So, the potentiometer and the sensor form a voltage divider to allow you to adjust the voltage to the gate's critical point at the junction. You would make that adjustment while you simulate the resistance of the sensor when there is a fire starting. When that has been done, the circuit will trip when the alarm is most helpful.

I hope this helps,
Steve