Middlesex County, Ontario | I'm fairly certain that the schematic as drawn will always activate the buzzer.
The sensor outputs an analog voltage. Kind of like a potentiometer. Something close by is 0 volts, something far away is 5v, something in between in distance is an in between in voltage. The scaling is 4.9mV/cm. The sensor output voltage is always between 0 and 5v.
The circuit as drawn compares that output voltage to ground. If that output voltage is anything higher than exactly ground, the buzzer sounds. The voltage is always higher than ground, that's just the nature of the sensor.So the buzzer always sounds.
I think that potentiometer that you removed, sets a bias on the negative terminal of the comparative. Say you set the potentiometer so the voltage at the comparator - terminal is 3v. The sensor outputs something like 2v because there is a truck ahead of you. The truck ahead of you pulls away, and the sensor voltage rises. As soon as the sensor voltage rises past 3v, the buzzer would sound.
If you want to remove the potentiometer you can, but you will need to replace it with a voltage divider (two resistors). If you tell us at what distance you want the buzzer to sound we can calculate those resistor values for you.
If you post the schematic that you were given, one that you havn't modified, that would be helpful.
The voltage regulator as drawn is missing a ground terminal. A real world voltage regulator will have a 12v input line, a 5v output line, and a ground line. If you buy a voltage regulator as a module it would have caps 1 and 2 inside of it already. You can still add them if you wish. If you use a chip type voltage regulator, like an lm7085, you should add the caps.
To specify the comparator you first need to specify the n channel transistor. Then you will have to look at the saturation values. You will have to make sure the output of your comparator can go high enough to saturate, and that the output of your comparator can go to 0. Basically an ideal comparator as drawn could output between 0 and 5v, but a real world comparator can only output from 0-4v. You could pay a little extra and use a rail to rail comparator which will go from 0-5v.
That's a $100 sensor. Have you looked around for anything else? I would think that you could find something cheaper.
Some people can do this stuff faster with circuits, some people can do this stuff faster with a digital microcontroller. Myself: I hate this circuit stuff. It takes too long to specify components and small projects like this take too long. I would buy the sensor, an arduino, and a 12v relay module. You will have very simple circuits with no extra components and you only need to write 15 lines of code. The output would be a 12v relay that you can use for a 12v buzzer, light, or whatever you want. With a microcontroller you could use a $20 sensor.
Edited by WildBuckwheat 3/20/2016 11:51
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Question on schmatic of truck movement sensor system.Pic
