Hi All!
I was studying about types of & uses for MOSFETs and came across a Water Alarm circuit that has me puzzled.
This is a circuit where the author of an article at https://www.cnet.com/how-to/how-to-build-your-own-flood-sensor/ used a dollar store window alarm to make a device that will sound an alarm when water touches the probes. (His circuit diagram is attached with a couple of my notes added.)
There were several things about the circuit design that I could not understand and couldn't find enough information on various references to help me figure out answers. So, if you don't mind some probably very basic questions, here we go...
A) Looking at the diagram, it seems to me that the wires I labeled "P-2" & "P-3" will end up making a direct connection between the positive and negative battery poles when the sensor probes are in conductive water. This is visible in the diagram where the source, the water probe and the negative terminal all come together. The other probe wire is on the positive battery terminal. Why does not the battery self-destruct when the two probe wires are 'joined' through the water?
B) I think I am correct that the purpose of the resistor is to protect the MOSFET device. Did I get that right?
C) How did the author choose the value for the resistor? (The text of his article shows it is 3.3 Meg Ohm.) The batteries are (3) 1.5 volt coin cells in series for 4.5 volts. I worked the numbers through Ohm's Law in various combinations and am still stumped.
D) Why did he use a MOSFET at all? Couldn't one take the alarm as-found and then interrupt one of the wires to the battery and send it to the water. Next one would take a wire from the water and send it back to the battery? Is the MOSFET to make it so a lower voltage (than the battery 4.5 volts) to make the alarm sound? My guess is that since water will have some resistance, a lower voltage trigger will allow the alarm to sound where the full 4.5 volt battery would not, thus the MOSFET. Did I get this correct?
Thank You all for helping me learn about the use of the MOSFET and the resistor calculations for this circuit. I appreciate the education.
Paul
I was studying about types of & uses for MOSFETs and came across a Water Alarm circuit that has me puzzled.
This is a circuit where the author of an article at https://www.cnet.com/how-to/how-to-build-your-own-flood-sensor/ used a dollar store window alarm to make a device that will sound an alarm when water touches the probes. (His circuit diagram is attached with a couple of my notes added.)
There were several things about the circuit design that I could not understand and couldn't find enough information on various references to help me figure out answers. So, if you don't mind some probably very basic questions, here we go...
A) Looking at the diagram, it seems to me that the wires I labeled "P-2" & "P-3" will end up making a direct connection between the positive and negative battery poles when the sensor probes are in conductive water. This is visible in the diagram where the source, the water probe and the negative terminal all come together. The other probe wire is on the positive battery terminal. Why does not the battery self-destruct when the two probe wires are 'joined' through the water?
B) I think I am correct that the purpose of the resistor is to protect the MOSFET device. Did I get that right?
C) How did the author choose the value for the resistor? (The text of his article shows it is 3.3 Meg Ohm.) The batteries are (3) 1.5 volt coin cells in series for 4.5 volts. I worked the numbers through Ohm's Law in various combinations and am still stumped.
D) Why did he use a MOSFET at all? Couldn't one take the alarm as-found and then interrupt one of the wires to the battery and send it to the water. Next one would take a wire from the water and send it back to the battery? Is the MOSFET to make it so a lower voltage (than the battery 4.5 volts) to make the alarm sound? My guess is that since water will have some resistance, a lower voltage trigger will allow the alarm to sound where the full 4.5 volt battery would not, thus the MOSFET. Did I get this correct?
Thank You all for helping me learn about the use of the MOSFET and the resistor calculations for this circuit. I appreciate the education.
Paul
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