simulate this circuit – Schematic created using CircuitLab

Thanks for looking at my project i'm designing a circuit that turns OFF when i have a resistance of 4k and ON when i have a resistance of 2k without the use of IC chips. So i grabbed a mosfet transistor and tried manipulate it with a voltage divider. when i saw that i was still getting voltage across VGS i thought that my gate voltage was too high so i adapted a photoresistor in the middle of the GATE. So that when the voltage across the seconds resistor triggers the led, the photoresistor would decrease in resistance and allow voltage to go through, and turn it "ON" (how i though a mosfet transistor works). Instead i'm getting different results than what i thought would happen.


For sure, this is never going to do anything useful !

The gate current into a MOSFET is zero, so what does that mean for the voltage across the photo resistor when it is dark and when it is light ?

Hint: use Ohm's law.

You write about a Wheatstone bridge but I do not see it.

Study the Wheatstone bridge and figure out how it works. Note that those with a variable resistor/sensor do not have that resistor sitting between the left and right branches.

Also not that to effectively use a Wheatstone bridge you'd need an opamp or comparator, which you do not want to use.

Think of a simple circuit with one fixed resistor, one variable resistor and the MOSFET (so no photo resistor !) and try to make the NMOS switch on/off when you change the value of the variable resistor.

When you have that think where the photo resistor should be added in that circuit.

  • \$\begingroup\$ i tried to make it "clearer" ,does that help? \$\endgroup\$ – Elvis Medina Sep 14 '17 at 13:33
  • \$\begingroup\$ Can you understand the question better now? \$\endgroup\$ – Elvis Medina Sep 14 '17 at 13:54
  • \$\begingroup\$ The matter is not me understanding the question, I understand what you're doing and why it does not work. Do you understand my answer and the questions in it ? \$\endgroup\$ – Bimpelrekkie Sep 14 '17 at 13:57
  • \$\begingroup\$ oh i see what your saying, first try to make it turn on or off before applying it? \$\endgroup\$ – Elvis Medina Sep 14 '17 at 14:53

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