Electrif field detector

[peterski11]

can someone please.. help me in my project " the electric field detector". i need to know its theory of operation.View attachment 64236

 

[unclejed613]

FT1 is a buffer amplifier, TR1 and TR2 are a differential amplifier. as a positive (note: this circuit only detects positive electrostatic charges) voltage is applied to the gate of FT1, FT1's source goes more positive, making TR1 conduct. the gain of TR1 is effectively doubled by TR2 acting in an opposite manner from TR1. increased conduction in TR1 lights the LED.
R1 is a very high value, because static voltages are in the multi-thousand volt range, and only 0.7V is needed to turn on TR1, so R1 becomes part of a voltage divider with either FT1's gate, or R8, reducing the voltage, and limiting the current across FT1's g-s junction. this circuit is very similar to what is called an "electrometer".

 

[Roff]

FT1 is a buffer amplifier, TR1 and TR2 are a differential amplifier. as a positive (note: this circuit only detects positive electrostatic charges) voltage is applied to the gate of FT1, FT1's source goes more positive, making TR1 conduct. the gain of TR1 is effectively doubled by TR2 acting in an opposite manner from TR1. increased conduction in TR1 lights the LED.
R1 is a very high value, because static voltages are in the multi-thousand volt range, and only 0.7V is needed to turn on TR1, so R1 becomes part of a voltage divider with either FT1's gate, or R8, reducing the voltage, and limiting the current across FT1's g-s junction. this circuit is very similar to what is called an "electrometer".

Nitpick:
Q2 does not increase the gain. There is no signal on its base, and no output is taken from its collector. Here is a good description of how this circuit works.

 

[MrAl]

Hi,

I'd like to add some more input as to the operation of this circuit.

First off, it is a linear circuit, not a digital circuit. I point this out because this means the LED does not light up suddenly but gets brighter and brighter the more input voltage is applied, unless of course we apply a higher voltage right away.

To start off say we have 0v input. The two resistors that bias TR2 hold the base at some predetermined voltage and so the voltage at the emitter is held at that voltage minus about 0.7 volts. This holds the emitter of TR1 at almost that same voltage because TR1 is not conducting yet. Note that TR1 only starts to conduct when it's base becomes more than the predetermined base voltage of TR2 because to forward bias TR1's base emitter the base has to be 0.7 volts higher than it's emitter, and the emitter voltage is set by TR2's emitter and that voltage is set by the two resistors R6 and R7. At this point the JFET is not conducting as much as it could because the gate is at 0v. Once it does start to conduct more heavily, it will increase the current through R2 and that will cause a greater voltage drop across R2 which means the base of TR1 becomes more and more positive. Eventually it will make TR1 start to conduct.
As we increase the input voltage, we reach a point where the JFET conducts enough to cause a voltage across R2 (which is the voltage as the base of TR1) that causes TR1 to conduct and that causes current through the LED even if it is a very small current. The LED may or may not light up a little depending on how high the input voltage is. Once the input voltage becomes high enough, the JFET conducts more heavily and that turns on TR1 much more and that means a lot more current through the LED which means it glows as bright as it can.

So with a 0v input we dont get any light out of the LED, and with a small voltage we can get some light out of the LED but it might be dim, and as the input goes up above some value the LED starts to glow bright. This would appear as a bright glow if a high voltage was applied, and no glow at all if no voltage was applied, so for this type of operation it would appear as a digital circuit.

Also, since the JFET has such a high input impedance it allows sensing voltages that dont have much power capability whereas a regular bipolar transistor would not work as well for that function.

 

[peterski11]

Thank You so much!! that really helps!! ^^

 

[peterski11]

Hello,^^, thank you for answering my problem ^^ that really helps