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So I have a remote switch to control a light bulb. The transmitter circuit basically consists of:

-transmitter IC ( RF 433M Hz)

-DC power supply

-Switch ( turn on/off by applying voltage, transistor maybe, or thyristor? )

-Mechanical Touch sensor that generates a pulse of 1.5-2.5 V within 0.5 s per press. The current generated here is almost negligible, around 20 nA.

(See the schematic)

All I need to do is to turn on the switch (transistor) when I press the mechanical sensor that is connected to the gate of the transistor to allow the DC power supply feed the "Data pin" and then transmit it through antenna.

The problem is, I don't know what type of transistor ( or any other electronic device) that can be used to detect such a small voltage/current pulse and turn ON.

Any suggestion? If you have totally different ideas to tranmist the data with that sensitive mechanical touch sensor you're most welcome to share with me and I'll appreciate it :)

enter image description here

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  • \$\begingroup\$ Try to find out how to use transistor as a basic switch. This would surely help and the good part, ull learn. \$\endgroup\$ – Jasser Feb 7 '17 at 8:36
  • \$\begingroup\$ I already know how to do it. But the thing is, all the transistors I found on Digikey website has high Vgs(threshold)... more than what my sensor can provide. \$\endgroup\$ – Fahad Feb 7 '17 at 10:18
  • \$\begingroup\$ If you have in mind any type of transistor that turns ON at low gate voltage ( 1V lets say) please share it with me and i'll give you my deep gratitude \$\endgroup\$ – Fahad Feb 7 '17 at 10:21
  • \$\begingroup\$ BSS138: Vgs(th) = 06V~1.4V. Seriously, the most basic search on mouser/digikey will give results. Now, you realize that you need two transistors, do you? Because you can't provide a digital signal of 0V or 5V by just putting the BSS138 the way you shown. You need an additional PNP or P-Channel fet (triggered by BSS138) to make 5V flow. \$\endgroup\$ – dim Feb 7 '17 at 10:29
  • \$\begingroup\$ Oh, even lower: BSS816: specifies a Vgs(th) of 0.3V min / 0.75V max for Id=3.7µA. Didn't even know there were general-purpose FETs specified at such low gate voltages. Thanks, learnt something. \$\endgroup\$ – dim Feb 7 '17 at 10:35
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Ok, we have the MOSFET choice solved: BSS138, for example, can easily conduct with a gate voltage as low as you request. Now, your circuit is missing a few things. First, you should have a pull-down at the data input pin, because when the FET isn't conducting, there is no voltage applied at all, and this isn't good. Here is what it would give:

schematic

But: A mosfet actually turns on when its gate voltage is above Vgs(th) with respect to the voltage at its source terminal. Not necessarily ground. So, look what happens when the gate is, for example, at GND+2.5V: The mosfet starts conducting because its source is at ground (due to the pullup). Fine. But when the output (source voltage) starts raising, the Vgs voltage actually reduces. And when it reaches 2.5V-Vgs(th), the mosfet can't conduct more. So you'll never reach the required 5V at the output. The highest you'll get is about 1.5V (2.5V - Vgsth).

In short: The above circuit doesn't work.

Here is what works:

schematic

simulate this circuit – Schematic created using CircuitLab

When the input is at 2.5V, M1 start conducting, which lowers the voltage at the gate of M2 (BSS84, it is a P-channel FET), and delivers the full 5V at the data input PIN. Otherwise, the pin is pulled down by R2.

This is a basic level shifter, actually.

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  • \$\begingroup\$ You're absolutely right, all what you said makes sense. I'll build and test this circuit as soon as possible and I will let you know what will happen to me. Many many many thanks for your effort man, your explanation is just perfect and you made it easy to grasp the idea for non-electrical engineer xD... thanks again <3 \$\endgroup\$ – Fahad Feb 7 '17 at 14:53
  • \$\begingroup\$ You're welcome. After a second thought, however, I'm afraid the 20nA current capability of the sensor may be a problem. Although MOSFETs have very low leakage at the gate, it is usually specified at ~100nA worst case. And you need to overcome the gate capacitance. This will need to be thoroughly tested. \$\endgroup\$ – dim Feb 7 '17 at 15:08
  • \$\begingroup\$ Yes, I will try to modify the touch sensor in a way it generates more current. But what I plan to do is to connect an LED diode instead of the Data pin just for testing, since it has very high resistance ( i believe in megas) u think it's ok to replace it with R2 or should I just connect it in parallel with it? \$\endgroup\$ – Fahad Feb 7 '17 at 15:13
  • \$\begingroup\$ For a LED, you can omit R2, there is no need to pull down. But you need a low value (470Ohm) resistor in series, however, or it will burn the LED (like if you put the LED directly across the supply). \$\endgroup\$ – dim Feb 7 '17 at 15:21
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You can use an NMOS transistor which turns on if its gate voltage is higher than 0.7V (in your case it is between 1.5V to 2.5V).

An NMOS doesn't require an input current to turn on as opposed to BJT.

enter image description here

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  • \$\begingroup\$ Yes, but I could not find such transistor. I do have an NMOS that has Vth= 2.35V, but in fact when I tried it, it only started conducting at 4V... \$\endgroup\$ – Fahad Feb 7 '17 at 10:22

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