Many Electronic circuit projects in simple ways of learning

JFET delay Time circuits using 2N3819

This is the simple time delay circuits using JFET. Which it will alarm with a buzzer sound. In circuits have a few parts. And an important component is 2N3819-FET—Small Signal N Channel Transistor. Also, using capacitor charge and discharged through resistors. Then, use voltage across them to control gate of 2N3819 is ON.

The buzzer will sound or the output is high for a long time as its capacitance. It is ideal to learn simple timer.

You can do it in two-time delay circuits using JFET—VHF/UHF Amplifier N–Channel,

  • Very Basic timer circuit
  • 10 minutes FET delay Time circuits

Basic Timer circuit using FET


Basic Timer using FET 2N3819

How it works

This electronic bell sounds output until the delay time for 1.5 minutes long. First of all, Push a power switch (S2) on. Then, toggle switch S1 at the Reset position cause an electric bell or buzzer will loud.

If we toggle S1 come back at a set time an electric bell will silent, until time complete setting. We will stop the electric bell at one time again.

If friends change the value capacitor C1 or resistor R1 for delay long ago go up. And reduce R2 while stay in a position Reset for Reset fast go up.

We hope friends have fun the very simple alarm timer Basic Timer Circuit.

Parts you will need
Q1: 2N3819 JFET 2N3819-JFET Small Signal N Channel Transistor, Quantity = 1
BZ1: Electronic Bell or 3V Piezo-Buzzer, Quantity = 1
R1: 1K 0.25W Resistors, Quantity = 1
R2: 2.2M Potentiometer, Quantity = 1
C1: 4700uF 16V Electrolytic capacitor, Quantity = 1
S1: SPDT Slider Switch, Quantity = 1

10 minutes FET delay Timer  using 2N381910 minute time delay using 2N3819 FET

This circuit uses basic function of the 2N3819 JFET.  Which it acts as a switch. In the conduction and not conduction. It runs contrary to the transistor.

The circuit can be used to delay the cycle is small. Such as delay, 9-volt light bulb, or motor delay to any DC 9 volts.

When raising the power supply circuit FET Q1 is running. D and S stand by the legs of furniture to the Tong Q1 will flow together, current flows through R2, so the flow from the legs through the leg and S. D, and through R4 to ground.

So no current flows into R3 to transistor Q2 does not conduct. Resulting in a voltage appearing at the output is 12 volt, but when I press the S1 switch current flows into the capacitor C1 fully.

Then go to the legs G of the FET. Q1 stopped working. Therefore, the current flowing through R2 is flowing to the R3 instead, and to bias transistor Q2 to work on the output transistor Q2 is not working out for 10 minute, because C1 is discharged.

The S2 is the reset switch. By acting shock C1 to discharge out soon. That the time delay depends on the C1. If it will delay much longer, if less it will have less delay.

Parts you will need
Q1: 2N3819 JFET 2N3819-JFET Small Signal N Channel Transistor, Quantity = 1
Q2: 2N3906_40V 0.2A BJT PNP Transistor, Quantity = 1
BZ1: Electronic Bell or 3V Piezo-Buzzer, Quantity = 1
R1: 3.3M 0.25W Resistors, Quantity = 1
R2,R3,R6: 10K 0.25W Resistors, Quantity = 1
R4: 470 ohms 0.25W Resistors, Quantity = 1
R5: 2.7K 0.25W Resistors, Quantity = 1
C1: 100uF 16V Electrolytic capacitor, Quantity = 1
S1: normally open pushbutton switch, Quantity = 1

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Close Menu