FET is an electronic device where the flow of current through the conducting region where the control lies in the electrostatic energy of the electromagnetic field, hence the name Field Effect Transistor. As current conduction is only by majority carriers, FET is said to be a unipolar device.
Based on the construction, the FET device can be classified into types as 1) Junction FET and 2) Metal Oxide Semiconductor FET (MOSFET)
Depending upon the majority carriers, JFET has been classified into two types named as 1) N-channel JFET with electrons as the majority carriers and 2) P-channel JFET with holes as the majority carriers
What is it composed of?
Construction:
It consists of an N-type bar which is made of silicon. Ohmic contacts (terminals) made at the two ends of the bar, are called sources and drain.
Source(S)- This terminal is connected to the negative pole of the battery. Electrons which which are the majority carriers in the N-type bar enter the bar through this terminal
Drain(D)- This terminal is connected to the positive to the positive pole of the battery. The (majority carriers) leave the bar through this terminal
Gate(G) – Heavily doped P-type silicon is diffused on both sides of the N-type silicon bar by which PN junctions are formed. These layers are joined together and the called GATE (G).
CHANNEL- The region BC of the N-type bar in the depletion region is called the channel. Majority carriers move from the source to drain when a potential difference Vds is applied between the source and drain.
How does it work?
When Vgs=0 and Vds=0.When no voltage is applied between drain and source, and gate and source, the thickness of the depletion region around the PN junction is uniform as shown below
In the event the Vds=0 and Vgs falls into the negative region below zero state. In this case PN junctions are reserve biased and hence the thickness of the depletion region also increases. As Vgs is decreased from zero, the reverse bias voltage across the PN junction is increased and hence the thickness of the depletion region in the channel increases until the two depletion regions make contact with each other. When this happens the entire channel is then cut off from the circuit. The value of Vgs which is required to cut off the channel is called the cutoff voltage Vc.
When Vgs=0 and Vps is increased from zero- Drain is positive with respect to the source with Vgs=0.Now the majority carriers (electrons) flow through the N-channel from source to drain. Therefore the conventional current Id flows from drain to source. The magnitude of the current will depend upon the following factors:
- The number of the majority carriers (electrons) available in the channel, i.e the conductivity of the channel.
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the length l of the channel.
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The cross sectional area A of the channel at B.
- The magnitude of the applied voltage Vds. Thus the channel acts as a resistor of the resistance R given by R=kl/A; Id= Vds/R= ((A*Vds/kl)…where k(= is the resistivity of the material)
Because the resistivity of the channel and the applied voltage Vds, there is a gradual increase of the positive potential along the channel from source to drain. Thus the reserve voltage across the PN junctions increases and hence the thickness of the depletion also increases.Therefore the channel is wedge shaped.
As Vds is increased, the cross-sectional area of the channel will be reduced. At a certain value Vp of Vds, the cross-sectional area at B becomes the minimum. At this voltage, the channel is said to be pinched off and the drain voltage Vp is called the pinch off voltage.
As a result of the decreasing cross-section of the channel with the increase of Vds, the following results are obtained.
- As Vds is increased from zero, Id increases along OP, and the rate of increase of Id with Vds decreases
- When Vds =Vp, Id becomes maximum. When Vds is increased beyond Vp, the length of the pinch off region increases. Hence there is no further increase of Id.
- At a certain voltage corresponding to the point B, Id suddenly increases. This effect is due to the avalanche multiplication of electrons caused by breaking of covalent bonds of silicon atoms in the depletion region between the gate and the drain. The drain voltage at which the breakdown occur is denoted by BVdgo. The variation of Id with Vds when Vgs=0
When Vgs is negative and Vds is increased
When the gate is maintained at a negative voltage less than the negative cutoff voltage, the reverse voltage across the junction is further increased. Hence for a negative value of Vp and BVdgo are lower.
From the curves, it is seen that above the pinch off voltage, at a constant value of Vds, Id increases with an increase in Vgs. Hence JFET is suitable for use as a voltage amplifier, similar to a transistor amplifier
“The drain current Id is controlled by the electric field that extends into the channel due to reverse biased voltage applied to the gate; hence this device has been given the name (FIELD EFFECT TRANSISTOR)”