How does a n channel jfet work




















Thus JFET operates on the principle that width and hence resistance of the conducting channel can be varied by changing the reverse voltage V GS.

When a voltage V DS is applied between drain and source terminals and voltage on the gate is zero as shown in fig. The electrons will flow from source to drain through a channel between the depletion layers. The size of the depletion layers determines the width of the channel and hence current conduction through the bar. When a reverse voltage V GS is applied between gate and source terminals, as shown in fig. This reduces the width of conducting channel, thereby increasing the resistance of n-type bar.

On the other hand, when the reverse bias on the gate is decreased, the width of the depletion layer also decreases.

This increases the width of the conducting channel and hence source to drain current. A JFET is a voltage controlled, constant current device in which variation in input voltage control the output current. The arrow denotes the types of JFET. This arrow also indicates the polarity of P-N junction, which is formed between the channel and the gate. The current flowing through the Drain and Source is dependable on the voltage applied to the Gate terminal.

In the above image, we can see the basic construction of a JFET. JFET is constructed using the long channel of semiconductor material. In the long channel of semiconductor material, Ohmic contacts at each end are created to form the Source and Drain connections. A P-N junction is formed in one or both side of the channel. One best example to understand the working of a JFET is to imagine the garden hose pipe.

Suppose a garden hose is providing a water flow through it. If we squeeze the hose the water flow will be less and at a certain point if we squeeze it completely there will be zero water flow. JFET works exactly in that way. If we interchange the hose with a JFET and the water flow with a current and then construct the current-carrying channel, we could control the current flow. When there is no voltage across gate and source, the channel becomes a smooth path which is wide open for electrons to flow.

But the reverse thing happens when a voltage is applied between gate and source in reverse polarity, that makes the P-N junction reversed biased and makes the channel narrower by increasing the depletion layer and could put the JFET in cut-off or pinch off region.

In the below image we can see the saturation mode and pinch off mode and we will be able to understand the depletion layer became wider and the current flow becomes less. These FET transistors are very useful in the chip designing due to their low power consumption behavior. Due to this high impedance values the FET transistors are very sensitive to small input voltages.

The JFET transistors are used as electronically controlled switches, Voltage controlled resistors and as amplifiers. This channel is formed due to the either of P-type or N-type semiconductor materials. In the N-channel JFET the channel is doped with the donor impurities due to this the current passing through the channel is negative i. The small voltage at the gate G terminal controls the current flow in the channel between drain and source of the JFET. The small voltage applied at the gate terminal controls the current flow in the channel between the drain and source of the JFET.

This is a transistor with N-type of channel and with P-type materials of the region. If the gate is diffused into the N-type channel, then a reverse biased PN-junction is formed which results a depletion region around the gate terminal when no external supply is applied to the transistor.

Generally the JFETs are called as depletion mode devices. This depletion region produces a potential gradient with the variation of thickness around the PN-junction. This PN-junction opposes the current flow through the channel by reducing the channel width and by increasing the channel resistance. Now the channel of JFET conducts with zero bias voltage applied as input. Because of the large portion of the depletion region formed between the gate-drain and the small portion of the depletion region between gate and source.

Now if we apply a small amount of negative voltage -V GS i.



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