Characteristics of JFET

CHARACTERISTICS OF JFET

The curves which relate device current and voltage are known as characteristics curves. There are two important characteristics of a JFET.

(i) Transfer characteristics

(ii) V-I characteristics (or) Drain characteristics

Transfer Characteristics

The curves shows the relationship between drain current (I_D) and gate to source voltage (V_GS) for the different values of drain to source (V_DS) voltage is known as Transfer characteristics.

These transfer characteristics curves are obtained by using the circuit shown in fig.3.63. The R₁ and R₂ are used to vary the voltages V_GS and V_DS respectively. The drain current is measured by ammeter (A) connected in series with the JFET and the supply voltage V_DD

The following steps are used to calculate the transfer characteristics

(i) Adjust the drain to source voltage, (V_DS) to some value.

(ii) Increases the gate to source voltage (V_GS) in small suitable steps.

(iii) Record the corresponding values of drain current (I_D) at each steps.

(iv) Plot a graph with V_GS in x-axis and drain current (I_D) in y-axis.

It is shown in fig.3.64.

The upper end of the curve is shown by the a rain current value equal to I_DSS' while the lower end is indicated by a voltage V_gs (or) V_p. The curve is part of a parabola and therefore may be expressed by the equation.

Drain Characteristics

The curve drawn between voltage of drain to source (V_DS) and drain current (I_D).The following steps are carried out to obtain the characteristics.

(i) Adjust the gate to source voltage (V_GS) to zero.

(ii) Increase the drain to source voltage (V_DS) is small suitable steps.

(iii) Record the corresponding values of drain current (I_D) at each step.

(iv) Plot the curve V_DS in x-axis and I_D in y-axis. It is shown in fig.3.65.

A similar procedure used to obtain curves for different values of gate to source voltage V_GS = 1,2,3 and 4 volts.

The drain characteristics of curve V_GS = 0 is shown in fig.3.66. The curve may be sub-divided into various regions as follows.

(i) Ohmic region:

This region is shown as a curve OA in fig.3.66. In this region, the drain current increases linearly with the increases in drain to source voltage, obeying ohm's law. The linear increase in drain current is due to the fact that N-type semiconductors bar acts like a simple resistor.

(ii) Curve AB:

In this region, the drain current increases at the reverse square law rate with the increase in drain to source voltage. The drain current increase slowly as compared to that in ohmic region. At the point B, the channel width is reduced to a minimum value and is known as pinch off. The drain to source voltage, at which the channel pinch-off occurs, is known as pinch-off voltage.

(iii) Pinch-off region:

This region is shown by the current BC. It is also called saturation region (or) constant current region. In this region, the drain current remains constant at its maximum value the drain, current in the pinch off region, depends upon the gate to source voltage (V_gs) and is given by the relation.

ID = I_DSS [1- V_GS/V_p]²

The above relation is known as shockly's equation. The pinch off region is the normal operating region of JFET, when used as an amplifier.

(iv) Breakdown region:

This region is shown by the curve CD. In this region, the drain current increases rapidly as the drain-to-source voltage is also increased. It happens because of the breakdown of gate-to-source junction due to avalanche effect. The drain to source voltage corresponding to point C is called breakdown voltage.

Example : 11

The data sheet of a certain JFET indicates that I_DSS equal to 15 mA and V_GS(off) equal to -5V. Determine the drain current for V_GS equal to 0V, -IV and -3V.

Solution:

I_DSS = 15 mA, V_GS(off) = -5 volts

I_D = I_DSS [1- V_GS/V_GS(off)]²

V_GS = 0 volts

I_D = 15 [1 – 0/-5]²

I_D = 15 mA

V_GS = -1 volts

I_D = 15[1- -1/-5]²

I_D = 9.6 mA

V_GS = - 3V

I_D = 15[1- -3/-5]²

I_D = 2.4 mA

Example: 12

A JFET has parameter of V_GS(off)  equal to -20 V and I_DSS equal to 12 mA. Plot the transconductance curve for the device using V_GS values of 0, -5, -10, -15 and -20V.

Solution:

V_{GS (off)} = -20 V, I_DSS = 12 mA

I_D = I_DSS [1- V_GS/V_{GS (off)}]²

V_GS = -5V

I_D = 12[1- -5/-20]²

I_D = 6.75 mA

V_GS = - 10 V

1_D = 12[1- -10/-20]²

I_D = 3 mA

V_GS = - 15 V

1_D = 12[1- -15/-20]²

I_D = 0.75 mA

V_GS  = -20 V

1_D = 12[1- -20/-20]²

I_D = 0 mA