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# FET Biasing

The Parameters of FET is temperature dependent .When temperature increases drain resistance also increases, thus reducing the drain current.

FET Biasing

The Parameters of FET is temperature dependent .When temperature increases drain resistance also increases, thus reducing the drain current.

Unlike BJTs, thermal runaway does not occur with FETs

However, the wide differences in maximum and minimum transfer characteristics make ID levels unpredictable with simple fixed-gate bias voltage.

Different biasing circuits of FET are

1.     Fixed bias circuits

2.     Self bias circuits

3.     Voltage bias circuits

Fixed bias circuits

DC bias of a FET device needs setting of gate-source voltage VGS to give desired drain current ID . For a JFET drain current is limited by the saturation current IDS. Since the FET has such a high input impedance that no gate current flows and the dc voltage of the gate set by a voltage divider or a fixed battery voltage is not affected or loaded by the FET. Fixed dc bias is obtained using a battery VQG. This battery ensures that the gate is always negative with respect to source and no current flows through resistor RG and gate terminal that is IG =0. The battery provides a voltage VGS to bias the N-channel JFET, but no resulting current is drawn from the battery VGG. Resistor RG is included to allow any ac signal applied through capacitor C to develop across RG. While any ac signal will develop across RG, the dc voltage drop across RG is equal to IG RGi.e. 0 volt.

Calculate VGS

For DC analysis IG =0., applying KVL to the input circuits

VGS+ VGG=0

VGS= - VGG

As VGS is a fixed dc supply, hence the name fixed bias circuit

Calculate IDQ

IDQ=IDss(1- VGS/VGp)2

Calculate VDS

This current IDQ then causes a voltage drop across the drain resistor RD and is given as

VDSQ = VDD – ID RD

The fixed bias circuit of FET requires two power supplies.

Self-Bias circuits

Self-Bias circuits is the most common method for biasing a JFET. Self-bias circuit for N-channel JFET is shown in figure The gate source junction of JFET must be always in reverse biased condition .No gate current flows through the reverse-biased gate-source, the gate current IG = 0 and, therefore,vG = iG RG = 0

With a drain current ID the            voltage at the S is

Vs= IDRs

1)The gate-source voltage is then

VGS = VG - Vs = 0 – ID RS = – ID RS

So voltage drop across resistance Rs provides the biasing voltage VGg and no external source is required for biasing and this is the reason that it is called self-biasing. 2)Calculate IDQ

ID=IDSS(1- VGS/ VP)2

Substituting the value of VGS

ID= IDSS (1+IDRS  / VP)2

3)The operating point (that is zero signal ID and VDS) can easily be determined from equation given below :

VDS = VDD – ID(RD + RS)

Self biasing of a JFET stabilizes its quiescent operating point against any change in its parameters like transconductance. Any increase in voltage drop across RS, therefore, gate-source voltage, VGS becomes more negative and thus increase in drain current is reduced.

Voltage -Divider Bias circuits The resistors RGl and RG2 form a potential divider across drain supply VDD. The voltage V2 across RG2 provides the necessary bias. The additional gate resistor RGl from gate to supply voltage facilitates in larger adjustment of the dc bias point and permits use of larger valued RS.

The coupling capacitors are assumed to be open circuit for DC analysis

1) The gate is reverse biased so that IG = 0 and gate voltage

VG =V2 = (VDD/R G1 + R G2 ) *RG2

2) Applying KVL to the input circuit we get

VGS= VG – VS = VG - ID RS

3) IDQ= IDSS(1- VGS/ VP)2

4) VDS = VDD – ID (RD + RS)

The operating point of a JFET amplifier using the Voltage -Divider Bias is determined by

IDQ= IDSS(1- VGS/ VP)2

VDSQ = VDD – ID (RD + RS)

VGSQ = VG – ID RS

Example Problems

1)Determine IDQ, VGSQ, VD, VS, VDS, and VDG         Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
Electronic Circuits - Biasing of Discrete BJT and MOSFET : FET Biasing |

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