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FET Biasing

Chapter7. FET Biasing

JFET Biasing configurations Fixed biasingSelf biasing & Common GateVoltage divider

MOSFET Biasing configurations Depletion-typeEnhancement-type

FET Biasing

JFET: Fixed BiasingExample 7.1:As shown in the figure, it is the fixed biasing configuration of n-channel JFET. Determine:VGSQ , IDQ , VDS , VD , VG and VS .

Solution: Mathematical approach

VGSQ = - VGG = - 2V.

FET Biasing

2

1

P

GSDSSDQ V

VII2

82110

VVmA

= 5.625mA

VDS = VDD – ID RD = 16V - (5.625mA) (2k) = 4.75VVD = VDS = 4.75V

VG = VGS = - 2V

VS = 0V

FET Biasing

Three points are sufficient to plot the curve.

Graphical approach

It is known that IDSS= 10mA and VP= -8V.

First, according to Shockley’s equation, we can sketch out the transfer curve.

(0 , IDSS) (0, 10mA)

(VP, 0) (-8V, 0)

(VP /2, IDSS/4) (-4V, 2.5mA)

FET Biasing

The fixed level of VGS has been superimposed as a vertical line at VGS= -VGG = -2V.The intersection of the two curves is the solution to the configuration, referred to as the quiescent or operating point.

By drawing a horizontal line from the Q-point to the vertical ID axis, we get

IDQ = 5.6mA

FET Biasing

VDS = VDD – ID RD = 16V - (5.6mA) (2k) = 4.8V

VD = VDS = 4.8V

VG = VGS = - 2V

VS = 0V

The remaining work is almost the same as previous approach:

FET Biasing

Figure: Example 7.1, FET fixed biasing

FET Biasing

Figure: Example 7.1, Graphical approach

FET Biasing

JFET: Self-BiasExample 7.2:As shown in the figure, it is the self biasing configuration of n-channel JFET. This network needs only one dc supply. Determine:VGSQ , IDQ , VDS , VD , VG and VS .

Solution:

This time, we only use graphical approach.

FET Biasing

From the dc circuit, we get

For dc analysis, all the capacitors are replaced with open circuit.

VGS = – ID RS

It is a straight line through the origin.The other point is (-4V, 4mA).So the load line is plotted through the two points.

Note that IG = 0mA .

FET Biasing

Three points are sufficient to plot the curve.

It is known that IDSS= 8mA and VP= -6V.

Then, according to Shockley’s equation, we can sketch out the transfer curve.

(0 , IDSS) (0, 8mA)

(VP, 0) (-6V, 0)

(VP /2, IDSS/4) (-3V, 2mA)

FET Biasing

The intersection of the two curves is the Q-point.So we get

IDQ = 2.6mA

VGSQ = -2.6V

VDS = VDD – ID (RD+RS)

= 20V - (2.6mA) (1k+ 3.3k)

= 8.82V

FET Biasing

VS = ID RS = (2.6mA) (1k) = 2.6V

VD = VDS +VS = 11.42V

or

VG = 0V

= 8.82V+ 2.6V

= 11.42V

VD = VDD – ID RD = 20V- (2.6mA) (3.3k)

FET Biasing

Figure: Example 7.2, FET self-bias

FET Biasing

Figure: Example 7.2, Q-point of self-bias

FET Biasing

JFET: Common GateExample 7.4:As shown in the figure, it is the common gate configuration of n-channel JFET. Determine:VGSQ , IDQ , VDS , VD , VG and VS .

Solution:This network is corresponding to common-base network of BJT.

FET Biasing

Three points are sufficient to plot the curve.

It is known that IDSS= 12mA and VP= -6V.

According to Shockley’s equation, we can sketch out the transfer curve.

(0 , IDSS) (0, 12mA)

(VP, 0) (-6V, 0)

(VP /2, IDSS/4) (-3V, 3mA)

FET Biasing

From the dc circuit, we getVGS = – ID RS

It is a straight line through the origin.The other point is (-4.08V, 6mA).So the load line is plotted through the two points.

= – ID (680 )

So from the Q-point, we getIDQ 3.8mAVGSQ -2.6V

FET Biasing

VS = ID RS

VG = 0V

= 6.3V

= 12V- 3.8mA 1.5k

VD = VDD – ID RD

= 3.8mA 680 = 2.58V

VDS = VD – VS

= 6.3V-2.58V = 3.72V

FET Biasing

Figure: Example 7.4, Common gate configuration

FET Biasing

Figure: Example 7.4, Transfer curve & load line

FET Biasing

JFET: Voltage DividerExample 7.5:As shown in the figure, it is the voltage divider configuration of n-channel JFET.Determine:VGSQ , IDQ , VD , VS , VDS , and VDG .

Solution:This network is the same as voltage divider network of BJT.

FET Biasing

Three points are sufficient to plot the curve.

It is known that IDSS= 8mA and VP= -4V.

According to Shockley’s equation, we can sketch out the transfer curve.

(0 , IDSS) (0, 8mA)

(VP, 0) (-4V, 0)

(VP /2, IDSS/4) (-2V, 2mA)

FET Biasing

For dc analysis, all the capacitors are replaced with open circuit. Note that IG = 0mA .

Also it’s obvious that

So we get

DDG VRR

RV21

2

V

kMk 162701.2

270

= 1.82V

VGS = VG – ID RS = 1.82V- ID (1.5k)

FET Biasing

(1.82V, 0mA) & (0V, 1.21mA)

So the load line is plotted through the two points.

Then from the Q-point, we getIDQ 2.4mAVGSQ -1.8V

= 10.24V

= 16V- 2.4mA 2.4kVD = VDD – ID RD

FET Biasing

VS = ID RS = 2.4mA 1.5k = 3.6V

= 6.64V= 16V- 2.4mA ( 2.4k + 1.5k )

VDS = VDD – ID (RD+RS)

Or VDS = VD – VS = 10.24V-3.6V = 6.64V

VDG = VD – VG

= 10.24V-1.82V = 8.42V

FET Biasing

Figure: Example 7.5, Voltage divider configuration

FET Biasing

Figure: Example 7.5, Transfer curve & load line

FET Biasing

Depletion-Type MOSFETExample 7.7:As shown in the figure, it is the n-channel depletion-type MOSFET configuration.Determine:VGSQ , IDQ , and VDS .

Solution:For the n-channel depletion-type MOSFET, VGS can be positive and ID can exceed IDSS.

FET Biasing

Three points are sufficient to plot the curve while VGS<0.

It is known that IDSS= 6mA and VP= -3V.

According to Shockley’s equation, we can sketch out the transfer curve.

(0 , IDSS) (0, 6mA) (VP, 0) (-3V, 0) (VP /2, IDSS/4) (-1.5V, 1.5mA)

FET Biasing

when VGS > 0, letting VGS =1V,

and according to Shockley’s equation, we get

So the transfer curve has been sketched out.

2

1

P

GSDSSD V

VII2

3116

VVmA

= 10.67mA

FET Biasing

For dc analysis, all the capacitors are replaced with open circuit. Note that IG = 0mA .

Also it’s obvious that

So we get

DDG VRR

RV21

2

V

MMM 18

1111011

= 1.5V

VGS = VG – ID RS = 1.5V- ID (750)

FET Biasing

(1.5V, 0mA) & (0V, 2mA)

So the load line is plotted through the two points.

Then from the Q-point, we get

IDQ 3.1mA VGSQ -0.8V

10.1V

= 18V- (2.4mA) (2.4k+750)

VDS = VDD – ID (RD + RS)

FET Biasing

Figure: Example 7.7, depletion-type MOSFET configuration

FET Biasing

Figure: Example 7.7, Transfer curve & load line

FET Biasing

Enhancement-Type MOSFETExample 7.11 :As shown in the figure, it is the n-channel enhancement-type MOSFET feedback biasingconfiguration.Determine: VGSQ and IDQ.

Solution:Due to the existence of VGS(Th), we need four points to obtain the transfer curve.

FET Biasing

The ID is defined byID = k (VGS – VT)2

Solving for k, we obtain

2)()(

)(

)( ThGSonGS

onD

VVI

k

2)38(

6VV

mA

= 0.2410-3 A/V2

Then, Letting VGS = 6V, we get

ID = 0.2410-3 (VGS – VT)2

= 0.2410-3 (6V – 3V)2 = 2.16mA

FET Biasing

Also, Letting VGS = 10V, we get

ID = 0.2410-3 (VGS – VT)2

= 0.2410-3 (10V – 3V)2

So four points are sufficient to plot the curve.

VGS(Th) (3V, 0mA)

(VGS(on) , ID(on) ) (8V, 6mA)

(6V, 2.16mA)

= 11.76mA

(10V, 11.76mA)

FET Biasing

From the dc circuit, we get

VGS = VDD – ID RD

(0V, 6mA) and (12V, 0mA)

So the load line is plotted through the two points.

= 12V– ID (2k)

So from the Q-point, we get

IDQ 2.75mA VGSQ 6.4V

FET Biasing

Figure: Example 7.11, enhancement-type MOSFET configuration

FET Biasing

Figure: Example 7.11, Transfer curve & load line

FET Biasing

Summary of Chapter 7

JFET Biasing configurations Fixed biasingSelf biasing & Common GateVoltage divider

MOSFET Biasing configurations Depletion-type: voltage dividerEnhancement-type: feedback biasing