Module – 5 UNIT -5 Field Effect Transistors

Review Questions:

1. Draw the structure of JFET and discuss its working. 2. What is ‘pinch off’ voltage? How to get its value experimentally? 3. An n-JFET is operated with negative gate voltage and not with positive one. Give reasons. 4. Give the structure of depletion MOSFET (D - MOSFET). How is D - MOSFET different from enhancement MOSFET (E - MOSFET)? 5. Draw and discuss drain characteristics for a D-MOSFET. 6. Discuss the formation of channel in E-MOSFET emphasizing the role of inversion layer. 7. Give self bias circuit for JFET and explain the biasing process. 8. How can we obtain negative or positive bias voltage with proper choice of resistors in a voltage divider bias? 9. Develop a simple small signal model/equivalent circuit for FET. 10. How are DMOSFET and EMOSFET connected in a circuit to work as resistors? 11. Derive expression for voltage gain for a common source amplifier. 12. Illustrate power efficiencies of CMOS devices through a CMOS inverter circuit.

Problems:-

5.1 The device parameters for an n-Channel JFET are: Maximum current IDSS = 10mA, Pinch off voltage,

Vp = - 4V

Calculate the drain current for

(a) VGS = 0

(b) VGS = - 1.0v

The expression for drain current ID, in the saturation region is,

2 VGS IIAD DSS 1 ...... ( ) VP

(a) When VGS = 0, from Eq(A) above,

I D I DSS 10mA

(b) When VGS = -1.0V, the drain current from Eq (A) is,

2 1 I 10 10 3 1 D 4 10mA 0.56

or I D 5.6 mA

or , I D 0

5.2 A JFET produces gate current of 2nA when gate is reverse biased with 8V. Determine The resistance between gate and source. Solution:-

Since reverse gate-source voltage, VGS, of 8v produces gate current, IG of 2nA, Therefore, gate-to-source resistance, RGS, is

VGS 8V RMGS 4000 IG 2 nA

RMGS 4000

5.3 The reverse gate voltage of JFET when changes from 4.4V to 4.2V, the drain current changes from 2.2 mA to 2.6 mA. Find out the value of transconductance of the transistor. Solution:-

The transconductance, gm is defined as

ID gm VGS

Where ΔID is change in drain current when change in gate-source voltage is ΔVGS.

In the given problem,

ΔID = (2.6 – 2.2) mA

= 0.4 mA

And,

ΔVGS = (4.4 – 4.2) V

= 0.2 V

Therefore,

0.4mA g m 0.2V or, gm 2.0 m mhos .

5.4 Find out the operating point current and voltage values (IDQ and VDSQ) for a self biased JFET having the supply voltage VDD = 20V and maximum value of drain current as 12 mA. Solution:-

We know that the value of drain current at Q-point may be taken as half of the maximum current, that is,

I 12mA IDSS 6.0 mA DQ 22

In the same way, the value of drain-source voltage at Q-point may be taken as half of supply voltage VDD. That is,

V 20V V DD DSQ 22 or, VDSQ 10 V

Therefore,

IDQ = 6.0 mA

VDSQ = 10.0 V

5.5 Calculate the value of source resistance RS required to self bias a n-JFET such that

VGSQ = - 3V. The n-JFET has maximum drain-source current IDSS = 12 mA, and pinch- off voltage, Vp = - 6V Solution:-

The drain current, ID, in a JFET, in the saturation region is,

2 VGS IID DSS 1 Vp

We have, IDSS = 12 mA, VGS = -3V and Vp = -6V, Therefore,

2 3 I12 mA 1 D 6 or, ID 9.0 mA

Since the voltage VGS is generated across the source resistor RS, we have,

VGS 3V RS  333 ID 9 mA

RS  333

5.6 For the DMOSFET circuit shown in fig., the device parameters are:

VGS(off) = -8V, IDSS = 10mA

Determine drain-to- source voltage VDS.

V = + 18V DD

RD 680Ω

+ VDS

- RG 10M

Solution:-

In a MOSFET, there is no gate current. Therefore, there is no voltage drop across resistor RG .

Thus, VGS = 0.

Further, when VGS =0, ID = IDSS, the maximum drain current.

Summing up voltages in the output loop and using ID = IDSS,

We have,

IDSS .RD + VDS = VDD or VDS = VDD – IDSS . RD

= 18 – 10 X 10-3 X 0.68 X 103

= 18 – 6.8 or VDS = 11.2 V

5.7 Data sheet of an EMOSFET specifies following parameters:

ID(on) = 50 mA at VGS = 6V and VT, the threshold voltage for EMOSFET, 2V.

Determine the drain current at VGS = 3V. Solution:-

We first determine the conductance parameter k for the device using the relation,

I D() ON k 2 VVGS T 50mA

(6 2)2 or, k 3.12 mA v2

Now, the drain current, ID is expressed as,

2 ID k V GS V T 3.12 1032 (3 2) or, ID 3.12 mA

5.8 The drain current changes from 5 mA to 7 mA when the gate voltage is changed from – 4.0V to – 3.7V in the amplifier circuit shown in fig.

Calculate the voltage gain of the amplifier.

+15V

6k RD v o vi

RL 6k RS

400Ω

Solution:-

In case, the source resistance RS is ac grounded as done in the circuit of fig. the gain of amplifier is,

AV = gm.rD

Where gm is transconductance of the transistor and rD is effective ac resistance seen by the drain terminal.

Now,

ID gm VGS VDS 2mA

0.3V or, gm 6.66 mS

And, rD = RD RL = 6k 6k = 3kΩ

Therefore,

-3 3 AV = gm X rD = 6.66 X 10 X 3 X 10 or, AV = 20

5.9 Find the drain-source voltage, VDS, for the NMOS transistor circuit shown in fig. The 2 device parameters are: conductance parameter, k = 600μA/v and VT = 2V.

+15V

RD 2k

R1 4M

VDS

2M R2

Solution:-

The gate current, IG, is zero in a MOSFET. Then, from voltage divider network,

R2 VVGS DD RR12 2M 15V 42MM or,5 VGS V

And, as we know

2 ID k V GS V T 62 600 10 (5 2) 5.4 mA or, ID 5.4 mA

Applying voltage summation in the output loop,

VDD = IDRD + VDS

-3 3 or, VDS = VDD – IDRD = 15 – (5.4 X 10 X 2 X 10 ) or, VDS = 4.2V

5.10 Calculate the voltage gain in the amplifier shown in fig. The transconductance of the transistor is 4000μs. If the 400Ω source resistance is by passed by an capacitor, how much is voltage gain now?

+20V

RD 6k vo

v i

R RG L 10k

5M 400Ω RS

Solution:-

When the source resistance RS is not by passed, the voltage gain is,

grmD AAV ...... ( ) 1 grms

Where rD and rS are effective (ac) resistance seen by the drain and source of the transistor.

And, rD = RD RL

= 6k 10 k = 3.75 kΩ

And, rS = RS = 400Ω

Therefore

4000 1063 3.75 10 A V 1 4000 106 400 4 3.75 5.7 2.6

AV 5.7

In case, RS is bypassed, the gain (rS = 0 in the Eq(A)),

-6 3 AV = gm rD = 4000 X 10 X 3.75 X 10 or, AV = 15