An ideal n-channel MOSFET as the following parameters: V_T = -0.35 V, mu_p = 210

Question

An ideal n-channel MOSFET as the following parameters: V_T = -0.35 V, mu_p = 210 cm^2/V-s, t_ox = 11 nm = 110 A, W = 35 mu m, L = 1.2 mu m. (a) Plot I_D versus V_SD for 0 greaterthanorequalto V_SD greaterthanorequalto 3 V and for V_SG = 0, 0.6, 1.2, 1.8, and 2.4 V. Indicate on each curve the V_SD (sat) point. (b) Plot squareroot I_D (sat) versus V_SG for 0 lessthanorequalto V_SG lessthanorequalto 2.4 V. (c) Plot I_D versus V_SG for 0 lessthanorequalto V_SG lessthanorequalto 2.4 V and for V_SD = 0.1 V.

Explanation / Answer

As par the question; up = 210 cm2/V-S, tox = 11 nm, w = 35 um , L = 1.2 um and Vt = -0.35 V.

Therefore,

up tox = 23 uA/V2 and (w/L) = 30 and Assume p = 0;

In this case, VSG < |Vt|; hence ID = 0 uA.

2.

Case 2: VSG = 0.6 V

Here, VSG > |Vt|

So current through ID follows the expression;

ID = up tox(w/L) [VSG+Vt – (VSD/2)]*VSD upto VSD <= (VSG + |Vt|) <= 0.95

Here,

ID = 690*[0.95 - (VSD/2)] * VSD

When VSD > 0.95; then drain current will be;

ID = up tox(w/2L) [VSG+Vt ]2

ID = (690/2)[0.952] = 311.36 uA.

VSD(in V)

ID(in uA)

0

0

0.2

117.3

0.4

207

0.6

269

0.95

311.36

2

311.36

3

311.36

Case 3: VSG = 1.2 V

Here ID will be in linear region when VSD <= 1.55V, after that it will be in saturation mode.

This case also has the similar characteristics as previous;

VSD(in V)

ID(in uA)

0

0

0.5

448.5

0.75

517.5

1.25

537.5

1.55

829

2

829

3

829

Case 4: VSG = 1.8 V

Here ID will be in linear region when VSD <= 2.15V, after that it will be in saturation mode.

This case also has the similar characteristics as previous;

VSD(in V)

ID(in uA)

0

0

1

1138.5

1.75

1540

2

1587

2.15

1594.7

2.5

1594.7

3

1594.7

Case 5: VSG = 2.4 V

Here ID will be in linear region when VSD <= 2.75V, after that it will be in saturation mode.

This case also has the similar characteristics as previous;

VSD(in V)

ID(in uA)

0

0

1

1552.5

2

2415

2.5

2587.5

2.75

2609

2.9

2609

3

2609

b) The squareroot(IDsat) vs VSG data is given below;

VSG(in V)

Squareroot(IDsat) (in uA)

0

0

0.5

15.8

1

25

1.5

34.36

2

43.69

2.4

51.09

The equation of the saturation drain current can be written as;

squareroot(ID) = squareroot(690/2) * (VGS + Vt)

c) In this case VSD <= VSG + |Vt|

ID = 690 * [VSG + 0.35 - 0.05] * 0.1 = 690*[VSG + 0.3] * 0.1

So the VGS vs ID chart is given below for VDS = 0.1V

VSG(in V)

ID (in uA)

0

0

0.5

55.2

1

89.7

1.5

124.2

2

158.7

2.4

186.3

VSD(in V)

ID(in uA)

0

0

0.2

117.3

0.4

207

0.6

269

0.95

311.36

2

311.36

3

311.36

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.