Field Effect Transistors: Important Example Solved Problems

Electron Devices

Chapter 5: Field Effect Transistors

Important Example Solved Problems

The Pinch‒off Voltage V_P

Ex. 1: For an n‒channel silicon FET, find the pinch‒off voltage and the channel half‒width.

Assume: a = 3 × 10^‒6 m, N_D = 10²¹ electrons/m², V_GS= 1/2 V_P, I_D = 0 and relative dielectric constant of silicon = 12.

Solution:

Since relative dielectric constant of silicon is 12 we have ε = 12ε₀. Using values of q and ε₀ we have

V_P = [ 1.6×10^‒19×10²¹×(3×10^‒6)² ] / [2×12×(8.85×10^‒12)] = 6.8 V

From equation (5.6.6) we have

 V_GS = (1 ‒ (b/a))² V_P

(1 ‒ (b/a))² = V_GS / V_P = 1/2

 1 ‒ (b/a) = 0.707

 b = (1 ‒ 0.707)a

= (0.293) (3 × 10⁻⁶) = 0.879 m

This result shows that the channel width is reduced to about one‒third its value for

V_GS = ½ V_P

Ex. 2: Assume that the p+n junction of a uniformly doped silicon n channel JFET at T = 300 K has doping concentrations of N_a =10¹⁸ cm and N_d = 10¹⁶ cm⁻³. Assume that the metallurgical channel thickness a is 0.7 μm Calculate the pinch off voltage. Also calculate pinch‒off voltage if the metallurgical channel thickness a is 0.75 μm.

Solution:

Here all quantities are in centimeters instead of meters. Channel thickness a = 0.7 μm = 0.7×10^‒4 cm. The internal pinch off voltage is given by

Characteristics Parameters of JFET (Transconductance)

Ex. 3: For JFET, if I_DSS = 20 mA, V_GS(off) = − 5 V, and g_mo = 4 mS or mA/V. Determine the transconductance for V_GS = ‒ 4 V, and find I_D at this point.

Solution:

From equation (5.7.2) we have,

 = 4×10^‒3 × 0.2 = 0.8 mS

We have,

 = 20×10^‒3 × 0.04 = 0.8 mA

n‒Channel E‒MOSFET - Current Equation

Ex. 4: For N‒EMOSFET V_T = 0.7 V, W= 30 μm, L=5 μm, μ_n = 650 cm²/V‒S, t_ox = 450 A (450 × 10^‒8). ε_ox = 3.9 × 8.85 × 10^‒14 F/cm

Assume transistor is biased in saturation region:

i) Determine drain current when V_GS = 2V_T. ii) If t_ox is doubled, find new K.

Solution:

i) W = 30 μm = 30 × 10^‒6 m = 30 × 10^‒4 cm

L = 5 μm = 5 × 10^‒6 m = 5 × 10^‒4 cm

K = Wμ_nε_ox / 2Lt_ox

= [ (30×10^‒4) (650) (3.9×8.85×10^‒14) ] / [ 2(5×10^‒4)(450×10⁻⁸) ]

= 0.1495 mA/V²

V_GS = 2 V_T = 2 × 0.7 = 1.4 V

i_D = K (V_GS ‒ V_T)² = (0.1495) (1.4 ‒ 0.7)²

= 0.0732 mA

ii) For t_ox = 2 × 450 × 10^‒8 = 900 × 10^‒8

K = Wμ_nε_ox / 2Lt_ox

=  [ (30×10^‒4) (650) (3.9×8.85×10^‒14) ] / [ 2(5×10^‒4)(900×10^‒8) ]

= 0.07475 mA/V²

Ex. 5: Evaluate the body effect coefficient (γ) in a MOSFET with N_a = 3×10¹⁶/cm³, ε_r = 11.6 and ε₀ = 8.854×10^‒12 F/m and C_ox = 1.726×10^‒7 F/cm².

Solution::

where ε_S = ε_r × ε₀

= 11.6×8.854×10^‒12 F/m

= 11.6×8.854×10^‒14 F/cm

 γ = { √[2×(1.6×10^‒19)×(3×10¹⁶)×(11.6×8.854×10⁻¹⁴) ] / [ 1.726×10^‒7 ]

= 0.5753