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1. An nMOS transistor is formed in a Si substrate doped with N_A=2x10¹⁵cm^-3. If the device is biased for operation in the inversion region, and has V_SB = 0:
a) What is the Fermi potential, f _f, in the semiconductor, and what is the electrostatic potential at the oxide-semiconductor interface (f _sur)?
b) What is the depletion layer thickness, x_d?
c) If we assume x_d =0.6m m, f _f=0.3 V, how much charge is in the depletion layer?

a) f _f = kT ln(N_A/n_i) = 0.026 ln(2x10¹⁵/1.5x10¹⁰) = 0.307 V

f_sur = = f_s- f_f = 2 f_f= 0.614 V

b) Depletion layer thickness x_d = (2e f _sur/qN_A)^1/2 = (2x1.04x10^-12x0.614/1.6x10^-19x2x10¹⁵)^1/2 = 0.632 m m.

c) When a transistor is in inversion, there is a depletion layer in the device channel (under the gate oxide) which provides a charge that must be balanced by the gate voltage. This charge is termed as ‘bulk layer depletion charge’ and must be considered for finding the threshold voltage of transistor.

Charge contained in the depletion region Q_b= -qN_AX_d = -1.6x10^-19x2x10¹⁵x0.6x10^-4 = -1.92x10^-8 C/cm²

This charge is negative for nMOS transistor as ions in the bulk are negatively charged.

2. An nMOS transistor has the following data: the oxide capacitance is 80nF/cm², the work function difference between the gate and the semiconductor is -0.1 V, f _F = 0.3 V, Q_b = -2.5x10^-8 coulombs/cm², and Qss = 10^-8 coulombs/cm².
a) Calculate the equilibrium threshold voltage, Vto,
b) If g = 0.25 V^½, what is the threshold voltage if V_SB is 1V?

C_ox = 80nF/cm², F _MS = -0.1V, Q_b=-2.5x10⁸ C/cm², Q_ss = 10^-8 C/cm², f _F=0.3V.

a)Equilibrium threshold voltage V_t0=F _MS+2f _F-(Q_b/C_ox)-Q_ss/C_ox

=-0.1+0.6-(-2.5x10^-8/80x10^-9)-(10^-8/80x10^-9) = 0.6875 V

b) Threshold voltage at V_SB=1V is given by V_t=V_t0+g ((2f _F+V_SB)^1/2 – (2f _F)^1/2) = 0.6875+0.25((0.6+1)^1/2-(0.6)^1/2) = 0.81V

3. An n-channel transistor has a substrate concentration of N_A = 1.4 x 10¹⁷ cm^-3, m _nC_ox = 188 m A/V², W=6m m, L=0.6 m m, and V_th=0.8V.
a) If the device is biased at V_GS= 1.2V and V_DS = V_eff (V_eff = V_GS – V_th), which region of operation (cutoff, triode, saturation) is the device operating in?
b) If we ignore channel length modulation, what is the drain current?
c) If V_DS increases by 0.5V, what is the new drain current if l = 0.01 V^-1?

a) V_DS-Sat = V_GS-V_TH = 1.2-0.8 =0.4V. Since V_DS-sat = V_DS, the device enters the saturation region.

b) Ignoring channel length modulation, drain current I_D = (m _nC_ox/2)W/L(V_GS-V_th)² = (188x10^-6/2)6/0.6(1.2-0.8)² = 0.15 mA.= 150m A

c) Considering l ( channel length modulation), the new I_D=(m _nC_ox/2)W/L(V_GS-V_th)²(1+l V_DS) =

150x10^-6(1+0.01x0.5) = 151m A

4. An nMOS transistor in the saturation region is measured to have the drain current of 20 m A when V_DS = V_eff. When V_DS is increased by 0.5V, I_D increases to 23 m A.
a) Calculate the channel length modulation factor, l , for this device.
b) What is the output resistance, r₀, of this transistor?

a)I_dsat = (m _nC_ox/2)W/L(V_GS-V_th)² = 20m A= I_D1

I_D2 = 23m A when V_DS increases by 0.5V

Thus, I_D2= I_D1(1+ l V_DS) => l =(23-20)x10^-6 /(20x10^-6x0.5) = 0.3V^-1.

b) Output resistance = r₀ = (l I_D2)^-1 = 144.93KW

5. An n-channel enhancement mode MOSFET has parameters: V_th=2V, K=0.4mA/V² (K=½m C_OXW/L)
a) If V_GS=4.3V, how large must V_DS be for constant current operation?
b) What value of I_D flows for V_GS=4.3V?
c) What region of operation is the device in if V_DS is reduced to 2V?
d) What value of I_D flows at V_DS=2V?

a) For constant current or operation under saturation region, V_DS must be >=V_GS-V_th i.e. V_DS must be greater than or equal to 4.3-2 = 2.3V.

b) I_D= K(V_GS-V_th)² = 0.4x10^-3 (2.3)² = 2.116 mA.

c) V_DSsat = V_GS-V_th = 2.3 V. since V_DS<V_DSsat, the transistor is in the triode(linear) region of operation.

d)Under these conditions, I_D=2K[(V_GS-V_th)V_DS-V_DS²/2] = 2x0.4x10^-3[2.3x2 – 2²/2] = 2.08 mA.

6. An nMOS transistor has parameters W=100m m, L=10m m, m C_OX = 20 m A/V², l =0.01V^-1, t_ox=0.12m m, f _f=0.3V, V_t0=1.1V, N_A=10¹⁵ cm^-3. For the following cases, you should prepare a table of values (at least 3-4 values for each case, appropriately

spread over the range of interest) and then plot those values

a) Sketch the I_D – V_DS characteristics for V_DS from 0 to 10V at V_GS=0.5,1.5 and 3V. Assume V_SB= 0V

b) Sketch the I_D – V_GS characteristics for V_GS from 0 to 3V at V_SB = 0, and 2 V. Assume V_DS=5V.

Used Expressions: Current in saturation region: I_D=m C_OXW/L(V_GS-V_t)²[1+l (V_DS-(V_GS-V_t))]

Current in triode region: I_D=m C_OXW/L[(V_GS-V_t)V_DS-V_DS²/2]

a) When VGS = 0.5V, VGS<VTH. Hence ID=0 for all VDS.

When VGS = 1.5, VDS-SAT= VGS-VTH = 1.5-1.1= 0.4V. The transistor will be in triode region when 0<VDS<0.4 and in constant current region for VDS>0.4V.

When V_GS=3V, V_DSSAT=3-1.1=1.9V. Hence, the transistor will be in triode region for 0<V_DS<1.9V and in saturation region for V_DS>1.9V

VGS=0.5V		VGS=1.5V	VGS=3V
VDS	ID	ID	VDS	ID
0	0	0	0	0
0.2	0	12	1	280
0.4	0	16	1.9	361
3.4	0	16.5	4.9	1957
6.4	0	17	7.9	2000

b) V_SB=0: For V_GS<1.1V, I_D=0. For V_GS>1.1V, the transistor will be in saturation region till V_DS=V_GS-V_t i.e. till V_GS=1.1+5 = 6.5V. Hence, for 1.1<V_GS<3, the transistor will be in saturation region.

VSB=2V.There is a Change in threshold voltage. Vt=Vt0+g ((VSB + 2f F)^½ – (2f F)^1/2)

C_ox = 3.9xe ₀/t_ox= 3.9x8.85x10^-14/(0.12x10^-4) = 287.6x10^-10 F per unit area

g =(2qNAe )1/2/Cox = (2x1.6x10-19x1015x1.04x10-12)1/2/287.6x10-10 = 0.636 V^1/2

f _F = KT/q ln(N_A/n_i) = 0.026 ln(10¹⁵/1.5x10¹⁰)=0.289V

V_t=1.1+0.636((2+0.578)^1/2-(0.578)^1/2) = 1.63V.

For V_SB=0, I_D=0 if V_GS<1.1V and for V_GS>1.1V,

I_D=100x10^-6(V_GS-1.1)²[1+0.01(5-(V_GS-1.1))]

For VSB=2V, I_D=0 for V_GS<1.6V and for V_GS>1.6V,

I_D=100x10^-6(V_GS-1.6)²[1+0.01(5-(V_GS-1.6))]

VSB=0		VSB=2V
VGS	ID	ID
0	0	0
1.2	1	0
1.7	37	1
2.6	233	104
3	372	203