Homework 5 - Solutions                                                                                EE562, Fall 2000

1. An n-channel enhancement mode MOSFET has it’s gate connected to drain.

a) Show that the device always operates in constant current region for V_GS>V_t, regardless of the value of V_DS.

b) Sketch the v-i characteristics (I_D vs. V_DS) for this case if m C_OX=1 mA/V² and V_t=2V, W/L=1.

a) When gate is connected to drain, V_DS=V_GS. Hence, V_DS is always greater than V_GS-V_t regardless the value of V_DS. This implies the transistor is in constant current region for V_GS > V_t.

b)In this case, I_D=mC_oxW/L (V_DS – V_t)²  since V_GS=V_DS.

Thus, I_D =1x10^-3 (V_DS-2)^2., if we assume W/L = 1

2. An n-channel enhancement mode MOSFET has parameters: V_t=2V, m C_OX=0.4mA/V², W/L = 1

a) If V_GS=4.3V, how large must V_DS be for constant current operation?

b) What value of I_D flows for this V_GS if W = 10L?

c) What value of I_D flows if V_DS is reduced to 2V?

a) For constant current operation, V_{DS >} (V_GS-V_t) i.e. V_DS  > 4.3-2 or V_DS >2.3V

b) I_D=mC_oxW/L(V_GS-V_t)²

        =0.4x10^-3x10 (2.3)² = 21.16 mA

c)If V_DS =2V, it’s less than (V_GS-V_t). The device will now be in triode region.

   Hence, I_D=mC_oxW/L[2(V_GS-V_t) V_DS – V_DS² ] = 0.4x10^-3x[2x2.3x2-(2)(2)]

                   = 2.08 mA.

3. MOSFET Small Signal Model

a) Calculate low frequency small signal model parameters for an nMOS device and draw the small signal model with the elements labeled appropriately. Assume the device is operated at I_D=600m A at V_GS=2V and V_SB=1V with Vth=0.8V, l =0.06V^-1, g =0.5V^½, N_A=5x10¹⁵cm^-3, m C_OX=100m A/V and W/L=12m m/1.2m m.

b) Redraw the small signal model to include capacitors Cgs, Cgd, Csb, and Cdb and calculate values for Cgs and Cgd if W=10m m, L=2m m, L_D=0.15m m and t_OX=100nm. For calculation of Cgs, use a value for Leff that accounts for lateral diffusion under the gate (L_D).

a) g_m= (2mC_oxW/LI_D)^1/2 = (2x100x10^-6x10x600x10^-6)^1/2 = 1.1mA/V

   f_F = KT/q ln(N_A/n_i) = 0.026 ln(5x10¹⁵/1.5x10¹⁰) = 0.763V

    g_s = gg_m/(2(V_SB+2f_F)^1/2) = 0.5x1.1x10^-3/(2(1+1.53)^1/2) = 0.173 mA/V

   r_ds = 1/(lI_D) = (0.06x600x10^-6)^-1 = 27.77 KW

b) L_eff = L-2L_D = 2-2(0.15) = 1.7mm

    C_ox = 3.9xe₀/t_ox = 3.9x8.85x10^-14/100x10^-7 = 34.5x10^-9 F/(mm)²

   C_gs = 2/3(Wl_effC_ox) = 2/3(10x1.7x34.5x10-9) = 0.39 mF

   C_gd = WC_oxL_D = 10x34.5x10^-9x0.15 = 0.052 mF

4. Calculate the small signal model parameters for a BJT in low frequency operation if the device has b =120, V_A=75V, and is operating in the forward active region with collector current of 80m A. Draw and label the model.

I_C = 80mA. b=120. Thus, I_B = I_C/b = 0.66mA.

  g_m = I_C /V_T = 80x10^-6/0.026 = 3.07 mA/V

  r_p = b/ g_m = 120/(3.07x10^-3) =  39.1 KW

  r_o = V_A/ I_C = 75/(80x10^-6) = 0.94 MW

  r_b is only significant at high frequencies and you were not required to calculate a value for it.  Although a value can be approximated, it actually depends largely on the physical design of the integrated device and is a function of layout (which we know nothing about in this problem).

5. Calculate SPICE model parameters KP, GAMMA, PHI, LAMBDA, and VTO for an nMOS transistor with a Length of 2m m and the data shown in the table below. Be sure to watch your units carefully, especially meters, cm, mm conversions.

  KP = Uo e₀e_ox/TOX = 700x8.85x10^-14x3.9/(4x10^-6) = 60.4 mA/V2

C_ox=  e₀e_ox /TOX  = 8.63x10^-8 F per unit area.

 GAMMA = ( (2qe_SiNSUB)^1/2) / C_ox =(2x1.6x10^-19x10¹⁶x1.04x10^-12)^1/2/(8.63x10^-8)

                 = 0.67 V^1/2

PHI = 2KT/q ln(NSUB/n_i) = 2x0.026 ln(10¹⁶/(1.5x10¹⁰) )= 0.697

Leff = L-2L_D = 2-2(0.2) = 1.6 mm.  LAMBDA = I/L_eff(  DL/DV_DS) = 1/1.6 (0.1) = 0.06 V^-1

VTO = F_MS – qNSS/C_ox + (2qNSUB½PHI½e)^1/2/C_ox +½ PHI½

          = -0.6 – 1.6x10^-19x5x10¹⁰/8.63x10^-8  + (2x1.6x10^-19x10¹⁶x0.697x1.04x10^-12)^1/2/( 8.63x10^-8 ) +  0.697

          = -0.6-0.0927+0.558 + 0.697 = 0.562 V

6. Use SPICE to plot the DC transfer characteristics (v_out vs. v_i, i.e. v_ds vs. v_gs) of an nMOS transistor with a 1kW resistor load. Let VDD = 5V and vary V_GS (input) from 0 to 5V. Use the model parameters given below and ignore capacitances. Set W=20m m and L=10m m.

.model cmosn nmos level = 1 vto = 0.77 kp = 7.7e-5 gamma = 0.71 phi = 0.73 lambda = 0.0625 tox = 3e-8 ld = 2e-7 u0 = 670 nsub = 2e16

The following circuit, netlist, and output plots describe the results of this SPICE analysis using B2Spice.

***** main circuit

M1 2  1  0  0 cmosn  l = 10u  w = 20u

R 3  2  1K

Vin 1  0 DC 0

VDD 3  0 DC 5

.model cmosn nmos  level = 1  vto = 0.77   kp = 7.7e-5

gamma = 0.71   phi = 0.73   lambda = 0.0625   tox = 3e-8 

ld = 2e-7   u0 = 670   nsub = 2e16 

.OPTIONS

.DC  Vin 0 5 .1

.END

This is a common-source amplifier configuration, and you will notice that the output decreases as the input increases.  Notices also that the output never drops to zero (or even below 3V) because the voltage drop across the load resistor is not large enough.  If we increases the value of R and/or increased the value of W/L to increase the drain current of the MOSFET, the output would look like the plot below (done with R = 10kW and W/L = 10/2mm), which is a more typical inverting amplifier response.