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"name": "",
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter11:Memory and Advanced\n",
"Digital Circuits"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.1:pg-1017"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# The problem is solved using Hit and Trial as well as approximation thus no programming is needed hence the example is skipped"
],
"language": "python",
"metadata": {},
"outputs": [],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.2:pg-1032"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 11.2 Design of two-stage CMOS op-amp \n",
"\n",
"uC_n=50*10**-6; # u_n*C_ox (A/V**2)\n",
"uC_p=20.0*10**-6; # u_p*C_ox (A/V**2)\n",
"V_tn0=1.0; # (V)\n",
"V_tp0=-1; # (V)\n",
"fie_f=0.6/2; # (V)\n",
"y=0.5; # (V**1/2)\n",
"V_DD=5; # (V)\n",
"W_n=4*10.0**-6; # (m)\n",
"L_n=2*10.0**-6; # (m)\n",
"W_p=10*10**-6; # (m)\n",
"L_p=2*10.0**-6; # (m)\n",
"W=10*10**-6; # (m)\n",
"L=10*10.0**-6; # (m)\n",
"C_B=1*10.0**-12; # bit line capacitance (F)\n",
"deltaV=0.2; # 0.2 V decrement\n",
"WbyL_eq=1/(L_p/W_p+L_n/W_n); # WbyL_eq=(W/L)_eq\n",
"# Equivalent transistor will operate in saturation\n",
"I=(uC_n*WbyL_eq*(V_DD-V_tn0)**2)/2\n",
"r_DS=1/(uC_n*(W_n/L_n)*(V_DD-V_tn0));\n",
"v_Q=r_DS*I; # v_Q=r_DS*I\n",
"I_5=0.5*10.0**-3; # (A) \n",
"deltat=C_B*deltaV/I_5;\n",
"print deltat*1e9, \"is The time (ns) required to develop an output voltage of 0.2V\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"0.4 is The time (ns) required to develop an output voltage of 0.2V\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11.3:pg-1041"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 11.3 : Time required for v_B to reach 4.5V\n",
"# Consider sense-amplifier circuit\n",
"uC_n=50*10**-6; #uC_n=u_n*C_ox (A/V**2)\n",
"uC_p=20*10**-6; #uC_p=u_p*C_ox (A/V**2)\n",
"W_n=12*10**-6; # (m)\n",
"L_n=4*10**-6; # (m)\n",
"W_p=30*10**-6; # (m)\n",
"L_p=4*10**-6; # (m)\n",
"v_B=4.5; # (V)\n",
"C_B=1*10**-12; # (F)\n",
"V_GS=2.5; # (V)\n",
"V_t=1; # (V)\n",
"deltaV=0.1; # (V)\n",
"g_mn=uC_n*(W_n/L_n)*(V_GS-V_t); # (A/V)\n",
"g_mp=uC_p*(W_p/L_p)*(V_GS-V_t); # (A/V)\n",
"G_m=g_mn+g_mp; # (A/V)\n",
"T=C_B/G_m; # (s)\n",
"deltat=T*(math.log(v_B/V_GS)-math.log(deltaV));\n",
"print round(deltat*1e9,1),\" is The time for v_B to reach 4.5V (s)\"\n",
"# The answer in the textbook is slightly different due to approximation"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"6.4 is The time for v_B to reach 4.5V (s)\n"
]
}
],
"prompt_number": 18
}
],
"metadata": {}
}
]
}
|
