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{
"metadata": {
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"signature": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Solved Examination Paper"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Exa 1.b - page 476"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"R1= 10 # in k\u03a9\n",
"R2= 10 # in k\u03a9\n",
"Rf= 50 # in k\u03a9\n",
"V= 2 # in V\n",
"V1= V*R1/(R1+R2) # in V\n",
"V01= -Rf/R1*V1 # in V\n",
"print \"The value of V1 = %0.f Volts\" %(V1)\n",
"print \"The value of V01 = %0.f Volts\" %(V01)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of V1 = 1 Volts\n",
"The value of V01 = -5 Volts\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Exa 2.a - page 479"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"V_P= -4 # in V\n",
"I_DSS= 10 # in mA\n",
"V_GS= 0 # in V\n",
"R_D= 1.8 # in k\u03a9\n",
"V_DD= 20 # in V\n",
"I_D= I_DSS*(1-V_GS/V_P)**2 # in mA\n",
"# Applying KVL to the circuit, we get V_DD= I_D*R_D+V_D\n",
"V_D= V_DD-I_D*R_D # in V\n",
"print \"The value of I_D = %0.f mA\" %(I_D)\n",
"print \"The value of V_D = %0.f Volts\" %V_D"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of I_D = 10 mA\n",
"The value of V_D = 2 Volts\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Exa 2.c - page 480"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"V_GS= 3 # in V\n",
"Vth= 1 # in V\n",
"unCox= 25 # in mA/V**2\n",
"unCox= unCox*10**-3 # in A/V**2\n",
"W=3 # in \u00b5m\n",
"L=1 # in \u00b5m\n",
"r_DS= 1/(unCox*W/L*(V_GS-Vth)) # in \u03a9\n",
"print \"The value of r_DS = %0.2f \u03a9 \" %r_DS"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of r_DS = 6.67 \u03a9 \n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Exa 3.b - page 481"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Given data\n",
"I_CQ= 10 # in mA\n",
"I_CQ= I_CQ*10**-3 # in A\n",
"V_CQ= 5 # in V\n",
"V_CC= 10 # in V\n",
"R_C= 0.4 # in k\u03a9\n",
"R_C= R_C*10**3 # in \u03a9\n",
"V_BE= 0.075 # in V\n",
"V_BB= 0.175 # in V\n",
"beta=100 \n",
"beta_max=120 \n",
"beta_min= 40 \n",
"# Applying KVL we get, V_CQ= V_CC-I_C*(R_C+R_E)\n",
"R_E= (V_CC-V_CQ)/I_CQ-R_C # in \u03a9\n",
"print \"The value of R_E = %0.f \u03a9\" %( R_E)\n",
"I_B= I_CQ/beta # in A\n",
"R_B= (V_BB-V_BE)/I_B # in \u03a9\n",
"print \"The value of R_B = %0.f k\u03a9\" %(R_B*10**-3)\n",
"I_Cmax= beta_max*I_B # in A\n",
"I_Cmin= beta_min*I_B # in A\n",
"delta_I_CQ= I_Cmax-I_Cmin # in A\n",
"print \"The value of delta_I_C = %0.f mA\" %(delta_I_CQ*10**3)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of R_E = 100 \u03a9\n",
"The value of R_B = 1 k\u03a9\n",
"The value of delta_I_C = 8 mA\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}
|
