{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 04 : Thyristors"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.3, Page No 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"P=.5 #P=V_g*I_g\n",
"s=130 #s=V_g/I_g\n",
"\n",
"#Calculations\n",
"I_g=math.sqrt(P/s)\n",
"V_g=s*I_g\n",
"E=15\n",
"R_s=(E-V_g)/I_g \n",
"\n",
"#Results\n",
"print(\"Gate source resistance=%.2f ohm\" %R_s)\n",
"#Answers have small variations from that in the book due to difference in the rounding off of digits."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Gate source resistance=111.87 ohm\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.4, Page No 149"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"R_s=120 #slope of load line is -120V/A. This gives gate source resistance\n",
"print(\"gate source resistance=%.0f ohm\" %R_s)\n",
"\n",
"P=.4 #P=V_g*I_g\n",
"E_s=15\n",
"\n",
"#Calculations\n",
" #E_s=I_g*R_s+V_g % after solving this\n",
" #120*I_g**2-15*I_g+0.4=0 so\n",
"a=120 \n",
"b=-15\n",
"c=0.4\n",
"D=math.sqrt((b**2)-4*a*c)\n",
"I_g=(-b+D)/(2*a) \n",
"V_g=P/I_g\n",
"\n",
"#Results\n",
"print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3)) \n",
"print(\"\\nthen trigger voltage=%.3f V\" %V_g)\n",
"I_g=(-b-D)/(2*a) \n",
"V_g=P/I_g\n",
"print(\"\\ntrigger current=%.2f mA\" %(I_g*10**3)) \n",
"print(\"\\nthen trigger voltage=%.2f V\" %V_g)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"gate source resistance=120 ohm\n",
"\n",
"trigger current=86.44 mA\n",
"\n",
"then trigger voltage=4.628 V\n",
"\n",
"trigger current=38.56 mA\n",
"\n",
"then trigger voltage=10.37 V\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.5 Page No 150"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"#V_g=1+10*I_g\n",
"P_gm=5 #P_gm=V_g*I_g\n",
"#after solving % eqn becomes 10*I_g**2+I_g-5=0\n",
"a=10.0 \n",
"b=1.0 \n",
"c=-5\n",
"\n",
"#Calculations\n",
"I_g=(-b+math.sqrt(b**2-4*a*c))/(2*a)\n",
"E_s=15\n",
"#using E_s=R_s*I_g+V_g\n",
"R_s=(E_s-1)/I_g-10 \n",
"P_gav=.3 #W\n",
"T=20*10**-6\n",
"f=P_gav/(P_gm*T)\n",
"dl=f*T\n",
"\n",
"#Results\n",
"print(\"Reistance=%.3f ohm\" %R_s)\n",
"print(\"Triggering freq=%.0f kHz\" %(f/1000))\n",
"print(\"Tduty cycle=%.2f\" %dl)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Reistance=11.248 ohm\n",
"Triggering freq=3 kHz\n",
"Tduty cycle=0.06\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.6, Page No 151"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"I=.1\n",
"E=200.0\n",
"L=0.2\n",
"\n",
"#Calculations\n",
"t=I*L/E \n",
"R=20.0\n",
"t1=(-L/R)*math.log(1-(R*I/E)) \n",
"L=2.0\n",
"t2=(-L/R)*math.log(1-(R*I/E)) \n",
"\n",
"#Results\n",
"print(\"in case load consists of (a)L=.2H\")\n",
"print(\"min gate pulse width=%.0f us\" %(t*10**6))\n",
"print(\"(b)R=20ohm in series with L=.2H\")\n",
"print(\"min gate pulse width=%.3f us\" %(t1*10**6))\n",
"print(\"(c)R=20ohm in series with L=2H\")\n",
"print(\"min gate pulse width=%.2f us\" %(t2*10**6))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"in case load consists of (a)L=.2H\n",
"min gate pulse width=100 us\n",
"(b)R=20ohm in series with L=.2H\n",
"min gate pulse width=100.503 us\n",
"(c)R=20ohm in series with L=2H\n",
"min gate pulse width=1005.03 us\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.9 Page No 163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"def theta(th):\n",
" I_m=1 #supposition\n",
" I_av=(I_m/(2*math.pi))*(1+math.cos(math.radians(th)))\n",
" I_rms=math.sqrt((I_m/(2*math.pi))*((180-th)*math.pi/360+.25*math.sin(math.radians(2*th))))\n",
" FF=I_rms/I_av\n",
" I_rms=35\n",
" I_TAV=I_rms/FF\n",
" return I_TAV\n",
"\n",
"#Calculations\n",
"print(\"when conduction angle=180\")\n",
"th=0\n",
"I_TAV=theta(th)\n",
"print(\"avg on current rating=%.3f A\" %I_TAV)\n",
"print(\"when conduction angle=90\")\n",
"th=90\n",
"I_TAV=theta(th)\n",
"\n",
"#Results\n",
"print(\"avg on current rating=%.3f A\" %I_TAV)\n",
"print(\"when conduction angle=30\")\n",
"th=150\n",
"I_TAV=theta(th)\n",
"print(\"avg on current rating=%.3f A\" %I_TAV)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"when conduction angle=180\n",
"avg on current rating=22.282 A\n",
"when conduction angle=90\n",
"avg on current rating=15.756 A\n",
"when conduction angle=30\n",
"avg on current rating=8.790 A\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.10, Page No 164"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"def theta(th):\n",
" n=360.0/th\n",
" I=1.0 #supposition\n",
" I_av=I/n\n",
" I_rms=I/math.sqrt(n)\n",
" FF=I_rms/I_av\n",
" I_rms=35\n",
" I_TAV=I_rms/FF\n",
" return I_TAV\n",
"\n",
"#Calculations\n",
"th=180.0\n",
"I_TAV1=theta(th)\n",
"th=90.0\n",
"I_TAV2=theta(th)\n",
"th=30.0\n",
"I_TAV3=theta(th)\n",
"\n",
"#Results\n",
"print(\"when conduction angle=180\")\n",
"print(\"avg on current rating=%.3f A\" %I_TAV)\n",
"print(\"when conduction angle=90\")\n",
"print(\"avg on current rating=%.1f A\" %I_TAV2)\n",
"print(\"when conduction angle=30\")\n",
"print(\"avg on current rating=%.4f A\" %I_TAV3)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"when conduction angle=180\n",
"avg on current rating=8.790 A\n",
"when conduction angle=90\n",
"avg on current rating=17.5 A\n",
"when conduction angle=30\n",
"avg on current rating=10.1036 A\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.11 Page No 165"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#initialisation of variables\n",
"f=50.0 #Hz\n",
"\n",
"#Calculations\n",
"I_sb=3000.0\n",
"t=1/(4*f)\n",
"T=1/(2*f)\n",
"I=math.sqrt(I_sb**2*t/T) \n",
"r=(I_sb/math.sqrt(2))**2*T \n",
"\n",
"#Results\n",
"print(\"surge current rating=%.2f A\" %I)\n",
"print(\"\\nI**2*t rating=%.0f A^2.s\" %r)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"surge current rating=2121.32 A\n",
"\n",
"I**2*t rating=45000 A^2.s\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.12 Page No 165"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"V_s=300.0 #V\n",
"R=60.0 #ohm\n",
"L=2.0 #H\n",
"\n",
"#Calculations\n",
"t=40*10**-6 #s\n",
"i_T=(V_s/R)*(1-math.exp(-R*t/L))\n",
"i=.036 #A\n",
"R1=V_s/(i-i_T)\n",
"\n",
"#Results\n",
"print(\"maximum value of remedial parameter=%.3f kilo-ohm\" %(R1/1000))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"maximum value of remedial parameter=9.999 kilo-ohm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.16 Page No 172"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"V_p=230.0*math.sqrt(2)\n",
"\n",
"#Calculations\n",
"R=1+((1)**-1+(10)**-1)**-1\n",
"A=V_p/R\n",
"s=1 #s\n",
"t_c=20*A**-2*s\n",
"\n",
"#Results\n",
"print(\"fault clearance time=%.4f ms\" %(t_c*10**3))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"fault clearance time=0.6890 ms\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.17, Page No 176"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"V_s=math.sqrt(2)*230 #V\n",
"L=15*10**-6 #H\n",
"I=V_s/L #I=(di/dt)_max\n",
"R_s=10 #ohm\n",
"v=I*R_s #v=(dv/dt)_max\n",
"\n",
"#Calculations\n",
"f=50 #Hz\n",
"X_L=L*2*math.pi*f\n",
"R=2\n",
"I_max=V_s/(R+X_L) \n",
"FF=math.pi/math.sqrt(2)\n",
"I_TAV1=I_max/FF \n",
"FF=3.98184\n",
"I_TAV2=I_max/FF \n",
"\n",
"\n",
"#RESULTS\n",
"print(\"(di/dt)_max=%.3f A/usec\" %(I/10**6))\n",
"print(\"\\n(dv/dt)_max=%.2f V/usec\" %(v/10**6))\n",
"print(\"\\nI_rms=%.3f A\" %I_max)\n",
"print(\"when conduction angle=90\")\n",
"print(\"I_TAV=%.3f A\" %I_TAV1)\n",
"print(\"when conduction angle=30\")\n",
"print(\"I_TAV=%.3f A\" %I_TAV2)\n",
"print(\"\\nvoltage rating=%.3f V\" %(2.75*V_s)) #rating is taken 2.75 times of peak working voltage unlike 2.5 to 3 times as mentioned int book."
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(di/dt)_max=21.685 A/usec\n",
"\n",
"(dv/dt)_max=216.85 V/usec\n",
"\n",
"I_rms=162.252 A\n",
"when conduction angle=90\n",
"I_TAV=73.039 A\n",
"when conduction angle=30\n",
"I_TAV=40.748 A\n",
"\n",
"voltage rating=894.490 V\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.19, Page No 186"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"T_jm=125\n",
"th_jc=.15 #degC/W\n",
"th_cs=0.075 #degC/W\n",
"\n",
"\n",
"#Calculations\n",
"dT=54 #dT=T_s-T_a\n",
"P_av=120\n",
"th_sa=dT/P_av\n",
"T_a=40 #ambient temp\n",
"P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
"if (P_av-120)<1 :\n",
" print(\"selection of heat sink is satisfactory\")\n",
"\n",
"dT=58 #dT=T_s-T_a\n",
"P_av=120\n",
"th_sa=dT/P_av\n",
"T_a=40 #ambient temp\n",
"P_av=(T_jm-T_a)/(th_sa+th_jc+th_cs)\n",
"if (P_av-120)<1 :\n",
" print(\"selection of heat sink is satisfactory\")\n",
"\n",
"V_m=math.sqrt(2)*230\n",
"R=2\n",
"I_TAV=V_m/(R*math.pi)\n",
"P_av=90\n",
"th_sa=(T_jm-T_a)/P_av-(th_jc+th_cs)\n",
"dT=P_av*th_sa\n",
"print(\"for heat sink\") \n",
"print(\"T_s-T_a=%.2f degC\" %dT) \n",
"print(\"\\nP_av=%.0f W\" %P_av)\n",
"P=(V_m/2)**2/R\n",
"eff=P/(P+P_av) \n",
"print(\"\\nckt efficiency=%.3f pu\" %eff)\n",
"a=60 #delay angle\n",
"I_TAV=(V_m/(2*math.pi*R))*(1+math.cos(math.radians(a)))\n",
"print(\"\\nI_TAV=%.2f A\" %I_TAV)\n",
"dT=46\n",
"T_s=dT+T_a\n",
"T_c=T_s+P_av*th_cs \n",
"T_j=T_c+P_av*th_jc \n",
"\n",
"#Results\n",
"print(\"\\ncase temp=%.2f degC\" %T_c)\n",
"print(\"\\njunction temp=%.2f degC\" %T_j)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"for heat sink\n",
"T_s-T_a=-20.25 degC\n",
"\n",
"P_av=90 W\n",
"\n",
"ckt efficiency=0.993 pu\n",
"\n",
"I_TAV=38.83 A\n",
"\n",
"case temp=92.75 degC\n",
"\n",
"junction temp=106.25 degC\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.20, Page No 187"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"T_j=125.0 #degC\n",
"T_s=70.0 #degC\n",
"th_jc=.16 #degC/W\n",
"th_cs=.08 #degC/W\n",
"\n",
"#Calculations\n",
"P_av1=(T_j-T_s)/(th_jc+th_cs) \n",
"\n",
"T_s=60 #degC\n",
"P_av2=(T_j-T_s)/(th_jc+th_cs) \n",
"inc=(math.sqrt(P_av2)-math.sqrt(P_av1))*100/math.sqrt(P_av1) \n",
"\n",
"#Results\n",
"print(\"Total avg power loss in thristor sink combination=%.2f W\" %P_av1)\n",
"print(\"Percentage inc in rating=%.2f\" %inc)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total avg power loss in thristor sink combination=229.17 W\n",
"Percentage inc in rating=8.71\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.21, Page No 197"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"R=25000.0\n",
"I_l1=.021 #I_l=leakage current\n",
"I_l2=.025\n",
"I_l3=.018\n",
"I_l4=.016\n",
" #V1=(I-I_l1)*R\n",
" #V2=(I-I_l2)*R\n",
" #V3=(I-I_l3)*R\n",
" #V4=(I-I_l4)*R\n",
" #V=V1+V2+V3+V4\n",
" \n",
"#Calculations\n",
"V=10000.0\n",
"I_l=I_l1+I_l2+I_l3+I_l4\n",
" #after solving\n",
"I=((V/R)+I_l)/4\n",
"R_c=40.0\n",
"V1=(I-I_l1)*R \n",
"\n",
"#Resluts\n",
"print(\"voltage across SCR1=%.0f V\" %V1)\n",
"V2=(I-I_l2)*R \n",
"print(\"\\nvoltage across SCR2=%.0f V\" %V2)\n",
"V3=(I-I_l3)*R \n",
"print(\"\\nvoltage across SCR3=%.0f V\" %V3)\n",
"V4=(I-I_l4)*R \n",
"print(\"\\nvoltage across SCR4=%.0f V\" %V4)\n",
"\n",
"I1=V1/R_c \n",
"print(\"\\ndischarge current through SCR1=%.3f A\" %I1)\n",
"I2=V2/R_c \n",
"print(\"\\ndischarge current through SCR2=%.3f A\" %I2)\n",
"I3=V3/R_c \n",
"print(\"\\ndischarge current through SCR3=%.3f A\" %I3)\n",
"I4=V4/R_c \n",
"print(\"\\ndischarge current through SCR4=%.3f A\" %I4)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"voltage across SCR1=2475 V\n",
"\n",
"voltage across SCR2=2375 V\n",
"\n",
"voltage across SCR3=2550 V\n",
"\n",
"voltage across SCR4=2600 V\n",
"\n",
"discharge current through SCR1=61.875 A\n",
"\n",
"discharge current through SCR2=59.375 A\n",
"\n",
"discharge current through SCR3=63.750 A\n",
"\n",
"discharge current through SCR4=65.000 A\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.22, Page No 198"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"V_r=1000 #rating of SCR\n",
"I_r=200 #rating of SCR\n",
"V_s=6000 #rating of String\n",
"I_s=1000 #rating of String\n",
"\n",
"#Calculations\n",
"print(\"when DRF=.1\")\n",
"DRF=.1\n",
"n_s=V_s/(V_r*(1-DRF)) \n",
"print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
"n_p=I_s/(I_r*(1-DRF)) \n",
"print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))\n",
"print(\"when DRF=.2\")\n",
"DRF=.2\n",
"\n",
"#Results\n",
"n_s=V_s/(V_r*(1-DRF)) \n",
"print(\"number of series units=%.0f\" %math.ceil(n_s))\n",
"n_p=I_s/(I_r*(1-DRF)) \n",
"print(\"\\nnumber of parrallel units=%.0f\" %math.ceil(n_p))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"when DRF=.1\n",
"number of series units=7\n",
"\n",
"number of parrallel units=6\n",
"when DRF=.2\n",
"number of series units=8\n",
"\n",
"number of parrallel units=7\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.23, Page No 198"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"V1=1.6 #on state voltage drop of SCR1\n",
"V2=1.2 #on state voltage drop of SCR2\n",
"I1=250.0 #current rating of SCR1\n",
"I2=350.0 #current rating of SCR2\n",
"\n",
"#Calculations\n",
"R1=V1/I1\n",
"R2=V2/I2\n",
"I=600.0 #current to be shared\n",
" #for SCR1 % I*(R1+R)/(total resistance)=k*I1 (1)\n",
" #for SCR2 % I*(R2+R)/(total resistance)=k*I2 (2)\n",
" #(1)/(2)\n",
"R=(R2*I2-R1*I1)/(I1-I2)\n",
"\n",
"\n",
"#Results\n",
"print(\"RSequired value of resistance=%.3f ohm\" %R)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"RSequired value of resistance=0.004 ohm\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.25, Page No 223"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"f=2000.0 #Hz\n",
"C=0.04*10**-6\n",
"n=.72\n",
"\n",
"#Calculations\n",
"R=1/(f*C*math.log(1/(1-n))) \n",
"V_p=18\n",
"V_BB=V_p/n\n",
"R2=10**4/(n*V_BB) \n",
"I=4.2*10**-3 #leakage current\n",
"R_BB=5000\n",
"R1=(V_BB/I)-R2-R_BB\n",
"\n",
"#Results\n",
"print(\"R=%.2f kilo-ohm\" %(R/1000))\n",
"print(\"\\nR2=%.2f ohm\" %R2)\n",
"print(\"\\nR1=%.0f ohm\" %R1)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"R=9.82 kilo-ohm\n",
"\n",
"R2=555.56 ohm\n",
"\n",
"R1=397 ohm\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.26, Page No 223"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"\n",
"V_p=18.0\n",
"n=.72\n",
"V_BB=V_p/n\n",
"I_p=.6*10**-3\n",
"I_v=2.5*10**-3\n",
"V_v=1\n",
"\n",
"#Calculations\n",
"R_max=V_BB*(1-n)/I_p \n",
"print(\"R_max=%.2f kilo-ohm\" %(R_max/1000))\n",
"R_min=(V_BB-V_v)/I_v \n",
"print(\"\\nR_min=%.2f kilo-ohm\" %(R_min/1000))\n",
"\n",
"C=.04*10**-6\n",
"f_min=1/(R_max*C*math.log(1/(1-n))) \n",
"print(\"\\nf_min=%.3f kHz\" %(f_min/1000))\n",
"f_max=1/(R_min*C*math.log(1/(1-n))) \n",
"print(\"\\nf_max=%.2f kHz\" %(f_max/1000))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"R_max=11.67 kilo-ohm\n",
"\n",
"R_min=9.60 kilo-ohm\n",
"\n",
"f_min=1.683 kHz\n",
"\n",
"f_max=2.05 kHz\n"
]
}
],
"prompt_number": 17
}
],
"metadata": {}
}
]
}