{
"metadata": {
"name": "",
"signature": "sha256:b6ca6f88473b518209629b5a50e034951050cc84cc7fdfcdcdfe2cbba83ff5c4"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter03:Diodes"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.1:pg-143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.1: Peak value of diode current and maximum reverse voltage\n",
"#v_s is sinusoidal input voltage with peak 24V\n",
"#battery charges to 12V\n",
"I_d=(24-12)/100.0\n",
"max_v_rev=24+12.0;\n",
"print round(I_d,2),\"peak value of diode current (A)\\n\",round(max_v_rev,2),\"maximum reverse voltage acrossthe diode (V)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"0.12 peak value of diode current (A)\n",
"36.0 maximum reverse voltage acrossthe diode (V)\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.2:pg-145"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.2 : Values of Iand V for the circuit given\n",
"print \"Consider fig 3.6(a). Assume both diodes are conducting\"\n",
"I_D2=(10-0)/10.0;\n",
"I=(0-(-10))/5.0-I_D2; # node eqution at B for fig 3.6(a)\n",
"V_B=0;\n",
"V=0;\n",
"print I,\"= I (mA)\\n\", V,\"= V (V)\\n\" ,\"D_1 is conducting as assumed originally\"\n",
"print \"Consider fig 3.6(a). Assume both diodes are conducting\"\n",
"I_D2=(10-0)/5.0;\n",
"I=(0-(-10.0))/10-2; # node eqution at B for fig 3.6(b)\n",
"print I,\"= I (mA)\\n \", V,\"=V (V)\"\n",
"print \"Implies assumption is wrong. lets assume D_1 is off and D_2 is on\"\n",
"I_D2=(10-(-10))/15.0;\n",
"V_B=-10+10.0*I_D2;\n",
"I=0;\n",
"print I,\"= I (mA)\\n\", round(V_B,1),\"= V (V)\\n D_1 is reverse biased\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Consider fig 3.6(a). Assume both diodes are conducting\n",
"1.0 = I (mA)\n",
"0 = V (V)\n",
"D_1 is conducting as assumed originally\n",
"Consider fig 3.6(a). Assume both diodes are conducting\n",
"-1.0 = I (mA)\n",
" 0 =V (V)\n",
"Implies assumption is wrong. lets assume D_1 is off and D_2 is on\n",
"0 = I (mA)\n",
"3.3 = V (V)\n",
" D_1 is reverse biased\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.3:pg-150"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.3 : Evaluating junction scaling constant\n",
"#i-I_S*exp(v/(n*V_T)) implies I_S=i*exp(-v/(n*V_T))\n",
"n=1;\n",
"i=10**-3; # (A)\n",
"v=700; # (V)\n",
"V_T=25; # (V)\n",
"I_S=i*exp(-v/(n*V_T))\n",
"print round(I_S,17),\"= I_S (A) for n=1\"\n",
"n=2;\n",
"I_S=i*exp(-v/(n*V_T))\n",
"print round(I_S,11),\"= I_S (A) for n=2\"\n",
"print \"These values implies I_S is 1000 times greater \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"6.9e-16 = I_S (A) for n=1\n",
"8.3e-10 = I_S (A) for n=2\n",
"These values implies I_S is 1000 times greater \n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.4:pg-154"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 3.4: To determine I_D and V_D\n",
"V_DD=5; # (V)\n",
"R=1000; # (ohm)\n",
"I_1=1*10**-3; # (A)\n",
"V_D=0.7; # (V)\n",
"V_1=V_D;\n",
"I_D=(V_DD-V_D)/R;\n",
"I_2=I_D;\n",
"V_2=V_1+0.1*log10(I_2/I_1);\n",
"I_D=(V_DD-V_2)/R;\n",
"print round(I_D*1000.0,3),\"= The diode current (mA)\"\n",
"V_D=V_2+0.1*log10(I_D/I_2)\n",
"print round(V_D,3),\"= The diode volage (V)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"4.237 = The diode current (mA)\n",
"0.763 = The diode volage (V)\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.5:pg-157"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 3.5 : Repeating example 3.4 using piecewise linear model\n",
"V_D0=0.65; # (V)\n",
"r_D=20; # (ohm)\n",
"R=1000; # (ohm)\n",
"V_DD=5; # (V)\n",
"I_D=(V_DD-V_D0)/(R+r_D);\n",
"print round(I_D*1000,2),\"= I_D (mA)\"\n",
"V_D=V_D0+I_D*r_D;\n",
"print round(V_D,3),\"= The diod voltage (V)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"4.26 = I_D (mA)\n",
"0.735 = The diod voltage (V)\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.6:pg-162"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 3.6 : Power supply ripple\n",
"V_S=10; # V_S=V_+\n",
"V_D=0.7; # (V)\n",
"R=10*10**3; # (ohm)\n",
"n=2;\n",
"V_T=25*10**-3; # (V)\n",
"V_s=1; # (V)\n",
"I_D=(V_S - V_D)/R;\n",
"r_D=n*V_T/I_D;\n",
"v_d=V_s*r_D/(R+r_D);\n",
"print round(v_d*1000,2),\"= v_d(peak (mV))\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"5.35 = v_d(peak (mV))\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.7:pg-163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 3.7 : Percentage change in regulated voltage\n",
"V_DD=10; # (V)\n",
"V_D=0.7*3; # string of 3 diodes provide this constant voltage\n",
"R=1*10**3;\n",
"I_D=(V_DD-V_D)/R;\n",
"n=2;\n",
"V_T= 25*10**-3; # (V)\n",
"r_d=n*V_T/I_D; # incremental resistance \n",
"r=3*r_d; # total incremental resistance\n",
"deltav_O=2*r/(r+R); # deltav is peak to peak change in output voltage\n",
"print round(deltav_O*1000,1),\"is Percentage change (mV) in regulated voltage caused by 10% change in power supply\"\n",
"I=2.1*10**-3; # The current drawn from the diode string\n",
"deltav_O=-I*r; # Decrease in voltage across diode string\n",
"print round(deltav_O*1000,1),\"is Decrease in voltage across diode string (mV)\"\n",
"# The answer in the textbook is slightly different due to approximation"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"37.3 is Percentage change (mV) in regulated voltage caused by 10% change in power supply\n",
"-39.9 is Decrease in voltage across diode string (mV)\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.8:pg-169"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 3.8 : line regulation load regulation\n",
"\n",
"V_Z=6.8; # (V)\n",
"I_Z=0.005; # (A)\n",
"r_Z=20; # (ohm)\n",
"V=10; # V=V_+\n",
"R=0.5*10**3; # (ohm)\n",
"\n",
"# 3.8a\n",
"V_ZO=V_Z-r_Z*I_Z;\n",
"I_Z=(V-V_ZO)/(R+r_Z)\n",
"V_O=V_ZO+I_Z*r_Z;\n",
"print round(V_O,2),\"= V_O (V)\"\n",
"\n",
"# 3.8b\n",
"deltaV=1; # change in V is +1 and -1\n",
"deltaV_O=deltaV*r_Z/(R+r_Z); # corresponding change in output voltage\n",
"print round(deltaV_O*1000,1),\"= Line regulation (mV/V)\"\n",
"\n",
"# 3.8c\n",
"I_L=1*10**-3; # load current\n",
"deltaI_L=1*10**-3;\n",
"deltaI_Z=-1*10**-3; # change in zener current\n",
"deltaV_O=r_Z*deltaI_Z;\n",
"print round(deltaV_O*1000),\"= Load regulation (mV/mA)\"\n",
"\n",
"# 3.8d\n",
"I_L=6.8/2000; # load current with load resistance of 2000\n",
"deltaI_Z=-I_L;\n",
"deltaV_O=r_Z*deltaI_Z;\n",
"print round(deltaV_O*1000,2),\"= Corresponding change in zener voltage (mV) for zener current change of -3.4mA\"\n",
"\n",
"# 3.8e\n",
"R_L=0.5*10**3; # (ohm)\n",
"V_O=V*R_L/(R+R_L);\n",
"print V_O,\"V_O (V) for R_L=0.5K ohm\"\n",
"\n",
"# 3.8f\n",
"I_Z=0.2*10**-3; # Zener t be at the edge of breakdown I_Z=I_ZK\n",
"V_Z=6.7; # V_Z=V_ZK\n",
"I_Lmin=(9-6.7)/0.5; # Lowest current supplied to R\n",
"I_L=I_Lmin-I_Z; # load current\n",
"R_L=V_Z/I_L;\n",
"print round(R_L,1),\"= R_L (Kohm)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"6.83 = V_O (V)\n",
"38.5 = Line regulation (mV/V)\n",
"-20.0 = Load regulation (mV/mA)\n",
"-68.0 = Corresponding change in zener voltage (V) for zener current change of -3.4mA\n",
"5.0 V_O (V) for R_L=0.5K ohm\n",
"1.5 = R_L (Kohm)\n"
]
}
],
"prompt_number": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3.9:pg-181"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 3.9 : Value of capacitance C that will result in peak-peak ripple of 2V\n",
"V_P=100.0; # (V)\n",
"V_r=2.0; # (V)\n",
"f=60.0; # (Hz)\n",
"R=10.0*10**3; # (ohm)\n",
"I_L=V_P/R;\n",
"C=V_P/(V_r*f*R);\n",
"print round(C*1000000,1),\"= C (microF)\"\n",
"wdeltat=math.sqrt(2*V_r/V_P);\n",
"print wdeltat,\"= Conduction angle (rad)\"\n",
"i_Dav=I_L*(1+math.pi*math.sqrt(2*V_P/V_r));\n",
"print round(i_Dav*1000),\"= i_Dav (A)\"\n",
"i_Dmax=I_L*(1+2*math.pi*math.sqrt(2*V_P/V_r));\n",
"print round(i_Dmax*1000),\"= i_Dmax (mA)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"83.3 = C (microF)\n",
"0.2 = Conduction angle (rad)\n",
"324.0 = i_Dav (A)\n",
"638.0 = i_Dmax (mA)\n"
]
}
],
"prompt_number": 35
}
],
"metadata": {}
}
]
}