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{
"metadata": {
"name": "",
"signature": "sha256:b8c93f0ed8e762e74a7e32d1ae1c9fc9e20a3917843706f0d24d06eedbf42b4e"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter01:INTRODUCTION TO ELECTRONICS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.1:pg-17"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example1.1: Amplifier gain, power and eficiency\n",
"# Amplifier operates at +10-V/-10-V power supply.\n",
"A_v=9/1.0; # sinusoidal voltage input of 1V peak and sinusoidal output voltage of 9V peak\n",
"I_o=9/1000.0; # 1 kilo ohms load\n",
"print A_v,\"= Voltage gain (V/V) \" \n",
"print round(20*log10(A_v),1),\"= Voltage gain (dB) \" \n",
"I_i=0.0001 # sinusoidal current input of 0.1mA peak\n",
"A_i=I_o/I_i;\n",
"print A_i,\"= Current gain (A/A) \"\n",
"print round(20*log10(A_i),1),\"= Current gain (dB)\"\n",
"V_orms = 9/math.sqrt(2);\n",
"I_orms = 9/math.sqrt(2);\n",
"P_L=V_orms*I_orms; # output power in mW\n",
"V_irms=1/math.sqrt(2);\n",
"I_irms=0.1/math.sqrt(2);\n",
"P_I=V_irms*I_irms; # input power in mW\n",
"A_p=P_L/P_I; \n",
"print A_p,\"= Power gain (W/W) \"\n",
"print round(10*log10(A_p),1),\"= Power gain (dB) \"\n",
"P_dc=10*9.5+10*9.5; # amplifier draws a current of 9.5mA from each of its two power supplies\n",
"print P_dc,\"= Power drawn from the dc supplies (mW)\"\n",
"P_dissipated=P_dc+P_I-P_L;\n",
"print round(P_dissipated,1),\"= Power dissipated in the amplifier (mW)\"\n",
"n=P_L/P_dc*100;\n",
"print round(n,1),\"= Amplifier efficiency in percentage\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"9.0 = Voltage gain (V/V) \n",
"19.1 = Voltage gain (dB) \n",
"90.0 = Current gain (A/A) \n",
"39.1 = Current gain (dB)\n",
"810.0 = Power gain (W/W) \n",
"29.1 = Power gain (dB) \n",
"190.0 = Power drawn from the dc supplies (mW)\n",
"149.6 = Power dissipated in the amplifier (mW)\n",
"21.3 = Amplifier efficiency in percentage\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.2:pg-21"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 1.2: Gain of transistor amplifier\n",
"# Amplifier has transfer characteristics v_O=10-(10**-11)*(exp**40*v_1) applies for v_1 is greater than or equal 0V and v_o is greater than or equal to 0.3V\n",
"L_l = 0.3; # limit L_-\n",
"print round(L_l,2),\"=The limit L_- (V) \"\n",
"v_I=1/40.0*math.log((10-0.3)/10**-11); # from the transfer characteristics and v_o=0.3V\n",
"print round(v_I,2),\"=v_I in volts \"\n",
"L_u=10-10**-11; # obtained by v_I=0 in transfer characteristics\n",
"print round(L_u,3),\"=the limit L_+ (V) \"\n",
"V_I=1/40.0*math.log((10-5)/10**-11); # V_O=5V\n",
"print round(V_I,3),\"=The value of the dc bias voltage that results in V_O=5V (V)\"\n",
"A_v=-10**-11*exp(40*V_I)*40; # A_v=dv_O/dv_I\n",
"print round(A_v,2),\"=Gain at the operating point (V/V) \"\n",
"print \"NOTE the gain is negative that implies the amplifier is an inverting amplifier\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"0.3 =The limit L_- (V) \n",
"0.69 =v_I in volts \n",
"10.0 =the limit L_+ (V) \n",
"0.673 =The value of the dc bias voltage that results in V_O=5V (V)\n",
"-200.0 =Gain at the operating point (V/V) \n",
"NOTE the gain is negative that implies the amplifier is an inverting amplifier\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.3:pg-25"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 1.3 : Overall voltage gain of cthree-stage amplifier\n",
"gainloss_in=10**6/(1*10**6+100.0*10**3); # fraction of input signal is obtained using voltage divider rule , gainloss_in= v_i1/v_s\n",
"A_v1=10*100000.0/(100000+1000); # A_v1 = v_i2/v_i1 is the voltage gain at first stage\n",
"A_v2=100*10000.0/(10000+1000); # A_v2 = v_i3/v_i2 is the voltage gain at second stage\n",
"A_v3=100/(100+10.0); # A_v3 = v_L/v_i3 is the voltage gain at the output stage\n",
"A_v=A_v1*A_v2*A_v3; # A_v is the total voltage gain \n",
"print round(A_v),\" = The overall voltage gain (V/V) \"\n",
"print round(20.0*log10(A_v),1),\"= The overall voltage gain (dB) \"\n",
"gain_src_ld=A_v*gainloss_in;\n",
"print round(gain_src_ld,2),\"= The voltage gain from source to gain (V/V) \"\n",
"print round(20.0*log10(gain_src_ld),1),\"= The voltage gain from source to load (dB) \"\n",
"A_i=10**4*A_v; # A_i=i_o/i_i=(v_L/100)/(v_i1/10**6)\n",
"print \"{:.2e}\".format(A_i),\" = The current gain (A/A)\"\n",
"print round(20.0*log10(A_i),1),\"= The current gain (dB) \"\n",
"A_p=818*818*10**4; # A_p=P_L/P_I=v_L*i_o/v_i1*i_i\n",
"print \"{:.2e}\".format(A_p),\"= The power gain (W/W) \"\n",
"print round(10*log10(A_p),1),\"= The power gain (dB) \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"818.0 = The overall voltage gain (V/V) \n",
"58.3 = The overall voltage gain (dB) \n",
"743.88 = The voltage gain from source to gain (V/V) \n",
"57.4 = The voltage gain from source to load (dB) \n",
"8.18e+06 = The current gain (A/A)\n",
"138.3 = The current gain (dB) \n",
"6.69e+09 = The power gain (W/W) \n",
"98.3 = The power gain (dB) \n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.4:pg-29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example1.4 : Bipolar junction transistor\n",
"\n",
"# 1,4a\n",
"# using voltage divider rule the fraction of input signal v_be=v_s*r_pi/(r_pi+R_s)\n",
"# output voltage v_o=-g_mv_be(R_L||r_o)\n",
"r_pi=2.5*10**3; # (ohm)\n",
"R_s=5*10**3; # (ohm)\n",
"R_L=5*10**3 # (ohm)\n",
"g_m=40*10**-3; # (mho)\n",
"r_o=100*10**3; # (ohm)\n",
"gain=-(r_pi*g_m*(R_L*r_o/(R_L+r_o)))/(r_pi+R_s); # gain=v_o/v_s\n",
"print round(gain,1),\"= The voltage gain (V/V) \"\n",
"gain_negl_r_o=-r_pi*g_m*R_L/(r_pi+R_s);\n",
"print round(gain_negl_r_o,1),\"= Gain neglecting the effect of r_o (V/V) \"\n",
"\n",
"# 1.4b\n",
"# Bi_b=g_m*v_be\n",
"# B is short circuit gain\n",
"B=g_m*r_pi;\n",
"print B,\"= The short circuit gain (A/A) \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"-63.5 = The voltage gain (V/V) \n",
"-66.7 = Gain neglecting the effect of r_o (V/V) \n",
"100.0 = The short circuit gain (A/A) \n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.5:pg-36"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 1.5 : DC gain, 3dB frequency and frequency at which gain=0 of voltage amplifier\n",
"\n",
"# 1.5b\n",
"R_s =20*10**3; # (ohm)\n",
"R_i =100.0*10**3; # (ohm)\n",
"C_i =60.0*10**-12; # (ohm)\n",
"u = 144.0; # (V/V)\n",
"R_o = 200.0; # (ohm)\n",
"R_L = 1000; # (ohm)\n",
"K=u/((1+R_s/R_i)*(1+R_o/R_L));\n",
"print K,\"= The dc gain (V/V)\"\n",
"print round(20*log10(K),2),\" = The dc gain (dB) \"\n",
"w_o=1/(C_i*R_s*R_i/(R_s+R_i));\n",
"print \"{:.0e}\".format(w_o),\" = The 3-dB frequency (rad/s) \"\n",
"f_o= w_o/2/math.pi;\n",
"print round(f_o/1000,1),\" = Frequency (KHz) \"\n",
"print \"{:.0e}\".format(100*w_o),\" = unity gain frequency (rad/s)\"\n",
"print round(100*f_o/1e6,2),\" = Unity gain frequency (MHz)\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"100.0 = The dc gain (V/V)\n",
"40.0 = The dc gain (dB) \n",
"1e+06 = The 3-dB frequency (rad/s) \n",
"159.2 = Frequency (KHz) \n",
"1e+08 = unity gain frequency (rad/s)\n",
"15.92 = Unity gain frequency (MHz)\n"
]
}
],
"prompt_number": 39
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1.6:pg-46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Example 1.6: Time for the output to reach (V_OH+V_OL)/2\n",
"V_DD=5; # (V)\n",
"R=1000.0; # (ohm)\n",
"R_on=100.0; # (ohm)\n",
"V_offset=0.1; # (V)\n",
"C=10.0*10**-12; # (F)\n",
"V_OH=5; # (V)\n",
"V_OL=V_offset+(V_DD-V_offset)*R_on/(R+R_on);\n",
"T=R*C;\n",
"v_o_t_PLH=(V_OH+V_OL)/2; #to find t_PLH \n",
"t_PLH=0.69*T;# t_PLH is low to high propogtion delay\n",
"print t_PLH/1e-9,\"= time required for the output to reach (V_OH+V_OL)/2 (miliseconds) \""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"6.9 = time required for the output to reach (V_OH+V_OL)/2 (miliseconds) \n"
]
}
],
"prompt_number": 46
}
],
"metadata": {}
}
]
}
|
