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 {
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	   "source": [
       "# Chapter 8: Applications of Operational Amplifier"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.1: Device_Temperature.sci"
		   ]
		  },
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"//Chapter 8_Applications of Operational Amplifier\n",
"//Caption : Device Temperature\n",
"//Example8.1:The Heat generated by a linear IC,uA 741 is 200 mW.If the thermal resistance is 150 degree Celsius/Watt and the ambient temperature is 25 degree celsius.calculate the device temperature.\n",
"//Solution:\n",
"clear;\n",
"clc;\n",
"Pd=200*10^-3;//heat generated\n",
"Rt=150;//thermal resistance\n",
"Ta=25;//ambient temperature in degree celsius\n",
"//assuming thermal equilibrium conditon\n",
"Td=Pd*Rt+Ta;\n",
"disp('degree celsius',Td,'The device temperature is:')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.2: Device_Temperature.sci"
		   ]
		  },
  {
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"//Chapter 8_Applications of Operational Amplifier\n",
"//Caption : Device Temperature \n",
"//Example8.2: For the device in Example8.1, Pdmax=500 mW. Determine the device temperature after equilibrium is attained for an ambient temperature of 75 degree celsius and if the device is subjected to maximum heat generation.Maximum allowable device temperature is 150 degree Celsius.\n",
"//Solution:\n",
"clear;\n",
"clc;\n",
"Pmax=500*10^-3;\n",
"Pd=Pmax;//since device is subjected to maximum heat generation\n",
"Rt=150;//thermal resitance\n",
"Ta=75;//ambient temperature\n",
"Td=Pd*Rt+Ta;\n",
"disp('degree celsius',Td,'device temperature is:')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.3: Device_Temperature.sci"
		   ]
		  },
  {
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"//Chapter 8_Applications of Operational Amplifier\n",
"//Caption : Device Temperature\n",
"//Example8.3:a)The ambient temperature of the device of Example8.2 rises above 90 degree celsius. What is the new value of Td if it still generates 500 mW?\n",
"//a)Solution:\n",
"clear;\n",
"clc;\n",
"Pd=500*10^-3;\n",
"Rt=150;//thermal resistance\n",
"Ta=90;//ambient temperature\n",
"Td=Pd*Rt+Ta;\n",
"disp('degree celsius',Td,'New value of device temperature is:')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.4: Device_Temperature.sci"
		   ]
		  },
  {
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"//Chapter 8_Applications of Operational Amplifier\n",
"//Caption : Device Temperature\n",
"//Example8.4: Forced air cooling provided for the device in Example8.3 lowers the ambient temperature at 60 degree celsius.What is temperature of the device?\n",
"//Solution:\n",
"clear;\n",
"clc;\n",
"Pd=500*10^-3;\n",
"Rt=150;//thermal resistance\n",
"Ta=60;//ambient temperature\n",
"Td=Pd*Rt+Ta;\n",
"disp('degree celsius',Td,'Temperature of the device is:')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.7: Output_Voltage.sci"
		   ]
		  },
  {
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"//Chapter 8_Applications of Operational Amplifier\n",
"//Caption : Output Voltage\n",
"//Example8.7: In the summing amplifier(inverting mode) the signals to be combined are V1=3V, V2=2v, and V3=1V.The input resistor are R1=R2=R3=3 kilo ohm.The feeddback resistor Rf=1 kilo ohm. Consider ideal Op-Amp,determine Vo.\n",
"//Solution:\n",
"clear;\n",
"clc;\n",
"V1=3;//input signal\n",
"V2=2;//input signal\n",
"V3=1;//input signal\n",
"Rf=1*10^3;//feedback resitor\n",
"R1=3*10^3;//input resistor in ohm\n",
"R2=R1;//input resistor in ohm\n",
"R3=R2;//input resistor in ohm\n",
"Vo=-(Rf/R1*V1+Rf/R2*V2+Rf/R3*V3);\n",
"disp('V',Vo,'Output Voltage of summing amplifier is:')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.8: Vp_and_Vo.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
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"//Chapter 8_Applications of Operational Amplifier\n",
"//Caption : Vp and Vo\n",
"//Example8.8: In the circuit of non-inverting summing Op-Amp, V1=+2V, V2=-4V, V3=+5V. input resistors for all the three input signal are same and are equal to 1 kilo Ohm.The feedback resistor Rf is 2 kilo ohm. Determine the voltage Vp at the noninverting pin of the Op-Amp and the output Vo. Assume ideal Op=Amp.\n",
"//Solution:\n",
"clear;\n",
"clc;\n",
"Rf=2*10^3;//feedback resistor\n",
"R1=1*10^3;\n",
"R2=R1;\n",
"R3=R2;\n",
"V1=2;\n",
"V2=-4;\n",
"V3=5;\n",
"n=3;//no of inputs\n",
"Vp=(Rf/R1*V1+Rf/R2*V2+Rf/R3*V3)/n;\n",
"Vo=(1+Rf/R1)*Vp;\n",
"disp('V',Vp,'voltage at noninverting pin is:')\n",
"disp('V',Vo,'output voltage voltage of noninverting summing Op-Amp is:')"
   ]
   }
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