Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_125

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Step 1 of 3 8.093P Refer to Figure P8.93 from the text book. The values of the currents and are equal. Thus, the voltages and are equal since the transistors and are matched. Also, values of the voltages at the sources of and are equal. That is, Hence, the base to emitter voltage of the transistor is equal to the voltage drop across the resistor That is, = Substitute for IR = In (1) Therefore, the values of the voltage drop across the resistor R is Step 2 of 3 Consider the expression for the emitter current Substitute 0.7 V for , 25 mV for and 1 mA for Therefore, the value of the current Iₛ is 6.91x10⁻¹⁶ Step 3 of 3 Recall equation (1), to determine the value of resistance. IR = Substitute 6.91x10⁻¹⁶ A for 200 for I and 25 mV for Therefore, the value of the resistance R is