1 (818)

Prévia do material em texto

Step of 7 7.035P Base emitter voltage is 0.700 V Collector current is 0.5 mA Collector resistance is 5 Power supply voltage is +5 V Determine the collector voltage Vc = = =5-2.5 = 2.5 V Thus, the collector voltage Vc is 2.5 V Step of 7 The signal current applied to base is raised to 0.705 V The total instantaneous base emitter voltage is 0.705 V 0.700 V Thus, the value of the signal is 5 mV Step of 7 Refer to the equation (7.56) in the textbook. Write the expression for the total instantaneous collector current Substitute 0.5 mA for Ic 25 mV for and 5 mV for (0.5 = 0.61 mA Thus, the total instantaneous collector current ic is 0.61 - mA Step of 7 Determine the total Collector voltage after base emitter voltage increased. icRc =1.95 V Thus, the collector voltage after base emitter voltage increased to 0.705 V is 1.95 V Step of 7 The increased voltage Vc across the collector resistance is Thus, the voltage is -0.55 V Determine the voltage gain -0.55 V = mV =-110 Thus, the voltage gain using exponential characteristic is -110 Step of 7 The dc collector current is 0.5 mA Refer to the equation (7.62) in the textbook. Write the expression for the transconductance 8m for Substitute 25 mV for and 0.5 mA for 0.5 mA 8m 25 mV = 20 mA/V V Thus, the transconductance for BJT amplifier is 20 mA/V Step of 7 Write the expression for the voltage gain using small signal approximation. Substitute - -20 mA/V for and for =(-20 = kΩ) = 100 V V Thus, the voltage gain is 100 V/V 100 V Hence, the voltage gain obtain is equal to the voltage gain in small signal approximation.