4(A) The transistor in the circuit shown in figure 6 is so biased that the DC collector current Ic is 1 mA and supply Vee is 5V. The network components have the following values Rc= 2 KQ, Rs=1.4 KN, RE=100 N, The transistor has specifications, current gain is 100, and assume base spreading resistance is 100 Q and VT = 26 mV. The readings were taken on room temperature. Calculate: (i) Voltage gain, (ii) Evaluate the input resistance R¡ for two cases at a frequency of 10 KHz, a. CE bypass capacitor across RE is 25 µF, b. Bypass capacitor is removed leaving RE unbypassed. Vcc = +5V Rc Rs Vs RE Figure 6
4(A) The transistor in the circuit shown in figure 6 is so biased that the DC collector current Ic is 1 mA and supply Vee is 5V. The network components have the following values Rc= 2 KQ, Rs=1.4 KN, RE=100 N, The transistor has specifications, current gain is 100, and assume base spreading resistance is 100 Q and VT = 26 mV. The readings were taken on room temperature. Calculate: (i) Voltage gain, (ii) Evaluate the input resistance R¡ for two cases at a frequency of 10 KHz, a. CE bypass capacitor across RE is 25 µF, b. Bypass capacitor is removed leaving RE unbypassed. Vcc = +5V Rc Rs Vs RE Figure 6
Chapter55: Ac Adjustable Frequency Drives
Section: Chapter Questions
Problem 4SQ: What is the advantage of an insulated gate bipolar transistor over a common junction transistor?
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