C2 R3 01 μF Rp R1 100 k2 4.3 k2 1 kQ 0.1 μF RG RS Сз 243 k2 10 μF 1.3 k2 Figure 17.11 ac Model for common-source amplifier.

A common emitter amplifier has a 15V power supply.  If we design for a 6.5V swing (peak voltage) at the output voltage what will the DC voltage drop be across RE? Group of answer choices 2V 10V 6.5 5V

Figure 2 shows a typical BJT amplifier, with its parasitic capacitances displayed. The current gain of the transistor is B=150 and the voltage gain of the amplifier is Am =-125. The small signal resistances of the transistor are re =16 ohm and r0 = infinity, respectively. The values of resistors and capacitors in the figure are: R1 =80 kohm, R2 =20 Kohm, Rc = 2 Kohm, RE = 2 Kohm, Rs =50 W, RI=5 kohm, Cs=2 uF, Cc=2 uF, Ce =10 uF, Cbc =4 pF, Cbe = 10 pF, Cce =1 pF, Cwi = 4 pF, CWO = 9 pF, and Vcc = 20 V. a) Sketch a simplified circuit diagram of Figure 2 for high frequency analysis. b) Using the concept of "Miller effect capacitance", calculate the input and Output Miller effect capacitances of Figure 2, respectively. C) Determine the upper cut-off frequency of Figure 2 that is imposed by its input network only. d) Explain briefly the possible ways to increase the upper cut-off frequencv of this amplifier.

A base bias method is used in the following circuit. For Bpc - 300, the value of emitter current is: -Vc. +15 V Rc 1.8 k2 RB 560 k2 7.82 mA O 6.84 mA O 7.27 mA