Answer ASAP, thank you. I'll upvote if correct. A common emitter amplifier, with the circuit shown in the image, is designed to have a Bode magnitude and phase frequency response as shown by the...


Answer ASAP, thank you. I'll upvote if correct. A common emitter amplifier, with the circuit shown in the image, is designed to have a Bode magnitude and phase frequency response as shown by the graphs. The BJT has the following parameters: β = 100, VA
→ ∞, Cπ
= 500fF, and Cμ
= 10fF. Use VT
= 26mV, RC
= 10kΩ and note that the capacitor CX
is an external capacitor used to achieve the desired frequency response. For the circuit shown, use β = 200, RB
= 1kΩ, RC
= 2kΩ, RE
= 4kΩ, VBE,on
= 0.7V, VCE,sat
= 0.2V, VCC
= 10V, VEE
= -10V. Write the generalized form of the amplifier’s transfer function, A(ω) = (Vout/Vin)(ω).


Vcc<br>J<br>VIN O<br>VOUT<br>Cx<br>Rc<br>

Extracted text: Vcc J VIN O VOUT Cx Rc
70<br>195<br>180<br>60<br>165<br>150<br>50<br>135<br>{ 120<br>105<br>90<br>30<br>75<br>60<br>20<br>45<br>10<br>30<br>15<br>1.E+05<br>1.E+06<br>1.E+07<br>1.E+08<br>1.E+09<br>1.E+10<br>1.E+11<br>1.E+12<br>1.E+13<br>1.E+05<br>1.E+06<br>1.E+07<br>1.E+08<br>1.E+09<br>1.E+10<br>1.E+11<br>1.E+12<br>1.E+13<br>Frequency, rad/sec<br>Frequency, rad/sec<br>gp ' (m)v|<br>ZA(@), degrees<br>

Extracted text: 70 195 180 60 165 150 50 135 { 120 105 90 30 75 60 20 45 10 30 15 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09 1.E+10 1.E+11 1.E+12 1.E+13 1.E+05 1.E+06 1.E+07 1.E+08 1.E+09 1.E+10 1.E+11 1.E+12 1.E+13 Frequency, rad/sec Frequency, rad/sec gp ' (m)v| ZA(@), degrees

Jun 11, 2022
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