Question

l and....Debate Club | Offic A) Theoretical Design Design a common emitter BJT amplifier with the following requirements: -Rin-10 K2, and Ro-45 ㏀ (Neglect the Early voltage Effect) Vo/Vsig- Gv-40 VIV or 32 dB " VCC-9 V V, IC-1mA, VCE-3.25V and β-100 RL-40 kQ, Rsige I ka, R 1-3R2, and C1-C2-1 μF Voc RC C2 R1 Rsig C1 RL R2 RE B) Verify your design using Orcad Capture Pspice by doing 1) AC sweep (frequency response): use a source VAC-1 ㎷, sweep the frequency from 1 Hz to 10 MHz. a) Obtain two magnitude plots for (Vo / Vsig) vs. frequency (one in dB and the other in VV). b) On the dB plot mark the values of the midband gain, the lower 3dB frequency fi and the upper 3 dB frequency f. Vin /i.-VB(QM(Rsig) c) Obtain a plot for the input resistance Rin 2) Transient Analysis (time domain): use a source voltage (Vsin) with, Vo 0, F I kHz. Show about 3 cycles on the plot (time 0 to 3 ms). Obtain a plot of (Vo vs. Time) for two cases: i) VAMP- 100 mV, ii) VAMP- 400 mV 222 AM 5/2/2019

please I need details

Answer 1

1. Choose transistor

                β =100

2. Calculate collector resistor

Here supply voltage used is 9V. The voltage need to be half the supply voltage.

So Vc= 4.5V, Ic= 1mA

Vc= IcRc

Rc= 4.5/1 =4.5kΩ.

3. Calculate the emitter resistor

The emitter voltage should be 10% of supply voltage.The emitter current should be same as collector current.

0.9= 1x Re

R_E= 0.9/1mA = 0.9kΩ.

4. Determine base current:

I_B= Ic/β

1mA/100 = 10 μA

5. Determine the base voltage:

The emitter voltage is 0.9V.The transistor taken is silicon transistor that has 0.6V base emitter junction voltage.

Base voltage= emitter voltage + 0.6

= 0.9+0.6 =1.5 V

6. Calculate R1 and R2

R1=3R2

choose R2= 1kΩ

R1= 3kΩ

7. Emitter bypass capacitor:

The gain of the circuit without a capacitor across the emitter resistor is approximately R3/R4. To increase the gain of AC signals,the emitter resistor bypass capacitor C3 is added.

C=1/(2πf)Xc

I am taking the frequency value as 95Mhz

c= 1/(2*3.14*95*10⁶ * 0.9*10³ ) =1.8*10^-12 = 1.8PF