Question

3. Practical Amplifier Design using Multisim [30 Marks] (10mVpk) (2kHz) (Av=95)/ In this section you need to build the circuits that you have designed in the theoretical part, you should have at least the following circuits: Circuit of the CE amplifier only with 47K Ohm Load - Circuit of the CE amplifier with 50 Ohms Load, Circuit of the cascaded Amplifier: Both CE & CC are connected with 50 Ohms Load. You should show the input voltage diagram, and the output voltage diagram for each circuit and the gain of each circuit.

Answer 1

a. Design with 47 K Ohm Load:

Design Specifications: Let V_CC= 10V , I_C = 2mA, V_BE = 0.7V (Silicon Transistor), $\beta =100$

Calculation RE: take. R VE = VCC for better bias stabili stability: lo IV VE Xlov = 502 VE 8. RE= VE Ic 2x03 IE take RE 4702 Standard value Calculation Re: we have gain for Single Stage amplifier as Av hclRL) ne 132 re= 26mv IF 2omv Ic 26mv 18.2mA Av= 95 given in data. RL=4 Tur Giren 45 = REXRL RCARL 응 13 95x13 - RC X 4780 RC= 6 58.045 XL 45.675 X6² to Rc +47x0² [RC= 1268.322 IRC = 1.5kr take standard vahe Page - 1

Base Circut : VB- Base to ground Voltage 1 9.44 ke from B VB VBE + VE AT + E VBE VB = 0.7+ 1 = 1.70 VB 1 + RE VE take I, 10 JB Rule I2 9 IB thumb. +Vcc 11 Riş 107V R2 Iz 9IB OL VB Where IB Ic 2mA ß 100 IB A R2= 107 zouxa take a Standard vahu 2 R2 = lour Rejsc VB R2 Ws2 {RE ZOMA = 8 Ri= Vec-VB V&B 10 lot 41.5kr 지 10 IB 10 X 20 x zou take Standard value as R = 3942 Capacitor Design: Input capacitor: Xiz Ze. 10 Ze= Rill Rall pre = 10K2 113962 ) vox 13) Ze=117.472 Page 2

1117.47 10 where f = 100 H2 ZTFC as, 14.25uf take C, = 15MF C2: output capacitor: XC Z zo Jo Zo= RCRL 1.5k 11 476 = 1453.62 1 1453.6 ZITAC2 10 Con 10.95 10.95MF take Valne Standend 12= 10MF CE: By pass Capacitor XCES RE 10 470 => CE> 33uF ZIFCE 10 take CE= 33MF Page 3

a. Multi-SIM circuit with 47 K ohm Load

Input waveform:

* CE Amplifier with 47 K Ohm Load I ? Transient Schematic Grapher st Split Item Document Transient 1 Name Transient 1 12.5m Type Transient Graph Q -PR1: V(6) 0 Plots [?] V1 10m Show (129.38 us, 9.9849 mV) 10V PR1: V(6) 7.5m Cursors } R1 47KQ RC 1.5K2 5m Axes C2 3 2 2.5m 10F 5 PR1 C1 æti fen- Ay el À 01 100A/A RL 47KO Voltage (V) 0 15F VIN +20 C3 10mV 2kHz 0° HH -2.5m R2 10KΩ wo R4 4702 33F 0 -5m 1 -7.5m - 10m -12.5m 0 100 200 300 400 600 700 800 900 1m 500 Time (s)

Output waveform:

* CE Amplifier with 47 K Ohm Load [ ? Transient Schematic Grapher sti Split Item Document Transient 1 Name Transient 1 0 1 Type Transient Graph Q — PR1: V(5) V1 10V Plots 1 [?] (363.75 us, 878.32 mV) 750m Show PR1: V(5) R1 500m 47KQ RC 1.5ΚΩ 3 C2 PR1 Cursors 2 Axes 5 10uF 250m C1 ti e- A + y el 6 HH Q1 100A/A RL 47ΚΩ 15 F 0 VIN 2+ 10mV 2kHz 0° Voltage (V) + R2 R4 47092 C3 33F -250m 10KO 0 -500m -750m -1 -1.25 0 100 200 300 400 600 700 800p 900 1m 500 Time (s)

Gain AV = 878.32mV / 9.98 mV = 88 ( against 95 due to the stand value resistances)

b. Circuit with 50 Ohms Load

For RL = 50 ohm, RL = 52.1 Ohm

Take RC = 50 Ohm resistor

Input waveform:

* CE Amplifier with 47 K Ohm Load I ? Transient Schematic Grapher sa Split Item Document Transient 1 Name Transient 1 12.5m Type Transient Graph Q -PR1: V(6) 0 Plots [?] V1 10m Show (129.38 us, 9.9849 mV) 10V PR1: V(6) 7.5m Cursors w } R1 47KQ RC 500 3 5m Axes C2 a ti e B A y © 2 2.5m 10F 5 PR1 C1 01 100A/A RL 509 Voltage (V) 0 15F VIN 10mV 2kHz 0° HEE C3 -2.5m R2 R4 47022 33F wo 10KΩ 0 -5m 1 -7.5m - 10m -12.5m 0 100 200 300 400 600 700 800 900u 1m 500 Time (s)

output waveform:

gain = Vo/Vi = 15.68mV/9.98mv = 1.57

*CE Amplifier with 47 K Ohm Load ? Transient Schematic Grapher st Split Item Document Transient 1 Name Transient 1 Transient Graph 20m Type -PR1: V(5) 0 Plots [?] < V1 10V 15m Show (348.13 us, 15.683 mV) PR1: V(5) 10m Cursors R1 47ΚΩ RC 509 3 Axes C2 PR1 V 5m 2 10uF 5 C1 ste + Q1 100A/A RL 5022 Voltage (V) 0 15 F w VIN 4 + C3 ro 10mV 2kHz 00 + R2 R4 4702 33uF -5m 10KO 1. 0 - 10m - 15m -20m 0 100 2004 300 400 600u 700 800 900 1m 500 Time (s)