Hi , please can somebody help me to solve my assignment . the question circuit is attached and the question is ;

1) for the curcuit design a two stage common emitter amplifier for for the following specifications ;

RE1=REA1+REA2 , RE2=REB1+REB2 , IE1=1.8mA , IE2=4mA
REA1=0.22RE1 , REB2=0.4RE2
FOR BOTH STAGES ;
β=200 , R2=0.1βRE1 , R4=0.1βRE1 , VA=100V , VE=0.1VCC , VC=0.5Vcc , VBE=0.65V

2) PERFORM AC ANALYSIS ( from 10 Hz to 100 MHz ) and print output for V of the both stages separately .

thanks a lot in advance for your support and so operation .

best regards

 

What kind of help do you ask for?

 

This belongs in the homework section, and keep in mind that normally people wont solve problems for you. They will however gladly help you if you show some initiative and show us what you have done so far and where you are stuck.
So where are you stuck with this assignment ?

 

This belongs in the homework section, and keep in mind that normally people wont solve problems for you. They will however gladly help you if you show some initiative and show us what you have done so far and where you are stuck.
So where are you stuck with this assignment ?

sorry for the mistake. i have solve the circuit and find all the values , bat i dont know where i do mistake . the values taht i find is ;

Re1= 1000ohm , Re2=450ohm , Rea1= 220ohm , Rea2=780ohm , Reb1=270ohm , Reb2=180ohm , R4=10kΩ, R2=10kΩ , Rc1=5kΩ , R1=10kΩ , Ic1=1.791mA , Vb1=1.15V , Rc2=2.2kΩ , Bb2=1.15V , R3=10kΩ

bat my instructor is said this values incorrect . so kindly can you help me and inform me the right values .

thanks

 

Hmm, for example
VE=0.1VCC and IE1=1.8mA , IE2=4mA
So
Re1=1.6V/1.8mA=888=820

Re2=1.6V/4mA=400=390

And Rc1=6.4V/1.79mA=3.5733=3.6K and soon

 

Hmm, for example
VE=0.1VCC and IE1=1.8mA , IE2=4mA
So
Re1=1.6V/1.8mA=888=820

Re2=1.6V/4mA=400=390

And Rc1=6.4V/1.79mA=3.5733=3.6K and soon

VE=0.1VCC so Ve= 0.1 X 18 = 1.8 V am i right ? why 1.6 V

 

If so I'm sorry because I assume 16V.
So for 18V
Re1=1.8V/1.8mA=1K; Rc1=9V/1.79mA=5K=5.1K;

Re2=1.8V/4mA=450=470; Rc2=9V/3.98mA=2.2K.

R2=0.1*200*1K=20K -->
R1=( Vcc - (Vbe+Ve) )/(IR2+Ib)

IR2=(Vbe+Ve)/R2=2.45V/20K=122.5uA

Ib≈1.8mA/β=9uA

R1=( 18V - (1.8V+0.65V) )/(122.5uA+9uA)=15.55V/131.5uA=118K=120K