hobbyist CE capacitor see form his terminal RE resistor and parallel connect resistance seen from emitter looking into the BJT ( R1||R2 / (β+1) + re )
So CE time constant is equal
T = CE * Rt
Rt = RE || ( R1||R2 / (β+1) + re ≈ re

So for 50Hz we get
Ce > 0.16/ ( 50Hz * 26mV/3mA ) = 369uF

OK I have applied the capacitance values which you gave me & simulate the circuit I can see difference but the output signal does not look normal!
Also can you please give the equations you used for calculating C1 & C2. One more thing I forgot is that my instructor gave me this value (Vt=2.6mv) am not sure what is this & did not use it on any of my calculations.

I have attached the ORCAD .dsn file for my circuit so you can have a look at it in a better way but please note that I renamed the file & added extension ".zip" so forum uploaded can accept it so please rename it & delete the (.zip) so u can open with pspice

 

Vt is the thermal voltage. It's value is 0.026V at room temperature (300K)

 

The only given values are:
Voltage Gain= 70
Lower cutoff frequency= 70 Hz
Load Resistance= 7k ohm
Vce= 7v
Ic= 3mA

For this parameters it is better to use this schematic



I assume Ve = 1V and Vce = VRc = 7V, and β = 100.

Vcc = 1V + 2*7V = 15V

Rc = 7V/3mA = 2.2kΩ

Re1 = 1V/3mA = 330Ω

Vb = 3mA * 330Ω + 0.65V = 1.65V

Ib = 3mA/100 = 30μA (base current)

R2 = Vb / ( 10 * Ib) = 5.1KΩ

R1 = ( Vcc - Vb) / ( 11 * Ib) = 39KΩ

So know if we want voltage gain 70V/V

Au = 70 = (Rc|| RL) / ( re + (Re1||Re2) )

( re + (Re1||Re2) ) = (Rc|| RL) / 70 = 1.6KΩ/70 = 22Ω

22Ω = ( 8.7Ω + (330||Re2) )

Re2 = ( 330 * 13.3 ) / (330 - 13.3) = 13.8 =[ B]15Ω[/B]

Lower cutoff frequency= 70 Hz, becaues we have three High Pass Filters connect in series.
We must design Fc = √ ((70^2) / 3)= 40Hz

C1 = 0.16/ ( 40Hz * R1||R2|| [ β+1 * ( re + (Re1||Re2) ) ) = 0.16/ ( 40 * 1.5K) = 2.6uF = 3.3uF

C2 = 0.16/(40Hz * (Rc+RL) ) = 470nF

Ce = 0.16/ ( 40Hz * Re2) = 333uF=470uF

 

this is a very professional way but sadly I got it late & I saw the need of adding one more resistor Re2 which made the circuit give accurate output.. but I have used some different method which gave me different components values but output was 70 as needed.

 

Ok, so without change the schematics

Vce = 7V
Ic = 3mA
RL = 7K
Au = 70 V/V

Au = Rc||RL/re = Rc||7K / ( 26mV/3mA) = Rc||7K/8.7Ω

Rc||RL = 70 * 8.7Ω = 609Ω

Rc = (7K * 609) / (7K - 609) = 667.031763Ω = 680Ω

Re = 1V/3mA = 330Ω

Vcc = 1V + 7V + Ic*Rc = 10V

R2 = Vb / ( 10 * Ib) = 5.1KΩ

R1 = ( Vcc - Vb) / ( 11 * Ib) = 24KΩ

C1 =0.16/(Fc * R1||R2||(β+1)*re≈ 0.16/ ( 40Hz * 101*8.7Ω) = 4.7uF

C2 = 0.16/(Fc*Rc+RL) = 680nF

Ce = 0.16/( Fc * re) = 470uF

 

This one is great but I have some questions:
(((Au = Rc||RL/re = Rc||7K / ( 26mV/3mA) = Rc||7K/8.7Ω)))
where did you got the 26mv from?
(((C1 =0.16/(Fc * R1||R2||(β+1)*re≈ 0.16/ ( 40Hz * 101*8.7Ω) = 4.7uF)))
I believe fc is the cutoff freq. which was given as 70Hz but you have substituted 40Hz!! Am I right?

 

26mV is a thermal voltage
http://en.wikipedia.org/wiki/Boltzmann_constant#Role_in_semiconductor_physics:_the_thermal_voltage

(((C1 =0.16/(Fc * R1||R2||(β+1)*re≈ 0.16/ ( 40Hz * 101*8.7Ω) = 4.7uF)))
I believe fc is the cutoff freq. which was given as 70Hz but you have substituted 40Hz!! Am I right?

Yes I substituted 40Hz
Becaues in you amplifier you have three high pas filters connect in series.
So the net (equivalent) cutoff freq. is equal:
Fc = √ (F1^2 + F2^2 + F3^3)
And you wont Fc= 70Hz, and if we assume F1=F2=F3 then

F1 = √(Fc^2)/3 = √(70^2)/3=√(4900/3) ≈ 40Hz

And we cheek the result

Fc = √ (F1^2 + F2^2 + F3^3) = √(40^2+40^2+40^2) =69Hz

 

It is difficult to calculate the voltage gain from the output to input ratio because the output has 20% to 30% distortion. Its positive-going peaks are compressed.

Here is an example of the distortion:

 

I did the simulation & added Vin/Vout as my Trace on pspice. It gave me almost 70 if we ignore distortions. I think it is close enough.

Finally am done with my project & submitted the final report today. Appreciate your help guys.