Help!! - Circuit Design: Voltage divider amplifier

[simplynew]

i am a first year student in electrical and computer engineering and my project is to design and build a single stage voltage divider amplifier with the following design specs:
1. voltage gain = 50
2. lower cut off frequency must be less than 100Hz
3. it must have maximum symmetrical swing
4. supply voltage = 15V
5. small signal transistor (2N3904/2N2222) is to be used

and i cannot get my values of Icq, etc by using the rule-of-thumbs


i can determine Icq by rule-of-thumbs, but without that i am not sure how to choose my value of Icq


i was thinking to get my Icq value from a datasheet of hfe against Ic and just choose a midway value for hfe for the specified transistor ( i am using the 2N3904).
is this a good way to go about choosing a value for Icq?

 

[simplynew]

Ok so here is one of my attempts:


Icq = 10mA
Vcc = 15V
A (Voltgae Gain) = 50
Vbe = 0.6 V (I’m not sure how to choose my Vbe because it is in a range)
Vc = Vcc/2 = 7.5 V (because then you get maximum symmetrical swing about Vc)

After all this is determined, I proceeded to calculate Rc and Re by Kirchhoff’s Voltage Law:
Vcc = IcRc + Vce + IeRe
Vcc = Vce + Ic(Rc + Re) {Assuming that IC ~ IE }
Now I get
Ic = (Vcc-Vce)/(Rc+RE) (1)
Icq = Vcc/((2*A*Re)) {where A = Volatage gain} (2)

→ From equation (2) :
10mA = 15/(2*50*Re)
Therefore , Re = 15 Ω

Now,
Volatge Gain, A = - Rc/Re (3)
So, 50 = Rc/15
Rc = 750Ω

Ve = Veq = Icq * Re
= 10 mA * 15Ω
= 0.15V

In general Vbe = 0.60V
Therefore, Vbe = 0.60V + 0.15 V
Vbe = 0.75 V

I am still not sure how i am supposed to go about choosing a value for Vbe, do i again look at the data sheet?

Sp from here i have to now calculate R1 and R2 but this is where i am stuck, i am not sure how to do this without using the rule-of-thumb.

 

[Laplace]

My datasheet for the 2N3904 has a graph showing Vbe as a function of collector current. Initially you can assume Vbe=0.6 to calculate the collector current, then read the datasheet graph to get a better value for Vbe. Then do another iteration for accuracy.

Having a lower cutoff frequency implies that you should be using an emitter bypass capacitor, or can you get sufficient mid-range amplification without the capacitor?

 

[simplynew]

Another attempt:
Designing a CE Voltage divider circuit using Vcc = 15V
Choose Icq = 1mA for good current gain and frequency response and without causing undue heating.
Vbe = 0.7Volts.

For good quiescent current stability, Vre = 1/10 Vcc = 1.5 volts.
Hence, Re = 1.5/1mA = 1.5kΩ

For maximum symmetrical swing, Rc = (15-1.5)/(1.5*1mA) = 9kΩ

Now, Vbq = Vre +Vbe = 2.2Volts

Current through R1 and R2, Ir = 1/10 Icq = 0.1 mA.
This gives R2 = 2.2Volts/0.1mA=22kΩ
And R1 = (15-2.2)/0.1mA=128kΩ

I can design my circuit this way, but i am not sure why Vre has to be 1/10 Vcc
And why Ir has to be 1/10 Icq
My teacher said this way is using the rule-of-thumb and I need to calculate my resistor values in a different way.