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.

i uploaded my design layout.
View attachment single stage voltage divider amplifier project.doc

 

[FvM]

There are some points missing in your calculation:

- The voltage gain would be reduced by the finite transistor transcondutance, you have to reduce Re to about 12 ohm to compensate for it.
- You can try to get an exact value for Vbe at the respective operation point from the transistor datasheet, but it's not much help, because the actual value will be affected by exemplar variations and temperature changes. The circuit can't work realiable without an additional DC feedback, but the solution can be accepted as an exercise. Vbe = 0.6 should be O.K. for the time being.
- For dimensioning of the base voltage divider, the transistor current gain respectively Ib should be known. R2 current has to be fixed in addition. Usually it's choosen e.g. with 10*Ib.

After fixing these parameters, you should be able to calculate all resistor values.

 

[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.