Transistor Problem

I've a circuit like that



I'm trying to determine R1,R2,Re and Rc
When Ic= 10mA and, Vce=8v and Bdc=75

I've done my best and I got 3 equations

Here is them

14= 10Rc+10Re
22-Vb= 10 Rc + 7.3
Vb= 10 Re + 0.7

and i cannot go on

Can anybody help,please?

 

Don't forget your units in the equations. Otherwise you'll get wrong answers. If you just insert a lone “10”, it is not the same as “10 mA”.

 

I see but I'm just wrinting my equations . I've 3 equations with 4 unknws so I'm tring to get the fourth one

 

What about Vcc ??

 

Try this this is the rule of thumb.

R1 = (Vcc - Vb) / (11 * Ib)

R2 = Vb / (10 * Ib)

For good thermal stability we choose Ve to be grater then Vbe.
So for good thermal stability we usually choose Re=(0.1 ... 0.4)*Vcc/Ic or Ve large then 1V (Ve>>Vbe)

Rc = (Vcc - Vce - Ve)/Ic

 

Vcc = 22v volt

 

One equation ( at least ) should include some parameter of the transistor. Think about it.

 

I already did

 

So for Vcc = 22V and beta = 72 ; Ic= 10mA; Vce=8V

Rc + Re = ( 22V - 8V) / 10mA = 1.4KΩ

I choose Rc = 1.2 KΩ and Re = 200Ω

Ve = Re * Ie = 200Ω * 10.138mA ≈ 200Ω * 10mA = 2V

Vb = Ve + Vbe = 2.65V

Ib = Ic/beta = 10mA/72 ≈ 139μA

R1 = (22V - 2.65V) / (11 * 139μA) ≈ 13KΩ

R2 = 2.65V / ( 10 * 139μA) = 2K

 

How come Rc=1.2 KΩ and Re= 200 KΩ!! isn't anyway but this i.e away from guessing ?

 

200Ω*** Sorry

 

From this

Vcc = Ic * Rc + Vce + Ie*Re

And if we assume Ic = Ie

Rc+Re = ( Vcc - Vce)/ Ic = ( 22V - 8V)/10mA = 1.4KΩ

So I choose Rc = 1.2K and Re = 1.4K - 1.2K = 200Ω

 

I know I'm asking about the latest stop why choosing Rc=1.2 based on what! and should RE=>10R2 in order for this Voltage divider to be stiff

 

Usually we choose Rc > Re for for several reasons.
The first one and perhaps most important reason is that Re reduces the amplifier output voltage swing. And the voltage gain is equal to Rc/Re.

I use some other condition to achieve the desired voltage divider stiffness.
I choose voltage divider current much larger than the base current (10 * Ib typical). This is almost the same condition as yours.

 

oh ok thanks