Look at this image.

 

Look at this image.
View attachment 194144

I did this oo..even used the spice file you provided with the My_3904 but yet can't see

But i give up..so many frustration for one night : go to bed

 

OK, get some sleep, tomorrow is a new day, we can then continue.
E

 

BF >< BetaDC

 

They cannot make a 2N3904 or any other transistor with a certain beta. Some that are made on a Friday have a gain of only 99 and some that are made on a Tuesday have a gain of 300 or more.
But a simulator program don't no dat.

 

If you want to find the DC operating point.
We start by assuming β = ∞ and apply a Thevenin theorem to RP and RB.

Vth = 5V*1kΩ/(6.2kΩ + 1kΩ) = 0.694V and Rth = RP||RB = 0.861kΩ

And the situation will look like this:


Therefore from KVL we can write:

Vcc - Ic1*Rc - Vbe2 - Ix*RF - Vbe1- Ic1*RE = 0 (1)
Vcc - Ic1*Rc - Vbe2 - Ix*(RF + Rth) - Vth = 0 (2)

And we can solve this for Ic1 and Ix

\[ I_C = \frac{R_F (V_{TH} - V_{BE}) + R_{TH}*(V_{CC} - 2 V_{BE})}{R_C R_{TH} + R_E (R_F + R_{TH})} \approx 187\mu A \]

And if we know Ic1 we can find all other voltages and current in the circuit.

Ie2 ≈ ((Vcc - Ic1*Rc - Vbe) - Vee)/(Re1 + Re2) ≈ (0.286V - ( -5V))/4.9kΩ ≈ 1mA

Also, this circuit is very Vbe sensitive, changing the Vbe to 0.7V changes the current very much Ic1 ≈ 135μA and Ie2 ≈ 1.3mA.
Therefore exact calculations are redundant due to the fact that we do not know the exact value of a Vbe and Vbe mismatch, and the same we can say about beta value.

 

If you want to find the DC operating point.
We start by assuming β = ∞ and apply a Thevenin theorem to RP and RB.

Vth = 5V*1kΩ/(6.2kΩ + 1kΩ) = 0.694V and Rth = RP||RB = 0.861kΩ

And the situation will look like this:
View attachment 194161

Therefore from KVL we can write:

Vcc - Ic1*Rc - Vbe2 - Ix*RF - Vbe1- Ic1*RE = 0 (1)
Vcc - Ic1*Rc - Vbe2 - Ix*(RF + Rth) - Vth = 0 (2)

And we can solve this for Ic1 and Ix

\[ I_C = \frac{R_F (V_{TH} - V_{BE}) + R_{TH}*(V_{CC} - 2 V_{BE})}{R_C R_{TH} + R_E (R_F + R_{TH})} \approx 187\mu A \]

And if we know Ic1 we can find all other voltages and current in the circuit.

Ie2 ≈ ((Vcc - Ic1*Rc - Vbe) - Vee)/(Re1 + Re2) ≈ (0.286V - ( -5V))/4.9kΩ ≈ 1mA

Also, this circuit is very Vbe sensitive, changing the Vbe to 0.7V changes the current very much Ic1 ≈ 135μA and Ie2 ≈ 1.3mA.
Therefore exact calculations are redundant due to the fact that we do not know the exact value of a Vbe and Vbe mismatch, and the same we can say about beta value.

Also did thevenin like you did but notice Vth = 0.694 but the VB1 is greater than this.. it's 754mv

 

So what? Vth is not the voltage at the T1 base.

 

For Vbe = 0.6V we have Ic1 ≈ 187μA and VB1 ≈ 187μA*330Ω + 0.6V ≈ 0.662V

And for Vbe = 0.7V ---> Ic1 ≈ 135μA ---> VB1 ≈ 135μA*330Ω + 0.7V ≈ 0.745V

All this assuming Ib1 = Ib2 = 0A

 

So what? Vth is not the voltage at the T1 base.

Thought the voltage will be less than Vth but see my mistake now

 

If you want to find the DC operating point.
We start by assuming β = ∞ and apply a Thevenin theorem to RP and RB.

Vth = 5V*1kΩ/(6.2kΩ + 1kΩ) = 0.694V and Rth = RP||RB = 0.861kΩ

And the situation will look like this:
View attachment 194161

Therefore from KVL we can write:

Vcc - Ic1*Rc - Vbe2 - Ix*RF - Vbe1- Ic1*RE = 0 (1)
Vcc - Ic1*Rc - Vbe2 - Ix*(RF + Rth) - Vth = 0 (2)

And we can solve this for Ic1 and Ix

\[ I_C = \frac{R_F (V_{TH} - V_{BE}) + R_{TH}*(V_{CC} - 2 V_{BE})}{R_C R_{TH} + R_E (R_F + R_{TH})} \approx 187\mu A \]

And if we know Ic1 we can find all other voltages and current in the circuit.

Ie2 ≈ ((Vcc - Ic1*Rc - Vbe) - Vee)/(Re1 + Re2) ≈ (0.286V - ( -5V))/4.9kΩ ≈ 1mA

Also, this circuit is very Vbe sensitive, changing the Vbe to 0.7V changes the current very much Ic1 ≈ 135μA and Ie2 ≈ 1.3mA.
Therefore exact calculations are redundant due to the fact that we do not know the exact value of a Vbe and Vbe mismatch, and the same we can say about beta value.

Thanks so much for helping with insight but solution is 6/10 but thanks. Ib = 0...