Transistor circuit

Audioguru again

Joined Oct 21, 2019
867
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.
 

Jony130

Joined Feb 17, 2009
5,145
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:
BJT2a.PNG

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.
 

Thread Starter

Zeeus

Joined Apr 17, 2019
508
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
 

Jony130

Joined Feb 17, 2009
5,145
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
 

Thread Starter

Zeeus

Joined Apr 17, 2019
508
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...
 
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