Electronic Devices - DC Biasing-BJTs - Discussion

Using nodal Analysis:

Let Eth = B (for simplicity)
[(B-20)/8200 ]+ [(B +10)/ 2200) = 0.
B = -3.65 V * answer.

Since βRE>>>10R2 ; 216k ohms >>> 22k ohms , approximate analysis can be used. Where Ib = 0, Ie=Ic & I1=I2 (R1&R2 in series). By KVL on left side, Vcc-I(R1+R2) +Vee = 0; I = (Vcc+Vee)/ (R1+R2) = 2.88 mA.


To get Eth it is usually the voltage drop in R2, but here the case is different since R2 is not connected to ground but is connected to a negative source, therefore, Eth = IR2+Vee is the total voltage drop on the R2 side.

Eth = (2.88mA)(2.2kohms)+ -10 =-3.65 V.

Vb = 20 - Ib(8200).

Also, Vb - Ib(2200) = -10 ------> Vb = -10 + Ib(2200).
20 - Ib(8200) = -10 + Ib(2200),
Ib = 2.8846mA,
Eth = Ib(R2) - VEE,
Eth = (2.88mA)(2200) - 10 = -3.6538 Volts.

Here, we can use Vcc (R2/(R1+R2)) - 0.7.

Current I = (Vcc + Vee)/(R1 + R2),
Then Eth = IR2 - Vee.

The formula isn't R2 * Vcc/R1 + R2 = Eth?

If it is, how that the answer is C?

Equivalent thevenin votage. But reference pts are not specified.

What is E(Th)? anyone please?