Solution
Transform the circuit on the left side of BJT into an equivalent circuit containing a voltage source Vth in series with a resistor Rth.
Cut the circuit on the left side from that on the right at the Base terminal and calculate the Vth at the cutting point.
V15kΩ = [15 V - (-15 V)][15kΩ/(15kΩ+15kΩ)] = (30)(1/2) = 15 V
Vth = -15V + V15kΩ = -15V + 15V = 0 V
To find Rth, make all DC power sources to zero (15 V to 0 V and -15 V to 0 V). At this point, both 15 kΩ resistors are in parallel.
Rth = 15kΩ||15kΩ = (15kΩ)(15kΩ)/(15kΩ+15kΩ) = 7.5 kΩ
Therefore the equivalent circuit for the left side of BJT is simply a ground in series with a 7.5 kΩ resistor.
Now, we assume that the BJT is in active mode (we will check whether this is true or not later). Therefore VBE = 0.7 V. With β = 100, IC = (100)IB and IE = (101)IB, we set up a KVL equation from -5 V to 0 V.
-5 V + (2kΩ)(101)IB + 0.7 V +(7.5kΩ)IB = 0 V
IB = 0.0205 mA, IC = 2.05 mA and IE = 2.07 mA
Finally, use KVL equations to find VB, VE, and VC as following.
VB + (7.5kΩ)(0.0205 mA) = 0 V; VB = - 0.15 V Ans
-5 V + (2kΩ)(2.07 mA) = VE; VE = - 0.86 V Ans
VC + (3kΩ)(2.05 mA) = 10 V; VC = 3.85 V Ans
At saturation,
VCE = 0 V, IC(sat) = [10 V- (- 5 V)] / [2kΩ + 3kΩ] = 3.0 mA
Both IC and IE are less than IC(sat). The BJT is therefore not saturated. Ans
Check the results;
VBE = (- 0.15 V) - (- 0.86 V) = 0.71 V ==> OK
VCE = (3.85 V) - (- 0.86 V) = 4.71 V
0 V < VCE < 15 V ==> BJT is not saturated (in active mode as we assumed at the beginning).
