Question

Determine the Q point of the circuit shown in fig. Also, draw the D.C load line. Given β=158
& RC = 1058 Ω.

Answer 1

According to KVL & KCL:

$$I_BR_B+0.7+I_ER_E=10,\\ I_E=I_C+I_B.$$

Since this is an amplifier:

$$V_B>V_E,\\ V_B<V_C.\\ I_C=\beta I_B\rightarrow I_E=(1+\beta)I_B,\\ I_BR_B+(1+\beta)I_BR_E=10-0.7,\\ I_B=\textbf{0.01171}\text{ mA},\\ I_E=(1+\beta)I_B=\textbf{1.862}\text{ mA},\\ I_C=I_E-I_B=\textbf{1.850}\text{ mA}.$$

Voltages:

$$V_{CC}=I_CR_C+V_C,\\ 10=I_CR_C+V_C,\\ V_C=10-I_CR_C=\textbf{8.043}\text{ V}.\\ V_E=I_ER_E-10=\textbf{-1.249}\text{ V}.\\ V_B=V_{BE}+V_E=0.7-1.249=\textbf{-0.549}\text{ V}.$$

Q point, therefore, is:

$$(V_{C},I_C)=(8.043\text{ V},1.850\text{ mA})$$

For DC load line the saturation current:

$$I{_C(sat)}=\frac{V_{CC}-V_E}{R_C}=0.0106\text{ A}.$$

Slope:

$$S=\frac{-0.0106}{10-(-1.249)}=-9.42\cdot10^{-4}\space\Omega^{-1}.$$