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# Answer to Question #136544 in Electrical Engineering for Rakibul

Question #136544
Determine the Q point of the circuit shown in fig. Also, draw the D.C load line. Given β=158
& RC = 1058 Ω.
1
2020-10-05T09:50:40-0400

According to KVL & KCL:

"I_BR_B+0.7+I_ER_E=10,\\\\\nI_E=I_C+I_B."

Since this is an amplifier:

"V_B>V_E,\\\\\nV_B<V_C.\\\\\nI_C=\\beta I_B\\rightarrow I_E=(1+\\beta)I_B,\\\\\nI_BR_B+(1+\\beta)I_BR_E=10-0.7,\\\\\nI_B=\\textbf{0.01171}\\text{ mA},\\\\\nI_E=(1+\\beta)I_B=\\textbf{1.862}\\text{ mA},\\\\\nI_C=I_E-I_B=\\textbf{1.850}\\text{ mA}."

Voltages:

"V_{CC}=I_CR_C+V_C,\\\\\n10=I_CR_C+V_C,\\\\\nV_C=10-I_CR_C=\\textbf{8.043}\\text{ V}.\\\\\nV_E=I_ER_E-10=\\textbf{-1.249}\\text{ V}.\\\\\nV_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}."

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