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Determine Q Point and draw the dc load line for the following circuit.
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Assuming $\beta=100$

$\mathrm{R_B=R_1||R_2}$

$\mathrm{R_B=\dfrac{15\times10^3\times5\times10^3}{15\times10^3+5\times10^3}}$

$\mathrm{R_B=3.75\ K\Omega}$

$\mathrm{V_{TH}=V_B=V_{CC}\times\dfrac{R_2}{R_1+R_2}}$

$\mathrm{V_B=20\times\dfrac{5\times10^3}{15\times10^3+5\times10^3}}$

$\mathrm{V_B=5\ V}$

Applying KVL to the Base - Emitter Loop

$\mathrm{V_B-I_BR_B-V_{BE}-I_ER_E}=0$

But, $\mathrm{I_E=\big(1+\beta\big)I_B}$

$\mathrm{V_B-V_{BE}-I_B\Big[R_B+\big(1+\beta\big)R_E\Big]}=0$

$\mathrm{I_B=\dfrac{V_B-V_{BE}}{R_B+\big(1+\beta\big)R_E}}$

$\mathrm{I_B}=\dfrac{5-0.7}{3750+\big(1+100\big)3000}$

$\mathrm{I_B=14.01\ \mu A}$

$\mathrm{I_C=\beta I_B}$

$\mathrm{I_C=100\times62.91\times10^{-6}}$

$\underline{\underline{\mathrm{I_C=1.4\ mA}}}$

$\mathrm{I_C\approx I_E}$

Applying KVL to Collector - Emitter Loop

$\mathrm{V_{CC}-I_CR_C-V_{CE}-I_CR_E}=0$

$\mathrm{V_{CE}=V_{CC}-I_C\big(R_C+R_E\big)}$

$\mathrm{V_{CE}}=20-1.4\times10^{-3}\big(2\times10^3+3\times10^3\big)$

$\underline{\underline{V_{CE}=13\ V}}$

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