Determine V_{CE} in the following silicon-bas | vedclass

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(D) The circuit is a voltage divider bias configuration. First, we find the Thevenin equivalent voltage at the base $(V_B)$:$V_B = V_{CC} 	imes \frac

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$2.4$

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(D) The circuit is a voltage divider bias configuration. First, we find the Thevenin equivalent voltage at the base $(V_B)$:$V_B = V_{CC} 	imes \frac{R_2}{R_1 + R_2} = 10 \ V 	imes \frac{5 \ k\Omega}{10 \ k\Omega + 5 \ k\Omega} = 10 	imes \frac{5}{15} = 3.33 \ V$.Applying Kirchhoff's Voltage Law $(KVL)$ to the base-emitter loop:$V_B = V_{BE} + I_E R_E$.For a silicon transistor, $V_{BE} = 0.7 \ V$.$3.33 \ V = 0.7 \ V + I_E (526 \ \Omega)$.$I_E = \frac{3.33 - 0.7}{526} \approx \frac{2.63}{526} \approx 0.005 \ A = 5 \ mA$.Assuming $I_C \approx I_E = 5 \ mA$, we apply $KVL$ to the collector-emitter loop:$V_{CC} = I_C R_C + V_{CE} + I_E R_E$.$10 \ V = (5 \ mA 	imes 1 \ k\Omega) + V_{CE} + (5 \ mA 	imes 526 \ \Omega)$.$10 \ V = 5 \ V + V_{CE} + 2.63 \ V$.$V_{CE} = 10 - 7.63 = 2.37 \ V \approx 2.4 \ V$.

Determine $V_{CE}$ in the following silicon-based transistor circuit. (in $V$)

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