Question

Consider a fixed biased circuit with a Q point (9.2 mA,3.4 V). The transistor has a DC common emitter current gain of 115. Given that Voltages,

\(V_{BE}= 0.7~ Volts ~ and ~ V_{CC} = 5\)

Then the value of base resistance \(R_B\) is ______\(k\Omega\)

Answer 1

The correct option is A 53.75

Redrawing the above circuit we get,

$I_B=\frac{I_C}{\beta }=\frac{9.2}{115}=0.08mA$

$\frac{V_{CC}-V_{BE}}{R_B}=I_B$

thus, $R_B=\frac{5-0.7}{0.08\times {10}^{-3}}=53.75k\Omega$