Question

3. a. Find the value of β if α=0.98 1 b. A transistor is connected in common emitter(CE) configuration in which collector supply is 8V and the voltage drop across resistance RC connected in the collector circuit is 0.5V.The value of RC=800Ω.If α=0.96 determine(i) collector-emitter voltage (ii) base current. 1+2=3 c. Calculate IE in a transistor for which β=50 and IB=20µA

Answer 1

Answer:

Explanation:

From given conditions IC = [(Voltage drop across 800Ω) / (800)] = [(0.5) / (800)] IC = [1 / (1600)]A given  α = 0.96 hence β = [α / (1 – α)] = [(0.96) / (1 – 0.96)] = 24 Now    IC = βIB hence IB = (IC / β) ∴        IB = [1 / (1600)] × (1 / 24) ∴        IB = 26μARead more on Sarthaks.com - https://www.sarthaks.com/327074/transistor-connected-common-emitter-configuration-collector-supply-voltage-resistor