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In the circuit shown in the figure, the ...

In the circuit shown in the figure, the input voltage `V_i = +5 V, V_(BE) = + 0.8 V` and `V_(CE) = +0.12 V`. Find the values of `I_B I_C` and `beta`.

Text Solution

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Given = `V_i = V_("BB") = + 5V`
`V_("BE") = + 0.08 V , V_("CE") = + 2.12 V, V_("CE") = 12V`
`R_B = 80 k Omega, R_C = 5 k Omega`
`I_B = ? , I_C = ? B =?`
(1) `V_("BE") = I_B R_B + V_("BE")`
`I_B = (V_(BB)- V_(BE))/R_B = (5-0*08)/(80xx10^3) = (4.92)/(80xx10^3) = 0*0615 mA`
(2) `V_("CC") = V_(CE) +I_C + R_C`
`I_C = (V_(C C)-V_(VE))/R_C = (12-0.12)/(5xx10^(3))=(11.88)/(5xx10^3)`
`I_C = 2.376 xx10^(-3) A`
`I_C = 2.376 mA`
`I_C = 2.376 mA`
(3) `B = I_C/I_B = (2.376)/(0.0615) = 39`.
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