An n-p-n transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The terminal of a 8 V battery is connected to the collector through a load resistance `R_L` and to the base through a resistance `R_B` The collector-emitter voltage `V_(CE) = 4 V`, base-emitter voltage `V_(BE) = 0.6 V `and base current amplification factor `B_(d.c)` = 100. Calculate the values of `R_L` and `R_B`

Explanation

Potential difference across ` R_L`
` (##AKS_NEO_CAO_PHY_XII_V02_C_C06_SLV_039_S01.png" width="80%">
` = 8 V V_(CE)`
`= 8 V - 4V `
`= 4 V `
Now` I_C R_L = 4V `
`R_L = (4)/( 4 x 10^(-3)) = 10^3 Omega = 11 k Omega `
Further for base emitter equation,
` V_(C C ) = I_B R_B +V_(BE)`
or ` I_B R_B ` = potential difference across `R_B `
` V_(C C ) = - V_(BE) = 8-06 = 7.4 V `
Again ` ,I_B = (I_C )/(beta) = ( 4 xx 10 ^(-3))/( 100 ) = 4 xx 10^(-5) A`
` therefore R_B = ( 7.4 )/( 4 xx 10^(-5)) = 1.85 xx 10^5 Omega = 185k Omega `