In the circuit shown in Fig. `E_(CE)=5.5V, R_(L)=1kOmega, R_=500kOmega`, the base current, `I_B` is `10muA` and collector current, `I_C=5.2mA`. Can this transistor circuit be used as an amplifier?
What happens if the resistance `R_(L)` is `500Omega` and `I_B, I_C` and `R_` remain the same as above.

Explanation

Here `E_(CE)=5.5V, R_(L)=1kOmega=10^(3)Omega`
`R_B=500kOmega =500xx10^(3)Omega=5xx10^(5)Omega`
`I_B=10muA=10xx10^(-6)A=10^(-5)A`
`I_C=5.2xx10^(-3)A`
In a battery, base-emitter circuit, we have `V_(BE)=E_(CE)-I_B R_B=5.5-10^(-5)xx(5xx10^(5))`
`=0.5V`
In battery, collector-emitter circuit, we have
`V_(CE)=E_(CE)-I_C R_(L)=5.5-(5.2xx10^(-3))xx10^(3)`
`=0.3V`
Hence, `V_(BC)=V_(BE)-V_(CE)=0.5-0.3=0.2V` It shows that base becomes positive by 0.2 V with respect to collector i.e., base collector junction becomes forward biased. As both the junctions become forward biased, the circuit can not work as an amplifier.
Here, `R_(L)=500Omega`
`V_(CE)=5.5-(5.2xx10^(-3))xx500=2.9V`
`V_(BC)=V_(BE)-V_(CE)=0.5-2.9=-2.4V`
Now collector is at 2.9V w.r.t. to emitter and base is at 0.5V w.r.t. emitter. So the base collector junction is reverse biased by -2.4 V. Hence the circuit can work as an amplifier.