Electronic Devices - BJT Amplifiers - Discussion

Based on the figure it is a Common Collector Bias since emitter (output) and base (input). With that info R_in(base) = β_ac*(r'e + Re).

At r'e = 25mv / Ie where Ie can be solved that will be equal to 6mA. Therefore r'e = 4.16
Rin(base) = 100(4.16+500) = 50K ohms

@Anele.

50k is correct if youre comparing it with the prev. question where Rin(base) is asked, it was state there that a bypass capacitor was installed @Re thus cancelling it from the formula

Rin(base) = (BETA)re' + (1+BETA)Re.

Thus the answer was 400+ohms.

But in this case, no bypass capacitor was added therefore the Re was not cancelled.

Please correct me if I am wrong.

What is the right answer?

50k is wrong because re=26mv/6mA=4.33=B*re=4.33*100=433.33.

@Junayd.

Your approach is correct, since he asked to find out the resistance seen back into the base of the transsistor.

If I am not taking it wrong then it is asked in question the resistance at base probably seeing resistance after the voltage divider circuit then it is almost 50k. Here it is how:

Rb = (beta)re x (beta+1)Re.

While re = 25mV/Ie and Ie after calculation is 6mA which will give re = 4.16 and putting all required values in above equation:

Rb = (100*4.16)*(101*0.5) = 50.916K.

Don't know if this is the thing which has been asked.

Yes its 4.07 k not 50 k.

How come 50k?

22K||5K = 4.07K?

Is there any other calculation?