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This is a picture of a Common-Emitter Configuration of an n-p-n BJT, taken from a book :-
For the BJT to be in active mode, the Base-Emitter junction should be forward biased, and the Collector-Base junction should be reverse biased.
Here, I am having difficulty understanding how the Base is at a higher potential than the Emitter, and how the Collector is at a higher potential than the base.
First things first. Voltage is the potential difference between two points. This means it is a relative term. Have clear concept about this first. We can proceed forward if you are clear what voltage means actually. Also your figure has a grounded emitter and a ground is referred as 0V.
In your figure, you have a battery with its -ve end connected to the emitter and +ve end connected to the base. Since you have higher potential at p-type semiconductor, you have a forward biased base-emitter junction.
Now another battery has its +ve end connected to the collector and the -ve end to the emitter. The potential at the collector is determined by the voltage rating of battery. And the Vcc should be greater than Vbb for reverse biasing of the collector-base junction.
Lets take a example: you select the value of Vbb to be 5V and the Vcc to be 12V. When you connect the +ve of 5V to the base and -ve to the emitter, you have base at 5V higher potential than emitter. You can see what this results to!
Now you connect the +ve of 12V battery to the collector and -ve to the emitter. Here, you have the collector at 12V higher potential than the emitter.
Look at the collector base junction now. The base is at 5V higher potential than emitter and the collector is at 12V higher potential than emitter. So which (base or collector) is at higher potential? And does it make the collector-base reversed biased?
Hint: You are 2 feets taller than your son. Your partner is 1.5 feets taller than your son. Who is taller among you and your partner?
I am having difficulty understanding how the Base is at a higher potential than the Emitter, and how the Collector is at a higher potential than the base.
You're using ideal voltage sources to bias your transistor so it's very simple:
The base is at a higher potential than the emitter whenever the voltage source \$V_{BB}\$ has a positive value.
The collector is at a higher potential than the base whenever the voltage source \$V_{CC}\$ has a higher output voltage than \$V_{BB}\$.
Note: In the real world, you'd need \$V_{BB}>0.6\$ or so and \$V_{CC}-V_{BB}>-0.3\$ or so (rather than just have these values be greater than 0) to get into forward active operation.
