How is the Collector-Base region reverse biased in Common Collector configuration of BJT?

Question

This is the circuit diagram of an n-p-n BJT in Common Collector configuration, taken from the book :-

I know that for the transistor to operate in active mode :-

• the Emitter-Base junction should be forward biased.

• the Collector-Base junction should be reverse biased.

By looking at the diagram, I can see that the Emitter-Base junction is forward biased. But I am unable to understand how the Collector-Base junction is reverse biased, since the Base is connected to the +ve terminal of VBB and the collector is at zero potential ( looks like forward biased to me ) .

Answer 1

You are right; the diagram is wrong. With the polarity of V_BB as shown, both junctions are forward biased — the BC junction by V_BB alone, and the BE junction by the sum of V_BB and V_EE. There is no "transistor action"; they're just wasting power as heat. Redrawing the schematic to show the relative voltages makes this a little clearer:

^{simulate this circuit – Schematic created using CircuitLab}

For the transistor to be in active mode, the sign of V_BB would have to be reversed. Note that this means that current would be flowing "backward" through the V_BB supply. If it really is a battery, it would be getting charged by the V_EE supply.

Normally, you would draw the common-collector circuit with the collector as being the most positive point in the circuit:

^{simulate this circuit}