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

enter image description here

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 ) .

  • \$\begingroup\$ Does the book mention that the NPN is actually working in active mode in this configuration? Depending on the values of VBB and VEE I can get the NPN to work in different modes. \$\endgroup\$ Commented Oct 1, 2019 at 8:15
  • \$\begingroup\$ No, it doesn't say that. \$\endgroup\$
    – anonymous
    Commented Oct 1, 2019 at 8:17
  • \$\begingroup\$ @Bimpelrekkie Can you explain a little about how can I make it work in Active mode ? (Basically, how can I make the Collector-Base junction reverse biased while keeping the Emitter-Base junction forward biased) \$\endgroup\$
    – anonymous
    Commented Oct 1, 2019 at 8:28
  • \$\begingroup\$ You would need to make VBB negative. Only then will the CB junction be in reverse mode. Make VBB such that the BE junction is in forward mode (for example when VEE = + 5 V making VBB = -4.4 V then: Vbe = +0.6 V, Vce = 5 V and the NPN is in active mode. But read in the book, what is the point of this circuit? It is a bit uncommon as the collector is grounded and CB appears to be in forward mode (that could hint at saturation mode though). The NPN doesn't care about this, voltages are just for our (human) reference. \$\endgroup\$ Commented Oct 1, 2019 at 8:47
  • \$\begingroup\$ I don't know what to say. This was taken way too seriously. Aside from the correct statement that the circuit shown is useless, the fact is that CC, CE, and CB modes are about knowing how the source is applied and how the load is applied and how the 3-terminal BJT fits in-between. That can't be determined from any picture like the OP shows, even if the BJT actually was in active mode. Any of the three choices would be equally valid by simply making different terminal assumptions about input and output. Without those assumptions stated, you can't say. \$\endgroup\$
    – jonk
    Commented Oct 1, 2019 at 11:43

1 Answer 1


You are right; the diagram is wrong. With the polarity of VBB as shown, both junctions are forward biased — the BC junction by VBB alone, and the BE junction by the sum of VBB and VEE. 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 VBB would have to be reversed. Note that this means that current would be flowing "backward" through the VBB supply. If it really is a battery, it would be getting charged by the VEE supply.

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


simulate this circuit


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.