This is question for your reference

Imagine we have an NPN transistor and the base voltage Vb = 0V. How can the transistor be in the cut-off region?

I've started by assuming it to be in the active region and then calculated the emitter voltage Ve = -0.7 by taking Vbe = 0.7. In the book it was directly mentioned that the transistor is in the cut-off region.

I think I'm missing some concept. Can someone help me with this simple thing?

  • 1
    \$\begingroup\$ You need a positive voltage (more than 0.5V) between base and emitter to open the NPN transistor. Thus, If VB is 0V how can BJT be in the active region? \$\endgroup\$
    – G36
    Commented Aug 14, 2021 at 12:27
  • 2
    \$\begingroup\$ The rule of thumb that Vbe is about 0.7 volts only applies when the base-emitter junction is conducting "normal" amounts of current, usually something a milliamp or maybe a bit more or less. It's a very useful approximation. But, when the base voltage is zero, and the emitter voltage is zero, the base current is zero and you can't use the approximation. \$\endgroup\$ Commented Aug 14, 2021 at 14:38
  • 4
    \$\begingroup\$ You do not have a negative voltage source, so the emitter cannot be at a negative voltage. \$\endgroup\$ Commented Aug 14, 2021 at 15:56

4 Answers 4


This is one of those cases where a teacher's desire to "keep things simple" instead turns into gross over-simplification that then just leads to confusion and disaster.

Let's focus on something else for a moment:


simulate this circuit – Schematic created using CircuitLab

Do you see anything in that circuit that goes outside the bounds of \$0\:\text{V}\to 6\:\text{V}\$? Is there anywhere, at all, that \$-0.7\:\text{V}\$ could arrive? Are you suggesting that the BJT itself, \$Q_1\$, might be able to generate it, somehow?

The answer is, of course, that there isn't a way for \$-0.7\:\text{V}\$ to show up in that schematic. You can't plug in a BJT and expect it to create a voltage that is outside its world. As you can see, it is nestled in a tiny world that lives between \$0\:\text{V}\$ and \$6\:\text{V}\$. There is no possibility for \$-0.7\:\text{V}\$.

As to the rest, once you accept that cannot happen then just ask yourself this question, "If the base voltage is \$0\:\text{V}\$ and the other end of \$R_2\$ is also at \$0\:\text{V}\$, and if the emitter must be in between these two voltages, then what voltage is the emitter at?"

It should be obvious.

Once you know the emitter voltage and also know the base voltage (which, by now, I'm sure you do) then you can see why the BJT is said to be in cut-off.



Well, can the emitter be at -0.7 V, based on the circuit you have it in? If not, then the assumption you started with must be false.

Hint: What were you measuring the base voltage relative to in the first place? Remember, a voltage measurement always represents the potential difference between two circuit nodes. The transistor is only "aware" of the voltages between its terminals.

  • \$\begingroup\$ Voltages are with respect to Ground \$\endgroup\$ Commented Aug 14, 2021 at 12:15
  • \$\begingroup\$ @AYUSHMEENA Ayush, I hope things are okay with you. You've not responded in a while. Best wishes. \$\endgroup\$
    – jonk
    Commented Aug 24, 2021 at 3:56

When Vb is equal to 0V, Vbe is equal also to 0V. With no voltage between base and emitter there can be no base current. Therefore the collector current is also 0mA. If there is no collector current then there can be no voltage drop across the collector's 10k resistor. Therefore both ends of the 10k resistor are at +6V and the transistor is operating in the cut-off region.


I've started by assuming it to be in active region and then calculated Emitter Voltage Ve = -0.7 by taking Vbe = 0.7

That's fine but, didn't your question title say this: -

How the Transistor is in Cut-off Region?

And then you said this: -

then in the book it was directly mentioned that Transistor is in Cut-off Region

So, why are you setting up the base-emitter voltage for the active region? Set base and emitter voltages to be the same (for example) for the cut-off region.

  • \$\begingroup\$ Actually this process or solution is mentioned in a book which confused me, thats why im asking \$\endgroup\$ Commented Aug 14, 2021 at 12:16
  • \$\begingroup\$ @AYUSHMEENA I'm really unsure what it is that you are actually asking. \$\endgroup\$
    – Andy aka
    Commented Aug 14, 2021 at 12:18
  • \$\begingroup\$ actually the question is about When Vb = 0V find the Emitter current and Vc \$\endgroup\$ Commented Aug 14, 2021 at 12:18
  • \$\begingroup\$ Andy i've attached the question also please have a look \$\endgroup\$ Commented Aug 14, 2021 at 12:22
  • \$\begingroup\$ @AYUSHMEENA for the currents listed in answers (a), (b) and (c) which one satisfies the situation that makes Vbe 0 volts? Then, when you have decided which one, what will Vc be and does that selected answer also match that or, do you go for (d)? \$\endgroup\$
    – Andy aka
    Commented Aug 14, 2021 at 12:35

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