1) An NPN transistor in cut off mode:

enter image description here

"To get a transistor into cutoff mode, the base voltage must be less than both the emitter and collector voltages. VBC and VBE must both be negative. In reality, VBE can be anywhere between 0V and Vth (~0.7V for silicon) to achieve cutoff mode."

a) I've also read that both the base- emitter and base- collector junction will be reverse biased in cut off mode. But how? If the emitter is earthed, then even if Vbe (=Vb?) is less than 0.7V, the base is at a more positive voltage than the collector and consequently it is forward biased, right?

b) Also, why do we need to ground the input and the base if Vbe is already less than 0.7V? (How is the base grounded in the figure? I can see only the input being grounded)

2) In Saturation mode

enter image description here

a) This, I can understand because the collector voltage becomes close to zero during saturation and thus both Vc and Ve are less than 0.7V (and are thus forward biased)

b) But, why do we connect the input and base to Vcc?

  • \$\begingroup\$ The subject of your question says "Transistor as a Sitch". That is supposed to be "Switch" right? I went ahead and edited it. I hope you don't mind. \$\endgroup\$ – mkeith Feb 27 '19 at 6:25
  • \$\begingroup\$ Yes.. Its switch. \$\endgroup\$ – Gokulakrishnan Shankar Feb 27 '19 at 7:15
  • \$\begingroup\$ If the emitter is earthed, then even if Vbe (=Vb?) is less than 0.7V, the base is at a more positive voltage than the collector .. why? \$\endgroup\$ – Huisman Feb 27 '19 at 9:41
  • \$\begingroup\$ Regarding Vbe (=Vb?) : How would you define Vb? \$\endgroup\$ – Huisman Feb 27 '19 at 9:45
  • \$\begingroup\$ Vb is the potential at the base. And Vbe= Vb - Ve. Since Ve= 0, I thought Vbe= Vb. \$\endgroup\$ – Gokulakrishnan Shankar Feb 27 '19 at 13:18

1a) I'd say both are correct. At any voltage Vbe < Vth the transistor will "cut off", where by "cut off" we mean "the current drops to a super-low value, and gets even lower as Vbe drops further". It's always an approximation -- do we say the transistor is "cut off" once the current is less than a milli-amp? Or less than a micro-amp? For a milli-amp, driving Vbe down to 0.1V might be good enough. For a micro-amp we might need to drive Vbe to 0.0V or even a negative voltage.

1b) Vbe is less than 0.7V because we have tied the input to ground. The base is "grounded" because it is tied to 0V through Rin.

2a) Not actually a question ;-)

2b) We are using Vcc as a convenient voltage to drive Vin high, so that we can drive Vbe to a nice positive voltage and turn the transistor on.

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  • \$\begingroup\$ 1a) So, driving Vbe to 0.1V is technically forward bias, right (since the emitter is grounded)? But because the current becomes super- low, it's equivalent to saying that it's in reverse bias, correct? Thanks :) \$\endgroup\$ – Gokulakrishnan Shankar Feb 27 '19 at 7:22
  • \$\begingroup\$ At any voltage Vbe < Vth the transistor will "cut off"? Seems you're mixing up with mosfets... \$\endgroup\$ – Huisman Feb 27 '19 at 9:34
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    \$\begingroup\$ @Huisman how so? Yes a MOSFET is more traditionally analyzed in terms of its gate voltage, whilst a BJT is typically analyzed in terms of its base current, but it's perfectly valid (and in some cases preferred) to analyze a BJT's response in terms of its base-emitter voltage instead of its base current. \$\endgroup\$ – Mr. Snrub Feb 27 '19 at 15:52
  • \$\begingroup\$ Ok, didn't know. Thanks for pointing out. \$\endgroup\$ – Huisman Feb 27 '19 at 17:48
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    \$\begingroup\$ That's exactly why my question of "How so?" should taken at face value :-) i.e. not as an attempt to be defensive, but rather as a request for clarification. The point of this site is to teach people and provide high-quality answers, not to prove how superior any one individual's knowledge might be ;-) \$\endgroup\$ – Mr. Snrub Mar 1 '19 at 5:48

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