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I'm reading a tutorial on basic transistor operation. It says for an npn transistor to operate such conditions are necessary:

  1. VC > VE, by at least a few X 0.1V
  2. VB > VE
  3. VC > VB
  4. We do not exceed maximum ratings for voltage differences or currents

Here I don't get why VC > VB is necessary. It looks to be if VC is bigger than VB, then the current will flow thorough the base. Here is the npn transistor in question: http://i.stack.imgur.com/m9rv6.png

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closed as unclear what you're asking by Olin Lathrop, Leon Heller, Daniel Grillo, Joe Hass, Matt Young Mar 31 '14 at 18:20

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ What do you mean by "operate"? As an amplifier? As a switch? As a follower? If you're using it as a switch there's no problem with VC<VB. If you're using it as an amplifier you want the collector voltage in a region where operation is linear with lots of headroom. \$\endgroup\$ – John D Mar 31 '14 at 15:16
  • \$\begingroup\$ Voltage is electricity potential between two points. Thus Vc means nothing by itself. You need to specify always between what two points you talk about. For transistors you can use nomenclature as "Vcb", "Vbe" etc. Simply Vc means nothing. It could be -100V, +3,3V or 0V at the same time with regard to different points of circuit or even transistor parts: Vce, Vcb. \$\endgroup\$ – zzz Mar 31 '14 at 15:20
  • \$\begingroup\$ NPN transistors often operate with Vcb being negative, so this question makes as much sense as "Why is the moon made of cream cheese?". \$\endgroup\$ – Olin Lathrop Mar 31 '14 at 15:27
  • \$\begingroup\$ @zzz in maths, X > Y is the same as X - Y > 0; saying 'VB > VE' is the same as saying 'the potential difference measured between at point B relative to point E is positive'. \$\endgroup\$ – Pete Kirkham Mar 31 '14 at 16:00
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    \$\begingroup\$ @Pete Kirkham math is a hoax, I don't believe in it. \$\endgroup\$ – zzz Mar 31 '14 at 16:46
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Here I don't get why VC > VB is necessary

Actually, it is not : If the transistor is saturated, you can have (for example) Vbe=0.6V and Vce =0.3V.

When NOT saturated, you will have Vc>Vb.

It looks to be if VC is bigger than VB, then the current will flow thorough the base.

No, it won't. In NPN transistors, you have a diode junction between base and collector that is reverse based when Vc>Vb.

Why is the collector voltage must be bigger than the base voltage to operate an npn transistor ?

Basically, this is what you want if the transistor acts as an amplifier : the voltage at the output (Vc) must be bigger than the voltage at the input (Vb). Note that this is often voltage variations (AC part of the voltage) that must be amplified.

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It takes a year to a year and a half of college-level classes to lay down the prerequisite groundwork: calculus through differential equations, chemistry, physics, basic electric circuit analysis. It takes about half a semester of semiconductor device theory to understand the processes in bulk silicon and PN junctions, leading up to the bipolar transistor.

When you have all that, what you have in a bipolar transistor (NPN or PNP) are sort of like two diodes connected in series, polarity opposing. The base terminal of the transistor is the common point. In an NPN transistor, the base is P-type material, and is hence the "anode" of both of the diodes.

For amplifying action, the Emitter-Base diode must be forward-biased. This means that, in an NPN transistor, Vb > Ve. For amplifying action, the Collector-Base diode must be reverse biased. This means that, in an NPN transistor, Vc > Vb. The rest is design.

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  • \$\begingroup\$ MY QUESTION I MEANT WHY MOST OF THE ELECTRONS FLOW THROUGH COLLECTOR INSTEAD OF BASE WHEN VC > VB. \$\endgroup\$ – user16307 Mar 31 '14 at 16:51

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