How Do Transistors Work?

To turn on an NPN transistor, a voltage is applied across the base and emitter terminals. This causes electrons in the Base wire to move away from the transistor itself and flow out towards the power supply. This in turn pulls electrons out of the P-type base region, leaving 'holes' behind, and the 'holes' act like positive charges which are pushed in the opposite direction from the direction of electron current. What SEEMS to happen is that the base wire injects positive charges into the base region. It spews holes. It injects charge.

(Note that I'm describing charge flow here, not positive-charge "conventional current.")

              
       ____________
                  
       COLLECTOR N
      _____________                  ELECTRONS ARE PULLED FROM THE
                     ----->          BASE REGION AND INTO THE WIRE,
       BASE      P ______________    WHICH CREATES POSITIVE "HOLES"
      _____________                +  WHICH SPEW OUT INTO THE BASE
                             ________   REGION.
       EMITTER  N              _____
      _____________          _________
                                _____
             _____________________  -

That's part of the conventional explanation. Why is all of this important to transistor operation? ***It's not!*** The base current is not important to transistor operation. It's just a byproduct of the REAL operation, which involves an insulating layer called the Depletion Region. By concentrating on the current in the Base lead, most authors go up a dead end in their explanations. To avoid this fate, we must start out ignoring the base current. Instead we look elsewhere for understanding. See the diagram below.

              
       ____________
                           \
       COLLECTOR          
                            > full of wandering electrons
        n-doped                        
      _____________        /             
                               \ 
       BASE                     
                   --           > full of wandering "holes"
        p-doped                 
      _____________            /
                         \    
       EMITTER              
                          > full of wandering electrons   
        n-doped         
      _____________      /
                       
             

The Depletion Region is an insulating layer existing between the base region and the emitter region. Why is it there? It exists because the Base region is p-doped silicon; it exists because p-type silicon is full of naturally-occurring movable "holes," and because the p-type silicon is touching n-type silicon.

              
       ____________
                  
       COLLECTOR N
      _____________                    
                                
       BASE      P --
      _____________        
       _____________   <-- insulating "depletion layer"           

                       
      EMITTER  N 
      ___________            
             
           

Electrons in p-type silicon act like the closely-packed beads of an abacus, and the "holes" are like gaps in the rows of beads. Move one bead, and a hole has moved the other way. Touch the p-type silicon against the n-type, and wandering electrons from the n-type silicon will fall into the holes. Also, holes in the p-type's Base region flow out among the movable electrons from the N-type Emitter region and many are cancelled. Holes swallow electrons, and this leaves a thin region between N and P sections which lacks movable charges.

Remember: a conductor is not a substance which allows charges to pass. (Don't forget #3 above!) Actually a conductor is any substance which contains charges which are movable. Anything that lacks movable charges is an insulator. Inside the depletion layer, all the opposite charges have fallen together and vanished. The gaps in the abacus beads are gone, so no beads can move anymore. Lacking mobile charges, the silicon has turned into an insulator. When there's no voltage applied across the base/emitter terminals, this insulating layer grows fairly thick, and the transistor acts like a switch which has been turned off.

I like to visualize that a transistor's silicon is normally like a shiny silver conductor (sort of like metal) ...except for this insulating layer between the P and N regions which acts more like a layer of insulating glass. Silicon is like a metal which can become glass!

              
       ____________
                           \
       COLLECTOR N        
      _____________         >  Shiny silver conductive     
                          
       BASE      P --      /
      _____________           
       _____________    <-- Glasslike insulating "depletion layer"                                      \         EMITTER  N             >  Shiny silver conductive
      _____________          /
                       
             

When voltage is applied between base and emitter, this insulating layer changes thickness. If (+)voltage is applied to the p-type, to the base wire, while a (-) voltage polarity is applied to the n-type, to the emitter wire, then electrons in the n-type are pushed towards the holes in the p-type. The insulating layer becomes so thin that the clouds of electrons and holes start meeting and combining. A current therefore exists in the base/emitter circuit. But this current is not important to transistor action. What's important to notice is that the *VOLTAGE* across the base/emitter has caused the insulating Depletion Layer to become so thin that the charges can now flow across it. It's as if the transistor contains a layer of glass whose thickness can be varied when we alter a voltage. The layer becomes thinner when base/emitter voltage is increased. This happens because the voltage pushes the holes and the electrons towards each other, reducing the size of the empty insulating region between the clouds of holes and electrons, and allowing the stragglers to jump across the insulator. The depletion layer is a voltage-controlled switch which "closes" when the right polarity of voltage is applied. It is also a proportional switch, since a small voltage can close it only partially. For silicon material, charges start jumping across when the voltage is around 0.3V. Raise the voltage to 0.7V and the current gets very high. (That's for silicon. Other materials have different turn-on voltages.) The larger the voltage, the thinner the insulating layer, so the higher the current in the entire transistor. By applying the right voltage, we can thicken or thin the depletion layer as desired, creating an open, closed, or partially open switch.