I've tried really hard to understand the basic operational principle of a transistor. I've referred to many books and been to forums but have never got a convincing answer.

Here are the things I want to understand:

A transistor is similar to a reverse biased diode unless a voltage is applied to the Base. Since the Emitter-Base junction is forward-biased, there will be conduction of - say - electrons (npn). What happens then? Is it true that these electrons from the Base break the barrier of the Collector-Base junction and then the combined current passes to the Emitter? (IB + IC = IE)

And why is that we are getting more current? Where is amplification? It can't be like creating something out of nothing. I know I'm missing some crucial point here. Can somebody explain me clearly in simple terms?

It has been a week I'm tryin to understand this. :(


3 Answers 3


When electrons flow through a forward-biased diode junction, such as the base-emitter junction of a transistor, it actually takes a non-zero amount of time for them to recombine with holes on the P side and be neutralized.

In an NPN transistor, the P-type base region is constructed so as to be so narrow that most of the electrons actually pass all of the way through it before this recombination occurs. Once they reach the depletion region of the reverse-biased base-collector junction, which has a strong electrical field across it, they are quickly swept away from the base region altogether, creating the collector current.

The total current through the base-emitter junction is controlled by the base-emitter voltage, which is independent of the collector voltage. This is described by the famous Ebers-Moll equation. If the collector is open-circuit, all of this current flows out the base connection. But as long as there's at least a small positive bias on the collector-base junction, most of the current is diverted to the collector and only a small fraction remains to flow out of the base.

In a high-gain transistor, fewer than 1% of the electrons actually recombine in the base region, where they remain as the base-emitter current, which means that the collector current can be 100× or more the base current. This process is optimized through careful control of both the geometry of the three regions and the specific doping levels used in each of them.

As long as the transistor is biased in this mode of operation, a tiny change in base-emitter voltage (and a correspondingly small change in base-emitter current) causes a much larger change in collector-emitter current. Depending on the external impedance connected to the collector, this can also cause a large change in collector voltage. The overall circuit exhibits power gain because the output power (ΔVC × ΔIC) is much greater than the input power (ΔVB × ΔIB). Depending on the specific circuit configuration, this power gain can be realized as either voltage gain, current gain, or a combination of both.

Essentially the same thing happens in a PNP transistor, but now you have to think of the holes (the absence of an electron) as being the carrier of a positive charge that drifts all of the way through the N-type base to the collector.

  • \$\begingroup\$ Alright. So because of narrow base and lesser time, recombination doesn't take place. And the electrons are drifted to collector region that forms collector current. But I don't understand where and why is amplification in this whole process. Since the collector current is nothing but a part of current in forward biased np junction that is passing from base to.collector, from where we are obtaining more current or current gain? Why and how does the variation in base current causing variation.in collector current. Please explain me! \$\endgroup\$
    – JGalt
    Commented Mar 3, 2014 at 7:39
  • 5
    \$\begingroup\$ Amplification doesn't happen inside the transistor; amplification is a concept related to the overall circuit in which it is found. The point is, the transistor is a device that can cause large changes in collector current from small changes in base current. This fact can be used to create circuits that have voltage amplification, current amplification, or both. In every case, the signal output power is greater than the signal input power. The extra power at the output comes from the circuit's power supply. \$\endgroup\$
    – Dave Tweed
    Commented Mar 3, 2014 at 13:18
  • \$\begingroup\$ Hi. I read all the above discussion which talks about DC currents in transistor when no external input signal is not applied. Now, suppose I apply a few mV signal between base-emitter junction.Will you please explain how this a few mV input signal is amplified in transistor? \$\endgroup\$
    – yuvi
    Commented Mar 3, 2014 at 13:45
  • \$\begingroup\$ @yuvi: No, it isn't possible to provide such an explanation without the context of a specific circuit. Besides, EE.SE isn't the place for such a discussion, which can (and does) fill entire books. \$\endgroup\$
    – Dave Tweed
    Commented Mar 3, 2014 at 14:46
  • \$\begingroup\$ @DaveTweed, sorry for the delay. Thank you so much for your wonderful answer. \$\endgroup\$
    – JGalt
    Commented Mar 25, 2016 at 14:29

Read and re-read Dave's excellent answer.

Then mentally reverse what's going on...

You have a forward-biased base-emitter junction, and external circuitry connected to the base demands a current Ib, which is supplied from electrons sourced by the emitter.

But when an electron enters the base region, it encounters a strong electric field pulling it towards the (positive) collector. The majority (a large and quite well defined proportion) of these electrons are lost (from the base current) and emerge as collector current, for the reasons explained so well in Dave's answer. So rather than an efficient amplifier, you could equally well view the transistor as a hopelessly inefficient supplier of base current!

From this viewpoint, the base circuit demands Ib and the emitter supplies it. But as a byproduct, a much larger current (Ic = 100Ib) is "lost" to the collector. Which is of course what we really want.

EDIT re: comment: Ultimately (most of, say 99%) the electrons from emitter enter the collector region.

Ultimately the collector current has to be (slightly) smaller than the supply emitter current.

Right to both of these.

What is the purpose?

1) A very small base current controls a large collector current, and the emitter current is the sum of these two.

2) The ratio Ic/Ib (hFE or current gain) is approximately independent of the collector voltage Vce (until Vce is low, say < 1V). This means that for a suitable choice of impedance in the collector circuit, a small change in Ib can result in a large change in Ic and a large change in Vce; this is where voltage gain comes from.

So the usual "common emitter" amplifier has the load in the collector circuit and has both high current gain and high voltage gain.

  • \$\begingroup\$ Thank you Brian. I think I've now quite understood the actual process. The definition of amplification is so confusing that I'd thought that some internal process actually produces more charge carriers in collector circuit. However, I have few more questions. Ultimately it is the electrons emitted by emitter are going to enter collector region? Then what is the benefit of doing all this? The emitter current is going to branch and a small part of it is base current and much of it is collector current. Ultimately the collector current has to be smaller than the supply emitter current right? \$\endgroup\$
    – JGalt
    Commented Mar 3, 2014 at 11:35
  • \$\begingroup\$ If that's so, then what is being amplified? Can.you give me example? \$\endgroup\$
    – JGalt
    Commented Mar 3, 2014 at 11:36
  • 1
    \$\begingroup\$ Right. The fraction of emitter current that reaches the collector is called \$\alpha\$, and is typically a number like 0.99 or more. The fraction of emitter current that actually comes out of the base is \$1 - \alpha\$, which would be 0.01 or less. The ratio of these two currents is the \$\beta\$, or current transfer ratio (current gain) of the transistor. \$\beta = \frac{\alpha}{1 - \alpha} = 99\$. \$\endgroup\$
    – Dave Tweed
    Commented Mar 3, 2014 at 12:18
  • \$\begingroup\$ What is being amplified? Base current. \$\endgroup\$
    – user16324
    Commented Mar 3, 2014 at 13:19
  • \$\begingroup\$ Hi. I read all the above discussion which talks about DC currents in transistor when no external input signal is not applied. Now, suppose I apply a few mV signal between base-emitter junction.Will you please explain how this a few mV input signal is amplified in transistor? \$\endgroup\$
    – yuvi
    Commented Mar 3, 2014 at 13:44

This is the way I see it, I hope it adds something useful to the discussion:



Let's think of a row of pennies laid out in a line, touching, across a table. Move the right hand end penny one penny's width to the right, leaving a gap. Then keep moving the penny to the left of the gap into the space. As you proceed all the pennies have moved to the right, and the gap has moved across the table to the left. Now picture the pennies as electrons, and you can see how electrons moving one way across a semiconductor causes holes to move the opposite way.

To stretch the analogy, we could use little piles of pennies, so a lot have to move right before a hole moves left. Or we could have a few pennies and a lot of space so that holes travel easily as the sparse pennies are moved across the wide gaps. These two cases model the two forms of doped Silicon, lot of electrons added and we have N-type, lots of holes (electrons removed) and we have P-type. The types are achieved by mixing (doping) the Silicon with small quantities of other metals.

With the electrons having to struggle through the atoms of a semiconductor, its resistivity is relatively high. Early semiconductors used Germanium, but, except for special cases, nowadays Silicon is the universal choice.

Copper wire can be visualised as having big piles of penny electrons, all close together, so a current is the movement of the few pennies at the tops of the piles, no holes are produced at all. With so many available for the current, resistivity, as we know, is low.


The commonest semiconductor diode (there are other specialised types) has a junction between N-type and P-type. If a voltage is applied to the diode, positive to the N-type end and negative to the other, the electrons are all pulled to the positive end, leaving holes at the negative end. With hardly any electrons in the middle, almost no current can flow. The diode is "reverse biassed"

When the voltage is applied the other way, negative to the N-type end and positive to the P-type, electrons are attracted to the middle and can cross over to cancel out the holes in the P-type, and flow out into the connecting wire. At the other, negative voltage, end, electrons are repelled into the middle of the diode, to be replaced by those flooding in from the wire, so overall a current can flow easily: the diode is forward biassed.

The connections to a diode are called the "Anode" which is the positive end when the diode is forward biassed, and the "Cathode" which is the negative end. I remember these by analogy with the same terms for valves, which need a high positive voltage (H.T. for "High Tension" -- keep your fingers off) at the anode for current to flow. A good mnemonic for the polarity of a forward biassed diode might be PPNN: "Positive, P-type, N-type, Negative".

A varactor diode exploits the fact that two separated charge areas, positive and negative, make a crude capacitor. So, specially designed diodes are made to exploit this, when reversed biassed. The applied voltage pulls the charges apart, forming a "depletion layer" between the contacts. Increasing the applied reverse voltage makes this layer thicker, so reducing the capacity, and vice versa. Varactor diodes are commonly used in tuned circuits to vary the frequency, replacing the vaned capacitors that were used in the days of valves.


A bipolar transistor is one whose operation depends on both electrons and holes. It comprises two diodes back to back sharing a common central layer. One of the outer terminals is the Collector C and the other is the Emitter E. The central connection is the Base B, and it is part of both the CB and BE diodes. So we have a three layered sandwich. In normal use the diode between C and B is reverse biassed, so, without the presence of the BE diode and its effect, no current would flow, because all the electrons are pulled up to one end of the CB section, and the holes to the other end, as in a diode, by the applied voltage.

The BE diode is forward biassed, so a current can flow and the external circuit is set up to limit this to a fairly small value, but there is still a lot of holes and electrons flowing through the Base and Emitter.

Now the clever bit. The common connection of the CB and BE diodes at the Base is made very thin, so the flood of electrons and holes in the BE part replaces those that the reverse Collector voltage has pulled away, and a current can now flow though this CB diode in the reverse direction, and then on through the forward biassed BE junction to the Emitter and out into the external circuit.

I think it is obvious that you can't make a transistor by soldering two diodes back to back, the action requires that intimate sharing of the thin layer inside the Silicon.

The Collector current depends on there being a Base current flowing, and the transistor is designed so that a small current in the BE diode opens the way for a much larger current in the CB junction. Thus we have current amplification. Using voltage drops across external resistors, this can be converted into voltage amplification.

These transistors are called "bipolar" because they effectively have two junctions.

I have carefully avoided mentioning the type of material in the CB and BE diodes, the ideas are the same for both, and we can have NPN or PNP as the possible layers. The arrow, on the emitter, in the symbol, which shows the direction of the conventional Collector current (the opposite of electron flow), points in the direction of the negative side of the applied CE voltage, so the current is "out of P or into N at the emitter".


There are lots of different designs of FET, and this is a very simplistic look at their basic principle.

These are "unipolar" transistors, although the term is not often used, because their operation depends only on electrons and electric fields, not holes.

Here we have a single block of doped silicon, the "channel", with lumps of the opposite type on the sides, or as an encircling ring. So we have only one diode junction, which is called the Gate G, between the lumps or ring and the channel. The channel acts as a resistor, with current flowing though from one end, the source S, to the other the Drain D. The junction between gate and channel is reverse biassed, so no current flows, but there is an electric field set up which pulls charges, electrons or holes, to the sides of the channel, leaving less available for the SD current. Thus we have the SD current controlled by the voltage on the gate.

Note this is a voltage controlled device, virtually no current flows into or out of the Gate. Think of Ohm's law: Resistance = Volts/Amps, and we see that a very low current means a very high Resistance, so the FET is said to have a very high input impedance -- its main advantage over Bi-Polar, where, by contrast, it takes little voltage to send the current through the base, giving it a low input impedance


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