In fact, this question has been asked on the EE site, but it's not well-answered. I suppose it might be more on-topic here.

According to this answer:

Note that the holes injected into Emitter are supplied from Base electrode (Base current), whereas the electrons injected into the Base are supplied from Emitter electrode (Emitter current). The ratio between these currents is what makes BJT a current amplifying device - small current at Base terminal can cause a much higher current at Emitter terminal. The conventional current amplification is defined as Collector-to-Base currents ratio, but it is the ratio between the above currents which makes any current amplification possible.

First off, Why collector current increases as base current increase? Is the former causes the later, or the later causes the former, or something else (voltage on electrodes, maybe) causes both?

And here is my question, Why collector current always increases more than the increment of base current? Say after something changes, a extra holes are "injected into" emitter region, and b extra electrons are injected into base region. Then why b is greater to a?

  • 5
    \$\begingroup\$ Hmm, beta changes a lot. You should probably rephrase this question, because your assumption is wrong. \$\endgroup\$
    – pipe
    Commented Mar 17, 2016 at 3:14
  • 3
    \$\begingroup\$ I second pipe's comment. Beta does change, and by a very large amount, as collector currents and voltages change. See any good transistor data sheet and you'll find different gains at different operating conditions. At the very least, a transistor with a beta of several hundred at low currents will display betas of 10 or lower when in saturation. \$\endgroup\$ Commented Mar 17, 2016 at 3:30
  • 2
    \$\begingroup\$ Beta also changes with temperature!! You can even totally destroy a transistor (i.e. beta likely = 1 or even 0) in exposure to a high energy EMP from something like a nuclear blast. \$\endgroup\$ Commented Mar 17, 2016 at 3:56
  • 2
    \$\begingroup\$ I can't answer either part properly for you but nobody will want a transistor that amplifies base current by < 1. It would be an attenuator and would be very little, if any, practical use. \$\endgroup\$
    – Transistor
    Commented Mar 17, 2016 at 8:54
  • \$\begingroup\$ The case of saturation (negligible Vce) is anomalous, because the emitter current is not removed via the collector; it just piles up unrecombined charges until the slow recombination in the base DOES happen, and the 'base-current-is-small' approximation doesn't work. The space charge that causes this base current causes a delay in any subsequent attempt to turn the transistor OFF, so avoiding saturation is a key element of speeding up switch transistors. Baker clamps and Schottky clamps help. \$\endgroup\$
    – Whit3rd
    Commented Apr 6, 2016 at 21:23

3 Answers 3


Answering questions:

@LvW Just out of curiosity: What if VCE = VBE? C-B pn junction won't be reverse-biased then, so it won't attract electrons in base region. Thus, IC will be zero, and IE will be equal to IB?

But the C-B diode is not forward biased. This is an application where the BJT is used as a diode and no "classical" amplification is possible (transition region between saturation and amplifying region).

IC and IE are controlled and only controlled by VBE; IB is just a side product; Once VCE is greater than VBE, its specific value does not matter, because E-B junction is reverse-biased. Am I right?

It does not matter too much - on the other hand: Look at the Ic=f(VCE) curves. Ic slowly rises with VCE because of the Early-effect.

Given VBE, IE is fixed, and as a result, the sum of IB and IC is fixed. When VCE < VBE, what IB and IC are depend on VCE. The greater VCE is, the greater IC/IB is. However, the value of IC/IB is capped by "beta", which is reached when VCE = VBE. " Is this right?

In this case (VCE < VBE) the C-B diode is open and there is a small current Ic which has a direction opposite to the "normal" Ic direction. Example: For VCE=0 we have a current Ic which is negative (The Ic=f(VCE) curves do NOT cross the origin!).

  • \$\begingroup\$ But my text book claims that with Ib fixed, The Ic = f(Vce) curves DO cross the origin. The graph on my book looks like this: postimg.org/image/jwhm8a6en (You can just ignore these Chinese characters) \$\endgroup\$
    – nalzok
    Commented Mar 17, 2016 at 15:55
  • \$\begingroup\$ Actually, I was confused with this graph; thus asked a series of questions. \$\endgroup\$
    – nalzok
    Commented Mar 17, 2016 at 15:59
  • \$\begingroup\$ I know that many graphical illustrations do not show that Ic<0 for VCE=0. In most cases, it is a matter of the used scale. Moreover, this property is not very important for practical applications. But - you can be sure that the Ic=f(VCE) curves do not touch the origin. For VCE=0 we have VBE=VBC and both diodes are open - hence there is a current from B to E (normal) and from B to C (opposite to the "normal current Ic"). \$\endgroup\$
    – LvW
    Commented Mar 17, 2016 at 16:02
  • \$\begingroup\$ Yes, absolutely. There is another thing that confuses me: What's the point of keeping Ib fixed? Does this guarantees Ube won't change or imply something else? \$\endgroup\$
    – nalzok
    Commented Mar 17, 2016 at 16:10
  • \$\begingroup\$ We have three quantities which matter: IC, VCE and VBE (or IB which is fixed related to VBE). Therefore, drawing IC=f(VCE) we must the third quantity keep constant. Both drawings are common: Either IB=const or VBE=const. In most cases. IB is held constant because the various curves look "nicer": They are equidistant. More than that, with IB=const. it is possible to define the EARLY voltage in the diagram. Furthermore, for IB=const (dIB=0), we can define the hybrid parameter h22 (output conductance for open signal input). However. VBE=const. is possible and can be found in some datasheets. \$\endgroup\$
    – LvW
    Commented Mar 17, 2016 at 16:36

The base-emitter diode carries current from both holes and electrons; for an NPN, the emitter (N type, electrons) current is dominant because the emitter is heavily doped compared to the base. There are lots of electrons in the emitter that move in response to the base-emitter voltage bias, and fewer holes in the base (moving in the opposite direction). The base current must make up the outflowing holes in order that the transistor not lose the base-emitter bias voltage (there aren't any holes in the collector or emitter). So, a roughly proportional base (hole) current must be supplied at the base wire, to the larger emitter (electron) current. A second contribution to base current is the recombination of electrons from the emitter, which also depletes holes from the base, but without moving them (an electron 'falls' into a hole). This contribution is also proportional to emitter current, and is minimized by keeping the base (P type, holes) region very thin; most electrons from the emitter travel through the region without recombining, and are then in the collector where they are... collected. Both cause loss of base charge, have to be 'replaced' by base current, or the emitter bias (and current) turns off.

To summarize: Base current is from holes-to-emitter diode current, plus some emitter-sourced electrons causing recombination events in the base. Collector current depends on base-emitter VOLTAGE, because that determines the dominant (emitter electrons) current source. The base current just restores the base-emitter voltage condition after charge carriers move in to stay, or out and never return.

  • \$\begingroup\$ Then can you tell me what does the corrector-emitter voltage control? \$\endgroup\$
    – nalzok
    Commented Mar 17, 2016 at 10:18
  • \$\begingroup\$ And what essentially causes the emitter current? Perhaps base-emitter voltage? \$\endgroup\$
    – nalzok
    Commented Mar 17, 2016 at 10:20
  • \$\begingroup\$ Also, can I say "Collector current only depends on base-emitter VOLTAGE"? (Sorry for asking so many questions!) \$\endgroup\$
    – nalzok
    Commented Mar 17, 2016 at 10:22
  • \$\begingroup\$ Collector-emitter voltage VCE (a) reversely biases thge C-B pn junction (necessary for amplification) and (b) causes the charged carriers emitted by the emitter to move through the base region (99.5%) because they are attracted by the VCE>VBE.. \$\endgroup\$
    – LvW
    Commented Mar 17, 2016 at 11:21
  • \$\begingroup\$ Yes - collector current IC only depends on VBE. There is a current IB, of course, but it has no controlling function (can 3 holes from the base cause 1000 electrons to leave the emitter?). \$\endgroup\$
    – LvW
    Commented Mar 17, 2016 at 11:23

Your Q apparently refers to a NPN transistor ('holes injected into Emitter').

In a bipolar transistor (NPN or PNP; referring to NPN in this answer), when the base-emitter junction is forward biased, current flows. This consists of holes injected from the base to the emitter, and electrons from the emitter to the base. Transistors are constructed (richer doping of Emitter than Base) so that most of the current is carried by electrons rather than by holes.

Now, the holes injected into the emitter will find a dense field of electrons (emitter is heavily doped), and so will recombine quickly. This requires replacement electrons to be supplied by the emitter terminal.

Electrons injected by the emitter into the base will find very few holes around -- the base is relatively lightly doped. So, a relatively small amount of recombination occurs, although this does require holes and consequent base current. As soon as these electrons arrive at the base end of the depletion region, they diffuse away from it. because the base is thin, this diffusion is 'fast'.

Any electrons that diffuse close to the collector-base junction will be swept across that junction (if the collector-base junction is reverse biased), because the field is such that it 'attracts' electrons from base to collector. These electrons form collector current.

Thus there are two significant components of base current -- holes injected from B to E, and holes to recombine with some of the electrons injected from emitter to base (there is a negligible additional component of reverse collector-base leakage). While not equal, these values are generally similar (recombination current is usually lower than injection current).

Emitter current consists of holes recombining and electrons injected. Because of the structure of the junction, the injection component dominates.

Collector current is primarily the injected emitter electron current, minus some small amount that is lost due to recombination.

So, because a) at the B-E junction electron injection is greater than hole injection, and b) electron recombination in the base is small, the collector current is a large (say 99 %) fraction of the emitter current -- therefore the base current (which is the difference) is about 1 % of the emitter current.

These parameters differ from device to device, with temperature, and with some imperfections and other defects in devices, but the basic principles are consistent.


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