# Why is Ib proporional to Ic in a bipolar transistor?

I am really puzzled why the collector current is considered proportional to the base current in a bipolar NPN transitor, ie

$$I_c = \beta I_b$$

Where I_c is the collector current, I_b the base current and below I_e the emitter current. I can't find any explanation in the books, it seems to be just stated as an empirical result. Is that how I should treat it, just an empirical result?

I know that it can be derived by assuming that the proportion of (conventional) current that enters the emitter from the base is a constant fraction of the collector current, ie

$$I_c = \alpha I_e$$ and that $$I_b + I_e = I_c$$

but that begs the question why is the fraction of base current entering the emitter stream always assumed to be a fixed proportion?

• You might want to get hold of this book: "Integrated Electronics", by Millman and Halkias or, better yet, "Electronic Devices and Circuits". Chapter 5 in the first one, and chapter 9 in the second one. Do not get a modern x-edition copy of "Microelectronics" or "Millman's Electronics". Get the old ones. Might even find a copy in some... Archive. – Sredni Vashtar May 11 '17 at 20:40
• Actually, this is an Engineering site. In Engineering, people rarely ask questions "why" fundamental relationships are such or such. The standard answer is that "there is a formula in Physics, where under reasonable assumptions about geometry of electrodes the relationship appears APPROXIMATELY as Ic = beta*Ib." In reality it is not, it is a convenient approximation. Ask this question on Physics forum. – Ale..chenski Jun 25 '17 at 0:05
• One issue: a transistor is also a PN diode, and in diodes, each metal terminal only emits new mobile carriers whenever other electrons or holes have collided/cancelled. The diode terminals "know" when carriers in the depletion zone have annihilated. Same with Ib: the metal Base terminal of NPN can only inject a new hole into the Base region if an existing hole has been eaten by an electron. This makes it obvious why, if Base region is doped lightly and made thinner, Ib must become smaller, while Ic does not. – wbeaty Dec 9 at 2:36

The collector current is only approximately proportional to the base current. The rate varies with collector current and can vary wildly between individual units of the same type.

It is just that both the collector current and the base current are exponential functions of the base voltage.

• An exponential function has two degrees of freedom, though. One accounts for the ratio of collector to base current ($\alpha$). The other, the coefficient inside the exponent, has to be the same, for the two quantities to remain proportional. Importantly, the Shockley equation says the coefficient is determined by temperature, and the two p-n junctions are in such proximity that the temperature (and thus the thermal voltage) is the same for both. – Ben Voigt May 11 '17 at 3:59
• Kevin white: I think your second paragraph is the best clue I've seen for the supposed linear relationship. I've been doing my own experiments and I've so far shown the exponential rise in collector current as a function of vbe. I should try ic vs ib to see how linear it really is. – rhody May 11 '17 at 5:30
• Thinking about it more I can agree now that the ratio of the two exponentials will be a constant and that explains the origins of the proportionality. I probably now need to pose a new question which is why do the currents respond exponentially to voltage. – rhody May 11 '17 at 5:49
• rhody, the exponential relationship goes back to the diode principle. The drift and diffusion currents are expressed using the hole and electron density as a function of the length coordinate x. This gives a first order diff. equation which can be solved if an external voltage is applied. As a result, we get an exponential function (Shockleys equation). – LvW May 11 '17 at 8:33
• Yes that makes sense. – rhody May 11 '17 at 16:26

You would need much more training to understand semiconductor physics.

It is only approximately constant and the tolerance is often a wide range over 3:1 range around nominal value unless binned or from the same batch.

For now try to learn the basic functions and variables to understand how things work from a logic and analog behaviour rather than the physics model.

• If you feed charges in a high electric field, they accelerate.
• The rate of change of charges is called current.
• The acceleration ratio of these charges in a high electric field , we call current gain.

It changes somewhat with many variables, so you only show the simple formula and the more accurate model is here.

• I went to the page you mention but it seems to assert again that the base current is proportional to the collector current, doesn't explain why (equ 3.45). For a given transistor I've assumed that the relationship is reasonably linear. – rhody May 11 '17 at 5:27
• re-examine the diode equations and model then examine Ic vs Ib vs Vce curves in datasheets. There is no formulae to derive beta but I assure you it is never constant except at one DC operating point in a constant environment. This is just a 1st order approximation. This is why we use negative feedback and R ratios and H biasing – Tony Stewart Sunnyskyguy EE75 May 11 '17 at 12:35
• Agreed, negative feedback will linearize. I'll go back the diode equations and figure it out. This helped however. – rhody May 11 '17 at 16:25

A device-physics guy explained BETA to me thusly: the base current charges enter the base region, and the emitter provides opposing charges that try to HIT and recombine with the base current charges. Fortunately, most of the emitter charges miss and continue on into the collector.

We thus benefit from those charges that happen to miss, which provides power gain and current multiplication.

• But why is the proportion of those that combine, approximately proportional to the collector current? – rhody May 11 '17 at 5:19
• Those charges injected from the emitter that DO NOT COMBINE become the collector current. Go read this explanation by James Early: ethw.org/Oral-History:James_Early – analogsystemsrf May 11 '17 at 7:15

Think of a vacuum tube: Electrons are emitted by the cathode, attracted by the grid, then pass through the grid and are accelerated to the anode. However, some small fraction of the cathode current will be "stolen" by the grid.

The same thing happens in a bipolar transistor. The base-emitter voltage sets the emitter current. In normal operation, most of that current continues to flow to the collector but the base steals some hopefully small fraction of the emitter current. The exact amount varies between devices and operating conditions. At a given operating condition, the base current is approximately proportional because each electron (or hole in a PNP transistor) acts nearly independently -- it has a certain probability to continue onto the collector, and a certain probability to recombine, in which case it won't be able to traverse the collector-base junction, and will instead leave via the base electrode.

• The question appears to be asking why it is a constant fraction, as opposed to say a constant absolute rate. – Ben Voigt May 11 '17 at 3:52
• Excactly, that's what I'm asking, why is the small fraction that goes to the base approximately proportional to the collector current? – rhody May 11 '17 at 5:22

Unfortunately the rationale quickly goes deep into semiconductor physics. The DC current gain $\beta$ is related by the forward transit time of a carrier to cross the base region and the average lifetime of a carrier in the base region. If you inject some base current $I_B$ into a transistor, a charge in the base region builds and eventually the collector current reaches approximately $I_C = \beta I_B$.